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614,801 | A rifle is fired, and the bullet moves much faster than the rifle, and momentum is conserved. My question is whether the kinetic energy of the bullet is greater than the kinetic energy of the rifle because the mass of it is smaller, therefore the force acting on it (for the same time it acts on the rifle) results in greater acceleration, and thus greater final velocity (basically, the change in momentum needs to be equal and in opposite direction to the rifle, which has greater mass). On the other hand, kinetic energy depends on work done, which is force times displacement. Therefore, we can say that the force acted on the bullet for a greater distance traveled. Are these explanations stating the same thing, or are they causally different? (From the expression of the momentum formula as m. dx) | Conservation of momentum implies $$mv+MV=0$$ where $m$ and $v$ are the mass and speed of the bullet, $M$ and $V$ of the rifle. Of course, it is $0$ in total as the system is at rest at the beginning. This implies $$|v|=MV/m$$ so the smaller the bullet, the bigger its speed. In terms of KE $$ke={1\over 2} m v^2={1\over 2} M^2V^2/m$$ for the bullet whereas $$KE={1\over 2} MV^2$$ for the rifle. Thus the ratio is given by $$ke/KE={M/m}$$ Now, a gun/rifle is in the 5-10 kg range whereas for a bullet we have 10 g (rounding up) so you - just by momentum conservation - get a bullet's kinetic energy 500-1000 times bigger than the rifle's. (Notice that from this perspective we can only extract the ratio between the two energies. The absolute values would be given by information about the energy generated by the gunpowder. But that energy is partitioned between bullet and rifle according to momentum's conservation.) Now, because - as you mention - the change in kinetic energy is the same as the work done, this indeed means that the work done on the bullet is greater than the work done on the rifle. To see it, assume the explosion generates a constant force $F$ on both the rifle and the bullet, but in different directions (because of action-reaction), and that the explosion lasts for $t_0$ seconds (this is of course an approximation because $F$ is in general a non-constant force but we approximate it as a mean constant force applied for a mean time $t_0$ )). The bullet position over time will be $$x={1\over 2}at^2={1\over 2}{F\over m}t^2$$ where I used $a=F/m$ . Thus at the end of the interaction ( $t=t_0$ ) the bullet will have travelled a distance $$d={1\over 2}{F\over m}t_0^2$$ If we do the same for the rifle, we get $$D={1\over 2}{F\over M}t_0^2$$ so you see that despite $F$ and $t_0$ being constants, the bullet moves more as it is smaller ( $m\ll M$ so $d\gg D$ ). Now, because work is force times distance, this means that the work done on the bullet ( $w=Fd$ ) is bigger than the work done on the rifle ( $W=FD$ ) by the same ignition. Their ratio is of course $$w/W=Fd/FD=d/D=M/m$$ which is the same ratio of the kinetic energy (and it has to be as work done = change in kinetic energy). (Again, we only know the ratio as we don't know the value of $F$ - we just know that because of action-reaction it will be the same in magnitude on both bullet and rifle.) So a simple mass difference makes it so that the bullet accelerates more hence gets more kinetic energy in the same time. To this you need to add the fact that: as you said, one usually holds the rifle fixed so indeed the bullet might get more of the ignition force transformed into momentum. bullets are small and their energy is "condensed" into a minimal surface making them very good at perforation. Also, they have all sort of structural details made to decrease air friction, etc. But from a purely momentum conservation point of view you can understand that the bullet, just by being smaller, absorbs a bigger part of the total energy and therefore moves at high speed with respect to the rifle. | {
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615,003 | Today, I've decided to observe my PC fan as I shut the computer down. The fan slowly lost angular momentum over time. What I've found really interesting is the fact that the momentum vector change did not stop at the zero vector, but instead flipped its orientation and "went to the negatives", albeit very small in the absolute value compared to the powered spin; this caused the fan's angle to deviate by a few degrees (opposite to the powered spin rotation) compared to the observed angle when momentum was equal to the zero vector. If I let $\overrightarrow{L}$ be the momentum vector, $\overrightarrow{L}_0$ be the momentum vector at $t_0$ (= poweroff time), and $\overrightarrow{L}(t) = y(t) * \overrightarrow{L}_0$ (with $y_0 = y(t_0) = 1$ ), then these are the plots of $y$ through the course of time. Expected fan poweroff behavior: Observed fan poweroff behavior: Can anyone explain why may this happen? | The overshoot behavior you noticed is called cogging and occurs when the magnet arrangement in the motor "catches" the rotating magnetic core of the motor during shutdown and jerks it back to one of the local strong spots in the field. You can demonstrate this yourself by carefully rotating the fan blade around with your finger when the motor is off. You will notice there are certain rotation angles where the fan wants to come to rest and others which it wants to avoid. If the motor passes one of the preferred spots but fails to rotate far enough to "climb the hill" and snap forward into the next cog spot, the motor will very briefly rotate backwards a fraction of a turn and go "boing-oing-oing-oing" as it settles into that cog position. | {
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615,013 | I am currently in high school and I came across a physics problem that asked to find whether the height of the mercury in a mercury barometer would change when it is kept in an elevator that accelerating upwards with a magnitude of 'a'. Here is how I approached the problem(demonstrated in the image file attached): 1)In the frame of the elevator, the acceleration due to gravity would be (g+a) downwards 2)I assumed the Mass of the air column above the liquid mercury exposed to the atmosphere to be 'M' 3)I assumed the area of the mercury in the container exposed to the atmosphere to be 'A'. 4)I assumed that a vacuum exists in the space between the tube and the mercury. I assumed the density of mercury to be 'rho' After you read my attempt at solving the problem, you see that at the end I get an equation for 'h' is not a function of either g or a. So that must imply that the height of the mercury inside the elevator accelerating upwards must be the same as in the case where the elevator was not accelerating. But after checking various solutions on the internet, which propose that the height of the mercury changes, I am confused and not able to determine what I assumed wrong or where I made a mistake. I would appreciate if someone could clarify my conceptual doubt. (P.S I do not know how to write mathematical equations in PhysicsStackExchange so please excuse any distastefulness in my handwriting or illustration skills) | The overshoot behavior you noticed is called cogging and occurs when the magnet arrangement in the motor "catches" the rotating magnetic core of the motor during shutdown and jerks it back to one of the local strong spots in the field. You can demonstrate this yourself by carefully rotating the fan blade around with your finger when the motor is off. You will notice there are certain rotation angles where the fan wants to come to rest and others which it wants to avoid. If the motor passes one of the preferred spots but fails to rotate far enough to "climb the hill" and snap forward into the next cog spot, the motor will very briefly rotate backwards a fraction of a turn and go "boing-oing-oing-oing" as it settles into that cog position. | {
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615,177 | I'm not sure where this question should go, but I think this site is as good as any. When humankind started out, all we had was sticks and stones. Today we have electron microscopes, gigapixel cameras and atomic clocks. These instruments are many orders of magnitude more precise than what we started out with and they required other precision instruments in their making. But how did we get here? The way I understand it, errors only accumulate. The more you measure things and add or multiply those measurements, the greater your errors will become. And if you have a novel precision tool and it's the first one of its kind - then there's nothing to calibrate it against. So how it is possible that the precision of humanity's tools keeps increasing? | I work with an old toolmaker who also worked as a metrologist who goes on about this all day. It seems to boil down to exploiting symmetries since the only way you can really check something is against itself. Squareness: For example, you can check a square by aligning one edge to the center of straight edge and tracing a right angle, then flip it over, re-align to the straight edge while also trying to align to the traced edge as best you can. Then trace it out again. They should overlap if the square is truly square. If it's not, there will be an angular deviation. The longer the arms, the more evident smaller errors will be and you can measure the linear deviation at the ends relative to the length of the arms to quantify squareness. Other Angles: A lot of other angles can be treated as integer divisions of 90 degree angle which you obtained via symmetry. For example, you know two 45 degrees should perfectly fill 90 degrees so you can trace out a 45 degree angle and move it around to make it sure it perfectly fills the remaining half. Or split 90 degrees into two and compare the two halves to make sure they match. You can also use knowledge of geometry and form a triangle using fixed lengths with particular ratios to obtain angles, such as the 3-4-5 triangle. Flat Surfaces: Similarly, you can produce flat surfaces by lapping two surfaces against each other and if you do it properly (it actually requires three surfaces and is known as the 3-plate method), the high points wear away first leaving two surfaces which must be symmetrical, aka flat. In this way, flat-surfaces have a self-referencing method of manufacture. This is supremely important because, as far as I know, they are the only things that do. I started talking about squares first since the symmetry is easier to describe for them, but it is the flatness of surface plates and their self-referencing manufacture that allow you to begin making the physical tools to actually apply the concept of symmetries to make the other measurements. You need straight edges to make squares and you can't make (or at least, check) straight edges without flat surface plates, nor can you check if something is round... "Roundness": After you've produced your surface plate, straight edges,and squares using the methods above, then you can check how round something is by rolling it along a surface plate and using a gauge block or indicator to check how much the height varies as it rolls. EDIT: As mentioned by a commenter, this only checks diameter and you can have non-circular lobed shapes (such as those made in centerless grinding and can be nearly imperceptibly non-circular) where the diameter is constant but the radius is not. Checking roundness via radius requires a lot more parts. Basically enough to make a lathe and indicators so you can mount the centers and turn it while directly measuring the radius. You can also place it in V-blocks on a surface-plate and measure but the V-block needs to be the correct angle relative to the number of lobes so they seat properly or the measurement will miss them. Fortunately lathes are rather basic and simple machinery and make circular shapes to begin with. You don't encounter lobed shapes until you have more advanced machinery like centerless grinders. I suppose you could also place it vertically on a turntable if it has a flat squared end and indicate it and slide it around as you turn it to see if you can't find a location where the radius measures constant all the way around. Parallel: You might have asked yourself " Why do you need a square to measure roundness above? " The answer is that squares don't just let you check if something is square. They also let you check indirectly check the opposite: whether something is parallel. You need the square to make sure the the gauge block's top and bottom surfaces are parallel to each other so that you can place the gauge block onto the surface plate, then place a straight edge onto the gauge block such that the straight edge runs parallel to the surface plate. Only then can you measure the height of the workpiece as it, hopefully , rolls. Incidentally, this also requires the straight edge to be square which you can't know without having a square. More On Squareness: You can also now measure squareness of a physical object by placing it on a surface plate, and fixing a straight edge with square sides to the side of the workpiece such that the straight edge extends horizontally away from the workpiece and cantilevers over the surface plate. You then measure the difference in height for which the straight edge sits above the surface plate at both ends. The longer the straight edge, the more resolution you have, so long as sagging doesn't become an issue. From these basic measurements (square, round, flat/straight), you get all the other mechanical measurements. The inherent symmetries which enable self-checking are what makes "straight", "flat", "round", and "square" special. It's why we use these properties and not random arcs, polygons, or angles as references when calibrating stuff. Actually making stuff rather than just measuring: Up until now I mainly talked about measurement. The only manufacturing I spoke about was the surface plate and its very important self-referencing nature which allows it to make itself. That's because so long as you have a way to make that first reference from which other references derive, you can very painstakingly free-hand workpieces and keep measuring until you get it straight, round or square. After which you can use the result to more easily make other things. Just think about free-hand filing a round AND straight hole in a wood wagon wheel, and then free-hand filing a round AND straight axle. It makes my brain glaze over too. It'd also be a waste since you would be much better off doing that for parts of a lathe which could be used to make more lathes and wagon wheels. It's tough enough to file a piece of steel into a square cube with file that is actually straight, let alone a not-so-straight file which they probably didn't always have in the past. But so long as you have a square to check it with, you just keep correcting it until you get it. It is apparently a common apprentice toolmaker task to teach one how to use a file. Spheres: To make a sphere you can start with a stick fixed at one end to draw an arc. Then you put some stock onto a lathe and then lathe out that arc. Then you take that work piece and turn it 90 degrees and put it back in the lathe using a special fixture and then lathe out another arc. That gives you a sphere-like thing. I don't know how sphericity is measured especially when lobed shapes exist (maybe you seat them in a ring like the end of a hollow tube and measure?). Or how really accurate spheres, especially gauge spheres, are made. It's secret, apparently. EDIT : Someone mentioned putting molten material into freefall and allow surface tension to pull it into a sphere and have it cool on the way down. Would work for low tech production of smaller spheres production and if you could control material volume as it was dropped you can control size. Still not sure how precisely manufactured spheres are made though or how they are ground. There doesn't seem to be an obvious way to use spheres to make more spheres unlike the other things. | {
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615,238 | For elements where 'liquid', is relatively easy to define, at which temperature are the most elements liquid, and which ones? Assume 1 atm | We take the natural elements (atomic numbers $1$ to $92$ ) to have well-defined melting and boiling points. Additionally, all figures are quoted at the standard $1\text{ atm}$ pressure. Here's some quick Python code that fetches the number of elements in the liquid state at various temperatures (note that it relies on the values of the melting and boiling points of elements defined in the mendeleev package , however it is straightforward to instead use your own dataset). from mendeleev import element
elements = [element(i) for i in range(1, 92 + 1)] # Hydrogen to Uranium
n_liquid_list = [sum(element.melting_point < temp < element.boiling_point for element in elements) for temp in range(0, 5000)]
n_liquid_max = max(n_liquid_list)
temperature = n_liquid_list.index(n_liquid_max)
print(n_liquid_max, "elements are in the liquid state at", temperature, "K") prints 38 elements are in the liquid state at 2161 K For the full list, just add in for elem in elements:
if elem.melting_point < temperature < elem.boiling_point:
print(elem.name, end=", ") which gives you Beryllium, Aluminum, Silicon, Scandium, Titanium, Vanadium, Chromium, Manganese, Iron, Cobalt, Nickel, Copper, Gallium, Germanium, Yttrium, Zirconium, Palladium, Silver, Indium, Tin, Lanthanum, Cerium, Praseodymium, Neodymium, Promethium, Gadolinium, Terbium, Dysprosium, Holmium, Erbium, Thulium, Lutetium, Platinum, Gold, Actinium, Thorium, Protactinium, Uranium which broadly fall under transition metals, lathanides and actinides. As noted by ChrisH, there is an additional such temperature range starting at $2584\text{ K}$ , in which the liquid elements are: Beryllium, Boron, Aluminum, Silicon, Scandium, Titanium, Vanadium, Chromium, Iron, Cobalt, Nickel, Copper, Gallium, Germanium,
Yttrium, Zirconium, Technetium, Ruthenium, Rhodium, Palladium, Lanthanum, Cerium, Praseodymium, Neodymium, Promethium, Gadolinium, Terbium, Dysprosium, Holmium, Erbium, Lutetium, Hafnium, Platinum, Gold, Actinium, Thorium, Protactinium, Uranium I also thought it might be interesting to plot the total number of elements in the liquid state at a given temperature as a function of temperature: The first peak corresponds to the temperature range $2161\text{ K} - 2219 \text{ K}$ while the second peak corresponds to $2584\text{ K} - 2627 \text{ K}$ . | {
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615,268 | I am having trouble understanding how the unsteady flow equation is formulated. Why does it say that the RHS of the equation(change of energy in system) is made up of internal energy, kinetic energy and potential energy? Where is the flow energy? Shouldn't enthalpy $(u+pv)$ be considered? Note i am talking about the RHS of equation specifically. I understand how the energy components from mass in/out comes about. I noticed some people mentioned that the flow work is 0 in the system as there is no boundary work on the system but pressure can change in unsteady flow which means that change in flow work $(pv)$ should be accounted for right? Where am i going wrong with my concept. I hope i am not blabbering too much and being too confusing. | We take the natural elements (atomic numbers $1$ to $92$ ) to have well-defined melting and boiling points. Additionally, all figures are quoted at the standard $1\text{ atm}$ pressure. Here's some quick Python code that fetches the number of elements in the liquid state at various temperatures (note that it relies on the values of the melting and boiling points of elements defined in the mendeleev package , however it is straightforward to instead use your own dataset). from mendeleev import element
elements = [element(i) for i in range(1, 92 + 1)] # Hydrogen to Uranium
n_liquid_list = [sum(element.melting_point < temp < element.boiling_point for element in elements) for temp in range(0, 5000)]
n_liquid_max = max(n_liquid_list)
temperature = n_liquid_list.index(n_liquid_max)
print(n_liquid_max, "elements are in the liquid state at", temperature, "K") prints 38 elements are in the liquid state at 2161 K For the full list, just add in for elem in elements:
if elem.melting_point < temperature < elem.boiling_point:
print(elem.name, end=", ") which gives you Beryllium, Aluminum, Silicon, Scandium, Titanium, Vanadium, Chromium, Manganese, Iron, Cobalt, Nickel, Copper, Gallium, Germanium, Yttrium, Zirconium, Palladium, Silver, Indium, Tin, Lanthanum, Cerium, Praseodymium, Neodymium, Promethium, Gadolinium, Terbium, Dysprosium, Holmium, Erbium, Thulium, Lutetium, Platinum, Gold, Actinium, Thorium, Protactinium, Uranium which broadly fall under transition metals, lathanides and actinides. As noted by ChrisH, there is an additional such temperature range starting at $2584\text{ K}$ , in which the liquid elements are: Beryllium, Boron, Aluminum, Silicon, Scandium, Titanium, Vanadium, Chromium, Iron, Cobalt, Nickel, Copper, Gallium, Germanium,
Yttrium, Zirconium, Technetium, Ruthenium, Rhodium, Palladium, Lanthanum, Cerium, Praseodymium, Neodymium, Promethium, Gadolinium, Terbium, Dysprosium, Holmium, Erbium, Lutetium, Hafnium, Platinum, Gold, Actinium, Thorium, Protactinium, Uranium I also thought it might be interesting to plot the total number of elements in the liquid state at a given temperature as a function of temperature: The first peak corresponds to the temperature range $2161\text{ K} - 2219 \text{ K}$ while the second peak corresponds to $2584\text{ K} - 2627 \text{ K}$ . | {
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615,304 | My textbook states, 'The sound box has a large area, it sets a large volume of air into vibration, the frequency of which is same as that of the string. So due to resonance a loud sound is produced.' My question is why isn't the air around the string (apart from the hollow body) resonating or resonating sufficiently enough to produce a loud sound without the need for a hollow body? Isn't the hollow body's volume of air negligible compared to 'room/area' where the string is played? | Yes, the room has a lot of air, but most of it isn't in direct contact with the vibrating string. In order for the string to make much sound, it needs to transfer some of its energy to the surrounding air, but there are a couple of issues which make that difficult. Firstly, the string has a relatively small surface area, so it simply doesn't directly contact very much air. Secondly, the string isn't very effective at transmitting its vibrations to the surrounding air. It's a bit like if you get a thin rod and wave it back and forth in water. Yes, it makes a few ripples, but the rod mostly just cuts through the water. But if you get a wide paddle and wave it around, then it's easy to make large waves in the water. Similarly, a vibrating string mostly just cuts through the air, rather than making useful pressure waves in it. With air, it's even harder to effectively induce waves than in water, because it's so compressible. Obviously, pressure waves do travel through the air, but it's a lot harder to transmit such waves through a gas than through a denser, stiffer medium. So stringed instruments generally have some kind of soundboard: a surface with a relatively large area that's in contact with a lot of air. The string can efficiently transmit its vibrations to the soundboard, and the soundboard can, in turn, transmit those vibrations to the air. On a guitar, the soundboard is the front of the body. Here's a diagram, courtesy of Wikipedia : The soundboard of a piano is inside the body: ( source ) An important concept here is acoustic impedance . Acoustic impedance and specific acoustic impedance are measures of the opposition that a system presents to the acoustic flow resulting from an acoustic pressure applied to the system. [...] There is a close analogy with electrical impedance, which measures the opposition that a system presents to the electrical flow resulting from an electrical voltage applied to the system. Unfortunately, that Wikipedia article is heavily technical and mathematical, without much physical detail. The soundboard of a musical instrument, or the membrane of a loudspeaker, doesn't need to have a lot of back and forth movement. It just needs to vibrate relatively gently. Large movement wastes energy by causing bulk movement of the air, rather than inducing the desired pressure waves. | {
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615,412 | In a Rutherford-atom, the electron classically emits EM radiation on an average rate of, $$
-\frac{dE}{dt}=\frac {\omega^4 e^2 R_0^2}{3c^3(4π\epsilon_0)}
$$ Where $\omega$ is the angular frequency, $R_0$ is the initial orbit-radius and the electron will ultimately spiral into the nucleus. Now, as far as I know, the entire phenomenon can be explained via Maxwell equations. Building on the same, and assuming a set of gravitoelectromagnetic equations to be valid (which is a weak field approximation) for masses, can't we say that the Earth itself should one day spiral into the sun, radiating continuously at a rate: $$
-\frac {dE}{dt} = \frac {\omega^4 G M_e M_s R_e^2}{3c^3}$$ Where symbols have their usual meanings? The mean decay time comes to be $4.75 × 10^{11}$ years. | Good question!
Indeed, there is an analogous phenomenon in GR to the emission by a Rutherford atom, but your formula does not describe it: as written, it would apply to dipolar emission of gravitational radiation, but
gravitational emission only starts at the quadrupole order of the perturbative expansion, which is suppressed by a factor $(v/c)^2$ compared to the dipole. The reason for this, in Newtonian terms, is that when differentiating the gravitational dipole you find the total momentum of the emitting system, which is conserved, so the second derivative of the gravitational dipole vanishes (see here ) --- this reflects the fact that while in electromagnetism we only have one conservation law for the charge, in GR the full four-momentum is conserved. The formula describing the energy loss of a binary system due to quadrupolar emission can be found in this Wikipedia article ; note the $c^5$ term in the denominator as opposed to your $c^3$ .
There is also a hidden factor of $\omega^6$ (as opposed to $\omega^4$ ) in the Wikipedia formula: an alternative way to express the radiated power (averaged over a period) is $$
- \left\langle \frac{\mathrm{d} E}{\mathrm{d} t} \right\rangle = \frac{32}{5} \frac{G \mu^2}{c^5} R^4 \omega^6\,,
$$ where $\mu = (1/M_s + 1/M_e)^{-1}$ is the reduced mass of the Earth-Sun system. This reflects the higher power of $v$ . To give a sense of the difference, I've plugged some numbers in and your expression would predict a rate of energy loss of about $3 \times 10^{14} \,\mathrm{W}$ in the current Sun-Earth configuration, while the correct GR expression predicts $2 \times 10^{2} \,\mathrm{W}$ . | {
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616,039 | There are two ways to orient the toilet paper: "over" (left image), "under" (right image). Each has it's pros and cons. For some reason, it's always easier to tear off the paper in the "over" orientation even though we apply the same force thus same torque. I assume these rolls are almost perfect cylinders (no excess of mass in one orientation compared to other) thus they have the same moment of inertia (degree of difficulty on changing the angular velocity of the object around the longitudinal axis of the roll) therefore both ways should be equivalently easy but it's not. Conjectures Me finding it easier to tear off the paper in "over" orientation doesn't really mean that it is, my perceptions might be deluding me. That's why I checked the statistics and most people seem to agree with me. It's due to the orientation of my hand: I can grab the paper better in "over" orientation, hence I can apply a uniformly distributed force which makes the paper to be torn off easier compared to "under" orientation. People finding "over" orientation easier are mostly right-handed people ( $70\%$ ) and, those who find "under" orientation easier are mostly left-handed people ( $30\%$ ). If this were the case, it may be the explanation for the results seen in a survey of Cottonelle in $1999$ , which states that $68\%$ of the respondents prefer "over" orientation over $25\%$ "under" orientation. I wonder if any of these above conjectures are true or if there's a more physical explanation why it's easier to tear off the toilet paper in "over" orientation compared to "under". | I'll propose a theory, and I'll describe an experiment I did to test it. Both suggest that the "over" configuration is better, at least if the goal is to make the squares easier to rip off with one hand without making the roll spin out of control. Terms of use Please don't use this post as ammunition to defend a preference. As rightly emphasized in several good comments (moved to chat), the specific issue addressed in this answer is not the only issue or even the most important one, and it assumes specific conditions that probably don't represent most real-world situations. I only posted it because I thought the experimental results were amusing. After seeing how much attention this answer has been getting, I decided to add this terms-of-use clause in case the answer's lighthearted intent wasn't already clear from the writing style. :) Toilet paper physics: theory The diameter of the cylindrical hole in the toilet paper is larger than the diameter of the axle on which it rotates, as illustrated in the pictures below. So when the toilet paper is at rest, only the top of the hole is in contact with the axle. The bottom and sides of the hole are not in contact with the axle. That matters because when the toilet paper is in the "over" configuration, you can tear off a square by yanking it straight down. Since the top of the hole is already in contact with the axle, the roll's resistance to the yank is immediate, and the square rips off easily. In contrast, when the toilet paper is in the "under" position, the loose square is hanging close to the wall. In that case, yanking straight down is hard to do without scraping your hand against the wall. We can get around that by yanking at an angle (or horizontally) instead, but then the roll's resistance is not so immediate because the opposite side of the hole is not in contact with the axle when the yank begins. The roll must move laterally to bring the opposite side of the hole into contact with the axle. That delays the rip and increases the tendency for the roll to spin, which tends to wind the paper up around the roll in addition to (or instead of!) ripping it off. Toilet paper physics: experiment I tested this by removing the axle from the fixture and holding it in one hand while using the other hand to yank on the loose end of the paper. I tried all combinations of the following options: Hold the axle with the left hand and yank with the right, or hold the axle with the right hand and yank with the left. Hold the roll in the "under" configuration, with the paper hanging on the opposite side from the hand I used to yank the square, or hold the roll in the "over" configuration, with the paper hanging on the same side as the hand I used to yank the square. Yank straight down, or yank at a near-horizontal angle. In both cases, the angled yank was toward the side with the hand I used to yank (toward the left when yanking with my left hand, and toward the right when yanking with my right hand). In the "under" configuration, this meant yanking toward the opposite side from the side on which the paper was hanging. In the "over" configuration, this meant yanking toward the same side from the side on which the paper was hanging. Here are the results: Didn't matter which hand I used to hold the axle or yank the paper, even though I'm not ambidextrous. In both the "over" and "under" configurations, yanking straight down almost always ripped off the square without spinning the roll significantly at all. (I tried a few times in each configuration, and there were only a couple of exceptions overall.) In the "over" configuration, yanking at an angle still almost always ripped of the square without spinning the roll significantly, and the number or exceptions was only slightly greater than when yanking straight down. In the "under" configuration, yanking at an angle almost always made the roll spin at least one full revolution, unwinding much more paper than intended, and it often failed to separate the square at all. Here's a graphic summary: These results seem mostly consistent with the theory described above, and they suggest a refinement. When yanking at an angle in the "over" configuration, the straight segment of the toilet paper is tangent to the upper part of the roll before the yank, close to the point where the hole is resting on the axle. When yanking at an angle in the "under" configuration, the straight segment of the toilet paper is tangent to the lower part of the roll before the yank, far from the point where the hole is resting on the axle. As a result of the longer lever-arm, yanking at an angle in the "under" configuration applies a larger torque (hence a stronger tendency to spin the roll) compared to yanking at an angle in the "over" configuration, as observed in the experiment. Other things to try What if the roll's hole fits snugly onto the axle, so the hole is in contact with the axle around its whole circumference? The theory predicts that this should eliminate the differences between the "over" and "under" configurations. I didn't test this. I always tried to yank the square orthogonal to the perforated line. I didn't try oblique yanks, to make the rip propagate more gradually from one side of the perforated line to the other. The proposed theory doesn't explain the correlation with right- and left-handedness cited in the question. This answer only considered one goal: ease of removing a square with one hand without causing the roll to spin out of control. Different goals might be better served by different configurations. Disclosure of conflict of interest I've always preferred the "over" configuration. I tried to conduct the tests fairly, because I really was curious, but I suppose my prior preference could have had some subconscious influence on how I did the tests. | {
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616,258 | I was walking outside one cold afternoon with my mask on and my glasses began fogging up. The mist was initially gray. I kept walking without cleaning my glasses and eventually enough mist collected that that it transformed into clear water droplets. This got me thinking: why is mist gray but water clear? Or perhaps more specifically, why are smaller water droplets gray and larger droplets clear? I couldn't find any explanation online. What is the physics behind such shenanigans? | Mist is a suspension of tiny water droplets in air . Light traveling through the mist gets randomly scattered, mainly by bouncing of the droplets. That makes mist far less transparent than bulk water. I don't think mist is literally gray in colour but the fact that mist is far less transparent than pure air (or bulk water) causes it to look the way it does. Other suspensions like smoke (a suspension of tiny solid particles in air ) look quite similar, due also to light scattering. Another example is very much diluted milk (an emulsion of fat droplets in water , mainly). | {
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616,276 | We are told that in a 2D CFT, all primary operators can be written in the form of \begin{equation} \tag{1}
\mathcal{O}(z, \bar{z} ) = \sum_{m,n \in \mathbb{Z} } \frac{\mathcal{O}_{m,n} }{z^{m + h} \bar{z}^{n + \bar h}}
\end{equation} Recently I have been wondering about what happens when $h$ and $\bar{h}$ are not integers . (For instance, I wonder how $\mathcal{O}(0,0)$ when acting on the ground state $|0\rangle$ is not trivially $0$ .) In order to look at a specific example, I tried to see what the mode expansion of the operator \begin{equation}
\mathcal{O}(z, \bar{z}) = \; :e^{ i k_\mu X^\mu (z, \bar z)} :
\end{equation} is in the free boson CFT. When this operator acts on the ground state $|0\rangle$ when $z = \bar{z} = 0$ , it creates a state of weight \begin{equation}
(h, \bar{h} ) = \left( \frac{\alpha' k^2}{4}, \frac{\alpha' k^2}{4} \right)
\end{equation} from equation 2.4.17 in Polchinski. As $h$ and $\bar{h}$ aren't integers, I therefore tried to see what the mode expansion of this operator is. However, it doesn't quite seem to take the form of $(1)$ , which makes me wonder if all operators really can be expressed as $(1)$ . I present my work below. Equation 2.7.4 in Polchinski reads \begin{equation}
X^\mu(z, \bar{z}) = x^\mu - i \frac{\alpha'}{2} p^\mu \ln(|z|^2)+ i \left( \frac{\alpha'}{2} \right)^{1/2} \sum_{m \in \mathbb{Z} - \{0\} } \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha_m^\mu} }{\bar{z}^m} \right).
