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535,257 | I think I may recall sometimes a rainbow going away when a cloud comes between myself and the sun. I know that the appearance of the rainbow is location dependent. But do we need the sun, rainbow, observer triangle for the effect to manifest? | In chapter 3, he says, if I understood this right, that electrons can only exist in specific quanta - that is they can only be in certain regions, and will perform a quantum leap over regions that are not stable energy levels. Electrons (that are confined by a potential well, such as when they are part of an atom) are indeed limited to certain quantized energy levels. And they will only be able to gain or lose energy only in quantized amounts. But no matter the energy level, the electron is not really restricted in what position in space it might be found in. I.e. there are no "regions" of space where it can't be found. There are, depending on the energy level, regions of space where it's spectacularly unlikely to find the electron (like you could take as many electrons as ever existed in the visible universe, put them into this state, and then observe their position and you probably wouldn't find one at that location). But in principle it's still possible to find the electron anywhere in space. For example, if you have an electron in a hydrogen atom, excited to one of the "2p" energy states, and you observe its location, you'll most likely find it in the locations indicated by these drawings: ( image source ) It's highly unlikely to find an electron in the 2p y state (for example) located far (a nanometer or so) up along the z axis (and a picometer off the xz plane). But it's not entirely impossible. There is actually, as pointed out in another answer, a locus of points on the xz plane, called a node of the wavefunction, where the probability of finding the electron is actually 0. what is the need for a quantum leap? The quantum leap is a leap in the energy of the electron, not its location in space. | {
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535,442 | If an object hits a soft surface it will bounce lower compared to the object hitting a hard surface, isn't the impulse in the first case equal to the impulse in the second case, so why does the object bounce lower when it hits a softer surface? | You need to be careful as to exactly what you mean by the terms "softer" and "harder". It is rather the degree of inelasticity of the collision that determines how high the object bounces. For example, you can have a spring with a low spring constant that you might consider "softer" than a spring with a higher spring constant which is "stiffer". If the two springs are ideal (no friction losses) an object bouncing off each of them will reach the same height because the collisions would be considered perfectly elastic. (Assumes the object itself is perfectly rigid). However, at the macroscopic level all real collisions are inelastic. Kinetic energy will be lost due to friction associated with the inelastic deformation of the colliding objects. Since most softer surfaces will undergo more deformation, it stands to reason that the collision with such surfaces will be more inelastic than harder surfaces as they potentially undergo more deformation. The degree of inelasticity of a collision is reflected in the coefficient of restitution (COR) of the colliding objects. That is the ratio of the final to initial relative velocity between two two objects after they collide, and is a number less than 1. All other things being equal the COR will likely be lower in a collision with a softer surface depending on how much it deforms and the degree to which the deformation is permanent. Shouldn't there be an equal force in magnitude and opposite in direction when the object hits the surface in both cases? therefore in both case shouldn't the object bounce back to the same height? Per Newton's third law the force the surface exerts on the object is equal and opposite to the force the object exerts on the surface. But the magnitude of the equal and opposite forces will not be the same when the object initially contacts the surface as when it comes off the surface after penetrating the surface. That's because energy is lost as heat when the object penetrates the surface. That means there is less kinetic energy of the object when it comes off the surface then when it initially hit the surface. Less kinetic energy means less velocity and a lower rebound height. Hope this helps. | {
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535,496 | Airplanes can fly because the pressure on the top of the wing is lower than on the bottom. But the difference in pressure must be huge in order to lift an enire airplane. Also, the difference in air pressure between, let's say, Mount Everest and sea level is not neglectable, thus, air pressure is decreasing the heigher you go. This is where I asked myself: Can we theoretically build a material which is light enough or high enough (or both), that can levitate just due to the difference in pressure on the top vs. the bottom like an on an airplane wing. For example imagine you're holding a piece of paper horizontally, then air pressure on the top is slightly lower due to the decreasing air pressure and now just make the piece of paper as light as it needs to be to stay in the air like a wing on a flying airplane Is this theoretically possible or are there other effects or simplifications I overlooked? | A hot air balloon, or a helium-filled balloon floats in air, so either might meet your criteria. If you're looking for a solid material, perhaps a sphere of very sparse aerogel, with its outside surface sealed with a thin layer of plastic then evacuated, could come close to what you have in mind. But whatever the "material" is, it would need to have a lower mass density than the ambient air. | {
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535,594 | I have read over a dozen questions about the speed of light -- "why it $c$ constant?", "why can't anything travel faster than light?", "how do we know this?" The responses are quite clear: The invariance of light speed is determined empirically (e.g. from the Michelson-Morley experiment). The speed of light is simply an axiom for physics and was discovered experimentally. The invariant value of $c$ is a fundamental property of the universe. My question is why can't the invariance of $c$ not be deduced theoretically with the following logic. As an object's velocity increases, its kinetic energy also increases. The kinetic energy growth is asymptotic, meaning it approaches infinity as the velocity approaches some value. This makes it impossible for anything with a mass to reach this velocity because it would require infinite kinetic energy. Therefore velocity must be have a limit. See, this makes much more sense to me than the claim that the invariance of $c$ is just a postulate from lab work and that there's no reason for it to be invariant other than "it's just the way things are". I suspect I've made a mistake. Perhaps the idea that an object's mass increases rapidly as speed increases comes from special relativity itself, which is derived from the assumption that $c$ is invariant. This doesn't, however, seem obvious to me since we should be able to observe this effect in experiments. | The kinetic energy growth is asymptotic, meaning it approaches infinity as the velocity approaches some value. Unfortunately, this result already assumes that you know that there is an invariant speed. Without the invariant speed the formula for KE is $KE=\frac{1}{2}mv^2$ which has no limiting speed. It is only after you already know about the invariant speed that you get the expression $KE=((1-v^2/c^2)^{-1/2}-1)mc^2$ which goes to infinity as $v$ approaches $c$ . photons don't have mass so they can This also requires already knowing about the invariant speed. Without the invariant speed there is no known relationship between all three of mass, energy, and momentum. With the invariant speed we learn $m^2 c^2=E^2/c^2-p^2$ which given the energy and momentum of light implies that light is massless. So yes, those things, if known independently somehow, would have led to the conclusion of an invariant speed. But how could they have been known? Perhaps they could have simply been measured and known experimentally first, but historically it didn’t happen that way. Historically the invariance of c was postulated prior to measurements experimentally showing those points. Furthermore, such measurements would have been considered violations of classical physics. | {
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535,784 | Consider a leading theory for a phenomenon, then how are new theories shown to be better while both remain unfalsified? | The very bare minimum that a theory has to obey to be considered as viable is: The theory is self-consistent (it cannot predict two different results for the same input) The theory is complete (within its domain of applicability, any statement can be decided by the theory) The theory agrees with all past experiments The falsifiability part is basically related to the third criteria: if a newer experiment cannot be explained by the theory, it is falsified. What to do if two theories on the same domain of applicability obey all three? Here are a few options! Don't do anything! For a while now, physics has overall tried to be about applying models more than trying to find some kind of metaphysical truth . With this in mind, there is nothing fundamentally wrong about having two theories for the same domain, especially if they have about the same predictions for any reasonable experiment. Things may not be ideal, of course. Competing theories may have wildly different predictions (although that is less of a problem in physics usually). Also there are finite resources for research, and it may be useful to determine which theory is more promising, more practical, or closer to the fundamental truth . There are many, many criteria I could bring up, stemming from made-up examples (it's a big hobby of philosophers of science), but I'll try to stick to realistic cases. Apply heuristics There are many broad bias in physics which have shown to be more likely to lead to better models than others. There are no fundamental reasons why these should lead to better theories, but so far they seem to be productive, if not metaphysically true : Try to go for the simpler theory. "Simpler" is of course somewhat vague, and different people may have different opinions on what constitutes a simpler theory. A few cases can be useful, though. Overall, people will pick general relativity over the Einstein-Cartan theory (general relativity + a torsion field), Brans-Dicke theory (general relativity + a scalar field), or other versions that are general relativity with additional terms, since any of those additional objects have such low impact on actual experiments. We went with quantum chromodynamics rather than the patchwork of formulas we had with the meson zoo of particles. Try to go for a local theory. It's generally assumed that causal effects are always local, and only propagate from their point of origin. Symmetries tend to play a big role in modern theories, and relatedly, conserved quantities that stem from them. We tend to favor theories if they stem from some underlying principle of symmetry, and we prefer if the big conservation laws hold, at least locally (conservation of energy, momentum, charge, etc). Fred Hoyle's gravitational theory was in part not liked too much for breaking local time translation symmetry, and therefore conservation of energy. The neutrino was inferred from conservation of energy, and the theories that went against it didn't pan out. The second law of thermodynamics looms large in a lot of physicists' minds, and overall, any theory that would imply a break of it is considered suspicious. Causality is generally considered very important. This is why naked singularities, closed timelike curves or time-symmetric theories are not considered to be likely to be physical. This is linked to another concept, determinism. While not all modern theories are deterministic, they are at least probabilistic. Theories that allow for events to happen with no assigned probability are generally considered pathological. Beware that many such heuristics existed in the past and have been shown to be wrong, or at least not necessarily best. Galilean invariance, celestial objects moving on circles, etc. Go for the practical theory A big part of the appeal of a theory is its ability to get results, and get them easily enough. If a theory seems promising, but the mathematics necessary to even obtain measurable outcomes isn't there, or fairly intractable, it may be best to go with the simpler theory. It's best to keep measurable quantities obtainable, in realistic scenarios, with reasonable computation times. Another practical aspect: ideally, a theory should be such that we can apply it without omniscience. If a theory requires the knowledge of too many initial conditions, it is not going to be terribly applicable in experiments. Ideally the influence of most of the universe should be either very small, or can be approximated by some general principles. This is common in most theories: inverse square laws, the cluster decomposition principle, mean fields, etc. Raw philosophical bias A lot of choices in theory on the individual level is simply due to their own ideas on ontology, causality, epistemology, and other metaphysical business. This is what drives a lot of adherents of fringe theories, the fights between competing theories for new physics, as well as some of the choices regarding what is established science. It's entirely possible to have theories without any ontology, where we simply input a list of measurements in formulas and get another list of measurements, but in addition to being impractical, most people believe in some kind of physical objects, and the theories tend to reflect those ontologies. A lot of people don't like the Copenhagen interpretation due to its lack of determinism, or its contextuality. Many of the heuristics I listed are rooted in philosophical positions, in addition to being seen as generally useful. There was some resistance to general relativity from the idea of Kant that Euclidian geometry was fundamental. A lot of science in the Soviet Union was constrained by the requirement that it had to obey dialectic materialism, which favored deterministic quantum theories and eternal cosmologies. Opinions on what constitutes a reasonable spacetime in general relativity can be rooted a lot in the author's opinion on causality or the nature of time (presentism, growing block, eternalism, whether time is cyclical, eternal or has a starting point, etc). | {
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535,790 | Do we ever talk about rate of change of momentum with respect to movement in spacetime? That should be a good quantity, given that spacetime is more fundamental than time. | Do we ever talk about rate of change of momentum with respect to movement in spacetime? Yes. In spacetime just like time and space are unified into one spacetime so also energy and momentum are unified into one concept called the four-momentum. The four-momentum is a four dimensional vector where the first element is energy and the remaining three elements are the momentum with energy having the same relationship to momentum as time has to space. Equipped with the concept of the four-momentum we can define the rate of change of the four-momentum. This quantity is called the four-force and has exactly the properties that you would expect in a relativistic generalization of force. The timelike component of the four-force is power. So in relativity the four-force unites power and force in the same way as spacetime unites space and time. | {
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536,391 | When a sphere of radius $r$ is placed in the path of a parallel beam of light of intensity $I$ , the force exerted by the beam on the sphere is given by: $$F=\frac{\pi r^2 I}{c}$$ I derived the above result by assuming the sphere to be perfectly reflecting. However, it turns out that the force exerted by a light beam of same intensity on a perfectly absorbing sphere of same radius is also given by the same formula. Further, even if the sphere partially reflects and partially absorbs the incident photons, the force exerted on it by the beam remains the same. I understood the final case (partially absorbing and reflecting) by imagining it to be a combination of the first two cases - totally reflecting and totally absorbing. In short, the force exerted by the light beam on a sphere depends only on the area obstructed by the object, here it's just the area of the biggest circle in a sphere ( $\pi r^2$ ). I understood the mathematics behind this result. But, this seems to be counter-intuitive for me because, the change in momentum in case of total reflection is twice that of the case when the light beam is totally absorbed. The force exerted on the object is nothing by the rate of change in momentum and therefore the force on the object which totally reflects is more compared to totally absorbing or partially absorbing objects. However, in case of spheres placed in the beam, the force on it remains the same irrespective of the amount of light absorbed or reflected. What is the intuitive reason behind this fact? Also, is this a property of only spherical objects or are there even more examples for this? | Consider a photon that strikes the center and is reflected straight back on itself. That photon gives the sphere twice its momentum. Consider a photon that strike the edge at a glancing angle and is only slightly deflected. It hardly affects the sphere at all. The momentum change is about $0$ . If you integrate over the sphere, you get an average momentum change in between the two extremes. If a photon is absorbed, it doesn't matter what the angle of the surface it. It gives all its momentum to the sphere. You have shown that the average value for reflection comes out that same as the uniform value for absorption. For other geometries, consider a cone where the surface is at 45 degrees. Light everywhere would reflect at 90 degrees. This would impart the same momentum as being absorbed. This would apply to a flat disk at 45 degrees too. | {
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536,407 | I know that centrifugal force only "exists" when the reference frame is rotating. In this case, if there is centripetal force, there must also be centrifugal force so that bodies on the rotating frame will be motionless. But if centrifugal force is a pseudo-force then why we can observe its effects outside of rotating frame? I mean if we start to spin a table with objects on it very fast, we can see the objects flung to different directions. Hence there must be acceleration outwards and we can observe it when our reference frame is outside of rotation. The centrifuge device is another example. | When objects fly off a spinning table, from an external reference frame they simply stop travelling in circles. At the point of "unsticking" they do not accelerate radially outwards from the centre of rotation but merely stop accelerating inwards; they continue in a straight line tangential to the circle and at right angles to the radial direction. Thus, no centrifugal force is present - indeed, it is the very absence of any force subsequent to breakaway which explains their behaviour. | {
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536,431 | How can I calculate the deflection of a Cantilever beam if there is a load at different points on the beam? My current scenario is the following → I have a ruler that I have modelled to be cantilever and I am moving a piece of blu-tac down the ruler to calculate different resonance frequencies. Is there a formula relating the position of blu tac and deflection? Or any relationship between this and its centre of gravity | When objects fly off a spinning table, from an external reference frame they simply stop travelling in circles. At the point of "unsticking" they do not accelerate radially outwards from the centre of rotation but merely stop accelerating inwards; they continue in a straight line tangential to the circle and at right angles to the radial direction. Thus, no centrifugal force is present - indeed, it is the very absence of any force subsequent to breakaway which explains their behaviour. | {
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536,688 | Consider an object being pushed 3/4 of the distance around a circular track. The work done on the object would be the distance of 3/4 the track’s circumference times the force applied to the object (given that it was pushed at a constant force). Since we are multiplying a vector by a scalar, why is work a scalar measurement? Or would the work done on the object actually just be force times displacement? Thanks. | Work is the dot product of a vector force and a vector displacement, hence a scalar. Knowing just the scalar distance isn’t enough to calculate work. That distance might be in the same direction as the force, but it might be perpendicular or even opposed. All of those would give different values for the work done. | {
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537,201 | When I was a kid I happened to encounter a solar eclipse. I was taught that I should not look at the Sun directly when it is undergoing an eclipse, but I was extremely curious to see it. Somebody suggested to me that if I created a pinhole in cardboard and place the cardboard in the Sun and managed to get the light passing through the pinhole on a screen inside the room then I could see the eclipse on the screen. I did that and I could see the eclipse on the screen. My question is, why was I not seeing a circular illumination on the screen? But to my surprise, as the eclipse was progressing on the Sun, the illumination I saw on the screen was also undergoing the same eclipse! It means the illumination on the screen was the image of the Sun! Why was it not a uniformly illuminated circular patch on the screen? Why was it undergoing an eclipse? The light was passing through all the portions of the hole, so why was an eclipse showing on the screen? In summary, how does a pinhole create an image of the Sun? And not always a circular illumination? Edit1: If we place a single point source of light in front of the pinhole then it creates a circular illumination on the screen, but if we put an extended object in front of the pinhole then it creates an inverted image of the object on the screen, how? An extended object can also be considered as a collection of infinite point sources of light. If one source produces a circular patch then infinite sources should also produce the same circular patch, just of greater intensity. The shape of the patch should not change. Why does the shape of the patch change to the shape of the object on the screen?
Kindly help. | Let us start from the basics. Consider a point source of light placed on the principal axis of the pin hole camera as shown in the diagram below: The point source produces a circular illumination on the screen. Now let's displace the point source towards $D$ from the centre as shown below: The circular illumination also moves away from the centre but in the opposite direction i.e., towards $d$ . For the time being let us assume the displacement of the object is small compared to its distance from the pinhole. So that we can still consider the illumination on the screen to be nearly circular for the sake of simplicity. I've shown the displacement along one direction. But similar phenomenon happens for displacements in all other directions perpendicular to the principal axis. I'll leave it to your imagination to play with the system. Now, let's consider an extended object which consists of four point sources of light as shown below: The circular illumination due to the central yellow point source is also at the centre. But for off-centred red, green and blue point sources, the illumination is also off-centred as per our previous result. The corresponding inverted image formed is also shown above. It's not necessary for the extended object to be made of point sources emitting different colours (wavelengths to be more precise). I've just coloured them differently to make the point clear. Sun is not a point source and is an extended body which contains infinitely many point sources. Similar arguments can be used to explain why we observe the image of eclipse instead of a circular patch of light. To witness the solar eclipse of December 26, 2019 , I too made a pin hole camera and the image of the eclipse is shown below: Don't get puzzled by the three images of the eclipse numbered one, two and three. I just made three circular holes each of different diameters ( $r_1<r_2<r_3$ ) to check which one gives the best result. As explained by Farcher in his answer , there exists an optimum pinhole diameter for a given wavelength of light and distance of the pinhole from the screen. If the pinhole is too small, then the diffraction effects would become significant. Also the intensity of the image decreases with the decrease in the pinhole size. When we increase the pinhole size, the intensity increases, but at the same time the image becomes more blurred as the circle of illumination grows in size. With the given order of pinhole sizes you could also verify this from the image above (although the difference between the second and third image is not that pronounced in this image). As per the Wikipedia article on pinhole camera the optimum diameter $d$ of the pinhole is given by the following expression: $$d=2\sqrt{f\lambda}$$ where $d$ is pinhole diameter, $f$ is focal length (distance from pinhole to image plane) and $\lambda$ is the wavelength of light. Image courtesy: My own work :) | {
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537,550 | A vector at any angle can be thought of as resultant of two vector components (namely sin and cos). But a vector can also be thought of resultant or sum of two vectors following Triangle Law of Addition or Parallelogram Law of Addition, as a vector in reality could be the sum of two vectors which are NOT 90°.The only difference here will be that it is not necessary that components will be at right angle. In other words why do we take components as perpendicular to each other and not any other angle (using Triangle Law and Parallelogram Law). | Indeed, any vector can be resolved in terms of two components (in $n$ -dimensional space in terms of $n$ components). For this being possible the components should be linearly independent, i.e. in your case they should not be parallel. The advantage of using two orthogonal/perpendicular components is that their scalar product is zero, which simplifies the math when calculating the coefficients: \begin{array}
\mathbf{F} = F_x\mathbf{e}_x + F_y\mathbf{e}_y \Longrightarrow F_x = \mathbf{e}_x\cdot \mathbf{F}, F_y = \mathbf{e}_y\cdot \mathbf{F}.
\end{array} Indeed, \begin{equation}
\mathbf{e}_x\cdot \mathbf{F} = \mathbf{e}_x (F_x\mathbf{e}_x + F_y\mathbf{e}_y) = F_x\mathbf{e}_x \cdot\mathbf{e}_x + F_y\mathbf{e}_x \cdot\mathbf{e}_y = F_x,
\end{equation} and similarly for $F_y$ , since \begin{equation}
\mathbf{e}_x \cdot\mathbf{e}_x = 1, \mathbf{e}_y \cdot\mathbf{e}_y = 1, \mathbf{e}_x \cdot\mathbf{e}_y = 0.
