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506,977 | This video from Brian Greene suggests this is so: https://www.youtube.com/watch?v=2sZUNud6rRw&list=PLj6DWzIvBi4PFDXCCV1bNhVUgDLTwVbFc&index=60 It shows if you stop a pole in the barn (ignoring all the obvious engineering challenges of doing so) it will end up permanently length contracted just like the returning twin will end up permanently younger than her earth bound twin in the twin paradox. Ignoring the practical problems with infinite deceleration, she stops when she turns around and that causes her permanent age difference but does she also end up permanently flatter? Again just consider the relativistic math and not all the physical impossibilities this example entails. Relativity allows a frame jump without deceleration, it's called a clock handoff in the twin paradox. Since a clock is used to measure length for length contraction, a clock handoff could also keep a record of both permanent age difference and permanent length contraction when the twin hands off her clock readings to a ship passing her to return to earth. There's no physical crunching of the pole in a clock handoff. So does relativity sanction permanent length contraction along with permanent age difference in the clock handoff twin paradox? | Does relativity sanction permanent length contraction along with permanent age difference in the twin paradox? No, it does not. However, given the many analogies between time and space this may seem disturbing. What makes time different from space in this context? The issue is that a clock does something different than a ruler does: it maintains a record. A ruler merely measures the distance between its endpoints, and as a moving ruler is brought (gently) to rest that measurement agrees with a permanently resting ruler. The device that most closely resembles a ruler for time is not a clock but rather a metronome. There is no permanent time dilation for a metronome, and as a moving metronome is brought (gently) to rest that measurement agrees with a permanently resting metronome. In this way it is symmetric with the impermanence of length contraction. If you want a device that resembles a clock for distance that would not be a ruler, but rather an odometer. An odometer maintains a record and will register permanent length contraction in the same manner as a clock. In this way the symmetry between time and space is recognized again. The difference was not due to differences in the physics of time and space, but rather differences in the measuring devices. We were comparing a memory-less device for space to a device with memory for time. With a proper comparison of similar devices the issue is resolved. | {
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506,980 | A box is launched upwards along a $30^o$ incline that is $1.0 m$ long. How far from the edge of the incline will the box reach if its initial speed is 3.55 m/s. The coefficients of of static and kinetic friction are $\mu_s = 0.30$ and $\mu_k = 0.20$ . I am having trouble with this type question in general, as I find it difficult to draw the free body diagram and then interpret the components in different directions. * UPDATE * This is what I have so far - I think I am 95% of the way there. I note that there is no force pushing the box up once it has been launched, so the only two forces acting on it are the force of friction (kinetic, $\vec{F}_k$ ) and the component of gravity that is pulling down in the defined x direction (the same direction as $\vec{F}_k$ ). This is where I struggled most. The normal force is perpendicular to the object, and is equal in magnitude, but points in the opposite direction to the force of gravity in the y-direction. Then, the components of gravity are as follows: $$\vec{F}_{G_y} = (m)(g)\cos \theta \implies N = (m)(g)\cos \theta$$ and $$\vec{F}_{G_x} = (m)(g)\sin \theta$$ From here, I understand that $\vec{F}_k = \mu_k(N) \implies \vec{F}_k = \mu_k (m)(g) \cos \theta$ . Using kinetic friction since the object in question is in motion. Then, $$\vec{F}_{net} = \vec{F}_{G_x} + \vec{F}_k = (m)(g)\sin \theta + \mu_k (m)(g) \cos \theta$$ Plugging in the numbers, I get $\vec{F}_{net} = (m)(-6.5974)$ . From here I used $F = ma$ to get an acceleration of $-6.5974 \frac{m}{s^2}$ . Plugging in $v_f = 0 \frac{m}{s}$ and $v_i = 3.55 \frac{m}{s}$ into $(v_f)^2 = (v_i)^2 + 2ad$ I get $d$ equal to $0.95 m$ . However, the answer in the book is $0.05 m$ (which is conveniently 1 - my answer), so can anyone tell me where I went wrong? Any help is appreciated. | Does relativity sanction permanent length contraction along with permanent age difference in the twin paradox? No, it does not. However, given the many analogies between time and space this may seem disturbing. What makes time different from space in this context? The issue is that a clock does something different than a ruler does: it maintains a record. A ruler merely measures the distance between its endpoints, and as a moving ruler is brought (gently) to rest that measurement agrees with a permanently resting ruler. The device that most closely resembles a ruler for time is not a clock but rather a metronome. There is no permanent time dilation for a metronome, and as a moving metronome is brought (gently) to rest that measurement agrees with a permanently resting metronome. In this way it is symmetric with the impermanence of length contraction. If you want a device that resembles a clock for distance that would not be a ruler, but rather an odometer. An odometer maintains a record and will register permanent length contraction in the same manner as a clock. In this way the symmetry between time and space is recognized again. The difference was not due to differences in the physics of time and space, but rather differences in the measuring devices. We were comparing a memory-less device for space to a device with memory for time. With a proper comparison of similar devices the issue is resolved. | {
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507,001 | I have a question pertaining to the concept of acceleration and it's formula - Both seem to give me different answers. I was asked: A train is moving at a velocity of $20\ \mathrm{m/s}$ . It hits the breaks and starts slowing down at the rate of $8\ \mathrm{m/s^2}$ . How long does it travel before coming to a stop? $a=-8\ \mathrm{m/s^2}$ $u=20\ \mathrm{m/s}$ $v=0$ $s=?$ (Displacement is asked) According to the formula of $s= \frac{(v^2-u^2)} {2a}$ , answer I got was 25 meters. BUT! My concept of acceleration explains me this : According to my intuition of acceleration and velocity, when an object is said to be accelerating at $-8\ \mathrm{m/s^2}$ , it means that with every passing second, it's velocity will reduce $8\ \mathrm{m/s}$ . So, At 1st second - Velocity = $20\ \mathrm{m/s}$ - It'll travel 20 meters. At 2nd second - $v = 20\ \mathrm{m/s}-8\ \mathrm{m/s}=12\ \mathrm{m/s}$ - Train travels 12 meters At 3rd, it travels 4 meters. At 4th, it halts. So total is 20+12+4 = 36 meters. It travels 36 meters before halt. Where am I going wrong in my conceptual understanding, I really don't understand! | Your mistake is in assuming that within those one second intervals the velocity is constant. The velocity as a function of time $v(t)$ for motion under constant acceleration $a$ is given by $$v(t)=at+v_0$$ where $v_0$ is the velocity at $t=0$ . So, you are right that $v(0)=20\,\mathrm{m/s}$ , $v(1)=12\,\mathrm{m/s}$ , and $v(2)=4\,\mathrm{m/s}$ . But, for example, we have $v(1.5)=8\,\mathrm{m/s}$ , and so on for any value of $t$ until coming to rest. The velocity is continuously decreasing. It's not decreasing in the step-like manner you propose. You are actually on the right track though. If you wanted to take your approach to the correct extreme, we would break our time, not into $1\,\mathrm s$ intervals, but instead into really small time intervals $\text dt$ such that the velocity can be considered constant. Then we can add up all of the changes in position $\text d x=v\,\text dt$ . This is where calculus becomes useful, and we get an equation you are probably familiar with $$\Delta x=\int_{x_0}^x\text dx=\int_0^tv(\tau)\,\text d\tau=\frac12at^2+v_0t$$ Combining this with our expression for $v(t)$ to eliminate $t$ gives us the expression you give your first correct method. | {
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507,372 | This question is inspired by a sign I saw at a lift, which said something like: Going up 1 floor or down 2 floors? Take the stairs, it's better for your health and for the environment. There's no doubt that taking the stairs is better for one's health, but the physicist in me wonders about the latter. An analysis: Presumably whoever wrote that sign was thinking about energy usage. By taking the stairs instead of the lift, I save on the electricity required to power the lift. However, gravity is a conservative field . The end result in both cases is that I move up one floor, so the amount of work (i.e. energy) required to get me up one floor is independent of the path I take. It should only depend on the initial and final states. Therefore, the electricity saved is compensated for by the fact that I need to eat more food (so my muscles can produce the energy required to move me up one floor). If I take the stairs, I do more than $mgh$ of work, because the stairs isn't completely vertical and there's some amount of horizontal work done. Therefore it's actually more energy-intensive overall to take the stairs instead of the lift. (There is also some kinetic energy I have to impart to my body to make it move upwards, but my body also has kinetic energy in the lift, and we assume they are equal.) However #3 seems superficial because the final state isn't exactly the same: the lift has also moved up one floor. The mass of the lift is obviously quite large, so to get it up one floor takes more energy. But it's exactly because of this that modern lifts use a counterweight that serves as a gravity battery . Whenever the electric motor moves the lift down, some amount of weight is lifted as a counterbalance. When the lift moves up, most of the energy comes from the potential energy stored in the gravity battery; the electric motor doesn't actually do much work. Besides, even if the electric motor does more work, the argument only holds in one direction. If it takes more energy to go up the building (compared to walking), it takes less energy to go down the building (compared to walking). If we assume that on average I move both up and down the building, then this difference cancels out, and we're left with the argument in #3. Conclusion: the sign is wrong. It's (slightly) worse for the environment to take the stairs instead of the lift. I'm wondering if this analysis is good or if I missed something. | I follow your analysis but not your conclusion. You are missing two key points, which as far as I can see might outweigh all the other points you mention: You must compare the efficiency of the electric engine running the lift with your body-"engine". You must compare the " fuel " that the electric engine and your body-"engine" use. To the first case, all your other points pale and are negligible, if an old inefficient engine is in use, or opposite if you as the person walking up the stairs have a hard an inefficient time doing so. To the second case, what if the electric engine is run by the roof solar panel, geothermal heating or a local wind turbine? Then the elevator's impact becomes negligible*. On the other hand, if you eat, drink and gain nutrition in an environmentally sustainable manner**, then your impact might be considered negligible. These thermodynamic and fuel-production factors*** are, as far as I can see, more important than your presented analysis, which is based purely on mechanical energy. * Unless we include an analysis of the lifetime, production and disposal of the solar cell, geothermal plant or wind turbine, in which case they will never win with current development technologies. We will then also have to consider other uses of these sources, and the math quickly becomes large and complicated. ** Home-grown vegetables etc. *** I'm sure we can find more such factors if we dig deep into the details of the lift technology and human physiology. The environmental-impact-math will also highly depend on how we estimate and evaluate indirect factors that may make all of your and my points moot, some of which are mentioned in comments and other answers, such as the energy storing capability of a lift's counterweight or such as the question of whether the lift will be implemented and in use regardless of what you choose to do due to regulation requirements and accessibility for the disabled/elderly. | {
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507,374 | I was learning about black body radiation and the explanations given by Max Planck and Albert Einstein when a thought crossed my mind. When we heat an iron piece, its color changes gradually from red, orange, yellow to bluish white. Yet such a change is not visible in a glowing piece of charcoal obtained from wood. Why is it that, wood charcoal is not able to glow in colors of higher frequencies? | Any black body emits radiation of all frequencies. So a hot charcoal definitely emits some blue light. It doesn't appear blue, because most of its light is red, orange, or yellow. To shift the spectrum toward perceiving the white light, you must increase the temperature to the temperature of the Sun (5,778K), but this would destroy your furnace, as we don't have materials capable of sustaining this temperature. The highest recorded melting point is 4,215K of tantalum hafnium carbide. To make charcoal blue, you would need to increase its temperature even more, which is unrealistic in any existing furnace. Hot iron doesn't become bluish white, but only yellowish, because it melts at 1,811K. You may see bluish colors on it after it cools down. They are caused by thin layers of iron oxide reflecting light differently for different wavelengths. | {
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507,626 | I'll be honest, I only have a Highschool education, so there might be something obvious I'm overlooking. However particle physics is of massive Interest to me. My question is, I know there are unstable quark-antiquark pairs that form Mesons, but are there any Baryons that are not wholly made up of quarks or antiquarks? i.e. "up, up, anti-down" If the answer is no, why not? | No, a three-quark baryon can not be be made out of two quarks and one anti-quark (and vice versa) as this would necessarily give the particle color. Each quark carries one of three colors (red, blue, green) and each anti-quark respectively carries anti-color. Color is an additive quantity when constructing particle and the result must be color-neutral, i.e. it either is made out of for example red + anti-red or red + blue + green quarks. Although we can not observe the "color" of the constituents of a particle directly (as all observable particles must be color-neutral), we can measure its effects indirectly via certain cross-sections. With this being said, we can now clearly see that there is no way in which we can construct particles of two-quarks and one anti-quark as any possible combination would not be color-neutral. Hence, color-neutrality forbids observable three-quark particles composed out of quarks and anti-quarks. Penta-quarks however can and must contain quarks and anti-quarks as mentioned in a previous answer . | {
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507,944 | While deriving the wave function why don't we take into the account of the probability density of the nucleus? My intuition says that the nucleus is also composed of subatomic particles so it will also have probability cloud like electrons have. Do we not do it for simplicity of the calculation, or is the nucleus fixed, or any other property of the nucleus? | Yes, the nucleus is composed of subatomic particles that have a probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.) Uncertainty principle requires that the nucleus as a whole has some spatial spread. The easy part is that the "probabilistic cloud" of a nucleus and its constituents are way smaller than the space electrons pretend to occupy. That's what makes the point approximation viable. | {
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508,066 | Let's say that we have 2 tennis balls in space, one being in motion (say, pushed by an astronaut), and the other one still. If we could take a snapshot of both tennis balls, would there be any evidence that could suggest that one is moving and the other one is still? Is there anything happening, at the atomic level or bigger, being responsible for the motion? If there isn't, and both balls are absolutely identical, then how come one is still and the other one moving? Where does the difference of motion come from? | According to classical physics: no. It is impossible to tell how fast something is moving from a snapshot. According to special relativity: yes. If we choose a frame of reference where one of the balls is at rest then only that ball will look normal. The other ball is moving in this frame so it will be length contracted. If its rest length is $L$ then its length will now be $L\sqrt{1-v^2/c^2}$ . Since $1-v^2/c^2<1$ the ball will be shorter in the direction it is moving. According to quantum mechanics: yes? In quantum mechanics particles are described by a wavefunction $\psi(x)$ which (handwavingly) says how much of the particle is present at a certain point. A tennis ball is also described by a wavefunction which you can get by combining all the wavefunctions of its atoms. The wavefunction actually contains all the information you can possibly know about an object, including its velocity. So if you could pause time and look at the wavefunction you would have enough information to know its (most likely) velocity. In real life you can't actually look at wavefunctions: you have to perform an experiment to extract information from the wavefunction. At this point you might wonder if that still counts as taking a snapshot. | {
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508,091 | After reheating cold about 1.5 oz. of Annie's Mac & Cheese shells for 15 seconds on high power in the microwave, the mac & cheese was burnt black only at certain points where the pasta is touching each other. Does anyone have an idea of what might be going on? | I get the same thing reheating some discs of glazed carrots. And there are several videos of folks doing this intentionally with grapes. An article published last year in PNAS says this will happen with almost any pair of similarly-sized object with sufficient water. The shape of the pairs appears to set up a resonance that concentrates the electric field at the point where they touch. The higher intensity field there heats the surroundings and burns the food. | {
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509,402 | To put it simply. We group our EM waves into groups such as X-rays, microwaves, visible light etc. I was wondering, if the properties of say, x-rays, slowly change into the properties of say, gamma rays, or as soon as the wavelength is larger than 10 picometers it's properties become that of an x-ray. If it's the former, how do we calulate the standards for identifying a wave. | The EM waves form a continuous spectrum, thus your first description of a gradual change in wave behaviour is correct. However, because of this; there is also no precise way of determining exactly at which frequency a category of EM wave starts or finishes at. In fact, sometimes the classification of waves can be so hand wavy that the typical description of an EM wave: $$c=f\lambda$$ is treated as a secondary determinant for which category an EM wave belongs to! One such example of this is with X rays and $\gamma$ rays. Their frequencies overlap one another at approximately $10^{18}\:\mathrm{Hz}$ so it is difficult to quantitatively differentiate higher frequency X rays and lower frequency $\gamma$ rays. Instead, we let the origin of the waves decide, $\gamma$ rays are emitted from the nucleus of the atom while X rays are emitted from outside the nucleus; by excited electrons. To conclude, the EM spectrum can in no way be easily divided into discrete categories and as such there is always some overlap between adjacent categories of wave. | {
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509,541 | Keeping it simple, what is the physics approach to a current in a wire connected to a battery? How does the electric field behave? I've read somewhere that, contrary to the common conception, electrons don't really flow as water in a hose so, what is really going on there? How is energy transported at the speed of light if electrons don't flow like very fast water? | Below is an illustration I made of this, based on a reasonably realistic simulation of a simple DC circuit with a rectangular shape. The energy does not flow in the same direction as the current -- if it did, then just as much energy would flow out of the resistor as into it. Actually, the battery is using up its chemical energy, and the resistor is putting out energy as heat, so the flow of energy from the battery to the resistor is always to the right. In electromagnetism, we measure the flow of energy using something called the Poynting vector. For example, inside a laser beam, the Poynting vector points in the direction of the beam. The Poynting vector is proportional to $\textbf{E}\times\textbf{B}$ , where $\textbf{E}$ is the electric field, $\textbf{B}$ is the magnetic field, and $\times$ is a thing called a vector cross product. The vector cross product is perpendicular to the two vectors being multiplied. For a wire that's a perfect conductor, the electric field inside the wire is zero, so the Poynting vector vanishes. Therefore no energy flows inside the wire. The flow of energy is outside the wire. The white arrows in the figure show the direction of the Poynting vectors. How is energy transported at the speed of light if electrons don't flow like very fast water? The electrons flow more like very slow water, and this is another way of seeing that energy is not transported along with the electrons. If you open or close a switch in a circuit like this, there are changes that ripple outward in the electromagnetic field. Those electromagnetic waves carry the energy and information at speeds that are usually close to the speed of light. | {
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509,650 | Newton's explanation of gravity as an attractive force seems to have been superseded by Einstein's explanation of gravity as warping of space-time. Was there any advances in math and science that was not known in Newton's time, that would have laid the foundation for Einstein to give a more accurate description of gravity in General Relativity? | A well-developed idea of a field theory . Newton thought of the force of gravitation to be operating with an action-at-a-distance mechanism. While this bothered him, it remained an unresolved question to him. However, by the time of Einstein, the idea of thinking of the force of gravitation in terms of a field theory had been developed. Lorentz invariance . While the shift from thinking of the theory of the force of gravitation in terms of a field theory was an important conceptual shift, nothing really changed in terms of the mathematical description of the force of gravitation. But, with the development of special relativity, Einstein had realized that the laws of physics should be Lorentz invariant, unlike the Newtonian law of gravitation which was Galilean invariant. Mass-energy equivalence . This is another aspect of the development of special relativity which was relevant to going beyond the Newtonian law of gravitation. Einstein had realized through special relativity that mass and energy are not distinct properties but are rather unified in a profound way. This led him to believe that if mass plays a role in causing gravitational attraction then so should energy. However, as I said, this is closely related to my previous point: Lorentz invariance. Riemannian geometry . Putting together all the physical axioms that Einstein had developed crucially required the use of Riemannian geometry. In fact, learning the tools of Riemannian geometry was the hardest part for Einstein in his journey of developing his theory of gravity. Finally, I would like to mention that two crucial elements that went into the development of general relativity (perhaps, the most crucial two elements) were already present at the time of Newton. One of them was the equality of the inertial and the gravitational mass (something that Newton also found curious) and the other was the question of what determines which frame is an inertial frame (to which, Einstein ultimately found the answer: the freely falling frame is the inertial frame). This is not to say that Newton should've developed general relativity had he been clever enough. Lorentz invariance and non-Euclidean geometry were absolutely indispensable in the development of general relativity and they were too way ahead in the future to be discovered at the time of Newton. | {
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509,660 | If Energy and mass are the same thing, then is it logical to look for analogous (duals) of properties of one in another? or there is there any conceptual framework that such questions make any sense? are there any properties from mass to energy or backwards that remain invariant? or just equivalence of energy and mass is where all the relationships in the intrinsic properties end? Not trying to make this a philosophy question, only wondering if this concepts have a place in physics or have been previously dealt with in a any topics of physics. | A well-developed idea of a field theory . Newton thought of the force of gravitation to be operating with an action-at-a-distance mechanism. While this bothered him, it remained an unresolved question to him. However, by the time of Einstein, the idea of thinking of the force of gravitation in terms of a field theory had been developed. Lorentz invariance . While the shift from thinking of the theory of the force of gravitation in terms of a field theory was an important conceptual shift, nothing really changed in terms of the mathematical description of the force of gravitation. But, with the development of special relativity, Einstein had realized that the laws of physics should be Lorentz invariant, unlike the Newtonian law of gravitation which was Galilean invariant. Mass-energy equivalence . This is another aspect of the development of special relativity which was relevant to going beyond the Newtonian law of gravitation. Einstein had realized through special relativity that mass and energy are not distinct properties but are rather unified in a profound way. This led him to believe that if mass plays a role in causing gravitational attraction then so should energy. However, as I said, this is closely related to my previous point: Lorentz invariance. Riemannian geometry . Putting together all the physical axioms that Einstein had developed crucially required the use of Riemannian geometry. In fact, learning the tools of Riemannian geometry was the hardest part for Einstein in his journey of developing his theory of gravity. Finally, I would like to mention that two crucial elements that went into the development of general relativity (perhaps, the most crucial two elements) were already present at the time of Newton. One of them was the equality of the inertial and the gravitational mass (something that Newton also found curious) and the other was the question of what determines which frame is an inertial frame (to which, Einstein ultimately found the answer: the freely falling frame is the inertial frame). This is not to say that Newton should've developed general relativity had he been clever enough. Lorentz invariance and non-Euclidean geometry were absolutely indispensable in the development of general relativity and they were too way ahead in the future to be discovered at the time of Newton. | {
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509,662 | What causes a potential drop in a resistor or load? Why a wire having neglible resistance have same potential across it? The given figure is of a stretched wire potentiometer.
My question is why point a and A have same potential after connection through a wire of negligible resistance.