\end{equation} Note that $\alpha^\mu_m$ is an annihilation operator for $m > 0$ and a creation operator for $m < 0$ . Normal ordering pulls all the annihilation operators to the right and all the creation operators to the left. In this procedure $p^\mu$ is regarded as an annihilation for this procedure and $x^\mu$ is regarded as a creation operator. (The normal ordering proscriptions are related by equation 2.7.12 in Polchinski.) We now find \begin{align}
&:e^{ i k_\mu X^\mu (z, \bar z)} : \\
&= e^{i k_\mu x^\mu} \exp(\sum_{m<0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) )
e^{ \alpha' k_\mu p^\mu \ln|z|}
\exp(\sum_{m>0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) )
\end{align} As brief check, note that \begin{align}
p^\mu |0\rangle &= 0 \\
\alpha^\mu |0\rangle &= 0 \hspace{0.5 cm} m > 0
\end{align} so \begin{equation}
: e^{i k_\mu X^\mu(0,0)} : |0\rangle = e^{i k_\mu x^\mu} |0\rangle = |k; 0\rangle
\end{equation} as expected. We now use the commutation relation (2.7.5b in Polchinski) \begin{equation}
[x^\mu, p^\nu] = i \eta^{\mu \nu}
\end{equation} and Baker-Campbell-Hausdorff to write \begin{align}
e^{i k_\mu x^\mu} e^{\alpha' k_\nu p^\nu \ln|z|} &= \exp( i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z| + \tfrac{1}{2} [i k_\mu x^\mu , \alpha' k_\nu p^\nu \ln|z| ] ) \\
&= \exp( i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z| - \tfrac{1}{2} \alpha' k^2 \ln|z| ) \\
&= \exp( i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z| ) |z|^{ - \tfrac{1}{2} \alpha' k^2 }
\end{align} making \begin{align}
&:e^{ i k_\mu X^\mu (z, \bar z)} : \\
&= e^{i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z|} \\
&\exp(\sum_{m<0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) )
\exp(\sum_{m>0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ) \frac{1}{z^{\alpha' k^2/4 }}\frac{1}{{\bar z}^{\alpha' k^2/4 }}
\end{align} This is a somewhat confusing answer. It almost looks to be of the form \begin{equation}
:e^{ i k_\mu X^\mu (z, \bar z)} : \; \stackrel{?}{=} \sum_{m \in \mathbb{Z} } \frac{\mathcal{O}_{m,n} }{z^{m + h} \bar{z}^{n + \bar h}}
\end{equation} but not quite, due to the $e^{i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z|} $ out front. Does anyone have any ideas? | Mist is a suspension of tiny water droplets in air . Light traveling through the mist gets randomly scattered, mainly by bouncing of the droplets. That makes mist far less transparent than bulk water. I don't think mist is literally gray in colour but the fact that mist is far less transparent than pure air (or bulk water) causes it to look the way it does. Other suspensions like smoke (a suspension of tiny solid particles in air ) look quite similar, due also to light scattering. Another example is very much diluted milk (an emulsion of fat droplets in water , mainly). | {
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616,307 | For example, when we put two objects of +10C and +20C together and then take them apart, each of them acquires a charge of +15C. In a nucleus, the protons and neutrons are stuck together. Why is it that then neutrons has no charge and why does 1/2 the charge of proton not be transferred to neutron? | If I put a red billiard ball and a blue billiard ball in a bag, leave them for a while, and then draw one out, I will find I am holding a red ball or a blue ball. Never a purple ball. At the level of fundamental particles, we know from experimental evidence that electric charge is a discrete quantity and behaves like the colour on the billiard balls, not like a continuous quantity such as temperature. This is shown in experiments such as Millikan's oil drop experiment . Why ? That’s just the way the world works. | {
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616,320 | I am doing project in blackbody radiation and i am about to analyse the spectral data of the incandescent lamp, hot metal, Sun and Sirius A. So i can find the displacement of the peak of curves using Wien's displacement I don't have the required apparatus to obtain the data but i heard that i can get such data from internet. So i searched. But unfortunately I am noob at this so i couldn't get any data. If anyone know anything about it please help me to get the data | If I put a red billiard ball and a blue billiard ball in a bag, leave them for a while, and then draw one out, I will find I am holding a red ball or a blue ball. Never a purple ball. At the level of fundamental particles, we know from experimental evidence that electric charge is a discrete quantity and behaves like the colour on the billiard balls, not like a continuous quantity such as temperature. This is shown in experiments such as Millikan's oil drop experiment . Why ? That’s just the way the world works. | {
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616,364 | If you pour water from a container, the flowing water stream seems to cast a shadow. I am not sure you can call it a shadow, but it definitely is not letting all light through it. How is this possible and what uses can it have? | A large amount of water (i.e. if the path of light through it is long) will simply start absorbing light, as it's not completely transparent. For smaller amounts, as when pouring it from one container to another, this is mostly negligible. However, there is also surface reflection. A small amount of the incident light will be reflected off by the surface. The much larger contribution, however, will come from refractive effects. If you look closely, there are not only areas which are darker than the uniformly lit surroundings, but some will also be brighter. The stream of water forms shapes that act similar to a lens and will divert light off its original path. The patches where the incident light would have gone without the disturbance will then be darker, the places where it is directed to instead will be brighter. | {
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617,113 | Whenever I wash my thermos, I put hot water and then some soap in; then I seal the one end with my hand or use the lid. After shaking it up, if I slowly remove the lid or my hand, it expels a little air. Why is that? Does it have something to do with increased surface area of soapy water? Or is it the fact that the air is heated by the water, even though the water must surely cool slightly? | When you pour the hot water in, the air inside the thermos is still quite cold (ambient temperature, approx.) But then when you shake it up the cold air is heated by the hot liquid. Gases expand considerably when heated, approximately acc. the Ideal Gas Law: $$pV=nRT$$ This causes a modest (and harmless) pressure increase in the flask, which is what you experience. | {
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617,514 | There are a few questions only on this site about this but none of them answer my question. Can cannonballs go through water? Why does a bullet bounce off water? I find it hard to understand why bullets shoot through water at longer distances but stop in sand almost right away: Both water and sand are made up of smaller droplets/grains and both are relatively heavy elements (sand is only 1.5 times heavier per volume). Water molecules are bound by Van der Waals force into droplets, sand molecules are bound by covalent bonding into crystals At slow speed, I can put my hand through water and sand both. The droplets and grains can roll over and accommodate an object easily. At high speed, an airplane crashing onto water will fall into pieces because water acts in this case like a solid, because the molecules and droplets don't have enough time to rearrange to accommodate the object. Same with sand. Now in the case of a bullet, this argument seems not to work. In air, bullets reach speeds over 1800 mph. Bullets penetrate water, and can keep high speeds up to 10feet. On the other hand, bullets can't penetrate sand at all, they stop completely almost with no real penetration. Bullets can keep high speeds up to 10 feet in water. https://mythresults.com/episode34 Bullets in sand are completely stopped after 6 inches. https://www.theboxotruth.com/the-box-o-truth-7-the-sands-o-truth/ Question: Why do bullets shoot through water but not through sand? | The sand particles interact on a macroscopic level different from water. The edges can lock together and more efficiently distribute force. Water molecules, being part of a liquid, do not distribute force this way. They tend to move out of the way instead. Mr Wizard has a great demonstration of this , using a plunger and salt rather than bullets and sand. I recommend this experiment because it takes these effects away from bullets (which can be difficult and dangerous to experiment with), and brings it down to Earth, demonstrating the effect with a hand plunger. It's easy to visualize that water would not stop the plunger from going through and piercing the tissue paper (probably have to switch to wax paper, given the water), and you can play with different variants safely until it makes more sense. And then, when it all makes sense, take a look at fluidizes sand , and the physics will go back to being magical again! | {
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618,165 | I want to compute the commutation relation between a function of momentum and a function of space in Quantum Mechanics. I know the commutation relation between momentum and a function of space but how can I compute the commutation relation between functions of momentum $p$ and space $x$ ? Some thing such as $[f(x),g(p)]$ | There are dozens of tricks to do this, but most of them reside in phase-space quantization, with which you are likely unfamiliar if you are asking this, so I'm staying out of it. Let's skip the hats for operators, set $\hbar=1$ for simplicity, non-dimensionalizing, and work in the coordinate representation, so, essentially, p is shorthand for $p=-i\partial_x$ . You are then seeking $[g(p),f(x)]$ .You already said you have $$
[p,f(x)]= -i f'(x),
$$ from which you may compute the following $$
[p^2,f(x)]=p[p,f]+ [p,f]p \\ = -ipf'(x) -i f'(x) p = -f''(x)-2if'(x)p,
$$ and likewise for all monomials in p , whence all polynomials, by linearity of the commutator. I assume you are familiar with the combinatoric tricks for monomials, which mathematicians like, but this is a bit barking up the wrong tree; most of them don't get their hands dirty in our messy world, nor have they any interest in that. The key point is to convince your self that, for a c-number y (not operator like p ), $$
[e^{iy p}, f(x)] = \bigl(f(x+y) -f(x)\bigr) e^{iyp},
$$ by virtue of the fundamental Lagrange shift , with which you are familiar from the translation operator. But every function g(p) is a linear combination of such exponentials, by virtue of its Fourier transform, $$
g(p) = {1\over \sqrt{2\pi}}\int\!\! dy ~~e^{iyp} \tilde g(y)~~~~\leadsto \\
\tilde g(y)= {1\over \sqrt{2\pi}}\int\!\! dp ~~e^{-iyp} g(p).
$$ Consequently $$
[g(p),f(x)]= {1\over \sqrt{2\pi}}\int\!\! dy ~~ \tilde g(y)~~[e^{iyp},f(x)]\\= {1\over \sqrt{2\pi}}\int\!\! dy ~\bigl (f(x+y)-f(x)\bigr )~ \tilde g(y) e^{iyp},
$$ a numerical integral of functions in coordinate space when acting on a constant or translationally invariant state. | {
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618,180 | An electromagnetic wave is typically shown with this schematic
(taken from https://en.wikipedia.org/wiki/Electromagnetic_radiation ) We see graphically from the schematic that the amplitude of E and B is evolving with position and time. Thus, should we consider that the electromagnetic wave has a lateral extension? Or is there no lateral extension: that is, we should consider the electromagnetic wave as a pure mathematical line (not as shown in the schematic) where the electric field has a different value depending on the position on this propagation line? | In the image included in your question, one should not view the arrows as literally extending through space or having a spatial length associated to them. The length of these arrows is the magnitude of the field, and they should each be thought of as living inside of their own vector space which is "internal" or "attached" to a given point in space. Your last statement is essentially correct, there is no "lateral extension". As a very simple analogy, imagine a particle moving through space along some trajectory with coordinates $\gamma(t) = \{x(t),y(t),z(t)\}$ . The trajectory traces out of a one-dimensional curve through space. At every point on this curve, we can talk about the velocity vector , $\mathbf{v}(t) = \dot{x}(t)\hat{\mathbf{x}}+\dot{y}(t)\hat{\mathbf{y}}+\dot{z}(t)\hat{\mathbf{z}}$ . We often draw a picture of the curve and its associated velocity vector as if the velocity vector had a length in space, for example like this: Does the velocity vector literally represent something with a spatial extent? Of course not. We should understand the velocity vector as living in the tangent space $T_{\gamma(t)}E$ at the point $\gamma(t)$ , where $E$ is the spatial manifold, usually taken to be Euclidean space. There is a separate tangent vector space for every point in space, and these are the spaces where each of the velocity vectors live for each point of the trajectory. Note that the direction that the arrows point has a meaning with respect to space, but their length only tells you the magnitude of the velocity (i.e. the speed). We don't think of the velocity vectors as representing spatial displacement vectors (they can't since a velocity vector has magnitude measured in units of length/time while a displacement vector has a magnitude measured in units of length). In fact, the electromagnetic field should properly be understood as living in the space of 2-forms at each point of space, $\bigwedge^2 T_pM$ for each point $p$ in $M$ , where $M$ is 4-dimensional Minkowski space. The basic idea is that the "fields" we talk about in physics (including quantum fields) take a value at every point in space(time), and the value they take lives in an "internal space" at each point. This notion is formalized in the idea of a fiber bundle , where the fibers are the "internal spaces" connected to each point of space. Addendum Remember that the field takes independent values at every point in space. I made a little animation of a plane wave to show this. Note that in the image below each little arrow represents the local field strength and direction, but these arrows each live in the internal space. I can only show a finite number of arrows, but there are more arrows filling all of space in the gaps. Try focusing on just a single arrow. | {
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618,248 | When I was studying motion, my teacher asked us to derive the equations of motion. I too ended up deriving the fourth equation of motion, but my teacher said this is not an equation. Is this derivation correct? \begin{align}
v^2-u^2 &= 2ax \\
(v+u)(v-u) &= 2 \left(\frac{v-u}{t}\right)x \\
(v+u)t &= 2x \\
vt+ut &= 2x \\
vt+(v-at)t &= 2x \\
2vt-at^2 &= 2x \\
x &= vt- \frac{at^2}{2}
\end{align} And why is this wrong to say that this is the fourth equation of motion? Given 3 equations of motion:- \begin{align}
v&=u+at \\
x&=ut+ \frac{at^2}{2} \\
v^2-u^2&=2ax
\end{align} | The problem of perception as to "What is a new Equation of Motion?" seems to originate with the dogmatic teaching of the Three Equations Of Motion as a Set of Results to be Learned. They are in fact three results derived from the distillation of Newton's Laws: $$\mathbf f = \dfrac {\mathrm d} {\mathrm d t} (m \mathbf v)$$ which differential equation is solved, where $\mathbf f$ is set to a constant (and $m$ is taken for granted as being constant also). This of course can't be done at the elementary level at which the SUVAT equations are initially introduced. So the three convenient "equations of motion" are introduced instead, in a way that the students can get their heads round them. Whether an equation is given an official Name to Be Remembered is not all that important. What is important is the ability to use them. Working out that fourth equation from the given three is actually a worthy exercise in its own right. Granted it is not a particularly profound equation, as it can be obtained from the other three. But -- get this -- each of the other three has also merely been derived from other equations. For your teacher to dismiss it as "not an equation" is appalling. It may be the case that the teacher is teaching from the book, and not from his or her own expertise in the subject. It can be disheartening to be taught by teachers who do not understand the subject they are teaching, but hang on in there, it gets better as you go on in your schooling. | {
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618,264 | I have a doubt in understanding the meaning of this statement written on my homework. The components $F_1$ and $F_3$ are equal along the u-axis Does the statement mean that the x and y components of those two forces are really equal? | The problem of perception as to "What is a new Equation of Motion?" seems to originate with the dogmatic teaching of the Three Equations Of Motion as a Set of Results to be Learned. They are in fact three results derived from the distillation of Newton's Laws: $$\mathbf f = \dfrac {\mathrm d} {\mathrm d t} (m \mathbf v)$$ which differential equation is solved, where $\mathbf f$ is set to a constant (and $m$ is taken for granted as being constant also). This of course can't be done at the elementary level at which the SUVAT equations are initially introduced. So the three convenient "equations of motion" are introduced instead, in a way that the students can get their heads round them. Whether an equation is given an official Name to Be Remembered is not all that important. What is important is the ability to use them. Working out that fourth equation from the given three is actually a worthy exercise in its own right. Granted it is not a particularly profound equation, as it can be obtained from the other three. But -- get this -- each of the other three has also merely been derived from other equations. For your teacher to dismiss it as "not an equation" is appalling. It may be the case that the teacher is teaching from the book, and not from his or her own expertise in the subject. It can be disheartening to be taught by teachers who do not understand the subject they are teaching, but hang on in there, it gets better as you go on in your schooling. | {
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618,266 | I came across $S^{ij}$ when studying GR and Riemann Curvature. $$S^{ij} = \oint_{\text{loop on the surface}} x^i dx^j - x^j dx^i \approx \text{Area Enclosed By The Loop.}$$ What is $S^{ij} S_{ij} ?$ And does this mean something like "area" in any coordinates or does this have to be done in cartesian coordinates to give "area". I used this to parameterize a circle and find the area. Then I calculated it in polar coordinates by doing the integral such that $x=r$ and $y=\theta$ and by transforming the tensor and got the same answer, $2\pi r$ . And that's not an area. | The problem of perception as to "What is a new Equation of Motion?" seems to originate with the dogmatic teaching of the Three Equations Of Motion as a Set of Results to be Learned. They are in fact three results derived from the distillation of Newton's Laws: $$\mathbf f = \dfrac {\mathrm d} {\mathrm d t} (m \mathbf v)$$ which differential equation is solved, where $\mathbf f$ is set to a constant (and $m$ is taken for granted as being constant also). This of course can't be done at the elementary level at which the SUVAT equations are initially introduced. So the three convenient "equations of motion" are introduced instead, in a way that the students can get their heads round them. Whether an equation is given an official Name to Be Remembered is not all that important. What is important is the ability to use them. Working out that fourth equation from the given three is actually a worthy exercise in its own right. Granted it is not a particularly profound equation, as it can be obtained from the other three. But -- get this -- each of the other three has also merely been derived from other equations. For your teacher to dismiss it as "not an equation" is appalling. It may be the case that the teacher is teaching from the book, and not from his or her own expertise in the subject. It can be disheartening to be taught by teachers who do not understand the subject they are teaching, but hang on in there, it gets better as you go on in your schooling. | {
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618,997 | I recently read about the property of terminal velocity for objects and I got a question when doing so. If from a very tall building I throw a ball faster than terminal velocity downwards, will the ball slow down, then continue with terminal velocity, or will it continue with the speed I threw it with? | The ball will slow down to terminal velocity. This is because the force of air drag increases with increasing speed. Terminal velocity is the speed where the force of air drag equals the force of gravity, so the total force is zero and the object travels at a constant speed. If the ball has a higher speed, then it will have an air drag force greater than gravity and slow down. | {
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619,000 | Apparently there is a lack of resources regarding inertia in the web. I have already asked 2 similar questions in this site, but still I have a problem with inertia. My question - What is value of inertia of a box placed on the floor of an accelerating bus? I know the standard explanations; but most are qualitative. Please provide with a quantitative solution. P.S.- As @G. smith suggests, my question is about inertial or fictitious forces. | The ball will slow down to terminal velocity. This is because the force of air drag increases with increasing speed. Terminal velocity is the speed where the force of air drag equals the force of gravity, so the total force is zero and the object travels at a constant speed. If the ball has a higher speed, then it will have an air drag force greater than gravity and slow down. | {
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619,222 | I want to begin my explanation using abstract mathematical explanation to repetition possibility by taking independent samples $X_n$ from some continuous probability distribution: https://math.stackexchange.com/q/1739927/ If we applied this same principle and its conclusion and assumed that the universe is homogeneous and were able to prove its infinity someway (along with matter inside it including stars and planets) then, statistically, it would mean that the chance of repetition of life again, at least and in the most conservative approach here, as a form of doppelganger extraterrestrial life would be zero? meaning that we are alone in this universe | No, quite the opposite. If the universe is truly infinite and approximately homogeneous, then I invite you to point in some direction in the sky at night, and if you travel far enough in that direction, you might have to point through several stars and planets and alien organisms to get there, but eventually you might well be pointing at a nearly exact copy of yourself pointing back at you. With infinite repetition things that are possible can become probable and things that are probable can become certain. Not everything works this way, and it does depend on how “identical” your setups are as you repeat. But the basic math is like this: imagine we model a bridge but we ignore continuous “damage” to it and treat its failure instead as an all-or-nothing thing which has an 0.01% chance of failing in any given day, independent of any other day. One can derive that this bridge will live about 10,000 days or 27 years, plus or minus another 27. It actually in the continuous limit has a 50/50 chance of failing apart after $10000~\ln2$ days or 19 years. So those independent little chances of something happening, under many many repetitions, eventually lead to this very unlikely thing (one in ten thousand) becoming probable at $10000~\ln2$ repetitions and becoming 99.995% likely after a hundred thousand repetitions. Very simple math. And all that the above statement is doing is the same argument, extrapolating an Earth-sized cylinder through space and saying “hey, that this Earth is roughly the way it is (up to your uncertainty) is some unbelievably tiny probability $p$ , but let’s chain those cylinders together in the direction you're pointing until we get $10/p$ repetitions and I suppose if you're right that everything is infinite and homogeneous then we'll not run out of universe before $10/p$ repetitions, even for small $p$ .” The “(up to your uncertainty)” is also very important here, because it is what makes the probability finite rather than infinitesimally zero. Mathematicians deal with say real numbers being perfectly specified, so 1.00125267... is very different from 1.00125264... but us physicists have to say “look my experiment has an 0.2% error and I can't resolve the difference between 1 and 1.002, so all of these are approximately 1.000 up to my error bars.” Actually, something that bugs some people is the idea of “Boltzmann brains.” This is the idea that generating a brain randomly is much easier than generating a whole body and Earth and brain all together, so that if the world really does proceed to a state of maximum entropy, a thermal equilibrium, then there is some sort of recurrence time (in the simplest case a Poincaré recurrence) where the thermal equilibrium randomly generates a tiny bit of brain like yours perceiving the things that you are perceiving but the vast vast majority of times this happens that brain is totally and utterly detached from the real world and vanishes after a moment. And the reason that this application of infinity bugs people is that it means that averaging over all the entities in our universe which are having the conscious experience that we are having of staring at our computer screens on a physics website pondering the universe, almost none of them are actually living on an Earth and almost none of them are going to last longer than the next five seconds and almost none of them are therefore actually seeing anything like the real world, they are just in a very temporary hallucination. And that brings up questions about why we are so sure that we are not. | {
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619,249 | Earth core temperature is range between 4,400° Celsius (7,952°
Fahrenheit) to about 6,000° Celsius (10,800° Fahrenheit). Source Why can't the Earth's core melt the whole planet? In other words, what is stopping Earth from being melted up to its surface? | First thing to notice is that the heat flow is limited, so the heat from the core does not flow to the surface instantaneously. Second point is that the surface of the Earth radiates energy to the space.
The combination of these effects makes it possible to have a molten core but a cold surface. | {
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619,987 | Imagine a system of two objects: a supermassive black hole mass of approximately $10^{36}$ kg and another $1$ kg object revolving around the black hole with an average mean distance of about $1.5×10^{8}$ m. The angular velocity of the smaller object is $2$ rad/s and correspondingly the linear velocity is equal to the speed of light. But how could this be? I know this might be wrong, but where? | A black hole of mass $10^{36}$ kg would have a Schwarzschild radius (the distance from the center to the event horizon) of about $1.5\times 10^{9}$ m. So your choice of "orbital" distance is inside the black hole, and it won't orbit, it will just fall inward. Outside the event horizon there is a minimum distance for a stable circular orbit which is three times the Schwarzschild radius . You cannot use Newtonian physics for these calculations; you need full blown general relativity. If you try mixing these two systems together you get nonsense. It is only at distances where the gravitational field is weak that you can apply Newtonian physics with good accuracy. Regarding why we are so confident that we know what happens inside a black hole (like everything falling inward) when we can never test that directly it comes down to the way the theory is designed. The theory for what happens outside the event horizon and inside the event horizon are the same theory and as we have verified that theory for outside the event horizon to a very, very high level of confidence so there is a very high level of confidence in it's predictions for inside the event horizon. The basic core the theory of general relativity is that physics must work the same way everywhere . The event horizon doesn't make any difference in this sense. | {
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620,137 | I admit this sounds like a dumb question. I know very little about physics, but I've always wondered that given the fact that: A unit of measurement is a very specific constant value. Then how can a formula like: $$E=mc^2$$ Be so precise and so clean when it has to use already pre-defined unit of measurements as values.
I assume (I'm googling this and I know almost no physics, just what I remember from high-school)
That this formula claims that $1 \text{joule} = 1 \text{kg} * (\text{speed of light}) ^ 2 \frac{m}{s}$ The meter, the kilogram, the second, the joule (composed of the base units) and so on are very very specific values defined by the International System of Units.
How can one find a law, that just so happens to find a relationship between these exact values, without needing a custom made constant value or set of values to multiply or add by? Edit: So I understand that my understanding of the equation itself was off. It does not claim that $1 \text{joule} = 1 \text{kg} * (\text{speed of light}) ^ 2 \frac{m}{s}$ but instead that a mass of $1 \text{kg} * (\text{speed of light}) ^ 2$ = A certain amount of energy expressed in joules. My understanding of the equation itself improved, but my original question still stands. How is such a clean relationship found between the kg and the speed of light without multiplying or summing by a custom constant or set of constants? If you were to change to a different system, wouldn't you be unable to neatly say that exactly 1 unit multiplied by the speed of light = a certain amount of energy? You would need to translate your metric for mass to 1 kg.