\end{equation} For non-orthogal vectors $\mathbf{e}_x \cdot\mathbf{e}_y \neq 0$ , the math becomes a bit more complicated and the interpretation of the projections as coordinates is less intuitive. There are however some cases where using non-orthogonal components is beneficial, notably when dealing with non-orthogonal crystal lattices, such that of graphene (a hexagonal lattice.) | {
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537,756 | Imagine you have a pull back toy car . Its back part is on $x_0$ . You push it down and move it in the back direction to the point $y$ (not marked): Then you leave the car to move away: Then you mark the final position by $x_1$ : Let's say, that the distance you have pushed the car is $d_1$ and the distance the car has travelled is $d_2$ . As you can see, the car has travelled more than before (same as $d_1 < d_2$ ). If you don't believe it, try it yourself. Why has that happened? The law of conservation of the energy tells us, that energy can't be created out of anything. What has happened? | Are you are expecting that if you roll the car backwards $30\ \mathrm{cm}$ then release it, it should move forward $30\ \mathrm{cm}$ ? Why? Most toy cars wouldn't move at all. If you put a stone in a catapult, pull it back $30\ \mathrm{cm}$ then release it, it goes forward much further than $30\ \mathrm{cm}$ . If you did this in empty space the stone would keep going indefinitely. Energy hasn't been created out of nothing. You have done work against the catapult, storing elastic energy. When you release the catapult the stored elastic energy is transformed into the kinetic energy of the stone, which is dissipated as heat and sound as the stone flies through the air and hits a target. If there is no air resistance or friction, and nothing impedes the stone, its kinetic energy remains constant forever – its speed doesn't change, it goes infinitely more than $30\ \mathrm{cm}$ . The toy car is the same. Instead of an elastic band, it contains a spring. Pushing down engages a gear wheel. As you push the toy car backwards you wind up the spring quickly using a relatively large force. You do work, elastic energy is stored in the spring. When the car is released, it springs back up and a different gear wheel is engaged . Now the spring unwinds itself slowly, supplying a much smaller force to the toy car. (See Note .) Elastic energy is transformed into the kinetic energy of the car, which is dissipated by friction. The car loses its kinetic energy gradually; it slows down and stops. If there was no friction the car would keep going indefinitely on a flat surface. It is not the distances which you need to compare but the work done , which is force times distance. You give the car elastic potential energy by pushing with a large force over a short distance. The much smaller force of friction takes that energy away over a much longer distance after it has been transformed into kinetic energy. Suppose the friction force is $0.1\ \mathrm N$ and you push the car backwards with a force of $5.1\ \mathrm N$ through a distance of $30\ \mathrm{cm}$ . Then you have done $5.1\ \mathrm N \times 0.3\ \mathrm m = 1.53\ \mathrm{Nm}$ of work. Friction works in both directions so $0.1\ \mathrm N \times 0.3\ \mathrm m = 0.03\ \mathrm{Nm}$ of the work you do is wasted pushing against friction. The remaining $1.50\ \mathrm{Nm}$ of energy gets stored in the spring. When the car is released it is transformed into the kinetic energy of the car. The friction force of $0.1\ \mathrm N$ slows the car. You can expect the car to go a distance of $15\ \mathrm m$ before stopping because $0.1\ \mathrm N \times 15\ \mathrm m = 1.5\ \mathrm{Nm}$ . The car goes $50$ times further forwards than you moved it backwards. But you haven't created any energy. In fact, some energy was lost pushing against friction. Only $1.50\ \mathrm{Nm}$ of the $1.53\ \mathrm{Nm}$ of energy which you supplied was used to move the car forward. Note: When the spring is fully unwound it is disengaged from the wheels so that the car rolls forward freely instead of winding the spring back up. That's like the catapult which releases the stone; otherwise the stone would stretch the elastic again and keep oscillating until its kinetic energy was used up. | {
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537,986 | I have heard it said that charged planets could not orbit a massless (low mass) oppositely-charged star based on electromagnetic attraction the same way they can with gravitational attraction, because Maxwell's laws dictate that accelerating (orbiting) charges produce electromagnetic waves and therefore lose energy which would lead to the planets slowdown and eventually crash. But it occurred to me that something similar would seem likely with gravitational waves in real-life gravity-based orbits. Is it true than that planets' orbits are decaying slowly and turning that energy into gravitational waves? If not, how can that be, given we know gravitational waves exist and surely expend energy in the same way the production of electromagnetic waves do? | Yes, but undetectably. The Earth-Sun system radiates a continuous power average of about 200 watts as gravitational radiation. As Wikipedia explains, “At this rate, it would take the Earth approximately $1\times 10^{13}$ times more than the current age of the Universe to spiral onto the Sun.” The Hulse-Taylor binary (two neutron stars, one a pulsar) was the first system in which the gravitational decay rate was measurable. It radiates $7.35\times 10^{24}$ watts as gravitational radiation, about 1.9% of the power radiated as light by the Sun. | {
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538,550 | I have been thinking about this and I know that other people have answered this on here, but there's one part that still baffles me, and it has to do with parallel circuits. If I connect a battery to a resistor, and connect another in parallel to it, and measure the current across both, there will be a current! That is, if I connect a $6\ \mathrm V$ battery to a $100\ \mathrm\Omega$ resistor ( $R_1$ ) and connect a $200\ \mathrm\Omega$ resistor in parallel to it ( $R_2$ ), I would still measure $6\ \mathrm V$ across both (voltage is preserved in parallel circuits, correct?) and my current (per Ohm's Law) is $$I=\frac VR\Rightarrow I=\frac{6\ \mathrm V}{100\ \mathrm\Omega}=0.06\ \mathrm A $$ $$I=\frac VR\Rightarrow I=\frac{6\ \mathrm V}{200\ \mathrm\Omega}=0.03\ \mathrm A $$ So that means one resistor has $6o\ \mathrm{mA}$ and the other has $30\ \mathrm{mA}$ . Well and good, but why doesn't this apply to a bird? That is, a bird dropping its feet on a wire isn't completing a circuit between two different potentials but it is making a parallel circuit. This is what confuses a lot of people I think including me. If the usual laws for parallel circuits apply, why doesn't it apply to birds on a wire? One explanation I hear is that birds aren't connecting two places of differing potential – but if that was the case then why does my parallel circuit work? One resistor should register no current (or very little) – and I know if I make the resistor large enough (the one in parallel, say, $R_2$ ) the current draw will be smaller. Is that what is happening? The resistance of the bird is large enough that the current drawn is small? Let's say a bird has $1\ \mathrm{M\Omega}$ of resistance. A $600\ \mathrm V$ wire would still put $0.6\ \mathrm{mA}$ through the animal. But that doesn't satisfy me because we are dealing with a $\mathrm{kV}$ scale wire a lot of the time. You'd need for the bird, which is effectively a bag of water and such, to have a lot of resistance for that to work, but maybe it does. I am always reading that in order for the circuit to be complete the bird (or person) must be grounded, but that doesn't make sense to me because then no parallel circuit would work from batteries! Or even the house current, which is basically a lot of circuits in parallel. I feel like I am missing something here, and if anyone can tell me what it is that would be greatly appreciated. | A birds legs are pretty close together. An electrical transmission wire has very little resistance. This means that the voltage as a function of distance barely changes. So the voltage difference between two birds feet is essentially 0, because the potential on each foot is practically the same. The potential difference between the wire and the ground might be large; but the bird isn't offering any pathway between the wire and anything at much lower voltage. It only offers a pathway between it's two legs, and so voltage difference remains small. To add on to that, the bird has a lot more relative resistance than the wire, since the wire is supposed to minimize voltage drop across it. This means that most of the current will also flow through the wire, and relatively little current would flow through the bird. The bird isn't really at risk unless it can connect the high voltage line to something of a significantly different potential, which the same line a few inches further down isn't. For an example of the numbers, Solomon Slow estimated in the comments : Suppose the wire is equivalent to 000-gauge copper, 0.0618 ohms per 1000 ft. Suppose it's carrying close to its rated capacity: 300A. Suppose a bird, maybe the size of a dove, with legs that grip the wire about 1 inch apart. According to my calculation, the potential difference between points 1 inch apart along the length of that wire will be about 1.6 millivolts . Emphasis mine. It should be pretty easy to check that estimation for yourself, but it really illustrates the problem. | {
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538,565 | In gravity (GR) apparently there are forces which occur which are related to the spin of two masses . For example if we had a rotating gravitational source and dropped, say, some particles into it with instrinsic spin, they would fall differently under gravity. From what we learned at school, every object no matter what it's made of must fall the same under gravity. It only depends on the mass of the source. (i.e. drop a feather and a rock they must fall the same on the moon). So was Galileo wrong? What's more, using this spin-spin interaction seems to suggest that one might use this interaction to create an anti-gravity device. (In the sense that one could create an object that fell every so slightly slower towards a spinning object such as the sun or Earth). In fact I've seen papers which suggest that two spinnning black holes could even repel each other. Since gravity is equivalent to acceleration, is there an equivalent accelerated frame that causes different spinning particles to behaving differently? I'm trying to reconcile this spin-spin interaction with what I was previously taught about how things must fall the same under gravity. | A birds legs are pretty close together. An electrical transmission wire has very little resistance. This means that the voltage as a function of distance barely changes. So the voltage difference between two birds feet is essentially 0, because the potential on each foot is practically the same. The potential difference between the wire and the ground might be large; but the bird isn't offering any pathway between the wire and anything at much lower voltage. It only offers a pathway between it's two legs, and so voltage difference remains small. To add on to that, the bird has a lot more relative resistance than the wire, since the wire is supposed to minimize voltage drop across it. This means that most of the current will also flow through the wire, and relatively little current would flow through the bird. The bird isn't really at risk unless it can connect the high voltage line to something of a significantly different potential, which the same line a few inches further down isn't. For an example of the numbers, Solomon Slow estimated in the comments : Suppose the wire is equivalent to 000-gauge copper, 0.0618 ohms per 1000 ft. Suppose it's carrying close to its rated capacity: 300A. Suppose a bird, maybe the size of a dove, with legs that grip the wire about 1 inch apart. According to my calculation, the potential difference between points 1 inch apart along the length of that wire will be about 1.6 millivolts . Emphasis mine. It should be pretty easy to check that estimation for yourself, but it really illustrates the problem. | {
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538,568 | In the figure, there are two blocks of masses m and 2m connected by a light string passing over a frictionless pulley. The mass m is placed on a smooth inclined plane of inclination 30 degrees and 2m hangs vertically. Now we gotta find the acceleration when the system is released. Solution: acceleration=net force/total mass = 2mg-mgsin30/3m Now I dont get why is it mgSin30. Like according to trigo sin30=mg/F where F is the force the block m is putting on the system. And since we have to put the net force in the answer. Shouldn't it be:
force 2m is applying - force m block is applying/3m so shouldnt it be F=mg/sin30
So i dont get why are we supposed to use mgsin30. | A birds legs are pretty close together. An electrical transmission wire has very little resistance. This means that the voltage as a function of distance barely changes. So the voltage difference between two birds feet is essentially 0, because the potential on each foot is practically the same. The potential difference between the wire and the ground might be large; but the bird isn't offering any pathway between the wire and anything at much lower voltage. It only offers a pathway between it's two legs, and so voltage difference remains small. To add on to that, the bird has a lot more relative resistance than the wire, since the wire is supposed to minimize voltage drop across it. This means that most of the current will also flow through the wire, and relatively little current would flow through the bird. The bird isn't really at risk unless it can connect the high voltage line to something of a significantly different potential, which the same line a few inches further down isn't. For an example of the numbers, Solomon Slow estimated in the comments : Suppose the wire is equivalent to 000-gauge copper, 0.0618 ohms per 1000 ft. Suppose it's carrying close to its rated capacity: 300A. Suppose a bird, maybe the size of a dove, with legs that grip the wire about 1 inch apart. According to my calculation, the potential difference between points 1 inch apart along the length of that wire will be about 1.6 millivolts . Emphasis mine. It should be pretty easy to check that estimation for yourself, but it really illustrates the problem. | {
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540,065 | I'm a laywoman in physics and recently found myself pondering about the matter reflected in the title of this post. To make my question more precise from the mathematical standpoint, let's suppose you are given a 3D image of the momentary positions of the nuclei of all atoms of an unknown monoatomic substance in a certain volume at a certain moment of time. Rotating the image in a 3D visualization program, you see that the positions look pretty chaotic from any angle, unlike a crystalline structure. You know neither the image's scale nor any of the parameters such as the pressure or temperature. The only information you are given is that the substance is not ionized and is in a thermodynamic equilibrium and either in the liquid state or in the gaseous state and that the pressure and the temperature are below the critical pressure and the critical temperature, respectively. You can extract the numerical XYZ positions and do any calculations with them, but, as stated above, you don't know the scale. How can you tell whether it's a liquid or a gas? What criterion can be used to reach that end? My first guess was that whilst a gas doesn't have any correlation between the positions, a liquid does, but then I realized it's a wrong answer because a gas is not necessarily an ideal gas, so it's unclear to me how I can tell whether it's a liquid or a gas if there's some correlation between the positions in the image. I tried to find the answer on the Internet and this SE, but did not succeed and humbly hope that physics experts on this SE can tell me the answer. UPDATE: Sure, the limiting cases of an ideal gas and a tightly packed liquid are easy, but what do I do in the general case? In other words, how can I deduce whether it's a liquid or a gas if the spread of distances between neighboring nuclei is moderate, that is, neither very small nor very large? | Everything you've said is correct, which is why the conclusion is: there is no fundamental difference! Under the modern classification, they're just the same fluid phase of matter. For example, consider the phase diagram of water . If you take water vapor, slowly heat it up, then pressurize it, and then slowly cool it down, you'll end up with liquid water. This entire process is completely smooth. There isn't any sharp point, like a phase transition, where the behavior qualitatively changes; thus we can't make a sharp distinction between liquids and gases. There are fluids that are "liquid-like" (densely packed, strong interactions between neighbors) and fluids that are "gas-like" (sparse, weak interactions between neighbors) but no dividing line, just like how there's no moment where a shade of grey changes from white to black. By contrast, ice really can be distinguished from liquid water or water vapor. You can't turn either of the two into ice without crossing a phase transition. At that point, the atoms will suddenly become ordered, and you can see this from a snapshot of their positions. Edit: in response to the 25 comments, I'm not saying there's no difference between liquids and gases, I'm saying that there are clearly liquid-like things, and clearly gas-like things, but a continuous spectrum between them. Here are some properties that characterize gases: large distance between molecules weak interactions large mean free path high compressibility very low surface tension upward density fluctuations at small separation The opposite properties characterize liquids. In the easy cases, you could use any of these to make the call. But all of these properties change continuously as you go from one to the other, as long as you go around the critical point. This isn't true for a solid/liquid or solid/gas phase transition. | {
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540,090 | For the Ch. 27 in book QFT by Srednicki, in the modified minimal-subtraction renormalization scheme( $\overline{MS}$ ), the residue for pole at $-m_{ph}^2$ is $R$ , instead of one. However, I can not understand why we need only change the external leg as: add a $R^{1/2}$ coefficient replace the Lagrangian mass $m$ as $m_{ph}$ but we don't need to change the internal legs. Also, renormalization scheme means to determine the coefficient for counterterms, or say, determine the coefficient for $Z_i$ in Largranian: $$L=-\frac{1}{2}Z_\phi(\partial \phi)^2-\frac{1}{2}Z_mm\phi^2+\frac{1}{4!}Z_g g \phi^4\\=-\frac{1}{2}(\partial \phi_0)^2-\frac{1}{2}m_0\phi_0^2+\frac{1}{4!} g_0 \phi_0^4$$ according to the answer in Is the effective Lagrangian the bare Lagrangian? , I know that renormalized $m$ is not the physical mass, but the pole of propagator is. But I am confused that: since it seems like $m_0$ is just pole according to the second line of equation above, does this mean bare mass $m_0$ is actually the physical mass(the mass what we detected)? I think so because Srednicki says "bare parameters" must be independent of $\mu$ " in Ch.28. But other book, such as p.323 Peskin, says bare mass is not not the values measured in experiments. | Everything you've said is correct, which is why the conclusion is: there is no fundamental difference! Under the modern classification, they're just the same fluid phase of matter. For example, consider the phase diagram of water . If you take water vapor, slowly heat it up, then pressurize it, and then slowly cool it down, you'll end up with liquid water. This entire process is completely smooth. There isn't any sharp point, like a phase transition, where the behavior qualitatively changes; thus we can't make a sharp distinction between liquids and gases. There are fluids that are "liquid-like" (densely packed, strong interactions between neighbors) and fluids that are "gas-like" (sparse, weak interactions between neighbors) but no dividing line, just like how there's no moment where a shade of grey changes from white to black. By contrast, ice really can be distinguished from liquid water or water vapor. You can't turn either of the two into ice without crossing a phase transition. At that point, the atoms will suddenly become ordered, and you can see this from a snapshot of their positions. Edit: in response to the 25 comments, I'm not saying there's no difference between liquids and gases, I'm saying that there are clearly liquid-like things, and clearly gas-like things, but a continuous spectrum between them. Here are some properties that characterize gases: large distance between molecules weak interactions large mean free path high compressibility very low surface tension upward density fluctuations at small separation The opposite properties characterize liquids. In the easy cases, you could use any of these to make the call. But all of these properties change continuously as you go from one to the other, as long as you go around the critical point. This isn't true for a solid/liquid or solid/gas phase transition. | {
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540,320 | I saw these images of how a telescope works and it seems like it is shrinking the image down to the size of the eye. I don’t understand how that makes the image bigger. My thinking is that shrinking an image makes the image obviously smaller and you lose details. What am I not getting here? | While a telescope can make an image larger, the diagram shown above doesn't really show that happening. Your eye perceives the size of an image based upon the angular extent it takes up in the visual field. A ball that takes up one degree of your field would look larger to you if it took up five degrees of your visual field instead. This could happen by moving it closer to you, or by using an optical magnifier. So a telescope (or similar) will magnify an image by changing the angles through which it is seen. Because only one set of parallel rays are drawn (and they are shown as parallel again when passing through the last lens), nothing about the magnification of this system is evident. The other thing a telescope will do is to collect light from a large area and concentrate that light into your pupil. This might not magnify the image, but it would make it brighter. Whether intentional or not, this is what is shown above. So don't think of the big mirror as "shrinking" an image, think of it as "concentrating" the light. Whether the image is reduced or enlarged depends on the angles at the end, not on the density of the lines. | {
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540,437 | The electron absorbs the energy of photon(with specific frequency)and re-emits the photon.The photon can be emitted in any direction. So why do they get re-emitted in a specific direction after reflection? On hitting normal to surface the photons follow path in reverse normal direction and hitting at an angle it follows V path. | Although a single photon can only be absorbed and emitted by a single electron, it leaves that electron in exactly its original state. There is no record, and no way of knowing, which electron absorbed and emitted the photon. According to quantum theory, to calculate the result when any electron could have absorbed and emitted the photon, we must form a superposition of all the processes which could have taken place. The calculation in quantum mechanics takes the form of wave mechanics -- this does not have to mean that there is actually a wave, only that the mathematical theory behaves as though there was a wave. Since a wave would reflect at a particular angle, that is also what happens for photons. The reasons for this strange quantum behaviour are deep and subtle. Quantum mechanics is actually a theory of probabilities, not a theory of physical waves. The mathematics tells us that the probability of a particular angle of reflection is actually a certainty; we can even show that this mathematical behaviour is necessary to a consistent probabilistic interpretation of quantum theory. It is much harder, probably impossible, to conceptualise exactly what is physically taking place at the level of elementary particles leading to these kinds of results. | {
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540,485 | How can photons have different energies if they have the same rest mass (zero) and same speed (speed of light)? | Some areas of physics are counter-intuitive. For them, your everyday experience is a poor guide to how the universe really works. This is one of those areas. Photons have no mass. They all have the same speed. Yet they have energy and momentum, and it isn't the same for all photons. If you are used to $p = mv$ , this doesn't make sense. The explanation is simple. $p = mv$ doesn't apply to photons. It applies to massive objects at low speeds, and photons are something different. One way to make sense out of photons is to treat them like the new thing they are. Before you encountered quantum mechanics, you never encountered anything that was sort of like a particle and sort of like a wave. So what are the properties of this new and different thing? An excited atom can drop to the ground state, and at the same time experience a recoil. A while later, another atom that was at rest with respect to the first atom can experience a recoil in the opposite direction and get promoted to an excited state. A photon is what happens in between. Experiments like this show that photon had enough energy to excite an atom and enough momentum to give it a recoil. They show a photon is something like a particle. Experiments with diffraction gratings show photons have frequency and wavelength, and higher frequency/shorter wavelength corresponds to higher energies and momenta. I am glossing over other counter-intuitive results, like uncertainty of momentum. Having said this much, I hope I don't muddy the waters by saying there isn't any such thing as a red or blue photon. This gets back to relativity. You do have some everyday experience with Galilean relativity, which isn't entirely different from special relativity. Suppose you are floating in space and you encounter a rock. If the rock isn't moving fast, it taps you gently. If it is moving fast, it does damage. But you can't really say how the rock is moving. You can only say how fast it is moving with respect to you. Two people could see the same rock. One could see it moving slowly, and the other fast. They would disagree on how much energy and momentum the rock has. Suppose you are sitting in a boat watching waves go by. You count peaks passing by per second to get frequency. If you move into the waves, you encounter peaks more often, and your value for the frequency goes up. You also see the waves moving faster with respect to the boat. The distance between peaks does not change. Photons don't have mass and their speed is always c. But their energy and momenta behave something like what you would expect from watching rocks. Their frequency behaves something like what you would expect from watching water waves or sound waves. There are differences in details, but your intuition can be something of a guide. Photons are like rocks in that different atoms will see different energy and momenta, depending on how they move. If we repeat the exited atom experiment with atoms that are approaching each other, we find the recoil is higher than for an atom at rest, the photon has an energy higher than is needed to excite the atom. The intuitive part is that the photon "hits harder" when you run upstream into it. The counter intuitive part is that photons always travel at c, so it hits at the same speed. You also get semi-sensible results when an atom and a diffraction grating are approaching each other. Like water waves, the diffraction grating encounters peaks more often and sees a higher frequency. The counter-intuitive part is that the speed doesn't change, but the distance between peaks gets shorter. The diffraction grating reflects the photons at a different angle. So there is no such thing as a red or blue photon because it matters how fast the thing it hits is moving. The thing it hits will see it as red or blue, and something else would see it differently. But again, this is counter-intuitive. Even though the photon always hits a speed c, there is a difference. It is more intuitive when you think of the relative velocity between the thing that was hit and the thing that emitted the photon. Quantum mechanics is often like this. There are two interactions, and everything adds up before and after. But what goes on in between can be murky. A photon or electron is emitted from a source. There is no trajectory it follows, only a wave that describes probabilities. Then it hits something. The recoil of the source and target match. Intuition has lead people to look for a deeper theory that explains more. If there is a cause, there must be a predictable effect. It turns out that this intuition leads down a wrong path. This is the way the universe works. The best thing to do is find ways to get used to it. | {
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541,598 | I have a rather naive question. In stars such as the Sun, what prevents the whole thing exploding at once? Why is the nuclear fusion happening slowly? I can only assume that something about the fusion is fighting the gravity and slowing the fusion down and when that process is done gravity starts the fusion process again. | The fusion that occurs in the core of the Sun occurs in nothing like the conditions you might be thinking of in a bomb, or a fusion reactor. In particular, it occurs at much lower temperatures and at a much lower rate. A cubic metre of material in the solar core is only releasing around 250 W of power by fusion. The fusion rate is set by the temperature (and to a lesser extent, density) of the core. This in turn is set by the need for a pressure gradient to balance the weight of material pressing down on it from above. The core pressure is set by the need to balance the weight above it; the core pressure is determined by the temperature and density of the core; and the temperature and density of the core set the fusion reaction rate. At 15 million kelvin (the Sun's core temperature, which is much lower than the temperatures in nuclear bombs or fusion reactors), the average proton has a lifetime of several billion years before being converted (with three others) into a helium nucleus. There are two reasons this is slow. First, you have to get protons, which repel each other electromagnetically, close enough together to feel the strong nuclear force. This is why high temperatures are needed. Second, because the diproton is unstable, one of the protons needs to change into a neutron via a weak force interaction, whilst it is in the unstable diproton state, to form a deuterium nucleus. This is just inherently unlikely and means the overall reaction chain to helium is very slow. The reason there is no bomb-like explosion is because there is no problem in shifting 250 W per cubic metre away from the core, in the same way that a compost heap, which generates about the same power density, does not spontaneously explode. In the case of a star any additional heat goes into more radiation that diffuses away and in work done in expanding the star. As a result, the temperature of the core is stable - the timescale for the structure of the star to change and maintain stability (millions of years) is much shorter than the timescale on which the nuclear fuel is burned (billions of years). Ultimately, any additional energy emerges as sunlight at the solar photosphere. If for some reason, the opacity to radiation in the core increased, then the temperature would rise and more energy would be generated by fusion. This is exactly what happens in the core as more hydrogen is turned into helium; the core temperature and luminosity do rise, but slowly, on timescales of billions of years. There isn't a runaway explosion because the star can easily adjust its size and luminosity to radiate away the rising fusion rate. | {
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542,298 | This morning I made myself some coffee. There was cream on it as how I liked my coffee. I lifted my mug and rotated it so I can sip. Then I realized the cream on coffee wasn't rotating with the mug. It was in rest, the mug was rotating but the pattern of cream on my coffee was looking exactly the same. So the coffee had to be in rest or it was rotating the opposite direction I rotated my mug but why? Can anyone explain this phenomena or say the name of it? | Suppose your cup was full of a lump of stuff that had no friction. Then as you rotate your cup there is no force between the cup and the stuff, and so it would not rotate. In the case of coffee - the friction is from the viscosity of the coffee. But, as your cup pulls some of the coffee around with it right at the edge, the rest of the coffee slips on that layer, rather like if you slip on a patch of coffee on the floor. Eventually, if you keep rotating the coffee, say on a turn table, the small friction will rotate all the coffee. Then, if you stop rotating the cup, the coffee will keep rotating from its inertia, until the small amount of friction has brought it to a halt. | {
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543,056 | Whenever I study the photoelectric effect and the Compton effect, I have always had a question about how a photon can possibly collide with an electron given their unmeasureably small size. Every textbook I've read says that the photo-electrons are emitted because the photons collided with them. But since the photons and electrons virtually have no size, how can they even collide? I have searched for the answer on the internet but I couldn't find any satisfying one. | This is an answer by a particle physicist that has been working with data for forty years: Photons and electrons are quantum mechanical entities, and to really really understand their interactions, quantum mechanics has to be invoked. When detected, the photon has a point particle footprint (as does the electron) consistent with the axiomatic particle table of the standard model . The leftmost frame shows the collision of a countable single photons on a screen (in a double-slit experiment) . The accumulation of photons (light emerges in a calculable manner from many photons), shows the wave nature's interference effects. It is the probability of landing on the (x,y) of the screen that displays a wave behavior. Not the individual photons . Here is another measurement of a photon The original picture is here . That the single photon (gamma) electron interaction is at a point is evident. Now let us see how what we call size for macroscopic particles in quantum mechanics appears. It is all dependent on probabilities of a particle being at an (x,y,z) to interact with another particle. Look what an electron around a hydrogen atom has as a probable location : This is what defines the macroscopic charge distribution, and the probability of an incoming gamma ray to interact with the electron is a mathematical combination of this, and the coupling constants of the quantum mechanical interactions. A free electron has a very small probability to be hit by a photon. That is why high density beams are used in high energy experiments. In general it will be the coupling constants which will give high probabilities the closer the two point particles are, and of course not to forget Heisenberg's uncertainty principle , which also will define a volume in space and momentum where interactions can happen. The photoelectric effect involves electrons that are in orbitals and a large number of atoms and molecules, and the fact that it exists means that there is enough probability for an incoming photon to hit an electron in the orbitals distributions of the specific solid. | {
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543,199 | According to the third law of Newton, every action has an equal and opposite reaction. This would (to me) suggest that every surface should put a force exactly opposite the weight of the body. If so, why does compression occur? Say you keep a weight on a sponge, then the sponge compresses, why is this? Does the sponge not give an equal reaction to the weight on it? Edit: If Newton's third law is true, why can we sink in sand? This question is different from this one because there the question is mainly about which bodies the forces act on according to third law while the question I asked is mainly about compression and how much force a body applies against a weight put placed on it. | Imagine you are in a crowd of people. A huge crowd, with everyone, almost squeezed together. Imagine you run into the crowd. In that case, you will have an almost rigid body, because, you apply a force on him, he applies a force back on you, you stop, but he will have a force, which will be balanced by the next person and next and so on till you reach a wall. But, if the crowd has a little less number of people, or more like people standing in rows holding hands to form a chain, it will be different.
