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415,409 | I was wondering, if a person climbs a tower and shouts at the top of his lungs how much distance would the sound travel ? Would it reach someone 1km far? Assuming that there are no tall buildings in the way and the wind is still. | Gabriel Golfetti's answer assumes no dissipation. In reality, atmospheric attenuation is quite important for this calculation. According to Engineering Acoustics/Outdoor Sound Propagation: Attenuation by atmospheric absorption (Wikibooks), dissipation in the atmosphere exponentially decreases the sound's intensity with distance, which leads to a linear reduction in the loudness of the sound in dB. Therefore, the loudness of the sound will actually be $$L=88\;\text{dB}-20\log_{10}\left(\frac{r}{0.3\;\text{m}}\right)-ar$$ where $a$ is the attenuation coefficient in dB/m. The table below gives the attenuation coefficient as a function of frequency and relative humidity for air at 20 degrees Celsius: For air at a pressure of 1 atm and sound at a frequency of 1 kHz (which is around the peak of the human vocal spectrum), for most values of relative humidity the attenuation coefficient is approximately $a\approx 1\;\text{dB}/100\;\text{m}$. So our equation for the loudness becomes $$L=88\;\text{dB}-20\log_{10}\left(\frac{r}{0.3\;\text{m}}\right)-\frac{r}{100\;\text{m}}$$ Solving for $L=-9\;\text{dB}$ gives $$r\approx 2\;\text{km}$$ which is drastically reduced from the original answer. Changing the attenuation coefficient by a factor of two (which is approximately how much it varies at that frequency for non-dry air) changes the maximum distance by a factor of 2, so the proper answer, accounting for this uncertainty, is a few km . | {
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415,531 | According to this article , a Foucault pendulum proved Earth was rotating. I'm not sure it really proved it. If Earth weren't rotating and a Foucault pendulum started in a state with zero velocity, it would keep swinging back and forth along the same line. If at its highest point, it has a tiny velocity in a direction perpendicular to the direction to the lowest point on the pendulum, then maybe it would have such a tiny deviation from moving exactly back and forth that a human can't see that tiny deviation with their own eyes, but that deviation would result in a slow precession of the pendulum and a day is so long that the direction it's swinging in would rotate a significant amount. If you have a system where a particle's acceleration is always equal to its displacement from a certain point multiplied by a negative constant, and it's not moving back and forth in a straight line, it will travel in an ellipse which does not precess at all. I think the same is not true about a Foucault pendulum. Its acceleration doesn't vary linearly with its distance along the sphere of where it can go to the bottom and the sphere of where it can go doesn't have Euclidean geometry. My question is can we really conclude from looking at a Foucault pendulum and the laws of physics that Earth is rotating? Maybe if its initial velocity is controlled to be very close to zero, we can tell from its precession that Earth is rotating. Also maybe if its arc is very small, the rate of precession for any deviation from going back and forth that's undetectably small to the human eye will be so much slower than the rotation of Earth that we can tell from watching it that Earth is rotating. That still might not prove Earth is rotating, because if the pendulum has a tiny charge in the presence of a weak magnetic field, the magnetic field could also cause a slow precession. | The thing that Foucault did was not just to predict that a pendulum would undergo precession, but also to predict which way the precession would go, and how rapidly , depending on the latitude of the observer.
(Although Cleonis points out in a comment and in another answer that ascribing all of this analysis to Foucault is a historical oversimplification.) If you imagine a Foucault pendulum set up at one of the poles, with the Earth rotating underneath it, you should be able to convince yourself that an observer standing on the Earth would see the pendulum complete one precession cycle every day. Likewise a pendulum swinging in the plane of the equator would have no tendency to precess, and one at the other pole would (relative to the ground) precess the other way. The precession period works out to be $\rm1\,day/\sin(latitude)$, which at Paris is about thirty-two hours. Many of the mechanical details of a Foucault pendulum are set up to reduce the contribution of the parasitic effects that you're thinking of. Foucault used a very large mass, to reduce the acceleration imparted by stray air currents, on a very long tether, so that individual swings of the pendulum are very slow and any intrinsic curvature or ellipticity is observable. He was careful that the material of the tether shouldn't "unwind" the way some multi-fiber ropes do, which would exert some extra torsion on the motion. And the pendulum was released by tying it to a horizontal fiber and then burning that fiber with a candle, rather than by cutting the fiber with a knife or having a person just give the pendulum a shove, to minimize exactly the sort of parasitic horizontal forces you're asking about. Even then you might be able to explain away a single demonstration as a lucky fluke. The real strength of the Foucault pendulum comes when you repeat the experiment so that all these tiny parasitic forces ought to be randomly different, but the precession period turns out to be exactly the same , and then you repeat it in a city at a different latitude and the precession period is different by the right amount . | {
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416,131 | I'm not a physicist, but I was wondering: Can ants walk on dense liquid like mercury and why? | Whether or not a small animal/insect can walk on a liquid is determined much more by surface tension than by density. To see why this is consider a dense liquid without any surface tension. You would float in it very well, but if you tried to walk on it you would step right through the surface and fall over, sinking until you were sufficiently submerged to float (the amount of liquid displaced is equal to your body weight). Now imagine a very high surface tension liquid that is not very dense. You would be able to walk across it as long as you don't break the surface. The second you break the surface you would sink like a stone. So now for the ant in mercury. The surface tension of mercury is much higher than that of water (~6x), so I think it's safe to say that an ant might succeed in walking across the mercury for a short while, as some insects of similar mass can walk along water. Though without some coating on the ants feet which repelled to mercury (like water-walking insects have for water), the ants feet would likely get covered in mercury and would start sticking the surface of the liquid. | {
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417,927 | I am not entirely sure this is an appropriate question for PSE, however since many of you have such diverse backgrounds, I'll give it a shot. I have noticed that when one takes a picture of a computer, TV screen with a phone, you can "see" what looks like a saddle phase portrait. Why does this happen? Does it have any relevance to differential equations or other areas of math? | What you're seeing is a moiré pattern caused by beating between the regular grid of the screen's pixels and the regular grid of your camera sensor's pixels. The departure of the pattern from straight lines is likely due to a slight deformation away from planarity in the screen. You can test for this by keeping the camera fixed on a tripod and mildly flexing the screen with your fingers, upon which the saddle pattern should shift slightly. Edit: As pointed out in the comments, the saddle pattern could also be caused by distortions caused by perspective as well as by the camera optics. The fact that similar hyperbolae appear in this related question , involving only straight grids at an angle, provides a good deal of support for the perspective-effect explanation. The paper Simulation and Analysis of Overall Raster Moire Patterns on Color CRTs. T. Inoue et al. Soc. Inf. Display: International Symposium, Digest of Technical Papers 33 , 256 (2012) seems to contain relevant results for your case. | {
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418,550 | Hydrogen is the most abundant element in nature. Does cosmological nucleosynthesis provide an explanation for why is this the case? Is the explanation quantitatively precise? | The short answer is that (i) protons (hydrogen nuclei) are produced abundantly in the early universe, but only a small fraction of these are able to engage in nuclear reactions leading to heavier elements, either during primordial nucleosynthesis or later inside stars. This means that hydrogen ends up being the most abundant element in the universe. (ii) Big Bang nucleosynthesis makes very precise predictions (to $<1$ % precision) about the fraction of baryons that end up as protons vs other nuclei. Further details In the early phases of the big bang only the constituent parts of nucleons (quarks and anti-quarks) plus leptons (e.g. electrons, positrons) and light existed. As the universe expanded and cooled, quarks were able to combine and form the basic building blocks of nuclei - the neutrons and protons. A proton is of course a hydrogen nucleus; any heavier elements needed to be built by fusing together protons. Protons are positively charged and repel each other strongly. In order to fuse protons and make helium requires high energies/temperatures and the intermediate stage of forming deuterons - a proton plus a neutron in a bound pair. The reason that most protons do not fuse in this way is threefold. The small excess mass of the neutron compared to the proton means that the equilibrium reactions that produce nucleons end up producing about six times as many protons as neutrons. Free neutrons then decay into protons on timescales of minutes, which increases this ratio further to about seven. The deuteron is weakly bound, so there is only a limited window of time between when it is too hot to remain bound (prior to 10s after the Big Bang) and when it is too cool to get deuterons to fuse to become helium (beyond about 10 minutes after the Big Bang). Of these things, by far the most important factor is #1 and this is fundamentally the answer to your question. It is because the neutron is more massive than the proton. All heavier nuclei must contain neutrons because the Coulomb repulsion between two or more protons is too strong to form stable nuclei without them. By the end of Big Bang nucleosynthesis, essentially all of the available free neutrons end up in helium nuclei (with traces of deuterons and lithium nuclei) and thus the H/He ratio ends up at around 12 protons for every He nucleus (2 protons, 2 neutrons). Is this quantitatively precise? Well, my figure of 12 protons for every He nucleus, or 75% by mass , is a bit handwaving. A detailed model must take into account in a more accurate way the temperature evolution of the universe, the various (more minor) nucleosynthetic pathways, photodisintegration and so-on, but yes, the models do provide a very precise prediction of the H/He ratio (and that of more minor species) at the end of cosmological nucleosynthesis. There are small uncertainties in the neutron lifetime, the various reaction cross-sections and so-on, but the most important uncertainty is that there is an important "free parameter" - the baryon to photon ratio - that must be fixed. This can be constrained by demanding that one consistent value of this number can explain all the measured primordial abundance ratios (He/H, Li/H, D/H, $^3$ He/H) or it can be found from other cosmological measurements (such as from the cosmic microwave background). The He/H ratio is not very sensitive to this free parameter and hydrogen is always by far the most dominant nucleus for the reasons I explained above.
Current levels of uncertainty on the mass fraction of hydrogen produced in the (standard) Big Bang are significantly smaller than 1% (e.g. Peimbert 2008 ). Coc et al. (2013) used the Planck cosmic microwave background constraints on the baryon-to-photon ratio and estimated a standard Big-Bang He/H mass ratio of $0.2463\pm 0.0003$ . One could consider additions to this standard model - e.g. change the number of neutrino families, have decaying dark matter particles in the early universe and so-on, but it seems hard to change the mass fraction of hydrogen by more than 1% without upsetting the concordance with other observations. A final point to make is that very little of this cosmological material has so far found its way into stars (perhaps 10%), and of that, much of it is still in the same (low-mass) stars that were formed. The amount of "processed" material made of heavier elements, fused from hydrogen in stars, that has enriched the cosmological material is therefore comparatively small - of order 1-2%. So the predominance of hydrogen has hardly decreased since the big bang. This latter property can be used to test the whole model. By looking at the He/H ratio as we go "back in time" we can see if the primordial ratio matches that predicted by the Big-Bang. In practice this can be done by estimating the He/H ratio in the oldest stars or by estimating He/H in the interstellar medium of the most metal-poor galaxies. These measurements are more uncertain than the predictions above, but are in reasonable agreement with them. An example would be Izotov & Thuan (2010) , who estimated a primordial He/H mass ratio of $0.2565 \pm 0.005$ from metal-poor galaxies - about two (small) error bars higher than the prediction above. | {
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418,552 | In a multiple choice question a simple circuit of only a battery and an inductor the length of the inductor is doubled. The question is how does the velocity of the electrons change. Big spoiler alert: the velocity decreases.
My reasoning was that in the equation $V=-\dot{I}L$ the inductance doubles so the change in current must halve in order for the voltage to remain constant. Thereby the current and the velocity of electrons decreases. Is this correct so far? Because the solutions say: inductance increases so the resistance increases and (by Ohm's law I guess) the current decreases. However, I do not understand why the resistance (there is no resistor) should change if the inductance increases? | The short answer is that (i) protons (hydrogen nuclei) are produced abundantly in the early universe, but only a small fraction of these are able to engage in nuclear reactions leading to heavier elements, either during primordial nucleosynthesis or later inside stars. This means that hydrogen ends up being the most abundant element in the universe. (ii) Big Bang nucleosynthesis makes very precise predictions (to $<1$ % precision) about the fraction of baryons that end up as protons vs other nuclei. Further details In the early phases of the big bang only the constituent parts of nucleons (quarks and anti-quarks) plus leptons (e.g. electrons, positrons) and light existed. As the universe expanded and cooled, quarks were able to combine and form the basic building blocks of nuclei - the neutrons and protons. A proton is of course a hydrogen nucleus; any heavier elements needed to be built by fusing together protons. Protons are positively charged and repel each other strongly. In order to fuse protons and make helium requires high energies/temperatures and the intermediate stage of forming deuterons - a proton plus a neutron in a bound pair. The reason that most protons do not fuse in this way is threefold. The small excess mass of the neutron compared to the proton means that the equilibrium reactions that produce nucleons end up producing about six times as many protons as neutrons. Free neutrons then decay into protons on timescales of minutes, which increases this ratio further to about seven. The deuteron is weakly bound, so there is only a limited window of time between when it is too hot to remain bound (prior to 10s after the Big Bang) and when it is too cool to get deuterons to fuse to become helium (beyond about 10 minutes after the Big Bang). Of these things, by far the most important factor is #1 and this is fundamentally the answer to your question. It is because the neutron is more massive than the proton. All heavier nuclei must contain neutrons because the Coulomb repulsion between two or more protons is too strong to form stable nuclei without them. By the end of Big Bang nucleosynthesis, essentially all of the available free neutrons end up in helium nuclei (with traces of deuterons and lithium nuclei) and thus the H/He ratio ends up at around 12 protons for every He nucleus (2 protons, 2 neutrons). Is this quantitatively precise? Well, my figure of 12 protons for every He nucleus, or 75% by mass , is a bit handwaving. A detailed model must take into account in a more accurate way the temperature evolution of the universe, the various (more minor) nucleosynthetic pathways, photodisintegration and so-on, but yes, the models do provide a very precise prediction of the H/He ratio (and that of more minor species) at the end of cosmological nucleosynthesis. There are small uncertainties in the neutron lifetime, the various reaction cross-sections and so-on, but the most important uncertainty is that there is an important "free parameter" - the baryon to photon ratio - that must be fixed. This can be constrained by demanding that one consistent value of this number can explain all the measured primordial abundance ratios (He/H, Li/H, D/H, $^3$ He/H) or it can be found from other cosmological measurements (such as from the cosmic microwave background). The He/H ratio is not very sensitive to this free parameter and hydrogen is always by far the most dominant nucleus for the reasons I explained above.
Current levels of uncertainty on the mass fraction of hydrogen produced in the (standard) Big Bang are significantly smaller than 1% (e.g. Peimbert 2008 ). Coc et al. (2013) used the Planck cosmic microwave background constraints on the baryon-to-photon ratio and estimated a standard Big-Bang He/H mass ratio of $0.2463\pm 0.0003$ . One could consider additions to this standard model - e.g. change the number of neutrino families, have decaying dark matter particles in the early universe and so-on, but it seems hard to change the mass fraction of hydrogen by more than 1% without upsetting the concordance with other observations. A final point to make is that very little of this cosmological material has so far found its way into stars (perhaps 10%), and of that, much of it is still in the same (low-mass) stars that were formed. The amount of "processed" material made of heavier elements, fused from hydrogen in stars, that has enriched the cosmological material is therefore comparatively small - of order 1-2%. So the predominance of hydrogen has hardly decreased since the big bang. This latter property can be used to test the whole model. By looking at the He/H ratio as we go "back in time" we can see if the primordial ratio matches that predicted by the Big-Bang. In practice this can be done by estimating the He/H ratio in the oldest stars or by estimating He/H in the interstellar medium of the most metal-poor galaxies. These measurements are more uncertain than the predictions above, but are in reasonable agreement with them. An example would be Izotov & Thuan (2010) , who estimated a primordial He/H mass ratio of $0.2565 \pm 0.005$ from metal-poor galaxies - about two (small) error bars higher than the prediction above. | {
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419,192 | It is well known that when water flows through a tube you can make it flow faster by making the tube narrow. Now consider what happens when a group of people are moving and the space becomes narrower. The opposite of what happens with water happens here. People start to move slower. I was wondering if there is any fluid that shows this kind of behaviour and what would cause that. | An incompressible (i.e. constant density, like water under most cirumstances) fluid has to satisfy the continuity equation $\nabla V = 0$, where $V$ is the velocity of the fluid. This means that because the same amount of mass per unit of time goes in at one end as goes out the other end and the volume per unit of mass stays constant, the velocity of the fluid has to increase as the cross-sectional area of the tube decreases along the flow direction. A compressible fluid on the other hand can change in density and therefore does not obey the same rules. If you take for example a supersonic gas flow like in a rocket nozzle or a jet fighter exhaust, the fluid will counterintuitively flow slower as the cross-sectional area decreases, and faster as the cross-sectional area of the flow increases. (Table from Introduction to compressible flow by Eric Pardyjak, University of Utah) A classic example is a laval nozzle, where the flow behind the critical cross-section (the narrowest part in the middle) is supersonic and will go faster (note the increasing V in the diagram) as the nozzle gets wider. (image taken from https://commons.wikimedia.org/wiki/File:Nozzle_de_Laval_diagram.png , public domain) | {
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419,200 | Suppose you have a glass of water at room temperature. Now suppose you put the glass of water on a weighting scale. What the quantity actually measures ? I have read that in fluids are basically lots of molecules in random motion and the intermolecular forces is really weak (I don't know if this is accurate or not). So when you put the glass of water on the scale, water particles are going in random motion. So is the weight of the particles which are not touching the surface of the water glass (but are rather freely floating at the time of weighting and at random motion inside the glass) at the time of weighting contributing to the weight of the water glass as measured by the scale? Or is the weight of the glass same as the only the weight of the particles touching the surface of the glass at the time of measuring the weight (I don't think this is true since if you cool down it will become a solid so the weight will be same)? Or is the weight which is measured is same as the water pressure at the bottom of the glass (I think this option is false too since you can create very high pressures at the bottom of the glass with some springs and pascal's theorem but that's not adding much weight ) | An incompressible (i.e. constant density, like water under most cirumstances) fluid has to satisfy the continuity equation $\nabla V = 0$, where $V$ is the velocity of the fluid. This means that because the same amount of mass per unit of time goes in at one end as goes out the other end and the volume per unit of mass stays constant, the velocity of the fluid has to increase as the cross-sectional area of the tube decreases along the flow direction. A compressible fluid on the other hand can change in density and therefore does not obey the same rules. If you take for example a supersonic gas flow like in a rocket nozzle or a jet fighter exhaust, the fluid will counterintuitively flow slower as the cross-sectional area decreases, and faster as the cross-sectional area of the flow increases. (Table from Introduction to compressible flow by Eric Pardyjak, University of Utah) A classic example is a laval nozzle, where the flow behind the critical cross-section (the narrowest part in the middle) is supersonic and will go faster (note the increasing V in the diagram) as the nozzle gets wider. (image taken from https://commons.wikimedia.org/wiki/File:Nozzle_de_Laval_diagram.png , public domain) | {
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419,353 | Note that my question is not why do you tilt your bike when on a curve . It's about the reduction in turning radius when one tilts the bike inwards. Short to-the-point answers are welcome. | It is actually not the tilting that causes more turning . You could in theory turn just as much while being straight up by just turning your steering wheel. But if you do that straight up, you fall. The torque produced by the friction from the turned wheel is unbalanced and will flip you over. When you turn on a bike, your body automatically leans in and tilts slightly - not because that tilting does the turning, but because it keeps your balance. By tilting your bike, you move your centre-of-mass sideways, so that gravity causes a larger and larger torque in you. When this gravity-torque counteracts the friction torque exactly, your turn is stable and you won't fall while turning. So, the tilt doesn't cause sharper turning, but it allows for sharper turning without you falling over. | {
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420,288 | IceCube, XENON, etc, keep yielding negative results. If dark matter exists, it doesn't interact with baryonic matter at the energy ranges they can detect. The response is to build even bigger detectors to search for even fainter energy signatures. Why? Is there evidence that dark matter is supposed to have weak interactions (instead of gravity-only)? Or is it just searching for your keys under the lamp post (i.e. it's the only possibility that we have a way to detect)? | The short answer is that they don't assume that. But among all the proposals that remain for what dark matter might be, weakly interacting stuff is the easiest to detect, 1 so that is what is getting the money right now. 2 And that is not unusual. The history of missing-mass/dark-matter is one of proposals being made and then ruled out one-by-one, in order of ease of accessibility. WIMPs are just the latest candidate to get top-billing. MACHOs were hot when I was in college but were largely disposed of in the nineties and naughties. Before that, decades were spent with ever improving telescopes in wider and wider bands just ruling out many of the ways that ordinary matter could be hiding in plain sight (gas and dust, mostly). And there are additional possible candidates in the theoretical catalog. I think that sterile neutrinos and/or axions will be next up if WIMPs are convincingly ruled out. 1 There is a caveat here in the form of sterile neutrinos which are not "detected" exactly but can be deduced by finding the three-flavor mixing matrix to be non-unitary. This is a hot topic again because MiniBooNe has recently announced an improved analysis of a larger data set in which the low-energy excess remains and the $\theta_{13}$ efforts have paid off in a big way so we've close to being able to quantify the unitarity (or lack thereof) of the matrix with some precision. 2 WIMPs in a particular mass range also offer the possibility of explaining additional features of the universe which makes them attractive for a second reason. | {
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420,291 | For an ideal gas, it is relatively easy to express the Euler equations in primitive form (variables $\rho$, $u$, $p$), starting from their expression in conservative variables ($\rho$, $\rho u$, $E$). I did not find any example of such derivation for a general real gas, governed by any equation of state. Is it possible to express the Euler equations in primitive form for any (unknown) real gas (involving the speed of sound somewhere)? | The short answer is that they don't assume that. But among all the proposals that remain for what dark matter might be, weakly interacting stuff is the easiest to detect, 1 so that is what is getting the money right now. 2 And that is not unusual. The history of missing-mass/dark-matter is one of proposals being made and then ruled out one-by-one, in order of ease of accessibility. WIMPs are just the latest candidate to get top-billing. MACHOs were hot when I was in college but were largely disposed of in the nineties and naughties. Before that, decades were spent with ever improving telescopes in wider and wider bands just ruling out many of the ways that ordinary matter could be hiding in plain sight (gas and dust, mostly). And there are additional possible candidates in the theoretical catalog. I think that sterile neutrinos and/or axions will be next up if WIMPs are convincingly ruled out. 1 There is a caveat here in the form of sterile neutrinos which are not "detected" exactly but can be deduced by finding the three-flavor mixing matrix to be non-unitary. This is a hot topic again because MiniBooNe has recently announced an improved analysis of a larger data set in which the low-energy excess remains and the $\theta_{13}$ efforts have paid off in a big way so we've close to being able to quantify the unitarity (or lack thereof) of the matrix with some precision. 2 WIMPs in a particular mass range also offer the possibility of explaining additional features of the universe which makes them attractive for a second reason. | {
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420,491 | Let's take the very simple problem of what happens if I drop a 1 kg ball from a height of 1 meter. Classically, $F = mg$ and $g \approx 10 \frac{\mathrm{m}}{\mathrm{s}^2}$, so the ball feels a force of $10 \, \mathrm{N}$. From $S = ut + 1/2 gt^2$, we get that the ball hits the ground in about $\sqrt{1/5} \, \mathrm{s}$. No problem. In principle, quantum mechanics should also be able to do this. So we start with the 3-dimensional time-independent Schrodinger equation with a potential of $V = -GM/r$:
$$
-\frac{\hbar}{2m}\nabla^2 \psi - \frac{GM}{r}\psi = E\psi
\,.$$ This is effectively the same equation as for solving the Hydrogen atom: a $1/r$ potential but with different coefficients. So the answer should take the same form, and we get all the energy levels, the n/l/m quantum numbers, etc. But qualitatively, this is clearly different from the classical result. It means the ball has many quasi-stable orbitals for example, and it doesn't give a precise prediction that the ball will hit the ground in $\sqrt{1/5} \, \mathrm{s}$. In the classical problem it's important that the ball was dropped from rest, otherwise the time taken to hit the ground will vary. But the quantum solution doesn't involve the initial conditions at all. What has gone wrong? My first thought was that I need the time-dependent Schrodinger equation instead of the time-independent one, but that doesn't lead anywhere - it just means the ball oscillates between solutions. Am I missing some way to get the classical result from the Schrodinger equation? Related but not quite the same: Is it possible to recover Classical Mechanics from Schrödinger's equation? | This question has been asked many times before. I last answered it here and there's an excellent answer here that explicitly constructs a solution that obeys Newtonian gravity. You're of course correct that if the ball were in a single stationary state, then it wouldn't do anything remotely like falling. It would occupy some standing wave about the Earth and not evolve in time. The reason we never see macroscopic objects in such states is because they are unstable in the same sense as Schrodinger's cat. Your proposed stationary states would delocalize the ball over the entire Earth, over a scale of thousands of miles. But if you just look at the ball, you can measure where it is, collapsing this superposition. Or, if you don't like the collapse phrasing, the interaction of any stray photon with the ball will entangle the ball's position with the photon's state. This decoheres the enormous position superposition you've constructed and gives the ball a somewhat definite position to the perspective of any external observer. That is to say, after a generic interaction with the environment the ball will quickly end up with its position peaked about a narrow value. (Not necessarily so narrow that the uncertainty principle comes into play, but effectively zero for all macroscopic purposes.) Such states exist in the Hilbert space, as complicated linear combinations of the energy eigenstates. To see that the peak in the wavefunction obeys Newton's laws, you can appeal to Ehrenfest's theorem,
$$m \frac{d^2 \langle x \rangle}{dt} = - \left\langle \frac{dV}{dx} \right\rangle$$
which immediately gives that result. You may still be troubled, because in classical mechanics we need to specify an initial position and initial velocity, while in quantum mechanics it seems we only need to specify the analogue of position. But you can't forget the wavefunction is complex. The "velocity" of a particle is encoded by how fast the phase winds around in position. For example, for the free particle, a constant wavefunction would give a totally stationary particle, while $e^{ikx}$ would give a moving particle. Since the ball is heavy, this property is stable under interaction with the environment, like the position. | {
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420,542 | XKCD usually has solid (and often contemporary) science behind it. Lightning Difference, #2027 one says: Q: What’s that trick for telling how many miles away lightning is? A: Just count the seconds between the visible flash and the radio wave burst, then multiply by 5 billion. Usually it's lightning versus thunder, and you divide the time by 5 (or thereabouts) to get the distance in miles. Here though, light time for 1 mile (about 1600 meters) would be about 5.3E-06 seconds, and if the difference between the visible light flash and the radio burst were one five-billionths of a second (2E-10 seconds), that suggests a velocity difference of about 38 ppm. What is the physics behind that 38 ppm difference? | I think it's fair to say that explainxkcd.com is the authoritative source for questions regarding xkcd. In this case, a detailed discussion (including formulas) is taking place on the page for xkcd 2027 . Here's a quote from its current text: According to Wikipedia and other sources ,
refractive index of air at 0°C is about 1.000277, which equates to
a speed of light around 299709.4 km/s (186230.8 miles/s). According to this paper ,
refractive index for radio waves in similar conditions is 1.000315,
which equates to a speed around 299698.1 km/s (186223.7 miles/s). This
means that to get the distance, the time difference in seconds between
visible flash and radio burst should be multiplied by about 4.9
billion for miles, or about 7.9 billion for kilometers. More details
for the calculations are in the comments below . As for why radio waves are slower in air than visible light - I don't know, and I didn't find any useful sources, but I guess it's because even in the troposphere some molecules are ionized, and the free electrons affect radio waves much more than waves of higher frequencies. What I read about the ionosphere and dispersion due to free electrons in the interstellar medium seems to support that idea. But it's just a guess - I may be completely wrong. | {
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420,552 | From Wikipedia : The SI base units and their physical quantities are the metre for measurement of length, the kilogram for mass, the second for time, the ampere for electric current, the kelvin for temperature, the candela for luminous intensity, and the mole for amount of substance. As far as I know, a base is a unit which cannot be broken down into units other than those from the above mentioned units. However, degrees (for angles) and 8-bit bytes (for digital data) cannot be expressed using one or more of these base units. So, why are these two units not considered base units? | The radian (not the degree) is the SI unit of angle, and it's defined in terms of lengths: it is that angle for which the length of a circular arc subtending that angle is equal to the radius of the circle. Since this definition refers to the relative ratio of two lengths, the SI considers it to be a "dimensionless derived unit", rather than a base unit. 1 As far as bytes go: Defining a unit amounts to specifying a certain amount of a quantity that we call "one unit". Physical quantities such as mass, length, time, etc., are (effectively) continuous quantities, and so there is no "natural" unit for us to use. We therefore have to make an arbitrary choice about how much of each quantity is equal to one unit. Digital information, on the other hand, is inherently discrete. All methods of quantifying data simply amount to counting bits; and you don't need to make an arbitrary choice of unit if you can simply count a quantity. There is therefore no need to define a unit for digital information, because there is already a natural unit (the bit). It's important to note that not every measurable quantity is inherently definable in terms of SI base units. If I count the number of people in my office building right now, and tell you that there are "12 people" in the building right now, then "people" is not expressible in terms of meters, kilograms, and seconds. But I don't need to worry that you're going to come along and use some different unit to count the people in this building, because a natural unit (1 person) exists. It's only when we are measuring a quantity that can take on any real-numbered value (e.g., the mass of all the people in this building) that it becomes important to define a unit; otherwise, you and I have no basis for comparison. Any system of units is essentially a set of these arbitrary choices; "natural" units of quantities that are inherently discrete are unnecessary simply because they're understood to be the obvious choice. 1 It's worth noting that the radian was officially a "supplementary unit" in SI until 1995, when they were reclassified as "dimensionless derived units". A bit of the discussion surrounding this change can be found on p. 210 of the Proceedings of the 20th Conférence Générale des Poids et Mesures (warning: large PDF). Reading between the lines, I suspect that the name "dimensionless derived unit" was something of a compromise between those who thought it should be thought of as a derived unit and those who didn't think it should be thought of as a unit at all; but I wouldn't want to speculate further than that. | {