In pounds for example I think you would have to say: E = $2.20462pounds * c^2 m/s$ Is this off? And so on.. If you want to keep it clean with 1 unit you would have to change the constant from the speed of light to a different value (a smaller value in the case of pounds at least). | The elegance of the equation $E = mc^2$ doesn't have much to do with the unit system. It's a great equation because the speed $c$ that shows up on the right-hand side is exactly the same as the speed of light, but this follows from the postulates of relativity, not the unit system used. But you might wonder: the unit system does matter when we're computing with numbers, right? That's right: unit systems are not completely arbitrary, and if you set them up differently then you can get worse results. Fortunately, we never do. For example, consider the formula for the area of a circle, $$A = \pi r^2.$$ In physics, we view this formula as a relation between physical quantities. The quantity $A$ is an area, and $r$ is a length. The purpose of a unit system is to convert these physical quantities to raw numbers , which we do by dividing both sides by a suitable unit, $$\frac{A}{1 \text{ square meter}} = \pi \, \left( \frac{r}{1 \text{ meter}} \right)^2.$$ This is now an equation relating two numbers. It says that the area of the circle (in square meters) is equal to $\pi$ times the radius (in meters) squared. The coefficient is still exactly $\pi$ , but we can screw it up if we use a different unit for area. For example, let's define the $\textit{Horace}$ to be a unit of area precisely equal to $1.39$ square meters. Then multiplying the left-hand side by $1 = (1.39 \text{ square meters}) / (1 \text{ Horace})$ gives $$\frac{A}{1 \text{ Horace}} = \frac{\pi}{1.39} \, \left( \frac{r}{1 \text{ meter}} \right)^2.$$ Now an ugly numeric constant appears, because the units for area are not compatible with those for length. The relation between $A$ and $r$ when those quantities are expressed in this unit system is no longer elegant, even though it's still true that $A = \pi r^2$ . Of course, in physics, we avoid this by defining all units to be products of powers of a set of base units, like how a square meter is the square of a meter. There's a subtlety that I swept under the rug, which doesn't apply to $E = mc^2$ but does to other equations. When we talk about a "unit system", we don't just mean a set of base units. We often also include an entire set of conventions about how to define physical quantities in the first place. Different conventions can lead to differences in the appearances of the equations themselves. For example, consider Faraday's law, $$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}.$$ I could define an alternative quantity $\tilde{\mathbf{B}} = c \mathbf{B}$ . The advantage of this new definition is that $\tilde{\mathbf{B}}$ has the same dimensions as $\mathbf{E}$ , which leads to a pleasing symmetry between Faraday's law and Ampere's law in vacuum, $$\nabla \times \mathbf{E} = - \frac{1}{c} \frac{\partial \tilde{\mathbf{B}}}{\partial t}, \quad \nabla \times \tilde{\mathbf{B}} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}.$$ But it also makes factors of $c$ pop up elsewhere. For example, the magnetic force becomes $$\mathbf{F} = q \mathbf{v} \times \mathbf{B} = \frac{q}{c} \mathbf{v} \times \tilde{\mathbf{B}}.$$ In the so-called Gaussian units in electromagnetism, the base units are the centimeter, gram, and second. That alone wouldn't change the forms of the equations, just the numbers you get when you express them in terms of the base units. However, in Gaussian units you additionally define the
"magnetic field" to be $\tilde{\mathbf{B}}$ , not $\mathbf{B}$ (along with a similar redefinition for electric charge and electric field). It's these conventions that change the forms of the equations. This isn't relevant to $E = mc^2$ because usually we only define energy, mass, and speed in a single way, even if we can use different units to quantify them. For example, the speed $c$ is defined as the distance traveled per time. You could define an alternative speed $\tilde{c}$ as, say, half the distance traveled per time, but this would be incredibly confusing, and require changing results as basic as the kinematics equations. However, it's not as confusing to redefine electric charge and electromagnetic fields, because we never measure them directly, only the forces induced by them. (And there's always an exception -- in natural units we define energy as $\tilde{E} = E/c^2$ , getting the simpler relation $\tilde{E} = m$ .) | {
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621,195 | First of all, I'm a layman to cosmology. So please excuse the possibly oversimplified picture I have in mind. I was wondering how we could know that the observable universe is only a fraction of the overall universe. If we imagine the universe like the surface of a balloon we could be seeing only a small part of the balloon or we could be seeing around the whole balloon so that one of the apparently distant galaxies is actually our own. In the example with the balloon one could measure the curvature of spacetime to estimate the size of the overall universe, but one could also think about something like a cube with periodic boundary conditions. Is it possible to tell the size of the overall universe? Artistic image of the observable universe by Pablo Carlos Budassi. | Yes, it's possible in principle that we see the same galaxy more than once due to light circling the universe. It wouldn't necessarily be easy to tell because each image would be from a different time in the galaxy's evolution. There is a way to test for this. The cosmic microwave background that we see is a 2D spherical part of the 3D plasma that filled the universe just before it became transparent. If there has been time for light to wrap around the universe since it became transparent, then that sphere intersects itself in one or more circles. Each circle appears in more than one place in the sky, and the images have the same pattern of light and dark patches. There have been searches for correlated circles in the CMB pattern (e.g. Cornish et al 2004 ), and none have been found. | {
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621,202 | So here's the setup. We have a spherical source. It emits a pulse of light in all directions with some wavelength $\lambda$ . It reflects off of a spherical mirror that is centered around this source. Now, when the light comes back, it bounces off of the source again (or some percentage, whatever). The source emits another pulse of light at the same time with exactly the same energy as the light that bounced off, but shifted back $\lambda$ / $2$ , so all the crests line up with all the troughs and the light completely destructively interferes. I can't see a way out of this. I've spent energy - but where did it go? | The energy went back into your spherical source. For the first pulse the source had to provide energy. For the second pulse the source had to absorb energy. The mistake is simply assuming that the energy required to drive the source is independent of the external fields acting on the source. | {
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621,237 | According to the fluctuation theorem , the second law of thermodynamics is a statistical law. Violations at the micro scale, therefore, certainly have a non-zero probability . However, the application of the theory, in particular the Jarzynski inequality indeed, in principle, extends to systems of particles comprising macro objects. ${\qquad\qquad\qquad\qquad\qquad}$ Does the fluctuation theorem imply that there is a non-zero probability that a broken egg may spontaneously reassemble itself, in effect implying that (at least some) other laws of physics (such as linear momentum conservation) are in fact statistical statements? | Up to the limits of our theoretical understanding, yes , there is nothing in principle wrong with seeing what happens in the video happen for real in the sense that you can formulate this entire scenario in principle in such theory without running into any contradiction, whereas something like a perpetual motion machine would (namely, that would require physical laws sensitive to the time at which something happens, by the contrapositive of Noether's theorem.). The actual problem is that it's stupendously super-duper unlikely for it to happen. Generally speaking, the probability for a fluctuation gets exponentially smaller the more atoms you want involved, and here we have on the order of some multiple of Avogadro's number, $6.022 \times 10^{23}$ , meaning that we can expect probabilities on the order of $10^{-10^{24}}$ to see this play out in real life. It is, thus, exceptionally unlikely it will happen before some mechanism - perhaps proton decay - dissolves all matter in the universe. Depending on your philosophical bent, you could then say that because of that, we cannot empirically verify physical theory to that point, and so maybe it's possible that in fact our theories are wrong on this and something does come up to intervene. But we have no way we can know that to be the case under all possible scenarios for which it could be. That said, one might also wonder about the possibility of creating this situation artificially, given that one can locally reverse entropy by doing some active work, i.e. expending energy processing information. That might be possible, but getting all the requisite matter into the right state that is exactly the time-reverse of the egg falling and breaking scenario, may be intractable for other reasons. It would require some impressive micro-control over all the movements of atoms to set up the right little "ripples" in the air and board that should all converge "just right" to have that "magic" reassembly power latent within them and "kick the egg back into wholeness". | {
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621,654 | I have many things to do during my day (work + study + projects + ... + ...) and seriously 24 hours a day is not enough to finish my tasks. I know that by Einstein's theories that time gets slower if travel fast. So I decided to travel fast ( not to the speed of light, but a closer speed maybe) All I want is 12 hours extra a day, so my day is 36 hours. So at what speed I should travel to make my 24 hours day (travelling hours) equals 36 hours (normal earth hours)? Edit Reading comments and answers, I realised that I need to have 36 hours travelling while earth only made 24 hours, if that possible! Which means I have to make earth goes close to the speed of light while I am still not moving!! but if I travelled to the speed of light ( from my perspective i am still and the earth is moving) I am confused now!. | I'm afraid you have it backwards. A stationary (inertial) clock takes the longest possible path through spacetime, and accelerating the clock will reduce the length of its path (I.e. the elapsed time for it). If you travel very fast then only a little time passes for you, while years may pass on Earth. This would not help you get more done in an Earth day! | {
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621,967 | I was recently told that in physics, if your equation contains an infinity, it usually suggests that something is wrong. As an example, the infinity around black holes (also) tells us that general relativity isn't the full picture of gravity. How true is this? And what logic underlies this methodology of verifying a theory? Edit to suffice this post The answers don't seem to explain why ; there is no analytical or empirical grounding as to why 'limits are(n't) okay'. | There's no rule against infinity, only a rule against being empirically wrong. Historically, the infinite has hinted we're missing something, but so have plenty of other things too. Let's discuss some historical examples. The ultraviolet catastrophe and its resolution respectively come down to physical reasons to care about $\int_0^\infty x^2dx=\infty$ and $\int_0^\infty\frac{x^3dx}{e^x-1}=\frac{\pi^4}{15}$ . To which is the power output of a glowing object (be it black or grey ) proportional? If it were the former, our power sources couldn't make anything hot enough to glow in the first place. The catastrophe was so-called not because mathematicians, philosophers or physicists are prejudiced against infinite quantities, but because it's an empirical fact that finite amounts of heat are convertible to light. Galilean relativity was at odds with Maxwell's equations until Einstein argued it was the former, not the latter, that had to change. The latter's Lorentz-invariance is the only alternative this group theory argument allows for. Long story short, a special speed appears, which is infinite in Galilean physics. No prejudice ruled against this, only the empirical success of a theory instead claiming that speed would be $c:=(\mu_0\epsilon_0)^{-1/2}$ . We could get around this by setting $\epsilon_0=0$ so $\rho=\epsilon_0\nabla\cdot E=0$ , i.e. electric charges cannot exist, or $\mu_0=0$ so currents can't generate $\nabla\times B$ , but those are also empirically untrue. Newton had been dissatisfied with the infinite-speed action-at-a-distance nature of his own theory of gravity, but he and many successors accepted it as a brute fact about the universe, and it seemed like they always would. Einstein spent a few years working out how a speed- $c$ gravity would work, not because nothing can be infinite, but because special relativity's implications seemed so far-reaching. Eventually, general relativity was not only theoretically decent, but empirically vindicated. But GR gave rise to a different source of infinite quantities: singularities . To this day, we don't know exactly how if at all post-GR gravitational physics will disprove that (but see also here ). I've discussed three problems so far. Quantum mechanics solved the first, but its unification with special relativity, quantum field theory, introduced infinities of its own, for each fundamental interaction in nature. The resolution to these refines an objection you might have had to my first example: what if $\int_0^\Lambda x^2dx$ is the integral that matters, i.e. a glowing object doesn't release arbitrarily high frequencies? The value of $\Lambda$ you'd get from that idea contradicts observed spectra, i.e. we really do need the integrand to taper off in a way $x^2$ doesn't. But it's not a bad idea in theory , which is why analogous ideas can be more successful. See here for a discussion of such techniques. For a reason the above link discusses (although it can also be explained in other ways), gravity's version of these problems has proven tougher. Quantum gravity isn't renormalizable (T & Cs apply, hence the multiple approaches or overlaps thereof to massaging the infinities we get this time). But issues of renormalization, regardless of the decade they're solved in, lead to a different, subtler "is it infinite & does it matter?" question: how many degrees of freedom does it take to nail down your theory? While infinitely many do present a problem (see e.g. Sec. 7.1 here ), not all useful theories are renormalizable , because the first few parameters can do most of the work in explaining or predicting empirical reality. As for examples where finite quantities are too large or too small in theory to match the data, we could be here with those all day. | {
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622,434 | Electrodynamics makes heavy use of vector calculus, which in turn is about differentiation and integration of scalar and vector fields in $\mathbb{R}^3$ . At this point everything seems fine to me, since the physical space is isomorphic to $\mathbb{R}^3$ . However, Maxwell's laws rely on mathematical abstractions such as supposing that electric charge is a continuous variable. For instance: $$\nabla \cdot \mathbf{D}=\rho$$ In the LHS, there is a local quantity, since the divergence of a vector field is defined at every point of space. On the other hand, in the RHS there is a global quantity, i.e. , it can only be defined as a mean charge density in volumes much greater than the dimensions of the charge carriers involved. This is because electric charge is quantized. All charged particles have charges that are integer multiples of $\frac{1}{3}e$ . Since the charge carriers are generally very small, macroscopically it seems that we can indeed define $\rho$ at every point of space, even though this statement does not make much mathematical sense. For instance, in a point of "empty" space (a point not belonging to the charge carriers), $\rho=0$ . Even though I am aware of the extraordinarily high predictive power of classical electrodynamics, the assumption of considering charges continuous instead of discrete seems pretty far-fetched. I'd like to know if, besides quantum applications, Maxwell's laws show some kind of error due to this assumption. | The title of the question asks if Maxwell's equations are mathematically precise . The answer is certainly yes. Maxwell's equations are well-posed differential equations. There is no ambiguity about the symbols, and one can unambiguously check if a given set of fields and charge/current distributions satisfy the equations. In fact, Maxwell's equations enjoy many additional nice mathematical features, such as existence and uniqueness theorems, as well as beautiful and non-obvious symmetry properties such as Lorentz invariance and electromagnetic duality. However the text asks a fundamentally different question: whether Maxwell's equations are a perfect model of the real world. The answer is surely no. Maxwell's equations do not describe many important phenomena, such as gravity. Even more to the point, Maxwell's equations describe a classical field theory, while the world we see around us is quantum . Having said that, it is essentially a triviality in physics to say that any given model does not perfectly describe reality. Even the Standard Model and General Relativity are not complete theories of reality. The interesting question is not a binary "yes or no" question about whether a given theory is "right or wrong", but rather to understand where a given theory breaks down (see Asimov's brilliant essay The Relativity of Wrong ). In the case of classical electromagnetism, it is precisely when dealing with small numbers of fundamental particles (a few photons, or a few electrons, say) that one needs to worry about quantum mechanics, and Maxwell's equations break down. While Maxwell's equations are not a perfect representation of reality, I don't accept the OP's explanation for why they are imperfect. Maxwell's equations can accommodate discrete charge distributions -- simply insert a sum of three dimensional delta functions for $\rho$ . Now, it is true that there is nothing in Maxwell's equations that says that the quantization of charge is logically required , which you may feel is a deficit in the theory. However, there is nothing in the entire Standard Model that requires charge quantization. So, even in our current best understanding of particle physics, the quantization of charge is an empirical fact, and not a consequence of other theoretical principles. Furthermore, mathematically it is a valid approximation to replace a dense sum of discrete point charges by a smooth charge distribution. This is not a particularly deep observation. It is quite common to approximate a sum containing a large number of terms by an integral. | {
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622,492 | I was wondering: If you let a small stone drop on a body of water, record it on film, and replay the scene in slow motion, will it be possible to see the difference with a huge rock that falls, in real-time, in a body of water? Let's take for the stone a spherical mass, and let's assume that the body of water has a flat surface. The stone hits the water perpendicularly. And let's assume that the rock is a spherical mass too with the same density as the small mass. Classical physics (mechanics, thermodynamics) can be applied. The big mass has a velocity of $n$ times that of the small one (upon hitting the water surface). $M=nm$ , where $M$ is the value of the big mass and m that of the second mass. We make the small mass fall into the water, record the whole process, and replay the record in slow-motion (suitably adjusted to the situation; let's say that we replay the record n times as slow, but maybe other paces are better). Is one able to see the difference with the big mass falling into the water in real-time? Is the process of a mass that falls in water scale-invariant wrt the slow-motion version of a smaller mass falling into the water? In the definition of scale-invariance, the variation of time isn't spoken about, though (see here ): In physics, scale-invariance is a feature of objects or laws that do not change if scales of length, energy, or other variables, are multiplied by a common factor, and thus represent a universality. Time can't be varied in reality (it's a parameter). So strictly speaking, we can't talk about scale-invariance. I think it's clear what I mean though. I make this edit because something has become clear to me. Of course, it's the splash that's problematic. In a slow-motion version of a small stone hitting the water, the water will have less time to evolve. And of course, there is less water to evolve. This means that one can tell if the splash is caused by a small mass or a big mass (as one can easily tell that a slow-motion trick is used in old disaster movies). And even though a slow-motion replay of a recorded droplet falling in water looks globally the same (initially; later on a high central uprise of water is to be seen and I'm not sure if water is uplifted in the same way for a big rock), one can tell the size is small. I don't think that changing the speed of impact changes much. Upon replaying the slowed-down speed must resemble the speed of the big mass when it hits the water (which can be done by placing (spacially) scaled down familiar objects around the water. When one tries to model ocean waves (like in MASK ) or in this real model of a rogue wave) the overall result looks good but nevertheless one can see immediately that the waves are scaled-down (like in the small droplet case ). So maybe we can use a liquid that's visibly the same as water but with different properties. The surface tension of water plays a role of significance in the development of the splashing water. Maybe using a liquid with a small surface tension renders a more faithful image of real water when the record is played in slow motion. For the rock, the water has more time to evolve than the water (transparent liquid) for the droplet. This clearly affects the water (liquid) development but I can't visualize exactly how. So maybe it's best to just go to a lake and let a rock fall into it while not forgetting to record it. Can this falling rock record on its turn, when replayed in slo-mo be used to mimic the impact of a 100 meter asteröid...? | In a crude way, the answer is yes, and this effect was commonly exploited in old school sci-fi films by shooting scaled-down models and playing back the action in slo-mo. This general effect is also exploited in the world of physics and engineering, where accurate scaling factors are used to model the behavior of full-scale systems by using models in which scale lengths, speeds, accelerations, forces and whatnot have all been scaled together through the use of nondimensional similitude parameters which will properly proportion all the pertinent effects. There are dozens of these, each with its own special area of application, typically named after the researcher who first described their use. Common examples include the Reynolds number, Mach number, Nusselt number, Froude number and my favorite one: the Number number , which tells you how many nondimensional scaling parameters are needed to solve a particular problem. This is an engineering joke, you may laugh now. | {
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622,595 | I saw a movie in which an object (a kind of steel wool in kitchens) was burning on the kitchen scale. When the burning process began, the weight of the object began to increase. Can anyone explain the cause of this phenomenon (increasing weight by burning) so that it is understandable to a high school student? I didn't find any related question in StackExchange. Edit ---> I didn't expect the following information to be necessary:
Steel wools are generally made of low-grade carbon steel wire, aluminum, bronze or stainless steel. The Steel itself is an alloy of iron and many other elements may be present or added to it depending on the industrial purposes. | For steel wool the combustion reaction is roughly: $$\require{mhchem} \ce{2 Fe (s) + 3/2 O2 (g) -> Fe2O3 (s)}$$ So the object 'absorbs' (and chemically binds ) air oxygen and thus gains weight. | {
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622,729 | This question about the speed of light prompted my own question. In the linked question it is asked if there is experimental proof that the speed of gravity equals the speed of light. I was surprised not to see the LIGO measurements mentioned. The experiment uncovered the arrival of a spacetime distortion coming from fast-spinning binary systems of black holes or neutron stars. Due to LIGO's extensive Nature (there is one observatory in Livingston and one in Hanford) it seems that upon arrival the gravitational wave (if it hits the Earth at a sharp angle) will hit one of both observatories first (which one depends obviously from the origin of the wave). So it should be possible to measure the speed of gravity. Or, at least, to measure if the speed is finite (or not). Has this been done? | Yes . In principle, the speed of gravitational waves can be measured using the data of LIGO. In fact, using a Bayesian approach, the first measurement of the speed of gravitational waves using time delay among the GW detectors was suggested/performed by Cornish, Blas and Nardini . By applying the Bayesian method, they found that the speed of gravitational waves is constrained to 90% confidence interval between $0.55c$ and $1.42c$ by use of the data of binary black hole mergers GW150914, GW151226, and GW170104. After that, a more precise measurement of the speed of gravitational waves was
performed by the measurement of the time delay between GW and electromagnetic observations of the same astrophysical source, as @Andrew nicely mentioned, by use of a binary neutron star inspiral GW170817 . They found the speed of gravitational wave signal is the same as the speed of the gamma rays to approximately one part in $10^{15}$ . Note that this study is primarily based on the difference between the speed of gravity and the speed of light . Recently, a new method has been introduced using a geographically separated network of detectors. As the authors mentioned, while this method is far less
precise, it provides an independent measurement of the speed of gravitational waves by combining ten binary black hole events and the binary neutron star event from the first and second observing runs of Advanced LIGO and Advanced Virgo. By combining the measurements of LIGO and Virgo, and assuming isotropic propagation, the authors have constrained the speed of gravitational waves to ( $0.97c$ , $1.01c$ ) which is within 3% of the speed of light in a vacuum. In my opinion, the best study is the second one (that @Andrew nicely mentioned), in which multiple measurements can be measured to produce a more accurate result, but the later (the third study) has its scientific significance. This is because the later method is an independent method of directly measuring the speed of gravity which is based solely on GW observations and so not reliant on multi-messenger observations , as the authors mentioned. Besides these achievements, there are other interesting results that one can extract from LIGO's data. For example, observations of LIGO have constrained a lower bound on the graviton Compton wavelength as $${{\lambda _{{\rm{graviton}}}} > 1.6 \times {{10}^{13}}{\rm{km}}},$$ which is really interesting. In fact, assuming that gravitons are dispersed in vacuum like massive particles, i.e. ${\lambda _{graviton}} = \frac{h}{{{m_{graviton}}\,c}}$ , one can find an upper bound for graviton's mass as ${{m_{{\rm{graviton}}}} \le 7.7 \times {{10}^{ - 23}}eV/{c^2} \sim {{10}^{ - 38}}g}$ , which is extremely small, beyond the technology of our detectors. | {
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623,146 | I read on Wikipedia that electrons can be created through beta decay of radioactive isotopes and in high-energy collisions, for instance when cosmic rays enter the atmosphere. Also, that they can be destroyed using pair annihilation. We also know that charge is a physical property which can be associated with electrons.