If you run into the crowd from outside, you will push a few people, and they will accelerate (inelastic collision) and you and the other person moves with the same velocity as you, but you are not falling into the person. He exerts an equal force, thus preventing you from sinking into him. But as there is no other person to provide an equal force on him, he starts falling in till he reaches another person because of the unbalanced forces on him But that does not mean that he is not moving, because he is, and if you were large enough it would be similar to the case you talk about. Several people moving in a localized region that would 'appear' as if the forces are not equal. But if you go a bit more closer, you will see that it is not violated. The sponge is not one body, rather it is like the crowd. The particles can move, to some extent, independent of the other particles. So, when you push a sponge, you are actually making some sponge mass to move, like crashing into the crowd. So, Newton's third law is not violated | {
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543,413 | I am new to physics and I have learned a little bit about gravity from Einstein's perspective. The gist is that heavy objects create curvature of spacetime, and free-falling objects move on the straight lines in the curvature. But I am failing to understand how this applies to objects on Earth. For example, why don't elephants make spacetime curvature and cause dust to go around them? Or simply, how does spacetime curvature work inside a planet? | The short answer: general relativistic effects are mostly not noticeable on such a small scale (except in a few cases). For example, one common way to characterize the strength of a gravitational field is through the dimensionless number: $$A=\frac{GM}{Rc^2}$$ where $G = 6.67 \times 10^{-11} \text{ }\mathrm{m^3 kg^{-1} s^{-2}}$ is the gravitational constant , $M$ is the mass, $R$ is the distance to the object, and $c=2.99\times 10^8\text{ m/s}$ is the speed of light. For being directly next to an African Bush Elephant: $$A \approx 10^{-24}$$ For the Earth: $A \approx 10^{-9}$ and even for the Sun, $A \approx 10^{-6}$ . Generally, when $A \ll 1$ the effects are practically negligible. In fact, most of the physics that describes planetary motion in our solar system can be accurately described with standard Newtonian physics. However, there are some extremely sensitive cases (e.g. orbit of Mercury / satellite communication) where we need to take into account general relativity. Just for comparison, a black hole (using the Schwarzschild radius) gives a value of $A=0.5$ , which is much stronger than our Sun's meaning we definitely will need to take into account general relativity. | {
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543,425 | I was wondering if at a given gradient temperature heat could flow in any direction satisfying moving from a higher to a lower temperature. Or it can only flow in the direction of gradient? Consider a point $(x_0,y_0,z_0)$ where $T_0=T(x_0,y_0,z_0)$ and $T_1$ , $T_2$ are the temperatures of two points very close to the $(x_0,y_0,z_0)$ . It is true that $T_0>T_1>T_2$ . Would some amount of heat flow to the point with $T_1$ and another amount to the point with $T_2$ or all the heat would flow to the point with the maximum temperature difference? What if $T_1=T_2$ ? | The short answer: general relativistic effects are mostly not noticeable on such a small scale (except in a few cases). For example, one common way to characterize the strength of a gravitational field is through the dimensionless number: $$A=\frac{GM}{Rc^2}$$ where $G = 6.67 \times 10^{-11} \text{ }\mathrm{m^3 kg^{-1} s^{-2}}$ is the gravitational constant , $M$ is the mass, $R$ is the distance to the object, and $c=2.99\times 10^8\text{ m/s}$ is the speed of light. For being directly next to an African Bush Elephant: $$A \approx 10^{-24}$$ For the Earth: $A \approx 10^{-9}$ and even for the Sun, $A \approx 10^{-6}$ . Generally, when $A \ll 1$ the effects are practically negligible. In fact, most of the physics that describes planetary motion in our solar system can be accurately described with standard Newtonian physics. However, there are some extremely sensitive cases (e.g. orbit of Mercury / satellite communication) where we need to take into account general relativity. Just for comparison, a black hole (using the Schwarzschild radius) gives a value of $A=0.5$ , which is much stronger than our Sun's meaning we definitely will need to take into account general relativity. | {
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543,434 | I am wondering why I have seen the covariant derivative for the first time in general relativity . Starting from the point that the covariant derivative generalise the concept of derivative in curved space (even if think it is better to consider it as the extension of the derivative such that it is covariant under a change of coordinates). To do that we introduce the Christoffel symbols $\Gamma^i_{jk}$ . In curved space-time we have globally non vanishing Christoffel symbols $\Gamma^i_{jk} \neq 0$ , but in general $\Gamma^i_{jk} \neq 0$ doesn't mean we are in curved space-time. For example, if I consider Minkowski space-time with Cartesian coordinates than, thanks to the Lorentz transformation, if the Gammas are zero in a reference frame they are zero in every reference of frame, but I could have $\Gamma^i_{jk} \neq 0$ even in flat space time with polar coordinates, as the Gammas don't transform like a tensor in this case due to the non tensorial part of the transformation law for $\Gamma^i_{jk}$ under a change of basis. If what I said before is true (a big if), then I'd interpret this in classical mechanics saying that in Cartesian coordinates, the basis vectors { $\hat{e}_x,\hat{e}_y$ }, solid to a point of a curve, are constant if the point is moved along the curve. While I think I can't say the same for { $\hat{e}_r, \hat{e}_{\theta}$ }, as moving a point along the curve in this case the tangent vectors to the coordinate lines aren't constant (they rotate while the point is moving). This is why I think I should see the Christoffel symbols even in classical mechanics, to reflect the property of the vectors { $\hat{e}_r, \hat{e}_{\theta}$ } that vary along the curve. | You don't see the covariant derivative as often because flat space has isometries that make Cartesian coordinates better, and in these coordinates there are no Christoffel symbols, so we use them as much as possible. But look at the formula for the divergence of a function $\mathbf{F} = F^\hat{r} \hat{\mathbf{r}} + F^\hat{\theta} \hat{\mathbf{\theta}}$ in polar coordinates: $$\nabla \cdot \mathbf{F} = \frac{1}{r} \frac{\partial(r F^\hat{r})}{\partial r} + \frac{1}{r} \frac{\partial F^\hat{\theta}}{\partial \theta} = \frac{\partial F^\hat{r}}{\partial r} + \frac{1}{r} F^\hat{r} + \frac{1}{r} \frac{\partial F^\hat{\theta}}{\partial \theta}.$$ That $1/r$ in the middle term with no derivatives comes from the Christoffel symbols! So the covariant derivative is definitely there, but instead of using the Christoffel symbols, we usually calculate it using the chain rule and the fact that the cartesian basis vectors have zero derivative. The derivatives of the basis vector are after all the Christoffel symbols, so the method is not that different. One final comment: the orthonormal basis vectors $\{\hat{\mathbf{r}}, \hat{\theta}\}$ in polar coordinates are not the basis vectors $\{\partial/\partial r, \partial/\partial\theta\}$ we know from differential geometry, because the latter are not orthonormal. The relation is simple: $$\begin{aligned}
\hat{\mathbf{r}} &= \frac{\partial}{\partial r} \\
\hat{\theta} &= \frac{1}{r} \frac{\partial}{\partial\theta},
\end{aligned}$$ so keep this in mind when applying the formulas. In differential geometry we tend to write the components of vectors with respect to the derivative basis, but the formulas we know from more basic calculus (like my divergence formula) are written in terms of the orthonormal basis. | {
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544,445 | I was stretching a pink colored rubber band, and I noticed that the longer I stretch it, the lighter the pink becomes. I haven't found answers to this question anywhere else. Is there a reason for this phenomenon? Why does this happen? | Colour can come from pigment particles embedded in the translucent rubber matrix absorbing light. When you pull the band the particles become separated by a longer distance, but being themselves inelastic remain the same size. Hence the amount of absorption per unit area decreases, and the band become lighter in color. Simulated rubber band with pigment particles embedded in the matrix. As it is extended it becomes more translucent Rubber bands are also incompressible ( $\nu=1/2$ ) so the volume is largely unchanged by pulling. This has the effect of reducing the cross section, further reducing absorption. | {
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544,453 | Suppose you place a point charge $+Q$ at the corner of an imaginary cube. Since electric field lines are radial, there is no flux through the three adjacent (adjacent to the charge) sides of the cube. However there is some amount of flux passing through the other three sides of the cube (flowing out of the cube). We can estimate that the flux through these three surfaces combined is equal to $Q/(8\epsilon)$ . As, if you consider $7$ other cubes having the charge at the corner, each of them would have equal flux flowing out by symmetry and since the total flux through the $8$ cubes is $Q/\epsilon$ , each cube would have a flux of $Q/(8\epsilon)$ . Now apply Gauss' law to the cube, and we find that the cube encloses a charge of $Q/8$ . This means that 1/8th of the charge belongs to this cube. But the charge we placed was a point charge with no dimensions. It cannot be split into parts. What is wrong? | Gauss's law applies to situations where there is charge contained within a surface, but it doesn't cover situations where there is a finite amount of charge actually on the surface - in other words, where the charge density has a singularity at a point that lies on the surface. For that, you need the " Generalized Gauss's Theorem " [PDF], which was published in 2011 in the conference proceedings of the Electrostatics Society of America. (I found out about this paper from Wikipedia .) The Generalized Gauss's Theorem as published in that paper says that $$\iint_S \vec{E}\cdot\mathrm{d}\vec{A} = \frac{1}{\epsilon_0}\biggl(Q_{\text{enc}} + \frac{1}{2}Q_{\text{con}} + \sum_{i}\frac{\Omega_i}{4\pi}q_{i}\biggr)$$ where $Q_{\text{enc}}$ is the amount of charge fully enclosed by the surface $S$ and not located on $S$ $Q_{\text{con}}$ is the amount of charge that lies on the surface $S$ at points where $S$ is smooth $q_i$ for each $i$ represents a point charge that is located on $S$ at a point where $S$ is not smooth (i.e. on a corner), and $\Omega_i$ represents the amount of solid angle around that that point charge that is directed into the region enclosed by $S$ . There are a few edge cases (haha) not handled by this formulation (although it should be straightforward to tweak the argument in the paper to cover those), but fortunately it does cover the case you're asking about, where a point charge is located at a corner of a cube. In that case, the amount of solid angle around the corner that is directed into the interior of the cube is $\Omega_0 = \frac{\pi}{2}$ . Plugging in that along with $q_0 = Q$ (the magnitude of the charge), you find that $$\iint_S \vec{E}\cdot\mathrm{d}\vec{A} = \frac{1}{\epsilon_0}\biggl(0 + \frac{1}{2}(0) + \frac{\pi/2}{4\pi}(Q)\biggr) = \frac{Q}{8\epsilon_0}$$ which agrees with what you've found intuitively. | {
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545,359 | I have just started studying about quantum mechanics, and I was studying the definition of the inner product for functions; I am also quite new to linear algebra. While studying I think I encountered a contradiction on the definition of the inner products between functions, and I can't resolve it. I am following the textbook "Mathematics for Physics by Frederick Byron". The book defines inner products as: (the function space is defined over the interval $[a, b]$ where $a,b \in \mathbb{R}$ ) $$ \langle f, g \rangle = \int_{a}^{b} f^*(x) g(x) dx $$ And of course the book proves the fact that the function space (the set of square integrable functions over some interval $[a, b]$ ) is in fact a vector space. As far as I know, as a consequence of the definition of vector spaces, the zero vector (or the zero function) has to be unique. As well, based on the definition of inner products the following condition should always be met: $$ \langle f,f \rangle = 0 \iff f=0 $$ However, in the textbook the authors note that $f$ could be a function which is non-zero at a set of points with a Lebesgue measure of 0, and $\langle f,f\rangle$ would still be $0$ . If the definition of the $0$ function is changed from a function which is $0$ for all $x \in [a, b]$ , to a function that is only non-zero at a set with a zero Lebesgue measure, then this issue will be resolved and the definition of inner products will be valid. But this also implies that the zero function is no longer unique, contradicting the fact that the function space is a vector space. What is my mistake? How can we satisfy both of these conditions (unique zero vector and the inner product property that only the zero function has a norm of 0) without arriving at a contradiction? I appreciate your help. I understand this question might be more of a mathematics question than a physics question, but considering the problem is relevant to the basis of quantum mechanics I think Physics Stack Exchange is the more appropriate place for this question to be asked. | This is precisely why the $L^2(\mathbb R)$ is not simply the space of square-integrable functions from $\mathbb R$ to $\mathbb C$ (which we might call $SI(\mathbb R)$ ). $SI(\mathbb R)$ consists of all functions $f:\mathbb R\rightarrow\mathbb C $ such that $\int_\mathbb{R} |f(x)|^2 dx$ exists and is finite. But as you note, if you try to make it into a Hilbert space, you run into problems. The solution is to define an equivalence relation $\sim$ on $SI(\mathbb R)$ , whereby $f \sim g$ if $f(x)\neq g(x)$ only on a set of Lebesgue measure zero - that is, $f\sim g$ if they agree almost everywhere . From there, define $L^2(\mathbb R)$ as the quotient set $SI(\mathbb R)/\sim$ , whose elements are equivalence classes of square-integrable functions under the equivalence relation $\sim$ . This resolves the ambiguity - the functions $f(x)=0$ and $g(x)=\begin{cases}1 & x=0\\ 0& x\neq 0\end{cases}$ are different elements of $SI(\mathbb R)$ , but they are two equivalent representatives of the same element of $L^2(\mathbb R)$ . | {
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546,845 | In air, when there is light propagating in a direction, we can still see it even when it is not primarily travelling in our direction, because a small part of the light hits the air molecules, and changes its direction; it travels towards us. Does this mean that, in a vacuum, you would not be able to see light which is not travelling towards you? | If the light has nothing to scatter off of to reach your eyes you won't see anything. | {
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547,371 | In school we learn that we can add velocities together, and then later on we learn that it's not correct and that there is a speed limit. Why create all this confusion when we could just use rapidity to begin with? Rapidity is defined as $w = \mathrm{arctanh}(v / c)$ , where $v$ is velocity and $c$ is the speed of light in a vacuum. Rapidities can be summed and have no upper bound. At non-relativistic speeds it acts proportional to velocity. In fact, at non-relativistic speeds, we could substitute $v$ for $wc$ (rapidity times speed of light), and one could hardly tell the difference. The ISS moves rather fast at a velocity of 7660 m/s (27,576 km/h), and has a $wc$ of about 7660.0000016667 m/s. Why can't we just substitute velocity for rapidity in real-world and classroom use, and end the confusion about why there is a speed limit once and for all? | In everyday life, we experience the universe in a non-relativistic classical way.
We are familiar with the concept of time and space. Defining the velocity as the ratio between a distance traveled in a given time interval is a much more natural choice instead of defining the rapidity. If we all lived at relativistic speeds, or close to the event horizon of a black hole, or if we were small as an atom, we would use other tools to describe the universe around us. However, in our case, the quantities of classical mechanics work quite well and we can have a direct intuitive grasp of their meaning. | {
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547,505 | This is a pretty basic question, but I haven't had to think about orbital mechanics since high school. So just to check - suppose a [classical] system of two massive objects in a vacuum. If the density of either object is the same at a given distance from the center, and both objects are spherical, then both objects can be treated as point-masses whose position is the [geometric] center of the original sphere. In the case that either object is not spherical or has an irregular distribution of mass (I'm looking at you, Phobos!), both objects can still be treated as point-masses but the center of mass rather than the geometric center must be used. Is this correct? | No. For example, the gravity of a cubical planet of uniform density, which can be computed analytically, is not directed towards its center (or any other single point). You can also imagine a dumbbell-shaped mass distribution where the two heavy ends are very far apart. If you drop an apple near one end it is going to fall toward that end, not toward the middle of the “neck”. | {
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547,512 | I believe I have a misunderstanding of some principles, but I have not, even through quite a bit of research, been able to understand this problem. My current understanding of transmission, reflection and absorption is as follows: transmission occurs when the energy of an incident photon does not correspond to any electron's energy transition within the material. Therefore, the photon does not interact with the atoms / electrons and is transmitted through. Absorption occurs when the incident photon's energy exactly equals that of an electron's energy transition. The photon is absorbed and excites an electron to a higher state. Reflection I feel like my understanding is flawed, since I have read multiple different views. I believe that a photon is absorbed by an atom, exciting an electron. The electron, however, almost immediately transitions back into a lower energy level, emitting a photon of identical wavelength. My question concerning reflection is: Why are some wavelengths absorbed and immediately re-emitted? I presume that it is because the electron is in a type of unstable state and therefore drops back to its previous energy level? Given a solid object that appears red to us (therefore reflects wavelengths somewhere between 625 and 740nm), how can it be possible that all other incident wavelengths are absorbed? They must be absorbed, since the only wavelength being reflected is in the "red" range, and I can clearly see there's no visible light being transmitted through the object. However, in my knowledge, the wavelengths can only be absorbed if they correspond to the energy transition of an electron, which is not the case for every wavelength in the visible spectrum. How is it then possible that they are absorbed?