Before connection there can be different potentials on poinrs a and A. So how the same potential achieved after connection. I wants it's machanism and explanation. | A well-developed idea of a field theory . Newton thought of the force of gravitation to be operating with an action-at-a-distance mechanism. While this bothered him, it remained an unresolved question to him. However, by the time of Einstein, the idea of thinking of the force of gravitation in terms of a field theory had been developed. Lorentz invariance . While the shift from thinking of the theory of the force of gravitation in terms of a field theory was an important conceptual shift, nothing really changed in terms of the mathematical description of the force of gravitation. But, with the development of special relativity, Einstein had realized that the laws of physics should be Lorentz invariant, unlike the Newtonian law of gravitation which was Galilean invariant. Mass-energy equivalence . This is another aspect of the development of special relativity which was relevant to going beyond the Newtonian law of gravitation. Einstein had realized through special relativity that mass and energy are not distinct properties but are rather unified in a profound way. This led him to believe that if mass plays a role in causing gravitational attraction then so should energy. However, as I said, this is closely related to my previous point: Lorentz invariance. Riemannian geometry . Putting together all the physical axioms that Einstein had developed crucially required the use of Riemannian geometry. In fact, learning the tools of Riemannian geometry was the hardest part for Einstein in his journey of developing his theory of gravity. Finally, I would like to mention that two crucial elements that went into the development of general relativity (perhaps, the most crucial two elements) were already present at the time of Newton. One of them was the equality of the inertial and the gravitational mass (something that Newton also found curious) and the other was the question of what determines which frame is an inertial frame (to which, Einstein ultimately found the answer: the freely falling frame is the inertial frame). This is not to say that Newton should've developed general relativity had he been clever enough. Lorentz invariance and non-Euclidean geometry were absolutely indispensable in the development of general relativity and they were too way ahead in the future to be discovered at the time of Newton. | {
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509,670 | Quote: The coherent states $|z\rangle$ is defined as $$|z\rangle
=e^{-|z^2|/2}\sum_{n=0}^\infty\frac{z^n}{\sqrt{n!}}|n\rangle
=e^{-|z^2|/2} e^{a^\dagger z}|n\rangle,$$ which was, very understandable and mathematically easy. However, I encountered an article that using translation operator to define the coherent states, See reference here page 28 . Quote: $$|\tilde{x}_0\rangle \equiv T_{x_0}|0\rangle=e^{-\frac{i}{\hbar}\hat{p}x_0}|0\rangle.\tag{4.14}$$ But this has suddenly be somewhat confusing, the former stated that coherent states is a superposition of infinite states of probability amplitude distribution of Gaussian wave pack. The later sates that coherent states is a spacial translation of $|0\rangle$ eigenstates. Further, the first definition involve only $\hat{a}^\dagger$ while the latter involved $\hat{p}$ , a linear combination of $\hat{a}$ and $\hat{a}^\dagger$ . Could you explain to me what's going on? Especially, what's the inituition between define coherent states from $T_{x_0}$ ? How to prove they are same or not the same? | A well-developed idea of a field theory . Newton thought of the force of gravitation to be operating with an action-at-a-distance mechanism. While this bothered him, it remained an unresolved question to him. However, by the time of Einstein, the idea of thinking of the force of gravitation in terms of a field theory had been developed. Lorentz invariance . While the shift from thinking of the theory of the force of gravitation in terms of a field theory was an important conceptual shift, nothing really changed in terms of the mathematical description of the force of gravitation. But, with the development of special relativity, Einstein had realized that the laws of physics should be Lorentz invariant, unlike the Newtonian law of gravitation which was Galilean invariant. Mass-energy equivalence . This is another aspect of the development of special relativity which was relevant to going beyond the Newtonian law of gravitation. Einstein had realized through special relativity that mass and energy are not distinct properties but are rather unified in a profound way. This led him to believe that if mass plays a role in causing gravitational attraction then so should energy. However, as I said, this is closely related to my previous point: Lorentz invariance. Riemannian geometry . Putting together all the physical axioms that Einstein had developed crucially required the use of Riemannian geometry. In fact, learning the tools of Riemannian geometry was the hardest part for Einstein in his journey of developing his theory of gravity. Finally, I would like to mention that two crucial elements that went into the development of general relativity (perhaps, the most crucial two elements) were already present at the time of Newton. One of them was the equality of the inertial and the gravitational mass (something that Newton also found curious) and the other was the question of what determines which frame is an inertial frame (to which, Einstein ultimately found the answer: the freely falling frame is the inertial frame). This is not to say that Newton should've developed general relativity had he been clever enough. Lorentz invariance and non-Euclidean geometry were absolutely indispensable in the development of general relativity and they were too way ahead in the future to be discovered at the time of Newton. | {
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509,803 | I have been reading/learning about the double slit experiment, its implications in quantum theory, and how it explained that “particles” can behave as both waves and particles. I know that the wave function is a probability of the location of the particle, and that shooting the electrons through the double slits causes an interference pattern associated with multiple waves. This, though not making intuitive sense (in relation to how anything can even exist as a wave), is something I can follow. However, I have read/heard that an “observer” collapses the wave function into a single point. This is what caused the electrons to actually show up on the wall behind the slit; however Feynman (admittedly, as a thought experiment) suggested that putting an “observer” prior to the slits would cause the electrons to fly through as particles, and leave no interference pattern on the back wall. What is an “observer”? How and why would the electron “know” it is being observed and therefore cause it to change behavior? | The other answers here, while technically correct, might not be presented at a level appropriate to your apparent background. When the electron interacts with any other system in such a way that the other system's behavior depends on the electron's (e.g., it records one thing if the electron went left and another if it went right), then the electron no longer has a wave function of its own: the electron+"detector" system has a joint state. The two are entangled . The electron doesn't have to "know" anything. The simple physical interaction results in a state vector which, by the laws of quantum mechanics, will preclude interference by any of the subsystems of this larger system. That said, the joint state can itself show a kind of "interference effect" (though not the kind you normally think of in the two-slit experiment). If this entanglement is well-controlled (as in a lab), then (a) showing this "joint interference" might be practical, and (b) undoing the entanglement is also possible, thus restoring the electron's sole superposition. This is how we know that it hasn't "collapsed." But if the entanglement is caused by stray photons, air molecules, etc., then any hope of controlling them becomes almost immediately dashed, and we can no longer exhibit interference in practice. From here on out, the system will appear to behave classically, with the different branches evolving independently. This fact is called decoherence . The superposition still hasn't "collapsed," but we no longer have the ability to show or exploit the superposition. You may notice that this still leaves open a crucial question: when do the many branches become one? This is called the measurement problem , and physicists don't agree on the answer even today. | {
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510,002 | I am aware that a constant force causes a constant acceleration but friction can counteract this. However, if I push something across a table, for example, it seems no matter how hard I push, the object travels at a constant velocity, even if I apply more force than the kinetic friction. The object seems to always travel at the same velocity as my hand, does this mean I am not actually applying a constant force? | It's not easy pushing something with by hand with a constant force greater than kinetic friction. Try using a rubber band and a ruler to pull something across the table with a constant force. I think if you focus on keeping the rubber band stretched a constant amount while you pull you will notice the object will accelerate. | {
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510,399 | We have all heard people saying that to lift an object of mass $m$ , you have to apply a force $F$ equal to its weight $mg$ . But isn't it getting the force equal to its weight from the surface to which it is attached to (normal force). Why it is willing to change that equilibrium state by getting the same force from us as from the surface? (Consider the situation devoid of any resistance) . I think we must be applying slightly more force to it in order to move it even with constant velocity at least at the start and balancing the force of gravity afterwards. | There are two points to be clarified here. The normal reaction force from the surface is a self-adjusting force. In particular, it can take any value so as to prevent the object in contact from penetrating. So, if an object resting on a surface has a weight $w$ then the normal reaction force would be $w$ in the upward direction. Now, if you apply an external upward force on the object (with your hand, say) of a magnitude $w/2$ then the normal reaction force from the surface would change its value to $w/2$ . Now, if you apply an external force of a magnitude $w$ in the upward direction then the normal reaction force from the surface would reduce to zero. However, as you correctly notice, when the upward external force is exactly the same as the weight in magnitude, the object is still in perfect equilibrium. And since the initial velocity of it was zero, its velocity would still remain zero because equilibrium means no acceleration. So, there would be no movement. So, in order to actually lift the object, you do need to provide an upward force which is at least slightly greater than the weight of the object. Once you apply such a force even for a tiny amount of time, the object would pick up an upward velocity because it would have been subjected to an upward acceleration for that tiny amount of time. Once this is accomplished, you can reduce the magnitude of the upward force to be exactly the same as the magnitude of weight and the object will continue to move in the upward direction, in equilibrium, but now, with a constant velocity (that it picked up during that tiny amount of time of acceleration). | {
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510,405 | Peoples ! Thanks you for your time. I have a question about acceleration equation. $$\begin{bmatrix}f_{x}\\ f_{y}\\ f_{z}\end{bmatrix}=\begin{bmatrix}\dot{u}\\\dot{v}\\ \dot{w}\end{bmatrix}+\begin{bmatrix}0 & w & -v \\ -w & 0 & u \\ v & -u & 0\end{bmatrix}\begin{bmatrix}p\\ q\\ r\end{bmatrix}+g\begin{bmatrix}\sin\theta \\-\cos\theta \sin\phi\\ -\cos\theta \cos\phi\end{bmatrix}$$ Here, why $uvw$ should be skew-symmetric matrix and
why skew-symmetric matrix of $uvw$ should be multiplied with $pqr$ ? Please, let me know. | There are two points to be clarified here. The normal reaction force from the surface is a self-adjusting force. In particular, it can take any value so as to prevent the object in contact from penetrating. So, if an object resting on a surface has a weight $w$ then the normal reaction force would be $w$ in the upward direction. Now, if you apply an external upward force on the object (with your hand, say) of a magnitude $w/2$ then the normal reaction force from the surface would change its value to $w/2$ . Now, if you apply an external force of a magnitude $w$ in the upward direction then the normal reaction force from the surface would reduce to zero. However, as you correctly notice, when the upward external force is exactly the same as the weight in magnitude, the object is still in perfect equilibrium. And since the initial velocity of it was zero, its velocity would still remain zero because equilibrium means no acceleration. So, there would be no movement. So, in order to actually lift the object, you do need to provide an upward force which is at least slightly greater than the weight of the object. Once you apply such a force even for a tiny amount of time, the object would pick up an upward velocity because it would have been subjected to an upward acceleration for that tiny amount of time. Once this is accomplished, you can reduce the magnitude of the upward force to be exactly the same as the magnitude of weight and the object will continue to move in the upward direction, in equilibrium, but now, with a constant velocity (that it picked up during that tiny amount of time of acceleration). | {
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510,417 | Does the launching of a satellite need the consideration of the general theory of relativity (GR)? | Newtonian physics would be sufficient to launch and orbit a satellite. But any equipment, sensors, or experiments on board that required exact timing might need to take into account relativistic time dilations from speed and gravity. | {
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510,579 | In cargo cult science Feynman writes: "Millikan measured the
charge on an electron by an experiment with falling oil drops, and
got an answer which we now know not to be quite right. It's a
little bit off, because he had the incorrect value for the
viscosity of air....Why didn't they discover that the new number was higher right away?
It's a thing that scientists are ashamed of--this history--because
it's apparent that people did things like this: When they got a
number that was too high above Millikan's, they thought something
must be wrong--and they would look for and find a reason why
something might be wrong. When they got a number closer to
Millikan's value they didn't look so hard. And so they eliminated
the numbers that were too far off, and did other things like that.
We've learned those tricks nowadays, and now we don't have that
kind of a disease." What tricks is he talking about specifically? In fact, in general, what tricks do physicists learn for doing experiments and avoiding fooling themselves? Second: it's my strong belief that Feynman is referring to something here which isn't in textbooks but instead built into the culture of physics, but I don't understand what exactly it is, and I suspect it's passed along in the culture of physics labs. If someone can explain with some stories, examples or general comments, what is that culture like? | There are lots of different strategies that are employed by the scientific community to counteract the kind of behavior Feynman talks about, including: Blind analyses : In many experiments, it is required for the data analysis procedure to be chosen before the experimenter actually sees the data. This "freezing" of methodology ensures that nothing about the data itself changes the way it's analyzed, and a methodology that changes once data starts coming in is a red flag that physicists check for in peer review. Statistical literacy : The more you know about statistics, the easier it is to spot data that has been manipulated. Much effort has been devoted to increasing the knowledge of proper statistical practices among physicists so that publishing a flawed analysis (whether intentional or not) is difficult. For example, most experimental physics courses nowadays include training on the basics of statistics and data analysis (for example, mine extensively used Data Reduction and Error Analysis for the Physical Sciences by Bevington and Robinson). Independent collaborations : It's common nowadays for multiple detectors to perform tests of the same hypotheses independently of each other. The procedure, data, and results are all carefully kept as separate as possible before the respective analyses are published. This increases the likelihood that bias or manipulation will be detected, since it will usually cause a difference in results between multiple studies of the same hypothesis. Verification of old results with new data : This is probably less common than it should be in science, due to a cultural preference for performing novel tests, but still occurs at a reasonable frequency in physics. Even in the huge detectors and giant collaborations of high-energy physics, new data is often cross-checked with old data as a side effect of some analyses. For example, an analysis trying to measure the mass of a new particle will often utilize already-known particles to detect its signature, and in the process of characterizing the dataset, will end up confirming earlier measurements of those particles as a "sanity check" of the integrity of the dataset. Incentivization of properly-done disproof : Typically the case where a measurement disagrees with existing hypotheses/theories is met with as much or even more excitement as one that confirms existing hypotheses/theories. This is especially true in high-energy physics, where the majority of the community is eagerly awaiting the first statistically-significant experimental disagreement with the Standard Model. This excitement also brings intense scrutiny of the experiments claiming to measure a disagreement, which helps filter out improperly-done analyses. This is, of course, a partial list. | {
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511,067 | In the Nature paper published by Google, they say, To demonstrate quantum supremacy, we compare our quantum processor against state-of-the-art classical computers in the task of sampling the output of a pseudo-random quantum circuit. Random circuits are a suitable choice for benchmarking because they do not possess structure and therefore allow for limited guarantees of computational hardness. We design the circuits to entangle a set of quantum bits (qubits) by repeated application of single-qubit and two-qubit logical operations. Sampling the quantum circuit’s output produces a set of bitstrings, for example {0000101, 1011100, …}. Owing to quantum interference, the probability distribution of the bitstrings resembles a speckled intensity pattern produced by light interference in laser scatter, such that some bitstrings are much more likely to occur than others. Classically computing this probability distribution becomes exponentially more difficult as the number of qubits (width) and number of gate cycles (depth) grow. So, from what I can tell, they configure their qubits into a pseudo-randomly generated circuit, which, when run, puts the qubits into a state vector that represents a probability distribution over $2^{53}$ possible states of the qubits, but that distribution is intractable to calculate, or even estimate via sampling using a classical computer simulation. But they sample it by "looking" at the state of the qubits after running the circuit many times. Isn't this just an example of creating a system whose output is intractable to calculate, and then "calculating" it by simply observing the output of the system? It sounds similar to saying: If I spill this pudding cup on the floor, the exact pattern it will form is very chaotic, and intractable for any supercomputer to calculate. But I just invented a new special type of computer: this pudding cup. And I'm going to do the calculation by spilling it on the floor and observing the result. I have achieved pudding supremacy. which clearly is not impressive at all. In my example, I'm doing a "calculation" that's intractable for any classical computer, but there's no obvious way to extrapolate this method towards anything actually useful. Why is Google's experiment different? EDIT: To elaborate on my intuition here, the thing I consider impressive about classical computers is their ability to simulate other systems, not just themselves. When setting up a classical circuit, the question we want to answer is not "which transistors will be lit up once we run a current through this?" We want to answer questions like "what's 4+1?" or "what happens when Andromeda collides with the Milky Way?" If I were shown a classical computer "predicting" which transistors will light up when a current is run through it, it wouldn't be obvious to me that we're any closer to answering the interesting questions. | The big difference between the quantum supremacy experiment and your pudding experiment is that the quantum supremacy experiment solved an unambiguous, well-posed mathematical problem. While people sometimes describe the computational task as "simulating the physical Sycamore computer", that's not right. The actual task was calculating the output of an abstract quantum logical circuit, of which the Sycamore computer was an approximate physical instantiation. The difference is subtle but crucial. From a computational perspective, the math came first and the physics came second. Crucially, the quantum supremacy problem was mathematically well-specified, and so it could be checked on a classical computer. The parallel classical computation wasn't just there to provide a time benchmark, but also - crucially - to check the quantum computation for accuracy . There's no such "slower but equivalent" computation to the pudding experiment. In the pudding experiment, you need to specify exact which of these two problems you're trying to solve: Simulate the pattern that will result if you knock a generic pudding cup off the table. Simulate the pattern that will result if you knock a particular pudding cup off the table [where you specify enough detail to nail down the initial conditions in enough detail to model its fall]. The first variant is obviously massively underspecified and doesn't have a unique answer. The second variant does in principle have a unique answer, but crucially, you can't actually capture all the necessary details about the initial condition in practice. So neither variant can actually be framed as a mathematically well-posed question. In the quantum supremacy experiment, the abstract problem to be solved (which was solving the abstract logical circuit, not simulating the physical hardware) was simple enough to pose that you could (very slowly) solve it exactly on a classical computer as well as on a quantum one. | {
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511,072 | for example when two objects undergo a collision and no external force is acting on them. Will both kinetic energy and momentum be conserved or just momentum? | The big difference between the quantum supremacy experiment and your pudding experiment is that the quantum supremacy experiment solved an unambiguous, well-posed mathematical problem. While people sometimes describe the computational task as "simulating the physical Sycamore computer", that's not right. The actual task was calculating the output of an abstract quantum logical circuit, of which the Sycamore computer was an approximate physical instantiation. The difference is subtle but crucial. From a computational perspective, the math came first and the physics came second. Crucially, the quantum supremacy problem was mathematically well-specified, and so it could be checked on a classical computer. The parallel classical computation wasn't just there to provide a time benchmark, but also - crucially - to check the quantum computation for accuracy . There's no such "slower but equivalent" computation to the pudding experiment. In the pudding experiment, you need to specify exact which of these two problems you're trying to solve: Simulate the pattern that will result if you knock a generic pudding cup off the table. Simulate the pattern that will result if you knock a particular pudding cup off the table [where you specify enough detail to nail down the initial conditions in enough detail to model its fall]. The first variant is obviously massively underspecified and doesn't have a unique answer. The second variant does in principle have a unique answer, but crucially, you can't actually capture all the necessary details about the initial condition in practice. So neither variant can actually be framed as a mathematically well-posed question. In the quantum supremacy experiment, the abstract problem to be solved (which was solving the abstract logical circuit, not simulating the physical hardware) was simple enough to pose that you could (very slowly) solve it exactly on a classical computer as well as on a quantum one. | {
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511,789 | I'm aware of a similar question being asked in Do rainbow shows spectral lines? : The response to this question is that the body producing the light is not the water droplet that merely diffracts it but rather the Sun, which acts as a black body and thus produces a complete spectrum. However, it is not entirely true that the sun is a black body: like all stars, it has an spectral class with absorption lines. I am wondering why these lines are not observable in the rainbow or when decompossing sunlight with a prism in otherwise uncontrolled conditions. Are they simply too narrow to be seen without optical instruments? Is it a product of the light source not being coherent? How is this consistent with the possibility of obtaining spectra from distant stars in which the spectral lines are visible? | First you must understand how a (primary) rainbow is formed . Here is a summary with some details that are not in the linked article: Whenever a beam of light encounters an air-water boundary, it is either reflected or refracted. To form a primary rainbow, we must first have enough small water droplets in the air, as these are close to spherical. Parallel light beams that reach such a water droplet must be refracted once as they enter, reflected off the inner surface of the droplet once, and refracted once more as they exit: (image from the linked article) There are 3 important points to note here: Water droplets in the air are never perfectly spherical. That is one source of fuzziness. Parallel beams entering the same water droplet can exit at different angles! This is the major reason why rainbows can never give sharp spectra even if you have perfectly spherical water droplets (say in outer space). Why then do we still see the rainbow? There are three reasons, which together result in the rainbow being seen roughly at 42°: Different incident light beams will have different amount reflected/refracted. In particular, the bottommost incident beam (in the diagram) will mostly pass through without being reflected at the back of the droplet, and the topmost incident beam will mostly be reflected rather than enter the droplet. The light beams emerging from the droplet after the above process are 'denser' along the so-called caustic ray, because the emergent angle does not vary monotonically with the distance of the incident beam from the central axis, and it reaches a maximum for the caustic ray, around which the emergent angle varies less. The incident light beams further from the central axis undergo greater refraction, hence resulting in greater separation of different wavelengths. In contrast, the rainbow rays from incident beams close to the central axis largely overlap one another and wash out. (See this webpage for an image illustrating this.) Light beams may encounter more than one droplet! This is another major reason why we cannot expect a sharp spectrum from a (natural) rainbow. Even if we assume that the water droplet is a point, the light beams from the sun will not be perfectly parallel. In fact, the sun subtends an angle of about 0.5° to an observer on Earth, so this leads to roughly that same amount of spreading of the rainbow as compared to one generated by a point light source. Still, this is a much less significant effect than point 2. ( jkien 's answer is incorrect but inexplicably has lots of upvotes.) | {
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512,834 | I hope this is the right place to ask this question. Suppose I found a small irregular shaped rock, and I wish to find the surface area of the rock experimentally. Unlike for volume, where I can simply use Archimedes principle, I cannot think of a way to find the surface area. I would prefer an accuracy to at least one hundredth of the stone size. How can I find the surface area experimentally? | The way I would do it is to first dip the rock in thinned fingernail polish. Let that dry, and then dip the rock into hot liquid wax. Let the wax cool. Peel the wax off the rock and measure the thickness of the wax layer. Melt the peeled-off wax and measure its volume. Divide the volume by the thickness, and you have the area. | {
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512,902 | The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus). Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, Proposition XIX). Here, Spinoza, following Descartes, denies that a body, the direction of which is changing, is at rest for some instant. https://archive.org/details/principlesdescar00spin/page/86 How is it solved by modern physics? If the object is at rest at some instant, one cannot understand how the movement starts again ( due to the inertia principle). If the object is not at rest at some instant, it seems necessary that there is some instant at which it goes in both directions ( for example, some moment at which a ball bouncing on the ground is both falling and going back up). In which false assumptions does this dilemma originate according to modern physics? | After the invention of modern calculus and notions like continuity and differentiability, the answer is quite trivial in Newton's formulation of mechanics assuming the body is moving along a line. The second derivative of the position should be always defined as it equals the total force acting on the body. Therefore the first derivative must be continuous. This derivative is the velocity. You are assuming that it changes its sign passing from time $t$ to time $t'$ . A continuous function defined on an interval which changes its sign at the endpoints of the interval must vanish somewhere in the interval. That is an elementary result of Calculus. In summary, the body must be at rest at some time, if accepting the modern version of Newton's formulation of classical mechanics. The objection that if the body stops at a certain instant, then the direction of the motion immediately after that instant cannot be decided is untenable. The direction is actually decided by the acceleration, that is by the total force acting on the body at the said instant. | {
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513,026 | I was wondering, there are several examples where energy comes in very handy. Many problems in kinematics which are really challenging if you could use only forces, can be solved effortlessly by converting into terms of energy. (Illustrative image) What makes most force calculations tedious and energy calculations dead simple? Intuitively, is it right to think of energy as a more fundamental quantity than force? | Intuitively, is it right to think of Energy as a more fundamental quantity than Force? The “fundamental-ness” of something is difficult to quantify so it is hard to say whether something is more or less fundamental. However, I would tend to agree with you, but for different reasons. There are two basic approaches to classical mechanics. The Newtonian approach is based on forces, and the Lagrangian/Hamiltonian approach is based on energy. In quantum mechanics only the Lagrangian/Hamiltonian approach is used, the Newtonian approach doesn’t apply, and forces can be difficult to define let alone calculate. So insofar as QM is considered more fundamental than Classical Mechanics I think you would consider energy more fundamental than force. One other point that may lead to the idea of energy being more fundamental than force is that energy is conserved while force is not. Again, fundamental-ness is fairly vague so it is not clear that conservation is a criterion, but it does seem reasonable to consider it briefly. You mention simplicity, which is not as clear cut as these other points. Some (few) problems are simpler in terms of forces, especially ones with friction. It is good to know forces also for those problems. You can apply Lagrangian methods to dissipative systems, but it is not simple. I think your overall impression is correct, that there are more problems which are simpler in the Lagrangian approach than are simpler in the Newtonian approach. | {
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513,999 | I was taught in school that clouds are white due to the scattering of light. Since all rays are reflected it appears as white. But I am wondering about rain clouds. Why are rain clouds darker? | Rain clouds are dark because the part of the cloud you see is in the shade . Clouds are white because they contain tiny water droplets that scatter light of all colors equally in all directions. "Scatters light of all colors equally in all directions" means "white". But if you put a layer of white stuff over another layer of white stuff, the top layer will scatter light from the Sun, reflecting a lot of it into space. That means there's less left to light up the layer underneath. Compared to the top layer, the bottom layer will look darker. For a cloud to produce rain, it needs to be fairly tall (thick). That means the upper parts of the cloud reflect away most of the sunlight, leaving the lower parts in the shade. If you're under the cloud, the lower part is all you see -- and it looks dark. | {
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514,009 | This is not a homework question because I do not want help with solving my homework. I would rather want an explanation of why the derivative of this seems to break math as I know it. Background information: The problem from book: Part of the solution from Slader: My question: When taking derivative of $x=r\sin(\theta)$ , shouldn't this become $x=r\cos(\theta)$ ? Apparently it does not, it should be $x=r\dot\theta\cos(\theta)$ . Why is that? I cannot understand where $\dot \theta$ comes from | Rain clouds are dark because the part of the cloud you see is in the shade . Clouds are white because they contain tiny water droplets that scatter light of all colors equally in all directions. "Scatters light of all colors equally in all directions" means "white". But if you put a layer of white stuff over another layer of white stuff, the top layer will scatter light from the Sun, reflecting a lot of it into space. That means there's less left to light up the layer underneath. Compared to the top layer, the bottom layer will look darker. For a cloud to produce rain, it needs to be fairly tall (thick). That means the upper parts of the cloud reflect away most of the sunlight, leaving the lower parts in the shade. If you're under the cloud, the lower part is all you see -- and it looks dark. | {
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514,262 | I am not asking about whether the photon goes through both slits, or why. I am not asking whether the photon is delocalized as it travels in space, or why. I have read this question: Do we really know which slit the photon passed through in Afshar's experiment? Which theory explains the path of a photon in Young's double-slit experiment? Shooting a single photon through a double slit Where John Rennie says: The photons do not have a well defined trajectory. The diagram shows them as if they were little balls travelling along a well defined path, however the photons are delocalised and don't have a specific position or direction of motion. The photon is basically a fuzzy sphere expanding away from the source and overlapping both slits. That's why it goes through both slits.