My question is why can't charges be created or destroyed if electrons can? | Electrons can only be created and destroyed in processes that keep electric charge constant. There are three Standard Model interactions involving the electron: $\rm W^-\to e^-\bar{\nu}_e$ and $\rm \gamma\text{ (or Z)}\to e^-e^+$ . $^1$ the first case, the W boson has the same charge as the electron, so no charge is created or destroyed. In the other cases, a neutral particle turns into two particles with equal and opposite charge. Again, no charge is created or destroyed. In a certain sense, there's no reason that electric charge, in particular, has to be conserved- our theories start with the fact that that charge cannot be created or destroyed as an assumption because we have never seen it be created or destroyed, and the universe as a whole appears to be electrically neutral. $\scriptsize^1\text{Also technically }{\rm H\to e^-e^+}\text{, but that coupling is very small and for practical purposes it can be ignored.}$ | {
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623,344 | Most thermodynamics textbooks introduce a notation for partial derivatives that seems redundant to students who have already studied multivariable calculus. Moreover, the authors do not dwell on the explanation of the notation, which leads to a poor conceptual intuition of the subject. For example, in maths, given a sufficiently well-behaved function $f: \mathbb{R}^3 \to \mathbb{R}$ , we can define its partial derivatives unambiguously by: $$\frac{\partial f}{\partial x}(x,y,z) \qquad\text{ or simply}\qquad\frac{\partial f}{\partial x} \text{ (in shorthand notation)}$$ Writing $$\left(\frac{\partial f}{\partial x} \right)_{y,z}$$ would be verbose. In thermodynamics, why do we have to specify the variables which are held constant by writing subscripts? | That's because in thermodynamics we sometimes use the same letter to represent different functions. For example, one can write the volume of a system as $V=f_1(P,T)$ (a function of the pressure and the temperature) or as $V=f_2(P,S)$ (a function of the pressure and the entropy). The functions $f_1$ and $f_2$ are distinct in the mathematical sense, since they take different inputs. However, they return the same value (the volume of the system). Thus, in thermodynamics it is convenient to symbolize $f_1$ and $f_2$ by the same letter (simply $V=V(P,T)$ or $V=V(P,S)$ ). The subtlety here is that there can be more than one rule that associates pressure (and other variable) to volume. Therefore, the notation $$\frac{\partial V}{\partial P}$$ is ambiguous, since it could represent either $$\frac{\partial V}{\partial P}(P,T)=\frac{\partial f_1}{\partial P} \qquad\text{or}\qquad\frac{\partial V}{\partial P}(P,S)=\frac{\partial f_2}{\partial P}$$ (Here, I am supposing a single component system. Due to Gibbs' phase rule , we need $F=C-P+2$ independent variables to completely specify the state of a system.) However, if we write $$\left(\frac{\partial V}{\partial P}\right)_{T}\qquad\text{or}\qquad\left(\frac{\partial V}{\partial P}\right)_{S}$$ there is no doubt about what we mean, hence the importance of the subscripts. You can indeed check that for a single component system, $\left(\frac{\partial V}{\partial P}\right)_{T}\neq\left(\frac{\partial V}{\partial P}\right)_{S}$ . $$\left(\frac{\partial V}{\partial P}\right)_{S}-\left(\frac{\partial V}{\partial P}\right)_{T}=\frac{TV^2 \alpha^2}{Nc_p}$$ If you want to read more about this, I suggest Representations of Partial Derivatives in Thermodynamics . | {
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623,353 | In thermodynamics, we talk of useful energy vs. useless energy, or heat. I want to know what is the formal definition of useful energy. It seems too subjective. So, can anyone clarify this for me. | That's because in thermodynamics we sometimes use the same letter to represent different functions. For example, one can write the volume of a system as $V=f_1(P,T)$ (a function of the pressure and the temperature) or as $V=f_2(P,S)$ (a function of the pressure and the entropy). The functions $f_1$ and $f_2$ are distinct in the mathematical sense, since they take different inputs. However, they return the same value (the volume of the system). Thus, in thermodynamics it is convenient to symbolize $f_1$ and $f_2$ by the same letter (simply $V=V(P,T)$ or $V=V(P,S)$ ). The subtlety here is that there can be more than one rule that associates pressure (and other variable) to volume. Therefore, the notation $$\frac{\partial V}{\partial P}$$ is ambiguous, since it could represent either $$\frac{\partial V}{\partial P}(P,T)=\frac{\partial f_1}{\partial P} \qquad\text{or}\qquad\frac{\partial V}{\partial P}(P,S)=\frac{\partial f_2}{\partial P}$$ (Here, I am supposing a single component system. Due to Gibbs' phase rule , we need $F=C-P+2$ independent variables to completely specify the state of a system.) However, if we write $$\left(\frac{\partial V}{\partial P}\right)_{T}\qquad\text{or}\qquad\left(\frac{\partial V}{\partial P}\right)_{S}$$ there is no doubt about what we mean, hence the importance of the subscripts. You can indeed check that for a single component system, $\left(\frac{\partial V}{\partial P}\right)_{T}\neq\left(\frac{\partial V}{\partial P}\right)_{S}$ . $$\left(\frac{\partial V}{\partial P}\right)_{S}-\left(\frac{\partial V}{\partial P}\right)_{T}=\frac{TV^2 \alpha^2}{Nc_p}$$ If you want to read more about this, I suggest Representations of Partial Derivatives in Thermodynamics . | {
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623,713 | Mirrors are able to reflect light but are not perfect and after a number of reflections, light loses intensity. However I wonder, during the reflection by a different type of mirror, could the light photons lose some energy and thus be red shifted instead of just losing intensity? I am not talking about light becoming more red, as a simple color filter on top of a mirror could do, but rather e.g. blue becoming green and green becomes red, etc. Does such a mirror exist? Also I am not looking for a digital system with a camera and screen. I am looking for physical processes that would shift light frequency. It would seem there is no natural pitch shifter for acoustics too (swallowing helium gas is not pitch shifting an existing sound). Addendum: I am looking for a static device, not a moving mirror. Simulation: Very simplistic simulation of how you would look like in such a mirror, assuming you are a brunette wearing a red top. Left picture is the original. Middle is shifted: blue becomes green, green becomes red, red is invisible (black). Right: one bigger shift, blue is now red and the rest is gone to infrared. This is just an illustration (or even dramatization, given the amount of shift), don't take it too literally in terms of redshifts I am interested in. Also this assumes there is no UV or higher energy photons in the original scene that would become visible after being shifted. | Yes, this is possible using nonlinear optics. This kind of frequency shift can be done using acousto-optic modulators and electro-optic modulators , and it is normally done in a transmission geometry. The basic idea is that you have a block of material whose refractive index depends on the acoustic pressure or on the local electric field, and then you make that pressure (resp. electric field) oscillate by driving it with an acoustic or radio wave. This induces an oscillation in the refractive index of the material, which then induces a frequency shift in the light that's transmitted. If you want to work in a reflection geometry, then it's probably possible to re-work the EOM principle of operation so that the frequency shift happens to a reflected component. The simplest idea there would be to set up the driving field to create a grating in the direction of propagation, so that the incoming light suffers a Bragg reflection from it $-$ and then you adjust your driver so that this Bragg grating propagates over time. In essence, you create an effective "moving mirror" without any matter getting transported. (... or, of course, you can just use a transmission modulator and put a mirror behind it.) Now, in the real world, this won't impart a huge frequency shift. With realistic devices, the largest shift you can get from an EOM is a few GHz (whereas, for comparison, visible light is in the hundreds of THz). But the principle is there, and there is no physical law that forbids this from happening at higher frequency shifts. (Though then again, if you push too hard in this direction, then it will just start looking like a difference-frequency-generation configuration, and those are also viable candidates for the functionality you're asking about.) Oh, and also $-$ note that these principles are equally applicable to acoustic waves, and it should also be possible to build equivalent devices there. (Just saying.) | {
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623,965 | High tides and low tides are caused by the Moon. The Moon's
gravitational pull generates something called the tidal force. The
tidal force causes Earth—and its water—to bulge out on the side
closest to the Moon and the side farthest from the Moon. ... When
you're in one of the bulges, you experience a high tide. If ocean water rises on full moon. And gravitational acceleration is not dependent on the mass of the attracted body. Just as a metal ball and feather falls at the same speed, why doesn't both bottle water and ocean water rise by same levels on a full moon? If air is the reason, then on an atmosphere less planet does bottle water and ocean water rise by same levels? | Does bottle water rise a little bit on full moon days? No. Tidal forces are about the difference in gravitational pull at different points in the same body. For oceans and other very large bodies of water, this difference causes water to flow from one region to another, which causes the rise in tides. For example, this is why, even though the sun's gravitational pull is much larger on the earth than the moon's, the moon dominates the tides because it is closer to the earth and therefore the difference in gravitational pull is larger. So for the bottle, the difference in gravitational pull from one side of the bottle to the other side of the bottle is extremely small because the distance is extremely small relative to the distance to the moon, and the tidal forces can not be observed. | {
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625,098 | If we have a chain of fixed length hanging from two points we know that it will form a curve that minimizes the chain's potential energy. If we imagine the chain as having many small segments, then the potential energy of each segment is $E_p = mgh$ . As the number of small segments approaches infinity, their masses equalize because the difference in mass between any two segments goes to $0$ as the number of segments goes to infinity. So if we want to minimize the total potential energy of the chain, we need to minimize the sum of potential energies of every segment as the number of segments goes to infinity. However, since we know that in the limiting case all small segments have the same mass, minimizing the sum of their potential energies is equivalent to minimizing the sum of their heights (as the number of segments goes to infinity). This limit of a sum is by definition the integral of the chain curve. Therefore, to minimize the potential energy of a hanging chain, the area under the chain must be minimal. However, this is not the case because the curve that minimizes that area is known to be a semicircle. This would mean that the shape of a hanging chain is a semicircle which is obviously false. What is wrong with my argument? I don't see a reason why minimizing the area doesn't minimize the potential energy. By the way, I don't know physics above high school level, so I would be very thankful if someone can answer this without super heavy math. Calculus is fine. | What is wrong with your argument is this paragraph: If we imagine the chain as having many small segments, then the potential energy of each segment is $E_p=mgh$ . As the number of small segments approaches infinity, their masses equalize because the difference in mass between any two segments goes to $0$ as the number of segments goes to infinity. Especially the part "their masses equalize". The shape of the curve is $y=\cosh(x)$ , but for a segment of chain with a small change in $x$ of $\mathrm{d}x$ in the middle, where the gradient is zero, the length of the segment is shorter by a significant fraction than for a part at the edge with the same $\mathrm{d}x$ . Due to the comments You said that the difference in the $\mathrm{d}x$ values would go to zero, that's true but it's the ratio, or fraction, of the $\mathrm{d}x$ values at different parts of the chain that matter as follows... 1cm of chain (with the 1cm being a change in $x$ ) near the edge might be made up of a million small $\mathrm{d}x$ parts, but these million would add up to a greater mass of chain than a million similar parts that make up a a 1cm section (in the $x$ direction) in the middle. | {
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625,503 | Does physics explain why the laws and behaviors observed in biology are as they are? I feel like biology and physics are completely separate and although physics determine what's possible in biology, we have no idea how physics determine every facets of biology. We know roughly how forces in physics may impact biological systems, but not every little connections and relations that exist between physics and biology. Am I wrong? | To get more insight into a question like this, you might like to ponder the relationship between logic gates and programming languages in the case of computers. This is a lot simpler than the physics—biology question, but begins to open up some of the issues. When a computer runs a program, certainly lots of logic gates and memory elements etc. are enacting the process described by the program. But the logic gates do not themselves tell you much about the structure and nature of a high-level programming language such as Java or Python. In a similar way, further study of atoms and molecules will not in itself reveal much about the immune system in mammals, or the social structure of an ant colony, and things like that. This "answer" is really a brief comment on what is, in the end, quite a deep issue concerning the whole nature and structure of scientific knowledge. Another useful thing to ponder is the relationship between the concepts involved when one moves from the equations of particle physics to many-body physics. There is every reason to consider that the motions of a non-linear many-body system are all consistent with the description offered by the Standard Model of particle physics for all the various fields and interactions. However, the low-level description does not in itself tell us how to formulate a field theory which correctly captures the main elements of the collective behaviour. It is a bit like the difference between knowing the rules of chess and knowing how the game is played to a high standard. For the latter one needs to appreciate some higher-level issues such as the importance of the central squares, pawn structure, open files and things like that. It is not that these are somehow operating without regard to the laws constraining the movement of the pieces, but rather the low-level laws (about how individual pieces may move) simply do not frame a language adequate to describe the higher-level issues. This analogy with chess is not perfect, of course, but it is apt nonetheless, and it illustrates why it is quite misleading to claim, as many do, that "physics explains chemistry". The situation is more subtle than that. For example, the behaviour of many chemical reaction networks has features which do not depend much if at all on the individual reactions, but on the global structure of the network. It is not that such networks fail to respect any law of physics, but the description at the level of individual components cannot frame a language adequate to express and thus grapple with the higher-level issues such as whether the network is stable overall, and things like that. And what is really telling is that it is very common for these higher-level languages to have internal consistency and a certain robustness, such that they can be supported by more than one underlying hardware. This is similar to the way a given computer program can run on different types of hardware as long as the operating system is in common. | {
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625,561 | For a lack of a better title and phrasing of my question, I am asking: do units have to make sense in physics? Let me give an example of what I mean. Let's say that I have some arbitrary units (for argument's sake): $$\frac{\mathrm{meters}^{2} \cdot \mathrm{seconds} \cdot \mathrm{henry}}{\mathrm{coulomb}^{3}}$$ And I want this to be equal to charge let's say, $$C = \frac{\mathrm{meters}^{2} \cdot \mathrm{seconds} \cdot \mathrm{henry}}{\mathrm{coulomb}^{3}}$$ Which has units of Coulombs. But the units don't add up to equal one coulomb as you can see. So my real question is that do you have to do some manipulation with the units in order for them to equal one coulomb? What are the rules and conventions in physics for this? I have seen in EE (electrical engineering) that if you can cancel out all units, then you can give it any units you want in the end. It is the same for the physics field? (EDIT: I may have misunderstood this part and it is probably not true). Or can you set the units to whatever you want even if they are not equal to one coulomb? I have heard that the constant in the EFE's is a conversion factor to get the right units: $$\frac{8\pi G}{c^4} T_{\mu \nu }$$ Where $\frac{8\pi G}{c^4}$ is the conversion factor. This is a real example that I found that I want to understand as well. In summary: Can you set the units to whatever you want even if they are not equal to one coulomb? If you can cancel out all units, then can you give it any units you want in the end? Do you have to do some kind of manipulation of units to get the right units? | I suppose the Wikipedia page on dimensional analysis might give you a better context. Anyway: Can you set the units to whatever you want even if they are not equal to one coulomb? - Absolutely no, unless see point #3. If you can cancel out all units, then can you give it any units you want in the end? - no you can't. I suppose you mean what happens if you construct an expression with no units, such as meter/meter. You just get a dimensionless quantity. Do you have to do some kind of manipulation of units to get the right units? - Not sure about the "manipulation" term, but you can use dimensional constants in some cases. For example, you're absolutely sure that in your problem where you are looking for a speed of a relativistic electron $v$ , it has to depend on a dimensionless parameter $\alpha$ . Then you can write $v=\alpha c$ , where $c$ is the speed of light. It still does not guarantee that your expression is correct (it can be $v=\alpha^2 c$ or $v=c/\alpha^4$ ) but at least this is a reasonable assumption. So in this example, we used a dimensional constant $c$ to create something with a dimension of a velocity. Of course this will not help you to identify numerical factors like $4\pi$ or so. | {
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625,643 | Suppose I stand on a friction less floor, and another object of finite nonzero mass stand in front of me. Can I push the mass, so that it has a nonzero acceleration? Also, where does this force arise, since I cannot push the ground, since that would make me slip on the floor. | Absolutely you can apply a force on some other object and make it accelerate. What will happen is that you too will accelerate - in the opposite direction.
You push the thing, the thing pushes you, it's classic Newton's Third Law. This is akin to rocket propulsion. | {
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626,151 | Apart from the fascinating mathematics of TQFT s, is there any reason that can convince a theoretical physicist to invest time and energy in it? What are/would be the implications of TQFT s? I mean is there at least any philosophical attitude behind it? | I have personally done original research in the field of TQFTs, so I can tell you the reasons I find TQFTs interesting. Some that come to mind are: Some "real life" theories are accurately described/approximated by TQFTs; for example, gauge theories such as QCD. If you have a regular QFT that is gapped, then its strongly-coupled regime is almost by definition a TQFT. The cleanest example of this phenomenon that I know is arXiv:1710.03258 , where the infrared structure of QCD $_3$ is argued to be a certain TQFT. At low energies (below the gap), this TQFT is an excellent approximation to the real dynamics of the theory. So TQFTs are a tool that allows you to describe the strongly-coupled phase of (some) gauge theories, a regime where no other method really works. TQFTs are exactly solvable, so they are a great toy model for more complex QFTs... ... but TQFTs exhibit all the usual features of regular QFTs. For example, TQFTs may have symmetries, which can be Continous or discrete, Old school zero-form or modern higher-form (cf. arXiv:1412.5148 ), Old-school invertible or modern non-invertible (cf. arXiv:2008.07567 ), Modern higher-categorical (where symmetries of different degree mix non-trivially, cf. arXiv:1802.04790 ), anomalous (in the sense of 't Hooft), non-perturbatively, classical or quantum (e.g., non-trivial dualities, cf. arXiv:1607.07457 ) etc. All these features are shared with conventional QFTs, but the advantage of TQFTs is that these are exactly solvable, so one can study these properties much more reliably and explicitly. Also, although TQFTs can be solved exactly, they can also be solved in perturbation theory via Feynman diagrams. So you can compare the perturbative expansion to the full answer, something you cannot do for other QFTs. So TQFTs allow you to understand non-perturbative aspects of QFTs in a controlled enviroment. Other aspects you can study in TQFTs is e.g. the gauging of (continuous or discrete) symmetries. The general principle is the same as in generic QFTs, but here you can do the computations explicitly and cleanly, so you can learn much better what it means for a symmetry to be gauged, and why you may have obstructions ('t Hooft anomalies) or ambiguities/choices (theta parameters). TQFTs classify conformal blocks in one lower dimension, so if you want to want to have a proper understanding of CFT $_2$ (i.e., String Theory) then you will have to learn about TQFT $_3$ sooner or later. Along similar lines, TQFTs give some examples of holography so I guess they are also interesting if you care about that. Finally, TQFTs sometimes appear in actual experiments, I believe. The Hall effect is an oft quoted example, although I never really cared for the real world so I don't know the details. | {
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627,141 | Why does the Sun appear white through clouds? It seems there should not be any absorption of, say, the reddish component, as this would not produce a white colour. So what is going on? Am I right that clouds are white due to direct yellow light from the Sun and scattered blue light arriving from all directions? | It's kind of a funny misconception that the sun is yellow. I mean, astronomically speaking it is indeed a yellow star , more precisely G-type main sequence / yellow dwarf ... but don't be fooled by the terminology: astronomically speaking, you'll also find that the Earth consists completely of metal ! Actually you should consider the sun as white . The main reason, strangely enough, why we think the sun is yellow is that we never look at it . That is, directly enough to judge its colour. When the sun is high in a cloudless sky, it's just too bright to see its colour (and evolution has trained us to not even try, because it would damage the eyes). Only near sunrise or sunset do we actually get to look at the sun, but then it's not so much the colour of the sun but the colour of the atmosphere we're noticing – and the atmosphere is, again counter to perception, yellow-orange-red in colour. Well, not quite – the point is that the atmosphere lets red / yellow light through in a straight line whereas bluer frequencies are more Rayleigh scattered . That's the reason why the sky is blue, and also adds to the perception of the sun being yellow: it's yellow-ish in comparison with the surrounding sky colour. When you see the sun through clouds, you get to see its actual colour more faithfully than usual, both because (as Mark Bell wrote ) Mie scattering doesn't have the colour-separating effect that Rayleigh scattering does, and because you then see it against a grey / white backdrop instead of against the blue sky. | {
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627,156 | Let's suppose i have a strong metal container 1 x 1 x 1 cm filled with water at room temperature (25 c). How much pressure would i need to apply from one side of the box container to raise temperature of contained water to > 100 degrees? And how much energy i would get from steam when i release the pressure and let water insta-boil? I'm trying to figure out what it would take to make water steam-explode using just pressure. | It's kind of a funny misconception that the sun is yellow. I mean, astronomically speaking it is indeed a yellow star , more precisely G-type main sequence / yellow dwarf ... but don't be fooled by the terminology: astronomically speaking, you'll also find that the Earth consists completely of metal ! Actually you should consider the sun as white . The main reason, strangely enough, why we think the sun is yellow is that we never look at it . That is, directly enough to judge its colour. When the sun is high in a cloudless sky, it's just too bright to see its colour (and evolution has trained us to not even try, because it would damage the eyes). Only near sunrise or sunset do we actually get to look at the sun, but then it's not so much the colour of the sun but the colour of the atmosphere we're noticing – and the atmosphere is, again counter to perception, yellow-orange-red in colour. Well, not quite – the point is that the atmosphere lets red / yellow light through in a straight line whereas bluer frequencies are more Rayleigh scattered . That's the reason why the sky is blue, and also adds to the perception of the sun being yellow: it's yellow-ish in comparison with the surrounding sky colour. When you see the sun through clouds, you get to see its actual colour more faithfully than usual, both because (as Mark Bell wrote ) Mie scattering doesn't have the colour-separating effect that Rayleigh scattering does, and because you then see it against a grey / white backdrop instead of against the blue sky. | {
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627,158 | $$\nabla \times A = B$$ $A$ is vector magnetic potential, $\mathrm{Wb/m}$ $B$ is magnetic field intensity, $\mathrm{Wb/m^2}$ Where does one more m come from for $B$ ? Is that from the gradient operator so it is in meter or something? | It's kind of a funny misconception that the sun is yellow. I mean, astronomically speaking it is indeed a yellow star , more precisely G-type main sequence / yellow dwarf ... but don't be fooled by the terminology: astronomically speaking, you'll also find that the Earth consists completely of metal ! Actually you should consider the sun as white . The main reason, strangely enough, why we think the sun is yellow is that we never look at it . That is, directly enough to judge its colour. When the sun is high in a cloudless sky, it's just too bright to see its colour (and evolution has trained us to not even try, because it would damage the eyes). Only near sunrise or sunset do we actually get to look at the sun, but then it's not so much the colour of the sun but the colour of the atmosphere we're noticing – and the atmosphere is, again counter to perception, yellow-orange-red in colour. Well, not quite – the point is that the atmosphere lets red / yellow light through in a straight line whereas bluer frequencies are more Rayleigh scattered . That's the reason why the sky is blue, and also adds to the perception of the sun being yellow: it's yellow-ish in comparison with the surrounding sky colour. When you see the sun through clouds, you get to see its actual colour more faithfully than usual, both because (as Mark Bell wrote ) Mie scattering doesn't have the colour-separating effect that Rayleigh scattering does, and because you then see it against a grey / white backdrop instead of against the blue sky. | {
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627,163 | For instance, since weight is given by mass x gravitational field strength and assuming both objects have the same gravitational field strength and the only changing variable is its mass, why do they fall at the same rate? | It's kind of a funny misconception that the sun is yellow. I mean, astronomically speaking it is indeed a yellow star , more precisely G-type main sequence / yellow dwarf ... but don't be fooled by the terminology: astronomically speaking, you'll also find that the Earth consists completely of metal ! Actually you should consider the sun as white . The main reason, strangely enough, why we think the sun is yellow is that we never look at it . That is, directly enough to judge its colour. When the sun is high in a cloudless sky, it's just too bright to see its colour (and evolution has trained us to not even try, because it would damage the eyes). Only near sunrise or sunset do we actually get to look at the sun, but then it's not so much the colour of the sun but the colour of the atmosphere we're noticing – and the atmosphere is, again counter to perception, yellow-orange-red in colour. Well, not quite – the point is that the atmosphere lets red / yellow light through in a straight line whereas bluer frequencies are more Rayleigh scattered . That's the reason why the sky is blue, and also adds to the perception of the sun being yellow: it's yellow-ish in comparison with the surrounding sky colour. When you see the sun through clouds, you get to see its actual colour more faithfully than usual, both because (as Mark Bell wrote ) Mie scattering doesn't have the colour-separating effect that Rayleigh scattering does, and because you then see it against a grey / white backdrop instead of against the blue sky. | {
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627,687 | Due to the forecasted frost last night, I placed yesterday evening, some 1.5l standard PET bottles filled up to 90% with warm tap water(+60°C) close to some vegetables that I wanted to protect in my garden. The temperature dropped to roughly -3 ~ -4°C last night. This morning I went to see how it went. Of the 10 bottles, one was filled half with water, half with ice. In all the other bottles, the water was still in a liquid state. So I decided to empty them. Here comes the interesting thing: I uncap the bottles, flip them upside down to empty the water, and give them a shake/twist so that it empties faster, and I noticed that some not well structured ice (it was more looking like melting snow actually) was forming almost instantaneously. Curious, I decided to give a strong shake on the next one while emptying it, and well, this mix of ice forming at that moment reminded me the texture of the icy fruit smoothies one can find during summertime. How do you explain that the water, when still, was 100% liquid, and that when I shake the bottle, ice was forming in no time? I mean, for me, shaking = adding energy, so it should warm the water, not cool it to the point it will form ice? From this experiment I guess not, and that instead, it more or less 'helps' the remaining energy of the 0°C water to dissipate, forming ice super quickly. Am I right in my reasoning? I'll redo the experiment the next night, trying to take a photo to add it here. Edit 1: I've placed the same bottles, with the same warm tap water in them at the same position as yesterday. The upcoming night might even be colder... I'll try to take some photos tomorrow morning. Edit 2: Okay, so this morning it wasn't as impressive as yesterday but it happened again: Fig.1 When you start to empty the bottle, only cold, clear water comes out of it. Shaking a little, then Fig.2 Ice particles have attached to the inside of the bottle as it is being emptied. Fig.3 Unstructured ice has accumulated on the ground. It's a totally "wild" and uncontrolled experiment so it's not as impressive as the videos linked by Philip hereunder. Here are the videos from which the screenshots were extracted: https://vimeo.com/534346291 https://vimeo.com/534347556 I also made a tiny additional observation but this is probably entirely due to chance: because I filled the bottles with warm water yesterday, they were a little depressurized this morning, having kind of a global concave shape. I have shaken them all before opening; but the water stayed clear. It's only once I opened them, and emptied them, that ice was formed. | Congratulations, it sounds to me like you've just observed supercooled water ! There are many videos on YouTube that describe this phenomenon, and explain it much better than I could, see here for a Veritasum video where this is discussed for example. The basic idea is this: when water freezes it forms ice, which is a nice regular crystalline structure. However, ice-crystals need a nucleation site , which is a point where the crystal can start to form, before they can actually start to form. In normal situations, water usually has some impurities which can serve as such nucleation sites, around which the crystal starts to grow and ice starts to form. However, if you use very pure water, there are no such "natural" nucleation points and so there is a chance that the water molecules want to form ice, but can't quite get around to it. As a result, the liquids are trapped in a "metastable" state well below their freezing point, but such a state has a precarious stability that can easily be disturbed. Shaking the bottle is one way to disturb this stability, as it gives a couple of the water molecules the chance to align in just to right way to start the crystallisation process, and once it's done, it is energetically favourable for the system to form ice, so all the other water molecules hop on as well. As a result, you would usually see the crystal "growing" in one direction until all the water becomes ice. Of course, you don't need to shake it, you could just introduce a different type of nucleation point as can be seen in this very pretty video and it would produce the same results, or alternatively, you could very carefully pour the supercooled water on top of an ice cube and form a sort of ice sculpture (see this video from The Action Lab ). It takes more energy to form supercooled water than ice, meaning that when the water transitions to ice, it actually releases some heat, so supercooled water is actually colder than ice. Incidentally, this process is also observed in other materials, notably sodium acetate which is used in making heat packs like this one. I've never actually managed to see supercooled water myself, though I've tried quite a few times. I'm quite surprised that you were able to get it from tap water, since it usually requires very pure water. I hope you're able to reproduce the experiment! | {
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628,632 | Why does thunder, that is heard about five or ten seconds after the lightning is seen, start as relatively quiet high pitched 'crackling' thunder which is, about five or ten seconds later than that, followed/replaced/drowned out by much louder lower pitched 'booming' thunder? I am under the impression that loud low frequency sound travels much faster than quiet mid range frequency sounds (by a factor of approximately two in the case of the sound/blast of a large nuclear explosion, meaning that the blast wave of a large nuclear explosion travels at twice the 'speed of sound' i.e about 600 meters per second). So I would have expected the loud low frequency sound from the lightning to reach me earlier and not later. But the low frequency sound does seem to reach me later. Edit: Another thing I find strange is that in many cases the later sounds (the loud booms of thunder) are remarkably distinct (or brief, being at most about 0.1 seconds in duration each, say), like a series of large bombs going off, as if coming from points or small volumes, rather than a continuous roar/rumble that gets louder and quieter at random. The latter might be what is expected if the loud later sound is coming from a randomly oriented line (with or without branches) with a length of five hundred meters to twenty kilometers. | I'm not an expert, but I spent some time with references 1 and 2 several years ago. This answer is based on some notes I took. Measurements using the radio waves produced by lightning indicate that lightning bolts inside thunderclouds (the ones that we can't see directly) are often mostly horizontal, and they can be anywhere from 1/2 km in length to 20 km in length, spanning a large portion of a large thundercloud. That says that the most distant part of the lightning bolt can be several kilometers farther away than the closest part. That is at least part of the reason why thunder lasts so much longer than the flash of lightning that produced it. Also, lightning typically has a jagged, branching shape. Experiments with smaller sparks have shown that the sound from a given segment is typically directional (louder in some directions than in others), so different segments of a jagged lightning bolt will tend to contribute different amounts to the overall sound because they're oriented differently. This is at least part of the reason for the sound's rich texture. Based on that information, here's a guess about why the sound from moderately distant lightning often starts with a quieter higher-pitched part. I'm picturing the shape of lightning as similar to the shape of a river with many smaller tributaries contributing to it. If we're closer to one of those small tributaries than to the main part of the river, then we'll hear the sound from that small tributary first (especially if its orientation is favorable, because the sound is directional), followed by the sounds from the more substantial parts that are farther away. This at least seems to explain the quiet-to-loud trend, and maybe it also helps explain the frequency trend: smaller sparks apparently don't produce as much lower-frequency sound as larger sparks, so most of the lower frequencies will come from the larger parts, which are farther away in this scenario. Frequency-dependent attenuation in the atmosphere and in the ground might also play a role (reference 3). References: Rakov and Uman (2003), Lightning: Physics and Effects (Cambridge University Press) Holmes et al (1971), "On the power spectrum and mechanism of thunder" ( https://doi.org/10.1029/JC076i009p02106 ) Lamancusa (2000), "Outdoor sound propagation" ( https://www.mne.psu.edu/lamancusa/me458/10_osp.pdf ) | {
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628,767 | On the whole, the entropy of the universe is always increasing. There are far more possible states of "high" entropy than there are of "low" entropy. The example I've seen most often is an egg $-$ there is only one way for an egg to be whole (the low-entropy state), but many possible high-entropy states. Now, this is only a "statistical" law, however, and, as I understand it, it's not impossible (only incredibly unlikely) for a metaphorical egg to "unbreak". My question, then, is: have we ever observed a natural example of a "spontaneous" reduction in entropy? Are there any physical processes which work to reduce entropy (and, if so, is there a more concrete way to define entropy? Obviously, a cracked egg is intuitively a higher entropy state than an "uncracked" egg... but what to we really mean by entropy, which is distinct from disorder)? | Yes , one can observe a decrease in entropy of an isolated system . The statistics of these observation are quantified by the fluctuation theorem . The logic of it is based on what you (the OP) suggest: since statistical mechanics is about statistical laws, one would expect there to be fluctuations. The first paragraph of wikipedia states this well, so I'll just copy it with some emphasis added: While the second law of thermodynamics predicts that the entropy of an isolated system should tend to increase until it reaches equilibrium, it became apparent after the discovery of statistical mechanics that the second law is only a statistical one, suggesting that there should always be some nonzero probability that the entropy of an isolated system might spontaneously decrease ; the fluctuation theorem precisely quantifies this probability. The paper in the first note (pdf) there gives access to a PRL with an experimental observation of a decrease in entropy in an isolated system. Of course, these observations were done in very small systems; as the systems get larger, the chance of observing such fluctuations decreases. | {
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629,128 | I've read many times that the fact that gravitational mass is equal to inertial mass (as far as we can tell) used to be a puzzle. I believe that Einstein explained this by showing that gravity is itself just an inertial force. When I first encountered this concept, I thought "isn't there just one property called $m$ and it just appears in different equations (e.g. Newton's second law and the law of gravitation)? In a similar way that (say) frequency appears in many different equations." Obviously I am thinking about this in the wrong way, but does anyone have a good way to explain why so that I can understand it? | "isn't there just one property called m and it just appears in
different equations (e.g. Newton's second law and the law of
gravitation)? In a similar way that (say) frequency appears in many
different equations." There IS indeed just one property called m which appears in both the equations.
The point is that there is no intuitive reason why this should be the case. Forget the term mass for a second and just think in terms of the properties of an object. One property of an object determines how strong is the gravity of the object.
The other property determines how much acceleration it experiences under a given force. There is no obvious reason why these two properties should be the same. But, we observe in daily life, that these two ARE the same. That is what Einstein was able to explain i.e. why these two are the same. EDIT: A good example to compare and contrast is to think about the forces between 2 electrically charged objects, as pointed out by Arthur's answer to this question. One property of the object (namely the charge) determines the amount of attractive/repulsive force. There is no reason why this property that determines the magnitude of a force would be the same as the property that determines how the object would move under a given force. And indeed these properties are not the same.
But in case of gravity, we observe, that these properties are the same. | {
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629,159 | NOT a duplicate of Maximum length stretch of vertical spring with a mass? , I am asking about a system with two connected springs, as shown in this diagram For a single spring, you can simply equate the forces in the vertical direction, i.e. $$
d_{\text{max}} = \frac{mg}{k}
$$ How do you do this for a double spring pendulum system as shown above with spring constants $k_1$ and $k_2$ , masses $m_1$ and $m_2$ , and equilibrium lengths $L_1$ and $L_2$ .