Additionally, if the electron is excited to a higher level, does it just store the energy? Does it take thermal form?? I assume that perhaps I cannot simply apply these principles of absorption, that I was taught only in relation to a single atom, to a complex body consisting of billions of atoms. Could someone elaborate on this and explain my questions about absorption and reflection? Thanks very much! | Your misunderstanding is very common and quite easy to make. Basically, what students are usually introduced to first is the thermodynamics of ideal monoatomic gasses. This is good because it is simple and easy to understand, but can be problematic because features specific to the simple substance can be misunderstood as general features of all substances. In an ideal monoatomic gas light can interact either by scattering or by absorbing an amount of energy corresponding to an atomic transition*. Note, in the latter case the photon is not absorbed by the electron but by the atom as a whole because the atom has different internal states corresponding to the absorbed energy. As a result ideal monoatomic gasses tend to be transparent except at a few narrow** frequencies. Now, consider a molecular gas. Just like an atom has internal states that an electron does not, similarly a molecule has internal states that an atom does not. Some states correspond to electron transitions in the molecule, but others correspond to rotational or vibrational modes. The molecular electronic transitions combined with the molecular vibrational and rotational transitions gives rise to a multitude of absorption lines, often forming continuous absorption bands, so many times these are visibly not transparent. Now, consider a solid. Just like a molecule has states that an atom does not, similarly a solid has states that a molecule does not. The rotational and vibrational modes gain additional degrees of freedom and can act over fairly large groups of molecules (e.g. phonons). These states can have energy levels that are so closely spaced they form continuous bands, and are called energy bands. Any energy in the band will be easily absorbed. This makes most solids opaque as they absorb broad bands of radiation. Finally, when a photon is absorbed it may be re-emitted at the same wavelength to fall back to the original energy state. However, if there are other energy states available then the energy can be emitted and retained at different energy levels. For example, a UV photon could be absorbed and a visible photon could be emitted along with an increase in a rotational degree of freedom. *Even for an ideal monoatomic gas there are other less common mechanisms like ionization and deep inelastic scattering, but for clarity these are neglected here. **Note that even for an ideal monoatomic gas the frequency bands are not infinitely narrow but have some breadth. This is caused by two factors. First, the width of the peaks is fundamentally limited by the time-energy uncertainty relation which says that $2 \Delta T \ \Delta E \ge \hbar$ where $\Delta E$ is the width of the energy band and $\Delta T$ is the lifetime of the transition. Second, random thermal movement of the gas will cause Doppler and pressure broadening of the frequency band. | {
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548,031 | I'm trying to help my kids why their elementary-school physics. Their lesson today says that "work" is done only when a change in position is accomplished by the applied force. I have absolutely no background or training in physics whatsoever. So my question is if two people were to stand facing each other with a book (or another object) between them, and they both applied equal force to opposite sides of the book (each person attempting to push the book toward the other person) would that be "work" in the physics sense? Since each person is accomplishing a change in the book's movement (stopping it from moving) would that not be considered work? I guess my question involves two forces instead of one. Maybe that makes a difference? | No. No work would be done in this case, at least not at the macroscopic level. Work is the product of force and displacement in the direction of the force and in this case there is no displacement. I disagree with you that each person is "accomplishing a change in the book's movement". The book wasn't moving initially, or at the end, or at any time in between. The situation is exactly the same as if two people had been trying to push the book through a solid wall. Be careful when using the human body in questions where you are asking how much work has been done. If you actually try what you propose you will find that you will get tired. Your muscles are losing chemical potential energy but you are not doing work on the book. At the microscopic level your muscle fibers are contracting and slipping and contracting again so at that level work is being done. I find it more instructive to think of replacing the people in examples like these with some sort of simple mechanical device. You could lean something up against the book, use a clamp or set up some other simple mechanical system which would continue to apply a force but would require no ongoing energy input, which makes it clearer that no work is being done. | {
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548,186 | I know that on cutting a spring into n equal pieces, spring constant becomes n times. But I have no idea why this happens. Please clarify the reasons | Equations are nice, but if you're looking for a conceptual answer : | {
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548,519 | Let's say I have these equal size objects (for now thinking in 2D) on a flat surface. At the center of those objects I add equal positive angular torque (just enough to make the square tire to move forward). Of course the round tire will move faster forward and even accelerate (I guess). But how can I mathematicaly prove/measure how better the round tire will perform? This for my advanced simulator I'm working on and I don't want to just Hardcode that rounds rolls better, square worse, etc. I know the answer could be very complex, but I'm all yours. | Center of Mass The center of mass of a uniform sphere/disc is located at the center of that sphere/disc (this might sound trivial, but this is only true for the cases where mass distribution is spherically symmetric). The center of mass can be viewed as a collective representation of the whole body by a single point (do note that this isn't strictly true, but for our purposes, it will help in building some basic intuition). Similarly, the center of mass of a uniform cube/square plate is at the center of the cube/square plate. Gravitational Potential Energy The gravitational potential energy of an object is given by $$U=mg(h_{\text{COM}})\tag{1}$$ where $m$ is the mass of the body, $g$ is the gravitational acceleration and $h_{\text{COM}}$ is the height of the center of mass. In equation $(1)$ , we have assumed the potential energy to be $0$ at the ground level i.e. $h_{\text{COM}}=0$ . Now, to lift a body such that its center of mass moves form a height $h_1$ to a height $h_2$ , we need to do some work which is equal to the change in body's potential energy: $$W=\Delta U=mg(h_2-h_1)$$ Rolling Square As you can see in the GIF below, the rolling square has a kind of wobbly rotational motion. Wobbly in the sense that its center of mass goes up and comes down, goes up and comes down, and on an on. Animation source So as we calculated above, we need to do some work to raise the height of the square's center of mass (There's a specific angle, $45^{\circ}$ in this case, up till which you need to rotate the square if you want it to roll. If you rotate the square by an angle less than that, then the square would fall back). And once the center of mass reaches the maximum height, it then falls to the other side on its own and the kinetic energy gained by the square during the fall gets dissipated as sound and heat energy due to the inelastic nature of collision of the square with the ground. Now you would again have to raise the square's center of mass up to make it roll. This process continuously involves giving energy to raise the center of mass and then losing the energy because of the square falling back to the ground. And this makes it really difficult for a square to roll. Why does it undergo inelastic collisions? The square is prone to lose more energy due to inelastic collisions when compared to a circular disc because of a larger area of the surface which is in contact with the ground. This is similar to the case of a bicycle tyre. When the tyre is inflated, it is spherical and thus has a lower are in contact with the ground resulting in lower energy loses, whereas a deflated tyre has a larger area in contact with the ground which makes it more prone to inelastic collisions. Rolling Circle When a circle/sphere rolls, the height of the center of mass stays the same throughout the motion because of the symmetry of the shape. You can also see this in the GIF below. Animation source This means that none of the energy that we provide, gets wasted in changing the height of the center of mass. And all of the energy is utilised in speeding up the sphere/circle, which makes us feel easy to roll it faster. Why does its center of mass stay at the same height? For the sake of rigor, let's prove that a circle is the only 2D shape which has the property that its center of mass stays at the same height when it rolls. First let's assume that there exists another shape (not a circle) which also has this property. This implies that no matter how you put that shape on the ground (of course, we can't just lay it flat), the center of mass will always remain at a constant height. Which means the distance between the ground and the center of mass will always be the same. Which then implies that the distance between the boundary point touching the ground and the center of mass will always be the same. However this is true for all boundary points, since all boundary points can be made to touch the ground (again we are assuming a convex shape). This implies that all the boundary points are at the same distance from the center of mass. This means that the boundary points lie on a circle which is centered at the center of mass of the body. And thus the desired shape can be nothing else than a circular disc. Moment of Inertia Moment of inertia also has a role to play here. It can be shown that for a given constant area of any 2D shape, a circular disc would have the lowest moment of inertia (assuming all the shapes to be made from same materials/density). This means it would be a bit easier to roll a circular disc than any other 2D shape. A similar argument holds for 3D shapes, however, here we would keep the volume (the 3D analogue of area) constant while varying the shape. But here, theoretically, a cylinder with an infinitesimally small radius and infinitely large length will have the lowest moment of inertia. Addendum For special surfaces, you can even make a square rotate like a sphere. See the GIF below. Animation source As you can see, if we use a surface which is made up of inverted Catenary curves , we can even make a square roll. To see why this is true, you can check out the derivation here . Also, as this answer suggested, curves of constant width are also good candidates when it comes to rolling. So strictly speaking, circle isn't the only shape which can roll on a flat surface. However it is way better than a square when it comes to rolling. | {
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548,955 | If temperature is just the average kinetic energy of particles, why would moving air feel colder rather than warmer? | If the air was still, body heat warms a thin layer of air next to the skin. This warm air would stay near the skin, separating it from the cold air. Wind, however, continuously blows away this warm bit of air, replacing it with the colder surrounding air. There's a similar effect on humidity. Evaporating sweat increases the humidity right next to the skin, decreasing the rate of evaporation. Wind removes this humid air and replaces it with the less humid surrounding air. This is why a fan can cool a person down by blowing hot air at them. I've also heard stories from soldiers driving tanks in the desert that remaining still can make 120 $^\circ$ F (49 $^\circ$ C) days more tolerable. Their bodies create a layer of 98 $^\circ$ F (37 $^\circ$ C) air next to their skin. | {
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549,522 | The van't Hoff law for osmotic pressure $\Pi$ is $$\Pi V=nRT$$ which looks similar to the ideal gas law $$PV = nRT.$$ Why is this? Also, in biology textbooks, the van't Hoff law is usually instead written as $$\Pi=CRT =\frac{NC_m RT}M$$ where $C_m$ is the mass concentration, $N$ the number of ions, and $R$ the ideal gas constant. Why? | The law $PV = n RT$ gives the pressure $P$ of $n$ moles of ideal gas in volume $V$ . Meanwhile, the law $\Pi V = n R T$ describes the osmotic pressure $\Pi$ due to $n$ moles of solute in volume $V$ . These are qualitatively very different situations, but there's a simple fundamental reason that they end up looking the same. Both of these laws are derived under the idealized assumption that the ideal gas/solute molecules don't interact with each other at all. So the expressions for the entropy of the ideal gas/solute are the same, and since the pressure of a system can be derived from the entropy , both situations yield the same pressure. The reason that you see $\Pi V = n RT$ expressed in such different units in biology textbooks is simply because they're using the units that are most convenient for them. | {
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549,928 | If you were to play a flute, you would want to blow sideways across it, kind of like a bottle, like in Fig 1 . Why would this be better than blowing straight in to the hole, like in Fig 2 ? It seems like more air would go into the flute, which seems like a good thing, so why is it that you blow sideways? | How to create a sound? To create sound, you would first need to create some kind of a wave, some kind of an oscillation. In the case of a flute, an oscillating column of air works very well, because the flute (its body) resonates at particular frequencies of oscillation and thus you can hear a sound. What happens when you blow sideways? Now to create a wave, we need to setup an oscillating jet of air. This is what exactly happens when we blow sideways into the flute. Why? Here's how: Image source As you can see in the above image, when we blow a jet of air sideways, the blown air "collides" with the surface of the flute on the other side. You might expect that the air jet would get divided into two parts after striking the surface of the hole, but that's not what really happens $^{\phi}$ . Instead, the jet of air rapidly fluctuates between going all into the hole and going all away from the hole. This sets up a rapid vibration at the end of the flute. Once this oscillating air column is set up, the flute's body does the job of amplifying the sound by resonating with this oscillating air column at particular frequencies. Also, you might have to change your blowing pressure for playing different notes (which is expected) because the blowing pressure corresponds to the speed of air jet, which in turn corresponds to the frequency of the oscillation of air jet (See $[1]$ for further information). Also see this question on PSE which delves deeper into the relation between air jet speed and frequency. What if I blow straight into the hole? If you blow straight into the hole, you will not be able to setup any kind of air oscillations, except the fluctuations of your blown air. In this scenario, the blown air would simply enter from one end of the flute and exit from the other, producing nothing but some noise (due to the fluctuations of the air you blowed). $^{\phi}$ Why does the air just not simply get divided into two parts instead of oscillating? As the breath is directed toward the edge of the hole, high-pressure sound waves pass through the tube and reach openings such as the end of the foot joint and the sound holes. These waves then bounce back and try to force the air in the vicinity of the embouchure hole back out through the embouchure hole. As this happens, the sound pressure in this section of the instrument falls, and air is sucked back in. Waves are then produced that cause the air around the edge of the hole to vibrate up and down, producing changes in the sound. $^{[3]}$ Further reading/References $[1] :$ https://newt.phys.unsw.edu.au/jw/fluteacoustics.html $[2] :$ http://www.markshep.com/flute/Acoustics.html $[3] :$ Yamaha's guide on how flutes work and more $[4] :$ Working of a whistle $[5] :$ Working of brass instruments | {
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549,932 | So, this might sound rather childish, but I was walking a dark mountain road yesterday night and started thinking a bit. My phone's battery is used to light up it's torchlight, which then is used to create light that helps me see around. Things might be a bit confusing in my head, but that torchlight emits both heat energy and light. The heat energy will somehow dissipate in the air, but the light energy will go around me to lit up the objects, reflecting said light into my eyes, but then, where is that energy? Is light energy absorbed by objects? I always remember the "everything is transformed, nothing is lost" when talking about energy, so what about that light that goes reflecting in the guardrail? In what it is transformed? | How to create a sound? To create sound, you would first need to create some kind of a wave, some kind of an oscillation. In the case of a flute, an oscillating column of air works very well, because the flute (its body) resonates at particular frequencies of oscillation and thus you can hear a sound. What happens when you blow sideways? Now to create a wave, we need to setup an oscillating jet of air. This is what exactly happens when we blow sideways into the flute. Why? Here's how: Image source As you can see in the above image, when we blow a jet of air sideways, the blown air "collides" with the surface of the flute on the other side. You might expect that the air jet would get divided into two parts after striking the surface of the hole, but that's not what really happens $^{\phi}$ . Instead, the jet of air rapidly fluctuates between going all into the hole and going all away from the hole. This sets up a rapid vibration at the end of the flute. Once this oscillating air column is set up, the flute's body does the job of amplifying the sound by resonating with this oscillating air column at particular frequencies. Also, you might have to change your blowing pressure for playing different notes (which is expected) because the blowing pressure corresponds to the speed of air jet, which in turn corresponds to the frequency of the oscillation of air jet (See $[1]$ for further information). Also see this question on PSE which delves deeper into the relation between air jet speed and frequency. What if I blow straight into the hole? If you blow straight into the hole, you will not be able to setup any kind of air oscillations, except the fluctuations of your blown air. In this scenario, the blown air would simply enter from one end of the flute and exit from the other, producing nothing but some noise (due to the fluctuations of the air you blowed). $^{\phi}$ Why does the air just not simply get divided into two parts instead of oscillating? As the breath is directed toward the edge of the hole, high-pressure sound waves pass through the tube and reach openings such as the end of the foot joint and the sound holes. These waves then bounce back and try to force the air in the vicinity of the embouchure hole back out through the embouchure hole. As this happens, the sound pressure in this section of the instrument falls, and air is sucked back in. Waves are then produced that cause the air around the edge of the hole to vibrate up and down, producing changes in the sound. $^{[3]}$ Further reading/References $[1] :$ https://newt.phys.unsw.edu.au/jw/fluteacoustics.html $[2] :$ http://www.markshep.com/flute/Acoustics.html $[3] :$ Yamaha's guide on how flutes work and more $[4] :$ Working of a whistle $[5] :$ Working of brass instruments | {
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550,239 | Work done on a rotating body is equal to the change in its kinetic energy. When an electric fan rotates with a constant angular velocity, then its kinetic energy doesn't change. Does it mean that it doesn't consume electrical energy? | It definitely does consume electrical energy. Why? Because there's some opposing force faced by it while it rotates, and this force is often known as air drag/air resistance. You can see the effect of air drag once you switch off the fan. The fan decelerates from its original angular velocity until it stops completely. This deceleration is due to the motion opposing air drag. And thus while rotating, the fan continually loses it's kinetic energy (due to the air drag) and this lost energy is primarily converted to heat energy Thus you don't need electricity to change the kinetic energy, rather you need electrical energy to compensate for the energy lost due to the air drag acting on the fan . Also the air drag is the most common, most general and easy to understand among all the losses experienced by a fan. However there are many other factors which also increase the loss of energy in a fan. Here's an extremely nice flowchart/Sankey diagram showing this: Source (PDF) | {
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550,820 | If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\varepsilon_0$ , which is same as that if the charge $q$ was placed at the center of a spherical shell. The area vector for each infinitesimal area of the shell is parallel to the electric field vector, arising from the point charge, which makes the cosine of the dot product unity, which is understandable. But for the cube, the electric field vector is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electric field vector changes, i.e., they are no more parallel, still the flux remains the same? To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube. | Consider the flux through a tiny segment of a sphere. Since the electric field is parallel to the normal of the surface at all points, the flux is simply the electric field at that distance multiplied by the area of the element. Now imagine tilting the top of the cone by an angle $\theta$ so that the corners still lie on the conical section, as seen below: The area increases by a factor $\frac{1}{\cos\theta}$ , however the electric field vector in the normal direction $E_n$ is decreased by a factor of $\cos\theta$ . Therefore the flux through this surface is unchanged since flux is the product of the normal electric field component and the area. Now imagine splitting the cube up into lots of these conical sections. Clearly the tilting of the top surfaces of these sections due to the fact it being a cube rather than a sphere does not affect the flux flowing through each area element. Therefore the total flux flowing through the cube is the same as a sphere. Note that this was a simplified adaptation from a chapter of The Feynman Lectures on Physics which explains why the images do not quite match my explanations since I was just talking about the top surface of the conical section being tilted. Feynman explains the effect of the flux through a closed surface in a more complete way. | {
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550,846 | I was just watching some videos and came across beta+ radiation (when a positron is emitted). It then occurred to me, how can the following be true, given that a positron and an electron have the same mass: neutron = proton + electron [eq 1, beta- decay] proton = neutron + positron [eq 2, beta+ decay] As this would mean, a neutron = neutron + positron + electron (substitution of eq 2 in eq 1), which seems impossible? | A neutron does not have the same mass as a proton plus an electron, and a proton does not have the same mass as a neutron plus a positron. Proton mass = 938.272 MeV Neutron mass = 939.565 MeV Electron or positron mass = 0.511 MeV https://physics.nist.gov/cgi-bin/cuu/Value?mpc2mev|search_for=atomnuc ! So $m_p+m_e = 938.783 \text{ MeV} \ne m_n$ and $m_n + m_e = 940.075 \text{ MeV} \ne m_p$ Now, your unstated question may be based on thinking that because a neutron can decay into a proton and an electron (and an anti-neutrino) that must mean that it contains a proton and an electron (and an anti-neutrino). That is not the case. When a subatomic particle decays into a different particle the new particles are created, they are not merely separated out of the previous particle. The masses may not balance and the difference is the total KE of the products. | {
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551,191 | Newton's second law states: $$F(\vec{x})=m\vec{\ddot{x}}$$ For $\vec{x}$ scaled by some arbitrary constant $s$ , we obtain: $$F(s\vec{x})=ms\vec{\ddot{x}} \Longleftrightarrow \frac{F(s\vec{x})}{s}=m\vec{\ddot{x}}$$ Which is clearly just $F(\vec{x})$ ! Therefore: $$F(\vec{x})=\frac{F(s\vec{x})}{s} $$ for any $s$ , which can only be satisfied by a quadratic potential. Therefore, if Newton's second law is to hold and be consistent, all potentials in the universe are quadratic! Is there a very obvious mistake here, or is this inconsistency related to the fact the classical mechanics is not a complete description of nature? Because this discrepancy does not seem to arise if we use Ehrenfest's theorem in QM. | While other answers are correct, they fail to address your specific issue. It looks like you are treating Newton's second law like it defines a single function, when it does not. For example, in algebra if I say a function is $f(x)=x^2 + 3$ , then I can "plug into" this function something like $sx$ so that $f(sx)=(sx)^2+3$ by how we defined the function. This is not what Newton's second law is doing. $F(x)=m\ddot x$ is not a function saying "whatever I plug into the function $F$ I take its second derivative with respect to time and multiply it by $m$ ." So, your statement of $F(sx)=ms\ddot x$ is not correct. Newton's law is a differential equation , not a function definition. $F(x)$ is defined by the forces acting on our system, and Newton's second law then states that the acceleration is proportional to this force. | {
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551,336 | Suppose there are $N$ radioactive atoms and the half life of decay is $t$ . Then after one half life the number of remaining atoms will be $\frac{N}{2}$ . And so after each half life the number will be halved. Which means, $1/2$ of the atoms will have a life of $t$ Half of the the remaining half or $1/4$ of the atoms will have a life of $2t$ and so on. Then if the mean time for decay is $\tau$ , then it should be: $\tau = \frac{(\frac{N}{2}t+\frac{N}{4}2t+\frac{N}{8}3t+...)}{N}$ or $\tau = t(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...)$ But this infinite series doesn't equal to $\frac{1}{ln2}$ .
And we know that, $\tau =\frac{t}{ln2}$ So obviously my calculation is wrong. Why is this way of calculating the mean time for decay wrong? | Your mistake is here: Which means, 1/2 of the atoms will have a life of t Half of the the remaining half or 1/4
of the atoms will have a life of 2t and so on. The corrected statement is: Which means, 1/2 of the atoms will have a life $\le t$ Half of the the remaining half or 1/4
of the atoms will have a life between $t$ and $2t$ and so on. | {
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551,349 | In this picture(all others on google are the same), the positive plate is at the top and the negative plate is on the bottom. This means for this experiment to make any sense, the oil drops must be negatively charged. If this is the case, then why are X-rays used to cause the oil droplet to be charged? Surely since X-rays are ionising, they would remove electrons, making the oil droplets positively charged. | Your mistake is here: Which means, 1/2 of the atoms will have a life of t Half of the the remaining half or 1/4
of the atoms will have a life of 2t and so on. The corrected statement is: Which means, 1/2 of the atoms will have a life $\le t$ Half of the the remaining half or 1/4
of the atoms will have a life between $t$ and $2t$ and so on. | {
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551,563 | Given that they can reach terrifying energies and temperatures, why isn’t fusion of protons a concern? After all, they start with a plasma and ram protons into each other.
At some point the strong force will overcome the proton-proton electric repulsion, no? And as a corollary, can they re-purpose CERN to become a fusion reactor? | Proton-proton fusion happens at energies around 15 keV. The LHC currently operates at an energy of 13 TeV, which is literally one billion times larger. Fusion is one of the lowest-energy processes that could occur at the LHC, and most of the interesting reactions that are studied there go way beyond nuclear fusion. A proton is made up of fundamental, pointlike bits of matter called quarks, held together by force carriers called gluons. These quarks are bound together with a certain binding energy ; the quarks are also relatively light, so this binding energy actually makes up most of the mass of the proton. In other words, the binding energy of the proton is roughly equal to the mass of the proton. When you collide protons at an energy that's far below its mass, the proton acts as a single object. This is the regime in which nuclear physics typically occurs; all of the reactions that break nuclei apart, and the energy that holds them together, are typically small enough that it works to treat the proton as a single unit most of the time. This includes nuclear fusion; the mass of the proton is 938 MeV, and the amount of energy required for nuclear fusion (15 keV, as we said) is tens of thousands of times smaller than that. If we instead decide to collide protons at energies that are higher than a few times the proton mass, then there is now enough energy in the reaction for the quarks inside the protons to interact with each other directly. We can no longer treat protons as single objects, since the energy involved in the reaction is enough to expose its inner components. As the collision energy becomes higher and higher, the binding energy of the proton becomes less and less relevant, and the picture of two dense clouds of quarks and gluons interacting becomes more and more accurate. These reactions have enough energy to produce all manner of exotic matter, particles that you could never find in a nucleus, nor could you create them by nuclear fusion. The current energy of the LHC is roughly ten thousand times greater than the mass of the proton; at those energies, the proton's mass and binding energy are completely negligible, and the interesting reactions are entirely driven by individual quarks and gluons interacting with each other. The proton is merely a vehicle for us to deliver bits of fundamental matter to the collision site. If there were a way to just collide free quarks and free gluons, without all the mess generated by the proton, many particle physicists would jump at the chance. (Unfortunately, this turns out to be impossible.) There's certainly lots of extremely interesting physics to do at the nuclear-physics scale, and our understanding of the dynamics of nuclei is the subject of a great deal of active research in other experiments. But that's simply not what the proton-proton collisions at the LHC are meant to explore. The proton-proton collisions are actually better thought of as quark-quark collisions, or quark-gluon collisions, or gluon-gluon collisions. They're not meant to study protons, which is why the protons are often totally destroyed in the collision, transformed into exotic matter which then decays back into ordinary matter. Proton-proton collisions are meant to make conclusions about the interactions of fundamental particles, and that requires acceleration to extremely high energies, far above what you would need, or want, for nuclear fusion. So, given this, why aren't LHC physicists worried about triggering nuclear fusion? The answer is fairly straightforward: though each individual proton has an energy a billion times larger than the fusion threshold, the total amount of energy that is released into the surrounding area is still rather manageable, on a macroscopic scale. After all, 13 TeV is still only about a microjoule of energy, which is around a billion times less than the amount of energy that the Sun imparts to a square meter of Earth every second. That said, there are around 600 million collisions per second happening, so you definitely don't want to be standing anywhere near the interaction point. This is especially true since the individual particles of radiation released are much higher in average energy, meaning they're much nastier in terms of damage to life and inanimate objects than the radiation from the Sun. Because of this, the detector electronics have to be specially designed to deal with this extreme high-radiation environment; human access to the experimental hardware is also very tightly controlled, and is completely forbidden when the accelerator is running. But ultimately we're talking about around a few kilowatts, at most, of radiation released into the environment at each collision site. That's a pretty human-sized amount of power, and is roughly equivalent to the heating power of a large space heater (but, again, in a much more damaging form than the heat released by a space heater). This was by design - the collision rate at the LHC was chosen partly so that it would be feasible to build a detector that could withstand the influx of radiation. Nuclear explosions require many, many reactions all occurring at once, which is why they have such destructive power. The LHC collides at most a few individual protons at a time. So, given all this, the LHC would make a terrible fusion reactor. Its energy is much too high to actually reliably trigger nuclear fusion, and the energy released in collisions is miniscule compared to the energy required to keep the beams running, so it would be incredibly inefficient. | {
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551,876 | Kirchhoff's law states that the sum of voltages around any closed loop sum to zero. The law is true as the electric field is conservative in circuits. Why can we not apply the law here? Why doesn't the law hold here despite the fact that the electric field is conservative and the voltages should add up to $0$ ? | Just to complement the other answers: This isn't really about Kirchhoff's law. Rather, it is about an idealised situation that does not have a solution at all. When you draw such a diagram, you can think of it in two ways: As a sketch of a real circuit. Then the voltage source is, e.g. a battery or a power supply, and the line is a wire. You can connect them this way, and something will happen (possibly, something will break or catch fire). As an idealised circuit. Then the voltage source maintains a fixed (presumably nonzero) voltage $V$ between the poles and supplies whatever current is necessary. The wire has no resistance, inductance or capacitance -- it will carry any current and produce zero voltage drop. You immediately see that you cannot satisfy both conditions. Hence, this idealised circuit does not admit a solution. UPDATE To extend this a bit: You can approximate the behaviour of real devices with combinations of ideal circuit element. For a battery, a common way is a series conection of an ideal volatge source and a resistor ( see e.g. wikipedia ), and a real wire would be an ideal wire with, again, a resistor (and possibly inductance and capacitance, see wikipedia again ). So in your case, you would have to include two resistors: An internal resistance $R_\text{int}$ , which you can think of as part of the battery, and a wire resistance $R_\text{w}$ , which really is distributed along all of the real wire and not a localised element. The you wil have a current $$I=\frac{V}{R_\text{int}+R_\text{w}}\,$$ and an "external voltage", i.e. the voltage aong voltage source and internal resistance, of $$U_\text{ext}=V-I\cdot R_\text{int}=V\left(1-\frac{R_\text{int}}{R_\text{int}+R_\text{w}}\right)\,.$$ In the fully idealised case $R_\text{int}=R_\text{w}=0$ , these expressions are ill-defined. You can look at two posible limiting cases: "Superconducting wire": If $R_\text{w}=0$ but $R_\text{int}\neq0$ , i.e. superconducting ideal wire shorting a real battery, current is limited by internal resistance and external voltage is zero (and the battery will likely overheat). "Real wire on ideal battery": If, on the other hand, $R_\text{int}=0$ but $R_\text{w}\neq0$ , current is limited by the wire resistance, and the external voltage is just $V$ . | {
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551,992 | So to get a proton beam for the LHC, CERN prob has to make a plasma and siphon off the moving protons with a magnet. Are the electrons stored somewhere? How? I don’t mean to sound stupid but when they turn off the LHC, all those protons are going to be looking for their electrons. And that’s going to make a really big spark. | You're right that CERN gets its protons by ionizing matter and collecting them. But the number of electrons & protons CERN deals with is far smaller than you might think. They get about 600 million collisions a second at CERN. So call it 1.2 billion protons used per second. $1.2 \times 10^9$ . That'd be a large number in dollars, but it's not much in Coulombs. For comparison, a wire carrying a 30 amp current has about $2 \times 10^{19}$ electrons flowing through it every second. That's a factor of 10 billion. So there's not really any issue disposing of CERN's unneeded electrons. You probably make a bigger spark when you rub your feet on the carpet. If memory serves, the LHC has been running for its whole history off of a single canister of hydrogen gas. | {
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552,611 | Suppose an inductor is connected to a source and then the source is disconnected. The inductor will have energy stored in the form of magnetic field. But there is no way/path to ground to discharge this energy? What will happen to the stored energy, current and voltage of the inductor in this case? | Suppose an inductor is connected to a source and then the source is disconnected.