The photon position is only well defined when we interact with it and collapse its wave function. This interaction would normally be with the detector. Lasers, Why doesn’t a photon go through the same slit every time? Where ThePhoton says: for example, if you put a detector after a two-slit aperture, the detector only tells you the photon got to the detector, it doesn't tell you which slit it went through to get there. And in fact there is no way to tell, nor does it even really make sense to say the photon went through one slit or the other. In classical terms, this question might be obvious, because a classical billiard ball cannot be at two places in space at the same time. But this is not a billiard ball, this is a photon, a QM phenomenon. And this is not classical terms, but QM. And if we truly accept that the photon travels through both slits, then it basically must exist in space at both places (both slits) at the same time. But as soon as we interact with it (the wave function collapses), the photon becomes spatially localized, but only at a single location (at a certain time). What is not obvious from QM, is how we can have these two things at the same time: the photon pass through both slits but we can only interact with it at one slit (not both) What is that basic thing in QM, that will disallow for the photon to pass through both slits and be interacted with at both slits too? Somehow the QM world underneath will change to classical as soon as we measure, and interact with the photon. This change from QM to classical is where the possibility of the photon being at both places (both slits) at the same time gets disallowed somehow. This could be decoherence, as the QM entity gets information from the environment (because of the measurement), or just the fact that the wavefunction collapses and that has to have a single spatial location for the photon when measured. So basically the photon goes through both slits, thus, it in some form exists at both slits at the same time. But when we try to interact with it, it will only be spatially localizable at one of the slits, not both at the same time. Question: If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)? | If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)? Alright, let's play some word games: This isn't a well-defined question. "Detect a particle" doesn't mean anything in quantum mechanics. Quantum mechanical measurements are always measurements of specific observables . There is no holistic act of "observing all properties of a system at once" like there is in classical mechanics - a measurement is always specific to the one observable it measures, and the measurement irrevocably alters the state of the system being measured. People often use "detect a particle" as shorthand for "perform a position measurement of a particle". By definition, a measurement of position has as its outcome a single position, and interacts with the state of the particle being measured such that it now really is in the state in which it is at that single position and nowhere else. So if you could perform position measurements that yielded both slits as the position of the particle, this would mean you have performed an impossible feat - there are now two particles, each in the state of being at one slit and that slit only. Quantum mechanics may be weird, but it is hopefully clear it is not this weird - we cannot duplicate a particle out of thin air just by measuring it. If you don't insist on "detect" meaning "performing a position measurement", then of course the standard double slit setup is a "detection" of the photon at both slits - the pattern on the screen is only explainable by the particle's wavefunction passing through both slits and interfering with itself. This is of course just indirect reasoning - there simply is no observable whose eigenstates would naively correspond to "we have detected the photon at both slits at once". Lastly, you seem to confuse "interacting" with "measuring" or "detecting". Of course we can interact with the particle at both slits - we just cannot perform position measurements (or other "which-way" measurements) at both slits and expect them to yield the impossible result of the particle split in two. But if you look at more sophisticated setups like the quantum erasers, there certainly is interaction with the particle at both slits - just carefully set up to not destroy the interference pattern, and hence no obtaining useable which-way information. | {
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514,570 | I have an idea that the space elevator can be build in the following conditions: Not a classic theory. No cable. No Geostatic orbit. Not equator platform position The possible platform / base position is Earth's south magnetic pole - Antarctica, or north magnetic pole in Arctic. The lifting force is Earth's magnetic field. The lift design is a huge ring (more than 100 m) with a magnetic coil. The wire of the coil is a superconductor for a high current. The surface of the ring is closed with parabolic solar energy collector with lens in the focus. The lens is the laser beam receiver from the ground platform. Near the focus are located payload spacecrafts with engines for changing the motion vector after dumping and during the fall. Let us clarify the question in the following way:
1) “What conditions and parameters of the electromagnet ring or coil and of other components in the assembled structure (elevator) will satisfy the condition that the Earth’s magnetic field will be enough to lift up this structure.” The parameters can be like the following: radius of the ring R (m), number of circuits in a coil N (pcs.), current I (A), weight M (kg) and others required for calculations. The additional questions:
2) “What height can be reached if the method is possible at all”
3) “Is it possible somehow to focus the Earth’s magnetic field of some circle area in the pole which will help to increase the chances for questions 1 and 2” Legends: Item 4 - ring with coil is red color Item 5 - wire is inside the coil inside the ring (4) Item 6 - solar panels - energy collector is transparent blue Item 6 - lens in the focus - laser beam receiver - purple Item 7 - payload - green cylinders Updates (20.11.19). I see one more reason that the elevator will not work out. While the electromagnetic field inside the ring will be oriented in the opposite direction to the Earth’s magnetic field creating the lifting force, the electromagnetic field outside the ring will goes the same direction as the Earth magnetic field and will pull the ring down. Only in case if the radius of the ring will be bigger than the radius of the Earth’s magnet we will not face this issue until some height is reached. The picture of 3dmodel is updated with better rendering (20.11.19) | Setting aside the technical details, the biggest problem with this idea would be Earnshaw's theorem , stating that it is impossible to stably levitate a magnetic dipole in another magnetic dipole's (Earth's) field. In other words, you could theoretically generate enough magnetic force for lift-off with your dipole (the coil), but you would have trouble stabilizing it so that it doesn't topple over and fly back into the Earth. Edit, after bringing up active stabilization in the comments: While in theory, yes you could actively stabilize the whole setup. And just for the sake of arguing, you could also stabilize the ring by (mechanically) spinning it around its axis. The torque that would act to topple over the dipole would now cause precession instead. Voila: the ring is stabilized. However, I still don't think that this whole idea would work in reality. The magnetic moment of the dipole is $m_{B}=IS$ , where $I$ is the current through the loop and $S=\pi R^2$ is the enclosed area by the loop of radius $R$ . Since we're talking about a setup on the magnetic axis of the earth, the geomagnetic field can be written as $B_{\rm E}=B_0(1+h/R_{\rm E})^{-3}$ , where $B_{0}\approx62\,$ µT is the field strength at the surface of the Earth, $R_{\rm E}\approx6400\,$ km is the radius of the Earth, and $h$ is the height above ground. The force generated by our dipole lifter would be : $$
F=-\frac{\rm d}{{\rm d}h}\left[m_{B}B_{\rm E}\right]\
=-\pi IR^2B_{0}\frac{\rm d}{{\rm d}h}\left[(1+h/R_{\rm E})^{-3}\right]
=\frac{3\pi IR^2B_{0}}{R_{\rm E}}(1+h/R_{\rm E})^{-4},
$$ We can also approximate that $h\ll R_{\rm E}$ , so that $F\approx3\pi IR^2B_{0}/R_{\rm E}$ . Now, then lets say that you build yourself a ring with radius $R=500\,$ m, and drive a current of $I=10\,$ MA through it. That would generate a lifting force of about $230\,$ N (newton) , or about $23\,$ kg ( $50\,$ lbs for the imperial entanglementalists). But how much would such a device weigh? We, of course, use superconductors in order to not require extremely thick copper or silver conductors. But the problems with superconductors is that you can't just drive however large current you want through them. At some critical current density, the supercondictivity will break down due to the high magnetic fields generated by them. Lets say we build that loop of niobium-nin , a superconducting material with a reportedly record-breaking current density. It's highest current density is $j_{\rm max}\approx2600\,$ A/mm $^{2}$ . We wanted to push $I=10\,$ MA through it, so that would require a cross-sectional area of $I/j_{\rm max}\approx4000\,{\rm mm}^{2}=0.004\,{\rm m}^{2}$ . If we multiply that by the length of conductor needed, $L=2\pi R\approx3000\,$ m, we get a volume of $12\,{\rm m}^3$ . I don't have the (mass) density of niobium-tin, but needless to say $12\,{\rm m}^3$ of it will weigh in the order of at least $20{-}100\,$ tons, and that's not counting all the cooling apparatus you would need in addition to the pure weight of the superconductors. (And I haven't even gotten to address the mechanical strength required for you to spin the whole ring at hundreds or thousands of rpm, in order to stabilize it.) Tl;dr You won't be able to generate enough force to even remotely lift your apparatus. I hope this can make you sleep well tonight. | {
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515,086 | When the train accelerates, the bob would also accelerate with the same magnitude and direction as the train.
From the free body diagram, only tension and weight are exerted on the bob.
I understand how to relate the "horizontal component" of the tension to the acceleration of the bob.
It moves "backward" and in fact its vertical position is different from when the train is not accelerating. I am not sure whether inertia could explain this situation. What is the force moving it backward/ lifting with the height? | The bob doesn't move backwards at all. The train is moving forwards (according to your reference frame) and if the bob wasn't attached to the train it would remain stationary. Since it is attached to the train there is a tension in the wire (as the top anchor point moves with train and the bottom is attached to the weight, so stretching the wire slightly). The vertical component of that tension (which is in the direction of the wire) counteracts gravity and allows the bob to rise whereas the horizontal component accelerates the weight in the direction of the train. | {
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515,259 | I was wondering about this question since I learned about rolling motion in the chapter on rotational mechanics. I was unable to come to a solid conclusion due to the reasons mentioned below. The following diagram shows a ball on a frictionless inclined plane and the forces acting on it: The forces acting on the ball are shown in red and are the normal contact force $N$ and the gravitational force of attraction $mg$ . I qualitatively determined the torque of these forces about two axes - one passing through the centre of mass of the ball of uniform density, and the other passing through the point of contact of the ball and the inclined plane. Both of these axes are perpendicular to the screen. When the axis passes through the centre of the ball, the torque exerted by $mg$ is zero as its line of action meets the axis. Further, the torque exerted by $N$ is also zero due to the same reason. There are no other forces. So, net torque about this axis is zero, and this tempts us to conclude the ball slides down the inclined plane. When the axis passes through the point of contact, the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll i.e., it rotates while moving down the inclined plane. This conclusion is contradictory to the former case. So, what exactly will happen to a ball kept on a frictionless inclined plane - will it slide or roll? The following diagram is a visual interpretation of my question (if the terms slide and roll confuses the reader) where the red arrow denotes the orientation of the ball: Image Courtesy: My own work :) Please Note: The question - Ball Rolling Down An Inclined Plane - Where does the torque come from? discusses the case of ball rolling on an inclined plane where friction is present . Since the question - Rolling in smooth inclined plane is marked as duplicate of the former, and has no sufficient details, I planned to ask a new question with additional information. | ...the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll... Actually, it means that the angular momentum about that axis must increase. That is not the same as rolling. If the axis is through the center of mass of the object then the only way for the angular momentum to increase is through rolling. However, if the axis does not pass through the center of mass then there is also angular momentum due to the linear motion. In other situations this is the difference between orbital angular momentum and spin angular momentum. So let's calculate the "orbital" angular momentum in this problem. The torque is $m g R \sin(\theta)$ where $R$ is the radius of the ball and $\theta$ is the angle of the incline. The magnitude of the "orbital" angular momentum is given by $R m v$ where $v$ is the linear velocity of the center of mass, so its time derivative is $R m a$ where $a$ is the linear acceleration of the center of mass. From Newton's laws the linear acceleration is the component of gravity which is down the slope. This is $ma=mg \sin(\theta)$ so $a=g \sin(\theta)$ . Substituting the linear acceleration into the time derivative of the orbital angular momentum gives $R m g \sin(\theta)$ which is equal to the torque. This means that the increase in angular momentum due to the torque is fully accounted for by the increase in the "orbital" angular momentum and there is no left over torque for increasing the "spin" angular momentum. Therefore, the ball does not spin/roll regardless of which axis you examine. | {
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515,304 | Everything, in theory, can have a smell, but that is not the whole point of this question. My main query is, since things do smell, does that mean that everything is slowly evaporating (or, sublimating, I suppose)? For example, if we perceive metal to have a scent, this must be because some of the molecules of metal are in the air or are at least affecting the gasses between the metal and our noses. What is happening in terms of physics that causes almost anything to be perceivable by a (biological or otherwise) molecule receptor? (There are a few similar questions closed as off-topic, but I would like to argue that this is a valid physics question.) | Not everything is evaporating. You raise the point that we can "smell" certain metals which, given how you smell most things would imply that the metal is evaporating somehow and entering your nose. You'd be right to think this is strange and in contradiction with the idea that metallic bonds tend to be strong and so unlikely to evaporate. This is a good answer explaining why we appear to be able to smell metals. | {
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516,203 | Question A uniform rod of mass $M$ is placed almost vertically on a frictionless floor. Since it is not perfectly vertical, it will begin to fall down when released from rest. I have seen solutions online for this problem and while solving this problem, it is assumed that the end point of the rod that is in contact with the floor will continue to stay in contact with the floor till the rod, in its entirety, hits the floor horizontally. It is this assumption that lets us determine the normal force from the floor. However, how does one show that this assumption is true? Or is it taken to be an additional constraint of the problem? Check the figure in D1 to verify if you've got the right setup in mind. Duplicates in SE: D1: Will a falling rod always remain in contact with the ground? D2: Rod Falling on Frictionless Surface D3: Equation of motion for a falling rod (with one end touching a frictionless surface) I believe the OP in D1 has asked the same question (along with other questions) but it has been closed as off-topic. Simon Robinson, one of the answerers in D2, has also expressed concerns about this. I ask this question because it hasn't been addressed properly on SE. I don't feel that the answer to this question is specific only to this vertical rod problem. Instead, I feel that this question is onto something basic that I don't yet understand regarding the necessary constraints that need to be specified in a physics problem. My Attempt The problem with this question is that I feel like I have given all the information that's necessary to predict the entire dynamics of the rod's motion after its release. I'm unable to accept the idea that "rod-cannot-lose-contact" constraint must be specified as an additional piece of information to solve this problem. If we accept that it's not an additional constraint, then we should be able to show that the rod's end point cannot lose contact. But, that's the problem. I've been thinking about it for days and I can't seem to find a way to show that. I'm unable to see anything "violated" if it loses contact at some point during its fall. After it loses contact, it simply rotates about the center of mass with constant angular velocity [See $(1)$ ] and the rod's COM falls down with acceleration $\mathbf{g}$ . $$\frac{d\mathbf{L}_{CM}}{dt} = \boldsymbol{\tau}_{CM} \Rightarrow \text{$L_{CM}=I_{CM}\omega\;$ is constant} \tag{1}$$ Thanks for taking the time to read this question. I apologize if I have violated any code of conduct. Any insight that addresses my question would be greatly appreciated. Further Clarification, If Needed Clarifications which will hopefully help PhySE users to better understand my question are made here. Reading the following information is not necessary to answer my question. It is important to note that even if the rod's bottom end point loses contact with the floor at some point during the fall, the centre of mass of the rod will continue to fall vertically straight down just as before (but now with acceleration $\mathbf{g}$ ). So, the fact that the COM falls vertically straight down cannot be used to prove that the rod's bottom end point doesn't lose contact with the floor. COM falls vertically straight down $\not\Rightarrow$ the rod's bottom end point doesn't lose contact with the floor | The technique to use in problems like this is to assume that the rod remains in contact with the table, and to then try to figure out whether the normal force ever switches sign for some angle $\theta$ as the rod falls. If it does, then the rod's lower tip will have to leave the table, as a "frictionless table" cannot pull the rod downward; it can only push it upwards. Similar techniques are used in the solution to the classic "puck slides down a frictionless hemisphere" problem, as well as the "toppling ruler" problem. Actually doing this is something of a mess, but here's a rough sketch. Let $L$ be the length of the rod and $m$ be its mass. Let $I = \frac{1}{4} \beta m L^2$ be the rod's moment of inertia about its center of mass; note that $\beta = \frac{1}{3}$ for a rod of uniform density, while $\beta = 1$ if the mass is concentrated at the tips. This is done to provide a bit more generality; I will assume, however, that the mass distribution is symmetric, so that the center of mass is at the geometric center of the rod. The ingredients you'll need are: Geometric constraints: The vertical position of the center of mass of the rod will be $z = \frac{1}{2} L \cos \theta$ (taking positive $z$ to be upwards.) Differentiating this twice, we obtain for the velocity and acceleration of the center of mass $$
v = - \frac{L}{2} \omega \sin \theta, \\
a = - \frac{L}{2} ( \alpha \sin \theta + \omega^2 \cos \theta),
$$ where $\alpha$ is the angular acceleration of the rod. Conservation of energy: Since the table does no work on the tip of the rod, the mechanical energy of the rod is conserved. This gives a relationship between $v$ and $\omega$ . Newton's Second Law (translational): Using Newton's second law, you can relate $a$ and $N$ . Newton's Second Law (rotational): Calculating the torque about the center of mass of the rod, you can find a relationship between $N$ and $\alpha$ . This gives us a system of five equations and five unknowns $\{N, v, a, \omega, \alpha \}$ which can be solved. After going through it, I find that the normal force as a function of $\theta$ is $$
N = \frac{mg \beta (\beta + (1- \cos \theta)^2)}{(\beta + \sin^2 \theta)^2}
$$ which is manifestly positive for any value of $\theta$ . Thus, the tip of the rod does not leave the table; the table continually maintains an upward normal force as it falls. | {
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516,511 | Articles, with very little detail, have made their rounds about an X17 boson (16.7 MeV) being observed in tests of decaying beryllium-8 and perhaps once in a test with helium. Most of the undiscovered particles that are searched for in CERN or other colliders or dark underground quiet dark pools filled with water were predicted prior to being searched for. (WIMPs, the Higgs boson, neutrinos, axions - all predicted first), after which, there was a long period of trying to find them, in some cases, still ongoing. And, likewise, sometimes objects are observed before they're predicted ( dark matter and dark energy comes to mind, probably more). I know that a 5th force has been theorized, so in a sense, a 5th boson has long been proposed, but the articles I've read imply only that beryllium decay products were studied and X17 has been observed as a bump in the chart of output data, but no indication of what they were looking for when studying beryllium decay. I realize the discovery isn't official yet, but some articles suggest pretty good certainty that the observation is legitimate. No mention of what they were looking for suggests to me that this was a surprise discovery, not a study where they expected to find something new, but I wanted to confirm, as the articles I've seen are all very short. If there was something specific they were looking for when studying beryllium decay, then identifying that would be appreciated. A paragraph or two in layman's terms is also appreciated. | Is the X17 predicted, or discovered? It depends upon what the meaning of the word "is", is. Prediction vs. Discovery In particle physics, phrases like "the Higgs boson" or "dark matter" actually stand in for mechanisms or effects, which could be realized in many different concrete models. And even when you specify a model, it's still ambiguous because you can vary how "fundamental" your model is. The Higgs boson itself is a good example of this. Before the LHC, we knew for sure that particles in the Standard Model had mass, that this was difficult to accommodate the standard way (by adding so-called "mass terms" to the Lagrangian), and that a very simple and elegant way to do it was the Higgs mechanism . By calling it a mechanism, we mean that it required only general ingredients, such as spontaneous symmetry breaking and new bosons, but that it wasn't specified how the symmetries were broken or what new bosons were added. What we call "the" Higgs boson is just the simplest set of ingredients you could supply to get the Higgs mechanism to function. There were many alternatives, such as the two-Higgs-doublet model , which would give you five Higgs bosons instead. And even once you fix "the" Higgs boson, you could go deeper. In the resulting model, the mass of the Higgs boson is put in by hand, and basic consistency requirements only fix it within a rather wide range. You could consider more complex models that predict more specific values of the mass, or even explain where the Higgs boson itself comes from, e.g. if it falls out of an elegant grand unified model or is itself made of other particles, like the proton is made of quarks. There were so many of these theories that just about any Higgs mass in the reasonable range could be accommodated by one or ten. So I would say it's fairly clear that a Higgs boson was predicted before discovery... but it's completely up for debate whether a Higgs boson of mass $125$ GeV was predicted or discovered! Fifth forces Like "dark matter" or "the Higgs boson", "the fifth force" is another one of those vague phrases that really stands for thousands of distinct models -- basically any model that consists of the Standard Model and any new gauge bosons. Fifth forces are interesting because they are extremely simple extensions of the Standard Model (e.g. you could just take one of the existing forces and exactly copy it), and many more complex/fundamental models give you them automatically. Another appealing feature is that many types of fifth forces give rise to striking experimental signatures, which we can look for with great sensitivity. So in that sense, a fifth force has been predicted for a long time. However, because of the avalanche of possible distinct fifth force models, experimentalists usually don't try to test specific models. Instead, they parametrize the effects of the fifth force in terms of a few quantities (the mass of the gauge boson, the gauge coupling, the coupling to electrons, the mixing with the photon, etc.). Then they run tests that try to capture as much of this parameter space as possible. Many such tests have been performed in the past and are ongoing now. The experimentalists behind the X17 boson were looking for a specific type of fifth force, called a " dark photon ". These are part of a relatively new mechanism to produce dark matter (DM), using an entire "dark sector" where the DM only affects normal matter through the dark photon. This mechanism has been gaining in popularity because it results in lighter DM, and the traditional heavy DM has been pretty conclusively tested by WIMP searches. However, "dark photons" are still not a specific model, but rather stand for a general idea that has a range of corresponding concrete models, and within each one the mass of the dark photon can lie in a wide range. So they were inspired by these theoretical ideas to look for MeV-scale bosons, but not directed to any specific mass; they just were trying to capture as big a slice of parameter space as possible. They indeed found evidence for a new MeV-scale boson, but it behaved in a rather strange way. In a later theory paper , it was explained that the boson could not have been a dark photon, as a dark photon with that mass and coupling would have already been discovered in earlier experiments. Dumping the dark photon interpretation, the theorists showed that the observations could be accommodated, without conflict with earlier experiments, if the new boson had the unusual property of being "protophobic", i.e. not coupling to protons. The experimentalists then did a follow-up work confirming their results, leading to the recent media frenzy, and that's where we are now. I hope it's apparent that your question does not have a yes or no answer! Fifth forces have been thought about for a long time, but they're a very general thing. The experimentalists were motivated to look by a recent, particular kind of fifth force (dark photons) -- but what they saw wasn't compatible with it. Science often progresses this way. Addendum I realize the discovery isn't official yet but some articles suggest pretty good certainty that the observation is legit. This needs to be said: I don't think any practicing particle physicist would give this observation more than a 5% chance of holding up. At any moment, there are about 50 distinct experimental anomalies, each of which could revolutionize physics if true. Historically, the vast majority don't pan out. That's because physics is subtle, experimental physics is even more subtle, and these measurements are often pushing the limits of what our equipment can do. (There is the further problem that the lab claiming the X17 has a history of reporting similar anomalies -- if I recall correctly, they already pulled this exact same thing twice in the past, with different boson masses, and never explained why those observations went away.) It just so happens that you've heard more about this anomaly because it blew up in the media. This happens naturally: the more prominent a story gets, the more each journalist wants to write their own take on it, and so you automatically get a winner-take-all situation where X17 gets more attention than the next ten most recent anomalies combined. I wouldn't recommend worrying about the X17. Science will go on, the experiments will be thoroughly scrutinized and replicated, and in ten years' time we'll know for sure if it's real or not. If it is, you'll hear about it everywhere. If you hear nothing, assume it disappeared. | {