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420,570 | A leaf is denser than water, but floats on top of it because of surface tension. But how to calculate the force holding up the leaf? Surface tension has units of force per length , not area, but clearly the area of the leaf matters. It's not just the edges of the leaf that are carrying the weight. The standard calculations with water surface tension concerns e.g. a needle floating on water, and the calculation proceeds by considering the surface tension along the two lines of the water-needle interface. If we extend this treatment to a disc, we would calculate the force as the water's surface tension times the circumference of the disc (neglecting the effect of the small contact angle). But this seems wrong to me - intuitively, it seems that the disc/leaf would experience a pressure from the surface, so that the force pushing it up against gravity would be proportional to the area, not the circumference. | The radian (not the degree) is the SI unit of angle, and it's defined in terms of lengths: it is that angle for which the length of a circular arc subtending that angle is equal to the radius of the circle. Since this definition refers to the relative ratio of two lengths, the SI considers it to be a "dimensionless derived unit", rather than a base unit. 1 As far as bytes go: Defining a unit amounts to specifying a certain amount of a quantity that we call "one unit". Physical quantities such as mass, length, time, etc., are (effectively) continuous quantities, and so there is no "natural" unit for us to use. We therefore have to make an arbitrary choice about how much of each quantity is equal to one unit. Digital information, on the other hand, is inherently discrete. All methods of quantifying data simply amount to counting bits; and you don't need to make an arbitrary choice of unit if you can simply count a quantity. There is therefore no need to define a unit for digital information, because there is already a natural unit (the bit). It's important to note that not every measurable quantity is inherently definable in terms of SI base units. If I count the number of people in my office building right now, and tell you that there are "12 people" in the building right now, then "people" is not expressible in terms of meters, kilograms, and seconds. But I don't need to worry that you're going to come along and use some different unit to count the people in this building, because a natural unit (1 person) exists. It's only when we are measuring a quantity that can take on any real-numbered value (e.g., the mass of all the people in this building) that it becomes important to define a unit; otherwise, you and I have no basis for comparison. Any system of units is essentially a set of these arbitrary choices; "natural" units of quantities that are inherently discrete are unnecessary simply because they're understood to be the obvious choice. 1 It's worth noting that the radian was officially a "supplementary unit" in SI until 1995, when they were reclassified as "dimensionless derived units". A bit of the discussion surrounding this change can be found on p. 210 of the Proceedings of the 20th Conférence Générale des Poids et Mesures (warning: large PDF). Reading between the lines, I suspect that the name "dimensionless derived unit" was something of a compromise between those who thought it should be thought of as a derived unit and those who didn't think it should be thought of as a unit at all; but I wouldn't want to speculate further than that. | {
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420,576 | I am a beginning physics student. I have a question (probably very simple for you all haha) about forces on an overhanging book, off a table. We all know that the max you can push it would be half the length of the book. The pivot of the book would also be the edge of the table. If the book were to be pushed to it's max, what forces would be enacted on the book? What forces would prevent it from falling/swiveling about the pivot? It would be best if you could provide a diagram, but an explanation is fine. Thanks! | The radian (not the degree) is the SI unit of angle, and it's defined in terms of lengths: it is that angle for which the length of a circular arc subtending that angle is equal to the radius of the circle. Since this definition refers to the relative ratio of two lengths, the SI considers it to be a "dimensionless derived unit", rather than a base unit. 1 As far as bytes go: Defining a unit amounts to specifying a certain amount of a quantity that we call "one unit". Physical quantities such as mass, length, time, etc., are (effectively) continuous quantities, and so there is no "natural" unit for us to use. We therefore have to make an arbitrary choice about how much of each quantity is equal to one unit. Digital information, on the other hand, is inherently discrete. All methods of quantifying data simply amount to counting bits; and you don't need to make an arbitrary choice of unit if you can simply count a quantity. There is therefore no need to define a unit for digital information, because there is already a natural unit (the bit). It's important to note that not every measurable quantity is inherently definable in terms of SI base units. If I count the number of people in my office building right now, and tell you that there are "12 people" in the building right now, then "people" is not expressible in terms of meters, kilograms, and seconds. But I don't need to worry that you're going to come along and use some different unit to count the people in this building, because a natural unit (1 person) exists. It's only when we are measuring a quantity that can take on any real-numbered value (e.g., the mass of all the people in this building) that it becomes important to define a unit; otherwise, you and I have no basis for comparison. Any system of units is essentially a set of these arbitrary choices; "natural" units of quantities that are inherently discrete are unnecessary simply because they're understood to be the obvious choice. 1 It's worth noting that the radian was officially a "supplementary unit" in SI until 1995, when they were reclassified as "dimensionless derived units". A bit of the discussion surrounding this change can be found on p. 210 of the Proceedings of the 20th Conférence Générale des Poids et Mesures (warning: large PDF). Reading between the lines, I suspect that the name "dimensionless derived unit" was something of a compromise between those who thought it should be thought of as a derived unit and those who didn't think it should be thought of as a unit at all; but I wouldn't want to speculate further than that. | {
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421,058 | Up to a certain time, I was told photons a.k.a. light was just a wave of energy. Then I was told, no, light is actually a particle. And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? (An electron would also be unstable by absorbing the energy and thus it re-emits it but in the first place why does it absorb it?) *Note:- A similar question was asked earlier ( How does an electron absorb or emit light? ) but my question is not the same. The earlier asked question was how does it happen and I ask why does it happen. | And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? There is a basic misunderstanding in your question. An electron is an elementary particle of fixed mass. It can scatter off a photon, (which is also an elementary particle); if accelerated it can emit a photon, but it does not absorb it, because the electron's mass is fixed, and if it were able to absorb a photon - at the electron's center of mass - the mass would have to change, which contradicts observations and special relativity for elementary particles. The terms absorption and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom when the atom absorbs a photon . So it is not the electron that absorbs the photon, but the atom. The atom has energy levels , and if the photon energy coincides (within a small $ΔE$ , the width of the energy level) with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon (not the electron). So the answer to "why" , above, is "because the photon has the appropriate energy to transfer the electron to an empty energy level" . If the photon energy does not coincide with transition energy of the atom, the photon may scatter with the spillover electric fields of the atom or molecule either elastically, or transferring energy and a lower energy photon continues on its way. The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can. | {
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421,063 | Object B is on top of object A and there is no friction between the two. A horizontal force acts on object A, will the force also act on object B? My intuition is that the force only acts on object A , and thus only object A will accelerate in the horizontal direction, and object B will remain in it's original position (not accelerated). However, when approaching this question "show that tanθ=F/W where W is the combined weight of object A and B" we can treat the two objects as one and the force thus acts on both objects. The block B does not move relative the wedge. (Friction between the wedge and block or between the wedge and the ground is negligible). Can someone explain when can two objects be treated as one? In the question I proposed, I thought the two objects can't be treated as one when the force only acts on object A. Thanks for any help. | And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? There is a basic misunderstanding in your question. An electron is an elementary particle of fixed mass. It can scatter off a photon, (which is also an elementary particle); if accelerated it can emit a photon, but it does not absorb it, because the electron's mass is fixed, and if it were able to absorb a photon - at the electron's center of mass - the mass would have to change, which contradicts observations and special relativity for elementary particles. The terms absorption and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom when the atom absorbs a photon . So it is not the electron that absorbs the photon, but the atom. The atom has energy levels , and if the photon energy coincides (within a small $ΔE$ , the width of the energy level) with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon (not the electron). So the answer to "why" , above, is "because the photon has the appropriate energy to transfer the electron to an empty energy level" . If the photon energy does not coincide with transition energy of the atom, the photon may scatter with the spillover electric fields of the atom or molecule either elastically, or transferring energy and a lower energy photon continues on its way. The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can. | {
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421,895 | Let's say a scientist attaches a 1 kg brick to a large helium inflated balloon, lets the balloon go, and then it reaches an altitude of 10 000 meters before it pops, dropping the brick. The brick falls and hits the ground with with a kinetic energy of approximately 100 000 joules. (Actually a bit less, it gives some of that energy as air resistance, but it still stored that much energy.) For reference, a rifle shot is about 2 000 joules. But where did this energy come from? The scientist just inflated a balloon and tied a string. | Your estimate that the brick would lose "a bit" of its energy to air resistance is incorrect. It would lose most of its energy. The terminal velocity of a brick ( http://physicsbuzz.physicscentral.com/2018/01/ask-physicist-which-falls-faster-brick.html ) is reported to be around 95 m/s. This will be its speed when it hits the ground, so the energy it will deposit into the ground will be 4,500 J, not 100,000 J. Assuming your math is correct, over 95 percent of its energy is lost to air resistance. When the balloon rises, what's making it rise is the atmosphere pushing up on it. The gain in potential energy of the brick is equal to the loss in energy of the atmosphere; as the balloon rises, the atmosphere, filling in the space below it, becomes, on average, very, very, slightly lower to the ground. So the energy comes from the gravitational potential energy of the atmosphere*. *In principle, the thermal energy of the atmosphere also has to be taken into account, since temperature decreases quickly with altitude in the troposphere. This means that the original gain in gravitational potential energy of the atmosphere also slightly decreased its thermal energy, while the loss of gravitational potential energy of the atmosphere is countered somewhat by a gain in thermal energy. These energy transfers are likely small compared to the changes in gravitational potential energy, though (and even if they weren't, the ability to do work using the temperature gradient is hampered by thermodynamic considerations). | {
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421,904 | Suppose a cylinder is slipping rigidly on a frictionless horizontal surface. Then, at $t=0$ it reaches a different ground. The coefficient of friction between the cylinder and this new type of ground is $\mu$. What happens?
- assuming usual high-school friction to be the only horizontal force and a homogeneous cylinder with moment of inertia $I=MR^2/2$. I have come up with this question myself, it is not homework. Once the cylinder starts slipping on the new ground, kinetic friction will dissipate energy, slowing it down. It will also produce torque and make it roll. Eventually, the velocity of the contact point will vanish and the cylinder will start rolling without slipping (RWS). This happens when $v=\omega R$ where $v$ is the velocity of the center of mass and $\omega$ is the angular velocity with respect to the center of mass. Equations of motion are $FR=I\dot{\omega}$ and $F=M\dot{v}$. With $v(0)=v_0$ and $\omega(0)=0$ this leads to the conclusion - using that friction is $F=\mu mg$ and if I made no mistake - that RWS will start at $t=v_0/(3\mu g)$. Now here comes the actual question. Once the cylinder is RWS, the friction force must actually be zero (otherwise, it would slow down the cylinder without performing work, which is impossible). However, the friction force is not zero immediately before RWS (it is $\mu mg$). Hence the question: is the friction force dicontinuous? Or have I made some mistake? | Your estimate that the brick would lose "a bit" of its energy to air resistance is incorrect. It would lose most of its energy. The terminal velocity of a brick ( http://physicsbuzz.physicscentral.com/2018/01/ask-physicist-which-falls-faster-brick.html ) is reported to be around 95 m/s. This will be its speed when it hits the ground, so the energy it will deposit into the ground will be 4,500 J, not 100,000 J. Assuming your math is correct, over 95 percent of its energy is lost to air resistance. When the balloon rises, what's making it rise is the atmosphere pushing up on it. The gain in potential energy of the brick is equal to the loss in energy of the atmosphere; as the balloon rises, the atmosphere, filling in the space below it, becomes, on average, very, very, slightly lower to the ground. So the energy comes from the gravitational potential energy of the atmosphere*. *In principle, the thermal energy of the atmosphere also has to be taken into account, since temperature decreases quickly with altitude in the troposphere. This means that the original gain in gravitational potential energy of the atmosphere also slightly decreased its thermal energy, while the loss of gravitational potential energy of the atmosphere is countered somewhat by a gain in thermal energy. These energy transfers are likely small compared to the changes in gravitational potential energy, though (and even if they weren't, the ability to do work using the temperature gradient is hampered by thermodynamic considerations). | {
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421,919 | Being a physics grad student, I got used to the weird concepts behind quantum mechanics ( used to doesn't mean I fully understand it though). What I mean is that I'm not surprised anymore by the fact that a quantum system might be in a superposition state, that a particle in a potential well has a discrete spectrum of energy, that some pairs of observables cannot both be measured with arbitrary precision, etc. However, the problem arises when I'm talking about my work to my (non-physicists) friends and family. They are not used to think "quantum mechanically", so when I'm saying that Schrödinger's cat is really, simultaneously, both "alive" and "dead", they don't get my point as it just seems crazy. So I have been thinking for a while of what would be the easiest way to introduce them to the core principles of QM. ( EDIT: This part seems to have upset people; I don't want to go into the details of distinguishing "simultaneously alive and dead until you measure" VS "superposition of alive and dead states", I mainly want them to understand that the outcome of the measure is indeterminate ) When they ask me "So what is QM all about?" , I'd like to give
them a short answer, a simple postulate/principle that, once accepted,
makes most key features of QM seem more natural. In my opinion, some of the "weird" ideas people find most difficult to accept (due to some popular pseudoscience) and/or that are central to QM are: Quantum superposition Complex amplitude/phase (as it leads to interference) Uncertainty principle Entanglement Quantization of physical quantities/particles My problem is that I can't explain how these ideas are related . If my friends accept that particles can be in two states at once, why couldn't they know both its position and momentum? Or if they accept that an electron has a super abstract property called a "complex phase", why should it lead to a discrete spectrum of energy? Of course, I could just tell them that they need to accept it all at once, but then the theory as a whole becomes hard to swallow. In my opinion, QM sometimes suffers from a bad reputation exactly because it requires to put all our classical intuitions to scrap. While actually, I think it could make perfect sense even for a non-physicist given only 1 or 2 "postulates" above. I tried looking back at my own learning of QM with little success since undergrad students are often precisely bombarded with poorly justified postulates until they just stop questioning them. A few more points: I don't want to get started on mathematical definitions, so nothing like "Quantum systems are in an infinite-dimensional Hilbert space" as it doesn't mean anything to my relatives. That includes pretty much all of the actual postulates of QM. I want my explanation to be more intuitive. I'd like to avoid any mumbo jumbo related to "wave-particle duality". I think (and I know I'm not the only one ) that most explanations regarding this are inacurrate and actually inefficient at giving a good description of quantum behaviour, since most non-physicists aren't familiar with the properties of a classical wave anyway. And it's mostly used when talking about the position of a quantum particle; I've never heard someone say that spin "behaves like a wave" or "like a particle". I'm looking for something more fundamental, more universal. That being said, I think it's very important to include a discussion about the relative phase between two superpositioned states as it leads to quantum interference (perhaps the only "wave-like" feature that applies throughout QM). If there were only quantum superposition, quantum computing would be pretty much useless: even if we perform one billion calculations simultaneously, when measuring the system we would get a single random value. Our goal is to manipulate the system, to create destructive interference between the unwanted results in order to increase the probability of measuring the desired value. And that requires the notion of quantum phase. Uncertainty principle might be the toughest challenge here, since we can get the basic idea without invoking QM at all. First, even in classical mechanics, it's impossible to measure something with an arbitrarily big precision simply because of instrumental limitations. Second of all, it makes sense to get different results if we measure the position before or after the momentum ($[x,p]\neq 0$) since we interact with the system when making the first observation, leaving it in a different state afterward. How could I explain that there are more fundamental implications from QM? I wasn't sure about entanglement and quantization. The weirdness of entanglement doesn't come from how we achieve it, but from how we interpret it, which is still an unresolved issue as far as I know (Many-world? Copenhagen interpretation? etc). As for the quantization of particle properties, it doesn't seem like an intrinsic feature of QM, more of a consequence of boundary conditions. A free electron in a vaccum has a continuous energy spectrum, it's only when we put it in a potential well that some states are "forbidden" (not exactly forbidden of course; even Bohr explained the hydrogen atom in a semi-classical model using destructive interference between the state of the electron). Then, there is also the quantization of matter itself (photons, electrons, quarks, etc.) which is another story. But I think most people are familiar with the idea that matter is made of atoms, so it isn't a far stretch to include light as well. My point is that these two principles don't appear quite as essential as the other ones to understand QM. I haven't found a satisfying answer in the similar posts of this forum ( here , here or here for instance) since I'm more interested in the overall description of QM - not just of wavefunctions or superposition - and I don't want to get into the motivations behind the theory. Also, I'd like to give an accurate, meaningful answer, not one that leads directly to pseudoscientific interpretations of QM (" Particles are waves, and vice versa ", " Everything not forbidden is mandatory ", " Consciousness affects reality "; that kind of thing). EDIT: Just to clarify my question perhaps, here what I would say concerning special relativity (SR). The basic assumption of SR is that everyone always sees light traveling at the same speed, no matter what circumstances. So far so good. Now imagine that I'm in a train going at 50 km/h and that I throw a ball in front of me at 10 km/h. From my perspective, the ball is going at 10 km/h. But for you, standing on the ground, the ball is going at 50+10 = 60 km/h with respect to you. It all makes perfect sense since you have to consider the velocity of the train. Good. Now, let's replace the ball with a photon. Oh oh, according to our previous assumption, we will both see the photon travelling at the same speed, even though I am moving at 50 km/h. We cannot add my speed to your measurement as we did earlier. This implies that we both perceive time and space differently . Boom. And from this you can get to pretty much all of SR (of course, one should also mention that $c$ is the maximum speed). This simple assumption implies the core of the theory. What would be the equivalent in QM, the little assumption that, when accepted, implies most of the theory? @Emilio Pisanty I saw the post in your comment (it's one of the links at the end of mine). As I said, I'm not interested in why we need QM, I want to know/explain what fundamental assumption is necessary to build the theory. However it's true both questions are linked in some ways I guess. EDIT 2: Again, I'm not interested in WHY we need QM. It's pretty obvious we'll turn to the theory that best describes our world. I want to explain WHAT is QM, basically its weird main features. But QM has a lot of weirdness, so if I simply describe the above phenomena one by one, it doesn't reflect the essence of the theory as a whole, it just looks like a "patchwork" theory where everyone added its magical part. And I KNOW all about the QM postulates, but those are more a convention of how to do QM, a mathematical formalism, (we'll use hermitian matrix for observables, we'll use wavefunction on a complex Hilbert space to represent a particle, and so on) rather than a description of what QM is about, what happens in the "quantum world", what phenomena it correctly predicts, no matter how counterintuitive to our classical minds. EDIT 3: Thank you all for your contributions. I haven't chosen an answer yet, I'm curious to see if someone else will give it a try. There's a little point I want to stress: many answers below give a good description of the weird concepts of QM, but my goal is to wrap it all up as much as possible, to give an explanation where most of these are connected with as little assumptions as possible. So far @knzhou's answer is the closest to what I'm looking for, but I'm not entirely satisfied (even though there's probably no answer good enough). Most answers are very close to the wave-particle duality - which I'm not a fan of because people understand this as "Electrons are particles on week days and waves during the weekend". However perhaps it's unavoidable as it truly does include most of the phenomena listed above. Then I thought perhaps if we were talking about quantum fields rather than "waves of matter", the discussion would be less likely to lead to misinterpretation? Also, for all those who said "Well QM is just incomprehensible for lay people" I think that's giving up too easily and it feels a little dishonest. QM is not a secret society for a special elite. I think it's important that researchers explain what is their goal and how they intend to achieve it to the general public. I think the 2nd question physicists are asked most often is "Why should my tax money pay for your work?". Do you really think an answer such as "Well... you couldn't understand it. Just trust me." is satisfying? QM is counterintuitive, but well explained it can make sense, and I'm just looking for the easiest way to make people realize this without simply saying "Well forget anything you know". Also I think it is a theory that would be much easier to accept if people were more often exposed to these ideas. After all, children and high school students learn that the world is made of atoms and they accept it even though they cannot see atoms. Why then couldn't QM be part of the common scientific culture? It's true it may sound ridiculous at first to hear about superposition and indeterminate states, but it isn't by giving up and by saying "Well you need a degree to understand it" that we are going to change it and make it more appreciable. | You're asking a tough question! I try to explain QM to nonphysicists all the time, and it's really hard; here are a few of the accessible explanations I've found. What I'll say below is not logically airtight and is even circular. This is actually a good thing: motivation for a physical theory must be circular, because the real justification is from experiment. If we could prove a theory right by just talking about it, we wouldn't have to do experiments. Similarly, you can never prove to a QM skeptic that QM must be right, you can only show them how QM is the simplest way to explain the data. Quantization $\rightarrow$ Superposition If I had to pick one thing that led to as much of QM as possible, I'd say quantum particles can exist in a superposition of the configurations that classical particles can have . But this is a profoundly unintuitive statement. It's easier to start with photons, because they correspond classically to a field. It's intuitive that fields can superpose; for example the sound waves of two musicians playing at once superpose, adding onto each other. When a light wave hits a partially transparent mirror, it turns into a superposition of a transmitted and reflected wave, each with half the energy. We know experimentally that light comes in little bullets called photons. So what is the state of a photon after hitting a partially transparent mirror? There are a few possibilities. The photon is split in half, with one half transmitted. This is wrong, because we observe all photons to have energy $E = hf$ , never $E = hf/2$ . The photon goes one way or the other. This is wrong, because we can recombine the two beams into one with a second partially transparent mirror (the would-be second beam destructively interferes). This could not happen if the photon merely bounced probabilistically; you would always get two beams. The photon goes one way or the other, but it interferes with other photons somehow. This is wrong because the effect above persists even with one photon in the apparatus at a time. The photon is something else entirely: in a superposition of the two states. It's like how you can have a superposition of waves, and it is not a logical "AND" or "OR". Possibility (4) is the simplest that explains the data. That is, experiment tells us that we should allow particles to be in superpositions of states which are classically incompatible. Then you can extend the same reasoning to "matter" particles like electrons, by the logic that everything in the universe should play by the same rules. (Of course such arguments are much easier in hindsight and the real justification is decades of experimental confirmation.) Superposition $\rightarrow$ Probability Allowing superposition quickly brings in probability. Suppose we measure the position of the photon after it hits one of these partially transparent mirrors. Its state is a superposition of the two possibilities, but you only ever see one or the other -- so which you see must be probabilistically determined. This is not a proof. It just shows that probabilistic measurement is the simplest way to explain what's going on. You can get rid of the probability with alternative formulations of QM, where the photon has an extra tag on it called a hidden variable telling it where it "really" is, but you really have to work at it. Such formulations are universally more complicated. Superposition $\rightarrow$ Entanglement This is an easy one. Just consider two particles, which can each have either spin up $|\uparrow \rangle$ or spin down $|\downarrow\rangle$ classically. Then the joint state of the particles, classically, is $|\uparrow \uparrow \rangle$ , $|\uparrow \downarrow \rangle$ , $|\downarrow \uparrow \rangle$ , or $|\downarrow \downarrow \rangle$ . By the superposition principle, the quantum state may be a superposition of these four states. But this immediately allows entanglement; for instance the state $$|\uparrow \uparrow \rangle + |\downarrow \downarrow \rangle$$ is entangled. The state of each individual particle is not defined, yet measurements of the two are correlated. Complex Numbers This is easier if you're speaking to an engineer. When we deal with classical waves, it's very convenient to "complexify", turning $\cos(\omega t)$ to $e^{i \omega t}$ , and using tools like the complex Fourier transform. Both here and in QM, the complex phase is just a "clock" that keeps track of the wave's phase. It's trivial to rewrite QM without complex numbers, by just expanding all of them as two real numbers, as argued here . Interference does not require complex numbers, but it's most conveniently expressed with them. Wave Mechanics The second crucial fact about quantum mechanics is momentum is the gradient of phase , which by special relativity means that energy is the rate of change of phase . This can be motivated by classical mechanics, appearing explicitly in the Hamilton-Jacobi equation, but I don't know how to motivate it without any math. In any case, these give you the de Broglie relations $$E = \hbar \omega, \quad p = \hbar k$$ which are all we'll need below. This postulate plus the superposition principle gives all of basic QM. Superposition $\rightarrow$ Uncertainty The uncertainty principle arises because some sets of questions can't have definite answers at once. This arises even classically. For instance, the questions "are you moving north or east" and "are you moving northeast or southeast" do not have definite answers. If you were moving northeast, then your velocity is a superposition of north and east, so the first question doesn't have a definite answer. I go into more detail about this here . The Heisenberg uncertainty principle is the specific application of this to position and momentum, and it follows because states with definite position (i.e. those that answer the question "are you here or there") are not states of definite momentum ("are you moving left or right"). As we said above, momentum is associated with the gradient of the wavefunction's phase; when a particle moves, it "corkscrews" along the direction of its phase like a rotating barber pole. So the position-definite states look like spikes, while the momentum-definite states look like infinite corkscrews. You simply can't be both. Long ago, this was understood using the idea that "measurement disturbs the system". The point is that in classical mechanics, you can always make harmless measurements by measuring more gently. But in quantum mechanics, there really is a minimum scale; you can't measure with light "more gently" because you can't get light less intense than single photons. Then you can show that a position measurement with a photon inevitably changes the momentum by enough to preserve the Heisenberg uncertainty principle. However, I don't like this argument because the uncertainty is really inherent to the states themselves, not to the way we measure them. As long as QM holds, it cannot be improved by better measurement technology. Superposition $\rightarrow$ Quantization Quantization is easy to understand for classical sound waves: a plucked string can only make discrete frequencies. It is not a property of QM, but rather a property of all confined waves. If you accept that the quantum state is a superposition of classical position states and is hence described by a wavefunction $\psi(\mathbf{x})$ , and that this wavefunction obeys a reasonable equation, then it's inevitable that you get discrete "frequencies" for the same reason: you need to fit an integer number of oscillations on your string (or around the atom, etc.). This yields discrete energies by $E = \hbar \omega$ . From here follows the quantization of atomic energy levels, the lack of quantization of energy levels for a free particle, as well as the quantization of particle number in quantum field theory that allow us to talk about particles at all. That takes us full circle. | {
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421,957 | From Wikipedia: To answer this question: "What is the shape of a chain suspended at both ends?" we can use the variational principle that the shape must minimize the gravitational potential energy. The gravitational potential energy is highest at the ends. Then why isn't the chain shaped like a V? | The V shape makes sense for an ideal chain with all of its mass concentrated at the midpoint (with the rest of the chain being massless). But now consider a chain with its mass distributed over 3 points, e.g. equal masses at the quarter-way points and the midpoint. The quarter point masses will pull the V out of shape, introducing their own bends into the chain. These new bends must change the position of the midpoint mass, and by symmetry the movement must be vertical. Before we added the new masses the midpoint mass was at the lowest possible position, so adding the quarter point masses must lift the midpoint mass. This is an easy experiment to perform, using some light thread and a few nuts (the metal ones, not the edible kind :) ). Now subdivide the chain into 8 sections, add nuts to the midpoints of the new sections, and we'll get new bends. Repeat the process indefinitely, and we get the catenary. Here are the first few steps of this process, calculated using Python. All of the chains in this diagram have a length of 400 units, the background squares are 10x10 units. You can see the original SVG version of this diagram on GitHub . The diagram below shows the forces acting on each of the masses. It also shows the vector sums for the points on the right side of the chain. The tension forces on each segment of the chain are directed along the chain. They must be equal and opposite, and the forces at each mass must sum to zero. By symmetry, the forces on the left side of the chain must mirror those on the right side. Note that the shape of the chain is unchanged if we scale all the forces by the same amount. The SVG version of this diagram is here . For $i \ge 0$, let $(x_i, y_i)$ be the force vector pointing upwards to the right from mass $i$. Let the weight of each mass be 2 units, so the weight vector at each point is (0, -2). At mass #$0$ we have
$$(-x_0, y_0) + (x_0, y_0) = (0, 2)$$
So $y_0 = 1$ At mass #$1$ we have
$$(-x_0, -y_0) + (x_1, y_1) = (0, 2)$$
So $x_1 = x_0$ and $y_1 = 2 + y_0 = 3$ For $i \gt 0$, at mass #$i$ we have
$$(-x_{i-1}, -y_{i-1}) + (x_i, y_i) = (0, 2)$$
So $x_i = x_{i-1}$ and $y_i = 2 + y_{i-1}$ Thus the $x$ components of each vector $(x_i, y_i)$ on the right side of the chain are identical, and the $y$ components form the arithmetic progression 1, 3, 5, 7... Let $r_i = \sqrt{x_i^2 + y_i^2}$, and let $s$ be the length of a chain segment. To calculate the coordinates of each mass first choose a location $P_0$ for mass #$0$.