The inductor will have energy stored in the form of magnetic field.
But there is no way/path to discharge this energy? Short answer: It will find a way/path to discharge this energy. Longer answer: Let's have this simple electric circuit consisting of a battery (voltage $V_0$ ),
a switch, a resistor (resistance $R$ ), and an inductor (inductance $L$ ). (image from build electronic circuits - What is an inductor? ,
slightly modified by me) After closing the switch there will be soon a steady state,
with a current $I=\frac{V_0}{R}$ flowing.
The magnetic energy stored in the inductor is $E=\frac{1}{2}LI^2$ . When opening the switch you obviously interrupt the current $I$ suddenly.
The differential equation between voltage $V_L$ and current $I$ through the inductor is $$V_L=L\frac{dI}{dt}$$ or for a finite time step $$V_L=L\frac{\Delta I}{\Delta t}.$$ Now in our case $I$ changes from $\frac{V_0}{R}$ to $0$ ,
and therefore $\Delta I=-\frac{V_0}{R}$ .
And for an ideal switch it is $\Delta t=0$ .
So we expect the inductor producing a voltage $$V_L=L\frac{\Delta I}{\Delta t}=-L\frac{V_0/R}{0}=-\infty.$$ Can this be correct? Well, almost. When the voltage across the opening switch reaches several 1000 volts,
the air between the contacts of the switch gets ionized
and becomes an electrical conductor. According to
" Electrical breakdown - Gases " air begins to break down at 3000 V/mm.
You will actually see and hear a spark in the switch. (image from build electronic circuits - What is an inductor? ,
slightly modified by me) What will happen to the stored energy, current and voltage of the inductor in this case? For some milliseconds the current continues to flow across the already opened switch,
passing through the ionized air of the spark.
The energy stored in the inductor is dissipated in this spark. Summary: An inductor doesn't "want" the current to be interrupted and
therefore induces a voltage high enough to make the current continuing. Side note: In many electric engineering applications
this kind of inductive spark is a highly undesirable feature.
It can be avoided by adding a flyback diode to the circuit. However, in some applications (like electric ignition
in petrol engines) the inductive spark is the wanted feature. | {
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552,617 | I get confused understanding Constant or uniform Velocity and acceleration when I understand constant speed which means that same distance traveled at same time. So constant or uniform velocity means same distance and same time which so no change in velocity or no change in speed. Hence if a say 4 m/s constant or uniform velocity it means the object is moving 4 m per second. So there is no increase or decrease of velocity / speed - is this true for what we mean uniform velocity? Then again constant or uniform acceleration is same like constant or uniform velocity? Hence Constant or uniform Velocity and acceleration would mean same as Constant or uniform speed then? Is my understanding correct? | Suppose an inductor is connected to a source and then the source is disconnected.
The inductor will have energy stored in the form of magnetic field.
But there is no way/path to discharge this energy? Short answer: It will find a way/path to discharge this energy. Longer answer: Let's have this simple electric circuit consisting of a battery (voltage $V_0$ ),
a switch, a resistor (resistance $R$ ), and an inductor (inductance $L$ ). (image from build electronic circuits - What is an inductor? ,
slightly modified by me) After closing the switch there will be soon a steady state,
with a current $I=\frac{V_0}{R}$ flowing.
The magnetic energy stored in the inductor is $E=\frac{1}{2}LI^2$ . When opening the switch you obviously interrupt the current $I$ suddenly.
The differential equation between voltage $V_L$ and current $I$ through the inductor is $$V_L=L\frac{dI}{dt}$$ or for a finite time step $$V_L=L\frac{\Delta I}{\Delta t}.$$ Now in our case $I$ changes from $\frac{V_0}{R}$ to $0$ ,
and therefore $\Delta I=-\frac{V_0}{R}$ .
And for an ideal switch it is $\Delta t=0$ .
So we expect the inductor producing a voltage $$V_L=L\frac{\Delta I}{\Delta t}=-L\frac{V_0/R}{0}=-\infty.$$ Can this be correct? Well, almost. When the voltage across the opening switch reaches several 1000 volts,
the air between the contacts of the switch gets ionized
and becomes an electrical conductor. According to
" Electrical breakdown - Gases " air begins to break down at 3000 V/mm.
You will actually see and hear a spark in the switch. (image from build electronic circuits - What is an inductor? ,
slightly modified by me) What will happen to the stored energy, current and voltage of the inductor in this case? For some milliseconds the current continues to flow across the already opened switch,
passing through the ionized air of the spark.
The energy stored in the inductor is dissipated in this spark. Summary: An inductor doesn't "want" the current to be interrupted and
therefore induces a voltage high enough to make the current continuing. Side note: In many electric engineering applications
this kind of inductive spark is a highly undesirable feature.
It can be avoided by adding a flyback diode to the circuit. However, in some applications (like electric ignition
in petrol engines) the inductive spark is the wanted feature. | {
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552,796 | In introductory quantum mechanics I have always heard the mantra The phase of a wave function doesn't have physical meaning. So the states $| \psi \rangle$ and $\lambda|\psi \rangle$ with $|\lambda| = 1$ are physically equivalent and indiscernible. In fact by this motivation it is said that the state space of a physical system shouldn't be a Hilbert space, but rather a projective Hilbert space, where vectors which only differ up to a multiplicative constant of magnitude 1 are identified. But I also heard that one of the defining "feature" of quantum mechanics is the superposition principle: We can combine states $| \psi_1 \rangle, |\psi_2 \rangle$ to a new state $| \psi_1 \rangle + | \psi_2 \rangle$ . This should for example explain the constructive / destructive interference we see in the double slit. But if two states with the same phase are physically equivalent, so should the states $| \psi \rangle, -|\psi \rangle$ . But their sum is zero. I have seen experiments which exploit this and measure the relative phase difference between two different states. But if relative phase difference is measurable, then surely the phase of a wave function does have physical meaning? This should mean that we can identify the phases of all states of a quantum system up to a $U(1)$ transformation by gauging some state to have phase $1$ . Is this correct? How can this be solidified with the above mantra? I have asked a second question here ("The superposition principle in quantum mechanics") regarding the superposition principle which is closely related to this question. | When people say that the phase doesn't matter, they mean the overall , "global" phase. In other words, the state $|0 \rangle$ is equivalent to $e^{i \theta} |0 \rangle$ , the state $|1\rangle$ is equivalent to $e^{i \theta'} |1 \rangle$ , and the state $|0\rangle + |1 \rangle$ is equivalent to $e^{i \theta''} (|0 \rangle + |1 \rangle)$ . Note that "equivalence" is not preserved under addition, since $e^{i \theta} |0 \rangle + e^{i \theta'} |1 \rangle$ is not equivalent to $|0 \rangle + |1 \rangle$ , because there can be a relative phase $e^{i (\theta - \theta')}$ . If we wanted to describe this very simple fact with unnecessarily big words, we could say something like "the complex projective Hilbert space of rays, the set of equivalence classes of nonzero vectors in the Hilbert space under multiplication by complex phase, cannot be endowed with the structure of a vector space". Because the equivalence doesn't play nicely with addition, it's best to just ignore the global phase ambiguity whenever you're doing real calculations. Finally, when you're done with the entire calculation, and arrive at a state, you are free to multiply that final result by an overall phase. | {
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552,840 | So, when someone is red-green colorblind, the colors appear the same to them, like this: Source: https://iristech.co/what-do-colorblind-people-see/ And if you're totally colorblind, then things presumably just appear like they would in a black-and-white movie. However, this isn't how ultraviolet patterns seem to work. Compare how we see this flower to the version where ultraviolet is visible: Source: Dr Klaus Schmitt The UV pattern is completely invisible here. However, unlike with the red and green, this isn't because yellow and UV are colors that appear identical when you can't see UV. Look at these flowers: Source: https://blog.zoo.org/2012/01/ultra-awesome-ultraviolet-eyesight-in.html This time the flowers are purple, but the UV pattern is still invisible. Why is that? Shouldn't the UV pattern still be apparent on at least one of the flowers, just in a different color? And on some other flowers, the UV does appear as a different color. So: Why is the UV invisible only sometimes? Does it have to do with the flower using iridescent structures to produce color, instead of a pigment? Can this happen with red and green, as well? Image sources: https://iristech.co/what-do-colorblind-people-see/ | Color is a double valued variable.For physics there is a one to one correspondence between frequency of light and the color assigned to visible frequencies. As far as the spectrum of colors (rainbow) ultraviolet frequencies are invisible to our eye. The eye is a biological entity, the retina of the eye has color receptors, and these receptors do see the spectrum . BUT there is also a color perception , that the same color can be accepted by the brain although it has many different frequencies. Color blindness is due to this biological mechanism being misaligned . . Why is the UV invisible only sometimes? . Does it have to do with the flower using iridescent structures to produce color, instead of a pigment? Now ultraviolet frequency reflecting from materials as in the photos you show, may interact with them and give the perception of "seeing" ultraviolet, and that will depend on the material, which explains the differences in seeing an ultraviolet effect or not in the visible. Can this happen with red and green, as well? It might, i.e. the frequency scattering off a material may be degraded in energy and change the frequency( color) a bit. One would have to shine a fixed frequency red or green to see if there is an effect on the particular material. | {
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552,845 | I am confused that whether the positive terminal in physics refers just to a higher potential or a positive charge. This also comes from the fact that anode and cathode are differently charged in Galvanic and Electrolytic cells. But electricity always flows from Anode to Cathode. So is the anode just an indicator of higher electric potential in physics high school circuit diagrams? | Color is a double valued variable.For physics there is a one to one correspondence between frequency of light and the color assigned to visible frequencies. As far as the spectrum of colors (rainbow) ultraviolet frequencies are invisible to our eye. The eye is a biological entity, the retina of the eye has color receptors, and these receptors do see the spectrum . BUT there is also a color perception , that the same color can be accepted by the brain although it has many different frequencies. Color blindness is due to this biological mechanism being misaligned . . Why is the UV invisible only sometimes? . Does it have to do with the flower using iridescent structures to produce color, instead of a pigment? Now ultraviolet frequency reflecting from materials as in the photos you show, may interact with them and give the perception of "seeing" ultraviolet, and that will depend on the material, which explains the differences in seeing an ultraviolet effect or not in the visible. Can this happen with red and green, as well? It might, i.e. the frequency scattering off a material may be degraded in energy and change the frequency( color) a bit. One would have to shine a fixed frequency red or green to see if there is an effect on the particular material. | {
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553,072 | I am a student and due to school closures I am reading ahead in physics. I have been learning about relativity and I made up a problem for myself. You can see a diagram that I drew below: The diagram shows car X travelling at $0.7c$ with respect to car B and $0.5c$ with respect to car A. Car A is travelling at $0.2c $ with respect to car B, where $c$ is the speed of light. This means, using Lorentz transform, 1 year on car X is equivalent to: $$
\frac{1}{\sqrt{1-0.5^2}}=1.15
$$ 1.15 years on car A. And 1 year on car X is equivalent to: $$
\frac{1}{\sqrt{1-0.7^2}}=1.40
$$ 1.40 years on car B. This means that 1.15 years on car A should be the same as 1.40 years on car B. So if we were to use the Lorentz transform we should get the same answer but instead we get: $$
\frac{1.15}{\sqrt{1-0.2^2}}=1.17
$$ We get 1.17 years on car B is the same as 1.15 years on car A. I don't understand what I have done wrong. Could you help me why this problem arises? It is also worth noting that if I was to use a different method it would work. If I was to convert the time for car X into the time for car A then convert the time for car A into the time for car B, it would work. But I don't understand why this method works. In fact, I believe that method shouldn't work. This is because car X will be moving at 0.7c with respects to B still so 1 years on car X should be the same as 1.40 years on car B. Could you help me with this? | The contradiction is in the setup. It is impossible that those three relative velocities can simultaneously be true. You can't just add relative velocities together directly in special relativity. If an observer moving at velocity $v$ measures, in their own frame, a velocity $v'$ of some passing object, then a stationary observer would measure the velocity $$u=\frac{v+v'}{1+\frac{vv'}{c^2}}$$ If car A is moving at $0.2c$ with respect to car B, and car X is moving at $0.5c$ in the same direction with respect to car A, then car X must be moving at roughly $0.64c$ with respect to car B, not $0.7c$ . | {
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554,148 | Consider some known physical fact, e.g. $\nabla \cdot \mathbf B = 0$ for the magnetic induction $\mathbf B$ . Now, is it possible that a mathematical theorem exists, which yields a wrong prediction? E.g. a hypothetical - correctly proven - theorem that goes: "If $\nabla\cdot\mathbf B=0$ then some new planet should be between Earth and Mars." If this theorem was right, and after deep research we were sure that no such planet exists, one obvious possibility is that the previously known fact was incorrect i.e. perhaps $\nabla \cdot \mathbf B \neq 0$ under some strange conditions. But is this the only possibility? In other words, is it possible that both the premise and the theorem were right, but the mathematically obtained prediction is not true for physics? Please note that I picked just a silly example to make myself clear about a question regarding the relationship between mathematics and physics, but of course it is not this particular example that I am interested in. Also, I am not looking to discuss the existence of planets between Earth and Mars and, lastly, I am certainly not questioning the truth of Gauss' law. | Here is a mathematical theorem: the internal angles of a triangle add up to 180 degrees (i.e. half a complete rotation). To be a little more thorough, let's define a triangle: it is a closed figure consisting of three straight lines, and a straight line is the line of shortest distance between two points. Ok so we have a nice mathematical theorem. Now we go out into the world and start measuring triangles. They all have internal angles adding up to 180 degrees, to the precision of our instruments, so we are reassured. But then we get more precise instruments and larger triangles, and something happens: the angles are no longer adding up right! Oh no! What has happened? Is it a contradiction? Or perhaps our lines were not straight? We check that the lines were indeed of minimum distance. Eventually we go back to our mathematical theorem and realise that it had a hidden assumption. It was an assumption lying in a subtle way right at the heart of geometry and it turns out that it is an assumption that need not necessarily hold. One to do with parallel lines, called Euclid's fifth postulate. Then we discover a more general way of doing geometry and we can make sense of our measurements again---using the theory of general relativity and the geometry of curved spaces. So, to answer your question, what happens when physical observations contradict a mathematical statement has, up to now, always turned out to be like the above. What happens is that we find the mathematical statement was true in its own proper context, with the assumptions underling the concepts it was using, but that context is not the one that applies to the physical world. So, up till now at least, physics has never contradicted mathematics, but it has repeatedly shown that certain mathematical ideas which were thought to apply to the physical world in fact do not, or only do in a restricted sense or in some limiting case. | {
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554,241 | A train consists of an engine and $n$ trucks. It is travelling along a straight horizontal section of track. The mass of the engine and of each truck is $M$ . The resistance to motion of the engine and of each truck is $R$ , which is constant. The maximum power at which the engine can work is $P$ . The train starts from rest with the engine working at maximum power. Obtain an expression for the time $t$ taken to reach a given speed $v$ . I wrote $$a(t)=\frac{P}{v(t)M(n+1)}-\frac{R}{M} \tag{1}$$ Putting $(1)$ into standard differential form: $$[M(n+1)v]dv+[(Rv(n+1)-P)]dt=0 \tag{2}$$ Since $(2)$ is non-exact, let $$M(n+1)v=f(v) \tag{3}$$ and $$(Rv(n+1)-P)=g(v) \tag{4}$$ Since $$\frac{1}{g}\left(\frac{\partial f}{\partial t}-\frac{\partial g}{\partial v}\right)=\frac{R(n+1)}{P-Rv(n+1)}=h(v) \tag{5}$$ i.e. a function of $v$ only. The integrating factor to $(2)$ is then given by: $$I(v)=e^{\int h(v)dv}=e^{-\ln(P-Rv(n+1))}=\frac{1}{P-Rv(n+1)} \tag{6}$$ The final solution then looks something like this $$M(n+1)\left[\frac{-Rv(n+1)-P\ln(P-Rv(n+1))+P\ln(P)}{R^2(n+1)^2}\right]-t=0 \tag{7}$$ But doesn't the argument of $\ln()$ have to be some dimensionless quantities for it to make sense? (I got $\ln(P)$ and $P$ is not dimensionless in this case.) Can someone please explain where my conceptual errors lie? | In equation (7) you have the expression $$−P\ln(P−Rv(n+1))+P\ln(P)$$ But since this a difference between two logarithms
you can rewrite the expression
(remember $\ln a - \ln b = \ln \frac ab$ ) as $$P\ln\left(\frac{P}{P−Rv(n+1)}\right)$$ Now you have the logarithm of a dimension-less quantity,
as it should be. | {
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554,243 | In QFT the propagator is divergent for the on-shell momentum of the particles. When e.g. calculating the amplitude for a box loop, the propagator diverges for on-shell particles running in the box-loop. Is there a physical interpretation for this divergent behavior? | In equation (7) you have the expression $$−P\ln(P−Rv(n+1))+P\ln(P)$$ But since this a difference between two logarithms
you can rewrite the expression
(remember $\ln a - \ln b = \ln \frac ab$ ) as $$P\ln\left(\frac{P}{P−Rv(n+1)}\right)$$ Now you have the logarithm of a dimension-less quantity,
as it should be. | {
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554,246 | Current is the flow of charges (many say electrons). Do these charges flow just between the terminals of the source/battery (terminal to terminal) ? Or Do these charges flow through/inside the source/battery as well? What I know is that the positive terminal is the lack of electrons as compared to the negative terminal which is the excess of electron. Current flows when electrons flow from the negative terminal of the battery to the positive terminal of the battery/source? My question is what happens when the electrons reach from the negative terminal to the positive terminal? Are electrons then "absorbed" by the positive charges or they continue to flow through the source/battery to move in cycles? | In equation (7) you have the expression $$−P\ln(P−Rv(n+1))+P\ln(P)$$ But since this a difference between two logarithms
you can rewrite the expression
(remember $\ln a - \ln b = \ln \frac ab$ ) as $$P\ln\left(\frac{P}{P−Rv(n+1)}\right)$$ Now you have the logarithm of a dimension-less quantity,
as it should be. | {
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554,932 | Consider this situation: A ball is moving forward and undergoing rotation. Assume that it is not slipping. Eventually, the velocity and rate of rotation of the ball decrease, and it comes to a halt. But if you observe the direction of friction (when the ball is rotating clockwise), you will see that the friction should have provided a clockwise torque to the ball and the angular velocity of the body should have increased. But this doesn't happen. Why? | First let us clear up some definitions of the terms "reaction force", "normal force", and "frictional force". Whenever there is a contact between two bodies, there is a reaction force on each body at every point of contact. A reaction force can be split into a normal component (sometimes called the "normal contact force"), and a tangential component (sometimes called the "force due to friction"). The direction of the force due to friction - the tangential component - is such that it opposes relative motion due to sliding / slipping. This means that a perfectly rigid cylinder can roll on a perfectly rigid and rough surface forever since there is no sliding so no friction to provide a torque. So why do balls we observe normally slow down? The answer lies in something called "rolling resistance" (sometimes confusingly referred to as "rolling friction", or just "friction"), and entirely explains why a football comes to a stop after rolling it along the ground. The key is that footballs and dirt are both compressible - they are not rigid bodies. On contact, the weight of the ball deforms both it and the dirt. This means that there are many points of contact between the ball and the dirt. Due to our definition, there are now many normal forces - to be precise, one per point of contact. We will ignore the tangential frictional components for now. The deformation of the ball and the surface means that the line of action of these normal components is not through the center of the ball (see diagram). As a result, the ball experiences two torques: a counter-clockwise one from the normal components to the right of the centerline, and a clockwise torque from the normal components to the left. Since the normal forces are larger on the right side, the counter-clockwise torques are greater, and thus there is a net counter-clockwise torque and the ball slows to a stop. Notice how we did not even consider any tangential frictional components at all. Just due to the normal components the ball is slowed. There are a couple of points I have not covered, such as what role the frictional components do actually play and what happens in different types of deformations (e.g. the surface doesn't deform). Also you may be wondering why the normal forces on the right are greater. The answer to all these can be found at: https://lockhaven.edu/~dsimanek/scenario/rolling.htm . This is also where my diagram came from and I credit for these explanations. The exact diagram from your question is also used here and cited as a "naive pictures of friction and a rolling cylinder." | {
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554,934 | Consider a sphere of radius $R$ on a rough surface. Let it be rotating with angular velocity $\vec{\omega}$ and let it be moving with velocity $\vec{v}$ . Then what happens after a very long time? What is the magnitude and direction of both its velocity vector and angular velocity vector after a long time? EDIT: Let coefficient of static friction be $\mu$ . | First let us clear up some definitions of the terms "reaction force", "normal force", and "frictional force". Whenever there is a contact between two bodies, there is a reaction force on each body at every point of contact. A reaction force can be split into a normal component (sometimes called the "normal contact force"), and a tangential component (sometimes called the "force due to friction"). The direction of the force due to friction - the tangential component - is such that it opposes relative motion due to sliding / slipping. This means that a perfectly rigid cylinder can roll on a perfectly rigid and rough surface forever since there is no sliding so no friction to provide a torque. So why do balls we observe normally slow down? The answer lies in something called "rolling resistance" (sometimes confusingly referred to as "rolling friction", or just "friction"), and entirely explains why a football comes to a stop after rolling it along the ground. The key is that footballs and dirt are both compressible - they are not rigid bodies. On contact, the weight of the ball deforms both it and the dirt. This means that there are many points of contact between the ball and the dirt. Due to our definition, there are now many normal forces - to be precise, one per point of contact. We will ignore the tangential frictional components for now. The deformation of the ball and the surface means that the line of action of these normal components is not through the center of the ball (see diagram). As a result, the ball experiences two torques: a counter-clockwise one from the normal components to the right of the centerline, and a clockwise torque from the normal components to the left. Since the normal forces are larger on the right side, the counter-clockwise torques are greater, and thus there is a net counter-clockwise torque and the ball slows to a stop. Notice how we did not even consider any tangential frictional components at all. Just due to the normal components the ball is slowed. There are a couple of points I have not covered, such as what role the frictional components do actually play and what happens in different types of deformations (e.g. the surface doesn't deform). Also you may be wondering why the normal forces on the right are greater. The answer to all these can be found at: https://lockhaven.edu/~dsimanek/scenario/rolling.htm . This is also where my diagram came from and I credit for these explanations. The exact diagram from your question is also used here and cited as a "naive pictures of friction and a rolling cylinder." | {