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516,609 | Long story short, my brother made a joke about how stupid it is to celebrate the Earth making one "trip" around the Sun: New Year's Eve. So I got curious and was wondering: how could the first physicists guess that a revolution of the Earth around the Sun takes 365.25xxx days? I've tried my luck on the internet but didn't find anything simple enough to understand. I've read about Kepler's law, and there are nice formulas, but how does he work this out? And how could the first people who knew that Earth was revolving around the Sun guess how much time an entire revolution took? I asked my brother and he told me if he'd have to do it, he would simply watch the stars and map them, do that every night, and count the days until he came back to the exact same position of the first night. Is that realistic? | In ancient civilizations, astronomy was a serious business (among other reasons, to accurately predict the seasons), so there were a lot of scientists making very careful measurements. Even with the naked eye, you can make quite accurate observations, and the ancients used these observations well. The first really accurate determination of the length of the year was made by Hipparchus, a Greek astronomer who lived circa 190-120 BC. He calculated a year to be 365 + 1/4 - 1/300 (365.24667) days long, which is 6 minutes/year longer than the current estimate. Before that time, most people used 365.25, which is also not too far off (11 minutes), but is likely also due to the lucky coincidence the actual value is so near the neat round 1/4 day. Hipparchus mostly used exact measuring of the equinoxes (the moment day and night are just as long, and when the Sun rises exactly in the east and sets exactly in the west). It's possible to measure this quite exactly, up to an hour. But Hipparchus also used the data gathered by earlier astronomers, back until 432 BC (~300 years before him). When you have this large time-span, and combine multiple measurements to cancel out any errors, it's thus possible to get very accurate results. Oh, and he used solar eclipses to refine his measurements as well, though the equinoxes were more important (there are more of them). Incidentally, he was also the first one to discredit your question ;-) That is, what most people think of as "a year", is actually not the time it takes the Earth to complete one orbit around the Sun. It's very close, but the Earth also precesses: it's top (the poles) very slightly turn around. It also means that Polaris (our current "northern pole star") will not remain the northern pole star forever (but it will for several thousand years). This also means that the time from equinox to equinox is not exactly the same as the time it takes the Earth to complete an orbit, it is shorter by 20 minutes and 24.5 seconds. The time it takes the Earth to complete an orbit is called a sidereal year , but the time it takes for the equinoxes to repeat (and thus the seasons), is called a tropical year . Strictly speaking, "a year" could refer to both of them, but most of the time a tropical year is meant. And a tropical year is also what our calendar is based on. At least, the western calendar uses a value of 365.2425 days, while the modern established value for a tropical year is 365.24219 days (27 seconds shorter), and a sidereal year is 365.256 36 days (1197.5 seconds longer)
These values are all assuming a day is exactly 86400 SI seconds, without any leap seconds. | {
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517,159 | I am having an argument with my friend about how a nearly-full soda bottle should be stored in the fridge, with the goal of keeping the soda from going flat (i.e. keeping as much of the gas dissolved in the liquid as possible). Assume the bottle is a standard 2L bottle and its walls are impermeable so this is a purely static problem. He thinks it doesn't make a difference whether the bottle is standing up vertically or lying flat horizontally. I think that standing up is better. My vague reasoning is that in either case, the plastic surface will feel the same amount of pressure $P$ from the contents; but when the bottle is lying horizontally, the shape of the air has a larger surface area and needs more stuff in it to attain the same pressure (since pressure if force over area). My statement about the shape of the air is likely true by the isoperimetric inequality (or at least some heuristic which says that among shapes with the same volume, the ones "closer" to a sphere have less surface area), since the air-shape at the top of the bottle in the vertical case is closer to a sphere, than the shape when it is lying horizontally (in this case the air is like a long prism). However I'm not sure it's true that the pressure felt by the plastic is the same in both cases. Who is right and why? EDIT: thank you for your explanations. I understand how to answer this in terms of constant volumes now, although it isn't very intuitive to me. I wonder if there is an explanation involving the sum of the forces acting on the surface of the liquid being zero, for both the horizontal and vertical configurations. | The volume of the bottle and the volume of the liquid are the same both ways. By subtraction, the means the gas volume is the same either way. The difference assumes in the question’s reasoning doesn’t exist. There is more area in the sideways case. That’ll let equilibrium be reached more quickly. But it’ll reach the same equilibrium either way. | {
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517,636 | Vectors give both magnitude and direction, whereas scalars can be thought of as magnitude without direction. So, velocity is a vector since it is speed with direction. Similarly, what is the scalar analog of acceleration? Velocity is to speed as acceleration is to ______. If there is nothing to fill that blank, is there a reason why velocity is so special? | In English, it seems that: Position is a vector. Distance/length is a name of its magnitude. Velocity is a vector. Speed is a name of its magnitude. Acceleration is a name of a vector and its magnitude. Force is a name of a vector and its magnitude. Momentum is a name of a vector and its magnitude. ... Velocity/speed as well as position/length seem to be exceptions. The general trend is to not have different names for the scalar-forms of vectors. In fact, I asked why this is the case on the History of Science and Math SE site a few months ago. The answer told me that Gibbs and Wilson formally defined the difference between speed/velocity in technical English in 1901 in their book Vector Analysis : Velocity is a vector quantity. Its direction is the direction of the tangent of the curve described by the particle. The term speed is used frequently to denote merely the scalar value of the velocity. This convention will be followed here. Since then, others continued this trend and it eventually got settled. Before then, the distinction was less clear. In other languages, there is not necessarily such a distinction. It is consensus in English, Spanish, my mother-tongue Danish and others, but not in Russian, German etc. | {
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518,380 | From this source , on average a rubber tyre loses $2$ atoms thickness per revolution due to the friction with the road. Do these atoms/molecules that were peeled off the tyre get absorbed by the road? Then would the road increase in thickness due to this? If so, why doesn't the road lose it's thickness to the tyre instead? Thus making the tyre increase in thickness over time... | Anyone who has lived near a major thoroughfare knows that tire wear creates rubber dust that gets spread around everywhere. In some parts of Los Angeles, the residents have to hose off the sidewalks every day or so in order to prevent that dust from being tracked into their homes. Some of it also gets friction-welded into the surface of the road itself, leaving skid marks that are plainly visible. Road surfaces themselves also lose material to the passage of tires. On concrete roads, the tire scrubbing action rubs away the mortar between the pebbles and after a while the pebbles become unsupported, and they too become dislodged. The road material does not end up accumulating on the tire surface because the tire surface is being continually worn away. Any road material that might get embedded in it is thereby shed loose along with the tire dross. | {
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518,423 | I have a bit of misconception about weight which I want to clarify. The air pressure is explained as the weight of the air column above our head acting per unit area. But since air is not continuous how can the weight of all the air molecules (above our head) be acting on our head? I mean we would only feel the weight of the molecules near the surface of our head (if not then why not?) but how do we feel the weight of molecules so far away? I have edited this question and asked this follow up question (so that it remains specific). | Imagine that the air in the atmosphere was just somehow sitting there unpressurised. What would happen? Well, Earth's gravity would be attracting all that air towards the centre. So the air would start to fall downwards. The very bottom layer of air would be prevented from falling through the solid surface, as the air molecules rebound off the molecules of the surface. But the layer above that doesn't stop. So Earth's gravity forces the air in the lower part of the atmosphere to accumulate against the surface of the planet, becoming more and more dense. As the air gets denser near the surface, it becomes more and more likely that air molecules collide. That's what air pressure is: the average force of all those air that would hit a surface you placed in the air. But the air pressure also acts on the air itself. So eventually the force of the air pressure at the bottom layer of air pushes up on the layer of air just above it enough to counteract the pull of Earth's gravity on that layer of air. And so you get another layer that is prevented from falling. But the air above that is still being pulled down, and so more air is being squashed down into this second layer above the surface. This increases the force that the bottom layer needs to provide to the next layer; the air molecule collisions not only need to provide enough force to counteract the weight of the air immediately above it, but also to provide those molecules with enough momentum that when they in turn collide with the air in the third-bottom layer it can support the weight of that layer as well. So more air squeezes down to the surface until the pressure at the bottom layer is sufficient to support the weight of the 2 layers above that. Obviously the atmosphere isn't actually split into discrete layers like this 1 , but hopefully it's a helpful way to think about it. You should be able to see how gravity squeezes the air down against the solid surface, until the pressure at the bottom is just enough to support the weight of all the air above it. This is why air pressure drops off at higher altitude. As you go up, there is less air above squeezing down, so equilibrium with gravity is reached at a lower pressure. So it's not literally that the air pressure you feel is the weight of the column of air above you. It's not that your head is somehow "holding up" a 100km column of air above it. But the air pressure of the air surrounding you must provide an equivalent force to the weight of all the air above it. If it did not then the weight of the air above would be partially unsupported, so gravity would squeeze it down further, increasing the pressure until it was equal to the weight of all of the air above. This is also why the top of your head doesn't feel any difference in air pressure to the side of your body. Air pressure is the same in all directions, because the air molecules are really just zipping around in countless different directions, uncoordinated with each other. Those molecules colliding with things must supply enough average force in the upward direction to support the weight of the atmosphere, but when the pressure increases due to gravity it can't cause a coordinated force that is only upwards, so there is just as much force from air pressure on the side of your body as there is on your head. 1 And if you actually had the atmosphere of Earth spread out in a diffuse low pressure cloud and let it all fall under gravity the results would be much more exciting than I have described. | {
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518,425 | We can define cross products mathematically like if we take two vectors, we can find another vector with certain properties but why do we use it in physics, if we consider a hypothetical physical quantity like force which is equal to cross product of certain vectors? For example, the force exerted on a charge in motion in an uniform magnetic field. Why is it so? Why does that force have to be a cross product of two vectors? Is it possible to come up with them when what we do is just observe the nature? | This is a great question. The dot and cross products seem very mysterious when they are first introduced to a new student. For example, why does the scalar (dot) product have a cosine in it and the vector (cross) product have a sine, rather than vice versa? And why do these same two very non-obvious ways of "multiplying" vectors together arise in so many different contexts? The fundamental answer (which unfortunately may not be very accessible if you're a new student) is that there are only two algebraically independent tensors that are invariant under arbitrary rotations in $n$ dimensions (we say that they are " $\mathrm{SO}(n)$ invariant"). These are the Kronecker delta $\delta_{ij}$ and the Levi-Civita symbol $\epsilon_{ijk \cdots}$ . Contracting two vectors with these symbols yields the dot and cross products, respectively (the latter only works in three dimensions). Since the laws of physics appear to be isotropic (i.e. rotationally invariant), it makes sense that any physically useful method for combining physical quantities like vectors together should be isotropic as well. The dot and cross products turn out to be the only two possible multilinear options. (Why multilinear maps are so useful in physics is an even deeper and more fundamental question, but which answers to that question are satisfying is probably inherently a matter of opinion.) | {
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519,352 | Imagine a standard $^{235}\text{U}$ pellets scaled to the scale of Earth, I wonder, would it explode or just meltdown? I believe despite overshooting the critical mass, the heat produced is still too small to start sustained chain reaction? Right? | It is worth noting that there is no such thing as a critical mass : there is a critical condition where there is enough mass, neutron reflection and other shape properties to make a chain reaction run to completion rather than produce some heat or a fizzle. Still, a huge sphere of U235 will not just sit there. 50 kg of uranium-235 is the bare sphere critical mass of uranium-235, that is, it will become critical even without reflectors (see the linked paper by Chyba and Milne for calculations). For a huge sphere the neutron mean free path $\lambda_f = 16.9$ cm will be far less than the radius and all neutrons in the core hit something. That some of the surface is too "cold" to react does not matter if much of the bulk does. There is a final issue: can the explosion overcome the gravitational binding energy of a huge lump of uranium? I we assume one earth mass of uranium at normal density I get $R=4210$ km and $E_{binding}=3.4\cdot 10^{32}$ J. U235 fission releases about 82 TJ per kg. So the nuclear energy that can be released is $4.9\cdot 10^{38}$ J, or 1.4 million times the binding energy. So, yes, it will explode even if the reaction is fairly inefficient: even a fizzle would turn it into plasma. | {
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519,504 | Is the weight of the aircraft flying in the sky transferred to the ground? Is the weight of people swimming in the pool also transferred to the ground? How can we prove it? | Is the weight of the aircraft flying in the sky transferred to the ground? Yes, at approximately the speed of sound. Is the weight of people swimming in the pool also transferred to the ground? Yes How to prove it? The swimming one can be quantitatively proven with a kitchen scale. Weigh an object that will float. Fill a bowl partway with water. Weigh it without and then with the object. The airplane one can be qualitatively seen in pictures of aircraft flying low over water. | {
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519,904 | It's my experience that most people (including myself) fall backward on their behind when trying to learn to walk forward on flat ice and fall forward when trying to run on it. When walking or running (in the now "normal" way: heel first, ball of the foot last, contrary to the aesthetic walking of a ballet dancer and the same we do when walking backward; in fact, the more Natural way to walk is like the ballet dancer does; see the picture below; only with the advent of shoes, especially the modern back shock-absorbing shoes; just try running while putting the front part of your feet first on the ground; feels pretty good!) on a flat non-slippery underground you mostly fall forward and mostly after tumbling (tripping sounds a bit inappropriate in this case). By the way, by walking and running I mean(in the now "normal" way: heel first, ball of the foot last, contrary to the aesthetic walking of a ballet dancer and the same we do when walking backward. In fact, the more Natural way to walk (which is now used to walk as silently as possible) is like the ballet dancer does (see the picture below). Only with the advent of shoes, especially the modern back shock-absorbing shoes we changed our walk. Just try running while putting the front part of your feet first on the ground. Feels pretty good! But let stick to the modern way. In can understand why, when walking on high friction underground you mostly fall forward after tumbling. For example, bumping with one foot against a curb you didn't see causes a torque (the force of which grips my foot in backward direction) making my center of mass rotating forward around my foot whereafter gravity makes me fall. When walking on flat ice, only slipping is involved though, without obstacles your feet can bump against. Of course, there has to be some friction (and there is some friction, though little) because without friction walking or running would be impossible. Now when you are an experienced ice walker, suppose your mass rests on your whole left foot, and the ice gives just enough friction not to slip (let's say in any fall of the normal walking on ice). Then you give your right foot forward motion. This forward motion is what gives you a forward motion in combination with the left foot that stays put but "rolls" over the ice and pushes you forward. (the movement of you right feet gives the left foot a backward force on the ice, which propels you forward. Meanwhile, the left foot changes from staying flat on the ice to standing on the front feet and when the heel of your right leg touches the ice it "rolls" along and your whole mass rests now on your right feet, after which the left leg starts moving and the cycle is complete. Nothing different from normal walking on a rough underground. But why do people learning to walk on the ice fall so often backward, one of their legs moving upwards in a forward direction? Where can't the friction stop the slipping? Of course, they also fall forwards but this is mostly the case when they try to run on the ice. | This is not theoretical question about physics as much as biomechanical question about technique of walking. When I run during winter, I always fell forward (assuming there is no slope). That is because I am putting my legs on the ground with backward motion to use the force to accelerate. So when they slip, they go backward, while my body center goes still forward. A lot of people during walking are putting their legs on the ground with forward motion, so when they slip, the legs go forward faster than their body center and they fall on their backs. This is mostly because nowadays we learned to walk through our heels instead of toes as a consequence of wearing shoes. | {
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519,916 | I recently posted an answer to Sustainability StackExchange where I claimed that energy consumption for cars (specifically, for electric cars) with no air conditioning drops when it's hot. The reason being that hot air is less dense. The answer resulted in lots of scepticism judging by the downvotes. Is this the case? Does hot air really equate to smaller resisting forces? What is the physical basis of this, and how can it be shown with laws of physics? (I'm too lazy to write an answer with details so I posted this question, hoping someone could write a detailed, clear, easy-to-understand answer.) How much does very hot air reduce resisting forces (and thus energy consumption) when compared with freezing temperatures? Let's assume the car has no air conditioning or that the air conditioning is turned off. Let's also ignore pressure of air in the tires, i.e. the driver adjusts the tire pressure to be always correct regardless of the temperature. | This is not theoretical question about physics as much as biomechanical question about technique of walking. When I run during winter, I always fell forward (assuming there is no slope). That is because I am putting my legs on the ground with backward motion to use the force to accelerate. So when they slip, they go backward, while my body center goes still forward. A lot of people during walking are putting their legs on the ground with forward motion, so when they slip, the legs go forward faster than their body center and they fall on their backs. This is mostly because nowadays we learned to walk through our heels instead of toes as a consequence of wearing shoes. | {
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520,161 | It is possible to conserve momentum without conserving kinetic energy, as in inelastic
collisions. Is it possible to conserve the total kinetic energy of a system, but not its momentum? How? To clarify, I am not necessarily talking about an isolated system. Is there any scenario which we could devise in which momentum is not conserved but kinetic energy is? | In order for momentum to be conserved, it must be the case that $$\mathbf F_\text{net}=\frac{\text d\mathbf p}{\text dt}=0$$ In order for kinetic energy to be conserved, it must be the case that $$\text dK=\text dW_\text{net}=\mathbf F_\text{net}\cdot\text d\mathbf x=0$$ at all instants in time. So, is there a case where the net work done on an object is $0$ , yet there is still a net force acting on the object? The answer is yes! We just need $\mathbf F_\text{net}\neq0$ to be perpendicular to the path of the object at all times. A simple example of this is an object undergoing uniform circular motion. The object's kinetic energy is not changing (as its speed remains constant), yet the momentum is constantly changing due to the non-zero net force. | {
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520,299 | As far as I understand a photon is produced, or "born", whenever an electron moves from a high energy state back to its normal energy state. The photon then travels at the speed of light across space in a straight line until it hits another atom, or rather, interacts with the electron shell(s) of that atom. The energy signature of the photon can change at this point. The photon can then bounce off that atom, and will continue to travel across space at the speed of light until it hits another atom. And so on. (Please correct me if any of my understanding here is off.) However what I want to know is what happens when light stops, and how this relates to the photon. I want to know what happens when a photon "dies" - not in a literal sense, just in the sense of when it has finished its journey of bouncing from atom to atom. If you stand in a huge and pitch-black cavern, and shine a torch, the light will only carry so far. Am I right in assuming that the photons produced by the torch eventually stop bouncing from atom to atom, or does the journey of the photon continue and its just undetectable to human eyes? Similarly, the colour black "absorbs light" - does this mean the colour black is "eating" photons? Does the energy get transferred to the electrons of the black material? What happens to this energy? And finally, does the same "photon death" happen when a photon hits the retina in a persons eye? In short, what happens when a photon dies? | This question is about the nature of the electromagnetic field. The electromagnetic field is a physical system that is most fully described by quantum field theory, and the results match those of classical field theory in certain limiting cases. The 'photon' is a physical picture which gives us a useful way to imagine certain aspects of this field. It is primarily a way to track energy movements. The main thing you need to know is that energy is conserved, but photons are not. When energy moves from some other form to an electromagnetic form, then photons are created. When energy moves from an electromagnetic form to other forms, then photons are destroyed. Another way of saying the same thing is to note that when an electron moves from a higher to a lower energy level in an atom, it does so through the way its charge pushes on the surrounding electromagnetic field, causing it to vibrate at a higher amplitude (the electric and magnetic parts both start to vibrate). This vibration, when it happens at a fixed frequency, can be conveniently modelled by saying it has a fixed amount of energy, equal to $h f$ where $h$ is Planck's constant and $f$ is the frequency. If this $h f$ is equal to the energy change $\Delta E$ in the atom, then we say one photon has been created. You can also find cases where two photons are produced, one at frequency $f_1$ and the other at $f_2$ , and then $h f_1 + h f_2 = \Delta E$ . This kind of process is much rarer but it illustrates that energy is conserved, but a given amount of energy can be expressed physically in more than one way. Eventually a photon may arrive at some other atom and be absorbed. What happens then is that the oscillating electromagnetic field pushes on the electrons inside the atom, until one of them gains some more energy. The field vibration then falls away as the energy is transferred. We summarise the process by saying that the photon has been absorbed. Or, if you like, the photon 'dies'. This is just another way to say that the field has stopped vibrating. | {
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520,935 | Reading the book Schaum's Outline of Engineering Mechanics: Statics I came across something that makes no sense to me considering the subject of significant figures: I have searched and saw that practically the same thing is said in another book ( Fluid Mechanics DeMYSTiFied ): So, my question is: Why if the leading digit in an answer is 1, it does not count as a significant figure? | Significant figures are a shorthand to express how precisely you know a number. For example, if a number has two significant figures, then you know its value to roughly $1\%$ . I say roughly, because it depends on the number. For example, if you report $$L = 89 \, \text{cm}$$ then this implies roughly that you know it's between $88.5$ and $89.5$ cm. That is, you know its value to one part in $89$ , which is roughly to $1\%$ . However, this gets less accurate the smaller the leading digit is. For example, for $$L = 34 \, \text{cm}$$ you only know it to one part in $34$ , which is about $3\%$ . And in the extreme case $$L = 11 \, \text{cm}$$ you only know it to one part in $11$ , which is about $10\%$ ! So if the leading digit is a $1$ , the relative uncertainty of your quantity is actually a lot higher than naively counting the significant figures would suggest. In fact, it's about the same as you would expect if you had one fewer significant figure. For that reason, $11$ has "one" significant figure. Yes, this rule is arbitrary, and it doesn't fully solve the problem. (Now instead of having a sharp cutoff between $L = 9$ cm and $L = 10$ cm, you have a sharp cutoff between $L = 19$ cm and $L = 20$ cm.) But significant figures are a bookkeeping tool, not something that really "exists". They're defined just so that they're useful for quick estimates. In physics, at least, when we start quibbling over this level of detail, we just abandon significant figures entirely and do proper error analysis from the start. | {
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521,224 | I think we could put satellite to orbit earth in such a way that it always see the sun. Which is orbiting along the path of earth orbiting the sun, like a wheel perpendicular to the sun I don't know the specific name of this type of orbit so I can't find detail about it, are there any satellite orbit earth this way? | There are satellites that constantly observe the sun. Among them are the STEREO pair of satellites . These satellites don't orbit Earth, but orbit the Sun–one a little faster than Earth, one a little slower. At first, I thought the Earth-centered orbit you suggest wouldn't work. I was thinking that if a satellite starts out in a polar orbit, the plane of the orbit would not rotate as the Earth orbited the Sun, leaving the orbit parallel to the sun a quarter year later and behind the Earth at times. However, it is possible. From Wikipedia on polar orbits : To retain the Sun-synchronous orbit as Earth revolves around the Sun during the year, the orbit of the satellite must precess at the same rate, which is not possible if the satellite were to pass directly over the pole. Because of Earth's equatorial bulge, an orbit inclined at a slight angle is subject to a torque, which causes precession. An angle of about 8° from the pole produces the desired precession in a 100-minute orbit. Another source cited by the wiki article . These orbits are generally used for Earth-observing satellites that track Earth conditions (like atmospheric temperatures) at the same times of day. Also, from a comment by cmaster - reinstate monica: Those sun-synchronous orbits are also used for satellites that don't care about day and night on earth, but need an unceasing supply of energy. Examples are satellites that use radar to image the earth's surface. Radar is an active technology that needs lots of power to work over 800-km distances, and that power is supplied by the sun. The sun-synchronous orbits puts the satellite into eternal sunlight, allowing it to work 24/7 with a minimal battery and solar cell array size. | {