Then $P_{i+1} = P_i + (x_i, y_i)s/r_i$ | {
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422,570 | I thought a vector field was a function $\vec{A}:\mathbb{R}^n\to\mathbb{R}^n$, which takes a vector to a vector. This seemed intuitive to me, but in a mathematical physics course I found the definition
$$X=\sum\limits_{i=1}^n X^i \frac{\partial}{\partial x^i},$$
where $X$ seems to take functions from $C^\infty$ as argument and $X^i:\mathbb{R}^n\to\mathbb{R}$. I don't understand this new way of talking about vector fields, especially when our professor said we should see the $\frac{\partial}{\partial x^i}$ as basis vectors of the tangent space. I also can't really translate any vector field which I know into this new definition. So it would be great if someone could maybe make an example of an $\vec{E}$-field or something similar and "translate" it into this new way of expressing vector fields. I'm also struggling to visualize the new definition. It seems that $X(f):\mathbb{R}^n\to \mathbb{R}$ (for $f\in C^\infty$), to which I really can't assign any physical meaning. TL;DR: Can someone explain to me why someone would want to express a vector field in this way? Ideally I'd like to see an example of the same vector field expressed in the two forms and an explanation how one would interpret the visualization of the new vector field. | The motivation goes like this. When we define things mathematically, we want to use as few separate objects as possible. We don't want to define a new object independently if it can be defined in terms of existing things. Suppose a particle moves so that when it is at position $\mathbf{r}$ , its velocity is $\mathbf{v}(\mathbf{r})$ , where $\mathbf{v}$ is a vector field. Then if there is some function $f(\mathbf{r})$ , then the particle sees $$\frac{df}{dt} = v^i \frac{\partial f}{\partial x^i}$$ by the chain rule. That is, if we interpret a vector field as a velocity field, and give it a function $f$ , then we can compute another function $df/dt$ , which is the rate of change of $f$ seen by a particle following the flow of the vector field, as if it were a velocity field. By glancing at the chain rule, you see that if you know $df/dt$ for every $f$ , then you know what the vector field is. Hence, when we work in the more general setting of a manifold, where it's not immediately clear how to define a vector field in the usual way ("an arrow at every point"), we can use this in reverse to define what a vector field is. That is, a vector field $v$ is a map of functions $f \mapsto v(f) = df/dt$ which obeys certain properties. Note that not every vector field should be physically regarded as a velocity field. We're just making mathematical definitions here. The definitions are chosen to make the formalism as clean and simple as possible, possibly at the expense of intuition. Translating between the two is very simple. For example, the field $$\mathbf{E}(\mathbf{r}) = x \hat{i} + xy \hat{j}$$ translates to $$\mathbf{E}(\mathbf{r}) = x \frac{\partial}{\partial x} + x y \frac{\partial}{\partial y}.$$ That is, whenever you see $\partial/\partial x$ you can just imagine it as a unit vector pointing in the $x$ direction. The intuition is the same, because the two definitions obey the same properties. There are many examples of this in mathematics. For example, you might think $\log(x)$ is defined as "the number of times you have to multiple by $e$ to get to $x$ ", but it's unclear how to rigorously define that. Then, through semi-rigorous manipulation you can show that $$\log(x) = \int_1^x \frac{dx'}{x'}.$$ Now the mathematician sees this and chooses to define $\log(x)$ as this integral. This is simpler, because it automatically works for any real $x$ and uses only the notion of an integral, which we already know. Then one can derive the "intuitive" properties of the logarithm, such as $\log(xy) = \log(x) + \log(y)$ . By using a less intuitive definition, the formalism becomes simpler. And once you show that this definition is equivalent to the intuitive one, you are "allowed" to just keep on using the same intuition you started with, so you get the best of both worlds! Edit: the OP asks for examples where this definition of a vector field is more practically useful. I can think of two off the top of my head. First, how do basis vectors transform when you change coordinates from $x^i$ to $y^j$ ? In the usual formalism you may have to memorize a formula, but with the derivatives it follows from the chain rule, $$\frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j} \equiv J^j_i \frac{\partial}{\partial y^j}$$ where $J$ is the Jacobian matrix. Next, suppose the vector field actually is a velocity field, and you want to calculate $f(x(t))$ given $x(0)$ , i.e. you want to know where you'll be if you follow the flow for time $t$ . In this formalism, that's a one-liner. It's just $$f(x(t)) = (e^{t v} f(x))|_{x = x(0)}.$$ To prove this, expand the exponential in a Taylor series. Of course, the great thing is that once you prove two formalisms are equivalent, you can use the intuition from either one interchangably, because you know they're both equally valid. So you gain intuition for a few new cases, without losing any intuition you had before. You can always change back and forth, much like the same software program can run on different hardware. | {
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422,591 | just a copy of https://physics.stackexchange.com/q/422286/203741 , maybe I'll get answers there I'm on the way to understand em waves, but I decided, that I need to completely understand a current flowing. Not basics, not in general, I need to understand every trifle. I think, for a lot of people it will be very useful. Post only applies to induced current. I will write, below, how do I understand things, if it wrong - correct me, and answer question during posting. How does current induces According to Faraday law, changing electric field generates changing magnetic field. How, and why does it occurs. Below, you can see fields of charge with $v=0, v=const;a=0,a!=0$ . The first case called electrostatics and describes by Coulomb law . If you confused, how does standing charge, that don't lose its energy, generates force, according to classic physics - well, I'm as well. The next one is "constant magnet field", if you here because of questions too, you probably remember about Lorentz force , and now think - where is there perpendicular components ? Good question, I don't know. The third is electromagnetic wave, it actually has a perpendicular component, you can see it, it means, that the force there directs perpendicular to "previous line", it clear. Why do line curves by the next wave? Well, it could be intuitively understand, if you see the animation(link in the bot). By the way, if we assume that free electrons in metal would be fixed, and you will wave the wire, it will emit a em wave. Considering, last case, if we put a charge in right or left, you understand it will move up-bottom. So, I think it explains in fundamental level the phenomena of induction. Lets go to macroscopic scales. How does current flows Well, there goes all confusing stuff, like voltage, potential difference, E.M.F, etc.. Lets consider the picture above. Okay, the AC wire in the bottom emits an em wave, the force actually directs parallel to wire near. So, now very attentively: electrons are uniquely distributed in wire - there at all, no places with higher or lower number of electrons. Now there occurs the force only in the bottom part of wire , and electrons in the bottom starts to move right or left. Why do electrons in another parts of wire start to move? Well, I think it happens by the chain reaction: Here goes specific questionS: Is AC current in circuits without parallel connections the same in all points of wire(in one time frame)? According to all stuff above, if it is true, explain this picture. Why current is maximum where voltage is minimum? Animation | The motivation goes like this. When we define things mathematically, we want to use as few separate objects as possible. We don't want to define a new object independently if it can be defined in terms of existing things. Suppose a particle moves so that when it is at position $\mathbf{r}$ , its velocity is $\mathbf{v}(\mathbf{r})$ , where $\mathbf{v}$ is a vector field. Then if there is some function $f(\mathbf{r})$ , then the particle sees $$\frac{df}{dt} = v^i \frac{\partial f}{\partial x^i}$$ by the chain rule. That is, if we interpret a vector field as a velocity field, and give it a function $f$ , then we can compute another function $df/dt$ , which is the rate of change of $f$ seen by a particle following the flow of the vector field, as if it were a velocity field. By glancing at the chain rule, you see that if you know $df/dt$ for every $f$ , then you know what the vector field is. Hence, when we work in the more general setting of a manifold, where it's not immediately clear how to define a vector field in the usual way ("an arrow at every point"), we can use this in reverse to define what a vector field is. That is, a vector field $v$ is a map of functions $f \mapsto v(f) = df/dt$ which obeys certain properties. Note that not every vector field should be physically regarded as a velocity field. We're just making mathematical definitions here. The definitions are chosen to make the formalism as clean and simple as possible, possibly at the expense of intuition. Translating between the two is very simple. For example, the field $$\mathbf{E}(\mathbf{r}) = x \hat{i} + xy \hat{j}$$ translates to $$\mathbf{E}(\mathbf{r}) = x \frac{\partial}{\partial x} + x y \frac{\partial}{\partial y}.$$ That is, whenever you see $\partial/\partial x$ you can just imagine it as a unit vector pointing in the $x$ direction. The intuition is the same, because the two definitions obey the same properties. There are many examples of this in mathematics. For example, you might think $\log(x)$ is defined as "the number of times you have to multiple by $e$ to get to $x$ ", but it's unclear how to rigorously define that. Then, through semi-rigorous manipulation you can show that $$\log(x) = \int_1^x \frac{dx'}{x'}.$$ Now the mathematician sees this and chooses to define $\log(x)$ as this integral. This is simpler, because it automatically works for any real $x$ and uses only the notion of an integral, which we already know. Then one can derive the "intuitive" properties of the logarithm, such as $\log(xy) = \log(x) + \log(y)$ . By using a less intuitive definition, the formalism becomes simpler. And once you show that this definition is equivalent to the intuitive one, you are "allowed" to just keep on using the same intuition you started with, so you get the best of both worlds! Edit: the OP asks for examples where this definition of a vector field is more practically useful. I can think of two off the top of my head. First, how do basis vectors transform when you change coordinates from $x^i$ to $y^j$ ? In the usual formalism you may have to memorize a formula, but with the derivatives it follows from the chain rule, $$\frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j} \equiv J^j_i \frac{\partial}{\partial y^j}$$ where $J$ is the Jacobian matrix. Next, suppose the vector field actually is a velocity field, and you want to calculate $f(x(t))$ given $x(0)$ , i.e. you want to know where you'll be if you follow the flow for time $t$ . In this formalism, that's a one-liner. It's just $$f(x(t)) = (e^{t v} f(x))|_{x = x(0)}.$$ To prove this, expand the exponential in a Taylor series. Of course, the great thing is that once you prove two formalisms are equivalent, you can use the intuition from either one interchangably, because you know they're both equally valid. So you gain intuition for a few new cases, without losing any intuition you had before. You can always change back and forth, much like the same software program can run on different hardware. | {
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423,406 | We know, via Wien's law, that a body at 6000K emits an electromagnetic wave at the peak wavelength in the visible spectrum. How come say the fluorescent tubes which also emit the EM waves that we can see as visible not 6000K? | Black body, by definition, produces thermal radiation only, which is an EM radiation caused by heat. For such radiation, the temperature of a body defines its radiation spectrum and its peak. The EM radiation in fluorescent tube is not due to heat, but due to fluorescence, which is a type luminescence, defined as emission of light not caused by heat , but by other processes. More specifically, in a fluorescent tube, UV photons are emitted by mercury vapor atoms, excited by fast moving charge carriers (sort of electroluminescence), and then visible light photons are emitted by phosphor coating atoms, excited by UV photons (fluorescence). Both steps here are forms of luminescence, not thermal radiation. Since fluorescent light is not due to thermal radiation, its temperature is not governed by black body radiation curves. Therefore, even though most of the EM radiation emitted by a fluorescent tube is in the visible light spectrum, its temperature is very low. | {
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423,409 | Half-life for certain radioactive element is
5 min. Four nuclei of that element are observed
a certain instant of time. After five minutes Statement-1: It can be definitely said that two
nuclei will be left undecayed. Statement-2: After half-life i.e.5minutes, half of
total nuclei will disintegrate. So only two nuclei
will be left undecayed. (A)Statement-1 is true, statement-2 is true and
statement-2 is correct explanation for
statement-1.
(B)Statement-1 is true, statement-2 is true and
statement-2 is NOT the correct explanation for
statement-1
(C)Statement-1 is true, statement-2 is false.
(D)Statement-1 is false, statement-2 is false The correct answer for this question (D) . What is the reason? I approached the problem in this way After one half life , exactly half of the undecayed atoms will be left and this only depends on intial number of undecayed nuclei . So the correct answer according to me must be (A) | Black body, by definition, produces thermal radiation only, which is an EM radiation caused by heat. For such radiation, the temperature of a body defines its radiation spectrum and its peak. The EM radiation in fluorescent tube is not due to heat, but due to fluorescence, which is a type luminescence, defined as emission of light not caused by heat , but by other processes. More specifically, in a fluorescent tube, UV photons are emitted by mercury vapor atoms, excited by fast moving charge carriers (sort of electroluminescence), and then visible light photons are emitted by phosphor coating atoms, excited by UV photons (fluorescence). Both steps here are forms of luminescence, not thermal radiation. Since fluorescent light is not due to thermal radiation, its temperature is not governed by black body radiation curves. Therefore, even though most of the EM radiation emitted by a fluorescent tube is in the visible light spectrum, its temperature is very low. | {
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423,782 | There's this Newsweek article titled "Ancient Egypt: Incredible Electromagnetic Discovery in Great Pyramid of Giza's Hidden Chambers" ( here ) that says that Now, an international team of physicists has found that, under the
right conditions, the Great Pyramid can concentrate electromagnetic
energy in its internal chambers and under its base. The results, which are published in the Journal of Applied Physics,
could help scientists to create new nanoparticles — particles between 1
and 100 nanometers in size — that could be used, for example, to develop
highly efficient solar cells or tiny sensors. Above article links to this article in the Journal of Applied Physics . Question: Would the scientist somehow use this pyramid to create the nanoparticles, or did the discovery tell them something about electromagnetism that was not already known? | The scientific interest is to use pyramids in nano technology. They just scaled up their calculation to the size of the pyramid at Gizeh to get media attention. The scientific interest of the Gizeh is zero, but it is great clickbait. | {
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424,499 | Suppose a ball is going straight along a fixed line without rotating. Now you consider a point which is not along the trajectory of the ball. If I want to consider the angular momentum about that point, as you can understand that the $r$ and $v$ vector aren't in the same direction, so $r\times v $ should have a value. So this object isn't rotating but has an angular momentum. Can I conclude that the ball has angular momentum without rotating ? | Yes, it does. It may seem a bit more intuitive if you imagine a line connecting your reference point and the centre of the ball: as the ball moves, the line sweeps out an angle across the ball, so there should be some angular momentum. If you think in polar coordinates instead of cartesian, the ball is rotating. | {
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424,870 | How does this formula work, from a dimensional analysis perspective? $$ v_\text{escape} = \sqrt{\frac{2GM}{R}}$$ The way I'm thinking about it is that $G$ is in units $\text{N} \cdot \text{m}^2/\text{kg}^2$. You multiply by a kilogram amount (the mass) to turn $G$ into units $N \cdot \text{m}^2/\text{kg}$. You then divide by the radius of the object to turn $G$ into units $N \cdot \text{m}/\text{kg}$. However, $v_\text{escape}$ is in units $\text{m}/\text{s}$. $\sqrt{N \cdot \text{m}/\text{kg}} \neq \text{m}/\text{s}$.
Therefore, how does the equation even work if the units on either side aren't equal? Or am I doing this all wrong? | Newton is not a fundamental SI unit: $$\mathrm N=\frac{\mathrm{kg}\cdot\mathrm m}{\mathrm s^2}.$$ So, in fact: $$\frac{\mathrm N\cdot\mathrm m}{\mathrm{kg}}=\frac{\mathrm m^2}{\mathrm s^2},$$ the square root of which has the units of velocity. | {
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425,013 | According to Wikipedia : A closed system is a physical system that does not allow certain types of transfers (such as transfer of mass and energy transfer) in or out of the system. According to my textbook , the principle of conservation of charge is: The algebraic sum of all the electric charges in any closed system is constant. Isn't this circular logic? In terms of charge, a "closed system" is one in which charge can neither exit nor enter. If the charge neither exits nor enters, then of course the sum thereof stays constant. Or is the principle saying that the only way for the sum of charge in a system to change is via transfer of charge in or out of the system? (In this case, wouldn't it make more sense to state the principle as "charge can neither be created nor destroyed"?) | You're right that it's a bit circular as stated. The more rigorous way to state a conservation law is something like: The rate of change of [quantity] in a bounded system is equal to minus the rate at which [quantity] leaves through boundaries of that system. A "closed system" is then a system for which both of these rates are zero, i.e., the [quantity] is not moving through the boundary of the system. The version you propose, "[quantity] can neither be created nor destroyed", is closer to this more rigorous statement. But the rigorous statement is a bit stronger than this. If a charge were to suddenly teleport across the room, without passing through the points in between, this would satisfy your version of the statement; but it would not satisfy the rigorous version of the conservation law above. Moreover, it's perfectly possible in particle physics for charges to be created or destroyed, so long as equal amounts of positive and negative charges are created or destroyed. Your version of the statement would seem to outlaw these events, but the rigorous version does not. | {
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425,018 | I can't understand one thing. A good flow of water from bath faucet requires a good velocity of water or a good pressure?