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555,885 | My mom uses frozen sausages for a soup. She defrosts frozen sausages by submerging them in water (room temperature I believe). She claims this makes them defrost faster. She learned this from some article in a magazine and now swears by it. She is of no scientific background, so no accurate experiments have been made. Is there any physics why this might be true? Why would they defrost quicker in water? I cannot think of a reason why this would happen. Moreover, she sometimes keeps the sausages wrapped in aluminum foil to keep them from absorbing moisture. Would this surely slow down the melting because of this extra layer? | I do exactly the same. It is a very effective way of defrosting food fast. Compared to air water has a much higher heat capacity and a much higher thermal conductivity. That means heat flows from the water into the sausages much faster than it would in air and the water cools less than air would as it heats the sausages. Aluminium foil has a much, much higher thermal conductivity that either water or sausage meat. Provided the foil is tightly wrapped around the food, so there isn't a large air gap between the food and the foil, wrapping the sausages in aluminium foil will have almost no effect of the rate of thawing. Any small gaps are quickly filled with water as the frost on the surface of the food melts. | {
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555,958 | At a high-level Wikipedia states : "A convolution between two functions produces a third expressing how the shape of one is modified by the other." But there are clearly many ways of combining functions to get a third one. A convolution is a specific type of such combination; one that requires reversing and shifting one of the operands and that combines them with a product and an integral to generate the output. While not very complex algebraically, the operation itself is somewhat "convoluted" (pun intended). Why are convolutions noteworthy? What physical phenomena can be explained mathematically as a convolution? | Preamble The lesson here is that graphical intuition isn't always the best choice. For example, you can say that the intuition for derivatives is that they're slopes, and for integrals is that they're areas. But why would "slopes" be that useful in physics? I mean, besides inclined planes, you don't see that many literal slopes in a physics class. And why "areas"? We live in 3D space, so shouldn't volumes be more important? In case that all sounds dumb, the point is that sometimes, you can actually make something less physically intuitive by explaining it visually, because usually the visual explanation is completely devoid of the dynamic context that would be present in a real physics problem. One of the main reasons derivatives and integrals appear so often in introductory physics is that they're with respect to time, so the derivative means "a rate of change" and the integral means "an accumulation over time". This is a distinct intuition from the geometric one. The point of the geometric intuition is to help you see what the derivative and integral is given a graph , but it doesn't really help you interpret what it physically does. Similarly, there is a complicated geometric intuition for the convolution, which can hypothetically help you eyeball what the convolution of two graphs of functions would look like. But in this case the "dynamic" intuition is much simpler. One piece of intuition Convolutions occur whenever you have a two-stage process where the stages combine linearly and independently. Suppose that I kick an initially still mass on a spring at time $t = 0$ , and the subsequent trajectory of the spring is $x(t)$ . If I apply the same kick at time $t = 1$ , then by time translational invariance, the subsequent trajectory is $x(t-1)$ . Now suppose that I kick both at $t = 0$ and $t = 1$ , with strengths $f(0)$ and $f(1)$ . Then by linearity, the subsequent trajectory is $$f(0) x(t) + f(1) x(t-1).$$ This is like the "reversing and shifting" structure of a convolution. So more generally, if I apply a continuous force $f(t)$ , then the trajectory is $$\int dt' \, f(t') x(t-t')$$ which is precisely a convolution. This is where a large fraction of the convolutions in physics and electrical engineering come from. (A large fraction of the remainder come from the fact that the Fourier transform of the convolution is the product of the Fourier transforms, and products are simple and ubiquitous.) | {
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556,043 | Perhaps it goes without saying, but according to Newton’s laws “every action has an equal and opposite reaction”. How do astronauts, especially those inside small spacecraft like the Crew Dragon, not “push” the spacecraft when they bounce and push off walls? In orbit, where even button-sized ion thrusters push spacecraft, how does an astronaut pushing against a wall not cause it to move or spin? | When an astronaut bumps against the wall of the spacecraft, the spacecraft does gain whatever momentum the astronaut transfers to the wall. However, the astronaut loses momentum-or gains it in the opposite direction. The net result is that the center of mass of astronaut- plus-spacecraft does not move, and the combined momentum does not change. It is worthy of note that, similarly, the combined angular momentum does not change. However, the orientation of the axes of the combined system can change, so there is not a perfect analogy between translation and rotational motion. See " How cats land on their feet ". | {
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556,222 | I have several thermal mugs, two of them by the same brand, have the same look, shape and size except that one is in steel (inside and outside) and the other one, plastic. Both have an insulating layer of air or vacuum. I do not need a thermometer to tell me that the steel one is much better at keeping liquids cool/hot than the plastic one. With water near 90°C inside the mugs, I cannot feel any heat by placing my hand on the steel one (same as a thermos), while the plastic one feels almost like burning. However steels conducts heat much better than plastic (about one order of magnitude greater thermal conductivity). Therefore intuition tells me that a metallic mug should be less efficient than a plastic one to keep things cool/hot, but the reverse seems true! What am I missing? I am adding pictures of the two mugs. On the transparent one, we can see the insulating gap. I think it is an enclosed volume so there shouldn't be any air transfer between it and the surroundings. Here's the other picture: OK, so far the two answers given mention two different reasons. One is that there might be air between the 2 plastic surfaces in the transparent mug, and vacuum in the steel one. Air has a thermal conductivity of about $3\times 10^{-2}\,\frac{\mathrm{W}}{\mathrm{Km}}$ , so the thermal conductance should be about $3\times 10^{-2}\,\frac{\mathrm{W}}{\mathrm{Km}} \times \text{surface area} / (5\times10^{-3} \mathrm{m})$ for an air gap of $5\,\mathrm{mm}$ . So $6A\,\frac{\mathrm{W}}{\mathrm{Km^2}}$ . For radiation, $P=A\sigma\varepsilon(T^4-T_\text{room}^4) \approx A\varepsilon\, 462\,\mathrm{W}/\mathrm{m}^2$ . That steel is not polished so its emissivity is probably not as low as 0.1, but let's take it as zero for simplicity. And 1 for plastic, to get orders of magnitude. It looks like radiation might play the bigger role (the answer should run the numbers!) at high temperatures. But at lower temperature differences (when the liquid has partly cooled), conduction may play the bigger role. Thus, plugging numbers for a temperature difference of 80C, radiation losses are about 460 W/m^2 while conduction through the air about 480 W/m^2. In the comments, I am told that air convection should also transfer more heat, hence I conclude that even at "high" temperatures, the conduction through air might play the bigger role. At temperature differences below 60C, the role of radiation should lower faster than the one of conduction, because of the 4th power dependence on temperature. I note that I took the air's thermal conductivity at 1 atm. If the air has a lower pressure I guess the thermal conductivity would be lower and in that case radiation may play the bigger role at high temperatures. | Note that double-wall plastic cups aren't strong enough to withstand a vacuum and so they usually just contain a thin layer of air in which convection currents (absent in the case of the vacuum) can easily get started. This will bleed the heat out of the contents of the cup much faster than in the vacuum case. | {
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556,226 | A projectile moving through two spatial dimensions has a resolution of two velocities, $$\vec i(V_{0}\cos\theta)+\vec j(V_0\sin\theta -gt)$$ each of which should have their own air drag(?) But also orthogonal to each other, just as the velocoties are. Is it correct to sum the two air perpendicular air drags as: $$\sum F_{drag}= \vec i(bV_0\cos\theta)+\vec j(bV_0\sin\theta-gtb)$$ where $b$ is supposed to represent the product of the drag coefficient, fluid density, etc. (using the Stokes model of proportionality) seems like sensible reasoning (even if I do say so myself hehe) but i still unsure. googling could not reveal :( so I beseech thee. | Note that double-wall plastic cups aren't strong enough to withstand a vacuum and so they usually just contain a thin layer of air in which convection currents (absent in the case of the vacuum) can easily get started. This will bleed the heat out of the contents of the cup much faster than in the vacuum case. | {
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556,234 | I enjoy the occasional hot drink, but place it below a small fan in order to cool it to a drinkable temperature. Unfortunately, as expected, I commonly forget about my drink, and it ends up very cold. In fact, it ends up so cold that it feels much colder than I would expect given my relatively warm room ( $\sim \mathrm{25^\circ C}$ ). However, this could be an illusion caused by room temperature still being less than body temperature, or the mug being fairly cold. I’m wondering if the fan is able to cool the liquid below room temperature. I’m aware that evaporation works on the warmest molecules, and leaves the remaining liquid cooler, but I’m not sure that it can ever become cooler than the surrounding air temperature? | Yes, as other answers have stated, the temperature could drop below room temperature through evaporative cooling. In fact it could get as cold as the wet-bulb temperature of the air in the room. If you know the temperature and humidity of the air, you can figure out the wet-bulb temperature by using a psychrometric chart : Find your room temperature on the green "Dry Bulb Temperature" scale and the room relative humidity on the red "Relative Humidity" scale. Locate the point where they meet, and read off that position on the light blue "Wet Bulb or Saturation Temperature" scale. For example, if the room temperature is 25°C and the relative humidity is 30% (which you indicated), the lowest temperature your drink could reach by evaporative cooling would be roughly 14°C. If the air was completely dry however (relative humidity 0%), it could reach about 8°C. This does not necessarily mean that your drink will reach that temperature. It may require a large air flow, a large evaporation surface, and considerable time to actually reach the wet bulb temperature. | {
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556,528 | We all know that the asymmetry between matter and antimatter is a big puzzle in physics. But I don't know why one expects matter-antimatter symmetry in the first place? As in, is there a fundamental principle which suggests that matter and antimatter should have been produced in the same number at the beginning of the universe? | Well, suppose that you're in the office and you go to the coffee machine. You notice that there is an incredibly tiny puddle of coffee left in the pot. The pot is almost completely empty, but it's not totally empty either. Why is this? One theory is that it could be a complete coincidence -- the pot was going to have some coffee level or other in it, and that's just what it happened to be. Another theory is that your lazy coworker Steve didn't want to finish the coffee and have to make a new pot, so he left a tiny bit in so the next person would have to do it. The benefit of the second theory is that it explains two weird features of the amount of coffee, while the first doesn't explain anything. But the drawback is that you could waste time thinking about something that turned out to be a total coincidence. Trying to explain the matter-antimatter imbalance is a lot like this. It's a very small imbalance: there are billions of photons out there per baryon. But it's not zero, either. (If it were zero, you actually would end up with some regions with only baryons just by chance, but there would be much fewer than we observe.) Why is this? Again, one theory is to just refuse to answer the question: say that the number is what it is, and it doesn't make sense to ask where it comes from. This could definitely turn out to be the right answer! But you don't see scientists talking about this option that much, because there really isn't anything else to say. The people who believe this just decide to work on other things. Now suppose you decide you do want to try to explain the quantity of baryons, i.e. explain why it's (1) not zero, and (2) much less than the quantity of photons. You might wonder why the first aspect is always emphasized. The reason is that if you believe the theory of inflation, then any imbalance that existed before inflation gets rapidly diluted away due to the expansion of the universe. Thus, under typical assumptions, you begin after inflation ends with precisely no imbalance, so you need a way to build it up. | {
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557,113 | The equivalence principle as I understand it goes something like this: Let's suppose you're in a black box in the middle of nowhere in space, and we accelerate this black box in some direction. You'd feel a force just like you would if you were in a gravitational field which was causing the same acceleration.
In fact there is no experiment that you can do to say for sure if you are in a gravitational field or just accelerating in some direction in a black box. OK, so now let's imagine this elevator is being accelerated. Surely it can't accelerate at a constant rate forever, right? It's limited by the speed of light. So at some point or another the elevator will start to slow down. This sort of deceleration just does not happen in a gravitational field. Where did I go wrong? | The point of the thought experiment isn't to say that the elevator can accelerate forever. The point is that acceleration is indistinguishable from being in a gravitational field. The acceleration doesn't have to exist for all time. However, just to address another issue... It's limited by the speed of light. So at some point or another the elevator will start to slow down. This is incorrect. The speed of light limit doesn't say you will start to slow down once you get closer to the speed of light. The issue is that you are thinking in terms of absolute velocity, not relative velocity. The elevator cannot accelerate to move faster than the speed of light relative to something else . This doesn't affect the acceleration you would feel in the elevator though. If we had some magical infinite fuel source, then indeed you would feel the same acceleration in the elevator forever. The speed limit applies to an outside observer who would see your speed relative to them approach but not reach the speed of light. | {
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557,164 | When we talk about current, we say electrons are "flowing" through a conductor. But if electrons are identical particles, how does it make sense to talk about them flowing? To expand on that: imagine the simplest wire, just a 1-D chain of copper atoms, each with one conduction electron. If we apply a potetntial across the wire, what happens? Of course, we say there is a current, and the electrons "flow". But what does that really mean? Suppose when the electrons "flow", each copper atom gives its electron to the next atom in the line. From a QM perspective, nothing has changed! The 'before' wave function is identical to the 'after' wave function, because all that we have done is exchange particles, and the wavefunction has to be symmetric upon particle exchange. The state of the system before and after the "flow" occured is exactly the same. So what does it really mean to say that there is a current flowing? | Perhaps you're visualizing the electron flow as if it were a series of snapshots, timed so that the snapshots all look identical. But it's more than that. The wavefunction of a moving electron is different from that of a stationary electron: it includes a nonzero velocity-associated component. It's that added component (which is always there, even in the "snapshots" of electrons in a current-carrying wire) that equates to charge motion and thus to current. | {
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557,234 | Imagine I'm in an infinitely large vacuum and have a special apparatus built into my body that allows me to breath, eat, pee/poo, etc. and never age. The vacuum is similar to deep space and has no heat source or visible light and is therefore quite close to absolute zero. Perhaps we should say just above absolute zero to exclude some strange phenomena that might take place at absolute zero? Let's just say it is cold. I don't want this question to be about phenomena associated with deep space that might cause issues here (e.g., gamma waves or something). Could I survive off my body heat alone if I had a very very large sweater? What if the sweater was 10 or 10 million miles thick? Or thicker? If not, what if my sweater was pre-heated to some temperature, would it work then forever? | A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil. The body loses heat through a handful of mechanisms: During conduction , your body transfers heat to the surrounding air which is in contact with your skin. This raises the temperature of the air, which (if the air is still) decreases the rate of heat loss. If the air is moving, then that energy is carried away by the breeze, and you're in contact with fresh, cool air basically the entire time. This is convection . Evaporation occurs when moisture in your skin is pulled from the liquid into the gas phase, taking energy along with it. This depends on the relative humidity of the air - see wet bulb temperature for more. At all times, your body emits radiation (primarily in the infrared), with a total power loss given by $P=\epsilon A\sigma T^4$ . Here, $\epsilon$ is the emissivity of your body, ( $\epsilon\approx 0.95$ if
you are naked) $A$ is the "effective radiation area" of your body
( $A\approx 0.7 (2\text{ m}^2)\approx1.4\text{ m}^2$ ) $\sigma$ is the
Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8}
\frac{\text{W}}{\text{m}^2\text{ K}^4}$ $T$ is the absolute temperature of your body in Kelvin. (Note that your body emits radiation, but also receives it, with the amount depending on your particular radiation environment.) Of these four mechanisms, the first two are irrelevant to your question because you are in a vacuum. Evaporation will definitely occur, especially around your nose, mouth, and eyes, but I think that the primary mode of heat loss here will be radiation, so let's focus on that. Your body generates heat at all times via your metabolism as well as internal friction. If you are relaxing in comfortable conditions, you are producing roughly 100 W - but this number increases if you start to exercise. In particular, when your body gets cold your brain activates the shiver reflex, which can cause your body's power output to jump to 200-300 W. Source (Note that $1 \text{ Cal/hr} \approx 1 \text{ W}$ ). Ignoring for a moment the effect of clothing, then your equilibrium body temperature can be roughly estimated by equating the power generated by your metabolic processes (and possibly movement) with the power loss via radiation, assuming that you are not absorbing radiation from anywhere else. I am assuming that the body is at a uniform temperature here. This would not be the case - the core of your body would be warmest and then a gradient would form to your skin - but this can be neglected because the gradient would not be very extreme. In this simplified model, this is the resulting equilibrium body temperature as a function of emissivity, assuming first 100 W and then 300 W of generated power. As you can see, the situation is rather bleak if you're facing the void in the nude. Your core temperature can't drop much below its normal 37 C before you enter a hypothermic state; even shivering ferociously , this requires an emissivity of something like $0.425$ , far below your body's typical value of $0.95$ . This is where clothing comes in. Textiles have a somewhat lower emissivity than naked humans do. The surface emissivity of wool is about 0.74, and most textiles are in that range or higher, which means that the surface of the garment would still equilibriate below 0 C. However, the thermal conductivity of wool is only about $0.03\frac{\text{W}}{\text{m K}}$ . For a garment of thickness $t$ covering your entire body, the temperature gradient from your body's surface to the surface of the garment would be $$\frac{\Delta T}{t} = - \frac{100\text{ W}}{2\text{ m}^2 \cdot 0.03 \text{W/mK}} \approx 1670 \frac{\text{K}}{\text m}$$ Starting from the temperature of the garment's exterior, this allows us to track back and find the corresponding body temperature as a function of thickness. I've performed the calculation for wool and for cotton, with the results shown below. The surface of a wool sweater would equilibriate at approximately -5 C, which would correspond to a 37 C body temperature if the thickness of the sweater were only about 3 cm. That's thick, certainly, but not absurdly so. For a cotton sweater, which would have both higher emissivity and higher thermal conductivity, the surface would equilibriate around -10 C and you would need a thickness of closer to 6 cm to keep you warm. On the other hand, you could consider wrapping yourself in a layer of extremely low-emissivity material, and that would be much more effective. Polished silver, for instance, has an emissivity of only $0.02$ , which would be problematic in the wrong direction. To radiate 100 W/m $^2$ , our layer would need to have a surface temperature of about 60 C, which would roast us alive. The sweet spot - at which our body would equilibriate at 37 C - appears to correspond to an emissivity of approximately $0.15$ . Based on this table of emissivities , it seems that alumel (an alloy of nickel, aluminum, manganese, and silicon) would do the trick. Further Reading: Convective and radiative heat transfer coefficients for individual human body segments The Relative Influences of Radiation and Convection on the Temperature Regulation of the Clothed Body | {
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557,237 | Suppose there is a free spring (there is no mass attached to it ). Next , I apply a force " $f$ " to it and extend it's length to say x from mean position. Now, I hold the spring at that position. At length $x$ the spring force (on me by spring) $= -Fx$ (-ve because directed towards mean position ) and by Newton's $3$ rd law I apply a force $Fx$ on the spring. Now as soon as I leave the spring the spring accelerates and return to mean position. What's the cause of that acceleration ? Now due to absence of my force does spring apply a force on itself (as I thought a particle can't apply a force on itself). | A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil. The body loses heat through a handful of mechanisms: During conduction , your body transfers heat to the surrounding air which is in contact with your skin. This raises the temperature of the air, which (if the air is still) decreases the rate of heat loss. If the air is moving, then that energy is carried away by the breeze, and you're in contact with fresh, cool air basically the entire time. This is convection . Evaporation occurs when moisture in your skin is pulled from the liquid into the gas phase, taking energy along with it. This depends on the relative humidity of the air - see wet bulb temperature for more. At all times, your body emits radiation (primarily in the infrared), with a total power loss given by $P=\epsilon A\sigma T^4$ . Here, $\epsilon$ is the emissivity of your body, ( $\epsilon\approx 0.95$ if
you are naked) $A$ is the "effective radiation area" of your body
( $A\approx 0.7 (2\text{ m}^2)\approx1.4\text{ m}^2$ ) $\sigma$ is the
Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8}