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521,400 | Suppose 2 objects are placed not too far away. These objects start moving towards each other due to gravitational pull but there is no application of external force. So why do These objects move? Where does this energy come from? | The answer to the question is in the question itself. You wrote: Suppose 2 objects are placed not too far away from themselves. These objects start moving towards each other due to gravity. Then asked: Where does the energy come from? The energy comes from whoever placed the two objects apart. They had to do work to get them into that position. You could just as well ask "Suppose I roll an object up a hill. Then I let go and the object rolls back down. Where did the energy come from?" It came from you! | {
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521,414 | I am learning electricity and quite often see loops that are connected in one node. This is a simplified example: But I found many cases, with more elements. The thing that makes me question their usefulness is that (so far) they have always behaved as independent circuits. So, I wonder if circuits are ever designed this way in real life and if this concept (two loops connected at one node) has its own name or is used at all with any purpose. I also wonder what would happen if one of the loops does not have any voltage source in it, like this: Would current behave as shown in A or in B or a mix of both and why? Teacher says: "current is smart and always takes the laziest road" (in this case A), but I don't see how that can be a solid reasoning. Maybe with a clear why . | The answer to the question is in the question itself. You wrote: Suppose 2 objects are placed not too far away from themselves. These objects start moving towards each other due to gravity. Then asked: Where does the energy come from? The energy comes from whoever placed the two objects apart. They had to do work to get them into that position. You could just as well ask "Suppose I roll an object up a hill. Then I let go and the object rolls back down. Where did the energy come from?" It came from you! | {
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521,433 | We define electric potential and gravitational potential and use them quite often to solve problems and explain stuff. But I have never encountered magnetic potential , neither during my study (I am a high-schooler) , nor during any discussion on physics. So, does magnetic potential even exist? According to me, it should because a magnetic field is a conservative one and so we can associate a potential with it? Also if it's defined, then why don't we encounter it as often as we do the others (electric potential, gravitational potential, etc.)? I have encountered magnetic potential energy only in the cases where a dipole is subjected to a magnetic field. Is magnetic potential limited only to this scenario, or is there a general expression for the magnetic potential? I should have said this earlier, but don't restrain your answers due to my scope. You can use vector calculus as I am quite familiar with it. Also, this question is meant for everyone, so even the answers which are out of my scope are appreciated. | Actually, we do! It's just that it's not the same " kind " of potential - and the reason for this is that magnetic forces work differently than electric forces. Magnetic fields, if you know, do not directly exert a force on charged particles, simply for being charged. (They would exert such on hypothetical "magnetically charged particles", but we've never found any to exist.) Rather, the force they exert does one thing only: to change the direction of motion of moving charged particles. The usual thing we call "potential" is what you can think of as a kind of "specific potential energy": it is the potential energy that a unit quantity of charge has from sitting at a certain place in an electric field, and if the particle moves between two areas of different potential, it gains or loses energy as a result of the shifting - but ever-present - pull of the electric force upon it. Magnetic fields, though, cause no changes in energy - changing something's direction of motion takes no energy, only speeding or slowing its motion does. Think about how that a bullet fired from a gun doesn't hurt more or less depending on the direction from which it comes, only from how powerful the gun is. In theory, to deflect the bullet from one to the other direction in flight, likewise, would not take any energy (though you would need a rather strong source of deflecting force). But nonetheless, that doesn't mean you can't still describe them using something like a potential, but it doesn't have quite the same meaning any more. As I just see you mentioned you've tried some vector calculus, I'll give this a shot. You see, there's a kind of "duality", if one will, between two operations one can do with at least three-dimensional vectors: the dot product and the cross product , which gives rise to related differential notions of the divergence and gradient , vs. curl , respectively. The usual ideal of a "potential", that is, for a field like the electric field (and also the Newtonian gravitational field), is based on the following result. "Under certain reasonable conditions", the following implication holds true. If $\mathbf{F}$ is some sort of force field, and $$\oint_C \mathbf{F} \cdot d\mathbf{l} = 0$$ for all closed paths $C$ , then there exists a scalar function $V$ (i.e. with 3 real spatial coordinate arguments and outputting one real number) such that $$\mathbf{F} = -\nabla V$$ Intuitively, the first equation is a kind of "conservation of energy": the left hand integral is, in effect, a work integral if $\mathbf{F}$ is serving as a force field, describing the amount of energy gained or lost (positive is gain, negative is loss) by a particle moving in a closed circuit through that field as it is pushed and pulled by the exerted force. The above implication, then, says that "if the force field conserves energy, we can describe it by a potential energy". This is how you get the usual electric potential, which is the "specific" potential energy: energy per unit of charge, which in SI units comes to be joules per coulomb, which we call as "volts". Moreover, the first equation, "under certain reasonable conditions", corresponds to the one $$\mathbf{\nabla} \times \mathbf{F} = \mathbf{0}$$ where the left-hand side is a differential operation called "curl", and intuitively represents the amount by which a vector field, thought of as force, locally *fails* to conserve energy . Now, it turns out that, however, there is another , analogous form but involving this integral: if $$\mathop{\vcenter{\huge\unicode{x222F}}}_S \mathbf{F} \cdot d\mathbf{S} = 0,$$ a surface integral over a closed surface $S$ , then it follows that another , vector field $\mathbf{A}$ exists such that $$\nabla \times \mathbf{A} = \mathbf{F}$$ which is very much like the relationship to potential and, indeed, we call this $\mathbf{A}$ a vector potential . Again, we should think about the intuitive meaning of the first integral: this integral now is a flux integral - in effect, if you imagine the field as representing the stream lines of some fluid, i.e. if the vectors returned by $\mathbf{F}$ are mass-flow, i.e. mass per time, with direction of flow, vectors, the flux integral would represent the net amount of fluid flowing into or out of that space - and to set it to zero says that, in effect the field "conserves fluid": no new fluid is destroyed or created at any point. Analogously, this corresponds to a similar "local" statement per the divergence : $$\nabla \cdot \mathbf{F} = 0$$ which, you may recognize, is exactly the equation satisfied by the magnetic field , $\mathbf{B}$ : $$\nabla \cdot \mathbf{B} = 0$$ and says there are "no magnetic sources", i.e. no magnetic charges. In a sense, magnetic "flux", which can be thought of as a kind of "fluid", swirls around magnetic objects but none is created or destroyed, and this conservation of flux gives rise to a magnetic vector potential , also typically denoted $\mathbf{A}$ . This "potential" is a vector , not scalar, quantity - and this is the answer to your question. It doesn't represent energy , but more like "specific flux", I'd suppose, though it's hard to answer and, moreover, interestingly, is much less unique. | {
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521,447 | The following text is from Concepts of Physics by Dr. H.C.Verma, from chapter "Geometrical Optics", page 387, topic "Relation between $u$ , $v$ and $R$ for Spherical Mirrors": If the point $A$ is close to $P$ , the angles $\alpha$ , $\beta$ and $\gamma$ are small and we can write $$\alpha\approx\frac{AP}{PO},\ \beta=\frac{AP}{PC}\ \ \ \text{and} \ \ \gamma \approx\frac{AP}{PI}$$ As $C$ is the centre of curvature, the equation for $\beta$ is exact whereas the remaining two are approximate. The terms on the R.H.S. of the equations for the angles $\alpha$ , $\beta$ and $\gamma$ , are the tangents of the respective angles. We know that, when the angle $\theta$ is small, then $\tan\theta\approx\theta$ . In the above case, this can be imagined as, when the angle becomes smaller, $AP$ becomes more and more perpendicular to the principal axis. And thus the formula for the tangent could be used. But, how can this approximation result in a better accuracy for $\beta$ when compared to $\alpha$ and $\gamma$ ? I don't understand the reasoning behind the statement: " As $C$ is the centre of curvature, the equation for $\beta$ is exact whereas the remaining two are approximate. " I can see the author has used " $=$ " instead of " $\approx$ " for $\beta$ and he supports this with that statement. But why is this so? Shouldn't the expression for $\beta$ be also an approximation over equality? Is the equation and the following statement really correct? | Actually, we do! It's just that it's not the same " kind " of potential - and the reason for this is that magnetic forces work differently than electric forces. Magnetic fields, if you know, do not directly exert a force on charged particles, simply for being charged. (They would exert such on hypothetical "magnetically charged particles", but we've never found any to exist.) Rather, the force they exert does one thing only: to change the direction of motion of moving charged particles. The usual thing we call "potential" is what you can think of as a kind of "specific potential energy": it is the potential energy that a unit quantity of charge has from sitting at a certain place in an electric field, and if the particle moves between two areas of different potential, it gains or loses energy as a result of the shifting - but ever-present - pull of the electric force upon it. Magnetic fields, though, cause no changes in energy - changing something's direction of motion takes no energy, only speeding or slowing its motion does. Think about how that a bullet fired from a gun doesn't hurt more or less depending on the direction from which it comes, only from how powerful the gun is. In theory, to deflect the bullet from one to the other direction in flight, likewise, would not take any energy (though you would need a rather strong source of deflecting force). But nonetheless, that doesn't mean you can't still describe them using something like a potential, but it doesn't have quite the same meaning any more. As I just see you mentioned you've tried some vector calculus, I'll give this a shot. You see, there's a kind of "duality", if one will, between two operations one can do with at least three-dimensional vectors: the dot product and the cross product , which gives rise to related differential notions of the divergence and gradient , vs. curl , respectively. The usual ideal of a "potential", that is, for a field like the electric field (and also the Newtonian gravitational field), is based on the following result. "Under certain reasonable conditions", the following implication holds true. If $\mathbf{F}$ is some sort of force field, and $$\oint_C \mathbf{F} \cdot d\mathbf{l} = 0$$ for all closed paths $C$ , then there exists a scalar function $V$ (i.e. with 3 real spatial coordinate arguments and outputting one real number) such that $$\mathbf{F} = -\nabla V$$ Intuitively, the first equation is a kind of "conservation of energy": the left hand integral is, in effect, a work integral if $\mathbf{F}$ is serving as a force field, describing the amount of energy gained or lost (positive is gain, negative is loss) by a particle moving in a closed circuit through that field as it is pushed and pulled by the exerted force. The above implication, then, says that "if the force field conserves energy, we can describe it by a potential energy". This is how you get the usual electric potential, which is the "specific" potential energy: energy per unit of charge, which in SI units comes to be joules per coulomb, which we call as "volts". Moreover, the first equation, "under certain reasonable conditions", corresponds to the one $$\mathbf{\nabla} \times \mathbf{F} = \mathbf{0}$$ where the left-hand side is a differential operation called "curl", and intuitively represents the amount by which a vector field, thought of as force, locally *fails* to conserve energy . Now, it turns out that, however, there is another , analogous form but involving this integral: if $$\mathop{\vcenter{\huge\unicode{x222F}}}_S \mathbf{F} \cdot d\mathbf{S} = 0,$$ a surface integral over a closed surface $S$ , then it follows that another , vector field $\mathbf{A}$ exists such that $$\nabla \times \mathbf{A} = \mathbf{F}$$ which is very much like the relationship to potential and, indeed, we call this $\mathbf{A}$ a vector potential . Again, we should think about the intuitive meaning of the first integral: this integral now is a flux integral - in effect, if you imagine the field as representing the stream lines of some fluid, i.e. if the vectors returned by $\mathbf{F}$ are mass-flow, i.e. mass per time, with direction of flow, vectors, the flux integral would represent the net amount of fluid flowing into or out of that space - and to set it to zero says that, in effect the field "conserves fluid": no new fluid is destroyed or created at any point. Analogously, this corresponds to a similar "local" statement per the divergence : $$\nabla \cdot \mathbf{F} = 0$$ which, you may recognize, is exactly the equation satisfied by the magnetic field , $\mathbf{B}$ : $$\nabla \cdot \mathbf{B} = 0$$ and says there are "no magnetic sources", i.e. no magnetic charges. In a sense, magnetic "flux", which can be thought of as a kind of "fluid", swirls around magnetic objects but none is created or destroyed, and this conservation of flux gives rise to a magnetic vector potential , also typically denoted $\mathbf{A}$ . This "potential" is a vector , not scalar, quantity - and this is the answer to your question. It doesn't represent energy , but more like "specific flux", I'd suppose, though it's hard to answer and, moreover, interestingly, is much less unique. | {
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521,453 | For example, if I add 2.50 five times, I get 12.50, keeping the number of decimals. However, if I multiply 2.50 by 5, I get 12.5. Why is this? Which rule should take precedence, and why? | The first thing to realize is that sig fig rules are just a rule of thumb. They're a quick and dirty way of doing error propagation, but if you want a totally correct error propagation, do real error propagation. In your example, however, the rules can be interpreted in a way that might make sense, depending on the context. In a computation like $2.50+2.50+2.50+2.50+2.50$ , the context could be that each 2.50 is an independent measurement, and could have its own positive or negative error. Suppose each one has error bars of about $\pm 0.01$ . Adding 5 of these errors is likely to give something fairly small, since there will be some cancellation. (If they're independent and identically distributed, then they make error bars of $0.01\times\sqrt{5}$ .) But in the calculation of $2.50\times5$ , suppose the context is that the 5 is an exact value, but the $2.50$ has error bars of $\pm 0.01$ . You only get one chance to measure it. If it's off by $0.01$ , then the result is off by $0.05$ , which is enough to make the digit in the hundredths' place pretty meaningless. | {
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521,524 | Let's say that I have a neighbor who has a big dog that barks at 80 dB. between him and me, there is a big wall, vegetations, etc. The sounds are absorbed, there is isolation so from my place what I will hear is a bark of only 60 dB. Now, I also have a dog, exactly the same and it barks at 80 dB too. If what I hear is 60 dB, does my neighbor hear 60 dB too from my dog? No matter what separates us? No matter the shape of the wall or its material ? is it always " symmetric "? I guess the answer is 'yes' but I'm wondering if it's so trivial PS: If you have a better idea to the title of that question, feel free to edit EDIT Since it's more a thought experiment, you should consider that the two dogs are twins | It is not always "symmetric" (technically, this property is called reciprocity )! However, assuming the medium between the two dogs is linear and microscopically reversible, your intuition is correct. Under these assumptions, it is actually a theorem, due to Lord Rayleigh, that if you exchange the source of the sound and the point at which you measure it, the sound level is unchanged. (“ Some general theorems relating to vibrations,” Proc. London Math. Soc. 4, 357–368. https://doi.org/10.1112/plms/s1-4.1.357 ). In electromagnetism, this is known as Lorentz reciprocity . Now if we break one of these assumptions, we can avoid the consequences. For instance, imagine there is a very strong wind going westward, going faster than the speed of sound. Then, barks emitted by the west dog will never reach the east dog, while the opposite is possible. Such a system is called non-reciprocal . This is an extreme, but straightforward example. You can get around reciprocity in subtler ways. Read https://science.sciencemag.org/content/343/6170/516 for a recent reference on the subject. | {
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521,538 | I was reading some notes from a second-semester quantum mechanics course and had trouble understanding why position states of a particle moving along the real line $\mathbb{R}$ are linearly independent. Here is an excerpt from the notes: Roughly, $\lvert x \rangle$ represents the state of the system where the particle is at the position $x$ . The full state
space requires position states $\lvert x \rangle$ $\forall x\in\mathbb{R}$ . Physically, we consider all of these
states to be linearly independent: the state of a particle at some point $x_0$ can’t be build by
superposition of states where the particle is elsewhere. Can we prove that any two position states $\lvert x \rangle$ and $\lvert x' \rangle$ with $x\neq x'$ are linearly independent or is there an intuitive way to see why this is true? | It is not always "symmetric" (technically, this property is called reciprocity )! However, assuming the medium between the two dogs is linear and microscopically reversible, your intuition is correct. Under these assumptions, it is actually a theorem, due to Lord Rayleigh, that if you exchange the source of the sound and the point at which you measure it, the sound level is unchanged. (“ Some general theorems relating to vibrations,” Proc. London Math. Soc. 4, 357–368. https://doi.org/10.1112/plms/s1-4.1.357 ). In electromagnetism, this is known as Lorentz reciprocity . Now if we break one of these assumptions, we can avoid the consequences. For instance, imagine there is a very strong wind going westward, going faster than the speed of sound. Then, barks emitted by the west dog will never reach the east dog, while the opposite is possible. Such a system is called non-reciprocal . This is an extreme, but straightforward example. You can get around reciprocity in subtler ways. Read https://science.sciencemag.org/content/343/6170/516 for a recent reference on the subject. | {
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521,843 | All the graphs shown below come from completely different fields of studies and still, they share a similar distribution pattern. Why most distribution curves Bell Shaped? Is there any physical law that leads the curve to take that shape? Is there any explanation in Quantum Mechanics for these various graphs to take that shape? Is there any intuitive explanation behind why these graphs are Bell Shaped? Following is Maxwell’s Distribution of Velocity Curve, in Kinetic Theory of Gases. Following is the Wein’s Displacement Law, in Thermal Radiations. Following is the Distribution of Kinetic Energy of Beta Particles in Radioactive Decays. | First, distributions are not always bell-shaped. A very important set of distributions decrease from a maximum at $x=0$ , such as the exponential distribution (delay times until a random event such as a radioactive decay) or power-laws (size distributions of randomly fragmenting objects, earthquakes, ore grade, and many other things). Stable distributions Still, there is a suspicious similarity between many distributions. These come about because of statistical laws that make them "attractors": various very different random processes go on, but their results tend to combine to form similar distributions. As Bob mentioned, the central limit theorem makes addition of independent random factors (of finite variance!) approach a Gaussian distribution (since it is so common it is called the normal distribution). Strictly speaking, there are a few other possibilities . If random factors are instead multiplied, the result is the log-normal distribution . If we take the maximum of some random things, the distribution will approach a Weibull distribution (or, a few others ). Basically, many repeated or complex processes tend to produce the same distributions over and over again, and many of those look like bell-shapes. Maximum entropy distributions Why is that? The deep answer is entropy maximization . These stable distributions tend to maximize the entropy of the random values they produce, subject to some constraint. If you have something positive and with a specified mean, you get the exponential distribution. If it is positive but there is no preferred scale, you get a power-law. Specified mean and variance: Gaussian. Maximal entropy in phase space for given mean energy: Maxwell-Boltzmann . Statistical mechanics This is where we get back to physics. A lot of physical processes obey statistical mechanics, which runs by the equal a priori probability postulate: For an isolated system with an exactly known energy and exactly known composition, the system can be found with equal probability in any microstate consistent with that knowledge. If we know the energy and number of particles exactly each allowed microstate is equally likely (maximizes entropy), but anything macroscopic we calculate or measure will be a function of these random microstates - so its distribution will be bunched up if there are a lot of microstates that can generate that macrostate. If it has fixed particles but we only know the average energy , each state has probability $(1/Z)e^{-E/k_B T}$ where $E$ is their energy, $Z$ is a normalizing constant and $T$ the temperature: this distribution, the Boltzmann distribution, maximizes entropy with the constraint that the average energy is fixed. Similar distributions work when the number of particles can change . Quantum mechanics Finally, this links to quantum mechanics: QM describes the set of possible microstates, and from that plus statistical mechanics one can calculate the statistical distributions of macroscopic things like emitted photons of different wavelengths, gas molecule speeds, or kinetic energy distributions. The number of states available affect what curves we get, and the constraints of the experiment fix parameters like energy or temperature, but since nature is entropy-maximizing we get the entropy-maximizing distributions that fit these inputs. They are often loosely bell-shaped since there are more states available for high energies (the curve grows from low values at low energy) but the system cannot put all particles into high energy states while keeping the (average) energy constant (the curve has to decline beyond a certain point). But this is the average of a myriad micro-events that all have more complex or discrete distributions. | {
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522,578 | When we look into the beautiful sky in the night, exclaiming how beautiful these shining stars are. My question is how could we tell, whether any of these shining "point" is a star or a galaxy? If indeed many of these are shining galaxies, then what roughly the percentage is of these shining galaxies in the sky (using the naked eye only without telescope)? (And why.) | With the naked eye, virtually every point you see is a star.* That's because there are very few galaxies that are visible with the naked eye. With telescopes, for many galaxies you'll be able to resolve multiple light sources (aka. stars). That lets you tell it's a galaxy. For those galaxies that are further away, you can still tell by taking a spectrum . Galaxies have fundamentally different spectra from stars, because they're composed of lots of different stars at different metallicities, temperature, etc (+ other stuff). The percentage of stars/galaxies you see depends on what you're using to observe. In the Hubble Deep Field for example, virtually every point is a galaxy. Comparatively, if you're using an ordinary pair of binoculars, virtually every point is a star. *Some of the brightest points will not be stars, but planets. | {
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523,021 | Like many others, I have been following the sad development of the bush/forest fires in Australia recently. A claim that gets repeated is that one of the contributors to this blaze is the ongoing >45°C heatwave. Now I am not really putting this in question here. This is probably a well established fact, but my immediate intuition of physics fails to grasp why. Here's my reasoning: Obviously the vegetation has to be dry for this to happen, but surely - and in my experience - a +30°C temperature is more than enough to make sure everything is bone dry. 45°C is not significantly closer than the aforementioned 30°C to the ignition temperature, which is several hundreds degrees. I am aware that some vegetation in Australia (eucalyptus probably) is remarkably easy to light, but still air temperatures are pretty far from that. Heat conduction flux is also directly proportional to the difference in temperature (first order dependence) and not for example the square of it, so the few extra degrees shouldn't make grass that much easier to ignite. So why does the heatwave matter? | This is more a question of chemistry and biology than physics. Solid objects don't burn (try dropping a lit match on a piece of structural lumber sometime -- it'll just go out). Instead, they release or decompose into flammable gasses on heating, and it's those gasses that burn. Some plants (eg. pines) produce volatile resins that provide an easy ignition source, and the amount and speed of vaporization increases with temperature. Eucalyptus is especially bad for this: eucalyptus oil is extremely volatile, extremely flammable, and produced in copious quantities. In hot weather, there's far more vaporized oil in the air than in more moderate temperatures, leading to easier fire starts and faster spreads. Eucalyptus has evolved not just to survive fires, but to encourage them, as a way to clear out competing plant species. | {
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523,349 | Is there any final model of an atom, of which we can say, “This is it”! Or is it still improving and physicists are not completely sure about it? I am particularly interested to know how exactly electrons move inside an atom. Do the physicists have any actual image or video of electron orbitals? | Electrons do not move inside atoms. If an electron is in a given energy level $E$ , the wavefunction is given by $\psi(x,y) = \phi(x)_{n\ell m} \,\mathrm{e}^{-\mathrm{i}E t/\hbar}$ . The time dependence is a pure phase factor, hence the real-space probability density of the electron is $|\psi(x)|^2 = |\phi(x)|^2 \neq f(t)$ , not a function of time. These are called stationary states , for this reason. The fact that electrons do not actually move in atoms is good , and it's the whole point quantum mechanics was invented. If they were to move, they would be accelerating charged particles and would thereby lose energy to radiation (bremsstrahlung) eventually collapsing. The instability of the atom was exactly the shortcoming of classical physics that led to the invention/discovery of quantum mechanics. Furthermore: Atomic orbitals are only "correct" $^\dagger$ wave functions in one-electron systems such as the hydrogen atom. In many-electron atoms, orbitals are a useful approximation, usually a basis used for perturbative calculations. For instance, for Helium you already have to take into account the indistinguishability of the two electrons, which leads to the linear combinations of the orbitals to work out correction terms. In the Hydrogen atom, the orbitals have been indirectly observed, see Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States , by recording the diffraction pattern of light radiating away from atomic transitions: these patterns related to the nodal structure of the atomic wavefunctions. Angular-resolved photoemission spectroscopy (APRES) can give information on the shape of molecular orbitals, see Exploring three-dimensional orbital imagingwith energy-dependent photoemission tomography . $\dagger$ : but only within the pure Coulomb Hamiltonian. With corrections such as fine structure, Lamb shift etc., there is no analytical solution for both eigenvalues and eigenstates. EDIT from comments . Given the attention this answer has got, let me add a few points raised in the long discussion that ensued in the comments. First and foremost, the above answer reflects my opinion and my interpretation of the matter. Indeed, as @my2cts points out: Whether electrons move or not is pure interpretation. What QM does
unequivocally say is that electrons have kinetic and potential energy.