For example, I have a crane that has a pipe diameter of 1 inch. The better the flow of water, the higher velocity of water. But the higher velocity of water, the lower pressure (bernoulli principle). So it seems that the flow is good, but the pressure is low. | You're right that it's a bit circular as stated. The more rigorous way to state a conservation law is something like: The rate of change of [quantity] in a bounded system is equal to minus the rate at which [quantity] leaves through boundaries of that system. A "closed system" is then a system for which both of these rates are zero, i.e., the [quantity] is not moving through the boundary of the system. The version you propose, "[quantity] can neither be created nor destroyed", is closer to this more rigorous statement. But the rigorous statement is a bit stronger than this. If a charge were to suddenly teleport across the room, without passing through the points in between, this would satisfy your version of the statement; but it would not satisfy the rigorous version of the conservation law above. Moreover, it's perfectly possible in particle physics for charges to be created or destroyed, so long as equal amounts of positive and negative charges are created or destroyed. Your version of the statement would seem to outlaw these events, but the rigorous version does not. | {
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425,649 | In computer science, we reference colours using the RGB system and TVs have pixels which consist of groups of red, green and blue lines which turn on and off to create colours. But how does this work? Why would certain amounts of red, blue and green light make something seem yellow? Is this a biological thing, where our brain performs some kind of averaging operation, or are the waves actually interacting to make light of a new wavelength? It seems RGB is a "universal triplet", as every colour within the visible spectrum can be created by combining the three in different intensities. Is RGB the only such triplet? If so, why? If not, what features must a triplet of colours have to be universal? | Color perception is entirely a biological (and psychological) response. The combination of red and green light looks indistinguishable, to human eyes, from certain yellow wavelengths of light, but that is because human eyes have the specific types of color photoreceptors that they do. The same won't be true for other species. A reasonable model for colour is that the eye takes the overlap of the wavelength spectrum of the incoming light against the response function of the thee types of photoreceptors, which look basically like this: Image source If the light has two sharp peaks on the green and the red, the output is that both the M and the L receptors are equally stimulated, so the brain interprets that as "well, the light must've been in the middle, then". But of course, if we had an extra receptor in the middle, we'd be able to tell the difference. There are two more rather interesting points in your question: every colour within the visible spectrum can somehow be created by "combining" the three in different intensities. This is false. There is a sizable chunk of color space that's not available to RGB combinations. The basic tool to map this is called a chromaticity plot, which looks like this: Image source The pure-wavelength colours are on the curved outside edge, labelled by their wavelength in nanometers. The core standard that RGB-combination devices aim to be able to display are the ones inside the triangle marked sRGB; depending on the device, it may fall short or it can go beyond this and cover a larger triangle (and if this larger triangle is big enough to cover, say, a good fraction of the Adobe RGB space, then it is typically prominently advertised) but it's still a fraction of the total color space available to human vision. (A cautionary note: if you're seeing chromaticity plots on a device with an RGB screen, then the colors outside your device's renderable space will not be rendered properly and they will seem flatter than the actual colors they represent. If you want the full difference, get a prism and a white-light source and form a full spectrum, and compare it to the edge of the diagram as displayed in your device.) Is RGB the only such triplet? No. There are plenty of possible number-triplet ways to encode color, known as color spaces , each with their own advantages and disadvantages. Some common alternatives to RGB are CMYK (cyan-magenta-yellow-black), HSV (hue-saturation-value) and HSL (hue-saturation-lightness), but there are also some more exotic choices like the CIE XYZ and LAB spaces. Depending on their ranges, they may be re-encodings of the RGB color space (or coincide with re-encodings of RGB on parts of their domains), but some color spaces use separate approaches to color perception (i.e. they may be additive , like RGB, subtractive , like CMYK, or a nonlinear re-encoding of color, like XYZ or HSV). | {
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425,834 | I've just started reading about photo electric effect here , and my high school level understanding goes something like this : 1) By 1900 we had Maxwell equations and treated light as a wave. 2) But Maxwell couldn't explain the graph of black body radiation : The electrons in a hot object can vibrate with a range of frequencies, ranging from very few vibrations per second to a huge number of vibrations per second. In fact, there is no limit to how great the frequency can be. Classical physics said that each frequency of vibration should have the same energy . Since there is no limit to how great the frequency can be, there is no limit to the energy of the vibrating electrons at high frequencies. May I know why/how classical physics came up with the conclusion that each frequency of vibration should have the same energy ? Intuitively, I feel the energy of a mechanical wave depends on amplitude/frequency. How could the classical physics conclude all frequencies have same energy ? | This classical prediction comes from the equipartition theorem of statistical mechanics, though I have some issues with exactly how the statement you quote is worded. The equipartition theorem is for describing how the energy gets distributed in a system with many degrees of freedom. For example, consider a mono-atomic ideal gas, like helium, that you've heated up to some temperature $T$ (in absolute units ). If you increase the energy stored in the gas --- maybe you compress the gas, doing work on it --- the only degree of freedom available to store that energy is that the velocities of the gas particles can change. The energy of each gas atom is $\frac12 m v^2 = \frac12 m (v_x^2 + v_y^2 + v_z^2)$, where the $v_i$ are the components of the velocity in some coordinate system. But if the container is symmetrical, then there shouldn't be any reason for, say, the $x$-components of the gas velocities to have systematically more energy than the $y$-components: the energy should be partitioned equally among all three components. The equipartition theorem predicts, among other things, that all monoatomic ideal gases should have the same molar heat capacity $C_V = \frac32R = 12\,\frac{\rm joule}{\rm mole\ kelvin}$, because each atom in the gas has (on average) kinetic energy $\frac12 kT$ in each of the three possible directions of travel. Furthermore because the gas speeds obey $\frac12 mv^2 = \frac32 kT$, then it's possible to make predictions about the speed of sound in different gases at different temperatures. The equipartition theorem isn't only about speeds, but about all sorts of degrees of freedom. For instance, if you have a diatomic gas, like carbon monoxide, there is an 1 additional way that the gas molecules can store energy: rotation. Each molecule has two possible axes of rotation perpendicular to the bond between the two atoms.
Equipartition predicts those two extra degrees of freedom should also each store average kinetic energy $\frac12 kT$. So diatomic ideal gases should have molar heat capacity $5R/2$. Which they do --- except at very low temperatures, when the heat capacity falls back to the monoatomic value of $3R/2$. This behavior was also a mystery at the beginning of the twentieth century. In the case of blackbody radiation, the oscillators in question aren't electrons, but standing waves in the electromagnetic field .
(Electrons may be involved, but the electromagnetic field can oscillated even if all the charges in the universe are far away.)
If you consider the electromagnetic field inside of a metal box, where the magnitude of the field always has to be zero in the walls of the box, then you can count these modes the same way we counted translational and rotational modes for the ideal gases. There's a mode where half a wave fits in the box, so there are nodes at the walls of the cavity; a mode where one wave fits in the box; a node where one-and-a-half waves fit in the box; and so on to infinity. And according to classical equipartition, each of these infinite possible oscillatory modes should contain, on average, an energy of $\frac12 kT$. You can sample these electromagnetic oscillators by building such a cavity, making it hot, and opening a small hole to look at the radiation that comes out (on the assumption that a small hole doesn't change what's happening inside the cavity very much). Your eye interprets the different wavelengths/frequencies of electromagnetic radiation as different colors.
And what you find is that, for long wavelengths/slow frequencies, equipartition gives a pretty good prediction for the spectrum of light emitted by a hot cavity. The trouble, as your text points out briefly, is that continuum equipartition doesn't have any mechanism for omitting the short wavelength/rapid frequency oscillations, and therefore predicts that empty space should have an infinite heat capacity. The ideal gas predictions were ... less wrong than this. In the literature, this misprediction is called the "ultraviolet catastrophe." Planck's suggestion was that the minimum energy you can add to an oscillator with angular frequency $\omega$ is $E = \hbar \omega$. Given that assumption, classical thermodynamics predicts that the probability of finding an oscillator with $n$ lumps of energy is proportional to $e^{-n\hbar\omega/kT}$. If the temperature is hot or the frequency is slow, these probabilities are all proportional to $e^{-\text{small}} \approx 1$, and there's no real restriction on adding or removing energy from that degree of freedom, and the equipartition theorem holds. But if the frequency is fast or the temperature is cold, then the probability of finding your oscillator with one lump of energy is much smaller than for finding it with zero, and we can say that the degree of freedom is "frozen out." Note that, so far as we know, Planck's assumption applies to all oscillators, not just the electromagnetic field in a hot cavity.
For instance, the reason that a cold diatomic gas has the same molar heat capacity as a monoatomic gas (the $\frac52\to\frac32$ business from eariler) is that the rotational degree of freedom freezes out. 1 " an additional way that gas molecules can store energy":
For simplicity, I'm ignoring gas vibrational degrees of freedom in this answer that's nominally about blackbody radiation.
Generally, vibration freezes out at a higher temperature than rotation does, and for many diatomic gases the vibrational modes aren't completely accessible before dissociation starts to be significant.
In the carbon monoxide example, the heat capacity is $\frac52R \pm 1\%$ from far below room temperature to about 400K; the vibration temperature is about $\hbar\omega/k = 3000\rm\,K$, about twice the dissociation temperature. | {
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426,674 | The speed of light is a universal constant, so we definitely know the speed of the photons. If we know the speed, then we should not have any information about their location, because of Heisenberg's uncertainty principle. But I'm one hundred percent sure when light goes through my window. Why is this so? | The Heisenberg Uncertainty Principle does not involve speed. It involves momentum , and this is one of the places where that distinction is very important. Photons all travel at the same speed, yes, but their momentum can take on any value. As such, the uncertainty in its position and the uncertainty in its momentum are still linked in the same way they would be for ordinary matter. In fact, due to the fact that photons are traveling at the speed of light, their energy and momentum are related by $E=pc$. Applying the usual Planck-Einstein relation $E=hf$, we can see that an uncertainty in the photon's momentum is also directly proportional to the uncertainty in the photon's frequency, which might be easier to picture (i.e. in the case of photons, the Heisenberg Uncertainty Principle links the uncertainty in their position and the uncertainty in their frequency). | {
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426,698 | Sound is air particles vibrating (thus hitting each other to make longitudinal waves) and heat is the vibration of air molecules. Because we can only assume that heat made from fire is a higher intensity of vibration than sound (because we don't burn ourselves when we speak) why doesn't it make an extremely loud sound? | The combustion reactions don't inherently make any sounds. But they release plenty of energy causing the nearby molecules to acquire higher random kinetic energy, which is theoretically detectable as Brownian noise. But it's not as easy as detecting a human's speaking because the noise is indistinct and the power at the eardrum is lower: a lot of the particles' velocities aren't towards the eardrum so a limited amount of energy is transferred. Unlike longitudinal sound waves, where groups of particles oscillate periodically with large amplitudes, the movement here is rather disorganized and isn't audible for all reasonable temperatures, so you're more likely to hear the sound of wood popping due to the expansion of fibres and escape of moisture and air, or sounds of wind blowing as air expands and rises rapidly. I decided to take a shot at finding the temperature needed for the Brownian noise to be about as loud as a conversation. For $60\ \rm dB (SPL)$ , with the usual reference of $2\times 10^{-5}\ \rm Pa$ and considering the ear's highest sensitivity region, we're looking for an eventual rms pressure of $2\times 10^{-2}\ \rm Pa$ . Using the equation from On Minimum Audible Sound Fields by Sivian and White $$\bar{P} =\left [ \frac{8 \pi \rho k_B T} {3c} ({f_2}^3-{f_1}^3)\right ]^{1/2}$$ where $\bar{P}$ is the rms ( root mean square ) pressure, $\rho$ is the density of air, $k_B$ is the Boltzmann constant, $T$ is the temperature, $c$ is the speed of sound in air, and $f_1$ and $f_2$ are the frequency range. Let's consider the frequency range of $0\ \rm Hz$ to $2\times 10^4\ \rm Hz$ , because according to the Brownian frequency distribution, higher frequencies are negligible. Throwing in common values for all the constants, we see that for $\bar{P}=2\times 10^{-2}\ \rm Pa$ , we need an unbelievable $10^8\ \rm K$ , to 1 significant figure, which is hotter than the Sun's core. It's important to not think that this implies that we're never going to hear Brownian noise. If you take some software like Audacity (or probably any sound editing tool), you can render Brownian noise and listen to it at $60\ \rm dB$ . But there we have a deliberate superimposition of several frequencies' waveforms (with amplitudes according to the distribution) being played. But we won't hear the noise caused by the random kinetic energy of air molecules at a loudness comparable to normal conversations . "Because we can only assume that heat made from fire is a higher intensity of vibration than sound" isn't really true: because of the direction of particles' oscillations/movement, longitudinal sound waves end up transferring far more energy to the ear drum. Eventually, we're not going to detect the noise; the flows due to convection and expansion of hot air are more likely to be audible. Sivian, L. J., and S. D. White. “On Minimum Audible Sound Fields.” The Journal of the Acoustical Society of America , vol. 4, no. 3, 1933, pp. 176–177., doi:10.1121/1.1901988. And some cool related stuff: Difference between sound and heat at particle level | {
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426,912 | It is well established that the light speed in a perfect vacuum is roughly $3\times 10^8 \:\rm m/s$. But it is also known that outer space is not a perfect vacuum, but a hard vacuum. So, is the speed limit theoretically faster than what we can measure empirically, because the hard vacuum slows the light down? Is this considered when measuring distances with light? | If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $\Delta c/c$ of $3 \times 10^{-4}$. The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about: $$ \frac{\Delta c}{c} = 3 \times 10^{-4} \, P $$ In high vacuum labs we can, without too much effort, get to $10^{-10}$ torr and this is around $10^{-13}$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error: $$ \frac{\Delta c}{c} \approx 3 \times 10^{-17} $$ And this is already smaller than the experimental errors in the measurement. So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible. | {
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426,920 | I recently came across a mathematical field called complex analysis. There was an important equation called Cauchy-Riemann equation . When I saw it at first, I recalled a book's sentence stating that A: Why did we substitute the Newtonian equations with that efforts? B: It's because it is beautiful in terms of mathematics. Hamilton's canonical equation is good because it is beautiful (in terms of mathematics) From a Quantum mechanics book. It was
\begin{align}
\frac {dq}{dt}& =\frac {\partial H}{\partial p} \\
-\frac {dp}{dt}& = \frac{\partial H}{\partial q}.
\end{align} Doesn't this look similar to the Cauchy-Riemann equation? (Integrating $u$ makes $v$ and vice versa.) I guess the reason is of something like the CR equation. So my question is: Is there any relationship between complex analysis and Hamilton's canonical equation? | If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $\Delta c/c$ of $3 \times 10^{-4}$. The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about: $$ \frac{\Delta c}{c} = 3 \times 10^{-4} \, P $$ In high vacuum labs we can, without too much effort, get to $10^{-10}$ torr and this is around $10^{-13}$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error: $$ \frac{\Delta c}{c} \approx 3 \times 10^{-17} $$ And this is already smaller than the experimental errors in the measurement. So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible. | {
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426,954 | Why don't we use infrared (IR) or even the far IR just to heat food in a microwave oven instead of, of course, the conventional 2.45 GHz microwaves? Don't people call IR heat waves? | We do use (near) infrared radiation to heat food – whenever we toast food or grill (UK)/broil (US) by beaming infrared downwards on to food! The point is that the infrared is strongly absorbed by the food we cook in this way, and doesn't penetrate significantly beyond about a millimetre. So the surface of the food is strongly heated – seared, toasted or scorched! What lies below the surface is cooked much more slowly, mainly by conduction of heat from the surface. Microwaves are not as strongly absorbed and penetrate much further, so the food is 'cooked from the inside'. The microwaves are mainly absorbed by water molecules that are sent into a vibratory/rotatory motion by the electric field of the microwaves acting on the (polarised) molecules. These are forced oscillations, but not at resonance; the frequency of the microwaves (about 2.4 GHz) is not a natural frequency for the molecule. If it were, the microwaves would be absorbed by the surface layer, and we'd have another grill or toaster! Edit (prompted by comment below). I don't mean to give the impression that water molecules are the only ones that absorb microwaves. Fats are also strong absorbers. | {
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427,095 | Either a large, constrained cylinder of water or a large volume of open water (perhaps in the ocean) could be persuaded to form a vortex by pumping energy in. The body of water would be acting as a fluid version of a flywheel, storing energy. If one then removed the input energy, would the vortex continue for a significant time or would it quickly lose all its energy by heating up the water? Would storing energy in the rotational kinetic energy of a fluid be horribly inefficient compared to pumped-hydro storage? | Yes, this would be horribly inefficient compared to pumped hydro or even a regular flywheel. With a rotating fluid, there's a lot of viscosity. This viscosity generates heat and slows the fluid down. You would be able to offset this somewhat if you kept the container for the fluid moving with the fluid itself; but even then I believe there would still be significant losses from internal viscous effects. You would then also just have what is essentially a fluid filled flywheel, at which point you have to ask yourself why you didn't go with a solid material of higher density and likely better strength/stability. The advantage of pumped hydro over this vortex system would be because a pumped hydro system doesn't require constant motion. It stores the energy as potential, so it only loses energy to viscosity when it is moving the fluid to the storage tanks. The less you have to move the fluid, the less you will convert energy to heat through viscosity. It's an interesting idea, but I don't think it would be practical. | {
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427,827 | Paper is an extremely flexible material, at least when it is in sheet form. It will deform significantly according to the pressure applied and it is easy to fold. Therefore, it's extremely counterintuitive that a sheet of paper could cut through human skin and probably through stiffer/harder materials, since when the skin applies a pressure on the paper, one would expect it to fold or bend. Yet it is easy to have a severe cut from paper, through both the epidermis and the dermis. How is that possible? Certainly the width of the sheet of paper plays a big role: the smaller it is, the sharper it is, but also the more flexible it becomes and the less it should sustain an applied pressure without folding up! I can think of other materials such as thin plastic films and aluminium foils. My intuition tells me the plastic foil would not cut through skin but the aluminium foil would, although I am not sure since I did not try the experiment. If this hold true, what determines whether a material would be able to cut through skin? A hair for example, which is flexible and thinner than a paper sheet, is unable to cut through the skin. What makes paper stand out? What is so different that makes it a good cutter? Maybe it has to do with its microscopic properties and that it contains many fibers, but I highly doubt it because the aluminium foil does not contain these and yet would probably also cut. | Paper, especially when freshly cut, might appear to have smooth edges, but in reality, its edges are serrated (i.e. having a jagged edge), making it more like a saw than a smooth blade. This enables the paper to tear through the skin fairly easily. The jagged edges greatly reduce contact area, and causes the pressure applied to be rather high. Thus, the skin can be easily punctured, and as the paper moves in a transverse direction, the jagged edge will tear the skin open. Paper may bend easily, but it's very resistant to lateral compression (along its surface). Try squeezing a few sheets of paper in a direction parallel to its surface (preferably by placing them flat on a table and attempting to "compress" it laterally), and you will see what I mean. This is analogous to cutting skin with a metal saw versus a rubber one. The paper is more like a metal one in this case. Paper is rather stiff in short lengths, such as a single piece of paper jutting out from a stack (which is what causes cuts a lot of the time). Most of the time, holding a single large piece of paper and pressing it against your skin won't do much more than bend the paper, but holding it such that only a small length is exposed will make it much harder to bend. The normal force from your skin and the downward force form what is known as a torque couple. There is a certain threshold torque before the paper gives way and bends instead. A shorter length of paper will have a shorter lever arm, which greatly increases the tolerance of the misalignment of the two forces. Holding the paper at a longer length away decreases this threshold (i.e. you have to press down much more precisely over the contact point for the paper to not bend). This is also an important factor in determining whether the paper presses into your skin or simply bends. Paper is made of cellulose short fibers/pulp, which are attached to each other through hydrogen bonding and possibly a finishing layer. When paper is bent or folded, fibers at the folding line separate and detach, making the paper much weaker. Even if we unfold the folded paper, those detached fibers do not re-attach to each other as before, so the folding line remains as a mechanically weak region and decreasing its stiffness. This is why freshly made, unfolded paper is also more likely to cause cuts. Lastly, whether a piece of paper cuts skin easily, of course depends on its stiffness. This is why office paper is much more likely to cut you than toilet paper. The paper's density (mass per unit area), also known as grammage , has a direct influence on its stiffness. | {
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427,922 | I have recently read that helium is going to be used as coolant in Generation IV nuclear reactors, because Helium is radiologically inert (i.e., it does not easily participate in nuclear processes and does not become radioactive). ( Source: Ch 3 of The Impact of Selling the Federal Helium Reserve (2000) ) Does that mean, if there would be some helium leaks, it would not pollute the air?
Also, how is that some gas "conduct" radioactivity more than others? How is that determined? I have read something about cross sectional area of nuclei with barns as unit, but I don't really understand the process. What are some other gases which are radiologically inert? | Chemically, helium is inert because it has a "filled valence shell" of electrons, which is very stable; it's extremely difficult to change this structure, as doing so requires a lot of energy and produces a system which is likely to quickly revert back to its ground state under normal conditions. The helium-4 nucleus is in a very similar situation: in a sense, it has "filled shells" of protons and neutrons. Relative to its neighbors on the nuclear chart, it's one of the most stable nuclear configurations we have measured. Changing this nuclear structure in any way is difficult, so helium-4 is unlikely to become radioactive in the first place, and the configurations that are created when it does happen are so unstable that they decay almost instantly. Neutron capture (far and away the primary cause of secondary radioactivity) creates helium-5, which decays with a half-life of $7\times 10^{-22}$ seconds, so it barely even exists, and certainly won't be found outside the reactor. It's also basically impossible to excite the helium-4 nucleus to a higher energy level using gamma radiation from a fission reactor, as the next energy level is 20 MeV above the ground state (for reference, most of the steps of the uranium decay chain have a total released energy of only 4-7 MeV). So it's safe to say that helium-4 is radiologically inert. The term "conduction" is probably* referring to the following process: a radioactive nucleus predisposed to emit neutrons decays, and the emitted neutrons are captured by another nucleus, which might make it unstable and therefore radioactive. In this sense, what determines how readily a substance "conducts" radioactivity is its willingness to capture neutrons (aka the neutron capture cross section ), which is heavily dependent on the specific nuclear structure. (There are other ways to induce radioactivity, like beta decay of one nucleus followed by electron capture by another, or gamma-ray emission and absorption, but the conditions required for those processes are rarer.) For other radiologically-inert substances, one might look for other nuclei that have "filled shells" of protons and neutrons. In nuclear structure, these are called "doubly magic" nuclei (having a "magic" number of protons and a "magic" number of neutrons), and do indeed have a reputation for stability, though none are quite so stable as helium-4. Doubly-magic nuclei include oxygen-16, calcium-40, and iron-56. *Let me stress that "conduction" is highly nonstandard terminology; the term for the process that I describe here is "induced radioactivity." | {
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428,127 | Witnesses of nuclear explosions have described their hands becoming transparent, and that they could see the bones. For example, see here. How does that happen? | Have you never seen the bones of your hand when covering a flash light at night ? Imo it was just a very bright light over a large area and trying to shield the eyes the bones were seen. | {
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428,131 | I'm must be missing something again but isn't there some contradiction in 2 statements below ? Edit: I've rewrote the post and added images for clarification. On one hand, PD depends on amount/density of charge separated. As example: distribution of voltage during transients, like on this image: Voltage here is directly correlate with amount/density of charge in certain place of conductor. Or when incident pulse of voltage and current reach an open end there will be short-term increase of voltage. This is explained due pile up of charges at the end since current have to stop. So here we again have relation: more charge - higher potential of conductor in this place. In this way, if source voltage doesn't change at all, charge distribution on conductor must be uniform to maintain same potential along it length. Note: actually there will be some potential drop due resistance and hence amount of surface charges is slowly decreasing but it doesn't matter here. But, on the other hand, if we see at DC steady state (with steady current), distribution of charges on surface of conductors, actually, is non-uniform. I.e. in some place amount of charge is pretty small, on bends it larger, and really huge accumulation of charge in the area of wire - resistor interface. As example, image from A semiquantitative treatment of surface charges in DC circuits : But potential of conductor (and voltage between them) is considered as equal everywhere along its length (again neglecting drop due resistance). I mean there is area at the end of conductor (at resistor boundary) with significantly more amount of charge than other parts of the wire. If remember transmission line example then potential of this area must be higher. It sound strange, but in this way voltage across resistor must be higher than across wires itselfs. But this is nonsense and here it obviously same as between other parts of the conductors. Why voltage across resistor in this case same as across wires despite the fact that there is much more charge accumulation than on wires itselfs ? Maybe it somehow related to capacitance, but I don't know. | Have you never seen the bones of your hand when covering a flash light at night ? Imo it was just a very bright light over a large area and trying to shield the eyes the bones were seen. | {
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428,402 | Mixing of different wavelengths of light results in white, but why is that when paint with different colors are mix results in black? | Mixing light does result in white, but the black mixture of paint happens due to how paint works. Paint has color not because it's emitting light, but because it's absorbing colors other than the one that's supposed to be the paint's color. As such, when you mix paints, they absorb more and more of the spectrum, resulting in black. | {
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428,414 | For a project, I was wondering whether or not it is possible to increase the net downwards force acting on an object so that you are able to manipulate its weight. | Mixing light does result in white, but the black mixture of paint happens due to how paint works. Paint has color not because it's emitting light, but because it's absorbing colors other than the one that's supposed to be the paint's color. As such, when you mix paints, they absorb more and more of the spectrum, resulting in black. | {
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428,525 | So the formula for work is$$
\left[\text{work}\right] ~=~ \left[\text{force}\right] \, \times \, \left[\text{distance}\right]
\,.
$$ I'm trying to get an understanding of how this represents energy. If I'm in a vacuum, and I push a block with a force of $1 \, \mathrm{N},$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block. I must be missing something, but I can't really pinpoint what it is. It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed. | You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body. | {
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428,538 | I have seen quite a bit online about how there are only 20 independent components for the (lowered) Riemann tensor $R_{abcd}$ for the Schwarzschild metric. I've been told this follows from the symmetries of the tensor, i.e.: $R_{abcd}=-R_{bacd}=-R_{abdc}=R_{badc}$ and $R_{abcd}=R_{cdab}$ Now if the indices in the tensor can all run from 1 to 4, then $R_{abcd}$ has 256 components. These symmetries seem to reduce our need to calculate components, but why only 20? Note: I am aware there are some similar questions on the stack exchange. I have read these, but none of them very clearly explained this specific point, so I decided to ask it directly so hopefully I can get my head around it. | You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body. | {
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428,952 | From this comment by orlp: If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives? | The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast? To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel). Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s. So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed? Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J. At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust. | {
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428,990 | The Euler-Lagrange equation gives the equations of motion of a system with Lagrangian $L$. Let $q^\alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time. The Lagrangian is a function of the state of a particle, i.e. the particle's position $q^\alpha$ and velocity $\dot q^\alpha$. The Euler-Lagrange equation is $$ \frac{d}{dt} \frac{\partial L}{\partial \dot q^\alpha } = \frac{\partial L}{\partial q^\alpha}$$ Why is this a law of physics and not a simple triviality for any function $L$ on the variables $q^\alpha$ and $\dot q^\alpha$? The following "proof" of the Lagrange Equation uses no physics, and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function. $$\begin{align}
\frac{d}{dt} \frac{\partial L}{\partial \dot q^\alpha} & = \frac{\partial}{\partial \dot q^\alpha} \frac{dL}{dt} &\text{commutativity of derivatives} \\ \ \\
&= \frac{\partial \dot L}{\partial \dot q^\alpha} \\ \ \\
&= \frac{\partial L}{\partial q^\alpha} & \text{cancellation of dots} \end{align}$$ This can't be right, or else nobody would give a hoot about this equation and it would be totally useless to solve any problem. What is wrong with the logical reasoning above? | Ah, what a tricky mistake you've made there. The problem is that you've simply confused some notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler-Lagrange equation is not trivial. Let's first take a step back. The Lagrangian for a particle moving in one dimension in an external potential energy $V(q)$ is
$$
L(q, \dot q) = \frac{1}{2}m \dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clearly $q$ and $\dot q$ are not independent variables. Once $q$ is specified for all times, $\dot q$ is also specified for all times. A better way to write the above Lagrangian might be
$$
L(a, b) = \frac{1}{2}m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
\frac{\partial L}{\partial a} = -V'(a) \hspace{1cm} \frac{\partial L}{\partial b} = m b.