I think the maximum extensions should occur when both springs/pendula are poiting directly downwards, i.e. $\theta_1 = \theta_2 = \frac{\pi}{2}$ , as then gravity pulls the pendula parallell to the spring extensions. | "isn't there just one property called m and it just appears in
different equations (e.g. Newton's second law and the law of
gravitation)? In a similar way that (say) frequency appears in many
different equations." There IS indeed just one property called m which appears in both the equations.
The point is that there is no intuitive reason why this should be the case. Forget the term mass for a second and just think in terms of the properties of an object. One property of an object determines how strong is the gravity of the object.
The other property determines how much acceleration it experiences under a given force. There is no obvious reason why these two properties should be the same. But, we observe in daily life, that these two ARE the same. That is what Einstein was able to explain i.e. why these two are the same. EDIT: A good example to compare and contrast is to think about the forces between 2 electrically charged objects, as pointed out by Arthur's answer to this question. One property of the object (namely the charge) determines the amount of attractive/repulsive force. There is no reason why this property that determines the magnitude of a force would be the same as the property that determines how the object would move under a given force. And indeed these properties are not the same.
But in case of gravity, we observe, that these properties are the same. | {
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629,972 | The recent publication in Phys.Rev.Lett. 126, 141801 Measurement of the Positive Muon Anomalous Magnetic Moment to 0.46 ppm says that A fast pulsed-kicker magnet deflects the muon bunch into a 9-cm-diameter storage aperture, resulting in $\approx 5000$ stored muons per fill. Considering that those muons are moving with almost the speed of light ( $\gamma\approx30)$ and at rest they would only last about $2.2\ \mu s$ on average before decaying, catching one would be no mean feat. What am I missing? | As the paper says, the muons are deflected with a magnet into a storage ring, where they are stored for a fraction of a second. The muons are still moving with nearly the speed of light, which extends their average lifetime by a factor of $\gamma$ . A small fraction of a second later, the muons have all decayed and the storage ring is ready to receive another fill of muons. It sounds like you might be under the impression that the muons are brought to rest or stored for a long period of time. Neither is true- they are stored at high speed and for a small fraction of a second. Granted, hundreds of microseconds is an eternity by the standards of many particle physics experiments. | {
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630,479 | If a force of $10\,\mathrm{N}$ is applied to different objects of different mass in empty space, in the absence of gravity, why do lighter objects accelerate faster than heavier objects? Why does mass cause inertia? | I do not think it is mass that "CAUSES" inertia. Rather mass is DEFINED to be the property of an object which gives it inertia For example, imagine that you were a theoretical observer who did not know of the concept of mass. There were just 3 objects in front of you. You applied $10 \mathrm{N}$ to each object and observed that the different objects accelerated at a different rates. One accelerated at $1 \mathrm{m}/\mathrm{s}^2$ . Others at $2\mathrm{m}/\mathrm{s}^2$ and $4\mathrm{m}/\mathrm{s}^2$ respectively. So, you can conclude that the different objects have some property that determines how much they accelerate under a given force.
You would also notice that there was a linear relationship between the force on a body and its acceleration. Now, you can calculate what would be the value of this property. And you can see that the body that accelerated at $1 \mathrm{m}/\mathrm{s}^2$ would have a higher value of this property than the one that accelerated at $4 \mathrm{m}/\mathrm{s}^2$ . The theoretical observer can see that this property is directly related to what people call "heavier" and "lighter" objects. So, it is safe to conclude that lighter objects accelerate faster than heavier objects. To summarise, the reason lighter objects accelerate faster than heavier objects is because that is how we have defined the terms "lighter" and "heavier". Mass "causes" inertia, because that is how we have defined "mass". If you want to further ask, why objects with different values of this "mass" property accelerate differently, then you are asking a more fundamental 'why' question. And like most fundamental why questions in physics, it boils down to , because those are the laws of physics in our universe. And if you want to ask why are the laws of physics like that, the answer is " because of the initial conditions at the Big Bang. And researchers are working to figure this out " EDIT : As some comments have pointed out, there is also the quantum mechanics aspect of this. All particles get their inertia or the "tendency to resist motion under a force", through the way the fundamental particles interact with the Higgs field. To understand why massive objects have inertia, we need to understand where the mass of objects comes from.
We know that all massive objects are made up of atoms. So, atoms give objects their mass. But where do the atoms get their mass from? Atoms are made up of protons and neutrons (and electrons but they have negligible mass). So, protons and neutrons give atoms their mass. But where do protons and neutrons get there mass from ? According to the standard model, protons and neutrons are made of quarks. So, quarks should be giving atoms their mass, right ? WRONG. Quarks have substantially smaller and lighter mass than the protons and neutrons that they comprise. It is estimated that the masses of the quarks, derived through their interaction with the Higgs field, account for only about 1% of the mass of a proton, for example. So, 99% of the mass of a proton is not to be found in its constituent quarks. Rather, it resides in the massless gluons that bind together the quarks inside the proton. These gluons are the carriers of the strong nuclear force, that pass between the quarks and bind them together inside the proton. These gluons are "massless" but their interaction with the Higgs field is what we experience as mass. So, at a quantum level , the Higgs field ‘drags’ on the gluons , as though the particle were moving through molasses, (Another analogy used is moving through a crowded dance floor) . In other words, the energy of this interaction is manifested as a resistance to acceleration. These interactions slow the particles down, giving rise to inertia which we interpret as mass. | {
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631,854 | While driving, an unlucky butterfly was about to hit my windshield. But instead of splattering, it sort of glided smoothly upwards across the surface of my windshield. The butterfly was clearly not skillfully dodging the car because the required speed to do so would be too much for the little fellow.
There was like a repulsive force between the butterfly and the screen. Though I don't have any video evidence, I guess at least some of you must have experienced this. Illustrative image How does this happen? I feel like the answer lies in the nature of airflow around the car, not sure how exactly, though. This is a Fiat punto, and it does have a fairly aerodynamic shape with the windshield about 45 degrees slanted. | While @Nick gave a good answer (“air flows up and around the car”), that answer by itself would mean no bugs ever hit the windshield - and we know that is false. So what’s the difference between a bug and a butterfly? If we look at the problem in the frame of reference of a stationary car, there is an airstream moving towards it, and in that airstream there is a small solid object (bug, butterfly). From the frame of reference of the object, it is in a body of air that suddenly moves up. The question then becomes - will the object move with the air stream? This depends on the size and strength of the wings and the mass of the object. If you are a bug with small wings that you have to beat very fast to stay in the air, then most of your “lift” is generated by the motion of your wings. If the air moves a bit faster, it won’t change the lift you experience by much (because your wings were moving so fast to begin with, the extra speed of air over the wings is small). So you will go splat. If you are a butterfly, you get enough lift without moving your wings much (because the wings are big). So if the air starts moving faster, it will tend to carry you with it. Lucky quirk of evolution - small body with big wings will avoid fast-moving objects (though I am pretty sure that was not the main reason why butterflies evolved to have large wings...) | {
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632,057 | When I put my hand on a hot metal (say) solid, I can feel my hand heating up. I suspect this is caused mostly by particles (electrons, atoms, ...?) from the solid colliding with the particles that make up my hand thereby transferring kinetic energy to it. But why does this lead to my hand heating up and not it (also?) being pushed? | Here is another scenario where the thing that you describe does happen: A tube is filled with a gas, for example plain air. The tube fits nicely around a finger. The fit is so precise that there is a sufficient seal, so the air cannot escape, but there is only just enough friction between the tube wall and your finger to prevent the tube from sliding off just like that. Gently increase the temperature of the gas. Now the molecules of the gas have a higher average velocity . The effect of that higher average velocity is that your finger is pushed out of the tube. The force on your finger arises from the accumulative effect of gas molecules bouncing against your skin. A gas doesn't have internal cohesion. When you give a gas opportunity to expand it will. Now consider a solid. A solid has internal cohesion. A solid does not expand like a gas at room temperature, and neither does it expand like a gas when you heat it up. (A solid will expand a little, but that's not visible to the naked eye.) When you heat a solid the molecules of the solid move back and forth faster than at colder temperature. Let's say a particular molecule has - just for an instant - a velocity away from the bulk of the solid. So the molecule is on its way to ascend out of the solid. But as that molecule ascends the forces of cohesion from the neighbouring molecules increase. As a consequence the ascending molecule is pulled back into the solid . The molecule now acquires a velocity back towards the bulk of the solid . This molecule will overshoot, and will very briefly create a local indentation of the solid. The motion of the molecules of the solid do transfer heat to your skin as you are touching the heated solid. And it's not just the outward punches that transfer heat. There is also an effect of interaction with the transient indentations from molecules overshooting on their way back into the bulk of the solid . You can think of that as a suction effect, if you will. As to your skin being pushed one way or the other: the combined effect of the "punches" and the "suctions" adds to zero. What remains is the transfer of heat. For that transfer the effect of the "punches" and the "suctions" do add up; that is the transfer of heat from a solid to your skin. | {
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632,078 | In the article Ship Traffic Increases Dramatically, to Oceans' Detriment , there is a quote (emphasis mine) : "I was surprised to see that in 20 years, the growth is almost fourfold, or almost four times larger," Tournadre said. " We are putting much more pressure on the ocean ." The speaker does not intend to use the phrase literally, but this makes me wonder if this is true.
Does putting more and more ships on the ocean increase the pressure on the ocean bed?
If it's true, it will imply that the pressure on a scuba diver (or any object under water) would increase because of the ship! On the one hand, adding more weight on top of anything must increase the force it exerts on the ground, so the pressure must also increase. But on the other hand, since the ocean area is not strictly bounded, the ocean spreads out, so the pressure must decrease; hence the dilemma. Though the effect may be negligibly small, I would still like to know if a ship can increase the ocean's pressure. If yes, by roughly how much? This question explains the case for bounded fluids, it does not explain unbounded fluids like the ocean. | Here is another scenario where the thing that you describe does happen: A tube is filled with a gas, for example plain air. The tube fits nicely around a finger. The fit is so precise that there is a sufficient seal, so the air cannot escape, but there is only just enough friction between the tube wall and your finger to prevent the tube from sliding off just like that. Gently increase the temperature of the gas. Now the molecules of the gas have a higher average velocity . The effect of that higher average velocity is that your finger is pushed out of the tube. The force on your finger arises from the accumulative effect of gas molecules bouncing against your skin. A gas doesn't have internal cohesion. When you give a gas opportunity to expand it will. Now consider a solid. A solid has internal cohesion. A solid does not expand like a gas at room temperature, and neither does it expand like a gas when you heat it up. (A solid will expand a little, but that's not visible to the naked eye.) When you heat a solid the molecules of the solid move back and forth faster than at colder temperature. Let's say a particular molecule has - just for an instant - a velocity away from the bulk of the solid. So the molecule is on its way to ascend out of the solid. But as that molecule ascends the forces of cohesion from the neighbouring molecules increase. As a consequence the ascending molecule is pulled back into the solid . The molecule now acquires a velocity back towards the bulk of the solid . This molecule will overshoot, and will very briefly create a local indentation of the solid. The motion of the molecules of the solid do transfer heat to your skin as you are touching the heated solid. And it's not just the outward punches that transfer heat. There is also an effect of interaction with the transient indentations from molecules overshooting on their way back into the bulk of the solid . You can think of that as a suction effect, if you will. As to your skin being pushed one way or the other: the combined effect of the "punches" and the "suctions" adds to zero. What remains is the transfer of heat. For that transfer the effect of the "punches" and the "suctions" do add up; that is the transfer of heat from a solid to your skin. | {
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632,189 | Let's say I have a strong metal bar. I pull it apart with a very small constant force -- obviously it doesn't break. However, this would disturb the internal configuration. If I let go, then eventually the internal configuration would return to what it was before I started pulling on the bar. However, if I keep pulling on the bar long enough, would the bar eventually break, no matter how small the force is? | Yes, the rod will ultimately break —barring any other failure mechanism that occurs first. (Depending on the material and conditions, you may need to wait a very, very long time, but your mentions of "long enough" and "ultimately" suggest that you're interested even in glacially slow processes, so to speak. Read on to hear about one such class of processes.) The reason for inevitable failure is that at any finite temperature—which is any actual temperature—there's a finite chance that any particular molecular bond will fail under a pulling force. This is called creep , or time-dependent deformation under an applied load. Creep is often assumed to be negligible at less than about a third (or a half, or two-thirds) the material's homologous temperature , as reflected in various answers and comments on this page, but is always active in every material around us. Creep is frequently characterized using a deformation mechanism map , which identifies which creep mechanism (e.g., viscous flow, dislocation creep) predominates under the specific temperature and applied stress and displays the predicted strain rate (e.g., $10^{-10}\,\mathrm{s}^{-1}$ , with approximately evenly spaced lines reading $10^{-5}$ , $10^{-6}$ , $10^{-7}$ , and so on indicating an exponential rate decrease with decreasing temperature): Copper, ice, and olivine are shown together here to emphasize the commonality of creep in material types that aren't often grouped together. Creep is universal. The following generic form is typical: In these maps, we find various regimes marked power-law creep in addition to boundary and lattice diffusional creep. Now, it's a common outcome when comparing various kinetic processes that because of the exponential dependence of rate on the activation energy , we see an interplay of predominance between processes with higher or lower reactant concentrations and higher or lower activation energies. Here's what I mean: The regime of grain boundary and surface diffusion typically dominates at lower temperatures and lower stresses because the bonding at interfaces is poor and allows more mobility, corresponding to a lower activation energy for physical rearrangement. This is the regime in which engineering materials are typically implemented; the creep rate is generally negligible when considering the lifetime of the system, structure, device, or object. However, when you hear about turbine blades being grown as single crystals at great effort and expense, it's to eliminate problematic grain boundary creep: F. L. VerSnyder and E. R. Thompson, Alloys for the 80’s, R. Q. Barr,
Ed., Climax Molybdenum Co., 1980, p. 69, reprinted in Hertzberg's Deformation and Fracture Mechanics . At higher temperatures, lattice or bulk diffusion (i.e., diffusion within the host material) often dominates because there are many more atoms in the bulk than along interfaces. In other words, these bulk atoms are better bonded and less likely to flow than interface atoms, but there are so many of them that sufficient thermal activation causes their effect to add up. Now, I realize your question focuses on metals, but I'll add that in polymers, long molecular chains may slide past each other at an appreciable rate given that the glass transition temperature or melting temperature may not be much higher than room temperature. Consider the creep of PVC resin at 20°C (i.e., room temperature), showing that creep in engineering materials doesn't require searing temperatures or geological time scales: Larger stresses tilt the energy landscape such that the effective barriers to dislocation movement are lowered. In this way, dislocations are driven to carry plasticity through the material via a combination of glide and climb (power-law creep, so named because the rate increases exponentially with increasing stress). At a certain stress in certain materials, dislocations glide easily and nearly instantaneously; we call this stress the yield strength and in engineering contexts often ignore the nuance of slow creep in favor of a yield–no yield dichotomy. In other words, we replace the fanning out of lines representing strain rates of, say, $1\,\mathrm{s}^{-1}$ (representing a fast rate) to, say, $10^{-10}\,\mathrm{s}^{-1}$ (representing a slow rate) with a single horizontal line called the yield strength. It's important to note that regardless of which mechanism dominates under a certain set of conditions, all valid creep mechanisms for a particular material are acting all the time, although the rate may be minuscule. Finally, a classic example of noticeable creep is the sagging of lead pipes over decades : Given enough time, this sagging arc will accentuate and fail by material pinching off, in the manner of any viscous fluid. Lead's low melting temperature means that room temperature corresponds to a quite-high homologous temperature, so prominent creep observations of a familiar metal become accessible over a human lifespan. Make no mistake, however: the pipes could be made of any other material, and such sagging would simply be a matter of time. In the long term, elasticity is an idealization , albeit a very reliable one for many metals and ceramics (and some polymers) at familiar temperatures and over familiar time scales. Many of the other existing answers appear to rely on this idealization, but it sounds like you're asking about the nuances of the very long term. This answer is intended to address one such nuance. | {
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632,205 | The following statement is from this article : The behavior of a dense atmosphere is driven by collisions between its
atoms and molecules. However, the moon's atmosphere is technically referred to as an exosphere because it’s so thin, its atoms rarely
collide. However, it doesn't explain what is meant by "rarely collide". I find this strange, especially since the word 'technically' is used. There doesn't seem to be anything technical about such a vague definition. I mean, someone might think that the atoms of the atmosphere rarely collide at an altitude of 400 km above the earth's surface. So he or she might think that that must be exosphere, but it's not. It's the thermosphere. The following statement is from this article : However the moon's atmosphere is so thin, atoms and molecules almost
never collide. Instead, they are free to follow arcing paths
determined by the energy they received from the processes described
above and by the gravitational pull of the moon. Much the same thing in other words. I would have thought that between collisions atoms and molecules follow arcing paths whenever there is gravity. The higher the pressure the shorter the arcs, but arcs nevertheless. So this doesn't clarify anything for me. So my question is: what does it mean to say that "its atoms rarely collide"? Does it mean they normally bounce off the moon rather than each other? | Any given gas molecule goes a long way on average without colliding with another gas molecule. The mean free path for a gas molecule is on the order $$ \lambda\sim \frac{k_BT}{d^2p} $$ where $k_B$ is the Boltzmann constant, $T$ is the temperature, $d$ is the kinetic diameter of the molecule in question and $p$ is the pressure. Plugging in some rough values for the lunar environment gives $$ \lambda\sim 10^5~\rm km $$ as a back of the envelope estimate. The moon's circumference is only $\sim10^4~\rm km$ , so most gas molecules probably escape the moon's atmosphere or collide with the moon without ever colliding with another gas molecule. | {
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633,206 | I'm doing some self reading on Lagrangian Mechanics and Special Relavivity. The following are two statements that seem to be taken as absolute fundamentals and yet I'm unable to reconcile one with the other. Principle of Least action states that the particle's trajectory under the influence of a potential is determined by minimising the action that is $\delta S =0$ . The universal speed limit of information propagation is $c$ . Question: Consider a potential that exists throughout space (like gravity perhaps). Now, I give a particle $x,p$ and set it motion. As per principle of least action the motion is now fully deterministic since the Lagrangian $L$ is known. We arrive at this claim by assuming that the particle is able to "calculate" the trajectory such that $\delta S = 0$ using Principle of Least Action. This is where my confusion is. The first principle is a global statement whereas the second principle is a local statement. What is the modern day physics answer to this? I hope I could get my questions across! Look forward to your valuable opinions! | The key to this is that the Lagrangian cannot be just any old function. It has to be a function such that, when the action is stationary, its solution describes the kinematics of the system. Thus, if we assert that there is a universal speed limit in the real world, then that says something about the Lagrangians which can be used. They must be chosen such that the motion of the particle perscribed by solving the Lagrangian Mechanics problem is such that the path of any particle is the same, regardless of the state of the universe outside of its light-cone. These are not mutually exclusive. From one perspective, you have a collection of every particle, and every particle obeys some rules. From another perspective, you have a global system whose time evolution ensures said rules for each particle are obeyed. Perhaps obvious: Physicists who use Lagrangian Mechanics to explore relativistic systems do indeed use Lagrangians that have this property. | {
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633,661 | I have been learning about the solar system from popular science shows. In these shows they suggest that, after having seeing around 2500 other solar systems, astronomers have concluded that our solar system is not the normal one. They see sun hugging hot jupiters and super earths close to their star. They find most systems have most of their matter closer to their star. They draw conclusions about our system, namely that our system is a freak. We know that the first exoplanets found were hot jupiters. And they found super earths close to their star. We also know that telescope technology is always improving. With increasing telescope quality, we can find dimmer objects. These would naturally be smaller planets orbiting distant stars at further distances from their stars. My question is this: how do we know that the apparent rare quality of our system is not an artifact of limited observing power that selects for larger objects close to their star? | The solar system cannot be said (yet) to be "rare" because we lack the ability to examine the planetary systems around other stars in detail. In particular, the census of low-mass planets and planets that are more than an astronomical unit from their star is very incomplete. Nevertheless, enough is known to say that the solar system is unusual in some respects. The main oddity about the solar system is that it doesn't contain any "super-Earths" or "sub-Neptunes" at all (i.e. planets intermediate in size between Earth and Neptune), despite them being common in other systems, and the close-in planets are all small and rocky. Most ( $\sim 70$ %) solar-type stars have at least one exoplanet larger than the Earth orbiting with a period of 100 days or less (e.g. Kunimoto & Matthews 2020 ). Most of these close-in planets are 2-5 times the size of the Earth. The presence of a Jupiter-sized planet at 3-7 au is also somewhat unusual - occurring in $<10$ % of solar-type stars Wittenmyer 2016 ). The lack of a hot Jupiter in the solar system is not unusual, since the occurrence rate of these is only of order 1%. These frequencies are corrected for the known and well-understood biases in detection sensitivity associated with system geometry, signal-to-noise ratio and observing cadence. These factors can easily be accounted for in a forward modelling approach . This is where you simulate your exoplanet population, then "observe" it, in software, including all the observation biases and detection thresholds, and then adjust the characteristics of the simulated population until the simulated observations match the real observations. Note for those wanting to discuss the anthropic principle. It seems to me that whether the planetary system around our Sun has an unusual architecture has nothing to do with our presence. What the anthropic principle may have a bearing on is explaining why we live in an unusual solar system or why we might expect the solar system to be unusual. | {
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633,675 | A cube of side 20 cm has its center at the origin and its one side is along the x-axis, so that one end is at x = +10cm and the other is at x = -10cm. The magnitude of electric field is 100 N/C and for x> 0 it is pointing in the +ve x-direction and for x<0 it is pointing in the -ve x-direction as shown. The sign and value of charges inside the box, are The actual solution of this question is to be solved using Gauss' Law. However, I solved for charge inside (Q) using the formula E=kQ/r^2, putting E=+100 N/C and r=+0.1 m but the value doesn't match. Can someone guide me as to why my approach is wrong? Why can't we use the formula E=kQ/r^2? | The solar system cannot be said (yet) to be "rare" because we lack the ability to examine the planetary systems around other stars in detail. In particular, the census of low-mass planets and planets that are more than an astronomical unit from their star is very incomplete. Nevertheless, enough is known to say that the solar system is unusual in some respects. The main oddity about the solar system is that it doesn't contain any "super-Earths" or "sub-Neptunes" at all (i.e. planets intermediate in size between Earth and Neptune), despite them being common in other systems, and the close-in planets are all small and rocky. Most ( $\sim 70$ %) solar-type stars have at least one exoplanet larger than the Earth orbiting with a period of 100 days or less (e.g. Kunimoto & Matthews 2020 ). Most of these close-in planets are 2-5 times the size of the Earth. The presence of a Jupiter-sized planet at 3-7 au is also somewhat unusual - occurring in $<10$ % of solar-type stars Wittenmyer 2016 ). The lack of a hot Jupiter in the solar system is not unusual, since the occurrence rate of these is only of order 1%. These frequencies are corrected for the known and well-understood biases in detection sensitivity associated with system geometry, signal-to-noise ratio and observing cadence. These factors can easily be accounted for in a forward modelling approach . This is where you simulate your exoplanet population, then "observe" it, in software, including all the observation biases and detection thresholds, and then adjust the characteristics of the simulated population until the simulated observations match the real observations. Note for those wanting to discuss the anthropic principle. It seems to me that whether the planetary system around our Sun has an unusual architecture has nothing to do with our presence. What the anthropic principle may have a bearing on is explaining why we live in an unusual solar system or why we might expect the solar system to be unusual. | {
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633,681 | In the given figure flux through surface $\,\rm S_1\,$ is $\,\Phi_1\,$ and through surface $\,\rm S_2\,$ is $\,\Phi_2$ . Which option is correct? (A) $\quad \Phi_1\boldsymbol{=}\Phi_2$ (B) $\quad \Phi_1\boldsymbol{>}\Phi_2$ (C) $\quad \Phi_1\boldsymbol{<}\Phi_2$ The answer to this question is option A. But according to inverse square law, the density of flux lines is inversely proportional to the square of the distance from the source. So according to this statement, shouldn't $\Phi_1>\Phi_2$ be the correct answer, since density of flux lines is greater in $S_1$ ?