Anyone is free to interpret this. Then, regarding motion , it is true that electrons possess momentum, kinetic energy, and, for $\ell \neq s$ , a probability current $\mathbf{J}$ that is however also stationary but in the tangential direction $\hat{\boldsymbol{\phi}}$ (derivation here ) like the velocity of a classically orbiting object. Particularly, @dmckee says: the electrons have a well defined energy which has to be interpreted
as including a kinetic component and a momentum distribution which may
include zero but also includes non-zero value with non-trivial
probability density. My idea of "electrons do not move" stems from the idea that "standing waves do not move", in that they don't go from A to B. But of course there is motion nonetheless. See nice discussion here . | {
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523,868 | I asked this question on Physics Stack Exchange - "Can Increment / maximization of Entropy be the “Cause” behind any phenomena?" To which I received a beautiful explanation with a bewildering example. I quote the example below. In the example, it seems the Second Law of Thermodynamics is more powerful than Newton's Laws and if there is a conflict between the two, then the second law wins! And this is where I have the question - Can Second Law of Thermodynamics / Entropy override Newton's Laws? Following are some specific questions: Can the Second Law of Thermodynamics / Entropy override Newton's Laws? From the example below, it seems that there is an underlying "Force" behind the second law of thermodynamics which drives it and which is more powerful than any other law and if there is a conflict between the second law of thermodynamics and any other law then second law will win. Is there really an underlying "force" behind the second law of thermodynamics, which drives the system in a particular direction? Following is the example: I remember witnessing a demonstration that the above abstract example
is a close analogy to. The demonstration involved two beakers, stacked, the openings facing
each other, initially a sheet of thin cardboard separated the two. In the bottom beaker a quantity of Nitrogen dioxide gas had been had
been added. The brown color of the gas was clearly visible. The top
beaker was filled with plain air. Nitrogen dioxide is denser than air. When the separator was removed we saw the brown color of the Nitrogen
dioxide rise to the top. In less than half a minute the combined space
was an even brown color. And then the teacher explained the significance: in the process of
filling the entire space the heavier Nitrogen dioxide molecules had
displaced lighter molecules. That is: a significant part of the
population of Nitrogen dioxide had moved against the pull of gravity.
This move against gravity is probability driven. Statistical mechanics provides the means to treat this process
quantitively. You quantify by counting numbers of states. Mixed states
outnumber separated states - by far. The climbing of the Nitrogen dioxide molecules goes at the expense of
the temperature of the combined gases. That is, if you make sure that
in the initial state the temperature in the two compartments is the
same then you can compare the final temperature with that. The final
temperature of the combined cases will be a bit lower than the
starting temperature. That is, some kinetic energy has been converted
to gravitational potential energy. I think the above example counts as a case of probability acting as a
causal agent. | Can the Second Law of Thermodynamics / Entropy override Newton's Laws? No. In the example given, every particle obeys Newton's laws. There is no particle that is not obeying $F=ma$ . From the example below, it seems that there is an underlying "Force" behind the second law of thermodynamics which drives it and which is more powerful than any other law and if there is a conflict between the second law of thermodynamics and any other law then second law will win. Is there really an underlying "force" behind the second law of thermodynamics, which drives the system in a particular direction? Again, no. The second law describes how Newton's laws combine with the laws of probability. There is no other "force" acting in the problem. Why is this demonstration not a violation of Newtonian mechanics? Well, consider this question: Does Newtonian mechanics say that heavy particles cannot move upwards? What is happening in the beaker is that the many molecules are jostling around. The heavier molecules, on average, will be lower than the lighter molecules, since they will have the same amount of total energy, on average, but are heavier. But there are many, many molecules. Sometimes, the $NO_2$ molecules will have more kinetic energy, and bounce upwards. The Second Law simply expresses that this random motion will result in a mixture of the gases. It does not contradict Newton's laws, but expresses the inevitable consequences of them. | {
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524,201 | In high school physics, I was taught that mass was just how much "stuff" or matter there is in an object. However, now that I am learning physics again in college, I am taught that mass of an object (inertial mass) is how resistant an object is to acceleration. Which one's the correct definition of what mass is? | This is a deep question. There are (at least) two definitions of mass: gravitational mass is how something is influenced by gravity, which is the $m$ in $F = Gm_1m_2/r^2$ , and is more-or-less 'how much stuff there is'; inertial mass is how resistant to acceleration something is, and it's the $m$ in $F = ma$ . If we call these two versions of mass $m_G$ and $m_I$ , then we can conduct experiments which ask whether they are the same (or to be more precise, whether there is a constant ratio between them, which ratio can be absorbed into $G$ ). It's easy to see how to set up such experiments in principle. Given the two equations above we can equate the forces to get $$m_{I,1} a = \frac{Gm_{G,1}m_{G,2}}{r^2}$$ or $$\frac{m_{I,1}}{m_{G,1}} = \frac{Gm_{G,2}}{a r^2}$$ Well we can measure all the quantities on the right hand side of this, and we expect the left-hand-side always to be $1$ if the two definitions of mass are equivalent. Even if we can't measure $G$ or $m_{G,2}$ very well, we can repeat the experiment with lots of objects on the left-hand-side and we should always get the same answer. The weak equivalence principle (WEP) says that they are the same, and experiment has so far borne this out. There are various stronger equivalence principles which matter in General Relativity in addition. I won't go into them here as I am always confused about exactly which is which. However it's fairly easy to see that if we want a theory of gravity which states that gravity is about the geometry of spacetime, then we really must have only one definition of mass, and so we need to claim rather strongly that all definitions of mass are equivalent. | {
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524,510 | Two black holes can collide and merge into one bigger black hole, but not split into two. Does this mean colliding black holes violate time reversal symmetry? Related: Do black holes violate T-symmetry? Based on the answer to that question, time-reversing a black hole yields a white hole. However, that seems to imply that white holes are very unstable because they can spontaneously split into two, which would then split into four, ad infinitum, and the universe would be covered with tiny white holes all over. | You missed something: the gravitational waves. A black hole merger spacetime contains gravitational waves leaving the merger at the speed of light. Time reversal reverses time across the entire spacetime, and this converts those escaping gravitational waves into a converging gravitational wave front, as well as the black holes into white holes. These waves converge in on the central (now-)white hole and get so strong at that central point of convergence as to be able to "buck" it apart into two separate white holes. Without those incoming gravitational waves, such a split would not occur. EDIT: As A.V.S. points out in the comments, in fact, a better answer to this question would be that the future evolution of white holes is in general undetermined , or better unrestricted, in the sense that multiple future trajectories from identical phase-space points will satisfy the dynamical equations, though of course that means still that we must highlight that a crucial element of the answer here is that the time reversal turns the black hole into a white hole. (Indeed, this is part of why they're called "white" - technically that's understating it: they can literally spit out anything - even unicorns, no seriously, it'd be entirely [though unlikely] consistent with the equations for a 1-horned ungulate to pop out, as much as literally anything else.) In a realistic black hole collision case, which is what I assumed in the answer above, then of course, yes, you will have the gravitational waves and so forth and you do have to take them into account in the reversal. But the situation is even more serious. Since the future evolution of white hole is unrestricted, you can build scenarios with a totally causeless, spontaneous split of the white hole, and have it be consistent with the dynamical equations. As it is a consistent evolution, it doesn't violate time reversal symmetry. The reason that the Universe isn't covered with tiny white holes is that they are next to impossible to form in the first place - and likely, general relativity is not the final description of these things. (I want to point out that there is actually an analogy for this within ordinary Newtonian mechanics called " Norton's dome ". It is not physically achievable, but is still a system within the mathematical theory which has a similar property of its present state being equiconsistent with multiple future evolution trajectories.) | {
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524,838 | Some Background I became fascinated with how the overtone series, or harmonics, relates to how brass instruments function. Most trumpet players (or brass, really) should notice that as they play higher notes, they become closer together on the staff. What fewer might notice is the pattern, where each octave that one goes up, there becomes twice as many notes (played without valves), and that each newly available note comes in between the notes of the previous octave. This led me to notice one last thing: trumpets can't play the fundamental frequency for their overtone series. The lowest note one can clearly produce (without using the valves) is a middle C, but the next note up is a G, clearly not the 2nd harmonic. This means that the lowest clear note for a trumpet is the 2nd harmonic, not the fundamental. However, I have also played the flugelhorn, which has the same tube length as a trumpet, and the same overtone series (the 2nd and 3rd harmonics are still the same frequency C and G as on a trumpet). On a flugelhorn, I found that you can also comfortably drop an octave below middle C, which is the fundamental frequency for the instrument. The question: Why can a flugelhorn clearly and easily drop to the fundamental, when a trumpet can not? The main functional difference between the two instruments is that a trumpet has a cylindrical bore while a flugelhorn has a conical bore . Why does this difference in bore have such an impact on the range that each instrument can play? I know that the cylindrical bore has a "warmer sound", but I don't particularly care about that for the purposes of this question. I did notice a section of this Wikipedia article mentions " whole tube vs. half tube " in relation to the ability of some brass instruments' ability to hit their fundamental, but I am interested in a more in-depth explanation of why that is important. Visuals of the instruments in question You can clearly see the difference in tubing between the two instruments.The trumpet's diameter only widens along its final approach to the bell, while a flugelhorn's diameter widens along the entire length of its tubing. Otherwise, they play nearly identically (outside of the thing that sparked this entire question). | In a simplified model, the tube is effectively closed by the player's lips at the mouthpiece end. For the theoretical case of a straight tube with completely cylindrical or conical bore, and no bell, the harmonics of the conical bore have frequency ratios 1:2:3:4.... but the cylindrical bore only has the odd harmonics 1:3:5:7.... Those facts are most obvious in woodwind reed instruments. The clarinet has a cylindrical bore and the first harmonic is at the 12th (3 times the fundamental frequency) but almost all other instruments (e.g. the saxophone) have conical bores and the first harmonic is at the octave (2 times the fundamental). It is also apparent in organ reed pipes, where the pipes designed to imitate trumpet tone are literally conical along their entire length, with no bell. In "cylindrical bore" brass instruments, the large bell reduces the frequency of each harmonic compared with the simple theory, so the actual frequencies change from 1:3:5:7:... to something close to (a number less than 1):2:4:6:... The actual shape of the "cylindrical" tube is carefully designed to make the higher ratios as close as possible to 4:6:8:... and so far as the player is concerned, these are equivalent to the 2:3:4:... of a conical tube of twice the length. Hence the description of "half tube" instruments which are (approximately) half the length of a "whole tube" instrument that from the players point of view sounds the same harmonics. However the fundamental frequency can not be accurately "modified" in the same way as the higher harmonics. It is always physically playable, but it may be so out of tune, and its frequency very unstable and depending on the exact tension in the player's lips when blowing, that it is musically useless and hence described as "unplayable" in practice. The embouchure for playing the fundamental may be quite different from playing the higher harmonics, so there is no practical use for learning how to play a note that is completely out of tune in any case! It's worth noting that the physicist Helmholz who founded the modern study of acoustics in the 19th century got his theory of brass instruments completely wrong, because he didn't realize that the player's lips effectively formed a closed end to the tube and not an open end. And his reputation as a scientist was so great that it was half a century or more before anyone seriously questioned his incorrect theory! Of course musical instrument makers had been constructing practical instruments for centuries before that, and musicians had been playing them, with no theory at all except the empirical observation of the frequencies of the different harmonics. | {
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524,846 | It seems that a weather vane will rotate in order to minimize energy and thus orient itself parallel to the wind. What I do not understand is why it is implied that the weather vane arrow should point in the direction of the wind. I do not understand why the arrow pointing in the opposite direction of the wind is also not a minimum energy solution. | The vane has to be designed so that it has a preference to point in the right direction. In the example that you included, this is implemented by the flag at the back providing a broader cross section than the arrow head and also by the rooster standing slightly to the back half of the arrow. You are correct that if the vane became perfectly anti-aligned to the wind, it might stay there for a bit. That solution, however, is an unstable equilibrium solution. If the wind shifts even a little, assuming the vane is well-designed, it should snap around to the proper direction for the reasons above. | {
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525,047 | According to Einstein’s equation $$E=mc^2$$ Matter can be converted into Energy. An example of this is a nuclear reaction. What happens to the matter in the process ? Do the atoms/subatomic particles just vanish? Any insights into this process are appreciated. | Much of mass is just binding energy, so in a chemical reaction the electrons rearrange themselves and energy is released and the total mass of the molecules goes down (in an exothermic reaction, for example). The same is true for common nuclear reactions like spontaneous fission of uranium, with the caveat that some important nuclear reactions do involve the changing of fundamental particles, e.g., beta decay: $$ n \rightarrow p + e^- + \bar\nu_e $$ Here the mass of the RHS is less than that of LHS, so the electron and antineutrino are energetic (in the neutron rest frame). The energy comes from the difference in the neutron (936.6 MeV) and (938.3 MeV) proton masses, which is about 1.3 MeV. That 1.3 MeV is not binding energy, rather it is the quark mass difference coming from: $$ d \rightarrow u + e^- + \bar\nu_e $$ where the up (down) quark mass is $2.3\pm 0.7 \pm 0.5$ MeV ( $4.8\pm 0.5 \pm 0.3$ MeV). Note that most of the nucleon mass is once again binding energy. So what happens to mass when the down decays into an up? Nothing. Mass is not stuff, mass is a quadratic coupling to the Higgs boson that leaves finite energy at zero momentum: $$ E = \sqrt{(pc)^2 + (mc^2)^2} \rightarrow_{|p=0}=mc^2$$ or in other terms free particle waves (divide by $\hbar$ ): $$ \omega = \sqrt{(kc)^2 + (mc^2/\hbar)^2}\rightarrow_{|p=0}=mc^2/\hbar$$ it is just finite finite frequency at zero wavenumber. How does it work that matter is not stuff and is just a quanta in the quantum field? While we are generally comfortable with the photon being a quanta in the electromagnetic field which can appear and disappear at will (provided at least energy and momentum are conserved), this is because we view the EM field as a fundamental object. Moreover, the photon is its own antiparticle, and is neutral, so the conservation of charge and other quantum number does not always come to mind. They are also boson, so we don't think of them as "stuff" because they can be in the same quantum state. Well so is the quark field also fundamental, and its quanta are up and down quarks (and more), and in the case of beta decay, the $W$ boson changes the quark's flavor (and charge, and other quantum numbers) such that at rest, the initial quanta ( $d$ ) has a stronger coupling to the Higgs than the final quanta ( $u$ ), which we see as more mass in the initial state and hence more kinetic energy in the final state. Of course, you may still see the quarks as "stuff" and consider the Higgs coupling as some sort of binding energy leading to mass. That's OK. So let's look at electron-positron annihilation: $$ e^+ + e^- \rightarrow 2\gamma $$ Here, matter (2 leptons) just disappear and turn into 2 gamma rays. The initial state really acts like stuff: they are fermions, they cannot be in the same quantum state, they have mass, they are have charge, and so on. Enough electrons make a lightning bolt: that is very real. But the electron and positron are both quanta in the electron-field. That is it. They have opposite charge and lepton-number (electron number), so they can annihilate without violating any conservation laws. The initial state mass is just energy at zero momentum, it is not something more "real" or fundamental than the electron field itself. The rest energy is now available for the 511 KeV gamma rays, which is just 2 quanta in the EM field (and it is charge that couples the two fields). So in summary: all matter is made from quanta in quantum fields, and the fields are the fundamental objects. Mass is just Higgs coupling. If particle and antiparticle meet, they can annihilate, in which case the mass disappears (or not, there can be massive particle in the final state). Since things like baryon number and electron number are conserved, the basic particles they make (atoms), in the absence of antimatter, appear stable and look like "stuff". But fundamentally: they are neither. | {
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525,262 | Honestly, I don't get it. People say it's because it's a postulate. But, I mean, I see people deriving the Schrödinger equation with the help of the wave function, $T+U$ and partial differentials in three space coordinates and one time coordinate. How is that not a derivation? And why can't the Schrödinger equation be derived? | A derivation means a series of logical steps that starts with some assumptions, and ends up at the result you want. Just about anything can be "derived", as long as you vary what the assumptions are. So when people say "X can't be derived", they mean "at your current level of understanding, there's no way to derive X that sheds more light on why X is true, over just assuming it is". For example, can you "derive" that momentum is $p = mv$ ? There are several possible answers. You ask this as a student in introductory physics. Some might say yes . For example, you can start from the kinetic energy $K = mv^2/2$ , and then assume $K = p^2/2m$ . Combining these equations and solving for $p$ gives $p = mv$ , so this is a derivation. You ask this as a student in introductory physics. Some might say no . The above derivation is just nonsense. Starting from $K = p^2/2m$ is basically the same thing as assuming the final result, and if you're allowed to do that, it's no better than just taking $p = mv$ by definition. It's like "deriving" $1 + 1 = 2$ by defining $2$ to be $1 + 1$ . You ask this as a student in advanced mechanics. Most would say yes . You start from the deeper idea that symmetries are related to conserved quantities, along with the definition that momentum should be the conserved quantity associated with translational symmetry. Putting these together gives the result. The point is, you can make up a derivation for anything -- but you might not be at a stage in your education where such a derivation is useful at all. If the derivation only works by making up ad-hoc assumptions that are basically as unmotivated as what you're trying to prove, then it doesn't aid understanding. Some people feel this is true for the Schrodinger equation, though I personally think its elementary derivations are quite useful. (The classic one is explained in a later answer here.) There is often confusion here because derivations in physics work very differently than proofs in mathematics. For example, in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X). Once you finish putting high school math on a rigorous foundation, undergraduate math generally builds upward . For example, you can't use Stokes' theorem to prove the fundamental theorem of calculus, even though it technically subsumes it as a special case, because its proof depends on the fundamental theorem of calculus in the first place. In other words, as long as your classes are being rigorous at all, it would be very strange to hear "we can't derive this important result now, but we'll derive it next year" -- that would be in danger of logical circularity. This isn't the case in physics: undergraduate physics generally builds downward . Every year, you learn a new theory that subsumes everything you previously learned as a special case, which is completely logically independent of those earlier theories. You don't actually need any results from classical mechanics to completely define quantum mechanics: it is a new layer constructed below classical mechanics rather than above it. That's why definitions now can turn into derived things later, once you learn the lower level. And it means that in practice, physicists have to guess the lower level given only access to the higher level; that's the fundamental reason why science is hard! | {
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525,645 | As we all know, force applied to a spring is directly proportional to the extension of spring as shown below: However, my experimental results for a simple spring from a school laboratory don't match this behaviour: Why is this happening? I don't think systematic error is the reason that is causing the experimental graph to move away from the origin. | The coils of the spring are touching one another and the spring is initially under self-compression so it takes a finite force to move all the coils away from one another and for the spring to behave as you expected. That initial part of the source vs stretch curve is real (you had obtained data whilst undertaking an experiment) and hence should not be ignored. | {
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525,904 | This is something anyone could easily verify. When we open a tap slowly, water bends inwards (towards the axis) while maintaining its laminar flow. After a certain height below the opening, the flow becomes turbulent. I've approximately illustrated the shape of water near the top portion in the following diagram: I tried to explain the above phenomenon based on my knowledge on fluid dynamics. Let us consider the following diagram: Here, $A_1$ and $A_2$ are the areas of cross-section and $v_1$ and $v_2$ are the speeds of water molecules at two different heights (indicated by dotted red lines). Since, the shape of water remains fairly constant and the flow is laminar, in a time interval $\Delta t$ , the volume of water passing through level 1 must be equal to the volume of water passing through level 2. Mathematically, we can say: $$A_1v_1\Delta t=A_2v_2\Delta t$$ $$A_1v_1=A_2v_2$$ Or in other words, the product of the area of cross section and the speed remains the same at all heights and this is known as the equation of continuity. Since water molecules are under the gravitation force of attraction, they are accelerated downwards. So, $v_1<v_2$ . As the product of area of cross section and the speed must be a constant, $A_1>A_2$ . This explains why water bends towards the axis while falling slowly from a tap. But the above explanation fails at much lower heights above the fluctuating flow zone (where flow fluctuates from laminar to turbulent). Let us consider another diagram: The area of cross-section remains almost constant at the intermediate heights above the red zone. It doesn't decrease in accordance to the equation of continuity. Further, my method of explanation involves a lot of assumptions and I've also neglected surface tension, viscosity etc. I'm unable to imagine how these forces would affect our results. Is this a correct reason for " Why does water falling slowly from a tap bend inwards? " or is there any better explanation for this phenomenon? Image Courtesy: My own work :) | You can actually predict the shape of the profile precisely using the arguments you mention above, which are by and large correct. To do so, you can make the following assumptions: Neglect viscosity (not a great assumption, but it's a start). The pressure is the same everywhere in the fluid—the edges are free surfaces, so this is reasonable. The flow is axially symmetric (i.e. the top-down cross section is always circular). If you do this, and take the location of the faucet as the origin, you can then state the relationship between the gravitational potential energy and the flow speed using Bernoulli's equation as: $$\rho g h + \rho \frac{1}{2}v^2 = \rho \frac{1}{2}v_0^2$$ where $v$ is the speed of the fluid as a function of height $h$ , $\rho$ is density, and $v_0$ is the speed at which the water leaves the faucet. Solving for $v$ , you'll find that: $$v = \sqrt{v_0^2 - 2gh}$$ As the fluid moves further down (i.e. as $h$ becomes further negative), the speed increases as you'd expect. Then you can use conservation of mass for the rest. Assuming steady flow, you'll find that $$A_1 v_1 = A_2 v_2$$ for any two cross-sections of the flow. Using the cross-sections at the faucet and another arbitrary cross-section, and declaring the faucet radius as $r_0$ , you'll find: $$\pi r_0^2 v_0 = \pi r^2 v$$ $$\pi r_0^2 v_0 = \pi r^2 \sqrt{v_0^2 - 2gh}$$ Solving for the radius $r$ , you find up getting the following expression: $$\boxed{r(h) = \frac{r_0 \sqrt{v_0}}{(v_0^2 - 2 g h)^{1/4}}}$$ This drop in the radius as the height decreases is consistent with your illustrations. For example, here is what I analytically determine as the flow profile when I use standard values for a bathroom sink faucet flow ( $r_0 = 1.5$ centimeters, $v_0 = 0.134$ meters per second, and $g = 9.81$ meters per second squared): Notice that the flow profile becomes effectively straight at distances observable in your common bathroom sink (4 inches or so). This is consistent with your observations. After a certain point, the stream becomes so thin that surface tension effects along with shearing at the air-water interface begin to destabilize the shape and cause it to break up into droplets. In addition, the flow becomes turbulent after a certain distance from the faucet, so this prediction is only accurate for the early stages of such a flow (i.e. for "small" $h$ ). | {
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526,095 | I'm a first year engineering student who is new to physics, so I apologize if my question is stupid. But in our statics course we are using the book "Engineering mechanics: statics" by R.C. Hibbeler and it contains the following image: The astronaut is weightless, for all practical purposes, since she is far removed from the gravitational field of the earth. Now this conflicts with my previous understanding of weightlessness. I always assumed it was because astronauts are orbiting the earth. But if it's because they are far away from the earth, why would spacecraft spend so much energy to accelerate to such large speeds if it's not really necessary to "stay afloat"? | The statement in the image is shockingly wrong. Given that the Earth is visible behind the astronaut, the reduction of the gravitational force due to distance must be fairly small; on the ISS for example the force of gravity is about 88% the value at the Earth's surface - very far from being negligible! The reason why the astronaut appears weightless is that she is in freefall, accelerating downwards at $g$ . | {