$$
Usually, most people write this as
$$
\frac{\partial L}{\partial q} = -V'(q) \hspace{1cm} \frac{\partial L}{\partial \dot q} = m \dot q.
$$ However, $q$ and $\dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $\dot q$ are too when they're being put into the Lagrangian. In other words, we could put any two numbers into $L$; we just decided to put in $q$ and $\dot q$. Furthermore, let's look at the total time derivative $\frac{d}{dt}$. How should we understand the following expression?
$$
\frac{d}{dt} L(q(t), \dot q(t))
$$
Both $q$ and $\dot q$ are functions of time. Therefore, $L(q(t), \dot q(t))$ depends on time simply because $q(t)$ and $\dot q(t)$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
\frac{d}{dt} L(q(t), \dot q(t)) = \frac{dq}{dt} \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \frac{d \dot q}{dt} \frac{\partial L}{\partial b}(q(t), \dot q(t)) = \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t))
$$ In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $\partial L / \partial a$ and $\partial L / \partial b$ by plugging in $(q, \dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$. In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't. EDIT: You can see for yourself that the Euler-Lagrange equation is not identically $0$. If you take the Lagrangian $L(q, \dot q)$ I've written above and plug it into the Euler Lagrange equation, you get
$$
m \ddot q(t) + V'(q(t)) = 0.
$$
This is not the same as $0 = 0$. It is a condition that a path $q(t)$ would have to satisfy in order to extremize the action. If it was $0 = 0$, then all paths would extremize the action. EDIT: As Arthur points out, this is also a good time to discuss the difference between $dL / dt$ and $\partial L / \partial t$. If we have a time dependent Lagrangian, $$
L(q, \dot q, t)
$$
then $L$ can depend on $t$ explicitly, as opposed to just through $q$ and $\dot q$. So, for example, where as we might have the Lagrangian for a particle in a constant gravitational field $g$ is
$$
L(a,b) = \frac{1}{2} mb^2 - m g a
$$
if we let allow $L$ to depend on $t$ explicitly, we could have the gravitational field get stronger as time goes on:
$$
L(a,b,t) = \frac{1}{2} mb^2 - m ( C t )a.
$$
($C$ is a constant such that $Ct$ has the same units as $g$.) The quantity
$$
\frac{\partial}{\partial t} L(a, b, t)
$$
should be understood as differentiating the "$t$-slot" of $L$. In the above example, we would have
$$
\frac{\partial}{\partial t} L(a,b,t) = - m C a.
$$
The quantity
$$
\frac{d}{d t} L(q(t), \dot q(t), t)
$$
should be understood as the full time derivative of $L$ due to the fact that $q$ and $\dot q$ also depend on $t$. For the above example,
\begin{align*}
\frac{d}{d t} L(q(t), \dot q(t), t) &= \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t),t) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t),t) + \frac{\partial L}{\partial t} (q(t), \dot q(t), t) \\
&= (\dot q) (-mC t ) + \ddot q(t) (m \dot q(t)) - mC q(t)
\end{align*} | {
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428,994 | I would like to know how many hours a single burner electric cooktop to be heated at 440 ºC (electric power 1,200W/220V) to ignite papers(tissue) or vinyl wrappers put on the same shelf at same level with 5 cm distance from the cooktop in a room teperature of 10 ºC basis 30% humidity.
The cooktop has a dimension of 280 x 350 mm.
Please advise of the title of the papers on similar subject, if any. With thanks and Kind Regards, E.S.Kang [email protected] | Ah, what a tricky mistake you've made there. The problem is that you've simply confused some notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler-Lagrange equation is not trivial. Let's first take a step back. The Lagrangian for a particle moving in one dimension in an external potential energy $V(q)$ is
$$
L(q, \dot q) = \frac{1}{2}m \dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clearly $q$ and $\dot q$ are not independent variables. Once $q$ is specified for all times, $\dot q$ is also specified for all times. A better way to write the above Lagrangian might be
$$
L(a, b) = \frac{1}{2}m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
\frac{\partial L}{\partial a} = -V'(a) \hspace{1cm} \frac{\partial L}{\partial b} = m b.
$$
Usually, most people write this as
$$
\frac{\partial L}{\partial q} = -V'(q) \hspace{1cm} \frac{\partial L}{\partial \dot q} = m \dot q.
$$ However, $q$ and $\dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $\dot q$ are too when they're being put into the Lagrangian. In other words, we could put any two numbers into $L$; we just decided to put in $q$ and $\dot q$. Furthermore, let's look at the total time derivative $\frac{d}{dt}$. How should we understand the following expression?
$$
\frac{d}{dt} L(q(t), \dot q(t))
$$
Both $q$ and $\dot q$ are functions of time. Therefore, $L(q(t), \dot q(t))$ depends on time simply because $q(t)$ and $\dot q(t)$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
\frac{d}{dt} L(q(t), \dot q(t)) = \frac{dq}{dt} \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \frac{d \dot q}{dt} \frac{\partial L}{\partial b}(q(t), \dot q(t)) = \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t))
$$ In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $\partial L / \partial a$ and $\partial L / \partial b$ by plugging in $(q, \dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$. In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't. EDIT: You can see for yourself that the Euler-Lagrange equation is not identically $0$. If you take the Lagrangian $L(q, \dot q)$ I've written above and plug it into the Euler Lagrange equation, you get
$$
m \ddot q(t) + V'(q(t)) = 0.
$$
This is not the same as $0 = 0$. It is a condition that a path $q(t)$ would have to satisfy in order to extremize the action. If it was $0 = 0$, then all paths would extremize the action. EDIT: As Arthur points out, this is also a good time to discuss the difference between $dL / dt$ and $\partial L / \partial t$. If we have a time dependent Lagrangian, $$
L(q, \dot q, t)
$$
then $L$ can depend on $t$ explicitly, as opposed to just through $q$ and $\dot q$. So, for example, where as we might have the Lagrangian for a particle in a constant gravitational field $g$ is
$$
L(a,b) = \frac{1}{2} mb^2 - m g a
$$
if we let allow $L$ to depend on $t$ explicitly, we could have the gravitational field get stronger as time goes on:
$$
L(a,b,t) = \frac{1}{2} mb^2 - m ( C t )a.
$$
($C$ is a constant such that $Ct$ has the same units as $g$.) The quantity
$$
\frac{\partial}{\partial t} L(a, b, t)
$$
should be understood as differentiating the "$t$-slot" of $L$. In the above example, we would have
$$
\frac{\partial}{\partial t} L(a,b,t) = - m C a.
$$
The quantity
$$
\frac{d}{d t} L(q(t), \dot q(t), t)
$$
should be understood as the full time derivative of $L$ due to the fact that $q$ and $\dot q$ also depend on $t$. For the above example,
\begin{align*}
\frac{d}{d t} L(q(t), \dot q(t), t) &= \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t),t) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t),t) + \frac{\partial L}{\partial t} (q(t), \dot q(t), t) \\
&= (\dot q) (-mC t ) + \ddot q(t) (m \dot q(t)) - mC q(t)
\end{align*} | {
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429,213 | How does a particle know it should behave in such and such manner? As a person, I can set mass is so and so, charge is so and so - then set up equation to solve its equation of motion but who computes that equation of motion for a particle in real life? I, as a person, employ smart 'tricks' such as principle of superposition to avoid having to calculate super complicated situation (calculating electrical force by a shape where large circle is hollowed out in the off-center) but if I were to calculate this in a brute force manner, this would take long time for me to calculate. However, nature doesn't seem to face these types of problems. Given a school of fish, the ones at the edge will sense threat and gives signal to those near them and so on but this analogy doesn't seem to make sense for physical objects generally considered in general physics problems. Am I asking the wrong type of question? Would appreciate input on this. | I think this question makes hidden, inarticulated assumptions about reality. In physics, we make observations and then try to find models that match them. The models, though, belong only to us and exist in our heads and textbooks. We perform the calculations required to make our predictions in our models. We cannot say whether nature makes similar calculations, and asking 'how' nature or particles perform calculations seems wrongheaded, to me. Jaynes called this the mind projection fallacy ; you are projecting things that exist in your mind - calculations and models - to reality. Responding to @knzhou, I would not, though, advocate answering every question about physics with the fact that 'physics is a bunch of models'. It is possible to construct models, answer questions about them and make calculations without projecting our calculations to reality. The question 'how does water know to boil at 100 C?' would be better phrased 'how does our model explain the observation that water boils?'. We could answer that with @knzhou's answer: 'bubbles of steam finally have enough energy to expand against the water pressure.' This is indeed a valuable explanation. | {
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429,363 | I got an egg boiler machine, on the instructions is stated: Less water is used when cooking more eggs. My thermodynamics understanding cannot figure this out yet. Why would I need less water for more eggs? The machine beeps when the water has evaporated, so the eggs are ready. In that case, why will less water cook more eggs? | Presumably, the rate of the steam escaping the cooker depends on the "resistance" of the steam path: from the opening in the bottom, where the steam enters the dome, to the opening on the side of the dome, from where the steam escapes. The more eggs in the cooker, the narrower the path, the slower the flow. Also, as relatively slowly moving steam makes contact with more eggs, it is more likely to condense and make its way back to the water at the bottom of the cooker, which further reduces its escape rate. So, with more eggs in the cooker, a smaller amount of water will last about as long as a greater amount of water with fewer eggs, resulting in a similar degree of cooking. | {
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429,390 | Traditionally for a free electron, we presume the expectation of its location (place of the center of mass) and the center of charge at the same place. Although this seemed to be reasonable for a classical approximation
(see: Why isn't there a centre of charge? by Lagerbaer), I wasn't sure if it's appropriate for quantum models, and especially for some extreme cases, such as high energy and quark models. My questions are: Is there any experimental evidence to support or suspect that the center of mass and charge of an electron must coincide? Is there any mathematical proof that says the center of mass and charge of an electron must coincide? Or are they permitted to be separated? (By electric field equation from EM, it didn't give enough evidence to separate $E$ field with $G$ field. But I don't think it's the same case in quantum or standard model, i.e. although electrons are leptons, consider $uud$ with $2/3,2/3,-1/3$ charges.) What's the implication for dynamics if the expectation of centers does not coincide? | Can the center of charge and center of mass of an electron differ in quantum mechanics? They can. Particle physics does allow for electrons (and other point particles) to have their centers of mass and charge in different locations, which would give them an intrinsic electric dipole moment. For the electron, this is unsurprisingly known as the electron electric dipole moment (eEDM) , and it is an important parameter in various theories. The basic picture to keep in mind is something like this: Image source Now, because of complicated reasons caused by quantum mechanics, this dipole moment (the vector between the center of mass and the center of charge) needs to be aligned with the spin, though the question of which point the dipole moment uses as a reference isn't all that trivial . (Apologies for how technical that second answer is - I raised a bounty to attract more accessible responses but none came.) Still, complications aside, it is a perfectly standard concept. That said, the presence of a nonzero electron electric dipole moment does have some important consequences, because this eEDM marks a violation of both parity and time-reversal symmetries. This is because the dipole moment $\mathbf d_e$ must be parallel to the spin $\mathbf S$, but the two behave differently under the two symmetries (i.e. $\mathbf d_e$ is a vector while $\mathbf S$ is a pseudovector; $\mathbf d_e$ is time-even while $\mathbf S$ is time-odd) which means that their projection $\mathbf d_e\cdot\mathbf S$ changes sign under both $P$ and $T$ symmetries, and that is only possible if the theory contains those symmetry violations from the outset. As luck would have it, the Standard Model of particle physics does contain violations of both of those symmetries, coming from the weak interaction, and this means that the SM does predict a nonzero value for the eEDM, which falls in at about $d_e \sim 10^{-40} e\cdot\mathrm m$. For comparison, the proton sizes in at about $10^{-15}\:\mathrm m$, a full 25 orders of magnitude bigger than that separation, which should be a hint at just how small the SM's prediction for the eEDM is (i.e. it is absolutely tiny). Because of this small size, this SM prediction has yet to be measured. On the other hand, there's multiple theories that extend the Standard Model in various directions, particularly to deal with things like baryogenesis where we observe the universe to have much more asymmetry (say, having much more matter than antimatter) than what the Standard Model predicts. And because things have consequences, those theories $-$ the various variants of supersymmetry, and their competitors $-$ generally predict much larger values for the eEDM than what the SM does: more on the order of $d_e \sim 10^{-30} e\cdot\mathrm m$, which do fall within the range that we can measure. How do you actually measure them? Basically, by forgetting about high-energy particle colliders (which would need much higher collision energies than they can currently achieve to detect those dipole moments), and turning instead to the precision spectroscopy of atoms and molecules, and how they respond to external electric fields. The main physics at play here is that an electric dipole $\mathbf d$ in the presence of an external electric field $\mathbf E$ acquires an energy
$$
U = -\mathbf d\cdot \mathbf E,
$$
and this produces a (minuscule) shift in the energies of the various quantum states of the electrons in atoms and molecules, which can then be detected using spectroscopy. (For a basic introduction, see this video ; for more technical material see e.g. this talk or this one .) The bottom line, though, as regards this, Is there any experimental evidence to support or suspect the center of mass and charge of an electron must coincide? is that the current experimental results provide bounds for the eEDM, which has been demonstrated to be no larger than $|d_e|<8.7\times 10^{−31}\: e \cdot\mathrm{m}$ (i.e. the current experimental results are consistent with $d_e=0$), but the experimental search continues. We know that there must be some spatial separation between the electron's centers of mass and charge, and there are several huge experimental campaigns currently running to try and measure it, but (as is often the case) the only results so far are constraints on the values that it doesn't have. | {
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429,495 | I read the following from wikipedia: When a string is plucked normally, the ear tends to hear the
fundamental frequency most prominently, but the overall sound is also
colored by the presence of various overtones (frequencies greater than
the fundamental frequency). When I pluck a string, I just notice a node at each end and an antinode at the middle. How can we have overtones in addition to the fundamental frequency? It seems counterintuitive for me. | The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular. If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones. | {
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429,505 | If a mass rests at the edge of a disk, where friction acts on it, and the disk is rotated with an angular velocity just enough for this mass to slide out, then the linear velocity of the mass $v$ is given by $$\dfrac{mv^2}{R} = \mu_{\rm static}mg$$ Well, I don't understand why. It doesn't make sense to me because shouldn't this be the force to just counter-act the friction? How can the mass move if there is no acceleration to make the mass go that minimal amount off the disk? | The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular. If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones. | {
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429,708 | It seems impossible, yet I'm thinking that maybe because the ball compresses against the bat a bit it acts a little like a spring, and DOES travel faster than the bat? EDIT :
This is just a clarification, and not really part of the question, but I think it may be valuable. For people saying momentum is conserved, I'm not sure what you are imagining, but take a moment to think about the equation you keep on mentioning: $$M_\textrm{bat}V_\textrm{bat} = M_\textrm{ball}V_\textrm{ball}$$ This is saying that the bat somehow transfers ALL of its momentum to the ball. The only way this can ever happen is if the bat comes to a dead stop when it hits the ball, somehow holds it in place while transferring ALL of its momentum to it (that phrase doesn't even make logical sense), and then the ball flies off at a much larger velocity. WERE the bat floating through space and struck a ball, the bat would not stop when it hits the ball, and there is no way it makes sense for the ball to go off at THAT much of a larger velocity. Just imagine a spaceship very slowly drifting through space, and an astronaut who suddenly touches it. Conservation of momentum DOES NOT mean $$M_\textrm{ship}V_\textrm{ship} = M_\textrm{astronaut}V_\textrm{astronaut}$$ According to that, the astronaut would shoot off at hundreds (maybe thousands) of kilometers per hour when being touched by a massive spaceship, and the spaceship would come to a stop, but obviously that doesn't happen. TOTAL momentum is conserved, so that $$M_\textrm{ship1}V_\textrm{ship1} + M_\textrm{astronaut1}V_\textrm{astronaut1} = M_\textrm{ship2}V_\textrm{ship2} + M_\textrm{astronaut2}V_\textrm{astronaut2}$$ but even THAT equation doesn't even apply in the case of the baseball strike since there is a human being providing a force EVEN as the bat hits the ball. I know this is ingrained deeply in the minds of physics students because we have conservation of momentum drilled into our heads as young students in introductory physics, but I encourage everyone to always think intuitively about physics scenarios before applying equations. Anyways, I hope that was valuable. Cheers! | For an ideal heavy bat, the ball moves faster than its point of contact with the bat. Here's why. Suppose you swing the bat with velocity $+w$ and the ball comes in with velocity $-v$. Work in the reference frame of the bat. In this frame the ball has velocity $-v-w$. Since the bat is much heavier than the ball, and assuming the collision is elastic, the ball simply bounces off the bat as if it were a brick wall, ending up with velocity $v+w$. Transforming back to your frame, the ball ends up with velocity $v+2w$. This is indeed always greater than the speed of the bat. For example, if you hit the ball from a tee, so $v = 0$, then the baseball ends up going precisely twice as fast as the bat. This can also be understood from a force perspective. If you think of the bat and ball as squishing during impact like tiny springs, then at the moment they're moving at the same speed $w$, there is a sizable amount of energy stored in the springs. As the collision ends, the springs release this energy, increasing the speed of the ball over that of the bat. | {
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429,874 | ( I'm repeating myself a lot here, but it's because I want to make my confusion clear. ) If 2 billiard balls are the same exact mass, and one hits another stationary one head on, I have heard that the hitting ball will often stop entirely while the one which got hit will go off at the original velocity of the first one ( ignoring friction and heat and other potential loss of energy ). I understand that this agrees with conservation of momentum. However, now that I am thinking about it, I am a little confused as to how it is possible. Consider: Let's say we have 2 balls, Ball 1 and Ball 2. The balls have equal mass. $$M_{1} = M_2$$ Ball 1 is the hitting ball, with an original velocity of $V_1$ , and Ball 2 is getting hit, originally stationary. Once Ball 1 hits Ball 2, it immediately starts accelerating it at the same rate that it decelerates. In other words, Ball 1 exerts the same force on Ball 2 that Ball 2 exerts on Ball 1. In this way, momentum is conserved, so that for any amount of momentum that Ball 2 gains, Ball 1 loses. $$F_{1 \to 2} = F_{2 \to 1}$$ That's just Newton's third law. Since they have the same mass, Ball 1 will decelerate at the same rate Ball 2 accelerates. However, after a certain amount of time, both balls will have the same amount of momentum in the same direction. That is, Ball 1 will have been decelerated to ${V_{i}/2}$ and ball 2 will have been accelerated to ${V_{i}/2}$ . Ball 1 and Ball 2 at this point of equal momentum must have the same velocity, since they have the same mass. Now, the only way for Ball 1 to exert a force on Ball 2 is for them to be in contact. However, the instant after they gain the same velocity, the 2 balls would no longer be in contact, as Ball 2 would now be moving away from Ball 1, or at least at the same rate as Ball 1. That being said, it seems impossible that Ball 2 would ever become faster than Ball 1, since Ball 2 would only be able to be accelerated up to the point where it is going at the same velocity as Ball 1. And it seems even more impossible for Ball 1 to stop completely and Ball 2 to go off at the original velocity of Ball 1. What am I missing? EDIT: After reading some scrupulous comments, (not in a bad way though, I love it when there are new variables introduced and things pointed out, and all the "what if" questions allways make for awesome discussion, thanks for the comments!) I've realized its better to imagine the billiard balls as floating through space than on a pool table, to get rid of the possibility that spin will affect the collision. Or we could just imagine a frictionless pool table. But better to imagine them floating through space, because everyone loves space | Your analysis is correct up to the point when the two speeds are equal. This is also the point when the relative velocity is zero.
But at this point the two balls are deformed (elastically). The forces acting on the two balls during approach is an elastic force. However small the deformation is, it is there. So what happens now is that indeed the two bodies start to fall apart (the distance between them increases) and they start to expand back to the original shape. The interaction force starts to decrease (does not drop to zero instantly). The work done during this second part of the process is the same as during the first part so the kinetic energies and velocities change by another on half.
This is the difference between perfect elastic and non-elastic collisions.
In the later case the two bodies do not expand back to the same shape (or not at all) so the exchange of velocities is not complete. It may help if imagine a spring between the two bodies. When the velocities are equal, the spring is still compressed. Actually it has maximum compression. So obviously the force exerted by the spring is not zero as it starts to expand. Note :A nice example is considering a rubber ball bouncing off a ground(suppose perfectly elastically), in that situation there is also a point where the relative velocity with the ground is zero(when the ball is momentarily at rest with the ground) but still the ball rises up due to elastic forces as it gains kinetic energy due to the energy stored through the deformation it had initially endured. | {
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429,998 | By definition pressure is the perpendicular force applied to a unit area. So it has a direction which is perpendicular to the area. So it should be a vector. But I did sone googling and found out that it is a scalar quantity. So it would really be helpful if someone could show me how pressure is scalar with some mathematical derivation. | Pressure is proportionality factor. Area is the one that gives you direction. You have to recall that pressure is defined everywhere in the bulk volume, not just at the surface. A volume of gas has pressure defined everywhere. And the force direction is determined by you - by the way you orient your surface that you put into the gas. $$\vec{F}=p\vec{A}$$ Here you see that the area is the vector. Quoting wikipedia : It is incorrect (although rather usual) to say "the pressure is
directed in such or such direction". The pressure, as a scalar, has no
direction. The force given by the previous relationship to the
quantity has a direction, but the pressure does not. If we change the
orientation of the surface element, the direction of the normal force
changes accordingly, but the pressure remains the same. I should clarify, that this calculates the force caused by the pressure, so it GIVES you the force perpendicular to the area, given the area vector. It's the defining equation and the only one that captures what pressure actually does, so it's always true, but needs to be understood as a formula for calculating the force from pressure. If $\vec{F}$ is just caused by the pressure, it can't be anything else but perpendicular to the area, otherwise you have other forces present in the system, or the liquid is not isotropic. That being said, assuming that $\vec{F}$ is only caused by the pressure, you could calculate $p$ by taking absolute values: $$p=\frac{|F|}{|A|}$$ Mathematically, you transformed a vector equation into a scalar equation assuming parallel vectors, so now you're not allowed to put in anything, but only lengths (or projections - similar argument) of F and A that are guaranteed to have been parallel, otherwise you get nonsense. You also lose the sign of pressure (for gasses, it can't happen, but for elastic solids or liquids, it can "pull" due to intermolecular forces). However, strictly speaking, pressure is a tensor , but for gasses, it's isotropic, so it acts as a scalar. Without going into details what a tensor is, imagine above that in $p\vec{A}$ , $p$ can also transform the direction, not just the magnitude of $\vec{A}$ , so the force does not have to point perpendicularly. This is true in elastic solids, where you can transmit sideways forces to the surface, and in viscous flowing liquids, where the viscous force is also just a stress (generalized pressure) transmitted to the surface. In this situation, $p$ has $6$ independent components, so you cannot measure it just by measuring a force on a single surface. You'd need to measure all components of force on 3 surfaces placed in different orientations. Only in gasses, you can rely on the force having the same magnitude no matter the orientation. Further reading: https://en.wikipedia.org/wiki/Cauchy_stress_tensor | {
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430,419 | In Newtonian gravity, an infinite volume filled with a uniform distribution of mass would be in perfect equilibrium. At every point, the gravitational forces contributed by masses in one direction would be exactly counterbalanced by those in the opposite direction. But when Einstein tried to apply General Relativity to possible cosmologies, he found it necessary to include the cosmological constant in order to get a static universe. In qualitative terms, it seems to me that the gravitational stresses that the masses would impose on the spacetime should all cancel out, and likewise, that the resulting flat spacetime should have no effect on the motion of the masses. However, the math of the situation is beyond my current skills, so I'm asking how it produces the nonequilibrium condition? (I realize that such an equilibrium solution might not be stable, and that there are many other very good reasons to believe in an expanding universe, so I'm not trying to promote any alternative theories. I'm just curious about this particular point. ) | This is a rather subtle question, which confused even Newton . It is very tempting to think that an initially static Newtonian universe with perfectly uniform mass density will not collapse, because the gravitational force cancels everywhere by symmetry. This is wrong. Here's an analogous question: suppose a function $f$ obeys $$f''(x) = 1$$ and we want to solve for $f(x)$ . Since every point on the real line is the same as every other point, we might think that by symmetry, $$f(x) = \text{constant}.$$ But this is completely wrong, because the second derivative of a constant is zero. And stepping back, the whole question doesn't make any sense, because there isn't enough information. To solve a general differential equation, you need boundary conditions. One possible boundary condition is that the solution looks approximately even at infinity. That's enough to specify the solution everywhere, as $$f(x) = \frac{x^2}{2} + \text{constant}.$$ But now the translational symmetry has been broken: not every point is equivalent anymore, because we have a minimum at $x = 0$ . This is inevitable. You can't solve the differential equation without boundary conditions, and any choice of boundary conditions breaks the symmetry . Similarly in Newton's infinite universe we have $$\nabla^2 \phi = \rho$$ where $\rho$ is the constant mass density and $\phi$ is the gravitational potential, corresponding to $f$ in the previous example. Just as in that example, we "obviously" have by symmetry $$\phi(x) = \text{constant}$$ which indicates that the force vanishes everywhere. But this is wrong. Without boundary conditions, the subsequent evolution is not defined; it is like asking to solve for $x$ given only that $x$ is even. With any set of boundary conditions, you will have a point towards which everything collapses. So the answer to your question is that both the Newtonian and relativistic universes immediately start to collapse; the symmetry argument does not work in either one, so there is nothing strange to explain. The reason this point isn't mentioned in most courses is that we often assume the gravitational potential goes to zero at infinity (in Newtonian gravity) or that the metric is asymptotically flat (in relativity). But this boundary condition doesn't work when the mass distribution extends to infinity as well, which leads to the pitfall here. The same point can lead to surprises in electrostatics . We reasoned above in terms of potentials. A slightly different, but physically equivalent way of coming to the same conclusion is to directly use fields, by integrating the gravitational field due to each mass. In this case, the problem is that the field at any point isn't well-defined because the integrals don't converge. The only way to ensure convergence is to introduce a "regulator", which makes distant masses contribute less by fiat. But any such regulator, by effectively replacing the infinite distribution with a finite one, introduces a center towards which everything collapses; just like the boundary conditions, any regulator breaks the symmetry. So, again, collapse immediately begins. In the end, both the Newtonian and relativistic universes immediately begin to collapse, and in both cases this can be prevented by adding a cosmological constant. In the Newtonian case, this is simply the trivial statement that $\nabla^2 \phi = \rho - \Lambda$ has constant solutions for $\phi$ when $\rho = \Lambda$ . However, in both cases the solution is unstable: collapse will begin upon the introduction of any perturbations. | {
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431,010 | I recently read this article which claims that last year’s LIGO observation of gravitational waves is proof that, at least on massive scales, there cannot be more than three spatial dimensions. I don’t understand the physics fully, so could someone please explain this to me? I know it’s been theorized that gravity is relatively weak when compared to other forces because it leeches into other dimensions, and I think I understand how these observations disprove that , but how does this prove that there must be three and only three spatial dimensions? | I’m the lead author of the paper. Thanks for being interested in the work! Your question is a good one. Really, our work can’t say anything about extra spatial dimensions if they’re not doing anything to gravity or light. As you correctly mention, we can only constrain higher dimensions where gravity is actually leaking into them. If there are higher dimensions, but our physics experiments can’t see or hear them, are they really there? :p (this isn’t to say there might not be other ways of detecting extra spatial dimensions — but really, if they aren’t affecting physics in any measurable way, there’s not much we can say) | {
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431,017 | Consider a system that takes excess air (assume it to be an ideal gas, $Cp = \frac{7}{2} R$ ) from a power plant stack at $370 \rm K$ (typo in figure) and $2$ bar and compresses it to 20 bar adiabatically and reversibly. Since the air will heat as it compresses, we can send the outlet to a Carnot engine, and produce enough work to power the compressor by rejecting the heat to the water below a frozen at $273 \rm K$ . The temperature of the expelled air is $275 \rm K$ We still have compressed air at $20$ bar coming out of the Carnot engine to use elsewhere! Free energy. This cannot be possible. So, which law is being violated in this system? Since the compressor is adiabatic and reversible, we know that $Q = 0 \implies \Delta S= 0$ . From the adiabatic relations, we know that the temperature in the outlet of the compressor is $714.36 \rm K$ . The work done by the compressor is a flow work given by $$\int V \rm d P = \frac{\gamma}{\gamma - 1} RT_1 \Bigg(\Big(\frac{P_2}{P_1} ^{\frac{\gamma -1}{\gamma}} -1\Big)\Bigg) = 4097.89 J$$ Onto the Carnot engine. We know that $T_{hot} = 614.36 \rm K$ and $T_{cold} = 275 \rm K$ .