| The solar system cannot be said (yet) to be "rare" because we lack the ability to examine the planetary systems around other stars in detail. In particular, the census of low-mass planets and planets that are more than an astronomical unit from their star is very incomplete. Nevertheless, enough is known to say that the solar system is unusual in some respects. The main oddity about the solar system is that it doesn't contain any "super-Earths" or "sub-Neptunes" at all (i.e. planets intermediate in size between Earth and Neptune), despite them being common in other systems, and the close-in planets are all small and rocky. Most ( $\sim 70$ %) solar-type stars have at least one exoplanet larger than the Earth orbiting with a period of 100 days or less (e.g. Kunimoto & Matthews 2020 ). Most of these close-in planets are 2-5 times the size of the Earth. The presence of a Jupiter-sized planet at 3-7 au is also somewhat unusual - occurring in $<10$ % of solar-type stars Wittenmyer 2016 ). The lack of a hot Jupiter in the solar system is not unusual, since the occurrence rate of these is only of order 1%. These frequencies are corrected for the known and well-understood biases in detection sensitivity associated with system geometry, signal-to-noise ratio and observing cadence. These factors can easily be accounted for in a forward modelling approach . This is where you simulate your exoplanet population, then "observe" it, in software, including all the observation biases and detection thresholds, and then adjust the characteristics of the simulated population until the simulated observations match the real observations. Note for those wanting to discuss the anthropic principle. It seems to me that whether the planetary system around our Sun has an unusual architecture has nothing to do with our presence. What the anthropic principle may have a bearing on is explaining why we live in an unusual solar system or why we might expect the solar system to be unusual. | {
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633,803 | Imagine a 1m cube of air in an insulated container, where you have heated the air in one half but not in the other. Intuitively, given molecules are moving so fast, you would expect the energetic particles from the hot side to distribute themselves throughout the box within milliseconds, equalising the temperature. Why does this instead take much longer? I thought of this question while waiting for a warm room to cool down after opening a window. Clearly my mental model of temperature is wrong because it seems to me that heat should move through a gas way faster than it actually does. | The mean-free path of a nitrogen molecule in air at STP is about 60 nm, so the molecules may travel quite quickly, but they do not go very far before the run into another molecule and get stopped. | {
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633,865 | For this quantum mechanical system, is there an operator that corresponds to "position" the same way that there are operators corresponding to angular momentum and energy? My first guess for what the operator would be is $$
\hat \Phi \big[ \psi(\phi) \big] = \phi \, \psi(\phi)
$$ which would be analogous to the position operator for the "particle in a box" system, but I am getting confused on the following issue: If the particle has a 50% chance of being in one place on the ring and a 50% chance of being in the place that is exactly opposite the first place, what should the "expectation value" of the position be? With this operator it seems like the expectation value would depend on where $\phi = 0$ is chosen to be, which makes me doubt the validity of this operator since its expectation value changes depending on which coordinate system you use. Also if the particle has some probability to be at $\phi = \epsilon$ and the rest of the probability to be at $\phi = 2 \pi - \epsilon$ then it seems like the "expectation value" should be around $\langle \hat \Phi \rangle = 0$ but this operator would put it near $\pi$ . | This question is a great setup for explaining a better way of describing particles in quantum theory, one that bridges the traditional gap between single-particle quantum mechanics and quantum field theory (QFT). I'll start with a little QFT, but don't let that scare you. It's easy, both conceptually and mathematically. In fact, it's easier than the traditional formulation of single-particle quantum-mechanics, both conceptually and mathematically! And it makes the question easy to answer, both conceptually and mathematically. To make things easier, here's a little QFT In QFT, observables are tied to space, not to particles. That's the most important thing to understand about QFT. Instead of assigning a "position observable" to each particle (which would be impossible in many models because particles can be created and destroyed), we assign detection observables to regions of space. Let $D(R)$ denote an detection observable associated with region $R$ . In nonrelativistic QFT, the eigenvalues of $D(R)$ are natural numbers ( $0$ , $1$ , $2$ , ...) representing the number of particles found in $R$ when $D(R)$ is measured. If the model has more than one species of particle, then we have different detection observables for each species. By the way, the traditional bit about "indistinguishable particles" is really just an obtuse way of saying what I said above: observables are tied to space, not to particles. Very simple. Now, the difference between nonrelativistic QFT and nonrelativistic single-particle quantum mechanics is almost trivial: in the latter, the eigenvalues of $D(R)$ are restricted to $0$ and $1$ . As an example, consider ordinary nonrelativistic single-particle quantum mechanics, in one-dimensional space for simplicity. The traditional formulation uses a position operator $X$ , whose measurement magically returns the particle's spatial coordinate. That formulation is convenient for some purposes, but it also causes trouble: First, it causes conceptual trouble, because this is not how real-world measurements work: there is no measuring device that magically tells us the particle's coordinates no matter where it is in the universe. Real-world measurements are localized — they correspond to observables that are tied to regions of space, not tied to particles. Second, it causes mathematical trouble, because $X$ has a continuous spectrum — it doesn't have any (normalizable) eigenstates, so it can't be perfectly measured even on paper. Fortunately, we don't really need $X$ . We can do better, both conceptually and mathematically, by using a collection of projection operators $D(R)$ instead. For any given spatial region $R$ , the definition of $D(R)$ is simple: If the particle's wavefunction is concentrated entirely within $R$ , then it's an eigenstate of $D(R)$ with eigenvalue $1$ . If the particle's wavefunction is concentrated entirely outside of $R$ , then it's an eigenstate of $D(R)$ with eigenvalue $0$ . In other words, $D(R)$ counts the number of particles in $R$ , and since the model only has one particle, the answer can only be $0$ or $1$ . If the particle's wavefunction is partly inside $R$ and partly outside $R$ , then the result of the measurement will be $1$ with probability $$
p = \frac{\langle\psi|D(R)|\psi\rangle}{\langle\psi|\psi\rangle},
\tag{1}
$$ and the result will be $0$ with the opposite probability $1-p$ . That's nicer conceptually, because it's closer to how we do things in the real world. It's also nicer mathematically, because $D(R)$ is a bounded operator with a discrete spectrum. Note that the observables $D(R)$ for different regions $R$ all commute with each other (this should be obvious from the definition), so we can measure them all simultaneously if we want to — just like we can use an array of detectors in the real world. The traditional position operator How are the operators $D(R)$ related to the usual position operator $X$ ? Simple: $$
X \approx \sum_n x_n D(R_n)
\tag{2}
$$ where the sum is over a set of regions $R_n$ that partition all of space into non-overlapping cells, and $x_n$ is the coordinate of some point in the $n$ -th cell. In the limit as the size of the cells approaches zero, this approaches the usual position operator $X$ . Conceptually, we're just filling space with an array of little detectors. We know where each one them is located, so if the $n$ -th one detects the particle, then we know where the particle is. To be fair, the traditional position operator (2) has some advantages over the detection observables $D(R)$ . One advantage is that it allows us to use summary-statistics like averages (think of Ehrenfest's theorem) and standard deviations (think of the traditional uncertainty principles). This is just like the usual situation in statistics: the information is in the distribution (the results of measuring lots of $D(R)$ s), but choosing a convenient labeling scheme lets us define things like averages and standard deviations to convey some incomplete but concise information about the distribution. The particle on a circle Now, the answer to the question should be obvious. Does a particle living on a circle have a position operator? Well, it certainly does have the operators $D(R)$ , where now $R$ is any portion of the circle. Those operators are defined just like before. We can also define something analogous to $X$ if we really want to, like this: $$
\Theta \approx \sum_n \theta_n D(R_n)
\tag{3}
$$ or like this: $$
U \approx \sum_n e^{i\theta_n} D(R_n),
\tag{4}
$$ where the sum is over a partition of the circle into non-overlapping intervals $R_n$ , and $\theta_n$ is an angular coordinate within the $n$ -th interval. We can take the limit as the size of the intervals goes to zero, if we want to. The operator (3) is most closely analogous to (2), and it's perfectly well-defined mathematically, but it's unnatural because the angular coordinate must have a discontinuity ( $2\pi$ jump) somewhere on the circle. That makes the operator (3) less useful for talking about things like averages and standard deviations, but that's not quantum theory's fault. It's just mundane statistics: averages and standard deviations are most useful when using a monotonically increasing coordinate, which we can't do everywhere on a circle if we want the coordinate to be single-valued. The operator (4) is more natural mathematically, because $e^{i\theta}$ is continuous everywhere around the circle. Beginners might complain that (4) is not hermitian, but observables do not need to be hermitian (see this question ), contrary to what some introductions say. Unitary operators like (4) can also be used as observables, because quantum theory doesn't care about the coefficients of the projection operators, it only cares about the projection operators. The coefficients are useful for talking about averages and standard deviations, but we don't really need to talk about those things. They're convenient, but they're not necessary. Again, this is just like in ordinary statistics. Labels can be convenient, but they're not necessary. So... does a particle on a circle have a position operator? It depends. What properties do you want the position operator to have? It does have the observables that really matter, which are the detection-observables $D(R)$ . And we can use those to construct operators like (3) and (4) that each resemble the position operator (2) in different ways, but of course we can't have an operator that is completely like (2) because a circle is not completely like the real line. So, whether the answer is "yes" or "no" depends on exactly what you want, but at least the conceptual obstacles should be gone now. | {
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634,080 | I know that a typical stellar black hole would spaghettify someone who crosses its event horizon. Is this also true for a hypothetical tiny black hole with a small mass (the mass of an apple)? Would someone touching such a black hole spiral into it and get dead? | This is just a quick calculation that shows what would happen to a black hole with a mass equal to the mass of an apple: It is shown that not only the builder of this black hole but also the whole city around that person will probably be destroyed. Let first compute the initial temperature of a hypothetical black hole with a mass equal to an apple, e.g., $M=0.2$ $\rm{kg}$ . According to the Hawking radiation formula , for the temperature of a static (Schwarzschild) black hole, we have $${T_{BH}} \approx 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{\rm{K} },$$ where $M_\odot \approx 2×10^{30} \rm{kg}$ is the solar mass (the mass of sun). For your black hole with the mass of an apple, one obtains $${T_{BH}} \approx 6 \times 10^{23} \, {\rm{K} }.$$ This is extremely huge. Now, let's compute the lifetime of this black hole . This can be calculated by the use of the Stefan-Boltzmann law and assuming that the black hole initial mass ( $M$ ) will eventually evaporate to nothing. For static black holes, this results in $${\tau _{{\rm{life}}}} \approx (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3}.$$ Therefore, for your hypothetical black hole, the lifetime is $${\tau _{{\rm{life}}}} \approx 10^{-19}-10^{-18} \, \rm{second},$$ which is extremely small. Now, let's evaluate the relativistic energy of this hypothetical black hole by use of the Einstein's mass-energy equivalence , $E=Mc^2$ , where $c$ is the speed of light. This amount of energy will be released during evaporation. It is straightforwardly obtained as $$E \approx 1.8 \times 10^{16} \rm{J}.$$ It is huge. Comparing this amount with a nuclear bomb is interesting: A nuclear bomb typically has an explosion of about $10-10000$ kilotons of TNT and one kiloton of TNT is equivalent to one trillion Joules ( $10^{12}$ $\rm{J}$ ) of energy (see WikiPedia ). So, our hypothetical black hole released an amount of energy as much as a nuclear weapon! These calculations show that if you could create a black hole with a mass equal to an apple, it will evaporate suddenly ( ${\tau _{{\rm{life}}}} \approx 10^{-19}-10^{-18} \, \rm{second}$ ) and release a big explosion equivalent to a nuclear bomb. So not only the builder of this black hole but also the whole city around that person will probably be destroyed. And, finally, I think the problem of gravity and the event horizon radius of this hypothetical black hole is not important here. The radius of the event horizon, in this case, is extremely small (about $10^{-28}$ $\rm{m}$ ), much much smaller than the radius of the proton which is $10^{-15}$ $\rm{m}$ ! In addition, the gravitational field outside the black hole's event horizon is still the same as the gravity of an apple (spaghettification due to the tidal forces needs a very strong non-homogeneous gravitational field). On the other hand, the tidal forces act at the level of elementary particles of your body and there is no chance for any spaghettification in comparison with the black hole's lifetime. So, it's not important here. Explosion of an apple-mass black hole Here, a precise comparison with a nuclear bomb is discussed. $1$ kilotons of TNT are exactly equivalent to $4.184 \times 10^{12}$ $\rm{J}$ (joules) of energy (See TNT equivalent convention ). So, an apple-mass black hole release $4300$ kilotons of energy (See the computation in this link , using Wolfram alpha). The first nuclear weapon which used in warfare exploded with an energy of approximately $15$ kilotons of energy (see WikiPedia ). This amount of energy is equivalent to $698$ milligrams relativistic mass $m$ from Einstein’s formula $E=mc^2$ (See this computation in this link using Wolfram alpha ). Comparing this ( $15$ kilotons of energy = $698$ milligrams relativistic mass) with an apple-mass black hole ( $4300$ kilotons of energy = $200$ grams) shows that the evaporation of apple-mass black hole in the laboratory would actually be pretty catastrophic, as it would release about as much energy as a nuclear bomb. Furthermore, in comparison with a nuclear explosion, the blast radius (and the other relevant data) of an apple-mass black hole explosion with 4300 kilotons of energy has been presented in this link , again, by use of Wolfram alpha. You may be interested in comparing the data of apple-mass black hole explosion with Tsar Bomba data as the most powerful nuclear weapon ever created and tested on our planet which yielded around $50$ megatons of TNT ( $2.092 \times 10^{17}$ joules). See this link that I prepared using Wolfram alpha. I have checked this calculation and improved them using Wolfram Alpha to be more accurate (thanks to everybody for your comments). The Hawking temperature and the lifetime formulas (including all the physical constants) can be found in Ref. [1]. This calculation is just an estimate and the order of magnitude is important here ... and I should emphasize that there may be other details that could probably change the final result and we have not considered them ... I try to discuss further another aspect in the next section. Further technical information You can skip this part. In this section, the relationships used in this answer together with three discussions (Hawking radiation contents, massive particle emission, and the graybody factor) are presented. The Hawking (radiation) temperature formula [1]: $${T_{BH}} = \frac{{\hbar {c^3}}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}} \tag{1}.$$ The black hole lifetime formula [1]: $${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3} \tag{2}.$$ The rate at which the static black hole radiates energy [1]: $$\,\frac{{dE}}{{dt}} = - \frac{{dM{c^2}}}{{dt}} = 4\pi r_s^2\sigma {\left( {\frac{{\hbar {c^3}}}{{8\pi GM{k_B}}}} \right)^4} \tag{3}.$$ $G=$ gravitational constant , $\hbar=$ Planck constant , $c=$ speed of light , $\sigma=$ Stefan–Boltzmann constant , $k_B=$ Boltzmann constant and ${r_s} = \frac{{2GM}}{{{c^2}}}=$ Schwarzschild radius . On the contents of Hawking radiation: The contents of Hawking radiation are fermions and bosons (both particles and anti-particles). By implementing the quantum field theory in the black hole background (i.e., by quantizing fermionic/bosonic fields in the curved black hole background), one finds the same Hawking temperature, Eq. $(1)$ , for each fermionic/bosonic quantum field. However, for bosonic fields, one always obtains the Bose-Einstein statics while the Fermi-Dirac statics is obtained for fermionic fields. So, the Hawking temperature is the same for every type of particle (e.g., according to the surface gravity definition , ${T_{BH}} = \frac{1 }{{2\pi }}\sqrt { - \frac{1}{2}({\nabla _\mu }{\xi _\nu })({\nabla ^\mu }{\xi ^\nu })}$ ), and the number of species only adds a numerical factor to the Stefan-Boltzmann law (and consequently to the black hole's lifetime) but, still, the order of magnitude of the final answer is approximately valid. On the emission of massive particles: In addition, at the final stages of black hole evaporation, the production of pair massive particles will become relevant. In fact, there is a threshold $E > mc^2 $ for the production of massive quanta and the requirements needed for this to happen are simply provided at the final stages of black hole evaporation. And, perhaps more interestingly, it's been shown that massive particles tunnel across the horizon at the same rate as massless particles of the same energy $E$ even with $E < mc^2$ , meaning that in such cases massive particles (with $E < m$ ) do not reach infinity, but they could be detectable at any finite distance ( see Ref. [4] ). So, in our example, we expect to see the emission of massive particles as well. On the graybody factor ( $\Gamma (\Omega)$ ): In a more realistic situation, only a fraction of the emitted radiation reaches the asymptotic observer located at infinity. This means the particle (or the associated wave) scaping from the black hole needs to pass through the black hole potential and this change the resulting spectrum by a greybody factor ( $Γ(Ω)<1$ ). Here, we assumed that $Γ(Ω)=1$ , which is reasonable since the lifetime of a black hole is too small so particles can scape the black hole's potential. References [1] T.A. Moore, A General Relativity Workbook , University Science Books (2012) [2] S.M. Carroll, Spacetime and geometry: an introduction to general relativity , Addison-Wesley, San Francisco, USA (2004) [3] Wolfram Alpha [4] G. Jannes, Hawking radiation of $E< m$ massive particles in the tunnelling formalism , JETP letters 94 (2011) 18-21. [5] V. Mukhanov and S. Winitzki, Introduction to quantum effects in gravity, Cambridge University Press (2007) [6] Thanks to @Andrew for his friendly suggestion about the radius of a proton. [7] A suggestion : This handy calculator (for computing the Hawking radiation of static black holes) is really interesting and it is easy to work with (Thank you, @PM2Ring). | {
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635,243 | Just a big ol' slice down the middle. Would it drift apart over time, or eventually fuse back together? | Assuming you don't separate the halves to any great extent, it would immediately fuse back together due to its self-gravity. The tidal forces exerted by the Earth on the Moon are only sufficient to overcome this gravity if it were much closer (about the radius of the Earth actually). This is actually how the Moon is held together, rather than atomic or molecular bonds. However a rough calculation would be to assume the "centres of gravity" of the two halves are separated by $\sim R_M/2$ . This gives a gravitational acceleration between them of $2GM_M/R_M^2$ . The tidal acceleration, assuming it is at its greatest, with the halves separated along the Earth-Moon line, would be $2GM_E R_M/d^3$ , where $d$ is the Earth-Moon distance. The ratio of self-gravity to tidal force is $$\frac{g}{g_{\rm tidal}} = \left(\frac{M_M}{M_E}\right) \left(\frac{d}{R_M}\right)^3\ .$$ Putting the numbers in, the ratio is $1.5 \times 10^5$ . EDIT: To distinguish my answers from others, let me point out that if you sliced the Moon in half and then separated the halves by a distance much greater than $R_M/2$ , let's call it $s$ for separation, then the ratio calculated above changes to something like $$\frac{g}{g_{\rm tidal}} = \left(\frac{M_M}{M_E}\right) \left(\frac{d^3}{2s^3}\right)\ .$$ Thus if you separated the halves by $\sim 40 R_M$ along the line from the Earth to the Moon, then tidal forces would overcome self-gravity. EDIT2:
This calculation should be regarded as very rough and is more accurately a condition for the fragments to form a stable, bound binary system. The problem of a low-mass binary orbiting a larger mass does have approximate criteria for long-term stability (e.g. Donnison 1988 ). It does depend a bit on eccentricity and on whether the two fragments orbit in retrograde or prograde fashion. For a binary separation $s$ and distance from the Earth $d$ , the results of Donnison suggest ratios of $d/s > 12$ (prograde) or $>22$ (retrograde), for something with the mass ratio of the Moon/Earth - so $s<20R_M$ or $<11 R_M$ respectively. Inevitably, what will happen and whether the two halves would fall back together, would depend on the initial conditions. If you just magically separate them along a radial line from the Earth but with each fragment keeping the original angular velocity, then that necessarily means you inject a large amount of angular momentum around the centre of mass and the fragments would just orbit each other.
Further speculation is not worth the effort without a specific set of initial conditions. | {
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635,640 | I have been reading about the exclusion principle a little bit, but I have some questions about it. How does the information about the state of electrons get "passed around" so that other electrons in similar state can not have that same state? Is there some kind've information carrier? Is there some unique force created by a set of quantum numbers for a fermion so that, the force then prevents another fermion with the same set of quantum numbers from being permitted? How do electrons know the states of other electrons to determine if their state is allowed? | Sometimes people say that the Pauli exclusion principle says that "two electrons can't be in the same state." This is not correct. It's not as though each particle has its own "state" that it keeps to itself. It's actually much deeper: the electron field itself has one state it's in (that's the whole point of quantum field theory) and crucially there are simply no states of the electron field corresponding to two electrons with the same spin, position, etc. There is nothing for it to "know," it's just what the quantum electron field is. Fermionic fields are different from bosonic fields. A boson field also has a single state it's in, but the difference between a bosonic field and a fermionic field is simply that the boson field does have states that correspond to boson particles with the same position, spin, etc. Bosonic fields have way more states than fermionic fields. (Actually, going a bit further, EVERY field together is really in a single universal state, but that's not really so important for the question at hand.) | {
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636,630 | Let us imagine you are in a vacuum and after having maintained a speed of 0 km/s (standing still) you accelerate to 297,000 km/s (99%). You know this is now your speed because you have a speedometer telling you so. You then decide to maintain that speed for a while. With the speed of light is always ~300,000 km/s faster than you, what is preventing you from (again in your reference frame) increasing your speed, as shown by a speedometer, an arbitrary amount faster than ~300,000 km/s? After all, the speed of light will always be always faster. I feel like length contraction even backs this since it will make your space wheels tinier. You're essentially scaled down and your tiny wheels would have to rotate many more times to go the distance just 1 rotation would have taken you with your non-contracted length. This then would cause the speedometer to relay speeds faster than the speed of light. | You know this is now your speed because you have a speedometer telling you so. This is precisely where you hit a (metaphorical) roadblock. A speedometer must use something outside of your reference frame to measure your speed, as speed inside your frame is either 0 or meaningless (take your pick). It's measuring the speed of your space wheels. Your space wheels will never spin faster than the speed of light (where the speed of your wheels is the linear velocity at the edge of the wheel). It's measuring your current work being put in and converting it to a speed. Then you are measuring kinetic energy and not speed. There is no upper limit on the kinetic energy an object can have! However when you solve backwards for speed infinite kinetic energy leads to your speed approaching (but never exceeding) c. $$ KE = mc^2 (\gamma - 1) \\
\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} = \frac{KE}{mc^2} + 1 \\
\sqrt{1 - (v/c)^2} = \frac{1}{\frac{KE}{mc^2} + 1} \\
1 - (v/c)^2 = \left( \frac{1}{\frac{KE}{mc^2} + 1} \right)^2 \\
v = c \cdot \sqrt{ 1 - \left( \frac{1}{\frac{KE}{mc^2} + 1} \right)^2} \\
$$ If you plug in $\infty$ for KE you should see that you recover $c$ . | {
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636,662 | According to this article : Imagine that all of space is uniformly filled with an invisible substance—now called the Higgs field—that exerts a drag force on particles when they accelerate through it. Push on a fundamental particle in an effort to increase its speed and, according to Higgs, you would feel this drag force as a resistance. Justifiably, you would interpret the resistance as the particle’s mass. For a mental toehold, think of a ping-pong ball submerged in water. When you push on the ping-pong ball, it will feel much more massive than it does outside of water. Its interaction with the watery environment has the effect of endowing it with mass. So with particles submerged in the Higgs field. So if the Higgs field is present in all space, why are there massless particles? Does that mean that they do not interact or go through the Higgs field? | The usual pop-sci explanation of "the Higgs field exerting a drag force on particles that move through it, sapping their kinetic energy" is unfortunately not very accurate. In technical terms, the mass generation for the weak gauge bosons is really due to the spontaneous symmetry breaking in the Higgs mechanism , and the associated Yukawa couplings generate mass terms for the fermions of the Standard Model (except neutrinos). There is additionally an important distinction to be made between the "Higgs boson" and the "Higgs field", though these are often conflated and even fused in pop-sci descriptions. I will attempt to explain this in a manner that is more accurate than the pop-sci description while still being accessible. The punchline is: Although the massive particles continuously interact with the Higgs boson which is present throughout space, this does not give them mass. The Higgs mechanism gives them mass once and for all at the electroweak transition scale. "Spontaneous symmetry breaking at the electroweak transition scale" roughly means this: at very high energies, the picture of particle physics and phenomenology is very different from ordinary energies. "Very high energies" is entirely equivalent to "very small distances", and we call the energy at which we are looking at the theory the "energy scale". If, starting from very high energies, you begin to decrease the energy scale, when you reach the electroweak scale at 160 GeV, the Higgs field "condenses". An almost perfect analogy for this phenomenon is how, starting at say 20 °C, you can decrease the temperature of water until it suddenly freezes into ice at 0 °C. Now 160 GeV roughly corresponds to a temperature of $1.85\times 10^{15}\ \mathrm K$ , which is really, really high, but still within the domain of the Large Hadron Collider. At this point, and this point alone , the gauge fields and fermion fields gain a mass term, and hence their quanta - the gauge bosons and fundamental fermions - are no longer massless and become massive. At energies below this scale, they do not have to interact with anything (no pop-sci-esque "Higgs boson fluid") to keep this mass. Indeed, massless and massive particles are fundamentally different in quantum field theory, so a "continuous" interaction type of approach to give mass to particles is doomed right from the get go. To make sure that e.g. photons do not gain a mass in this process, we encode the exact way that the symmetry breaking takes place into the model. This choice of symmetry breaking is not arbitrary, nor is it predicted by theory. The whole idea of electroweak spontaneous symmetry breaking is to explain the masses of the $W^\pm$ and $Z$ gauge bosons (and fermions), all with the underlying assumption that the photon is massless, and this is what we put into our model by hand. We could very well have built a similar model where the photon is massive - but we don't, because we don't observe such a thing. Everything need not interact with everything else - the electromagnetic field too permeates throughout spacetime, but does not interact with uncharged particles. So the real reason that the photon does not gain a mass by the Higgs mechanism is because we don't want it to, otherwise our model would be inaccurate. Nevertheless, if you are interested in probing deeper, here are The specifics (but not too technical) For now, focus only on the nature of the $W^\pm$ , $Z$ and $\gamma$ (photon). Ignore the gluons, since they play a mere spectator role in the mass generation - they are unaffected (we say that "the $\mathrm{SU}(3)$ color group remains unbroken"). This is due to the nature of our model building - we should be able to declare that certain particles are massless, and dictate which particles should interact with each other, provided that the resulting model is consistent with the theoretical framework. Unfortunately, the simplest model that we can build to accommodate the massive nature of the weak force bosons and fermions fails immediately for a straightforward reason: the existence of a mass term violates the gauge symmetry of the fundamental forces, and so our theory is mathematically inconsistent. The Higgs mechanism is the simplest (and only viable) method to explain this, but curiously the photon is not merely a spectator like the gluons and is present in the mechanism. Prior to spontaneous symmetry breaking , i.e. at energies higher than the electroweak scale, there are four massless gauge fields living peacefully - call them the $W^1, W^2, W^3$ and $B$ . In other words, there are four electromagnetic-like forces above 160 GeV, and associated to each one is its own "charge" determining the strength and nature of interactions, exactly the way the photon/electromagnetic field at ordinary energies couples to the electric charge. These gauge fields coexist with the Higgs field - which, most importantly, has a non-zero charge under all of these fields. At the electroweak transition scale, the Higgs field undergoes spontaneous symmetry breaking. We say that the $\mathrm{SU}(2)_L\times \mathrm U(1)_Y$ symmetry is broken down to $\mathrm U(1)_\mathrm{EM}$ . What this means, roughly, is that the original four degrees of freedom (DOF) of the Higgs field break up into a vacuum expectation value with 0 DOF, a Higgs boson with 1 DOF and 3 " Goldstone bosons " with 1 DOF each. This is roughly like how water loses its freedom to flow in different directions after the phase transition, with the DOF "freezing out". So why don't we observe these Goldstone bosons as physical particles at regular energy scales? Well, owing to the nature of the symmetry breaking, three linear combinations of the four originally massless gauge fields "eat" one Goldstone boson each and become massive, forming the $W^+, W^-$ and $Z$ bosons. The final linear combination remains massless - there are no more Goldstone bosons left - and forms the photon. The upshot is that after spontaneous symmetry breaking, there are now 3 massive gauge bosons, 1 massless one (that is distinct from each of the above massless gauge fields) and a Higgs boson, a "remnant" of the symmetry breaking. To stress again, this mass generation occurs at exactly one point - the electroweak transition scale. It is not a process of a massless particle having to continuously "bombard" against the Higgs boson to gain mass or anything of that sort. So if this "mass generation" supposedly happens at a single scale, why can processes like pair production take place at any energy? It's because these are two completely distinct phenomena. In the Higgs mechanism, we are fundamentally altering the particle content of the theory. Above the electroweak scale, there simply aren't any massive fundamental particles, and below it, there are. Pair production of mass on the other hand is simply a consequence of the mass-energy equivalence of special relativity. If there did exist massive particles above the electroweak scale, then pair production would also be possible there - they are unrelated processes. "Mass generation happens at a single scale" should not be interpreted as "all mass is created forever, and no mechanisms like pair creation and annihilation can change this". It means that the quanta of the post-SSB fundamental fields now have a mass, whereas the quanta of the pre-SSB fields did not. These are very distinct statements. It is seen that while only the massive fields end up interacting with the residual Higgs boson post-SSB, these interactions don't play any role in generating masses. The existence of the Higgs boson does however serve as an important verification of the Higgs mechanism. There are of course no a priori reasons that photons and gluons should be massless gauge bosons (at ordinary energies) - these are experimentally validated through the nature of the forces that they produce. That is, the massless nature of photons makes accurate predictions of the resulting Coulomb-like electromagnetic force that we observe in nature. Indeed, there is no 100% experimental confirmation of the masslessness of the photon - we can only impose tighter and tighter upper mass bounds, as mentioned here . The massless model for gluons makes similarly accurate predictions, though the subtlety of non-abelian gauge theories means that the mediated force does not manifest as a Coulomb-like force. | {
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637,196 | Is there any way to know whether a group of particles is generated from the sun rather from an artificial source? | Photons are identical particles characterized by energy and a direction of propagation .