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526,869 | I've been wondering about how sunken objects would behave if you could instantaneously flip their container with water. Like if you had a bucket filled with normal tap water and you dropped a ball in it. Then SOMEHOW you instantly flip the bucket over such that for just a moment, the water is suspended in midair and the ball is still touching the base of the bucket. What I want to know is, first of all what would happen? I know that if you drop a bowling ball and a bouncy ball from the same height they'd just about reach the ground at the same time, but what about this scenario? If you "drop" water and an object inside the water at the same height, do they reach the ground simultaneously? Or does the object "float" down the falling water? Second of all, under what conditions could the object fall first? That's what I'm looking for- what kind of object could fall out of the water and reach the ground before the water does? What difference would there be between the object being a sunken wooden ship and a rock? Third of all, say we take this experiment to the ocean, where the water pressure gets higher as you go down. If there's an object at the bottom of the ocean and we flip the ocean, how does the pressure affect things? | I want to supplement Emilio's intuitive answer discussing what would happen with some thoughts as to why what you propose in your second part cannot happen. What kind of object could fall out of the water and reach the ground before the water does? Let's assume the water is a single entity. In order for the object to accelerate faster than the water, the object needs to have a larger downward acceleration than the water does. This would need to come from a downward net force that is larger than the weight of the object, as in free fall both objects will have the same downward acceleration. Where would this force come from? There is no buoyant force in a free-falling fluid, but even if there were the buoyant force would act upwards on the object, not downwards. Therefore, the best your could even hope for is that the object and the water move together, and this is indeed what happens. Perhaps the confusion comes from why certain objects normally sink. They aren't pulled down by the water, they are pulled down by gravity. In your everyday scenarios it is just that the fluid impedes this "falling". You could even think of us as all sinking in the Earth's atmosphere. Therefore, in your scenario, it is not the case that because the object is in water suddenly means it wants to move down through the water. The water itself is not the mechanism for why objects sink. | {
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527,390 | I know that this is very similar to How important is mathematical proof in physics? as well as Is physics rigorous in the mathematical sense? and The Role of Rigor . However, none of the answers to those questions really resolved my own question : Is there a case where mathematical proof can replace experimentation? Most of the answers I read seem to be saying that you can mathematically prove facts about a model, but not that reality corresponds to the model. You have to experimentally validate the assumptions of the proof which demand the conclusion as true. But what if the assumptions have already been experimentally validated? For example, if I show that if certain physical laws or accepted theories are true, a model must be (I'm not aware of such a proof, or if one exists), since the assumptions have been validated, do I still need to go through the trouble of experimentation? If we've shown it would be logically inconsistent for a conclusion to be false, and we take data that seems to be contradicting it, what's more likely to be false or mistaken - our logic, or our tools/experiment? I imagine that if scientists ever claimed to have found a right triangle in nature that violates Pythagorean's theorem, it would be more logical to assume they made a mistake. The reason I ask this is because most, if not just about all of the ToEs in theoretical physics pretty much only have their mathematics going for them. The one most infamous for this is string theory. If string theory could be mathematically proven in the way I presented, and this proof was independently replicated and stood the test of time in the same way the Pythagorean theorem has, do we need to go through all the trouble of actually making an experiment? | No. Physics remains an experimental science and so it is not possible to replace experiment by a proof. Descartes tried this when he proposed his theory of propagation of light - very elegant - but it predicted incorrectly that the angle would increase for light passing into an optically denser medium. Indeed the story goes he refused to attend a demonstration that showed him wrong A rigorous proof is essential to properly understand and extend some aspects (and possibly some limits) of a theory, and to shed light on how phenomena can be linked and explained, but has no physical applications if it predicts something that contradicts experiment. ————— Edit: There is a related discussion in this paper by David Mermin: Mermin ND. What’s bad about this habit. Physics today. 2009 May 1;62(5):8-9. | {
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527,875 | Where is the mass coming from when neutrons are produced from protons in the Sun? If a positron is made, will it possibly annihilate with an nearby electron? | While a free neutron does have more mass than a free proton, a bound helium-4 nucleus has less mass than two free protons and two free neutrons. In fact, the helium-4 nucleus has less mass than four free protons. The difference goes into the binding energy of the nucleus. Therefore, as the other answers state correctly, stars are constantly losing mass, not gaining it, through their fusion reactions. | {
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528,549 | I know this question sounds dumb, but please bear with me. This question came into my mind while I was looking at the photos in an astronomy book. How is it possible that IR and UV photos of stars and nebulae are taken if our eyes could not detect them? | The images are taken by UV/IR cameras. But the frequencies are mapped down/up to visible region using some scheme. If you want to preserve the ratios of frequencies you do a linear scaling. Generally the scaling is done in way to strike balance between aesthetics and informativity. In light of the overwhelming attention to this question, I have decided to add the recent image of a black hole as taken by the Event Horizon Telescope. This image was captured in radio waves by array of radio telescopes in eight different places on earth. And the data was combined and represented in the following way. A point that I forgot to mention which was pointed out by @thegreatemu in the comments below is that the black hole image data was all collected at just one radio wavelength (of $1.3$ mm). The colour in this image signifies the intensity of the radio wave. Brighter implying brighter signal. | {
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529,076 | I noticed that after stirring, a bubble in the centre of my mug of tea changed the speed it was rotating at periodically. Speeding up, then slowing down, then speeding up again, etc. Almost like when a ballerina pulls in her arms to increase her speed. Tea after stirring Edit:
I've repeated this with room temperature water to try and rule out any temperature-related effects and the same effect is present. | Just looking at the video, it appears that the shape of the surface is varying quasi-periodically, as if the liquid is moving outward (and upward) toward the cup walls, then moving inward and rising in the center of the cup. This can be expected, if in the beginning the shape is not a perfect equilibrium shape (e.g., like a parabolic surface in a rotating cup). But when the the liquid moves toward the center, the rotation necessarily speeds up due to conservation of angular momentum; and when it moves outward the opposite happens. A crude analogy: If you rolled a marble in a large wok with smooth spherical curvature, in such a way that it looped near the center/bottom, then out near the edge, you would see that its angular velocity increases when it approaches the center/bottom, and decreases when it recedes from the center/bottom. You can think of the volume of liquid doing the same thing as the surface shape changes from a shallow curve to a deep curve. | {
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529,432 | Coulumb's law states that $$F \propto q_1 \cdot q_2 \tag{1} $$ and $$ F \propto \frac{1}{r^2} \tag{2} $$ Why can we combine these two proportions into $$ F \propto \frac{q_1.q_2}{r^2}?$$ What property allows that? If we combine them by the transitive property, shouldn't the final relation be $ q_1 \cdot q_2 \propto \frac{1}{r^2}$ ? | Let us say we want to define electrostatic force between two charges. What would it be a function of? We make the simplest guess that it must be related to the charges $q_1,q_2$ and the distance $r$ between the two, $F\left(q_1,q_2,r\right)$ . Now to get any functional form we need a controlled experiment where we measure the force by varying any one of the three parameters at a time while keeping the others fixed. In doing these experiments we find the following, $F\left(q_1,q_2,r\right)\propto q_1q_2\big|_{\text{fixed r}}$ $F\left(q_1,q_2,r\right)\propto \frac{1}{r^2} \big|_{\text{fixed }q_1q_2} $ Due to the fact that these are true only when the other parameters are fixed, we can not directly “combine” the two proportionalities in any meaningful way. To do so, we need to do the following. Notice that the first proportionality is true for a fixed $r$ . Thus in general the constant of proportionality must be some number $\alpha_1$ multiplied by some function of $r$ , say $f(r)$ . Similarly the second constant of proportionality must be $\alpha_2~g(q_1q_2)$ . Now that we have converted our proportionalities into equalities, we can use transitivity to obtain $$\alpha_1f(r)q_1q_2=F=\frac{\alpha_2g(q_1q_2)}{r^2}\\
\alpha_1f(r)r^2= \frac{\alpha_2g(q_1q_2)}{q_1q_2}
$$ Since $\alpha$ ’s are numbers, the LHS is purely a function of $r$ and the RHS purely a function of $q_1q_2$ , the only way the equation will hold true is if both sides equal a constant (can be set to 1 without loss of generality). This can only happen if $f(r)=1/r^2$ and $g(q_1q_2)=q_1q_2$ and $\alpha_1=\alpha_2$ . This finally gives us (using either of the proportionalities) $$F\propto \frac{q_1q_2}{r^2}$$ | {
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529,799 | If space is the measured distance between 2 objects, then saying the space expanded is nonsensical unless we have a measuring stick outside of the space fabric to measure the expansion. 2 objects moving away from themselves isn't the same as space expanding. What measuring stick is outside space? Doppler redshift shows the relative speed of a star in relationship to us - but would have no effect on a space expansion. Edit : There seems to have been some confusion (very possibly on my part) on the nature of my question. I'm not talking about an explosion-type situation where matter in the universe is moving apart (with nuclear forces holding stars/planets and gravitation forces holding solar systems, and star clusters together). This is intuitive. I'm talking about statements made by physicists that "space itself" is expanding. An excerpt from a scientific American article: "the spacetime containing matter cannot remain stationary and must either expand or contract" https://www.scientificamerican.com/article/where-is-the-universe-exp/ Edit after related question presentation : So according to the answers on that related question, we should be able to measure exactly that extra energy spent holding matter together against the accelerating space-time expansion. Would radioactive decay would be speeding up too? Would it get easier and easier to split an atom? So thats how we'd measure spacetime expansion? | NB: The nature of the question has changed since I placed this answer. This answer does not address the current version of the question. This answer addresses how we distinguish expansion of space from a model where galaxies move away from us through space (which is not straightforward). At the moment it is Hubble's law, some indirect measurements and a bit of philosophy. We observe that galaxies appear to move away from us, in an isotropic fashion, at a rate that is proportional to their distance from us. Whilst one could argue that we are at (or near) the centre of this very uniform expansion it begs the question as to why Hubble's law should exist and why the universe appears isotropic to us, but wouldn't from a different position in the universe. The simplest explanation is that General Relativity applies (as we observe in a number of other cases) and we live in an expanding universe - this then means we do not need to occupy some privileged position in the universe (an erroneous assumption that has proved wrong every other time it has been made). In such a universe, the redshift of distant galaxies is not caused by relative motion, but by the expansion of space. At high redshifts, these phenomena become distinct in that the relationship between "velocity" and redshift is different, for instance allowing "faster than light" (apparent) speeds. So basically at present, expansion fits the facts (far) better and more simply than any of the alternatives. A further piece of indirect evidence comes from a careful analysis of the physical conditions of gas at high redshifts, illuminated by background quasars and subtle alterations to the cosmic microwave background (CMB) spectrum, caused by the Sunyaev-Zel'dovich effect, towards galaxy clusters at low redshifts. Both of these methods give the temperature of the CMB at those locations. In the expanding universe model, the temperature should increase as $1+z$ , where $z$ is the redshift. If one instead has a non-expanding universe, and explain the CMB as due to some expanding shell of material, then the average temperature wouldn't change for distant galaxies unless the shell gas has been uniformly cooling by an amount that just happens to agree with redshift of that galaxy. Avgoustidis et al. (2015) review the evidence for the temperature evolution of the CMB and conclude that it agrees with an adiabatic expansion to better than 1%. Direct evidence for the expansion is on the horizon though. In an expanding universe, the speed at which galaxies move away from us can change slowly with time (and with distance) by of order 10 cm/s per year, despite their being no force on them. This is known as the redshift drift . There are plans to measure this tiny effect with the European Extremely Large Telescope over the course of a decade. | {
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530,030 | Obviously the iron filings can be seen aligning themselves along the virtual magnetic field lines produced by the permanent magnet, the virtual magnetic field line is made of electromagnetic field due to the alignment of electrons in the magnet but why the patterns, why lines? Do these lines have thickness? Are they due to interference pattern? | Here's a map of the barometric pressure in the United States. The map contains isobars, which are lines of constant pressure. These are constructed by starting from an arbitrary point, and following the direction where the pressure doesn't change. Isobars don't "exist", in the sense that there isn't literally a big white line in the sky hovering over New York City at this moment. Isobars aren't made of anything, and whether or not an isobar goes through a point on a map is decided entirely by how the map maker decided to draw them. But they help you visualize pressure, which is very real. When people say that magnetic field lines don't "exist", they mean they're like isobars: a completely arbitrary visualization tool that doesn't exist outside of diagrams. But like pressure, the magnetic field itself is as real as it gets. Iron filings follow its direction. | {
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530,057 | Basically im asking if there's anything special about visible light other than the fact that we use it to see colors. If we saw in another wavelength, would it still be possible to see colors like we do now? Does visible light have something special about it that lets us see a variety of different colors? | Something special about the visible range is that water has low absorption in this range. It’s a rather sharp dip near the visible region. Since we know that life began in water , the beings that were receptive to these wavelengths had a significant advantage over the others. Thus natural selection would have favoured these life forms over the others. This maybe the reason why we are primarily receptive to the “visible” range. | {
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530,205 | Imagine we have a magnet (red side is the north pole, blue side is the south pole), and imagine two ways to split it. The first way: When we split it by separating the north pole from the south pole, we see that the two pieces are themselves dipole magnets. The two broken ends will be opposite poles and they will attract each other. The second way: When we split it along the two poles, the two resulting magnet pieces will repel each other. Note that the colors are just representations of the poles of the magnet, and when the colors change, they don't represent any change in the pieces of the magnet itself. My question: If there is a repulsive force within the magnet that is pushing the magnet horizontally apart (in my picture), then what is keeping it together? It seems to me that the stability of the magnet can't be explained with electromagnetic forces alone, but that doesn't seem quite correct. What keeps the domains of the magnet together? And at the atomic level, what keeps the atoms together? | The chemical bonds of the material keep it together. If the magnets you're thinking of are made of metal, then the chemical bond is the metallic bond , which is quite strong. You can get a sense of how strong it is if you try to rip a metal bar into two. Unless you are exceptionally strong, you probably won't manage – but you are probably able to pull a bar magnet off a refrigerator door, for example. The force of the metallic bond is much larger than the magnetic force. Ironically, the source of the metallic bond is also the electromagnetic force. For typical values, the electric force is much larger than the magnetic force. You can get a sense of this by examining the force between two spheres with charge $1\ \mathrm C$ . If the two spheres are separated by a vacuum and are at a distance of $1\ \mathrm m$ , the force between them is approximately $9\times10^9\ \mathrm N$ . Meanwhile, if one of the spheres is moving at a speed of $1\ \mathrm{m/s}$ in Earth's magnetic field, the magnetic force it experiences is approximately $3.2\times10^{-5}\ \mathrm N$ . This large discrepancy is what keeps the magnet (and atoms) together. | {
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530,369 | If the pressure of the Earth is keeping the inner core solid, keeping it rigid to take up the least space, and temperature is dependent on how much the atoms are moving, why isn’t the inner core cold? If the pressure is so high that it’s forcing the inner core to be solid then the atoms can’t move around and thus they can’t have temperature. | Your argument would require that all solids must be cold, because all solids have constituent atoms that are constrained to remain in their solid lattice positions. But clearly not all solids are cold, so there is something wrong with your argument. That thing is that atoms or molecules constrained within a solid structure can still vibrate and oscillate around their equilibrium positions. So they do have an internal energy and this is where the heat is stored. | {
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530,411 | In his book Gravitation and Cosmology , Steven Weinberg says that the Cosmic Microwave Background (CMB) makes it "difficult to doubt that the universe has evolved from a hotter, denser early stage". In my understanding, CMB is just a peculiar isotropic radiation representing a black body at ~ 2.7K. How and why does the CMB point to the early Universe being hotter and denser? | Beyond the fact that the cosmic microwave background (CMB) is a direct prediction of the big bang model, there is the question of how you would produce it in any other way. It is remarkably close to being isotropic and remarkably close to being a blackbody spectrum - i.e. it is almost a perfect blackbody radiation field. A blackbody radiation field is emitted by material in complete thermodynamic equilibrium (CTE). An example would be the interior of a star. A requirement for (CTE) is that the matter and radiation field are characterised by the same temperature and that the material is "optically thick" - meaning that it is opaque to that radiation at basically all wavelengths. Given that the universe is mainly made up of hydrogen, helium and (presently) traces of heavier elements, we can ask how is it possible to produce a perfect blackbody radiation field? Cold hydrogen and helium are transparent to microwaves. To make them opaque they need to be ionised, so that the free electrons can be a source of opacity at all wavelengths via Thomson scattering. But this requires much higher temperatures - about 3000 K. How do we uniformly raise the temperature of a gas (adiabatically)? By squeezing it. A smaller, denser universe would be hot enough to have ionised hydrogen and would be opaque to the radiation within it. As it expanded and cooled, the electrons combined with protons to form atoms and the universe becomes transparent, but filled with a perfect blackbody radiation spectrum. The light, originally at a temperature of 3000 K and mainly in the visible and infrared, has had its wavelengths stretched by a factor of 1100 by expansion of the universe, meaning we now see it mainly as microwaves. Additional evidence for this model is that the radiation field is not absolutely isotropic. These small ripples encode information such as the expansion rate of the universe at the time of (re)combination and the density of matter. When inferred from measurements, these parameters agree very closely with other determinations that are independent of the CMB, such as the Hubble redshift distance relationship and estimates of the primordial abundance of Deuterium and Helium. There is now direct evidence that the CMB was hotter in the past and by exactly the amount predicted by an adiabatic expansion. The source of this evidence is measurements of the frequency-independence of the Sunyaev-Zel'dovich effect towards galaxy clusters (e.g. Luzzi et al. 2009 ); or more precisely by probing the excitation conditions in gas clouds at high redshift using even more distant quasars as probes (e.g. Srianand et al. 2008 . New results have been published by Li et al. (2021) . They describe measurements of the Sunyaev-Zel'dovich effect to hundreds of galaxy clusters in the redshift range $0.07<z<1.4$ and show that, the temperature of the CMB goes as $T_0(1 + z)^{0.983^{+0.032}_{-0.029}}$ , consistent with an adiabatic expansion to 3%. | {
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531,441 | Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period? | An example that might help: Start with a big pile of coins. Flip them. Remove the heads. About half remain. Take the remainder and flip them. Remove the heads. About half remain. Take the remainder and flip them. Remove the heads. About half remain. The analogy: An atom has a 50% chance of decaying in some particular interval $T_{1/2}$ . After each of those intervals, half are left. | {
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532,708 | This statement is repeated so often that it has become somewhat of a cliche: 'billions of neutrinos pass through your body every second'. For example see 1 , 2 , 3 , 4 , 5 , 6 . What is the evidence for it, especially considering that we have never detected even a hundred neutrinos in a second through one detector? | Those neutrinos come from the Sun. Fusion converts protons to neutrons, so that must produce neutrinos. One can calculate the number of nuclear reactions necessary for the power output, and get a number for the neutrino flux. One can also estimated the flux from the cross section of the detector. The two rates differ by a factor of about three. That was resolved by the neutrino oscillations between the three flavors (electron, muon and tauon neutrinos). | {
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532,726 | Consider a rigid bar (infinitely thin and with uniform mass density) of length $L$ with $x_1(t), x_2(t) \in \mathbb{R}^3$ each describing the positions of an endpoint of the bar in some fixed inertial frame at time $t$ . The vector $q(t) := \frac{x_2(t) - x_1(t)}{L}$ is a curve on the unit sphere $S^2$ describing the orientation of the bar. The vectors $\dot{q}(t)$ and $\omega(t) := q(t) \times \dot{q}(t)$ describe the velocity and angular velocity of the bar with respect to the inertial frame. My question is, what is the rotational kinetic energy of the bar in terms of these variables? From the formulas I know, I would like to say it's just: $$KE_{rot} = \frac13 mL^2 ||\omega||^2$$ But I haven't worked with this material in a number of years, and am quite rusty. | Those neutrinos come from the Sun. Fusion converts protons to neutrons, so that must produce neutrinos. One can calculate the number of nuclear reactions necessary for the power output, and get a number for the neutrino flux. One can also estimated the flux from the cross section of the detector. The two rates differ by a factor of about three. That was resolved by the neutrino oscillations between the three flavors (electron, muon and tauon neutrinos). | {
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532,731 | I just discovered in my parents' basement a Sprite can from 1995* and also a Coca-Cola can probably from the same year. Both cans are unopened and have no visible damage or holes. The Coca-Cola can feels "normal", but the Sprite can is empty! You can hear in my video that there is no liquid sloshing around in it. I can't find any way that the soda could have escaped. We also don't see any mess near where the cans were, but I don't know for certain that the cans stayed in the same place for 25 years, so maybe there could have been a mess of liquid leaked out somewhere else if the cans had been stored somewhere else earlier. But nothing feels sticky or looks like there has been a leak of any kind. What are possible explanations for how a carbonated drink could disappear from a sealed aluminum can? *I had kept it as a "collector's item" from when the Houston Rockets won their second championship. | The soda can't escape through a sealed can, which leaves two options: the can wasn't sealed or the soda didn't escape. Soda didn't escape: This is the less likely and less interesting case: the can was produced defective and never had any soda in it. Issue on the manufacturing line, and whoever saved it (you/parents?) did so because that was novel, and subsequently forgot. Can wasn't sealed: Again, a manufacturing defect is certainly possible, and in the intervening years most of the contents escaped through a defect you can't see, probably around the lid or the pop tab itself. This requires a serious molar flux through the hole. I needed to prove to myself that it was possible. Using Fick's First Law of diffusion, $$ J = -D\times \frac{\delta C}{\delta x}$$ A temperature of $20\ \mathrm{^\circ C}$ Which leads to a density of air of $1.204\ \mathrm{kg/m^3}$ , or $0.1204 \ \mathrm{g/cm^3}$ And a partial pressure of water of $0.0231$ And a diffusion coefficient of water vapor in air of $0.242\ \mathrm{cm^2/s}$ About $6\ \mathrm{cm}$ as the distance, given that the can is on average half full and the storage space was very dry We get a flux of $0.004856\ \mathrm{g/(cm^2\ s)}$ Given that $12\ \text{fl.oz.}$ is roughly $355\ \mathrm{cm^3}$ (and the soda is mostly water which is roughly $1\ \mathrm{g/cm^3}$ at that temperature), that means that a water vapor leak through the top over a period of $25$ years ( $7.88\times10^8$ seconds) would only need to be a $0.1\ \mathrm{mm}$ square hole. That's $1{-}2$ times the thickness of a human hair, on average. I couldn't see that hole, especially in the region under the pop tab. We could say the can was nearly fully empty the whole time and that the room wasn't that dry, with $50%$ relative humidity, and the hole would still only have to be $0.2\ \mathrm{mm}$ on a side. Can wasn't sealed: (part 2) The other answer, which I admit I was going to present when the above answer required too big of a hole, is that the fluid escaped as a liquid rather than a gas. Water has strong adhesion and cohesion, and will tend to "wet" any surface it is in contact with. If there was the tiniest hole or crack for water to flow through, it would have done so and evaporated long ago. As an aside: Interestingly, colas generally contain phosphoric acid, while sprite has citric acid, which is much weaker (and both have carbonic acid, weakest yet). That means that coke is more likely to corrode a metal container, enlargening a hole. However, aluminum is wonderfully inert in this capacity, and rapidly forms a persistent, impervious layer of aluminum oxide only a couple of atoms thick which prevents any further corrosion, so the sodas should not be chemically etching the can in any appreciable way. Testing: As mentioned in a comment above, you can do the bubble test yourself fairly easily. You can even enhance the test by precooling the can (and air inside) and placing it in warm water, creating a pressure difference. You might be able to see bubbles coming up. For a really definitive leak test, you can have the container tested using a radioactive tracer gas. Businesses exist to do just this, usually connected with the semiconductor industry. They can calculate the exact size of the hole, but such a service isn't free. | {