Using the theorem of Clausius for a Carnot engine, $\frac{Q_{hot}}{T_{hot}} + \frac{Q_{cold}}{T_{cold}} = 0$ . We know that $Q_{cold} = C_p\Delta T = 12784.9 J$ . Plugging this value into the theorem of Clausius, $Q_{hot} = 33451.9 J$ . Using $Q_{hot} = Q_{cold} + W_{C}$ , we get the work done by the system is $20667.9J$ . Hence, the reversible adiabatic compressor is getting more work done on it and it is doing less work than it is getting. So this violates the first law, right?
Does it also violate the second law? Any advice would be appreciated. | I’m the lead author of the paper. Thanks for being interested in the work! Your question is a good one. Really, our work can’t say anything about extra spatial dimensions if they’re not doing anything to gravity or light. As you correctly mention, we can only constrain higher dimensions where gravity is actually leaking into them. If there are higher dimensions, but our physics experiments can’t see or hear them, are they really there? :p (this isn’t to say there might not be other ways of detecting extra spatial dimensions — but really, if they aren’t affecting physics in any measurable way, there’s not much we can say) | {
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431,861 | Quoting from the Feynman Lectures on Physics - Vol I : The atoms are 1 or $2 \times 10^{−8}\ \rm cm$ in radius. Now $10^{−8}\ \rm cm$ is called an angstrom (just as another name), so we say they are 1 or 2 angstroms (Å) in radius. Another way to remember their size is this: if an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple. How does this hold true? Let us assume the radius of an average apple is about $6\ \rm cm$ ( $0.06\ \rm m$ ). The radius of earth is about $6371\ \rm km$ ( $6371000\ \rm m$ ). Therefore, a $\frac{6371000\ \mathrm{m}}{0.06\ \mathrm{m}} = 106183333.33$ magnification, i.e., a magnification of about $10^{-8}$ times is required to magnify an apple to the size of the earth. If we magnify an atom of size say 1 angstrom ( $10^{-10}\ \rm m$ ) by $106183333.33$ times, we get $0.0106\ \rm m$ or $1.06\ \rm cm$ only. The atom has not been magnified to the size of the original apple. How does the quoted statement in the book hold good? | In " back of the envelope " calculations like this, all you can really do is look at orders of magnitude. As others have pointed out, not all apples have the same size, and not all atoms have the same "size". All we can then work with is orders of magnitude, so $1\ \rm cm$ and $6\ \rm cm$ (although many people in the comments are saying $3\ \rm cm$ ) should be considered to be the "same", since we could be dealing with apples and atoms of various sizes. Indeed, if an apple's radius is on the order of $10^{-2}\ \rm m$ , and if the Earth's radius is on the order of $10^6\ \rm m$ , then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$ . If our atom radius is on the order of one Angstrom, or $10^{-10}\ \rm m$ , then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^{-2}\ \rm m$ , which is what we were trying to show in the first place. You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the estimated size of an atom is on the order of $10^{-10}\ \rm m$ . $^*$ Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so if you take all of this into account I would say the book is correct. $^*$ Although I have to admit, I tend to remember the size of an atom, and I can't seem to keep in my head the order of magnitude of the Earth's radius. So I might use this memory tool the other way around to remember that the size of the earth is about 8 orders of magnitude larger than an apple. | {
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432,137 | The 2018 Nobel Prize in Physics was awarded recently, with half going to Arthur Ashkin for his work on optical tweezers and half going to Gérard Mourou and Donna Strickland for developing a technique called "Chirped Pulse Amplification". In general, optical tweezers are relatively well known, but Chirped Pulse Amplification is less well understood on a broader physics or optics context. While normally the Wikipedia page is a reasonable place to turn to, in this case it's pretty technical and flat, and not particularly informative. So: What is Chirped Pulse Amplification? What is the core of the method that really makes it tick? What pre-existing problems did its introduction solve? What technologies does it enable, and what research fields have become possible because of it? | The problem Lasers do all sorts of cool things in research and in applications, and there are many good reasons for it, including their coherence, frequency stability, and controllability, but for some applications, the thing that really matters is raw power. As a simple example, it had long been understood that if the intensity of light gets high enough, then the assumption of linearity that underpins much of classical optics would break down, and nonlinear optical phenomena like second-harmonic generation would become available, getting light to do all sorts of interesting things. Using incoherent light sources, the required intensities are prohibitively high, but once the laser was invented, it took only one year until the first demonstration of second-harmonic generation , and a few short years after that until third-harmonic generation , a third-order nonlinear process that requires even higher intensities. Put another way, power matters, and the more intensity you have available, the wider a range of nonlinear optical phenomena will be open for exploration. Because of this, a large fraction of laser science has been focused on increasing the available intensities, generally using pulsed lasers to achieve this and with notable milestones being Q-switching and mode-locking . However, if you try to push onward with a bigger laser amplifier and more and more power, you are basically destined sooner or later to hit a brick wall, rather brusquely, in the form of catastrophic self-focusing . This is a consequence of yet another nonlinear effect, the Kerr effect , happening inside the laser medium itself. At face value, the Kerr effect looks harmless enough: basically, it says that if the intensity is high enough, the refractive index of the material will rise slightly, in proportion to the intensity: $$
n(I) = n_0 + n_2\: I.
$$ So, what's the big deal? In short, if you have a laser beam propagating through such a medium, then the intensity of the light will be higher in the center, which means that the refractive index will be higher in the center. In other words, the material's optical properties will look like those of a convex lens, and it will tend to focus the beam. This will tend to make the beam sharper, which will increase the intensity at the center, which will raise the refractive index at the center even higher... ... which will then focus the beam even more tightly, leading to higher and higher intensities. This makes up a positive feedback loop, and if the initial intensity is high enough, the medium is long enough, and there isn't enough initial diffraction to counteract it, then it will spiral out of control and cause catastrophic laser-induced damage in the very medium that you're trying to use to amplify that laser beam. (Moreover, it is quite common, particularly in air, that the laser will diffract on the damaged spot and then re-self-focus a bit further down the line, a phenomenon known as laser filamentation . If you get things just right wrong, this can propagate a failure in the gain medium up to the destruction of an entire beamline.) Image source This sounds like a funky mechanism, but it was a huge roadblock for a very long time. If you plot the highest laser intensity available at different times since the invention of the laser, it climbs quickly up during the sixties, and then it hits a wall and stays put for some ten to fifteen years: Image source This represents the barrier of Kerr-lens self-focusing, and at the time the only way to overcome it was to build a laser which was physically bigger, to dilute the intensity over more gain medium to try to prevent the problem. Until, that is, Chirped Pulse Amplification came around to solve the problem. The solution At its core, Chirped Pulse Amplification (CPA) works by diluting the light, so that it can be amplified to a larger total power without reaching a dangerous intensity, but it does this stretching in time , i.e. longitudinally along the laser pulse. The basic sequence consists of four steps: First of all, you start with a short laser pulse that you want to amplify You then stretch it in time, by introducing chirp into the signal: that is, you use some sort of dispersive element, like a prism or a diffraction grating, which decomposes the pulse into all of its constituent colors and sends the longer wavelengths first and the shorter wavelengths last. This will naturally reduce the intensity of the pulse. (Why "chirp"? because the upward (or downward) sweep of frequencies over the pulse is precisely what gives bird chirps their characteristic sound.) You then pass this lower-intensity pulse through your laser amplifier, which is safe because the instantaneous intensity is below the self-focusing damage threshold of your gain medium. Finally, you pass your pulse through a reversed set of gratings which will undo the relative delay between the longer- and shorter-wavelengths of your pulse, putting them all together into a single pulse of the same shape and length as your original pulse... ... but at the much higher amplified power, and at intensities which would be impossible to achieve safely using direct amplification of the pulse. The core feature that makes the method tick is the fact that, when done correctly, the stretching of the pulse will completely conserve the coherence between the different frequency components, which means that it is fully reversible and when you add a cancelling chirp the pulse will go back to its initial shape. Furthermore, the method relies on the fact that stimulated emission will completely duplicate, in a coherent way, the photons that it is amplifying, which means that the photons that are introduced by the amplification will have the same frequency and phase characteristics as the initial pulse, which means that when you remove the chirp from the amplified pulse the added-in photons will also compress into a tight envelope. Applications Like I said at the beginning, CPA is particularly useful in places where raw laser power, and particularly concentrated laser power, is of paramount importance. Here are some examples: In the same way that lasers gave us nonlinear optics, CPA has been integral in the development of high-order harmonic generation which has pushed past the second- or third-order harmonics to happily produce tens or hundreds of harmonics. (The current record goes all the way to harmonic 5,000 .) This isn't only 'more', it's qualitatively different: it pushes nonlinear optics to regimes where the usual perturbative expansion completely breaks down, and where it needs to be replaced with a completely new set of tools, which revolve around the so-called three-step model , and which involve a nice and quite particular new interface between classical and quantum mechanics, where trajectories do (sort of) exist but over complex-valued time and space, due to the presence of quantum tunnelling . It has also helped push the study of light-matter interaction past that same perturbative limit, giving us the tools to extract electrons from molecules and control them in very precise ways, thereby allowing for the creation of tools like e.g. laser-driven electron diffraction , which can be used to image the shapes of molecules as they undergo bending and other vibrations. CPA also underpins several breakthrough measurements which have been touched on previously on this site, including the observation of the time-dependent waveform of a light pulse , itself done using high-order harmonic radiation; the observation of charge oscillations when atoms are placed in excited states , again using HHG; or performing electron holography from an atomic target using electrons pulled from that same atom. Of course, all the cool laser-driven QED stuff at the top of that second diagram: if your laser is strong enough that, if you release an electron into the focus, the kinetic energy of its oscillations will exceed $m_e c^2$ , then you can start to have things like laser-driven pair creation, and all sorts of fun stuff. Some of it is already on the table, some of it is in achievable plans for the future, and all of it is made possible by CPA. CPA is also extremely useful in delivering sharply controlled bursts of power to materials. This is extremely useful in laser micromachining , for example, where it is routinely used in e.g. using short laser pulses to etch waveguides into dielectrics, which are then extremely useful for chip-based quantum computation and quantum information processing. Similarly, the ability to deliver sharply controlled bursts of power is extremely useful in laser microsurgery, and there are several types of eye surgery that exclusively use CPA pulses to provide sharp 'kicks' of power which perform cleaner incisions. On a much larger scale, when you really turn up the power to the maximum, CPA is a vital component of laser wakefield acceleration , which uses the ionized pocket left behind by an intense laser pulse as it travels through a gas to accelerate electrons to energies that would otherwise require an extremely large particle accelerator, but which are now available using a much more modest table-top laser system. Further reading Some additional resources for further reading: The Nobel Prize's scientific background and popular information documents make excellent reading, and people don't go looking at those sources anywhere near enough. Go check them out! The original paper: Compression of amplified chirped optical pulses. D. Strickland and G. Mourou. Optics Comms. 55 , 447 (1985) . What power limitations was chirped radar designed to overcome? , a previous question of mine on similar radar technologies that predated CPA. This is a nice tutorial on U Michigan . | {
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432,798 | Assume you wanted to build a particle accelerator in a non-commerical/non-residential area. It costs more money the deeper you want to build it, therefore you want to build it as close to ground level as possible. Why aren't particle accelerators built on ground level? What is the shallowest depth at which particle accelerators can be feasibly built, and what are the equations such as synchrotron radiation or luminosity interference (or, at least, the phenomena, and not necessarily the equations behind them) that determine this? My speculation: Some situation (forgot where and when) in which a stray particle from an accelerator hit someone ended in them from the effects of being by a high-energy (hadron?). Also, a Fermilab researcher who taught one of my classes told us about one of the times in which some loose particles found their way out of the accelerator and shot an inches-wide hole through a steel beam in a fraction of a fraction of a second. Now, I doubt the particle-accelerator engineers sat down in a conference and said 'we must build them below ground or else particles beams could tear through people' but this is the only drawback I know of that comes with a ground-level accelerator; it can accidentally release fairly high-energy particles that can hit things. Solar radiation might also have notable effects, but I am not sure. | The main reason for going underground is that the earth above provides some radiation shielding. An accelerator where everything is working properly is (outside the beam pipe) a relatively low-radiation environment. However if you have a steering or focusing magnet malfunction, so that the beam spills out of the pipe, you can briefly generate lots of prompt radiation. The amount of shielding that you need depends on the energy of the accelerator. For example, The 12 GeV electron accelerator at JLab is seven or eight meters underground --- just a couple of flights of stairs. The 1 GeV proton machine at the Spallation Neutron Source is actually at ground level, but there's an earthen berm above it. The (shuttered) 25 MV tandem accelerator at ORNL actually did most of its acceleration in a tower aboveground, and the various beam pathways are in a single above-ground building. The lower the energy of your accelerator is, the less you need earthen shielding for safety reasons. Another answer points out that background-limited experiments go underground to reduce cosmic ray backgrounds. This is a reason to put your detectors underground, but not necessarily a reason to put your accelerator underground. | {
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432,807 | Let's say you wanted a top-of-the-line, impenetrable suit to withstand high levels of radiation. How much does one of these suits cost? | The main reason for going underground is that the earth above provides some radiation shielding. An accelerator where everything is working properly is (outside the beam pipe) a relatively low-radiation environment. However if you have a steering or focusing magnet malfunction, so that the beam spills out of the pipe, you can briefly generate lots of prompt radiation. The amount of shielding that you need depends on the energy of the accelerator. For example, The 12 GeV electron accelerator at JLab is seven or eight meters underground --- just a couple of flights of stairs. The 1 GeV proton machine at the Spallation Neutron Source is actually at ground level, but there's an earthen berm above it. The (shuttered) 25 MV tandem accelerator at ORNL actually did most of its acceleration in a tower aboveground, and the various beam pathways are in a single above-ground building. The lower the energy of your accelerator is, the less you need earthen shielding for safety reasons. Another answer points out that background-limited experiments go underground to reduce cosmic ray backgrounds. This is a reason to put your detectors underground, but not necessarily a reason to put your accelerator underground. | {
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433,427 | I've been wondering this for a while but I have not yet encountered an explanation. This is from my understanding of physics, which is by no means expert, so sorry for my crude explanation: Energy within earth can be considered a closed system; it transforms but cannot be created or destroyed -- and from what I understand, heat seems to be its most natural form, so it will always end up like that in some way. Two things affect the total sum of energy on earth: radiation into space will drain energy (and is limited because of the presence of atmosphere). Radiation from the sun adds energy to the system. Basically the sun is our only real source of energy (and we can consider it limitless, since when the sun is exhausted, we're over anyway). The way I understand it, solar panels increase the efficiency of how we 'harvest' this solar energy, reflecting less of it back into space, and turning more of it into en energy (in this case, electrical). So we take more energy from the sun by putting solar panels in place. But the amount of energy that is removed from the system stays the same. Hence the total sum of energy on earth increases (more) when we use solar panels. So how come we consider them to be a way to counter global warming, instead of a contributing factor? edit: since all answers are about comparing solar cells to fossil fuels, let me clarify a bit more. I understand that fossil fuels will contribute more to climate change than solar cells -- but I just wanted to clarify that is seems to me that both are a net negative (not if you replace one by the other). In other words, that the idea that solar cells are 100% clean (apart from production cost), is not really true, then. Wind of hydro however, would be, since they use energy that is already present in the earth system. (and of course what we really need to do is require less energy) | The purpose of solar cells is to generate electricity. This can replace the electricity generated by burning fossil fuels for electricity. The fact that it's becoming practical to run vehicles on electricity means we can also replace the fossil fuels burned to power vehicles, which makes things even better. But let's concentrated on generating electricity. Your analysis ignores two things, one minor and one absolutely crucial: The minor point: Generating electricity by burning fossil fuels also adds heat to the planet. For example, only about 1/3 of the energy liberated by burning coal in a coal power plant is turned into electricity; the rest is waste heat. The major point: Fossil-fuel power plants continually produce CO2. This post on RealClimate does an excellent job of going through the details. To take an unrealistic extreme case, they assume that solar cells are perfectly black (albedo = 0), and they ignore the fact that real solar cells are sometimes installed on already dark surfaces (such as roofs). In order to generate the current world electricity supply of 2 trillion watts, perfectly black solar cells would add about 6.7 trillion watts due to waste heat. As they point out, the efficiency of fossil fuel plants means 2 trillion watts of electrical power would be accompanied by about 6 trillion watts of waste heat. So if you replace fossil fuel power plants with solar-cell power plants, you don't really change the waste heat production. But you do change the CO2 production, and that's crucial, because the heat added to the atmosphere by adding CO2 is orders of magnitude larger than the waste heat from the power-generation process itself. (This is a continuing process: every second you run the fossil-fuel power plants, you add more CO2 to the atmosphere.) ... by the time a hundred years have passed, the heat trapped each
year from the CO2 emitted by using coal instead of solar energy to
produce electricity is 125 times the effect of the fossil fuel waste
heat. And remember that the incremental waste heat from switching to
solar cells is even smaller than the fossil fuel waste heat. What’s
more, because each passing year sees more CO2 accumulate in the
atmosphere, the heat trapping by CO2 continues to go up, while the
effect of the waste heat from the fossil fuels or solar cells needed
to produce a given amount of electricity stays fixed. (You can, if you like, argue that getting rid of electricity generation entirely -- closing all power plants, solar or fossil-fuel-powered -- would be marginally better than converting electricity generation to solar. But that's a very small difference, and not really an option if you want to continue to have some kind of human civilization on the planet.) | {
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433,863 | In 3 spatial dimensions, $$[E] = [ML^2 T^{-2}]$$ Would it change in higher dimensions? If yes, then what would be the dimensions for 4 spatial dimensions? | Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=\frac{1}{2}m\vec{v}^2$ . I include the vector notation for the velocity here because it's useful to keep in mind that $\vec{v}^2=\vec{v}\cdot\vec{v}$ . If the particle is moving in two dimensions, then $\vec{v}=v_x\hat{x}+v_y\hat{y}$ and $\vec{v}\cdot\vec{v}=v_x^2+v_y^2$ , which has units of velocity squared. In three dimensions, $\vec{v} = v_x\hat{x}+v_y\hat{y}+v_z\hat{z}$ and $\vec{v}\cdot\vec{v} = v_x^2+v_y^2+v_z^2$ , which still has the units of velocity squared. In four dimensions, $\vec{v} = v_x\hat{x}+v_y\hat{y}+v_z\hat{z}+v_w\hat{w}$ , and $\vec{v}\cdot\vec{v}=v_x^2+v_y^2+v_z^2+v_w^2$ , which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged. We can also see this more generally by the formal definition of work, which is the definition of change in energy: $$W=\int_C \vec{F}\cdot d\vec{s}$$ for an object being acted on by a force $\vec{F}$ moving along a path $C$ with arclength parameter $d\vec{s}$ . This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same. | {
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434,634 | I have been on a date recently, and everything went fine until the moment the girl has told me that the Earth is flat. After realizing she was not trolling me, and trying to provide her with a couple of suggestions why that may not be the case, I've faced arguments of the like "well, you have not been to the space yourself". That made me think of the following: I myself am certain that the Earth is ball-shaped, and I trust the school physics, but being a kind of a scientist, I could not help but agree with her that some of the arguments that I had in mind were taken by me for granted. Hence, I have asked myself - how can I prove to myself that the earth is indeed ball-shaped, as opposed to being a flat circle (around which the moon and the sun rotate in a convenient for this girl manner). Question: Ideally I want to have a proof that would not require travelling more than a couple of kilometers, but I am fine with using any convenient day (if e.g. we need to wait for some eclipse or a moon phase). For example, "jump an a plane and fly around the Earth" would not work for me, whereas "look at the moon what it is in phase X, and check the shape of the shade" would. Trick is, I know that it is rather easy to verify local curvature of the Earth by moving away from a tall object in the field/sitting on the beach and watching some big ship going to the horizon. However, to me that does not prove immediately that globally the Earth has same/similar curvature. For example, maybe it's just a shape of a hemisphere. So, I want to prove to myself that the Earth is ball-shaped globally, and I don't want to move much to do this. Help me, or tell me that this is not possible and why, please. As an example, most of the answers in this popular thread only focus on showing the local curvature. P.S. I think, asking how to use physics to derive global characteristics of an object from observing things only locally (with the help of the Sun and the Moon, of course) is a valid question, but if something can be improved in it, feel free to tell me. Thanks. Update: I have not expected such a great and strong feedback when asking this question, even though it is indeed different from the linked ones. Them are still very similar, which was not grasped by all those who replied. I will thoroughly go over all the answers to make sure which one fits the best, but in the meantime if you would like to contribute, please let me clarify a couple of things regarding this question: they were in the OP, but perhaps can be made more obvious. I do not have a goal of proving something to this date. I see that mentioning her might have been confusing. Yet, before this meeting I was certain about the shape of the earth - but her words (even though I think she's incorrect in her beliefs) made me realize that my certainty was based on the assumption I have not really questioned. So sitting on a beach with another friend of mine (both being ball-believers) we thought of a simple check to confirm our certainty, rather than to convince anyone else in us being right. I am only looking for the check that would confirm the GLOBAL shape of the earth being ball-like. There were several brilliant answers to another question that worked as a local curvature proof, and I am not interested in them. I am looking for the the answer that will show that the Earth is ball-shaped (or rather an ellipsoid), not that it is not flat. There are many other great shapes being neither ball/ellipsoid nor flat. I do still have an assumption that this shape is convex, otherwise things can go too wild and e.g. projections on the Moon would not help us. I think point 1. shows why is that a valid physics/astronomy question, rather than playing a devil's advocate defending the flat Earth hypothesis, and I would also happily accept the answer like you cannot show this by not moving for 20k kilometers because A, B, C if there's indeed no simple proof. At the same time, points 2 and 3 should distinguish this question from the linked ones. | I love this question, because it's a very simple demonstration of how to do science. While it's true that in science one should never accept anything 'without evidence', it's also true that blind skepticism of everything and anything gets one nowhere - skepticism has to be combined with rational inquiry. Your date has gotten the 'skepticism' part of science, but she's failed to grasp the equally-crucial part where one looks at the evidence and thinks about what the evidence implies. You cannot just refuse to think or accept evidence. If your goal is to learn nothing, then nothing is what you'll learn. There are many, many ways of verifying that the Earth is not flat, and most of them are easy to think about and verify. You certainly do not need to go to space to realize the Earth is round! If the Earth is flat, why can't you see Mt. Kilimanjaro from your house? Mt. Kilimanjaro is tall, probably taller than anything in your immediate neighborhood (unless you live in a very deep valley) and so the question is why wouldn't you be able to see it from anywhere on Earth? Or, for that matter, why can't you see it from even closer? You have to be really close, in planetary terms, to be able to see it. This wouldn't be true if the Earth were flat! One might argue that this is just because of the scattering of the atmosphere. Distant objects appear paler, so probably after some distance you can't see anything at all. So then let's think about things that are closer. Stand on the ground and the horizon appears only a few km away. Go to the top of a hill, or a large tower, and suddenly you can see things much farther away. Why is this the case if the Earth is flat? Why would your height above ground have anything to do with it? If I raise or lower my eyes with respect to a flat table, I can still see everything on that table. The 'horizon' of the table never appears closer. If the Earth is flat, why do time zones exist? Hopefully your date realizes that time zones exist. If not, it's pretty easy to verify by doing a video call with someone in a distant location. The reason for time zones, of course, is that the sun sets and rises at different times at different parts of the globe. Why would this be the case? On a flat Earth, the sun would rise and set at the same time everywhere. If the Earth is flat, why is the Moon round? The moon is round and not a flat disc, as you can see by the librations of the moon. What makes the Earth special, then? Further, all the planets are round, although to verify this you need a good telescope. Again, what makes the Earth special? If the Earth is flat, then what is on its 'underside'? Hanging dirt and leaves? A large tree? Turtles? Those who reject the roundness of Earth either have no explanation or their explanation is based on much less solid grounding than the pro-round arguments (which, of course, is because the Earth is not flat). If there is 'nothing' under the Earth, then lunar eclipses would make no sense as the Earth needs to be between the Moon and the Sun. EDIT: As to the question of whether the Earth is round or some weird hemisphere/pear/donut shape, among other things those would all lead to a situation where gravity is wrong. For a hemisphere for example, gravity would not point down (towards the Earth) at any point on the Earth's surface unless if you were sitting right at the top of the hemisphere. Similar arguments can be made for the other shapes. Sure, it's possible to make it 'work' by doing even stranger things like altering the distribution of mass and so on, but at that point you've gone very far into violating Occam's razor. | {