\frac{\text{W}}{\text{m}^2\text{ K}^4}$ $T$ is the absolute temperature of your body in Kelvin. (Note that your body emits radiation, but also receives it, with the amount depending on your particular radiation environment.) Of these four mechanisms, the first two are irrelevant to your question because you are in a vacuum. Evaporation will definitely occur, especially around your nose, mouth, and eyes, but I think that the primary mode of heat loss here will be radiation, so let's focus on that. Your body generates heat at all times via your metabolism as well as internal friction. If you are relaxing in comfortable conditions, you are producing roughly 100 W - but this number increases if you start to exercise. In particular, when your body gets cold your brain activates the shiver reflex, which can cause your body's power output to jump to 200-300 W. Source (Note that $1 \text{ Cal/hr} \approx 1 \text{ W}$ ). Ignoring for a moment the effect of clothing, then your equilibrium body temperature can be roughly estimated by equating the power generated by your metabolic processes (and possibly movement) with the power loss via radiation, assuming that you are not absorbing radiation from anywhere else. I am assuming that the body is at a uniform temperature here. This would not be the case - the core of your body would be warmest and then a gradient would form to your skin - but this can be neglected because the gradient would not be very extreme. In this simplified model, this is the resulting equilibrium body temperature as a function of emissivity, assuming first 100 W and then 300 W of generated power. As you can see, the situation is rather bleak if you're facing the void in the nude. Your core temperature can't drop much below its normal 37 C before you enter a hypothermic state; even shivering ferociously , this requires an emissivity of something like $0.425$ , far below your body's typical value of $0.95$ . This is where clothing comes in. Textiles have a somewhat lower emissivity than naked humans do. The surface emissivity of wool is about 0.74, and most textiles are in that range or higher, which means that the surface of the garment would still equilibriate below 0 C. However, the thermal conductivity of wool is only about $0.03\frac{\text{W}}{\text{m K}}$ . For a garment of thickness $t$ covering your entire body, the temperature gradient from your body's surface to the surface of the garment would be $$\frac{\Delta T}{t} = - \frac{100\text{ W}}{2\text{ m}^2 \cdot 0.03 \text{W/mK}} \approx 1670 \frac{\text{K}}{\text m}$$ Starting from the temperature of the garment's exterior, this allows us to track back and find the corresponding body temperature as a function of thickness. I've performed the calculation for wool and for cotton, with the results shown below. The surface of a wool sweater would equilibriate at approximately -5 C, which would correspond to a 37 C body temperature if the thickness of the sweater were only about 3 cm. That's thick, certainly, but not absurdly so. For a cotton sweater, which would have both higher emissivity and higher thermal conductivity, the surface would equilibriate around -10 C and you would need a thickness of closer to 6 cm to keep you warm. On the other hand, you could consider wrapping yourself in a layer of extremely low-emissivity material, and that would be much more effective. Polished silver, for instance, has an emissivity of only $0.02$ , which would be problematic in the wrong direction. To radiate 100 W/m $^2$ , our layer would need to have a surface temperature of about 60 C, which would roast us alive. The sweet spot - at which our body would equilibriate at 37 C - appears to correspond to an emissivity of approximately $0.15$ . Based on this table of emissivities , it seems that alumel (an alloy of nickel, aluminum, manganese, and silicon) would do the trick. Further Reading: Convective and radiative heat transfer coefficients for individual human body segments The Relative Influences of Radiation and Convection on the Temperature Regulation of the Clothed Body | {
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557,240 | I have a naive model of action potential energy use and I’m unsure where the model is wrong. Clearly the model is wrong because its conclusion is wrong: When an action potential moves along an axon, it moves along the entire surface area (circumference) of the axon. Hence, in order to maintain the desired amount of voltage along the axon, to a first approximation, it requires an amount of energy proportional to the circumference of the axon, i.e. proportional to its diameter squared $d^2$ . Since this applies for each spike, the energy per second $E$ is proportional to the spike rate $R$ and energy per spike which is proportional to $d^2$ , so that $E\propto R\cdot d^2$ . In order to supply the sufficient amount of energy to allow for a certain spike rate over a longer timespan, there need to be an amount of mitochondria present in the axon proportional to the energy needs per second. Moreover, since the mitochondria per segment of the axon take space that is proportional to a first approximation to the circumference $d^2$ of the axon, we also have $E\propto d^2$ . However, clearly these equations are simultaneously possible only if $R$ is a constant: if $R$ rises above a certain level, the amount of energy needed to maintain an action potential explodes: we need more and more energy to maintain the action potential across the entire circumference of the axon, and we need more and more axon circumference to store all the mitochondria, and because these effects increase in the same amount, there can't ever be a mean firing rate above this threshold, and there is a unique energy-efficient firing rate. What is wrong about this? | A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil. The body loses heat through a handful of mechanisms: During conduction , your body transfers heat to the surrounding air which is in contact with your skin. This raises the temperature of the air, which (if the air is still) decreases the rate of heat loss. If the air is moving, then that energy is carried away by the breeze, and you're in contact with fresh, cool air basically the entire time. This is convection . Evaporation occurs when moisture in your skin is pulled from the liquid into the gas phase, taking energy along with it. This depends on the relative humidity of the air - see wet bulb temperature for more. At all times, your body emits radiation (primarily in the infrared), with a total power loss given by $P=\epsilon A\sigma T^4$ . Here, $\epsilon$ is the emissivity of your body, ( $\epsilon\approx 0.95$ if
you are naked) $A$ is the "effective radiation area" of your body
( $A\approx 0.7 (2\text{ m}^2)\approx1.4\text{ m}^2$ ) $\sigma$ is the
Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8}
\frac{\text{W}}{\text{m}^2\text{ K}^4}$ $T$ is the absolute temperature of your body in Kelvin. (Note that your body emits radiation, but also receives it, with the amount depending on your particular radiation environment.) Of these four mechanisms, the first two are irrelevant to your question because you are in a vacuum. Evaporation will definitely occur, especially around your nose, mouth, and eyes, but I think that the primary mode of heat loss here will be radiation, so let's focus on that. Your body generates heat at all times via your metabolism as well as internal friction. If you are relaxing in comfortable conditions, you are producing roughly 100 W - but this number increases if you start to exercise. In particular, when your body gets cold your brain activates the shiver reflex, which can cause your body's power output to jump to 200-300 W. Source (Note that $1 \text{ Cal/hr} \approx 1 \text{ W}$ ). Ignoring for a moment the effect of clothing, then your equilibrium body temperature can be roughly estimated by equating the power generated by your metabolic processes (and possibly movement) with the power loss via radiation, assuming that you are not absorbing radiation from anywhere else. I am assuming that the body is at a uniform temperature here. This would not be the case - the core of your body would be warmest and then a gradient would form to your skin - but this can be neglected because the gradient would not be very extreme. In this simplified model, this is the resulting equilibrium body temperature as a function of emissivity, assuming first 100 W and then 300 W of generated power. As you can see, the situation is rather bleak if you're facing the void in the nude. Your core temperature can't drop much below its normal 37 C before you enter a hypothermic state; even shivering ferociously , this requires an emissivity of something like $0.425$ , far below your body's typical value of $0.95$ . This is where clothing comes in. Textiles have a somewhat lower emissivity than naked humans do. The surface emissivity of wool is about 0.74, and most textiles are in that range or higher, which means that the surface of the garment would still equilibriate below 0 C. However, the thermal conductivity of wool is only about $0.03\frac{\text{W}}{\text{m K}}$ . For a garment of thickness $t$ covering your entire body, the temperature gradient from your body's surface to the surface of the garment would be $$\frac{\Delta T}{t} = - \frac{100\text{ W}}{2\text{ m}^2 \cdot 0.03 \text{W/mK}} \approx 1670 \frac{\text{K}}{\text m}$$ Starting from the temperature of the garment's exterior, this allows us to track back and find the corresponding body temperature as a function of thickness. I've performed the calculation for wool and for cotton, with the results shown below. The surface of a wool sweater would equilibriate at approximately -5 C, which would correspond to a 37 C body temperature if the thickness of the sweater were only about 3 cm. That's thick, certainly, but not absurdly so. For a cotton sweater, which would have both higher emissivity and higher thermal conductivity, the surface would equilibriate around -10 C and you would need a thickness of closer to 6 cm to keep you warm. On the other hand, you could consider wrapping yourself in a layer of extremely low-emissivity material, and that would be much more effective. Polished silver, for instance, has an emissivity of only $0.02$ , which would be problematic in the wrong direction. To radiate 100 W/m $^2$ , our layer would need to have a surface temperature of about 60 C, which would roast us alive. The sweet spot - at which our body would equilibriate at 37 C - appears to correspond to an emissivity of approximately $0.15$ . Based on this table of emissivities , it seems that alumel (an alloy of nickel, aluminum, manganese, and silicon) would do the trick. Further Reading: Convective and radiative heat transfer coefficients for individual human body segments The Relative Influences of Radiation and Convection on the Temperature Regulation of the Clothed Body | {
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557,256 | How are angular velocity, speed and radius related to each other? | A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil. The body loses heat through a handful of mechanisms: During conduction , your body transfers heat to the surrounding air which is in contact with your skin. This raises the temperature of the air, which (if the air is still) decreases the rate of heat loss. If the air is moving, then that energy is carried away by the breeze, and you're in contact with fresh, cool air basically the entire time. This is convection . Evaporation occurs when moisture in your skin is pulled from the liquid into the gas phase, taking energy along with it. This depends on the relative humidity of the air - see wet bulb temperature for more. At all times, your body emits radiation (primarily in the infrared), with a total power loss given by $P=\epsilon A\sigma T^4$ . Here, $\epsilon$ is the emissivity of your body, ( $\epsilon\approx 0.95$ if
you are naked) $A$ is the "effective radiation area" of your body
( $A\approx 0.7 (2\text{ m}^2)\approx1.4\text{ m}^2$ ) $\sigma$ is the
Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8}
\frac{\text{W}}{\text{m}^2\text{ K}^4}$ $T$ is the absolute temperature of your body in Kelvin. (Note that your body emits radiation, but also receives it, with the amount depending on your particular radiation environment.) Of these four mechanisms, the first two are irrelevant to your question because you are in a vacuum. Evaporation will definitely occur, especially around your nose, mouth, and eyes, but I think that the primary mode of heat loss here will be radiation, so let's focus on that. Your body generates heat at all times via your metabolism as well as internal friction. If you are relaxing in comfortable conditions, you are producing roughly 100 W - but this number increases if you start to exercise. In particular, when your body gets cold your brain activates the shiver reflex, which can cause your body's power output to jump to 200-300 W. Source (Note that $1 \text{ Cal/hr} \approx 1 \text{ W}$ ). Ignoring for a moment the effect of clothing, then your equilibrium body temperature can be roughly estimated by equating the power generated by your metabolic processes (and possibly movement) with the power loss via radiation, assuming that you are not absorbing radiation from anywhere else. I am assuming that the body is at a uniform temperature here. This would not be the case - the core of your body would be warmest and then a gradient would form to your skin - but this can be neglected because the gradient would not be very extreme. In this simplified model, this is the resulting equilibrium body temperature as a function of emissivity, assuming first 100 W and then 300 W of generated power. As you can see, the situation is rather bleak if you're facing the void in the nude. Your core temperature can't drop much below its normal 37 C before you enter a hypothermic state; even shivering ferociously , this requires an emissivity of something like $0.425$ , far below your body's typical value of $0.95$ . This is where clothing comes in. Textiles have a somewhat lower emissivity than naked humans do. The surface emissivity of wool is about 0.74, and most textiles are in that range or higher, which means that the surface of the garment would still equilibriate below 0 C. However, the thermal conductivity of wool is only about $0.03\frac{\text{W}}{\text{m K}}$ . For a garment of thickness $t$ covering your entire body, the temperature gradient from your body's surface to the surface of the garment would be $$\frac{\Delta T}{t} = - \frac{100\text{ W}}{2\text{ m}^2 \cdot 0.03 \text{W/mK}} \approx 1670 \frac{\text{K}}{\text m}$$ Starting from the temperature of the garment's exterior, this allows us to track back and find the corresponding body temperature as a function of thickness. I've performed the calculation for wool and for cotton, with the results shown below. The surface of a wool sweater would equilibriate at approximately -5 C, which would correspond to a 37 C body temperature if the thickness of the sweater were only about 3 cm. That's thick, certainly, but not absurdly so. For a cotton sweater, which would have both higher emissivity and higher thermal conductivity, the surface would equilibriate around -10 C and you would need a thickness of closer to 6 cm to keep you warm. On the other hand, you could consider wrapping yourself in a layer of extremely low-emissivity material, and that would be much more effective. Polished silver, for instance, has an emissivity of only $0.02$ , which would be problematic in the wrong direction. To radiate 100 W/m $^2$ , our layer would need to have a surface temperature of about 60 C, which would roast us alive. The sweet spot - at which our body would equilibriate at 37 C - appears to correspond to an emissivity of approximately $0.15$ . Based on this table of emissivities , it seems that alumel (an alloy of nickel, aluminum, manganese, and silicon) would do the trick. Further Reading: Convective and radiative heat transfer coefficients for individual human body segments The Relative Influences of Radiation and Convection on the Temperature Regulation of the Clothed Body | {
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557,265 | Imagine a Pulse of light traveling through space at $c$ , coming towards an observer on Earth, while at the same time, Space-time fabric (metric) is continuously changing (expanding), then why is the speed of light constant throughout space-time, since the separation of the very two-points in space, light is traveling in between, is not constant? My guess to that was $c=\lambda\nu$ but, how does Frequency of light and its Wavelength changes in just the right way so there product gives the speed of light and not a speed lesser than that of light(because space is Expanding)? A few articles also argued about its effect on Sommerfield's constant, but I've read that String theory allows Sommerfield's constant to change over time. I am not a GR head (yet) so, this post is bound to have a lot of things wrong (or maybe, all of them) so kindly keep your explanations as descriptive as possible. It'll be really helpful if you could provide some intuitions or examples for the same. | A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil. The body loses heat through a handful of mechanisms: During conduction , your body transfers heat to the surrounding air which is in contact with your skin. This raises the temperature of the air, which (if the air is still) decreases the rate of heat loss. If the air is moving, then that energy is carried away by the breeze, and you're in contact with fresh, cool air basically the entire time. This is convection . Evaporation occurs when moisture in your skin is pulled from the liquid into the gas phase, taking energy along with it. This depends on the relative humidity of the air - see wet bulb temperature for more. At all times, your body emits radiation (primarily in the infrared), with a total power loss given by $P=\epsilon A\sigma T^4$ . Here, $\epsilon$ is the emissivity of your body, ( $\epsilon\approx 0.95$ if
you are naked) $A$ is the "effective radiation area" of your body
( $A\approx 0.7 (2\text{ m}^2)\approx1.4\text{ m}^2$ ) $\sigma$ is the
Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8}
\frac{\text{W}}{\text{m}^2\text{ K}^4}$ $T$ is the absolute temperature of your body in Kelvin. (Note that your body emits radiation, but also receives it, with the amount depending on your particular radiation environment.) Of these four mechanisms, the first two are irrelevant to your question because you are in a vacuum. Evaporation will definitely occur, especially around your nose, mouth, and eyes, but I think that the primary mode of heat loss here will be radiation, so let's focus on that. Your body generates heat at all times via your metabolism as well as internal friction. If you are relaxing in comfortable conditions, you are producing roughly 100 W - but this number increases if you start to exercise. In particular, when your body gets cold your brain activates the shiver reflex, which can cause your body's power output to jump to 200-300 W. Source (Note that $1 \text{ Cal/hr} \approx 1 \text{ W}$ ). Ignoring for a moment the effect of clothing, then your equilibrium body temperature can be roughly estimated by equating the power generated by your metabolic processes (and possibly movement) with the power loss via radiation, assuming that you are not absorbing radiation from anywhere else. I am assuming that the body is at a uniform temperature here. This would not be the case - the core of your body would be warmest and then a gradient would form to your skin - but this can be neglected because the gradient would not be very extreme. In this simplified model, this is the resulting equilibrium body temperature as a function of emissivity, assuming first 100 W and then 300 W of generated power. As you can see, the situation is rather bleak if you're facing the void in the nude. Your core temperature can't drop much below its normal 37 C before you enter a hypothermic state; even shivering ferociously , this requires an emissivity of something like $0.425$ , far below your body's typical value of $0.95$ . This is where clothing comes in. Textiles have a somewhat lower emissivity than naked humans do. The surface emissivity of wool is about 0.74, and most textiles are in that range or higher, which means that the surface of the garment would still equilibriate below 0 C. However, the thermal conductivity of wool is only about $0.03\frac{\text{W}}{\text{m K}}$ . For a garment of thickness $t$ covering your entire body, the temperature gradient from your body's surface to the surface of the garment would be $$\frac{\Delta T}{t} = - \frac{100\text{ W}}{2\text{ m}^2 \cdot 0.03 \text{W/mK}} \approx 1670 \frac{\text{K}}{\text m}$$ Starting from the temperature of the garment's exterior, this allows us to track back and find the corresponding body temperature as a function of thickness. I've performed the calculation for wool and for cotton, with the results shown below. The surface of a wool sweater would equilibriate at approximately -5 C, which would correspond to a 37 C body temperature if the thickness of the sweater were only about 3 cm. That's thick, certainly, but not absurdly so. For a cotton sweater, which would have both higher emissivity and higher thermal conductivity, the surface would equilibriate around -10 C and you would need a thickness of closer to 6 cm to keep you warm. On the other hand, you could consider wrapping yourself in a layer of extremely low-emissivity material, and that would be much more effective. Polished silver, for instance, has an emissivity of only $0.02$ , which would be problematic in the wrong direction. To radiate 100 W/m $^2$ , our layer would need to have a surface temperature of about 60 C, which would roast us alive. The sweet spot - at which our body would equilibriate at 37 C - appears to correspond to an emissivity of approximately $0.15$ . Based on this table of emissivities , it seems that alumel (an alloy of nickel, aluminum, manganese, and silicon) would do the trick. Further Reading: Convective and radiative heat transfer coefficients for individual human body segments The Relative Influences of Radiation and Convection on the Temperature Regulation of the Clothed Body | {
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557,812 | I put a pot of water in the oven at $\mathrm{500^\circ F}$ ( $\mathrm{260^\circ C}$ , $\mathrm{533 K}$ ). Over time most of the water evaporated away but it never boiled. Why doesn't it boil? | The "roiling boil" is a mechanism for moving heat from the bottom of the pot to the top. You see it on the stovetop because most of the heat generally enters the liquid from a superheated surface below the pot. But in a convection oven, whether the heat enters from above, from below, or from both equally depends on how much material you are cooking and the thermal conductivity of its container. I had an argument about this fifteen years ago which I settled with a great kitchen experiment. I put equal amounts of water in a black cast-iron skillet and a glass baking dish with similar horizontal areas, and put them in the same oven. (Glass is a pretty good thermal insulator; the relative thermal conductivities and heat capacities of aluminum, stainless steel, and cast iron surprise me whenever I look them up.) After some time, the water in the iron skillet was boiling like gangbusters, but the water in the glass was totally still. A slight tilt of the glass dish, so that the water touched a dry surface, was met with a vigorous sizzle: the water was keeping the glass temperature below the boiling point where there was contact, but couldn't do the same for the iron. When I pulled the two pans out of the oven, the glass pan was missing about half as much water as the iron skillet. I interpreted this to mean that boiling had taken place from the top surface only of the glass pan, but from both the top and bottom surfaces of the iron skillet. Note that it is totally possible to get a bubbling boil from an insulating glass dish in a hot oven; the bubbles are how you know when the lasagna is ready. (A commenter reminds me that I used the "broiler" element at the top of the oven rather than the "baking" element at the bottom of the oven, to increase the degree to which the heat came "from above." That's probably why I chose black cast iron, was to capture more of the radiant heat.) | {
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559,289 | Say there is a liquid which behaves like an incompressible fluid and is flowing steadily through a pipe which is moving from a cross section of area $A_1$ to the cross section of area $A_2$ , where $A_2$ is less than $A_1$ . As per the continuity equation, $v_2>v_1$ and so the liquid seems to be accelerating. What force is causing this acceleration? | You are right.
From continuity of the incompressible fluid you have $$A_1 v_1 = A_2 v_2.$$ So obviously the velocity is changing.
Thus the fluid is accelerated, and therefore there must be a force causing this acceleration.
In this case the force comes from the pressure difference
between the wide and the narrow part of the pipe. (image from ResearchGate - Diagram of the Bernoulli principle ) This can be described by Bernoulli's equation ( $p$ is pressure, $\rho$ is density) $$\frac{1}{2}\rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$ | {
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559,298 | In the NS, a well known expression of the convective term is $$\bf v \times (\nabla\times \bf v) = \bf v\cdot \nabla v - \frac{1}{2}\nabla v^2 $$ In order to derive it I use the commute rule of the vectors cross products: $$\bf a\times (b\times c) = (a\cdot c)b - (a\cdot b)c$$ hence I got $$\bf v \times (\nabla\times \bf v) = \bf v\cdot \nabla v - v^2 \nabla$$ This then puzzled me that, can you explain why $$v^2 \nabla = \frac{1}{2}\nabla v^2 $$ | You are right.
From continuity of the incompressible fluid you have $$A_1 v_1 = A_2 v_2.$$ So obviously the velocity is changing.
Thus the fluid is accelerated, and therefore there must be a force causing this acceleration.