If you see just a photon, without any other information, from these two properties, you cannot distinguish a solar photon from one coming, let's say, from a tungsten filament at a temperature of $5800\,\mbox{K}$ (the surface temperature of the Sun). On the other hand, if you see many photons coming from the big glowing spot in the sky, you can reliably tell that they come from the Sun. | {
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637,536 | On many mornings I get this cool light pattern on my ceiling: It's light coming in through the blinds, but there is this rippling/wavy/moving effect. Its intensity varies (as can be seen a little in the GIF). I am basically just curious as to what it is and why it occurs. My partner has seen this in other houses but I never have. The light is coming in after bouncing off an AC unit This ended up not being really correct; see edit below : So I was wondering if that has something to do with it (e.g. some sort of mirage?). Otherwise, I was having trouble finding out other information online. Has anyone seen this before? Any potential explanations for what is happening? First time poster here; please help me improve the post if needed! More information (and answer): I have now verified that this is actually light bouncing off water (caustic reflections) ; I will accept the answerer who suggested/explained this possibility. Here are the fancy lights happening again this morning after the AC had been off for a few minutes : Sure enough, it rained yesterday, and there is a puddle outside which is redirecting direct sunlight directly into the window: So, I apologize for my original post being misleading. I said this was light "bouncing off the AC unit" - this is I guess true (sort of trivially), but not as significant as the direct sunlight reflecting off the puddle. I realize this incorrect/incomplete information introduced a bias. That being said, kudos to the answerer who proposed the correct explanation! It is kind of a particular set of prerequisites that have to occur for this light effect to take place (i.e. it rains enough to create the puddle, but sunny for the ~hour when the angle is correct). Just as an aside, there is this funny sort of confounding variable thing going on. The asphalt shown in the puddle photo was recently all torn out; I think the puddle only started appearing since then, concurrent with about when we started running the AC more. One final note is that the pattern is much less striking when the blinds are more open: There is some movement, but not nearly the amount seen when the blinds are closed (and the boundaries of the light don't grow and shrink as much). Anyway cheers to all who have given input on this post! | These are probably Reflection Caustics . The video Taming light reflection to create images discusses engineered caustics (refractive rather than reflective in this case). See also this post from the EPFL Geometric Computing Laboratory: I don't think the explanations of refraction from density variations in air actually work in this case. Schlieren photography is a real thing, but the variation of index of refraction of air at different temperatures is very small, of order $10^{-6}$ per degree C, so you need special circumstances, highly collimated light and/or long distances to see such an effect. See this work for example. You might need one of these (from Window Air Conditioner and Rain ): The light is coming in after bouncing off an AC unit I saw a similar effect years ago and looked out the window to find that sunlight was reflecting off of a puddle of water on top of a slightly concave top panel of the AC unit. Either somebody set something heavy on it to try to stop noisy vibrations, or sat on it, or it perhaps deformed over time for other reasons. These normally have a matte finish, AC window units are not normally shiny polished metal that can reflect so nicely. So I suspect there is a puddle of water on top of a slightly concave metal panel, and we are seeing caustic effects from Fresnel reflections from ripples on the water induced by wind blowing or AC compressor vibrations. From James Gurney's blogpost Caustic Reflections From forums.sketchup.com's Water Reflection on Surfaces : Concave AC unit top with matte finish: These don't reflect by themselves, there's probably water on top. Source: Shutterstock | {
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637,557 | My daughter was doing a physics problem where she had to calculate the gravitational force of an object that was 250 km above the earth... She used the equation: $$F=\frac{G\cdot m_1 \cdot m_2}{r^2}$$ For $r$ , she used the distance between the two centers of gravity: 250 km + the radius of the earth (6371 km) But I don't think this is correct. While it is true that the center of gravity of the earth is in the middle of earth, I would imagine that the effective center of gravity is actually closer... Consider the same problem where your weight is a 1 kg weight 250 km above 'earth', but replace earth with two 1 kg weights spaced 6371*2 km apart, which are inline with the object floating above it. Note that the new earth has the exact same center of gravity as the old earth. So, now using the equation: $$F=\frac{G\cdot m_1 \cdot m_2}{r^2}$$ we get: $$F=\frac{G\cdot 1\,\mathrm{kg}\cdot 2\,\mathrm{kg}}{(6371\,\mathrm{km}+250\,\mathrm{km})^2}$$ which resolves to: $3.04423\times 10^{-24}\,\mathrm{N}$ But, if instead we sum the forces of the two weights individually, we get: $$ F = \frac{G\cdot 1\,\mathrm{kg}\cdot 1\,\mathrm{kg}}{250\,\mathrm{km}^2} + \frac{G\cdot 1\,\mathrm{kg}\cdot 1\,\mathrm{kg}}{(12742\,\mathrm{km}+250\,\mathrm{km})^2} $$ which gives you: $2.06801 \times 10^{-21}\,\mathrm{N}$ , which is not the same thing at all. So, I'm wondering how one would actually calculate the force of gravity on a 1 kg (point object) 250 km above the earth... | In general, you cannot treat arbitrary mass distributions as if they were concentrated at the center of mass for the sake of calculating the gravitational force, as your example shows. You can, however, do this for spherically symmetric mass distributions if you're outside that mass distribution (even if the radial mass distribution is not uniform) by the shell theorem 1 . So for a point outside the (roughly) spherically symmetric earth your daughter's calculation is correct, but for a more general setup, one would need to do it more explicitly. 1. The same theorem also includes the interesting fact that if we have a hollow shell, the gravity inside that shell caused by the shell's mass is zero, while outside of it, it acts the same as if all its mass was concentrated in the center. | {
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638,194 | I'm starting my studies now, and wondering why a lot of things that at the beginning I don't know if the motivation is based on the intuition of concrete thinking or the logic of abstract thinking, and this is the case with the definition of vectors. I understood that the vectors are composed by direction and magnitude, which represent the difference between two points, which mathematically speaking makes intuitive sense to be a line. But why are they defined exactly as a line if not always when applied to physical concepts, for example, not necessarily what happens between these two is a line? | A vector is not a line. Nor is it "something that has magnitude and direction". A vector is instead a type of mathematical object that can be used to represent things with magnitude and direction, or even just direction. That is, vectors come first , "magnitude and direction" come second. In fact, in the most general sense of vectors, vectors do not have "magnitude" in an absolute sense, and even their sense of "direction" is fairly weak. What a vector "is" is an element of a vector space. That may seem really unhelpful, but one can say vector spaces are designed specifically to make working with quantities where we need to encode magnitude and direction information into a single object easy and useful. A slightly less abstract way to look at it is vector spaces generalize the idea of a list of numbers: $$\langle v_1, v_2, \cdots, v_n\rangle$$ where we can add those numbers in an elementwise fashion, as well as multiply them all by a single specific number (called a "scalar"). This lets us encode magnitude and direction in a neat way, because if we interpret that list of numbers as a list of Cartesian coordinates, we can say the direction represented is that of an arrow from the origin to the point given by the numbers we have so taken as coordinates (i.e. the point $(v_1, v_2, \cdots, v_n)$ , where we have used different bracketing styles to distinguish the point from the vector), and the distance from the origin to the point is the magnitude. That's what you're seeing when you see "lines" or "arrows". But the vector itself is just this list of numbers, not the line or arrow, which is a pictorial representation of how the magnitude and direction information are encoded. The relation between these two ideas is that when you go through the essential properties that "adding and multiplying in elementwise fashion" entail, you can prove that the only spaces that have all those properties can be considered "equivalent to" such a space of number lists, possibly infinitely long (though things get tricky in that case). (This is described in linear algebra texts as "there is one vector space of each dimension up to isomorphism", where "isomorphism" here means the equivalence in question.) So when the vectors get used in physics, we are not using lines, we are using a purely abstract object which in some cases it is useful to interpret as describing a line, but in other circumstances, it may not be. And we may draw it using a little arrow or similar symbol when we want to convey or visualize the encoded magnitude and direction information. | {
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638,529 | The rainbow! What magical "thing". And even if you see the droplets of rain move in a sunlit storm, she's steady. I have been trying to understand but there are so many drops involved! And they are moving in turbulent ways on top. So what's going on? I know that each droplet sends a "rain circle" cone towards us, and that our eyes are sprayed with these cones. These cones all have the same orientation, no matter how the droplets move. Somehow this must be the key, but I don't see how. | The color depends on the relative angle between you, the drop, and the sun. If you were to track an individual drop it would change color as it falls "through" the rainbow. That would be cool to see! In more detail: Consider a cartoon similar to the one in @John Hunter's answer . There is a viewer at left looking right. There is a sun behind the viewer. For every point to the right of the viewer we can draw lines from that point to both the viewer and the sun. So we correspond an angle to EVERY point in space. We can ask "what are the surfaces of constant angle?" Well, imagine a line through the sun and the viewer. All points along this line will be angle zero. Larger angles will be cones coaxial with this center line. These are the surfaces of constant angle. A rainbow works because, for certain angles, the droplet will reflect the various spectral components of the sun to the viewer. So what is fixed in space, given the position of the viewer and sun, are the cones of constant angle. These are present whether or not there is water. A rainbow arises when there are water droplets occupying the appropriate cones. As a droplet falls, it passes through the different cones, meaning the component of the sun that it reflects changes in time as it passes through the cones of differing angles. | {
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638,530 | I don't have much experience in classical field theory and have been trying to study it for the past week. However, I don't know if my understanding of the equations of motion for the fields are correct. Specifically, the Hamiltonian of the Klein-Gordon field is given by: $$ H[\pi,\phi] = \int\frac{1}{2} (\pi^2 + |\nabla \phi|^2+m^2\phi^2) \text{ } d^3x, $$ where $\pi$ is the conjugate momentum to the field. And we have the usual Hamiltonian equations: $$\begin{align}\dot\pi &= -\frac{\delta H}{\delta \phi}\cr
\dot\phi&= \frac{\delta H}{\delta \pi}.\end{align}$$ This seems pretty straight forward from here but the Hamiltonian is supposed to be a function of $\phi$ and $\pi$ but we also have terms that depend on the spatial derivatives of the fields. If my memory is accurate, I don't remember having this problem in classical mechanics. Should the spatial derivatives of the fields be treated as being dependent on the on the field itself? If so, how do we compute the term $\frac{\delta H}{\delta\phi}$ ? | The color depends on the relative angle between you, the drop, and the sun. If you were to track an individual drop it would change color as it falls "through" the rainbow. That would be cool to see! In more detail: Consider a cartoon similar to the one in @John Hunter's answer . There is a viewer at left looking right. There is a sun behind the viewer. For every point to the right of the viewer we can draw lines from that point to both the viewer and the sun. So we correspond an angle to EVERY point in space. We can ask "what are the surfaces of constant angle?" Well, imagine a line through the sun and the viewer. All points along this line will be angle zero. Larger angles will be cones coaxial with this center line. These are the surfaces of constant angle. A rainbow works because, for certain angles, the droplet will reflect the various spectral components of the sun to the viewer. So what is fixed in space, given the position of the viewer and sun, are the cones of constant angle. These are present whether or not there is water. A rainbow arises when there are water droplets occupying the appropriate cones. As a droplet falls, it passes through the different cones, meaning the component of the sun that it reflects changes in time as it passes through the cones of differing angles. | {
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638,699 | After a star becomes a White dwarf, it resists gravitational collapse mainly due to the electron degeneracy pressure. If the mass of the white dwarf is greater than the Chandrasekhar limit, the degeneracy pressure cannot resist the collapse any longer and is doomed to become a neutron star or a black hole. Why can't the degeneracy pressure keep on self-adjusting itself to resist collapse forever? | The basic problem is that for a sufficiently massive star, the electrons become relativistic. The fine details of this calculation are rather complicated, but you can get a qualitative sense of the argument as follows: For non-relativistic fermions at zero temperature, it is possible to show that the total energy of $N$ particles in a box of volume $V$ is proportional to $N^{5/3}/V^{2/3}$ . This can be done via counting the density of states, and using the fact that the energy of a non-relativistic particle obeys $E \propto |\vec{p}|^{2}$ .
For a spherical volume of radius $R$ , we have $R \propto V^{1/3}$ , and the number of fermions present is proportional to the mass. This means that the total energy of the fermions is proportional to $M^{5/3}/R^2$ . This energy is positive. On the other hand, the gravitational energy of a solid sphere is negative and proportional to $M^2/R$ . This means that the total energy is the sum of a negative $R^{-1}$ term and a positive $R^{-2}$ term, and such a function will have a minimum somewhere. This will be the equilibrium point. At smaller radii, the energy of the degeneracy grows faster than the binding energy decreases, pushing the radius back to larger values. At larger radii, the reverse occurs. This means that the star will be stable. This argument doesn't hold up to arbitrarily large energies, though, because eventually the Fermi energy of the electrons exceeds the rest energy of the electron; in other words, the electrons become relativistic. This changes the relationship between energy and momentum of the electrons. For highly relativistic electrons, we have $E \propto |\vec{p}|$ instead; and going through the same calculations (neglecting the electron mass entirely), we find that the total energy of a relativistic fermion gas is proportional to $N^{4/3}/V^{1/3} \propto M^{4/3}/R$ . The gravitational binding energy, on the other hand, remains negative and proportional to $M^2/R$ . This implies that the overall energy is itself proportional to $1/R$ , and there is no extremum of the total energy of the system. Since the fermion energy and the binding energy always increase or decrease at exactly the same rate, there will be no stable equilibrium radius. The star will either blow itself apart or collapse in on itself, depending on whether the kinetic energy of the fermions or the gravitational binding energy wins out. | {
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638,704 | I've recently started studying the concept of space-time diagrams in special relativity, and I came across the concept of representing the time axis using $ct$ , with units being that of length. Now I'm told that this is done, first of all, to keep the speed of light = 1, and give a common unit to both the axis, so that we can show, that space and time are inherently the same thing. However, I'm still not being able to understand, how can I intuitively think of time in meters or any unit of length. Also, doesn't plugging $ct$ and $ct'$ into Lorentz transformations, make it dimensionally inconsistent.
For example : $x' = \gamma(x-vt)$ if $x$ and $x'$ are in terms of distance, and so is $t$ , then the term inside the bracket becomes dimensionally inconsistent. So, how can I intuitively measure time in meters, or solve Lorentz transformations this way? What exactly does setting the speed of light $c = 1$ mean? It would be really helpful, if someone can explain to me the motivation, and the intuition behind expressing both axis in the same units. Moreover, what would have happened, if we had kept them in separate units? | The basic problem is that for a sufficiently massive star, the electrons become relativistic. The fine details of this calculation are rather complicated, but you can get a qualitative sense of the argument as follows: For non-relativistic fermions at zero temperature, it is possible to show that the total energy of $N$ particles in a box of volume $V$ is proportional to $N^{5/3}/V^{2/3}$ . This can be done via counting the density of states, and using the fact that the energy of a non-relativistic particle obeys $E \propto |\vec{p}|^{2}$ .
For a spherical volume of radius $R$ , we have $R \propto V^{1/3}$ , and the number of fermions present is proportional to the mass. This means that the total energy of the fermions is proportional to $M^{5/3}/R^2$ . This energy is positive. On the other hand, the gravitational energy of a solid sphere is negative and proportional to $M^2/R$ . This means that the total energy is the sum of a negative $R^{-1}$ term and a positive $R^{-2}$ term, and such a function will have a minimum somewhere. This will be the equilibrium point. At smaller radii, the energy of the degeneracy grows faster than the binding energy decreases, pushing the radius back to larger values. At larger radii, the reverse occurs. This means that the star will be stable. This argument doesn't hold up to arbitrarily large energies, though, because eventually the Fermi energy of the electrons exceeds the rest energy of the electron; in other words, the electrons become relativistic. This changes the relationship between energy and momentum of the electrons. For highly relativistic electrons, we have $E \propto |\vec{p}|$ instead; and going through the same calculations (neglecting the electron mass entirely), we find that the total energy of a relativistic fermion gas is proportional to $N^{4/3}/V^{1/3} \propto M^{4/3}/R$ . The gravitational binding energy, on the other hand, remains negative and proportional to $M^2/R$ . This implies that the overall energy is itself proportional to $1/R$ , and there is no extremum of the total energy of the system. Since the fermion energy and the binding energy always increase or decrease at exactly the same rate, there will be no stable equilibrium radius. The star will either blow itself apart or collapse in on itself, depending on whether the kinetic energy of the fermions or the gravitational binding energy wins out. | {
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638,847 | My mom told me that one can check whether an egg is rotten by sinking it in a glass of water. If it floats, then it is rotten. I didn't find any explanation for this phenomenon. If anyone knows one, please answer. | The domestic chicken's egg shell has about 7000 pores that allow the embryo to breathe. When an egg rots the yolk and surrounding materials decompose and they give off gasses which can pass through the shell. This allows mass to leave the interior of the egg resulting in less density for the volume of the egg making it more buoyant. | {
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638,848 | I cannot comprehend how the increasing of amplitude of an electromagnetic wave increases the number of photons. How does this even happen. I am also not able able to make sense of that fact that, when you increase frequency, you increase the energy of each individual photon. How does this happen? Is there a better way to visualize this | The domestic chicken's egg shell has about 7000 pores that allow the embryo to breathe. When an egg rots the yolk and surrounding materials decompose and they give off gasses which can pass through the shell. This allows mass to leave the interior of the egg resulting in less density for the volume of the egg making it more buoyant. | {
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638,988 | Can you please explain the thing below? When we add a battery in circuit then, it gives out some electric field that moves through the circuit and gives a force on electrons in conductor to produce current. When we connect two batteries parallel with each other in a circuit then the electric fields coming out of both batteries moves through the conductor which should give more flux than before. As there is the double number of fields in the same area. But the voltage doesn't double. What have I done wrong here? Are there any increasing flux/electric field magnitude when two batteries are connected parallel in a series? New Edit: I think I have a got three of the very close answers. Let us think about what is going on on the positive terminal and then we will apply the same to the negative terminals just to make it easy to understand. "Because the batteries were originally designed for V volts. But on adding batteries in parallel makes a large amount of repulsion in positive terminals that is felt by both batteries and reduces the amount of charge on each terminal which also decreases the voltage of each battery. The total charge of both positive terminals is the same as we get when using battery." "But there are people also saying that it depends on concentration of charge not amount of charge" "The one another explanation was: first assume that E is the electric field released by each battery, that half of electric field(E/2) from one positive terminal moves toward another positive terminal therefore reducing its half strength and the another half field (E/2) moves in the circuit (not toward the positive terminal) to increase the strength by (E/2), therefore making the total electric in through the resistor = E " | tl;dr Batteries do not create electric fields to move charges. They move charges, which creates electric fields. a battery [...] gives out some electric field that moves through the circuit and gives a force on electrons in conductor to produce current. This description is, if not completely wrong, at least misleading. A battery is not a source of electric field , it is a source of electric potential . Imagine a battery with terminals in the shape of a pair of parallel conductive plates with an air gap in between (this is a capacitor): as you move the plates toward each other, the field strength (between the plates) increases, and as you pull them apart, the field strength decreases. There is no upper limit to how strong the field can be (well, until it reaches the breakdown voltage and begins to arc), and no lower limit either -- the strength of the field is not determined by the battery. So the battery itself does not directly create an electric field between its terminals. Moreover, electric field does not "move through" a circuit; charges do. In a simple DC setup like a battery driving current through a resistor, all electric fields are stable over time -- the charges move through the circuit, but the field itself does not. Saying that electric field moves through a circuit is a bit like saying that gravity moves through a rollercoaster. Moving electric fields do come into play with AC circuits and devices with moving parts, such as electric motors; but even a non-moving electric field causes charges to move through a conductor (this is, after all, essentially the definition of "conductor"). So how does current work? Batteries are a source of electric potential , which is measured in volts. Potential is a kind of pressure and in a typical battery this pressure is caused by chemical reactions inside the battery pumping electrons from one terminal (+) to the opposite one (-). The potential difference between the terminals does create an electric field. In the experiment above where the terminals of the battery are parallel flat plates separated by a distance $d$ , you could calculate the field strength between them as simply $E = {V \over d}$ . But the field is just a way of observing the potential difference between the terminals: it's not the reason the charges are moving in the first place (which is, again, the chemical reactions happening inside the battery). Pressure is relative, and electric potential is no different. To be strictly accurate, what a battery provides is an electric potential difference ("voltage") between its two terminals. This is basically a measurement of how strong an electron-pump the battery has inside it. This is why connecting two identical batteries in parallel does not add their voltages: because they both provide the same potential difference between the (+) and (-) terminals, there is essentially no difference between two batteries connected in parallel and two isolated, disconnected batteries. The electric potential between the (+) terminals and (-) terminals is the same, and both batteries "agree" on it, so there is no reason for current to flow between them. (Connecting two mismatched batteries in parallel will cause current to flow, in a direction determined by which battery has higher voltage.) What connecting two batteries in parallel does do is change how the system behaves when under load. If you connect a load, say a 1kΩ resistor, across the terminals of a single 1.5-volt battery, the current through the resistor will be 1.5 mA, all of which is supplied by the same battery. If you connect the same load across the terminals of two 1.5-volt batteries connected in parallel, the current through the resistor will still be 1.5 mA, but now each battery only has to supply 0.75 mA of current. This means that each individual battery is under less load than before, because the electrochemical pumps inside it only have to move half as many electrons to maintain the same voltage. These batteries may last longer and behave better under a variety of loading conditions. (This assumes ideal conditions and perfectly matched batteries. In real-world scenarios, batteries are never perfectly matched, and so you may need a load-balancing circuit to protect the batteries from each other.) | {
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639,320 | We use a 780 nm laser in our lab, and that makes it in the near infrared (IR) range. The majority of people can not see this wavelength of light. However, when the beam reflects off of an object (see image), the light becomes visible. This image has been taken with an iPhone camera which has a poor (or non existent) IR filter, though the light is visible with the eye. Some questions have been asked which are related. The answer to one of them suggests that when an object is stationary, the reflected beam should lose energy. So why is the reflected beam experiencing an increase in energy here? | You can never see any light beam from the side. You only see light (of whatever wavelength) propagating directly into your eye. When laser beams sometimes appear as a visible line through the air, what is happening is that dust (and to some extent molecules too) in the air are scattering the light, sending some of it towards your eye. When the beam hits a solid object, then unless the surface is extremely flat (such as a precise and clean mirror) there will be scattering at all angles, so some will go towards your eye. It is this light which you are seeing. In the case of infra-red radiation, the human eye sensitivity does not drop off immediately for wavelengths above 700 nm; it is low but non-zero, and the scattered radiation from a laser beam is often bright enough to be seen (obviously it depends on the intensity of the original beam). I have in this way seen 852 nm, for example. However, when you can see a wavelength such as this, you should take care: the radiation entering your eye is brighter than you may think, because your eye's sensitivity is low but you are seeing it. Eye protection is for this reason especially important with wavelengths outside the normal visible range. | {
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639,328 | This may sound trivial but I am having a hard time linking different statements of second law. What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy. Also from second law, entropy of any isolated system never decreases. How are the two statements similar? Please use simple layman terms, I am simply trying to grasp the idea. | You can never see any light beam from the side. You only see light (of whatever wavelength) propagating directly into your eye. When laser beams sometimes appear as a visible line through the air, what is happening is that dust (and to some extent molecules too) in the air are scattering the light, sending some of it towards your eye. When the beam hits a solid object, then unless the surface is extremely flat (such as a precise and clean mirror) there will be scattering at all angles, so some will go towards your eye. It is this light which you are seeing. In the case of infra-red radiation, the human eye sensitivity does not drop off immediately for wavelengths above 700 nm; it is low but non-zero, and the scattered radiation from a laser beam is often bright enough to be seen (obviously it depends on the intensity of the original beam). I have in this way seen 852 nm, for example. However, when you can see a wavelength such as this, you should take care: the radiation entering your eye is brighter than you may think, because your eye's sensitivity is low but you are seeing it. Eye protection is for this reason especially important with wavelengths outside the normal visible range. | {
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639,350 | I know this question was asked in some way or another, but my familiarity of concepts in electricity are very basic so I ask you to give an explanation using simple terms. I'm learning about Voltage, and I found this video which compares Electrical potential difference to gravitational potential difference . He explains it using the following example: The point in which the ball stands in has a gravitational potential , and a point in the height of the table has a gravtiational potential . So we say that there's a gravitational potential difference between the two points. And so with electricity etc. My question is: What is the difference between potential and potential difference? Where the ball stands now, at 2.0 meters, this point has a potential since this ball can fall to the height of 0.5 meters. So it seems to me that this height (1.5 meters) is embedded in the potential of the point, or we can say - of this situation - i.e., the 2.0 meters above the 0.5 meters. But no, we're assigning a potential difference between the two points. So what is the meaning of the potential of the 2.0 meters point of its own? And if it doesn't have a meaning of its own, why don't we call the difference a potential ? This feels pretty circular. I can go on lol. Please clarify with simple terms | You can never see any light beam from the side. You only see light (of whatever wavelength) propagating directly into your eye. When laser beams sometimes appear as a visible line through the air, what is happening is that dust (and to some extent molecules too) in the air are scattering the light, sending some of it towards your eye. When the beam hits a solid object, then unless the surface is extremely flat (such as a precise and clean mirror) there will be scattering at all angles, so some will go towards your eye. It is this light which you are seeing. In the case of infra-red radiation, the human eye sensitivity does not drop off immediately for wavelengths above 700 nm; it is low but non-zero, and the scattered radiation from a laser beam is often bright enough to be seen (obviously it depends on the intensity of the original beam). I have in this way seen 852 nm, for example. However, when you can see a wavelength such as this, you should take care: the radiation entering your eye is brighter than you may think, because your eye's sensitivity is low but you are seeing it. Eye protection is for this reason especially important with wavelengths outside the normal visible range. | {
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639,577 | Suppose I have an upside down glass of water that I somehow brought in the configuration shown below (without any air between the glass and water). Now the water will obviously fall down but my question is why exactly? The forces on the water column are $\rho Vg$ downwards due to gravity, $P_aA$ upwards due to the atmosphere and $F_g$ downwards due to the glass. If $P_aA>\rho Vg$ (which can be easily achieved by reducing $V$ ), $F_g$ could simply be equal to $P_aA-\rho Vg$ and the water would be in equilibrium and would have no reason to fall down. | You are absolutely correct that the picture shown is in equilibrium. The pressure at the top and sides need not be 0, but it will be less than the pressure on the bottom by the amount of the weight of the fluid. So the net force and net torque on the fluid is zero and there is no tendency to accelerate or rotate. This indeed means that the fluid in this configuration is in equilibrium. There are two types of equilibrium: stable and unstable. Although this configuration is an equilibrium it is an unstable equilibrium. Specifically, this configuration is subject to Rayleigh-Taylor instability Basically, if a small parcel of the water descends and is replaced by an equal volume of air going up, the potential energy of the system is reduced. This means that the system will not tend to return to the original configuration. So any deviation from the perfect equilibrium state will grow exponentially, regardless of how minuscule* it is initially. Since there is always some small deviation, the fluid deforms, forms drops, and falls down as expected from common experience. *Surface tension can actually stabilize very small deviations in some fluid interfaces. | {
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640,059 | I was going through the definition of stress. I came to know that stress is a tensor quantity. When I checked the net about - 'What is a tensor quantity', I came to know that it is neither a scalar quantity nor a vector quantity. So how can a quantity be neither vector nor scalar? Physicists on this site, please explain your answers in simple ways too, that would be just enough for others (who are school students, etc.) to understand. You can do so by keeping the heading "SIMPLER WAY" on top of your answers. | The best way to settle all the problems with vectors, tensors, and other related concepts, is to forget about the sloppy definitions quite widespread among physicists and start with mathematically clean definitions. Only after this first step can it make sense of most of the specialized (and sometimes confusing) physical language. I'll try to provide some reasons for this introductory statement. You start with the question " How can objects be neither vectors nor scalar? ". To provide a direct answer, it would be useful to start with the definition of a vector. The mathematical definition is that a (real) vector is a member of a set of objects such that a binary composition ( sum ) is defined for every pair of vectors. Moreover, for each vector, the concept of multiple by a real number of that vector is defined as an element of the set. The sum of vectors and multiplication by a real number must obey a set of rules (see, for instance, the definition section on the Wikipedia page for vector spaces ). In brief, the sum must be commutative, associative, equipped with a neutral element (the zero vector), and for each vector, an (additive) inverse vector must exist. Multiplication by a scalar (any real number) must also be associative, multiplication by $1$ must be the identity, and multiplication must be distributive with respect to the sum of vectors and scalars. This definition looks much more complicated than the usual physical definition of a vector as a quantity characterized by magnitude and direction. Still, it is more complete (it explicitly contains the relevant information about how to combine vectors), and it is more general (it applies even to vectors for which the concept of direction would be unclear, like the case of functions). Therefore a first partial answer to the original question appears: yes, there are physical quantities that are neither scalar nor vectors. Interestingly, this situation is not confined to the tensors. The state of a fluid at some point, as described by the local velocity, density, and internal energy, is a tuple of physical quantities, which is neither a scalar nor a vector, but instead the combination of one vector and two scalars. Even more important, finite rotations in three dimensions, even if they are characterized by magnitude and direction, fail to be vectors because finite rotations do not commute. Let's go to the tensors. In linear algebra, they emerge naturally from a combination of vectors different from the sum. A tensor space is a vector space made out of two or more starting vector spaces. As such, a tensor is a vector belonging to a set made by the original vector spaces. Therefore, a tensor shares with any other vector all the basic axioms. Tensors of the same tensor space can be summed and multiplied by a scalar. The meaning of the sentence "tensor is not a vector" is that a tensor does not belong to any of the vector spaces used to build the tensor product. It is a vector of a different vector space. To add complexity to this picture, tensor spaces can be used to describe linear transformations of vector spaces into vector spaces. That is the case of the stress or the strain tensor. The stress tensor maps a vector (describing the orientation of a surface element) into another vector (belonging to a different vector space): the force on that surface element. The meaning of the statement that the stress tensor is not a vector refers uniquely to the fact that it is neither a direction nor a force. However, the set of all possible stress tensors can be equipped with a sum and a scalar multiplication obeying all the axioms of a vector space. At a first glance, it may sound confusing, but some investment of time to understand the mathematical definition of a vector space is highly rewarding in terms of physical understanding. | {
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640,065 | Why current decreases when length of resistor increases and How the speed of electricity is almost $c$ (speed of light)? | The best way to settle all the problems with vectors, tensors, and other related concepts, is to forget about the sloppy definitions quite widespread among physicists and start with mathematically clean definitions. Only after this first step can it make sense of most of the specialized (and sometimes confusing) physical language. I'll try to provide some reasons for this introductory statement. You start with the question " How can objects be neither vectors nor scalar? ". To provide a direct answer, it would be useful to start with the definition of a vector. The mathematical definition is that a (real) vector is a member of a set of objects such that a binary composition ( sum ) is defined for every pair of vectors. Moreover, for each vector, the concept of multiple by a real number of that vector is defined as an element of the set. The sum of vectors and multiplication by a real number must obey a set of rules (see, for instance, the definition section on the Wikipedia page for vector spaces ). In brief, the sum must be commutative, associative, equipped with a neutral element (the zero vector), and for each vector, an (additive) inverse vector must exist. Multiplication by a scalar (any real number) must also be associative, multiplication by $1$ must be the identity, and multiplication must be distributive with respect to the sum of vectors and scalars. This definition looks much more complicated than the usual physical definition of a vector as a quantity characterized by magnitude and direction. Still, it is more complete (it explicitly contains the relevant information about how to combine vectors), and it is more general (it applies even to vectors for which the concept of direction would be unclear, like the case of functions). Therefore a first partial answer to the original question appears: yes, there are physical quantities that are neither scalar nor vectors. Interestingly, this situation is not confined to the tensors. The state of a fluid at some point, as described by the local velocity, density, and internal energy, is a tuple of physical quantities, which is neither a scalar nor a vector, but instead the combination of one vector and two scalars. Even more important, finite rotations in three dimensions, even if they are characterized by magnitude and direction, fail to be vectors because finite rotations do not commute. Let's go to the tensors. In linear algebra, they emerge naturally from a combination of vectors different from the sum. A tensor space is a vector space made out of two or more starting vector spaces. As such, a tensor is a vector belonging to a set made by the original vector spaces. Therefore, a tensor shares with any other vector all the basic axioms. Tensors of the same tensor space can be summed and multiplied by a scalar. The meaning of the sentence "tensor is not a vector" is that a tensor does not belong to any of the vector spaces used to build the tensor product. It is a vector of a different vector space. To add complexity to this picture, tensor spaces can be used to describe linear transformations of vector spaces into vector spaces. That is the case of the stress or the strain tensor. The stress tensor maps a vector (describing the orientation of a surface element) into another vector (belonging to a different vector space): the force on that surface element. The meaning of the statement that the stress tensor is not a vector refers uniquely to the fact that it is neither a direction nor a force. However, the set of all possible stress tensors can be equipped with a sum and a scalar multiplication obeying all the axioms of a vector space. At a first glance, it may sound confusing, but some investment of time to understand the mathematical definition of a vector space is highly rewarding in terms of physical understanding. | {
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640,247 | A block of ice as big as a room will have more thermal energy than a cup of hot tea.