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434,985 | What is the experimental evidence that the nucleons are made up of three quarks? What is the point of saying that nucleons are made of quarks when there are also gluons inside it? | What is the experimental evidence that the nucleons are made up of
three quarks? Some strong pieces of evidence for the quark model of the proton and the neutron, not stated in another answer, are the magnetic moment of the proton and the magnetic moment of the neutron , which are consistent with the quark model and are inconsistent with the magnetic moment that would be predicted by quantum electrodynamics in a point particle model. What is the point of saying that nucleons are made of quarks when
there are also gluons inside it? The reasons this is done is called in the field of science communication " lies to children ". Complex topics are often initially taught in a manner the oversimplifies the reality in order to develop key salient points. Emphasizing the quark composition of the nucleons while ignoring the gluon contribution allows one to explain many key conclusions of the quark-gluon model including: The charge of all of the hadrons Beta decay The list of all possible baryons and of all possible pseudoscalar and vector mesons The full list of Standard Model fermions The magnetic moment of hadrons Deep inelastic scattering of hadrons A formula for hadron spin The existence of gluons as a constituent doesn't have to be explained in detail to reach these results. Also, while the valence quark content of a hadron is specific to a particular kind of hadron, the gluon content of a hadron is not useful for hadron taxonomy. | {
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434,993 | I'm solving a question out of the textbook and it reduces to the following. a particle of mass 2m with velocity $v_0$ collides with a particle of mass m at rest. The collision is elastic. So using conservation of momentum we can deduce that $$2m\vec{v_0 }= 2m \vec{u_1} + m\vec{u_2} $$ Here $u_1$ is the final velocity in the same reference frame of the 2m particle, etc. We also know that energy is conserved: $$\frac{1}{2}2m\vec{v_0 }^2= \frac{1}{2}2m \vec{u_1}^2 + \frac{1}{2}m\vec{u_2}^2 $$ So we are given the system, after simplifying $$2\vec{v_0 }= 2 \vec{u_1} + \vec{u_2} $$ $$2\vec{v_0 }^2= 2\vec{u_1}^2 + \vec{u_2}^2 $$ I would like to get the $u_1$ and $u_2$ in terms of each other so I did the following: $$2\vec{v_0 }= 2 \vec{u_1} + \vec{u_2} $$ Dot both sides with itselves $$4\vec{v_0 }^2= 4 \vec{u_1}^2 + \vec{u_2}^2 +4 \vec{u_1}\cdot\vec{u_2} $$ Substitute into the energy equation $$\frac{1}{2}(4 \vec{u_1}^2 + \vec{u_2}^2 +4 \vec{u_1}\cdot\vec{u_2}) = 2\vec{u_1}^2 + \vec{u_2}^2$$ $$(2 \vec{u_1}^2 + \frac{1}{2}\vec{u_2}^2 +2 \vec{u_1}\cdot\vec{u_2}) = 2\vec{u_1}^2 + \vec{u_2}^2$$ Subtract $u_1$ squared terms $$(\frac{1}{2}\vec{u_2}^2 +2 \vec{u_1}\cdot\vec{u_2}) = \vec{u_2}^2$$ Subtract $u_2^2$ terms and multiply by 2 $$4 \vec{u_1}\cdot\vec{u_2} = \vec{u_2}^2$$ Now here is the suspect step: can I "antidot" and do the following: if yes, why is this phyiscally correct, if not, what properties makes it incorrect: $$4\vec{u_1} = \vec{u_2}$$ | What is the experimental evidence that the nucleons are made up of
three quarks? Some strong pieces of evidence for the quark model of the proton and the neutron, not stated in another answer, are the magnetic moment of the proton and the magnetic moment of the neutron , which are consistent with the quark model and are inconsistent with the magnetic moment that would be predicted by quantum electrodynamics in a point particle model. What is the point of saying that nucleons are made of quarks when
there are also gluons inside it? The reasons this is done is called in the field of science communication " lies to children ". Complex topics are often initially taught in a manner the oversimplifies the reality in order to develop key salient points. Emphasizing the quark composition of the nucleons while ignoring the gluon contribution allows one to explain many key conclusions of the quark-gluon model including: The charge of all of the hadrons Beta decay The list of all possible baryons and of all possible pseudoscalar and vector mesons The full list of Standard Model fermions The magnetic moment of hadrons Deep inelastic scattering of hadrons A formula for hadron spin The existence of gluons as a constituent doesn't have to be explained in detail to reach these results. Also, while the valence quark content of a hadron is specific to a particular kind of hadron, the gluon content of a hadron is not useful for hadron taxonomy. | {
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435,404 | If we have two lasers based at the equator and we direct one of them east and the other west on to two screens each 10 km away, will light emitted synchronously at the lasers arrive at the same time as each other on thier respective screens? My interest is that one laser moves with the rotation of the earth and the other one the opposite. From what I know the light shouldn't be affected by the rotation of the earth so the laser on the west side will arrive faster than the east side since the screen on the west side will come to the laser reference point.
Correct me if I'm wrong. (with mathematics if you can) | It depends on your frame of reference. If you are positioned by the laser on the Earth's surface you would observe the two laser beams to have the same speed and your would observe the two targets to be stationary and you would observe the light to take the same amount of time to reach each target. On the other hand if you were not at rest compared to the laser and screens, say for example you are observing from the moon, you would still observe the laser beams having the same speed but you would see the screen as moving at about 460 m/s. From this frame of reference you would see the light hit the screen to the west first. What is simultaneous in one frame of reference is not necessarily simultaneous in a different frame of reference. | {
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437,355 | Why doesn't current pass through a resistance if there is another path without resistance? How does it know there is resistance on that path? Some clarification: I understand that some current will flow through the resistance. A depiction of this phenomenon through Ohm's laws does not answer the 'why' part of the question as it does not illuminate the process behind this phenomenon. There seems to be disagreement among the answers on whether the charge repulsions are enough to cause this effect. Cited research would be appreciated to resolve this issue. To refine what I mean by this question, I think I should give an example of why I am curious as to the causal background of this phenomenon. If charge repulsions do in fact repel 'current' in the example of the short circuit, then why does the same phenomenon not happen in other parallel circuits. (Again, I know that not all circuits behave ideally) I believe the question is expressed in the most straightforward way possible. This should, however, not affect the scope of the answers. I do not wish to see or think on the level of an electron as circuit behavior always includes a cumulative effect due to the nature of electrostatic interactions. | The basic circuit theory "rules" you imply, are high level simplifications applicable at a large scale and at slow speeds. If you look at it close and fast enough, you could say that a current really starts to go into the obstructed path, but the electric field in front of the obstruction would build up gradually and current will start to repartition into the free path where it can start to flow. Naively you could say that the electric field will "sniff out" the paths. Actually in reality the current will also bounce off the obstructions, reflect and go back and forth etc. This is a real mess in practical electrical engineering at high frequencies. | {
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438,636 | If gravity disappeared, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards? | As other answers explain, Newton's third law wouldn't push you upwards, because reaction disappears as soon of action (gravity) vanishes. However, we need to keep in mind that we are siting on several thousand kilometres of rock heavily compressed by its own weight. If weight suddenly disappears, that rock will react like a spring and project itself and anything in the surface at very high speed to space. In fact, even the most conservative ballpark estimates of the elastic deformation of Earth in its present state are in the order of several kilometres, so that's the quite instant rebound we can expect. | {
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438,700 | Silicon has a bandgap of 1.1 eV, whereas germanium has 0.65 eV. Silicon has an indirect bandgap, whereas gallium arsenide has a direct bandgap. Still silicon is mainly used for making solar cells. Why? | Si is among the most abundant materials on Earth and widely used for processors as well. There are very few other materials that can even theoretically compete with that. Germanium and GaAs won't be ever able to. Organic solar cells were promising due to low fabrication cost (just ask bacteria or whatever to make your solar cells), but failed. Now perovskites and especially perovskite-silicon tandems are the hot topic in research. Back in the day, thin film technologies like CdTe, CIGS etc also looked promising and started gaining meaningful market share - the highest was about 13% or so, and many believed they will reach 20%+ of the market as they almost caught silicon efficiency. But then Chinese entered the market and killed other tech by dramatically lowering Si prices. GaAs and closely related other III-V technology is used where mass or area efficiency matters the most as this tech offers the highest efficiency - therefore, it is used for satellites and other space craft. However, ISS still uses silicon (even though GaAs had higher efficiency even back then). From Tristan's comment, they will be upgrading to the state of the art GaAs tandems fairly soon - the tandems here will be GaInP/GaAs/Ge. This tandem is the most typical, but various other configurations are possible. Such tandems are usually (but not always) lattice matched and combine Ga/In with N/P/As in various ratios to achieve variable bandgap. Now more specifically for the mentioned technologies in the question: GaAs is crazy expensive. A single wafer costs several hundred euros, while even floatzone silicon wafer costs tens, and typical solar cells are made of dirt cheap silicon, costing far below 1€ for the wafer (unprocessed wafer costs). Add tons of Si tech from CPU industry. Getting tools that can do magic on silicon is easy and cheap, tooling for III-V is expensive and much more problematic, so processing again favors Si. Germanium alone isn't a good solar cell material - too low bandgap. But great for tandems. Sure, it will collect tons of photons, but all photon energy beyond the bandgap will get wasted and you won't end up with a lot of energy. Unless you try (and eventually fail) to make viable downconverters to split the high energy photons in 2, each with half the energy.
Si is actually pretty good regarding bandgap, only ~1% (absolute) below the maximum. Indirect bandgap just means that your absorption coefficient severely drops near the bandgap. This has a single consequence optically: you require a thicker layer of absorber to get the same absorption. But, as silicon is cheap, it isn't much of a problem. As it turns out, move to say 100 μm wafers was postponed due to wafer handling more than the efficiency - it turns out you can break them easily unlike the robust 180 μm ones. Plus, even as thin as 1 μm of silicon would still absorb surprisingly large amount of light if it gets heavily scattered on each interface. | {
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439,235 | It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others. Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs". Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian? I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism. | The answer is yes, the essence of Noether's theorem for linear and angular momentum can be understood without using the Lagrangian (or Hamiltonian) formulation, at least if we're willing to focus on models in which the equations of motion have the form $$
m_n\mathbf{\ddot x}_n = \mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...)
\tag{1}
$$ where $m_n$ and $\mathbf{x}_n$ are the mass and location of the $n$ -th object, overhead dots denote time-derivatives, and $\mathbf{F}_n$ is the force on the $n$ -th object, which depends on the locations of all of the objects. (This answer still uses math, but it doesn't use Lagrangians or Hamiltonians. An answer that doesn't use math is also possible, but it would be wordier and less convincing.) The inputs to Noether's theorem are the action principle together with a (continuous) symmetry. For a system like (1), the action principle can be expressed like this: $$
\mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...)
= -\nabla_n V(\mathbf{x}_1,\mathbf{x}_2,...).
\tag{2}
$$ The key point of this equation is that the forces are all derived from the same function $V$ .
Loosely translated, this says that if the force on object $A$ depends on the location of object $B$ , then the force on object $B$ must also depend (in a special way) on the location of object $A$ . First consider linear momentum. Suppose that the model is invariant under translations in space. In the context of Noether's theorem, this is a statement about the function $V$ . This is important! If we merely assume that the system of equations (1) is invariant under translations in space, then conservation of momentum would not be implied. (To see this, consider a system with only one object subject to a location-independent force.) What we need to do is assume that $V$ is invariant under translations in space. This means $$
V(\mathbf{x}_1+\mathbf{c},\mathbf{x}_2+\mathbf{c},...) =
V(\mathbf{x}_1,\mathbf{x}_2,...)
\tag{3}
$$ for any $\mathbf{c}$ . The same condition may also be expressed like this: $$
\frac{\partial}{\partial\mathbf{c}}V(\mathbf{x}_1+\mathbf{c},\mathbf{x}_2+\mathbf{c},...) = 0,
\tag{4}
$$ where $\partial/\partial\mathbf{c}$ denotes the gradient with respect to $\mathbf{c}$ . Equation (4), in turn, may also be written like this: $$
\sum_n\nabla_n V(\mathbf{x}_1\,\mathbf{x}_2,\,...) = 0.
\tag{5}
$$ Combine equations (1), (2), and (5) to get $$
\sum_n m_n\mathbf{\ddot x}_n = 0,
\tag{6}
$$ which can also be written $$
\frac{d}{dt}\sum_n m_n\mathbf{\dot x}_n = 0.
$$ This is conservation of (total) linear momentum. Now consider angular momentum. For this, we need to assume that $V$ is invariant under rotations. To be specific, assume that $V$ is invariant under rotations about the origin; this will lead to conservation of angular momentum about the origin.
The analogue of equation (5) is $$
\sum_n\mathbf{x}_n\wedge \nabla_n V(\mathbf{x}_1\,\mathbf{x}_2,\,...) = 0
\tag{7}
$$ where the components of $\mathbf{x}\wedge\nabla$ are $x_j\nabla_k-x_k\nabla_j$ . (For three-dimensional space, this is usually expressed using the "cross product", but I prefer a formulation that works in any number of dimensions so that it can be applied without hesitation to easier cases like two-dimensional space.) Equation (7) expresses the assumption that $V$ is invariant under rotations about the origin. As before, combine equations (1), (2), and (7) to get $$
\sum_n \mathbf{x}_n\wedge m_n\mathbf{\ddot x}_n = 0,
\tag{8}
$$ and use the trivial identity $$
\mathbf{\dot x}_n\wedge \mathbf{\dot x}_n = 0
\tag{9}
$$ (because $\mathbf{a}\wedge\mathbf{b}$ has components $a_jb_k-a_kb_j$ )
to see that equation (8) can also be written $$
\frac{d}{dt}\sum_n \mathbf{x}_n\wedge m_n\mathbf{\dot x}_n = 0.
\tag{10}
$$ This is conservation of (total) angular momentum about the origin. | {
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439,450 | It is said quantum entanglement works regardless of distance. 2 particles can be entangled and information is shared instantaneously, even if they are lightyears away from each other. But how do we know this still works with such a vast distance between both particles?
I can image experiments in a lab, or even on opposites sides of the planet, but not with light years between them.
So how do we know? | Distance is not a relativistic invariant. Let A be the event marked by my beginning to write this answer, and B the event where I finish writing it. In the frame of reference of my desk, the distance between A and B is zero, but in the frame of reference of an observer moving at $0.9999999c$ relative to the earth, the distance between A and B is millions of kilometers. If the separation between two events is spacelike, then all we can say is that in other frames, they are also spacelike-separated. The distance between them can be made arbitrarily large by picking an appropriate frame of reference. So all we need is one experiment showing that entanglement works over spacelike intervals, and it follows that it works over arbitrarily large distances. One such experiment is Guerreiro et al., http://arxiv.org/abs/1204.1712 . | {
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439,475 | I'm writing image processing software and my goal here is to take an image of a projectile moving away from the camera and determine the launch angle. What I already know is: The actual size of the object The distance from the camera to point a The distance from point a to point b The speed of the projectile The elapsed time between points The height from the camera to the ground All technical specs of the camera Is it possible to calculate this angle? I realize that if point a to point b is the hypotenuse then I could make the calculation if I had either the opposite or the adjacent but I'm struggling to find those measurements. I'm not trying to model the flight path here. I need to determine the angle the object left the ground very soon (milliseconds) after launch. This angle will then feed into my flight model. | Distance is not a relativistic invariant. Let A be the event marked by my beginning to write this answer, and B the event where I finish writing it. In the frame of reference of my desk, the distance between A and B is zero, but in the frame of reference of an observer moving at $0.9999999c$ relative to the earth, the distance between A and B is millions of kilometers. If the separation between two events is spacelike, then all we can say is that in other frames, they are also spacelike-separated. The distance between them can be made arbitrarily large by picking an appropriate frame of reference. So all we need is one experiment showing that entanglement works over spacelike intervals, and it follows that it works over arbitrarily large distances. One such experiment is Guerreiro et al., http://arxiv.org/abs/1204.1712 . | {
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439,511 | It's definitely possible to propel your body using muscle to move towards the ground quicker than gravity. But is it possible to squat quicker than gravity?
Assuming squatting from your knees and not you bending down from your back like a lever, and that you're not gripping the ground with your feet and pulling your self down. If you dropped a basketball from the same height as your head,
would it be possible to squat below the basket ball before the basket ball hit the ground? | Your head would be able to beat the basketball, but your center of mass wouldn't. The reason for this is that your head is subject to the downward force due to the rest of your body, but your body is only subject to the force due to gravity and $\mathbf{F}^\textrm{ext} = M\frac{d^2}{dt^2}\mathbf{R}$ where $\mathbf{R}$ is your center of mass (which follows from Newton's second law for a particle). | {
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440,007 | The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values. Why? Why are the vacuum permeability and permittivity non-zero and non-infinite? What would a universe in which the vacuum permeability and permittivity were zero look like? What would a universe in which the vacuum permeability and permittivity were infinite look like? | I will assume throughout this answer that we fix the value of $c$ independently of $\varepsilon_0$ or $\mu_0$ . The vacuum permittivity and permeability are related to one another by $\varepsilon_0\mu_0 = 1/c^2$ , so they're not independent constants — as we should expect given that electricity and magnetism are both manifestations of the same fundamental force. The permittivity is related to the dimensionless fine structure constant $\alpha$ by $\alpha = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{\hbar c}$ . The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $\varepsilon_0$ . If we take $\alpha \rightarrow 0$ ( $\varepsilon_0 \rightarrow \infty$ ), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $\alpha \rightarrow \infty$ ( $\varepsilon_0 \rightarrow 0$ ), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case. We can see a little physics of both of these limits from Coulomb's law, \begin{equation}
F = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2},
\end{equation} where the former limit gives $F\rightarrow 0$ and the latter $F \rightarrow \infty$ for finite charges and distances. | {
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440,014 | In this picture note that only the branches tangent to the light are lit up, so concentric circles of lit branches appear around the light: I've noticed I similar effect with the peaks of water waves in a rainy puddle with a streetlight reflected in it. What process(es) underlie(s) this phenomenon? | I will assume throughout this answer that we fix the value of $c$ independently of $\varepsilon_0$ or $\mu_0$ . The vacuum permittivity and permeability are related to one another by $\varepsilon_0\mu_0 = 1/c^2$ , so they're not independent constants — as we should expect given that electricity and magnetism are both manifestations of the same fundamental force. The permittivity is related to the dimensionless fine structure constant $\alpha$ by $\alpha = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{\hbar c}$ . The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $\varepsilon_0$ . If we take $\alpha \rightarrow 0$ ( $\varepsilon_0 \rightarrow \infty$ ), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $\alpha \rightarrow \infty$ ( $\varepsilon_0 \rightarrow 0$ ), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case. We can see a little physics of both of these limits from Coulomb's law, \begin{equation}
F = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2},
\end{equation} where the former limit gives $F\rightarrow 0$ and the latter $F \rightarrow \infty$ for finite charges and distances. | {
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440,060 | In the water tank analogy below, the smaller diameter pipe (B2)
will drain tank A slower than the larger diameter pipe (B1) We are told that resistance has this same effect on current
the bigger the resistance, the smaller the current We are also told that superconductors have zero resistance So using the same water analogy, and connecting two superconducting cables of different diameter to batteries holding the same amount of charge, Would they both drain the batteries at the same speed, having the same current regardless of the cable diameter ? | I will assume throughout this answer that we fix the value of $c$ independently of $\varepsilon_0$ or $\mu_0$ . The vacuum permittivity and permeability are related to one another by $\varepsilon_0\mu_0 = 1/c^2$ , so they're not independent constants — as we should expect given that electricity and magnetism are both manifestations of the same fundamental force. The permittivity is related to the dimensionless fine structure constant $\alpha$ by $\alpha = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{\hbar c}$ . The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $\varepsilon_0$ . If we take $\alpha \rightarrow 0$ ( $\varepsilon_0 \rightarrow \infty$ ), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $\alpha \rightarrow \infty$ ( $\varepsilon_0 \rightarrow 0$ ), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case. We can see a little physics of both of these limits from Coulomb's law, \begin{equation}
F = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2},
\end{equation} where the former limit gives $F\rightarrow 0$ and the latter $F \rightarrow \infty$ for finite charges and distances. | {
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440,275 | I am not a physicist, just a curious mind. I was reading a novel by Iain Banks where it was mentioned, that shifting from artificial rotational "gravity" (in space, on a rotating space craft) to real gravity caused some level of discomfort. And this has me thinking; is there any truth to that? I mean I am aware that reading a science fiction novel does not science make; however it also strikes me as an unlikely story line to inject in there if it was not founded on at least some real theory or actual reality. So I guess it boils down to this. From the perspective of the individual experiencing it, is there any notable difference from being rotated and thereby experiencing a sensation of gravity, to a person experiencing real gravity (from the attraction of mass)? | I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside. It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort. | {
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440,399 | If a person is sitting on a chair his momentum is zero and his uncertainty in position should be infinite. But we can obviously position him at most within few chair lengths. What am I missing? Do we have to invoke earth's motion, motion of the galaxy etc. to resolve the issue? | If a person is sitting on a chair his momentum is zero... How close to zero? The uncertainty principle says that if $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum, then $\Delta x\,\Delta p\sim \hbar$ . So, consider an object with the mass of a person, say $M = 70\ \mathrm{kg}$ . Suppose the uncertainty in this object's position is roughly the size of a proton, say $\Delta x = 10^{-15}\ \mathrm m$ . The uncertainty principle says that the uncertainty in momentum must be $$
\Delta p\sim\frac{\hbar}{\Delta x}\approx\frac{1 \times 10^{-34}\ \mathrm m^2\ \mathrm{kg/s}}{10^{-15}\ \mathrm m}\approx 1\times 10^{-19}\ \mathrm{m\ kg/s},
$$ so the uncertainty in the object's velocity is $$
\Delta v=\frac{\Delta p}{M}\approx \frac{\approx 1\times 10^{-19}\ \mathrm{m\ kg/s}}{70\ \mathrm{kg}}\sim 1\times 10^{-21}\ \mathrm{m/s}.
$$ In other words, the uncertainty in the person's velocity would be roughly one proton-radius per month. This shows that the uncertainties in a person's position and momentum can both be zero as far as we can ever hope to tell , and this is not at all in conflict with the uncertainty principle. | {
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440,510 | When driving my car over a bump or a quick change in gradient (from flat to uphill or downhill to flat), if I don't drive slowly and the bump/change in gradient is large enough, I'll end up with my front bumper dragging on the road/bump. But if I go slowly, there's no such issue. I would expect that the wheels are still taking the same path regardless of the speed, so the car's body should also act the same regardless. Why does it make a difference when speed is changed? Why doesn't the car behave the same way regardless? | Because your car has a suspension. A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface. As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all. But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface. | {
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441,019 | When abruptly braking with the rear wheel on a bike, it tends to skid pretty easily. Doing the same with the front wheel is a very different experience. Instead of skidding, the bike lifts the rear wheel. I've never seen the front wheel of a bike skid. According to this answer the torque generated on a bike during breaking is set around the front wheel contact point. While this explains why the rear wheel is lifted, it doesn't really explain why the wheel almost never slips. What causes the front wheel to behave like it had more friction/grip with the ground? | Because, when you brake, your weight is being shifted towards the front wheel. The inertia coupled with gravity puts your weight and that of the bike onto the front wheel. More weight ⇒ more pressure ⇒ more friction/grip with the ground. | {
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441,648 | Just like this guy's , the color of my stove's flames were affected by the humidifier as well. Why does this happen?