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532,748 | I was reading a great answer by John Rennie earlier (and took a mental note of the question which I must have mislaid). I wanted clarification by comment so I thought I could look it up amongst John's answers but he has 191 pages of answers. Needle in a haystack. John explains that in relation to inflation and expansion if we plotted out a grid of points 1 light-year apart in present time it would be a grid of infinite size - and if we went back in time to the Big Bang and plotted out a grid of the same points less than a nanometer apart it would still be of infinite size. I really wanted to ask why we assume the Universe is infinite in this way? | The soda can't escape through a sealed can, which leaves two options: the can wasn't sealed or the soda didn't escape. Soda didn't escape: This is the less likely and less interesting case: the can was produced defective and never had any soda in it. Issue on the manufacturing line, and whoever saved it (you/parents?) did so because that was novel, and subsequently forgot. Can wasn't sealed: Again, a manufacturing defect is certainly possible, and in the intervening years most of the contents escaped through a defect you can't see, probably around the lid or the pop tab itself. This requires a serious molar flux through the hole. I needed to prove to myself that it was possible. Using Fick's First Law of diffusion, $$ J = -D\times \frac{\delta C}{\delta x}$$ A temperature of $20\ \mathrm{^\circ C}$ Which leads to a density of air of $1.204\ \mathrm{kg/m^3}$ , or $0.1204 \ \mathrm{g/cm^3}$ And a partial pressure of water of $0.0231$ And a diffusion coefficient of water vapor in air of $0.242\ \mathrm{cm^2/s}$ About $6\ \mathrm{cm}$ as the distance, given that the can is on average half full and the storage space was very dry We get a flux of $0.004856\ \mathrm{g/(cm^2\ s)}$ Given that $12\ \text{fl.oz.}$ is roughly $355\ \mathrm{cm^3}$ (and the soda is mostly water which is roughly $1\ \mathrm{g/cm^3}$ at that temperature), that means that a water vapor leak through the top over a period of $25$ years ( $7.88\times10^8$ seconds) would only need to be a $0.1\ \mathrm{mm}$ square hole. That's $1{-}2$ times the thickness of a human hair, on average. I couldn't see that hole, especially in the region under the pop tab. We could say the can was nearly fully empty the whole time and that the room wasn't that dry, with $50%$ relative humidity, and the hole would still only have to be $0.2\ \mathrm{mm}$ on a side. Can wasn't sealed: (part 2) The other answer, which I admit I was going to present when the above answer required too big of a hole, is that the fluid escaped as a liquid rather than a gas. Water has strong adhesion and cohesion, and will tend to "wet" any surface it is in contact with. If there was the tiniest hole or crack for water to flow through, it would have done so and evaporated long ago. As an aside: Interestingly, colas generally contain phosphoric acid, while sprite has citric acid, which is much weaker (and both have carbonic acid, weakest yet). That means that coke is more likely to corrode a metal container, enlargening a hole. However, aluminum is wonderfully inert in this capacity, and rapidly forms a persistent, impervious layer of aluminum oxide only a couple of atoms thick which prevents any further corrosion, so the sodas should not be chemically etching the can in any appreciable way. Testing: As mentioned in a comment above, you can do the bubble test yourself fairly easily. You can even enhance the test by precooling the can (and air inside) and placing it in warm water, creating a pressure difference. You might be able to see bubbles coming up. For a really definitive leak test, you can have the container tested using a radioactive tracer gas. Businesses exist to do just this, usually connected with the semiconductor industry. They can calculate the exact size of the hole, but such a service isn't free. | {
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532,772 | When you play recorder or whistle, the pitch depends on how hard you blow into the tube. E.g. when you blow a whistle, initially the pitch is slightly lower when there is less air flow. This seems counter intuitive since the airflow should only affect the amplitude of the sound waves (like in many other instruments and tubes) and the frequencies which the resonating cavity choose to amplify should depend only on its length, which is constant. So why would the dominant sound we hear be affected by the air speed? | I don't believe the other answers are correct. FGSUZ describes pushing air out of a tube, which sort of plays a little part, but not the whole story. The way woodwind instruments produce sound, is they cause a column of air within the instrument to vibrate. This is done by splitting the air stream . Instruments such as the sax or clarinet use a reed to do this. A concert flute or a wine bottle blows air across a sharp edge, and a recorder or a whistle uses something called a fipple. In any case, that splitting of the air causes a pressure differential in the stream. One side of the split goes into free air, the other side goes into the body of the instrument. Additionally, virtually all of the air you blow goes out into free air, very little goes into the body*. We know from Bernoulli's principle that the moving air is at a slightly lower pressure. In an attempt to equalize, the column of air in the body will begin to move to fill the low pressure zone. Because the air has some mass and momentum, it will overshoot, and a newly-created high pressure zone will push the column of air back the other way, and the process will repeat. Pressing keys or (un)covering different holes will change the effective length of that air column, which you can think of as changing its mass**, which results in different pitches sounding. So when you blow with a greater airspeed, you will create a slightly more intense pressure differential, and so will create a little bit more relative energy to oscillate the air column. Blow a little slower, and the pitch will go down a little bit. Smoothly alternate between and you may have a nice vibrato. What's really important here, is it's not the volume of air that is important, but the air's speed . This phenomenon is also why many wind instruments tend to sound sharp at high notes, and flat on low notes, and the player needs to correct by varying their airspeed, as the keys or holes on the instrument alone are not enough to get the right pitch. In the case of a concert flute, which uses a sharp edge, rather than a fipple or reed, the player can aim their air, and directly control that pressure relationship , by varying the proportion of how much goes into the embouchure hole and how much goes over it. As a result, a skilled flutist can bend notes often more than a whole step up or down, based on air stream control alone, without changing anything about the flute itself, or without changing airstream velocity. Lastly, if you produce enough power in your airstream, you can overblow and play 1 or more octaves above the note as fingered. When playing in the upper registers, the tendencies for the instruments to sound increasingly sharp as it goes higher becomes more dramatic. Edit: I want to mention, but couldn't figure out where to work it into the answer above, but air speed is really important. Especially on the concert flute, it is important to the extent of massive frustration to newcomers. A fishing-line sized stream of air over the mouthpiece at the right speed will speak louder and clearer than 100 times more air if it's uncontrolled and slower. New flute players are often taught to think about "hot" vs "cold" air when learning to control their air stream. And, ultimately, when a player has attained sufficient skill, they can play quiet notes, by carefully blowing very small amounts of air, at very high speeds, and sound out even the highest notes quietly. If the physics of the instrument was about pushing air out of the body of the instrument, this would be impossible. It's not, because that tiny bit of air at the right speed is still enough to create that pressure differential, no matter how small. *Not true for reed instruments; the air-splitting behavior is caused by the reed itself, but the rest of the concepts are still true. **Massive oversimplification that borders on being completely wrong, but frankly it doesn't really matter. | {
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533,051 | A laser is built on quantum mechanics to create a beam of photons with the same frequency and phase. Someone told me a free electron laser is a based on classical electrodynamics. Is that true? | is classical laser possible? IMO, the question doesn't make sense. "Classical" and "quantum" are not different options for how a thing can work. They are different options for us to try to understand how it works. LASERs aren't "built on" quantum mechanics, but rather, quantum mechanics is an appropriate tool for understanding stimulated emission. A free electron laser isn't built on classical electrodynamics, but classical electrodynamics offers a sufficient explanation for why it emits light. | {
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533,291 | A few years ago my school physics teacher told us, his students, that there is an elegant physics problem in which all given numbers are unity, but the answer is $e$ in the power of $π$ , that is, $e^\pi$ , where e = 2.718... is the base of the natural logarithm and π = 3.14... is pi, the constant commonly defined as the ratio of a circle's circumference to its diameter. Fast forward to now, I am a university student studying something unrelated to physics, and I recently mentioned the above recollection of mine in a conversation with a physicist whom I had helped improve his English in his physics articles. Long story short, he highly doubts that such a physics problem exists. He believes that either I misunderstood my school teacher or the formulation of the problem is too lengthy or unnatural. But I clearly remember my teacher's words that the formulation of the problem is simple and that the solution is elegant. Unfortunately, I do not know what that problem is about. I tried googling, but did not get any lead. What is that mysterious problem, or is there any physics problem matching the description above? | $e^\pi$ is the square root of the ratio of the load tension to the hold tension for a rope wrapped $1$ full turn around a capstan when the coefficient of friction is $1$ . If I am allowed to say “ $1$ half-turn” without violating the “unity” requirement then taking the square root is unnecessary. The relationship is $$T_\text{load}=T_\text{hold}\,e^{\mu\phi}$$ where $\mu$ is the coefficient of friction and $\phi$ is the wrapping angle, as explained in the Wikipedia article “ Capstan equation ”. | {
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533,927 | So my textbook says that the reason cables that are suppose to carry high currents, are thicker that those that are meant to carry lesser current, is that "less resistance (of the wire) means less heating..."? Is this even true? Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating! | They’re describing the situation where the wires are carrying power to a load. It’s the load that (mostly) determines the current in the wires leading to it. A $1200$ W oven on $120$ V needs $10$ A. Once the load has determined the current, the heat in the wires is given by their resistance via $I^2 R_{wire}$ . A $0.02$ ohm wire to the oven will have $2$ W of heat; a $0.01$ ohm will have less: $1$ W. That difference in wire resistance doesn’t change the current much because the current is really controlled by the ~ $10$ ohm heater resistance. But it changes the wire heat a lot. | {
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533,982 | If you fire a gun or rifle into the air, whether straight up or at an angle, as I understand physics, a metal projectile will gain surprising momentum on its way down again, more than capable of killing a grown human being, not to mention small animals and children, property, etc. I have never fired a gun, but I assume that one of the first things they tell you is to never point the gun at a living being unless you intend to kill them, and next after that, to never fire up in the sky . However, people "skeet shoot" all the time, and seem to not apply this basic safety measure whatsoever when it comes to shotguns. I understand that shotguns work in a different manner from a normal "bullet", instead causing tons of small particles to spread out, but still, won't those small particles also come back to Earth in the same manner as the lethal metal bullet? | One of the other rules of firearm safety is "Know your target and what is beyond it." A firearm (of any sort) can be safely fired if there is nothing of value between the muzzle and the point where the projectile loses the last of its kinetic energy. Some of these safe places are called "firing ranges". "Firearms, The Law, and Forensic Ballistics" by Margaret-Ann Armour provides formulae for the maximum range of a shotgun pellet based on the pellet diameter ( $PD$ ): $$r_\mathrm{yards} = 2200 \tfrac{\mathrm{yd}}{\mathrm{in.}} \times PD_\mathrm{inches}$$ or $$r_\mathrm{meters} = 100 \tfrac{\mathrm{m}}{\mathrm{mm}} \times PD_\mathrm{millimeters}$$ When shooting skeet as in your question, you might select a shot between .110 inch (2.79 mm) and .080 inch (2.03 mm) . Doing the math, we come up with approximately 175-275 meters total projectile distance. That range (of values) is the maximum range (in meters) you need to make a (firing) range...and most shots will fall harmlessly to Earth much, much closer than the maximum. You can see this when you look at a satellite image of a shotgun range. There is nothing of value within 300 meters of the firing line, but most of the shot falls to Earth within 50m. Oh, and just to drive home the point that we are really, really certain there won't be any risk beyond those distances, what could that be in the top-left corner of the image? It's an airport! | {
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533,993 | I'm going to ask* about what should be the effect/interaction, if any, when a gamma-ray burst crashes or hits against the magnetic bubbles at Solar System's edge. These magnetic bubbles are explained in [1] from YouTube. I add also as reference this Wikipedia Gamma-ray burst . Question. Is it possible to say anything about the interaction, the physics, of an impact of a great gamma-ray burst against/going through the magnetic bubbles of a solar system similar than ours? And, additionally, as a secondary question, what happens if the gamma-ray burst does not collide/impact, but passes near the magnetic bubbles, let's say tangentially? Many thanks. I hope that it is possible say something about it (I evoke what work can be done to elucidate something about my questions). If you know references about what should to be the effect of the expected phenomenom or physics after a
collisions of a gamma-ray burst and this kind magnetic bubbles of planetary systems similar than ours, feel free to refer it and I try to search and read it from the literature. References: [1] NASA |Voyager Finds Magnetic Bubbles at Solar System's Edge , from the official channel NASA Goddard of YouTube (June, 9th 2011). | One of the other rules of firearm safety is "Know your target and what is beyond it." A firearm (of any sort) can be safely fired if there is nothing of value between the muzzle and the point where the projectile loses the last of its kinetic energy. Some of these safe places are called "firing ranges". "Firearms, The Law, and Forensic Ballistics" by Margaret-Ann Armour provides formulae for the maximum range of a shotgun pellet based on the pellet diameter ( $PD$ ): $$r_\mathrm{yards} = 2200 \tfrac{\mathrm{yd}}{\mathrm{in.}} \times PD_\mathrm{inches}$$ or $$r_\mathrm{meters} = 100 \tfrac{\mathrm{m}}{\mathrm{mm}} \times PD_\mathrm{millimeters}$$ When shooting skeet as in your question, you might select a shot between .110 inch (2.79 mm) and .080 inch (2.03 mm) . Doing the math, we come up with approximately 175-275 meters total projectile distance. That range (of values) is the maximum range (in meters) you need to make a (firing) range...and most shots will fall harmlessly to Earth much, much closer than the maximum. You can see this when you look at a satellite image of a shotgun range. There is nothing of value within 300 meters of the firing line, but most of the shot falls to Earth within 50m. Oh, and just to drive home the point that we are really, really certain there won't be any risk beyond those distances, what could that be in the top-left corner of the image? It's an airport! | {
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534,036 | I understand that a binary orbiting around one another will circularize due to the emission of GWs due to Peters equations and that highly eccentric binaries evolve faster.
But GW emission also removes energy and angular momentum (wouldn’t the latter increase the eccentricity from the relation between the eccentricity and the angular momentum?).
What is the physical picture behind this? | The shape of a Keplerian orbit can be characterized geometrically by its semimajor axis $a$ and eccentricity $e$ or dynamically by its energy $E$ and angular momentum $L$ . The latter are expressible in terms of the former as $$E=-\frac{Gm_1m_2}{2}\frac{1}{a}\tag{1}$$ and $$L^2=\frac{Gm_1^2m_2^2}{m_1+m_2}a(1-e^2)\tag{2}.$$ The former are expressible in terms of the latter as $$a=-\frac{Gm_1m_2}{2}\frac{1}{E}\tag{3}$$ and $$e^2=1+2\frac{m_1+m_2}{G^2m_1^3m_2^3}EL^2\tag{4}.$$ Note that $E$ is negative for a bound orbit. Gravitational radiation carries energy and angular momentum (and also linear momentum) off to infinity. $E$ decreases and becomes more negative, so its absolute value increases; $L^2$ decreases and becomes less positive, so its absolute value decreases. Whether the absolute value of the negative number $EL^2$ increases or decreases, and thus what happens to the eccentricity, is not obvious. One must do the calculation! Here is what Peters did. He first derives/rederives the formulas $$\frac{dE^\text{rad}}{dt}=\frac{G}{c^5}\left(\frac15\dddot{Q}_{ij}\dddot{Q}_{ij}\right)\tag{5}$$ and $$\frac{dL_i^\text{rad}}{dt}=\frac{G}{c^5}\left(\frac25\epsilon_{ijk}\ddot{Q}_{jl}\dddot{Q}_{kl})\right)\tag{6}$$ for the rate at which energy and angular momentum are carried to infinity by gravitational waves, in the leading order of a multipole expansion. Here $$Q_{ij}=\sum_n m^{(n)}\left(x_i^{(n)}x_j^{(n)}-\frac13\delta_{ij}x_k^{(n)}x_k^{(n)}\right)\tag{7}$$ is the system's traceless mass quadrupole moment tensor when the system is considered as $n$ point masses. He then applies this to a Keplerian binary and averages over one elliptical orbit. Using the conservation of energy and angular momentum, he finds that the binary's energy and momentum decrease at the average rate $$\left\langle\frac{dE}{dt}\right\rangle=-\frac{32}{5}\frac{G^4}{c^5}\frac{m_1^2m_2^2(m_1+m_2)}{a^5}\frac{1+\frac{73}{24}e^2+\frac{37}{96}e^4}{(1-e^2)^{7/2}}\tag{8}$$ and $$\left\langle\frac{dL}{dt}\right\rangle=-\frac{32}{5}\frac{G^{7/2}}{c^5}\frac{m_1^2m_2^2(m_1+m_2)^{1/2}}{a^{7/2}}\frac{1+\frac{7}{8}e^2}{(1-e^2)^2}\tag{9}.$$ Differentiating (3) gives $$\frac{da}{dt}=\frac{Gm_1m_2}{2}\frac{1}{E^2}\frac{dE}{dt}\tag{10}$$ and differentiating (4) gives $$e\frac{de}{dt}=\frac{m_1+m_2}{G^2m_1^3m_2^3}\left(L^2\frac{dE}{dt}+2EL\frac{dL}{dt}\right)\tag{11}.$$ Substituting (1), (2), (8), and (9) into (10) and (11) gives $$\left\langle\frac{da}{dt}\right\rangle=-\frac{64}{5}\frac{G^3}{c^5}\frac{m_1m_2(m_1+m_2)}{a^3}\frac{1+\frac{73}{24}e^2+\frac{37}{96}e^4}{(1-e^2)^{7/2}}\tag{12}$$ and $$\left\langle\frac{de}{dt}\right\rangle=-\frac{304}{15}\frac{G^3}{c^5}\frac{m_1m_2(m_1+m_2)}{a^4}\frac{e(1+\frac{121}{304}e^2)}{(1-e^2)^{5/2}}\tag{13}.$$ You can see that the rate of decrease of the eccentricity is very rapid for a highly eccentric orbit with $e$ near $1$ , due to the $(1-e^2)^{5/2}$ in the denominator. In others words, the orbit rapidly circularizes. From these equations, Peters proceeds to find $a$ as a function of $e$ (with two unusual exponents, $12/19$ and $870/2299$ ) and a differential equation for $e(t)$ from which the lifetime of the binary can be found. | {
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534,289 | I have a simple question regarding matter-antimatter gravity interaction. Consider the following though experiment: If we imagine a mass $m$ and an antimass $m^-$ , revolving around a large mass $M$ the potential energy of mass $m$ should be: $$ U_1=-\frac{GmM}{R} $$ and the potential energy of mass $m^-$ should be: $$ U_2=-\frac{GmM}{R} $$ or: $$ U_2=\frac{GmM}{R} $$ depending on the sign of the gravity interaction between matter and antimatter. If the two particles annihilate to energy, then the gravitational field of $M$ will interact with the emitted photons and will change their frequency. But, as the interaction between gravity and the photons has nothing to do with the question of the gravity between matter and antimatter, can't we simply use the interaction between gravity and photons, and the energy conservation to establish the nature of the gravity interaction between matter and antimatter? | This is a perfectly good argument and one of the reasons that all the physicists I know believe that antimatter behaves just like matter in a gravitational field. It is important to distinguish between antimatter, which is well understood from countless collider experiments, and negative matter (also known as exotic matter ), which has never been observed. Antimatter does not have a negative mass. Indeed antimatter is just perfectly ordinary matter - we think it special only because we are made from matter and therefore biased. Negative/exotic matter is very different. If it existed it would cause all sorts of problems with conservation of energy and the stability of the universe. | {
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534,348 | Our hands are the most common source for measuring heat transfer $^{\dagger}$ in our day-to-day life (they do not measure temperature, though they do indicate relative temperature). Now consider the following: When we rub our hands their temperature rises due to the friction. Now if there is equal increment in the temperature of both the hands then there should be no flow of heat (zeroth law of thermodynamics). Hence we should feel no difference than usual as there is no heat, rather due to there being a temperature gradient heat should flow out of our hands to surrounding and our hands should feel cooler (assuming that surrounding has same temperature as our normal body temperature). So why do our hands feel warm when we rub them? One possible explanation is that one of the hand gets slightly warmer than the other and hence there is a heat flow from one to other and hence we feel hot. But there is a problem that if that were the case then we would get an alternating hot and cold sensation, which isn't what is observed. $\dagger$ It is a common misconception that our hands measure the temperature. Rather they measure the heat flow (which is an indicator of relative temperature). Related: Veritasium: Misconceptions About Heat . Also note that I know that the increment in temperature is due to the friction so no explanations needed for that. | The frictional heat generated in the contact area between our hands flows roughly equally to each hand. The interface gets hotter, and there is a heat flow by conduction away from the interface in both directions. At the interface itself, there is a discontinuity in the heat flux, with opposite signs to the two fluxes. The difference between the two heat fluxes is the frictional heat generation rate per unit area. So we have $$\left[-k\frac{dT}{dx}\right]^+-\left[-k\frac{dT}{dx}\right]^-=q_{gen}$$ So, $$\left[-k\frac{dT}{dx}\right]^+=\frac{q_{gen}}{2}$$ and $$\left[k\frac{dT}{dx}\right]^-=\frac{q_{gen}}{2}$$ As a result of all this, the temperature is highest at the interface (higher than the bulk of your hands), and this is what you sense as your skin temperature. | {
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534,614 | I agree with the fact that the principle points out to the inaccuracy in the measurement of the two quantities of the particles (momentum and position). But measurements apart, does it explain anything about how nature works, in general? As in, I think the particle would have some exact value of momentum at that point in space (if not, please explain why). So why not just tell that 'okay it does possess some momentum at that position, but I can't tell what that exact value is'? Edit: I understood that the principle points out at nature as a whole, in general, and does not just point out at measurements | Heisenberg uncertainty is not a measurement effect - it's a fundamental property of objects in the physical universe . Historically, the uncertainty principle has been confused with a related effect in physics, called the observer effect, which notes that measurements of certain systems cannot be made without affecting the system, that is, without changing something in a system. Heisenberg utilized such an observer effect at the quantum level (see below) as a physical "explanation" of quantum uncertainty. It has since become clearer, however, that the uncertainty principle is inherent in the properties of all wave-like systems , and that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects. Thus, the uncertainty principle actually states a fundamental property of quantum systems and is not a statement about the observational success of current technology . (Emphasis mine) Therefore you cannot say "it does possess some momentum, I just don't know what it is". If it did possess that, it would be a so-called hidden variable , most versions of which have been excluded by experiment. | {
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534,847 | I've come across a couple of videos where some interesting sounds are produced using ice. (Click on the images to see the video.) Here, they drop a block of ice into a deep crevice and as the block falls, you can hear some strange sounds. These sounds remind me of some sound effects in old cartoons and sci-fi movies. The pitch of the sounds start high and end up low. Is this some kind of doppler effect? Here, there's a man performing ' nordic skating ', making the same sounds as he skates on top of a thin ice sheet. Maybe they are caused by the cracking ice. But why does that sound like this? There are some other videos too, where they skip stones on frozen lakes. Here the ice makes some high-frequency sounds which correspond to the chirping of birds. What are these sounds and why do they sound the way they sound? | The pitch of the sounds start high and end up low. Is this some kind of doppler effect? No. This is chirp induced by dispersion , which is the acoustic version of the same phenomenon for light . This refers to the fact that, generically, sound waves of different frequencies travel at different speeds. This can be caused by the material properties of the medium, and it can also occur if the sound wave is confined in some kind of waveguide (like a hole or a sheet of ice). For each individual sample a full analysis is necessary to understand which bit of the circumstances caused the dispersion (so it's hard-to-impossible to tell you any specifics about the individual recordings you've given) but the phenomenon is fairly generic so it really doesn't matter much. If you have short, sharp sounds, the waveform will typically have a very high harmonic content, i.e., if you decompose it as a superposition of monochromatic waves at different frequencies, it will cover a broad band of such frequencies. For the sounds you've given, the higher frequencies travel faster, and you get that descending chirp as the sound transitions to the lower frequencies which take longer to arrive. The reason this sounds so similar to a sci-fi 'blaster' sound is because this is how blaster sounds were first recorded: Star Wars sound engineer Ben Burtt created (recorded) that sound by recording the guy wire of a tower when he tapped it with a spanner: Image source Under tension, the transverse waves have exactly the same dispersion as the ice, so the physics is the same, and the sound is the same. For more discussion of this, see this previous question on this site . It's also worth mentioning that the equivalent phenomenon for light ─ taking a short pulse and passing it through a dispersive element so that the higher and lower frequencies arrive at different times ─ is the heart of what makes Chirped-Pulse Amplification (the dominant technology for creating intense pulses of light) work. | {
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535,240 | So recently I've been reading "How to teach Quantum Mechanics to your Dog" by Chad Orzel. In chapter 3, he says, if I understood this right, that electrons can only exist in specific quanta - that is they can only be in certain regions, and will perform a quantum leap over regions that are not stable energy levels. However, I've also seen stuff online about how electrons can be "anywhere" and that the further away they are from the nucleus, the less probability they have to be there. Is the probability of an electron being somewhere zero? Can electrons be anywhere? And if so, how and what is the need for a quantum leap? | In chapter 3, he says, if I understood this right, that electrons can only exist in specific quanta - that is they can only be in certain regions, and will perform a quantum leap over regions that are not stable energy levels. Electrons (that are confined by a potential well, such as when they are part of an atom) are indeed limited to certain quantized energy levels. And they will only be able to gain or lose energy only in quantized amounts. But no matter the energy level, the electron is not really restricted in what position in space it might be found in. I.e. there are no "regions" of space where it can't be found. There are, depending on the energy level, regions of space where it's spectacularly unlikely to find the electron (like you could take as many electrons as ever existed in the visible universe, put them into this state, and then observe their position and you probably wouldn't find one at that location). But in principle it's still possible to find the electron anywhere in space. For example, if you have an electron in a hydrogen atom, excited to one of the "2p" energy states, and you observe its location, you'll most likely find it in the locations indicated by these drawings: ( image source ) It's highly unlikely to find an electron in the 2p y state (for example) located far (a nanometer or so) up along the z axis (and a picometer off the xz plane). But it's not entirely impossible. There is actually, as pointed out in another answer, a locus of points on the xz plane, called a node of the wavefunction, where the probability of finding the electron is actually 0. what is the need for a quantum leap? The quantum leap is a leap in the energy of the electron, not its location in space. | {
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