But the tea feels hotter because the average kinetic energy (temperature) is higher in case of tea.
I conclude from this that total energy doesn't matter when it comes to what feels hotter. But it is said that steam at 100 °C feels hotter than water at 100 °C because steam has more energy, which is contradictory to my conclusion. Please explain where have I gone wrong. To resolve the confusions: When we heat water, its temperature keep on increasing. Boiling starts when it reaches 100 degree Celsius and bulk vaporization takes place. These vaporised molecules, possessing same energy of water (is at same temperature as it) plus latent heat of vaporisation, is what I am referring to as steam. So, it has more energy but same temperature.
I expected no confusion regarding such steam causing more severe effects. But now that it has, let me mention that this is a common secondary textbook fact, that is taught and studied in India. Here is a link to a related school material: https://byjus.com/questions/what-produces-more-severe-burns-boiling-water-or-steam/ . | What we define as "hot" or "cold" is the transfer of energy -- how much (quantity) and how fast (rate of transfer) -- and how it raises our temperature. The more energy that is transferred from the object quickly, the hotter the object feels. First, steam is in a vaporized phase -- which is why it has more energy. At 100 Celsius, water can exist both in gaseous and liquid phases. However, to vaporize liquid water, an energy input is required. This energy (called vaporization energy) is specific to each material, but if added, won't raise the temperature, but will simply vaporize the liquid into a gas. So, by vaporizing 100C water, you have water vapor at 100 degrees. Similarly, you can condense this vapor, by removing that same amount of energy required to vaporize it. In that case, you'd recover water at 100 degrees. When you touch something hot, it will transfer heat to you until the temperatures have equalized. So when you touch hot water, the water will simply transfer whatever energy it needs to reach your hand's surface temperature (which won't happen, you'll take your hand out much sooner). However, when you touch steam, it will also transfer the condensation energy to you -- which is actually a lot of energy. This energy drastically raises your hand's the temperature, and you feel it as "hot." Consider the simple heat transfer equation: the heat transfer rate $H$ is $$H=kA \dfrac{T_\text{hot}-T_\text{cold}}L$$ $L$ is unimportant for our case. What is important is $k$ , the thermal conductivity constant -- this constant depends on the material. The higher this constant, the faster heat gets transferred, so more heat gets transferred, and your hand's temperature increases. Next, $A$ represents the contact area between the surfaces. As @Wrzlprmft points out, steam can more easily enter skin pores. This will ensure more heat is transferred, since the total contact area is greater. We can also maximize heat transfer by increasing the temperature difference, $T_\text{hot}-T_\text{cold}$ . The greater this difference, the greater the heat flow. Note that as heat flows, this difference will shrink. In the case of water, $T_\text{hot}$ gets lowered and $T_\text{cold}$ gets higher. However, with vapor at 100C, the condensation energy leaves the vapor first, without changing the gas's temperature, so $T_\text{hot}-T_\text{cold}$ shrinks more slowly; $T_\text{hot}$ does not change, and thus, heat transfer is faster. Furthermore, the condensation energy is, for lack of a better word, quite large, which means that a lot will be transferred at that high rate. TLDR: The reason steam feels hotter, is that it can transfer more energy to us faster (that is, without decreasing its temperature by transferring condensation energy), whereas water cannot. Our feeling of what's hot is determined by how much energy and how quickly an object transfers that energy to raise our temperature. Edit: I forgot to mention that unlike water, steam can be packed tightly because it is a gas. Depending on how compressed the steam is in a given volume, you may experience 100C steam to feel warmer or colder than 100C water. For the purpose of my answer, I assumed the steam to be dense and tightly packed -- which can eventually make up for steam's lower thermal conductivity constant. | {
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640,875 | The writings of Rupert Sheldrake tend to provoke strong emotions, be they ridicule, curiosity, outrage, sympathy, disgust, or otherwise. While Physics SE is not an appropriate forum in which either to debunk or to promote his general worldview, it does strike me as an appropriate place in which to examine some of the more specific claims he has made. One recent claim which struck me as interesting - I lack the expertise either to support or to denigrate it any further than that - was that various celestial bodies could be conscious - note the conditional - on the grounds that the physical structures within, for example, our sun are at least as complex as those found within the human brain . (The paper in which Sheldrake sets out these ideas can be read here .) Is there any evidence in support of the specific claim about the complexity of structures within our sun? Is there any evidence to the contrary? Is it even meaningful, within the language of physics, to talk about Structure A being more complex than Structure B? | The structure of the interior of the sun has been extensively studied by the technique of helioseismology and the findings are consistent with a model in which that structure is relatively simple. Helioseismology yields structural information on length scales of order ~tens of thousands of kilometers and thus cannot reveal any fine structure which might be present in the interior. However, observations of the gaseous surface of the sun with powerful telescopes and radar show that there is lots of structure on relatively short (~thousands of kilometers) length scales in the form of convection cells that "boil" hot gas up to the surface and suck cooler gas down, plus bunched-up zones of strong magnetic field activity. Almost all of that short-scale structure is transient and does not persist for more than timescales of order ~hours to days. But the complexity of those structures is of a fundamentally different sort than that of the human brain, in that the brain is interconnected with nerve pathways that convey phenomenally complicated signals from one part of it to other parts, that those nerve pathway structures persist on timescales of order ~tens of years, and the nerve pathways consist of protein molecules of equally phenomenally complex molecular structure. Nothing even remotely like that is known to exist in the sun. Finally, note that the solar structure models account well for the observed behavior of the sun with no assumptions at all about short-scale "Sheldrake" structure. If there were such structure present, it is unlikely that the models would work properly since they do not include them. Sheldrake's implication that astronomical bodies are therefore capable of consciousness and emotions is without any experimental evidence. Note that there are phenomenal amounts of "structure" in a sand dune viewed through a microscope, but nothing that would support anything like consciousness. | {
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640,902 | In every book or pdf I read about the path integral, I see: $$
\begin{array}{ccl}
\langle q_2,t_2|\hat{q}(t_1)|q_0,t_0\rangle & = & \displaystyle{\iint} dq_3dq_4\langle q_2,t_2|q_3,t_1\rangle\langle q_3,t_1|\hat{q}(t_1)|q_4,t_1\rangle\langle q_4,t_1|q_0,t_0\rangle\\
& = &\displaystyle{\iint}dq_3dq_4\langle q_2,t_2|q_3,t_1\rangle q_3\delta(q_3-q_4)\langle q_4,t_1|q_0,t_0\rangle\\
&=&\displaystyle{\int}dq_4\langle q_2,t_2|q_4,t_1\rangle q_4\langle q_4,t_1|q_0,t_0\rangle\\
&=&\displaystyle{\int dq_4\left(q_4\int_{q(t_0)=q_0}^{q(t_1)=q_4}\mathcal{D}[q]e^{iS[q]} \int_{q(t_1)=q_4}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]}\right)}
\end{array}
\tag{1}$$ and then, they directly write: $$\langle q_2,t_2|\hat{q}(t_1)|q_0,t_0\rangle = \int_{q(t_0)=q_0}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]}q(t_1)\tag 2$$ I guess they use: $${\displaystyle \langle q_{2},t_{2}|q_{0},t_{0}\rangle = \int d q_4 \langle q_{2},t_{2}|q_{4},t_{1}\rangle \langle q_{4},t_{1}|q_{0},t_{0}\rangle}\tag 3$$ Since it's not explained where I looked, I suppose this is trivial, but I can't manage to find (2) from (1)... I tried integration by part, but it doesn't seem to work. | The structure of the interior of the sun has been extensively studied by the technique of helioseismology and the findings are consistent with a model in which that structure is relatively simple. Helioseismology yields structural information on length scales of order ~tens of thousands of kilometers and thus cannot reveal any fine structure which might be present in the interior. However, observations of the gaseous surface of the sun with powerful telescopes and radar show that there is lots of structure on relatively short (~thousands of kilometers) length scales in the form of convection cells that "boil" hot gas up to the surface and suck cooler gas down, plus bunched-up zones of strong magnetic field activity. Almost all of that short-scale structure is transient and does not persist for more than timescales of order ~hours to days. But the complexity of those structures is of a fundamentally different sort than that of the human brain, in that the brain is interconnected with nerve pathways that convey phenomenally complicated signals from one part of it to other parts, that those nerve pathway structures persist on timescales of order ~tens of years, and the nerve pathways consist of protein molecules of equally phenomenally complex molecular structure. Nothing even remotely like that is known to exist in the sun. Finally, note that the solar structure models account well for the observed behavior of the sun with no assumptions at all about short-scale "Sheldrake" structure. If there were such structure present, it is unlikely that the models would work properly since they do not include them. Sheldrake's implication that astronomical bodies are therefore capable of consciousness and emotions is without any experimental evidence. Note that there are phenomenal amounts of "structure" in a sand dune viewed through a microscope, but nothing that would support anything like consciousness. | {
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643,252 | In the given circuit, find the equivalent resistance between points A and B. I solved this question in two different ways, out of which one gave the correct answer. The correct method is the most popular method: assuming the potential of given points and then calculating the others using the fact (or assumption) that in the absence of resistance, the potential difference is zero. This gives the correct answer: $R_{eq} = \frac{R}{3}$ . I was confused regarding why the other method did not work. Current follows the least resistance path when available. That is why you should not connect a conducting wire (R = 0) parallel to a resistor as it might lead to short circuit. Using this, I can conclude that the current can flow in two possible paths: (1) from $V_A$ (leftmost) to $V_A$ (second from the right) to $V_B$ (rightmost) (2) from $V_A$ (leftmost) to $V_B$ (second from the left) to $V_B$ (rightmost). Each of the paths has resistance R, so $R_{eq} = \frac{R}{2}$ . But this is incorrect. Where is the mistake? Update Current follows the least resistance path when available. The proof for this lies in the current divider rule. $$I_1 = I\frac{R_2}{R_1+R_2}$$ Putting $R_1 = 0$ we find that the entire current flows through $R_1$ . | I am used to smoothing out badly shaped circuits by pulling the wires: Then I get a better circuit by cutting the extra wires: So there are three resistors in parallel, indicating that the current flows through three possible paths. | {
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643,286 | I am so confused about the interaction between the photon and the atoms especially when we see it and perceive the color. I think there are many questions like this before but I still don't understand its answer since I can't picture what happened in the atom when we see things. As it is mentioned that the atom is almost empty so the photon should pass through it easily but things are opaque and I want to understand why. There are some explain that when photon has exact energy to excite an electron in the atom to jump into the higher state then electron absorbed it and no light in that frequency passes through the atom. The exact value means greater than or equal so that the electron can be excited into the higher level? If not why the greater energy cannot excite the electron and just let the excess energy go somewhere? The above statement still is unclear to me because the atom filled with space so why photon can interact with the electron? The probability for them to meet should not be high? Or is it because of the cloud atomic model that they somehow can have a high probability to interact with each other? If it is the case that electron is excited into the higher state, assume that the object appears blue, then all except blue wavelength photon will be absorbed each by corresponding electrons? Does it mean that many electrons get excited and jump into the higher state (corresponding to their exact energy photon that they absorbed)? Why doesn't electron jump back to its original state? Because if it jumps back to its original state it should emit energy the same amount it absorbed, which then can be combined with all other electrons to be appeared almost white(except blue)? If 3. is the case then the atom that electrons jump to the lower positions (but higher than its original) will have to be excited (next time) by the photon with higher energy than before (since it's in the higher state than original) and hence we should see the object with the different color than before? I think maybe this fails at the very first question so someone please explain the interaction step by step for this (color perception) phenomena? | I am used to smoothing out badly shaped circuits by pulling the wires: Then I get a better circuit by cutting the extra wires: So there are three resistors in parallel, indicating that the current flows through three possible paths. | {
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643,615 | In the series Crisis of Infinite Earths , the whole story is that there is a wave of antimatter rampaging through the multiverse. So I got interested and googled "how to create antimatter", and I found out that when an antiproton and a positron are present in an atom, it creates an anti-hydrogen atom. My question is, what is anti-hydrogen? What can be its properties? We all know that hydrogen is flammable, so is anti-hydrogen a non-flammable material? Because it has opposite properties, right? | In particle physics, every type of particle is associated with an antiparticle with the same mass but with opposite physical charges (such as electric charge) Bold mine. Note physical charges . There are specific numbers associated with elementary particles that define them, and those are the ones reversed, as in the proton antiproton case, or the electron positron case. A world made completely out of antimatter would have the same chemical and electric etc interactions. We all know that hydrogen is flammable, so is anti-hydrogen a non-flammable material? Because it has opposite properties, right? The "opposite" does not go to interactions, only to quantum numbers. So yes antihydrogen it would be flammable in an antimatter world. There is an ongoing search to see whether antimatter exists in bulk in our universe . | {
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643,765 | Let's say that there is a large mass $M$ a light-year or so away from a black hole merger, which causes a very large gravitational wave to be produced. When the gravitational wave reaches $M$ , does it bend like light bends when it comes into a curved spacetime? If you were to think of this situation in terms of quantum gravity, where the gravitational wave is a wave of gravitons, then if gravity does bend gravity, then gravity would be a self-interacting quantum field like the Higgs field? | The short answer is yes, it does. You can find a whole chapter on this phenomenon in Misner, Thorne, and Wheeler's text "Gravitation". It is a messy thing to distinguish because general relativity is a non-linear theory and in principle the gravitational wave and the gravitational field are one and the same thing. When we describe gravitational waves mathematically we assume that the metric tensor consists of a static background field (that may be very strong, like the field close to a black hole) and a small perturbation. Working out the equations, one finds that the small perturbation describes a gravitational wave that propagates on the strongly curved background. Just as gravity bends light, it will bend, or deflect, the path of the gravity wave. This is very similar to our description of linear acoustics in a moving fluid medium. The equations of fluid dynamics are non-linear and what we choose to call "acoustics" or sound is a small perturbation. At some point it will become difficult to separate the wind from the sound, and for high amplitude sound fields we need to include non-linear corrections. Numerical models have been developed that directly solve the non-linear equations for a vibrating source just to validate our assumptions. I would expect that researchers in the field of GR and GW have thought of the same situation here. A strong, high amplitude, GW might be hard to separate from the background. | {
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643,783 | I was recently solving a problem which had rotation of rotating axis. The answer was pretty simple: just vector addition of their angular velocities. After searching up a bit about rotation of axis of rotation, I came across Euler rotational theorem. But I was a bit confused with the statement that the resultant axis of rotation is not necessarily fixed and can change with time . What does that even mean? The theorem clearly states that there can be only one axis of rotation, but if that axis is itself changing, won't it result in some new axis of rotation and so on? I know I am missing something but have no idea what. (Please try to keep the language simple as I am a high school student.) | The short answer is yes, it does. You can find a whole chapter on this phenomenon in Misner, Thorne, and Wheeler's text "Gravitation". It is a messy thing to distinguish because general relativity is a non-linear theory and in principle the gravitational wave and the gravitational field are one and the same thing. When we describe gravitational waves mathematically we assume that the metric tensor consists of a static background field (that may be very strong, like the field close to a black hole) and a small perturbation. Working out the equations, one finds that the small perturbation describes a gravitational wave that propagates on the strongly curved background. Just as gravity bends light, it will bend, or deflect, the path of the gravity wave. This is very similar to our description of linear acoustics in a moving fluid medium. The equations of fluid dynamics are non-linear and what we choose to call "acoustics" or sound is a small perturbation. At some point it will become difficult to separate the wind from the sound, and for high amplitude sound fields we need to include non-linear corrections. Numerical models have been developed that directly solve the non-linear equations for a vibrating source just to validate our assumptions. I would expect that researchers in the field of GR and GW have thought of the same situation here. A strong, high amplitude, GW might be hard to separate from the background. | {
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643,788 | A collection of accepted answers on Physics Exchange paint a seemingly inconsistent picture of Unruh radiation and it's isotropy. One of the most sophisticated answers, from Lubosz Motl, to the question Is Hawking radiation really the same as Unruh radiation states that Unruh radiation is the flat-space limit of Hawking radiation. Explicitly, an observer maintaining a fixed position in Schwarzschild coordinates is necessarily accelerating; that observer experiences Unruh radiation, and if the observer is at infinity, that radiation is interpreted as Hawking radiation. The answer to Is the Unruh radiation isotropic? suggests that the radiation is coming "from the horizon", and that it could be blocked by a thermal blanket, which seems to contradict the idea that one is sitting in a thermal bath. (That is, the vacuum is "everywhere", so when I accelerate, my guts and muscles should experience Unruh radiation, and not just my skin. It's conventionally said to be isotropic. This is not at all the same as saying that my skin is feeling warmth coming from the direction of the horizon.) Can Hawking radiation be blocked by a thermal blanket (or a lead brick) or is it experienced as a bath? It is conventionally said that a freely falling observer experiences no Unruh radiation. Yet, near a BH, there is still an event horizon for the freely falling observer, and thus the observer would necessarily observe Hawking radiation (but not Unruh radiation). Is that correct? That Hawking radiation is then perceived as a flux from a direction; can it be blocked and shielded against? The answer to Direction of Unruh radiation invokes the idea of a "Rindler wedge" and states that "Rindler particles (are) without physical direct meaning." I find this confusing: surely if I accelerate and carry a thermometer with me, it will quickly reach thermal equilibrium, and report a non-zero temperature? It is sometimes said that spacetime itself is falling into a black hole (that is certainly the interpretation invited by the Christodolou and Rovelli work at arXiv:1411.2854) and so, in that question, it is imagined that an observer accelerating above the horizon would experience a "headwind" much as a bicyclist might, yet curiously in the form of a thermal bath. Again, this seems to contradict the notion of Hawking radiation coming "from" the event horizon (a "tailwind"). I believe that the correct resolution is that one should not think of Hawking radiation as originating from some thin shell of distance epsilon above the horizon, but rather as momentum eigenstates of indeterminate position: this is why its not directional. One last: in Why is there a flux of radiation in the Hawking effect but not in the Unruh effect? (and other questions) , the answer provided by Lubosz states that "the static observer sitting on a heavy star doesn't see any radiation!" Yet other discussion suggests that an observer riding a rocket near that very same star would perceive Unruh radiation, because there is a horizon that forms due to acceleration. If there's a horizon, then surely it matters not if one is sitting on a chair on the surface of the heavy star, or is riding a rocket just slightly above it: either way, there is a horizon! Insofar as Hawking radiation is a flux, one could, in principle, ride it on a solar sail, as, after all, I should feel a momentum transfer from it. Which should provide an acceleration. If I deploy my solar sail just above the event horizon, is that flux sufficient to nudge me, err, um, I'm not sure what I'm asking, but, uh, how would that work right at the horizon? Is it no longer a flux, there? The core confusion appears to have something to with the Bogoliubov transformation, which "feels like it should be" a "local" transformation applied at the (accelerated) space-like tangent space of a manifold. Why does it "feel like that"? Because this tangent space is what provides the coordinate frame for Hamiltonians to be written in: a local symplectic manifold, over there, where one is accelerating, providing the coordinates, and then using those coords to write conventional 2nd-quantized (yet covariant!) QFT in that frame. That all sounds peachy-keen, except that those local coords need to be connected to, or transported to other, distant, inertial frames, in such a way that those field excitations which "fell and disappeared" behind the horizon are no longer present, leaving behind a real particle flux (at the temperature of the horizon), i.e. the Hawking radiation. So maybe my real question is: is there a recommended review article? One that treats the topic from a geometric view, i.e. showing how to take a Hamiltonian/symplectic frame of reference, in the accelerated frame, and transport/connect the tangent vectors and 1-forms over to an inertial frame, such that those vectors and 1-forms that "fell behind the horizon" are no longer present in the inertial frame? (Maybe I need to work with jets, and not 1-forms, to get it right? some non-commutative thingy on a 2-jet?) Bonus points for doing it on generic pseudo-Riemannian spaces (won't work on Riemannian spaces, because it seems that at least one time dimension is needed) and super-duper bonus points for working with spin structures? (why? Because spin structures are cough cough "where fermions come from", roughly speaking, when one 2nd quantizes them.) | The short answer is yes, it does. You can find a whole chapter on this phenomenon in Misner, Thorne, and Wheeler's text "Gravitation". It is a messy thing to distinguish because general relativity is a non-linear theory and in principle the gravitational wave and the gravitational field are one and the same thing. When we describe gravitational waves mathematically we assume that the metric tensor consists of a static background field (that may be very strong, like the field close to a black hole) and a small perturbation. Working out the equations, one finds that the small perturbation describes a gravitational wave that propagates on the strongly curved background. Just as gravity bends light, it will bend, or deflect, the path of the gravity wave. This is very similar to our description of linear acoustics in a moving fluid medium. The equations of fluid dynamics are non-linear and what we choose to call "acoustics" or sound is a small perturbation. At some point it will become difficult to separate the wind from the sound, and for high amplitude sound fields we need to include non-linear corrections. Numerical models have been developed that directly solve the non-linear equations for a vibrating source just to validate our assumptions. I would expect that researchers in the field of GR and GW have thought of the same situation here. A strong, high amplitude, GW might be hard to separate from the background. | {
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643,794 | Ok I'll start my question with laying some background: (Correct me if I'm getting things wrong - but don't be picky). Put electro-magnetism aside for this discussion - an electric field is some space in which an electric charge will experience force. What generates an electric field is an electric charge. Electric charge is a property of matter, and it can be either positive or negative. When some charged material is put in an electric field, it will experience force rejecting or attracting it, depending on the direction of the field and the type of the charge. Now, let's talk about Voltage - electric potential difference. Every video or article I watched or read about this explained the concept of electrical potential difference using the analogy to gravitational potential difference this way: Gravitational potential energy (PEg) is the potential energy stored in an object, with respect to some other point lower than where it is currently placed. It is $m * g * h$ where h is the delta between the two points. Gravitational potential (Pg) is a property of position: It is the PEg (in Joules) per kg of mass. So every kg you place at this point will hold this amount of Joules.
Eventually, what we really care about when talking about the potential of a position is the potential difference between this position and some other position. Now let's see how this translate to electricity: In electricity, much like in gravity, we have a field exerting a force on objects. This time we talk about charges rather than mass. Electrical potential energy (PEe) is the amount of energy (Joules) that a certain amount of charge possesses when placed at a given position in the field. But again: that amount of energy is only relevant compared to some other position. I mean - the PEe is how much work is needed to bring this charge from some other position to this position, and therefore how much energy it stores when placed at that position.
Same as with gravity - Potential (Pe) is a property of position in the field, and defines how much PEe one Coulomb of charge will hold at that position. So far so good, but there are 2 basic things I don't quite understand in all this: When we talk about gravity - we are talking about one system. It's true that the potential of a position is only relevant compared to another position. But eventually, all the positions in the world can be referred to one position - let is be the earth's center or sea level. How does this translate to electric fields? I don't quite understand how this model is applied in a 9V battery for instance. All the videos I watched are showing this big floating somewhere in space, and then talk about the little q pushed towards the big Q how does this model fit in a battery (or any other DC power supply)? What is the source of the electric field? And - as far as I understand - voltage is the mere difference in electric potential between two points. So if we take every two points in the field we can calculate the Voltage between them. But if we won't place any charges there, there won't be any current. But with batteries - we say that the mere presence of voltage meaning that there'll be current flowing if we close the circuit. EDIT I'll try to make the question more clear:
When we talk about gravitational potential energy, every point we'll choose has some height. We can always take 2 points (let's leave outer space out foe this discussion) and tell the difference between them. My question is: is it the same with electricity? Is there any reference point we can all agree upon and measure every two positions in the world and tell the difference between them? | The short answer is yes, it does. You can find a whole chapter on this phenomenon in Misner, Thorne, and Wheeler's text "Gravitation". It is a messy thing to distinguish because general relativity is a non-linear theory and in principle the gravitational wave and the gravitational field are one and the same thing. When we describe gravitational waves mathematically we assume that the metric tensor consists of a static background field (that may be very strong, like the field close to a black hole) and a small perturbation. Working out the equations, one finds that the small perturbation describes a gravitational wave that propagates on the strongly curved background. Just as gravity bends light, it will bend, or deflect, the path of the gravity wave. This is very similar to our description of linear acoustics in a moving fluid medium. The equations of fluid dynamics are non-linear and what we choose to call "acoustics" or sound is a small perturbation. At some point it will become difficult to separate the wind from the sound, and for high amplitude sound fields we need to include non-linear corrections. Numerical models have been developed that directly solve the non-linear equations for a vibrating source just to validate our assumptions. I would expect that researchers in the field of GR and GW have thought of the same situation here. A strong, high amplitude, GW might be hard to separate from the background. | {
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