Is it a good thing or a bad thing ? | The explanation I furnish below will stand or fall on the outcome of an experiment I and others here have suggested which is also outlined in my response. I promise to edit or delete my answer per the recommendations of the moderators here if that experiment shows it to be wrong. Humidifiers that operate on the "cold" principle- mixing tiny droplets of water thrown from the blades of a fan with a blast of air- produce a mist of water vapor-enriched air mixed with the partially-dried remains of water droplets that are enriched in salts by evaporative attrition. Those salt-enriched specks, when drawn into a hot gas flame, then emit light at frequencies corresponding to the line spectra of the salt constituents. In the case of sodium chloride (the most common salt in tap water), the sodium produces a yellow-orange glow when it hits the flame. This phenomenon forms the basis of a chemical analysis technique called flame spectroscopy, in which a platinum wire is dipped into a solution containing an unknown mixture of salts, and then stuck into a hot flame. The colors emitted as the salts in the solution are heated are then used to identify the chemical constituents of those salts. (Since sodium is ubiquitous, and this test is so sensitive to it, the platinum wire must be dipped in hydrochloric acid, heated to redness, quenched in the acid again and reheated several times to rid it of sodium before running the test on the sample.) This mechanism can be ruled in our out by observing the flame through a grating that separates out the primary sodium line and I invite anyone here who has a gas range (which I do not) and a grating (which I also do not, sorry) to perform the experiment and report back to us here. Since any dust in the kitchen would likely have salt in it, if the humidifier fan is blowing dust into the flame it would make the flame yellow as well. This can be tested by running the humidifier without water in it. | {
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442,271 | The GPS uses the flat space light propagation formula to calculate the distance from the source (the satellite) to the receiver (observer on Earth): $$ d=c \cdot \Delta t$$ where $c$ is the speed of light in Minkowski vacuum, $\Delta t$ is the difference between the times of emission and absorption of the signal ( corrected for relativistic time dilations ) and $d$ is the Euclidean distance. This formula is fed with the data beamed from 4 satellites to solve for the location of the receiver. My questions are : what is the rationale for using this formula? Shouldn't the distance be calculated in the curved geometry setting, e.g. using the Schwarzschild metric? What are the errors in using the Euclidean version $ d=c \cdot \Delta t$ ? N.B.: The time difference $\Delta t$ contains relativistic corrections to times . However, it is not clear to me why it is correct to use the flat space (Minkowski) formula for light propagation with just the value of $\Delta t$ amended to account for gravity. Please, try to be as clear as possible and support your statements with calculations/derivations. ADDENDUM: I found really good papers discussing in depth all the relativistic details and effects to GPS(-like) navigation in spacetime.
They are Thomas B. Bahder's Navigation in Curved Space-Time , Clock Synchronization and Navigation in the Vicinity of the Earth and Relativity of GPS Measurement . | The general-relativistic corrections are too small to matter. The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$ . Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light. At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude. The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $\Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system. The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric $$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 d\theta^2+r^2 \sin^2{\theta}\;d\phi^2$$ in geometrical units where $G$ and $c$ are 1. The signal follows a null geodesic where $ds=0$ . A radial null geodesic satisfies $$(1-2M/r)\;dt^2=(1-2M/r)^{-1}\;dr^2$$ which is a differential equation from which we can obtain $t(r)$ as $$t=r_0-r+2M\log\frac{r_0-2M}{r-2M}.$$ The initial conditions are that at $t=0$ the signal starts at $r=r_0$ , the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$ . For calculating $t_E$ in seconds, restore $G$ and $c$ to get $$ct_E=r_0-r_E+R_s\log\frac{r_0-R_s}{r_E-R_s}$$ where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm. Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is $$\Delta d=R_s \log\frac{r_0}{r_E}.$$ Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $\Delta t$ of 43 picoseconds and a $\Delta d$ of 1.3 cm. | {
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442,275 | Suppose we are moving two electrons in opposite direction both having a speed of 1.6e8 meter per sec.When they will cross each other then their relative velocity will be 3.2e8 ms_1.Isn't it faster than speed of light? | The general-relativistic corrections are too small to matter. The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$ . Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light. At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude. The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $\Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system. The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric $$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 d\theta^2+r^2 \sin^2{\theta}\;d\phi^2$$ in geometrical units where $G$ and $c$ are 1. The signal follows a null geodesic where $ds=0$ . A radial null geodesic satisfies $$(1-2M/r)\;dt^2=(1-2M/r)^{-1}\;dr^2$$ which is a differential equation from which we can obtain $t(r)$ as $$t=r_0-r+2M\log\frac{r_0-2M}{r-2M}.$$ The initial conditions are that at $t=0$ the signal starts at $r=r_0$ , the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$ . For calculating $t_E$ in seconds, restore $G$ and $c$ to get $$ct_E=r_0-r_E+R_s\log\frac{r_0-R_s}{r_E-R_s}$$ where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm. Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is $$\Delta d=R_s \log\frac{r_0}{r_E}.$$ Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $\Delta t$ of 43 picoseconds and a $\Delta d$ of 1.3 cm. | {
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442,764 | When water is poured out in space, why does it always take a spherical ball-like shape? | Minimizing energy. If there is a small amount of water, then surface tension wants to try and minimize the surface area of it, and the minimum surface area for a given volume material is a sphere. For really large volumes of water (if you, for instance, sucked all the water out of the oceans and placed it somewhere far away in space in the standard mad-scientist way), then you also get a sphere, but for a different reason: the mass of water wants to minimize its (self-)gravitational potential energy and this is also done when it is spherical. If such a volume is in the presence of external gravitational field (for instance if it was orbiting the Earth) then it would not be completely spherical: this is one of the reasons the Moon has a slightly odd shape, for instance. In between these two regimes -- if you had a few thousand gallons of water for instance, then although it would eventually end up spherical in the absence of other influences, this would take a very long time. Quantifying the effects It is interesting to try to quantify the differences between the effects. One way to do this is to consider a spherical ball of water (or anything else, but I'll stick to water because numbers are easy to come by) and consider what force you would need to bisect the sphere and move the two halves apart. Then we can compute the force that would be needed to break the surface tension, and that needed to overcome the gravitational attraction of the two halves. Surface tension Let the radius of the ball be $R$ , and surface tension be $T$ : $T$ has units of force per length. So the total force we need to exert when splitting the sphere is simply the total force exerted by surface tension around a circumference of the sphere, and we can see immediately that this goes like $R$ . $$F_T = 2\pi R T\tag{T}$$ For water, $T = 7.3\times 10^{-2}\,\mathrm{N/m}$ approximately. Gravity This is more complicated. First of all we can say something about the behaviour of the force: the masses of the two hemispheres go like $R^3$ , and the separation like $R$ , so it's immediately obvious that the force is going to go like $R^3\times R^3 / R^2$ : like $R^4$ in other words. Gravity is going to win as $R$ gets big! But we can actually get a number, even though the hemispheres are not, well, spheres, and thus hard to treat gravitationally: if you think about the surface which cuts the the ball into two hemispheres, then what is preventing the ball collapsing inwards across this surface is pressure. So the gravitational force between the two halves of the ball, when they are touching, must be equal to the integral of the pressure over that surface (it took me ages to realise this trick!). Let's assume the density is uniform, which it won't be for really large objects but it will be for reasonably small ones. Call the density $\rho$ . Then we can work out the gravitational acceleration at radius $r$ from the centre, relying on the shell theorem and knowing the mass inside $r$ is $m(r) = 4/3 \pi r^3$ . $$g(r) = \frac{4\pi}{3}G\rho r$$ And this gives us the pressure at $r$ , just by integrating $g$ from $r$ to $R$ : $$
\begin{align}
p(r_0) &= \int\limits_{r_0}^R \rho g(r)\,dr\\
&= \frac{4\pi}{3} G \rho^2 \int_{r_0}^R r\,dr\\
&= \frac{2\pi}{3} G \rho^2 \left[R^2 - r_0^2\right]
\end{align}
$$ or $$p(r) = \frac{2\pi}{3} G \rho^2 \left[R^2 - r^2\right]$$ And finally we can integrate this over the surface to get the total force: $$
\begin{align}
F_G &= \int\limits_0^R 2\pi r p(r) \,dr\\
&= \frac{4\pi^2}{3}G\rho^2 \int\limits_0^R R^2r - r^3 \,dr\\
F_G &= \frac{\pi^2}{3} G \rho^2 R^4\tag{G}
\end{align}
$$ (I hope this is right: it is dimensionally OK but I might have missed factors somewhere.) Compared So, then, given $\rho = 10^3\,\mathrm{kg/m^3}$ , $G = 6.7\times 10^{-11}\,\mathrm{m^3/(kg s^2)}$ , we can solve for the radius $R$ where $F_T$ = $F_G$ , and the answer is about $12.8\,\mathrm{m}$ . I was surprised how small this is (and I am worried I have made a mistake therefore). So if this is right, it means that gravity starts beating surface tension for a ball of water which is about $13\,\mathrm{m}$ in radius, and beyond that it wins rather rapidly due to the dependence on $R^4$ . What this does not tell you is anything about how long it takes something to become spherical: I think that would be a bunch harder to work out. | {
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443,260 | In The Lord of the Rings it is claimed that no man has the strength to pierce the skin of the giant spider Shelob. In a dramatic turn of events, Shelob lunges at Sam with such force that she pierces herself with Sam's sword. This seems physically impossible to me, assuming the real-world laws of physics are in effect. Wouldn't Sam need the strength to hold the sword in place in order to actually pierce, and by Newton's third law and the above claim, require impossible strength? Maybe this has something to do with impulse, which I never fully understood. | There are three ways ways this might work, all of which using Shelob's own strength. First, as noted in other answers, the sword could have been driven down until the hilt struck the floor and was braced that way. The sword could have been held straight up with the arms and legs locked and the crosspiece of the sword pressed against the hands. In this stance, a body could withstand more force than the muscles could actively exert, and thus act as sufficient brace against Shelob's own superior strength. The text seems to support that Shelob's body was hard. Piercing thus is a matter not of force, per se, but of impulse: mv = Ft. The harder the shell, likely the more a quick impulse will break it. Shelob could have moved with such speed against a thrust that the impulse time was shortened, and the strength of Sam's own thrust multiplied because of the shortened interval, since that equation rearranges to F = mv/t. A combination of 2 and 3 are also possible. As any good martial artist knows, you can deliver more power by timing your blows so that your joints are locking just as contact is made. This multiplies the speed and strength of, say, a punch with the power and weight of the body following through behind the punch. You can multiply the force of a punch by putting your whole torso into it and achieve more power than you can muster using your arm alone. Shelob throwing her own strength against Sam just as an upward thrust was reaching home and his arms locking combines the added impulse from the speed of contact with the force resistance of Sam's locked joints. It is perfectly reasonable to assume that this may have ultimately driven Sam down, but still have pierced her. | {
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443,358 | In depictions of the double-slit experiment that model the photon or electron as a particle, i.e. when attempting to measure which slit the particle passes through, it always shows the particle entering one of the two slits. Why is it that the particle can't hit the space between or outside the slits, i.e. never even make it through? Is it implied that the experiment is just repeated until a particle makes it through, i.e. shows up on the film or detector on the other side? I see how, modeled as a wave, the wave always makes it through. But the illustrations of particles kind of don't make sense to me. Is it because they are just simplified illustrations? I realize this sounds like a silly question, but I'm trying to go back and question everything I've taken for granted. (For example, I wondered what if there is some weird, hidden interaction between the ones that didn't make it through and the ones that did ?) | "Is it because they are just simplified illustrations?" you ask. The answer is simply: yes it is because they are simplified illustrations. Furthermore, not only can the particle hit the barrier outside or between the slits, typically most of the particles do that. Only a small fraction make it through. I say 'typically' because in such experiments we don't normally bother to set up the optics (whether for photons or electrons) so as to restrict illumination to only the two slits and not the surrounding area. But in principle it could be done, and then only a few particles would miss the slits. It is quite common, in experimental physics in this area, to do what is called 'post-selection'. That is the name for the practice of selecting from your dataset only those outcomes triggered by some signal, such as, in this case, the signal that a dot appeared somewhere on the final screen. Then after that the discussion is really saying not 'this is what happened in every run' but 'of those runs where something made it to the detector, this is what happened'. One can regard the simplified pictures as showing what is understood to have happened for those runs which were singled out by this 'post-selection'. | {
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443,586 | Why can't you see a reflected image on a piece of paper? Say you put a pen in front of the paper, even when light rays are coming from other sources, hitting the pen, reflecting back, and hitting the paper, there is no reflection. What's wrong with the following "ray diagram" and why such even don't happen and the image of the pen don't form on the paper (right side is a paper)? When then can you see the image of a torch when you shine it on the paper? When you put a convex lens in front of the pen, why you can now see the image of the pen on the paper? | Because the real situation looks a lot more like this: Your pen is (presumably) not made of mirror-like polished metal, but rather of something like wood or plastic that reflects light diffusely . This means that the light from each part of the pen is scattered all over the paper (and, of course, in other directions too), so it won't project a clear image onto the paper. (And since the paper itself is also a diffuse reflector, all the light that hits it gets scattered in all directions again, and some of it ends up hitting your eye. If you replaced the paper with a mirror, then only those rays that were coming from just the right direction would have a chance of getting reflected towards your eye, and so you'd see a sharp reflected image.) OK, so why won't the pen at least form a blurred image on the paper, then? Well, actually it does. Here's a photo I just took with my cellphone. Sorry that it's a bit dark, I wanted to make sure I didn't burn out any highlights. If you look closely and carefully, you can see just the slightest hint of yellowish color on the paper near the pencil, and a hint of red near the pen. Those faint colors are caused by light reflecting off the pen and the pencil onto the paper. But they're not very bright, because most of the light hitting the paper is still coming from other directions (primarily, from the lamp illuminating the scene), and not very distinct, because the light that does reflect off the pen and the pencil gets spread in all directions. Just in case you can't see it clearly on your screen, here's the same photo with the color levels adjusted for maximum contrast: There, I bet you can see it now. It almost looks as if the pen and the pencil were glowing... which, of course, technically they are. Not with their own light, of course, but with light reflected from the lamp that illuminates them, just like e.g. the moon glows with reflected sunlight. As a bonus, if you look carefully, you can also see some brighter spots on the paper near the pencil. Those are specular reflections from the glossy print on the pencil. So we actually have both diffuse and specular reflection demonstrated in the same picture. | {
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444,307 | A rock sitting on land, the ocean floor, or floating in space maintains its shape somehow. Gravity isn't keeping it together because it is too small, so I'm assuming it is chemical or nuclear bonds keeping it together as a solid. If not it would simply crumble apart. So, what type of energy maintains the shape of a rock, where did this energy come from, and is it slowly dissipating? As a corollary, if a large rock is placed on top of a small rock, is the energy required to maintain the shape of the small rock 'used' at a greater rate? | No, the exact opposite is true. The molecules in a rock don't stay together because they're spending energy. They stay together because of attractive chemical bonds. The molecules have lower energy when they're together than when they're not, so you have to spend energy to break the rock apart, not to keep it together. | {
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444,387 | In music, when two or more pitches are played together at the same time, they form a chord. If each pitch has a corresponding wave frequency (a pure, or fundamental, tone), the pitches played together make a superposition waveform , which is obtained by simple addition. This wave is no longer a pure sinusoidal wave. For example, when you play a low note and a high note on a piano, the resulting sound has a wave that is the mathematical sum of the waves of each note. The same is true for light: when you shine a 500nm wavelength (green light) and a 700nm wavelength (red light) at the same spot on a white surface, the reflection will be a superposition waveform that is the sum of green and red. My question is about our perception of these combinations. When we hear a chord on a piano, we’re able to discern the pitches that comprise that chord. We’re able to “pick out” that there are two (or three, etc) notes in the chord, and some of us who are musically inclined are even able to sing back each note, and even name it. It could be said that we’re able to decompose a Fourier Series of sound. But it seems we cannot do this with light. When you shine green and red light together, the reflection appears to be yellow, a “pure hue” of 600nm, rather than an overlay of red and green. We can’t “pick out” the individual colors that were combined. Why is this? Why can’t we see two hues of light in the same way we’re able to hear two pitches of sound? Is this a characteristic of human psychology? Animal physiology? Or is this due to a fundamental characteristic of electromagnetism? | This is because of the physiological differences in the functioning of the cochlea (for hearing) and the retina (for color perception). The cochlea separates out a single channel of complex audio signals into their component frequencies and produces an output signal that represents that decomposition. The retina instead exhibits what is called metamerism, in which only three sensor types (for R/G/B) are used to encode an output signal that represents the entire spectrum of possible colors as variable combinations of those RGB levels. | {
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444,894 | Why do waves that were traveling in a straight direction change direction when passing through an opening? I thought that the waves (red arrow) when colliding with the wall bounce in the opposite direction (green arrow). And the waves that pass through the aperture follow its path normally as shown in the image on the right. The waves that go in a straight direction should follow traveling straight line like a car that goes under a bridge the car is straight on the road.
But this is not so. Why does the direction of the waves change? How is the direction of the waves calculated? | For the full math, you can look up 'diffraction' and 'Huygens Principle' but here I will just post a quick observation that is enough to get a good physical intuition. Suppose we are considering water waves, and imagine yourself sitting behind the barrier in the 'harbour' (at the lower part of your diagram), watching the waves approaching from 'out at sea' (i.e. the top of your diagram). As the waves reach the 'harbour mouth' (i.e. the small opening in your diagram) the water there is caused to go up and down. So there is this water bobbing up and down in the small opening. Now the surface of the water nearby is going to bob up and down too, isn't it? And the ripples will spread out from there. It doesn't really matter in what direction you consider: the waves will spread out into the 'harbour' because the water at the harbour mouth is moving. From this way of thinking, you begin to wonder why the waves out at sea are so straight! Ultimately it is because in that case you have oscillating water all along a long line, and so the water all along that long line is caused to move in synchrony. As I say, this is not a full mathematical answer, just an attempt to give you some intuition about the physics. | {
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445,062 | I saw this arrangement for tensioning overhead cables from my train window (schematic below). Why not just have one pulley wheel leading directly to the weights? What function do the additional pulleys serve? For that matter, what are the cables for? They're clearly not power lines. | Having more pulleys increases the mechanical advantage of the system. In this case the mechanical advantage is 3. This means that the weights involved need to be a third as massive and the cables passing over the pulleys need to have a third the strength. This makes everything cheaper, smaller, and more tractable: it means, for instance, that you can use cheap, rather low-density, materials for the weights (they are often piles of concrete disks around a central metal rod). Cheap weights are both, well, cheap, but also less interesting to thieves: no-one wants to steal concrete disks, a lot of people want to steal lead, say, and metal theft is a big problem for many railways (obviously this part of the reason has no physics content, but it's important). Another reason for reducing the mass of the weights may be to do with how hard it is to install and maintain things: the lighter the weights are the less heavy machinery you need to get close to them. I don't know to what extent this is a consideration, and it's also not really a physics issues. Finally the pulleys can be a lot smaller as well as thin cables are more flexible. In answer to the second question: yes these probably are the ends of the power lines (although there may be some big insulator out of the shot). I'm not an expert on railway power systems but I think what they tend to do is have overlapping sections of power cable, so at a mast one set terminates while the other set carries on (assuming anything ends at all there: a lot of the masts are just for support I think). | {
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445,158 | How/why do green screens work? What's so special about the color green that lets us seamlessly replace the background with another image and keep the human intact? Are there other colors that work similarly? | It's partly about how human colour vision works, partly about avoiding colours you want to keep, such as those of the actors. Colour cameras record concentrations of red, green and blue light to mimic human colour vision. Before digital techniques, blue screens were preferred because, of the three primary colours, that's the one rarest in human skintones. When digital cameras were invented, they were given greater sensitivity to green light to mimic a bias in human vision. Green screen doesn't require as much illumination of the screen as blue screen does, which prevents the risk of chroma spill onto the foreground subject's edge, which can cause a special effects failure called a chroma halo. In the pre-digital era, when the foreground-background distinction had to be much larger than is required today (because of the complicated optical process involved in achieving chroma key), it was almost impossible to get away with any colour beyond blue. Nowadays both colours are very common, with green almost the new default; but, unlike the blue-only era of the past, typically both colours are now on standby. | {
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445,164 | Consider a bottle partially filled with water and it is sealed from everywhere so that no air can enter or exit from bottle.Now make a small hole at the bottom of the bottle and hang it vertically so that hole faces downward direction. As we know bottle is sealed from everywhere else so the hole is the only place from where entry or exit of air or water is possible.Now two events will occur one after another repeatedly a drop of water came out from hole due to gravity and falls down on ground a bubble of air (or air)wil enter in bottle through hole and moves upward against the gravity and mixes with the air present at the top of liquid in bottle. Now i know the reason for these two events to occur just because to maintain atmospheric pressure of air in bottle and due to gravity.But the thing in which i get confused is that which of the two events will occur first? I tried this experiment myself but not able to figure it out.I personally feel that event1should occur first | It's partly about how human colour vision works, partly about avoiding colours you want to keep, such as those of the actors. Colour cameras record concentrations of red, green and blue light to mimic human colour vision. Before digital techniques, blue screens were preferred because, of the three primary colours, that's the one rarest in human skintones. When digital cameras were invented, they were given greater sensitivity to green light to mimic a bias in human vision. Green screen doesn't require as much illumination of the screen as blue screen does, which prevents the risk of chroma spill onto the foreground subject's edge, which can cause a special effects failure called a chroma halo. In the pre-digital era, when the foreground-background distinction had to be much larger than is required today (because of the complicated optical process involved in achieving chroma key), it was almost impossible to get away with any colour beyond blue. Nowadays both colours are very common, with green almost the new default; but, unlike the blue-only era of the past, typically both colours are now on standby. | {
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445,200 | The net force on an object is equal to the mass times the acceleration, $F = ma$ When I brake on a (moving) car, the net force is negative, therefore causing the resulting acceleration to also be negative. This all makes sense, but if the acceleration of the car is negative, why does it not keep moving backward? I know cars in real life come to a stop, but I am having trouble explaining why the car does not continue to accelerate backward while the brakes are applied, with physics, so to speak. Where is the logic incorrect? | A notable property of frictional forces is that they resist motion (as opposed to other types of forces, which might resist displacement, for example, which is how a spring behaves). As a result, the brakes on your car slow down the motion of your wheels that produces forward movement of your car—but they also slow down motion that produces reverse movement. If instead you used another type of force-applying system to slow down your car (e.g., a giant spring), then your car would slow down, then stop, and then start moving backwards. | {
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446,840 | I was given an interesting dilemma today. A co-worker saw me adding a liquid (Diisopropyl ethylamine AKA DIPEA) to a flask filled with another liquid (Tetrahydrofuran AKA THF). I needed to weigh out exactly 5 grams of DIPEA into the THF and so I zero'd the scale with the flask+THF on it, then proceeded to add the DIPEA until the scale said 5.000g. Since masses are additive I assumed this was fine. My co-worker, however, stopped and told me that although masses of two liquids are additive, the combined weights would not be, and since the scale measures weight as opposed to mass I had apparently just added an incorrect amount of DIPEA. He explained the reasoning to me but I'm a chemist, not a physicist and certainly not skilled in fluid mechanics, so I would like someone to dumb it down for me a bit or tell me if I'm way off. From what I understand, the scale measures weight which is a function of gravitational force. But gravitational force is a function of buoyant force (its less if the buoyant force is greater since the buoyant force pushed a liquid up). Finally, buoyant force is a function of density. This means that my THF (which had a density of .9 g/ml) had a greater buoyant force than my THF/DIPEA solution (DIPEA density is only .74 g/ml so the solution would be somewhere between .74 and .90). And this means that technically as I'm adding DIPEA, the added mass is not the only thing causing the weight to increase; but rather the decreased buoyant force is also causing that. And so, when the scale finally read 5.000g, I had possibly only added 4.950 or maybe 4.990 etc (something less than 5.000). Is my reasoning correct? Any help is appreciated. | Of course, by common sense, if you put together two objects with masses $m_1$ and $m_2$ , and nothing comes out, then you end up with mass $m_1 + m_2$ . Weights are a little more complicated because of buoyant forces. All objects on Earth continuously experience a buoyant force from the volume of the air they displace. This doesn't matter as long as volume is conserved: if you stack two solid blocks their weights add because the total buoyant force is the same as before. But when you mix two liquids the total buoyant force can change, because the volume of the mixed liquid might not be equal to the sum of the individual volumes. To estimate this effect, let's say (generously) that mixing two liquids might result in a change of total volume of $10\%$ . The density of air is about $0.1\%$ that of a typical liquid. So the error of this effect will be, at most, around $0.01\%$ , which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second. | {
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