In this case the force comes from the pressure difference
between the wide and the narrow part of the pipe. (image from ResearchGate - Diagram of the Bernoulli principle ) This can be described by Bernoulli's equation ( $p$ is pressure, $\rho$ is density) $$\frac{1}{2}\rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$ | {
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559,741 | In discussions of sun spots and auroras on Earth, magnetic field lines are often described as "snapping" or "breaking", with the result of releasing charged particles very energetically. My understanding is that field lines are just a visualization tool. I don't understand, intuitively, how a field line could snap or break, or why that would result in a release of energy. I'm having trouble even framing this question because the concept of a field line breaking just doesn't make sense to me. What is happening when a magnetic field "snaps"? | Consider the following bar magnet, with the unphysical field lines drawn around it. The real magnetic vector field is tangent to these lines and is represented by black triangles (a magnetic vector field always emanates from the north pole to end up at the south pole, though it continues inside the magnet): Now consider the following picture of two of equivalent magnetic bars and the associated field lines (in which for every field line only one direction of the magnetic vector field vector is shown by a very small triangle): Field lines are always closed lines. This is easy to see in the single magnet (the lines continue inside the magnet). All field lines between the two magnets are connected (via the lines inside the magnets) with the lines on the far- left and far-right directed away from the magnets (which makes them closed, though that's hard to visualize). Now when we pull the magnets away from each other (to form two separate bar magnets), the field lines between the magnets (which are actually not separate, but you can't draw an infinity of field lines) move away from each other too, like the lines on the left and right of the double bar magnet arrangement. The field lines on the far left bend up (forming closed lines with the right ones bending up, which makes their closed Nature visible, like the single lines already bending inward are closed) to connect with the field lines on the left of the field lines in the middle (wrt to a vertical line in the middle of the two magnets). So these lines in the middle seem to "snap", just as the closed lines emerging from the left and entering on the right after which they reconnect to form two closed loops in each magnet. The reversed process, i.e. two closed lines forming one closed line (which is also a form of snapping), occurs, as you might have guessed when bringing two bar magnets together, in the same arrangement as depicted, to form one bar magnet. Because we pull the magnets apart the potential energy contained in the magnetic fields of two bar magnets is bigger than in a single one (if the two bar magnets were made from a single one by cutting it in two). You can imagine pulling them away from each other, and "snap!", two magnets with higher energy will emerge (actually the energy increases by infinite snaps in a continuous way, but separating them very fast will feel like a single snap). The magnetic fields around the bar magnets are produced by the spins of unpaired electrons in the outer shell of the atoms. Each spin produces a tiny magnetic field and in ferromagnets (which are the ones we consider here), if the temperature is not too high, all these tiny fields are permanently aligned, which minimizes the internal energy of the ferromagnet. Now, these kinds of processes (in very distorted ways and on much bigger scales) also take place on the surface of the Sun, but the (closed) magnetic vector fields are produced by huge plasma currents and the magnetic field lines are closed lines around these plasma currents. These plasma currents constantly change and thus the magnetic field lines. This induces electric fields, which accelerate charged particles, mainly protons, electrons, and a relatively small fraction helium nuclei (solar cosmic rays). When two or more closed field lines emerge from one closed field line (for example when one plasma current splits in two or more), the induced electric field becomes suddenly higher and this sudden increase of the induced electric field gives a burst of high energy protons, electrons, and helium (alpha particle). So just as in the case of two magnets that are separated very fast, thereby increasing the magnetic field energy in a snap, the sudden increase in the magnetic field's energy is converted into a burst of cosmic radiation, which reduces the energy contained in the magnetic fields around the two (or more) emerging plasma currents (because of which the plasma currents are reduced in strength as a reaction). The difference with the case of the two magnets is that the increased energy in the magnetic field of the two magnets stays (approximately) the same, without imparting the increased energy to other stuff. You can compare it with the lines of equal pressure in the weather developing. These lines are always closed too and they can merge or split to form new closed lines of equal pressure. The associated energies are contained in the winds. When one closed low-pressure line "snaps" into two closed lines, more wind energy will be released than in the case of the one closed low-pressure line. | {
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559,744 | I have been told that few models in statistical mechanics can be solved exactly. In general, is this because the solutions are too difficult to obtain, or is our mathematics not sufficiently advanced and we don't know how to solve many of those models yet, or because an exact solution genuinely does not exist, i.e. it can be proven that a model does not admit an exact solution? | Exact (non-)solvability is an issue that pops up in every area of physics. The fact that this is surprising is, I believe, a failure of the didactics of mathematics and science. Why? Consider the following: You solve a simple physical problem and the answer is $\sqrt{2}$ meters. So what is the answer? How many meters? Have you solved the problem? If I do not give you a calculator or allow you to use the internet, you will probably not be able to actually give me a very precise answer because " $\sqrt{2}$ " does not refer to an exact number we are able to magically evaluate in our mind. It refers to a computation procedure by which we are able to get the number with high precision, we could use e.g. the iterative Babylonian method $$a_0=1, a_{n+1} = \frac{a_n}{2} + \frac{1}{a_n}$$ after three iterations ( $n=3$ ) you get an approximation valid to six significant digits. But have you solved the problem exactly? No, you have not. Will you ever solve it exactly? No, you will not. Does it matter? No, it does not, since you can solve the problem extremely quickly to a degree that is way more precise than any possible application of the model itself will be. So when people refer to exact solvability they really mean "expressible as a closed-form reference to a standard core of functions with well-known properties and quickly-converging computational approximations". This "standard core" includes rational functions, fractional powers, exponentials, logarithms, sines, cosines,... Much of them can either be understood as natural extensions of integer addition, division and multiplication (rational functions), to solutions of simple geometrical problems (sines, cosines), and solutions of particular parametrized limits/simple differential equations (exponential). But there are other functions known as special functions such as elliptic integrals and Bessel functions that are sometimes understood as part of the "standard core" and sometimes not. If I express the solution of a problem as an exponential, it is an exact solution, but if it is an elliptic integral , it is not? Why is the reference to the circle and certain lengths within it (sine, cosine) more important than those within the ellipse (elliptic integral)? When you dig deeper, you find out that the notion of exact solvability is largely conventional, and trying to formalize it will typically either exclude or include many systems that either are or are not considered solvable. So you can understand your question as "Why are most problems in physics not possible to express as solutions of a rather arbitrarily chosen set of simple geometrical problems?" And the reason is because, well, there is no reason why to believe they should be. EDIT There have been quite a few criticisms of the original answer, so I would like to clarify. Ultimately, this is a soft question where there is no rigorous and definite answer (you would lie to yourself if you were to pretend that there was), and every answer will be open to controversy (and that is fine). The examples I gave were not supposed to be a definite judgment on the topic but rather serve the purpose of challenging the concept of "exact solvability". Since I demonstrated that the concept is conventional, I wanted to finish on that point and not go into gnarly details. But perhaps I can address some of the issues raised in the comments. The issue with $\sqrt{2}$ : I take the perspective of a physicist. If you give me a prediction that a phenomenon will have the answer $\sqrt{2}$ meters, then I take a measurement device and check that prediction. I check the prediction with a ruler, tape meter, or a laser ranging device. Of course, you can contruct $\sqrt{2}$ meters by the diagonal of a square with the tape meter, but if your measurement device is precise enough (such as the laser rangefinder), I can guarantee you that the approximate decimal representation is ultimately the better choice. The $\sqrt{2}$ factor can also be replaced by any constant outside the class of straightedge and compass constructions such as $\pi$ or $e$ to make the argument clearer. The point is that once you think about "door to door", "end to end" approaches, not abstract mathematical notions and "magical black boxes" such as calculators, the practical difference between "exactly solvable" and "accessible even though not exactly solvable" is not qualitative, it is quantitative. Understood vs. understandable vs. accessible vs. exactly solvable : I would like to stress that while there are significant overlaps between "understood", "understandable", "accessible", and "exactly solvable", these are certainly not the same. For instance, I would argue that the trajectories corresponding to a smooth Hamiltonian on a system with a low number of degrees of freedom are "accessible" and "fully understandable", at least if the functions in the Hamiltonian and the corresponding equations of motion do not take long to evaluate. On the other hand, such trajectories will rarely be "exactly solvable" in any sense we usually consider. And yes, the understandability and accessibility applies even when the Hamiltonian system in question is chaotic (actually, weakly chaotic Hamiltonians are "almost everywhere" in terms of measure on functional space). The reason is that the shadowing theorem guarantees us that we are recovering some trajectory of the system by numerical integration and by fine sampling of the phase space we are able to recover all the scenarios it can undergo. Again, from the perspective of the physicist you are able to understand where the system is unstable and what is the time-scale of the divergence of scenarios given your uncertainty about the initial data. The only difference between this and an exactly solvable system with an unstable manifold in phase space is that in the (weakly) chaotic system the instability plagues a non-zero volume in phase space and the instability of a chaotic orbit is a persistent property throughout its evolution. But consider Liouville-integrable Hamiltonians , which one would usually put in the bin "exactly solvable". Now let me construct a Hamiltonian such that it is integrable but its trajectories become "quite inaccessible" and certainly "not globally understandable" at some point. Consider the set $\{p_i\}_{i=1}^{N+3}$ of the first $N+3$ prime numbers ordered by size. Now consider the set of functions $\xi_i(x)$ defined by the recursive relation $$\xi_1(x) = F(p_3/p_2,p_2/p_1,p_1;x), \; \xi_{i+1} = F(p_{i+3}/p_{i+2},p_{i+2}/p_{i+1},p_{i+1}/p_i;\xi_i)$$ where $F(,,;)$ is the hypergeometric function . Now consider the Hamiltonian with $N$ degrees of freedom $$H_N(p_i,x^i) = \frac{1}{\sum_{i=1}^N \xi_i(x^i) } \left(\sum_{i} p_i^2 + \sum_{i=1}^N \xi_i(x^i)^2 \right)$$ For all $N$ the Hamiltonian can be considered exactly solvable (it has a parametrized separable solution), but there is a certain $N$ where the solution for a generic trajectory $x^i(t)$ becomes practically inaccessible and even ill-understood (my guess this point would be $N\sim500$ ). On the other hand, if this problem was very important, we would probably develop tools for better access to its solutions and for a better understanding. This is also why this question should be considered to be soft, what is considered undestandable and accessible is also a question of what the scientific community has been considering a priority. The case of the Ising model and statistical physics: The description of the OP asks about the solvability of models in statistical physics. The Ising model in 3 dimensions is one of the famous "unsolvable/unsolved" models in that field, which was mentioned by Kai in the comments and which also has an entire heritage question here at Physics SE. Amongst the answers I really like the statement of Ron Maimon: "The only precise meaning I can see to the statement that a
statistical model is solvable is saying that the computation of the
correlation functions can be reduced in complexity from doing a full
Monte-Carlo simulation." This being said, the 3D Ising model can be considered "partially solved", since conformal bootstrap methods provide a less computationally demanding (and thus ultimately more precise) method of computing its critical exponents. But as I demonstrated in the paragraphs above, "understandability" and "accessibility" do not need to be strictly related with "exact solvability". The point about the 3D Ising model is that the generic line of attack for a numerical solution of the problem (direct Monte Carlo simulation) is largely inaccessible AND the computational problem cannot be greatly reduced by "exact solvability". This also brings us to an interesting realization: exact solvability, in its most generous definition, is simply "the ability to reduce the computation problem considerably as compared to the generic case". In that sense it is tautological that exact solvability is non-generic. However, we should also ask why are certain classes of problems "generically computationally inaccessible" so that a lack of solvability becomes a huge issue. We do not know the full answer, but a part of it definitely has to do with the number of degrees of freedom. More complex problems are harder to model. Systems with more degrees of freedom allow for a higher degree of complexity. Why should we assume that as the number of degrees of freedom becomes large the computation of certain statistical properties of the system becomes simple? Of course, the answer is that we should not assume that the large- $N$ limit will become simple, we should instead assume that problems with computational complexity of large- $N$ systems will be generic and simplifications special. | {
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559,895 | In electrostatics we say that charge is quantised. Then my question is how can we integrate them? | You may also wonder why we use the concept of density of a material despite that material is made of molecules, or atoms, or in general quantized entities. This is the basis of hydrodynamics and solid mechanics (we use differential equations and integrals even if there are atoms). Electrostatics is just another branch of the " physics of continuous media ". Second, it is also worth to remember that we discovered that the charge is quantized after we developed electrostatics (the Millikan oil-drop experiment has been made in 1909, the Coulomb law dates 1785): in fact,
in our macroscopic world almost everything can be described as a continuum. In classical electrodynamics you do not need to know that charge is quantized, simply because you do not see the effects of quantization (for the same reason why you do not see the single grains of sand when looking at a beach from a certain distance). The basic concept is that of using a "fluid element", namely a small piece of matter containing many "atoms" that is big with respect to the microscopic length scale (so that its average properties like "density" are well defined) but that can be considered a point from the macroscopic perspective. Clearly, the "microscopic length scale" is the average (typical) distance between the discrete entities that compose your fluid element. In general all the classical mechanics of continuous media is made in this way. After all, also car traffic in cities can be described (to some extent) in terms of differential equations and integrals, despite cars are quantized (see e.g. " traffic flow "). | {
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560,067 | We know quantum mechanics gives a random result when we observe a particle that's in a superposition, but why is it random? One of the explanations I've heard is that because light comes with those discrete energy packets called photons, when a photon is passing through a polarized filter, it must either all pass through or all be blocked. You can't let a fraction of the photon pass through while others are blocked. Is it correct? It seems reasonable, but I couldn't find any proper source about this statement. | If it helps, it's not that the nature of the universe is random, it's that we model it as random in Quantum Mechanics. There are many cases in science where we cannot model the actual behavior of a system, due to all sorts of effects like measurement errors or chaotic behaviors. However, in many cases, we don't need to care about exactly how a system behaves. We only need to worry about the statistical behavior of the system. Consider this. We are going to roll a die. If it lands 1, 2, or 3, I give you \$1. If it lands 4, 5, or 6, you give me \$1. It is theoretically very difficult for you to predict whether any one roll is going to result in you giving me \$1 or me giving you \$1. However, if we roll this die 100 times, we can start to talk about expectations. We can start to talk about whether this die is a fair die, or if I have a weighted die. We can model the behavior of this die using statistics. We can do this until it becomes useful to know more. There are famous stories of people making money on roulette using computers to predict where the ball is expected to stop. We take some of the randomness out of the model, replacing it with knowledge about the system. Quantum Mechanics asserts that the fundamental behavior of the world is random, and we back that up with statistical studies showing that it's impossible to distinguish the behavior of the universe from random. That's not to say the universe is random. There may be some hidden logic to it all and we find that it was deterministic after all. However, after decades of experimentation, we're quite confident in a whole slew of ways the universe can't be deterministic. We've put together experiment after experiment, like the quantum eraser, for which nobody has been able to predict the behavior of the experiment better than the randomness of QM. Indeed, the ways the universe can be deterministic are so extraordinary that we choose to believe the universe cannot be that fantastic. For example, there's plenty of ways for the universe to be deterministic as long as some specific information can travel instantaneously (faster than light). As we have not observed any way to transfer information faster than light in a normal sense, we are hesitant to accept these deterministic descriptions of quantum behavior (like the Pilot Wave interpretation). And in the end, this is all science ever does. It can never tell us that something is truly random. It can never tell us what something truly is. What it tells us is that the observed behaviors of the system can be indistinguishable from those of the scientific models, and many of those models have random variables in them. | {
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560,714 | My teacher told me that an air bubble has 2 surfaces and a liquid drop has only 1 which is the reason for the air bubble having twice the pressure difference as a liquid drop with same surface tension. But I couldn't get it how an air bubble has 2 surfaces isn't it just 1 spherical surface ? Please help | Think about what a microscopic observer would experience as they moved from somewhere outside the bubble to its centre. With an air bubble they go from air (outside) to water (the "shell" of the bubble) to air again (centre). So they pass through two surfaces. With a water drop they go from air (outside) to water (all the way from the surface of the drop to its centre). So they only pass through one surface. | {
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560,722 | With respect to both gravity and electromagnetism, to the best of my understanding potential energy is added or subtracted from a system based on the distance between two objects such as charged or massive objects. Would this then imply that when oppositely charged objects or two massive objects are brought together within a small enough radius that the drop in potential energy would be greater than the combined mass of the two objects making the net mass/energy of the system negative? | Yes, potential energy can be negative: consider Newton’s law of gravitation $$V = -\frac{GMm}{r}$$ Where $G$ is Newton’s constant, $M$ and $m$ are masses, and $r$ is the distance between them. It can clearly be seen that this is always negative. The key thing is that the absolute value of potential energy is not observable; there is no measurement that can determine it. The only thing that can be measured is differences in potential energy. So actually there is a redundancy in the equation above: if I add any constant to it, the difference in potential energy for two given separations is the same. The common form of Newton’s law of gravitation is set by the convention that two objects an infinite distance apart have zero gravitational potential energy, but this is purely a convention. The idea of redundancies in physical descriptions is very important in theoretical physics, and is known as gauge invariance. EDIT: following some comments by the original poster, I've added some more to this answer to explain the effect on total energy of a system of attracting objects at very short distances. Let's consider two equal point masses $M$ separated by some distance $r$ : the total energy of the system, using the above definition of potential energy, is $$E = 2Mc^2 - \frac{GM^2}{r}.$$ If the total energy is negative, $E < 0$ . We can rearrange this inequality to give a condition on the radius for negative total energy: $$r < \frac{GM}{2c^2}.$$ Compare this to the Schwarschild radius $r_\mathrm{s} = 2GM/c^2$ . The distance at which the Newtonian energy becmes negative is less than the Schwarzschild radius---if two point masses were this close they would be a black hole. In reality we should be using GR to describe this system; the negative energy is a symptom of the breakdown of our theory. One can do the same calculation with two opposite charges $\pm e$ and find $$r < \frac{e^2}{8 \pi M c^2 \varepsilon_0}.$$ We can then compare this to the classical electron radius $r_\mathrm{e}$ and similiarly find that $r < r_\mathrm{e}$ for a negative total energy. The classical electron radius is the scale at which quantum fluctuations must be taken into account, so again the negative energy is a symptom of the breakdown of the theory. | {
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560,853 | I am new to the physics category of the Stack Exchange site. I apologize if my question is wrong, too broad, simple, or worded incorrectly. I am just trying to figure out what is true and false when it comes to electricity and its vast world. I want to have the right resources to learn from, however, I have come across many things that say electricity is the flow of electrons; and then there are people that contradict this statement. For example, this guy said, "First we must realize that "electricity" does not exist. There is no
single thing named "electricity." We must accept the fact that, while
several different things do exist inside wires, people wrongly call
all of them by a single name." I looked at some more of his information on his website, here are a few links: http://amasci.com/miscon/whatis.html http://amasci.com/amateur/elecdir.html http://amasci.com/miscon/eleca.html I am not asking anyone by any means to read everything, if someone could just browse it briefly - a few sentences or something, to let me know if these articles align with the truth and facts of electricity - then I would really like to study the articles and learn | The idea that electricity "does not exist" is just verbal sophistry along the same lines as "matter does not exist, it is frozen energy" or, "you do not exist, you are a figment of your own imagination". At best these are all just over-dramatic and misleading ways of saying that what these things actually are is not what you probably think they are. At worst, misguided eccentrics create "straw" definitions of such well-known words just so they can burn them and trump them with their own untenable notions. This last, sadly, is what is happening with the pages you link to. Although basically sound at an experimental and phenomenological level (and that has to be clearly acknowledged), he argues for his own wacky definitions of words. The guy claims that the scientific definition "means only one thing: quantities of electricity are measured in Coulombs". In fact it is electric charge which is measured in coulombs, not "electricity" per se ; his claim is a classic example of giving a straw definition so that he can debunk it. He concludes that "Because there are *two* things flowing, we cannot call them both by the name 'electricity.'" This is typical eccentric pedantry; of course we can. We can simply say that the phenomenon of electricity comprises electric charges and electric fields. And we do. For example "static electricity" is a buildup of charge, which creates an electric field capable of making your hair stand up or making molecules of glue stick together. Electricity as found in domestic wiring has three main properties; its flow is measured in (electron) charges per second or Amps, its "pressure" in Volts and the power (flow of energy) carried by the charged electrons in Joules per second or Watts. All these are aspects of electricity. There are many others. All the sophistry in the world will not change these facts, only what we choose to call such things. | {
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560,860 | Due to quantum randomicity, it's impossible to determine the future knowing the present. But is it possible to determine the past knowing the present? As far as I understand, it is impossible because of symmetry of quantum laws regarding time. But I am not sure. | The idea that electricity "does not exist" is just verbal sophistry along the same lines as "matter does not exist, it is frozen energy" or, "you do not exist, you are a figment of your own imagination". At best these are all just over-dramatic and misleading ways of saying that what these things actually are is not what you probably think they are. At worst, misguided eccentrics create "straw" definitions of such well-known words just so they can burn them and trump them with their own untenable notions. This last, sadly, is what is happening with the pages you link to. Although basically sound at an experimental and phenomenological level (and that has to be clearly acknowledged), he argues for his own wacky definitions of words. The guy claims that the scientific definition "means only one thing: quantities of electricity are measured in Coulombs". In fact it is electric charge which is measured in coulombs, not "electricity" per se ; his claim is a classic example of giving a straw definition so that he can debunk it. He concludes that "Because there are *two* things flowing, we cannot call them both by the name 'electricity.'" This is typical eccentric pedantry; of course we can. We can simply say that the phenomenon of electricity comprises electric charges and electric fields. And we do. For example "static electricity" is a buildup of charge, which creates an electric field capable of making your hair stand up or making molecules of glue stick together. Electricity as found in domestic wiring has three main properties; its flow is measured in (electron) charges per second or Amps, its "pressure" in Volts and the power (flow of energy) carried by the charged electrons in Joules per second or Watts. All these are aspects of electricity. There are many others. All the sophistry in the world will not change these facts, only what we choose to call such things. | {
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561,101 | I've read a lot of answers to the questions why the sky is blue. However all the answers I found contain mostly qualitative analysis: Rayleigh scattering is changing the direction of blue light, so there is more blue light coming to the eye along the line of sight than the red one. However these explanations raise additional questions. First of all, the scheme of only single scattering seems to be an oversimplification: the light direction should be changed more than once. Can we prove that this is negligible by calculation, or is it not negligible? Does this change the analysis? Further, the explanation says nothing about the exact amount of the blue light being scattered when looking into a particular direction. Assuming the sun is in zenith, it follows from symmetry that the color of the sky in the directions having the same zenith angle must be the same, but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith — so is it possible to derive theoretically a formula which would predict the color of sky given the azimuth angle and the position of the Sun (at least in a simple geometrical setup when the Sun is in zenith)? It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position: after all, the atmosphere is thicker along the lines going closer to horizon! The sky seems to be more uniformly blue than the typical explanation suggests. Further, it follows from the usual explanation that blue light is partially reflected back into the space. Due to this, about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light, which seems to contradict the observable reality. Does it? I'm mainly interested in quantitative analysis, not the observations or qualitative considerations. I've read the answers to this question and know that the physiology of the eye comes additionally into play, but let's neglect this for the sake of simplicity. | First of all, the scheme of only single scattering seems to be an oversimplification: the light direction should be changed more than once. Can we prove that this is negligible by calculation, or is it not negligible? This is an oversimplification, but for a clear sky at daytime it's not too wrong. See the following comparison of an atmosphere model computed with only single scattering and that including 4 orders of scattering (basically, 4 direction switches per light ray). The projection here is equirectangular, so you can see all the directions in one picture. This becomes a much more problematic simplification when the sun is under the horizon, particularly noticeable under the belt of Venus , where the Earth's shadow is located: Assuming the sun is in zenith, it follows from symmetry that the color of the sky in the directions having the same zenith angle must be the same, but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith — so is it possible to derive theoretically a formula which would predict the color of sky given the azimuth angle and the position of the Sun (at least in a simple geometrical setup when the Sun is in zenith)? If we neglect non-uniformity of the atmosphere with latitude and longitude, this scenario will lead to the colors independent of azimuth. It's not quite clear what you mean by "position of the Sun" though, if you already put it into zenith. Also, if by "derive theoretically a formula" you mean some closed-form expression, then it's unlikely, given that the atmosphere is not a simple distribution of gases and aerosols. But it's possible to calculate the colors numerically, and the above pictures demonstrate this calculation done by my (work in progress) software, CalcMySky . It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position: after all, the atmosphere is thicker along the lines going closer to horizon! It shouldn't be bluer at the horizon than at the zenith. After all, you have relatively small thickness near zenith, which makes most of the light scattered to you not too extincted due to Beer-Lambert law , while near the horizon the thickness is much larger, and the light scattered into the observer, in addition to becoming bluer due to Rayleigh scattering depending on wavelength, becomes also redder due to extinction along this long path. The combination of this bluing and reddening effects gives a color closer to white (which you can see in the daytime simulation above), or reddish-orange (in the twilight). Further, it follows from the usual explanation that blue light is partially reflected back into the space. Due to this, about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light, which seems to contradict the observable reality. Yes, the Earth indeed looks bluish from space, so the total radiation incoming from above should be redder at the ground level than at the top of atmosphere. But this is modified by the ozone layer, without which we'd have sandy color of twilight instead of blue. See for details the question Why is there a “blue hour” after the “golden hour”? | {
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561,102 | This fantastic question essentially asks what is the noise floor of air? Both the answer given on that thread and the value stated by Microsoft are around -23 or -24 dBSPL. However, overall loudness is only one metric. What does the amplitude of the noise in dBSPL look like when graphed out as a function of frequency in the audible range? How does the shape and level of the curve defined by that graph compare to the threshold of human hearing as described by the equal loudness contour ? | First of all, the scheme of only single scattering seems to be an oversimplification: the light direction should be changed more than once. Can we prove that this is negligible by calculation, or is it not negligible? This is an oversimplification, but for a clear sky at daytime it's not too wrong. See the following comparison of an atmosphere model computed with only single scattering and that including 4 orders of scattering (basically, 4 direction switches per light ray). The projection here is equirectangular, so you can see all the directions in one picture. This becomes a much more problematic simplification when the sun is under the horizon, particularly noticeable under the belt of Venus , where the Earth's shadow is located: Assuming the sun is in zenith, it follows from symmetry that the color of the sky in the directions having the same zenith angle must be the same, but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith — so is it possible to derive theoretically a formula which would predict the color of sky given the azimuth angle and the position of the Sun (at least in a simple geometrical setup when the Sun is in zenith)? If we neglect non-uniformity of the atmosphere with latitude and longitude, this scenario will lead to the colors independent of azimuth. It's not quite clear what you mean by "position of the Sun" though, if you already put it into zenith. Also, if by "derive theoretically a formula" you mean some closed-form expression, then it's unlikely, given that the atmosphere is not a simple distribution of gases and aerosols. But it's possible to calculate the colors numerically, and the above pictures demonstrate this calculation done by my (work in progress) software, CalcMySky . It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position: after all, the atmosphere is thicker along the lines going closer to horizon! It shouldn't be bluer at the horizon than at the zenith. After all, you have relatively small thickness near zenith, which makes most of the light scattered to you not too extincted due to Beer-Lambert law , while near the horizon the thickness is much larger, and the light scattered into the observer, in addition to becoming bluer due to Rayleigh scattering depending on wavelength, becomes also redder due to extinction along this long path. The combination of this bluing and reddening effects gives a color closer to white (which you can see in the daytime simulation above), or reddish-orange (in the twilight). Further, it follows from the usual explanation that blue light is partially reflected back into the space. Due to this, about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light, which seems to contradict the observable reality. Yes, the Earth indeed looks bluish from space, so the total radiation incoming from above should be redder at the ground level than at the top of atmosphere. But this is modified by the ozone layer, without which we'd have sandy color of twilight instead of blue. See for details the question Why is there a “blue hour” after the “golden hour”? | {
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561,192 | So I'm aware of this and this , but the question is Are Hard X-rays and Gamma-rays the same thing? If not, then what would be the key difference between them. Moreover, How much would the properties of each type differ from each other? I'd appreciate a simple answer without much technicalities. | It can be a little confusing because there are two conventions. The modern convention is to distinguish x-rays from gamma rays by how they are produced. X-rays are produced by electron energy transitions, typically inner orbital transitions, whereas gamma rays are produced by electromagnetic transitions in the nucleus. Usually, gamma rays have shorter wavelength (and therefore higher frequency and energy) than x-rays, but not always. Some radioactive processes release gamma rays with frequencies in the ultraviolet portion of the electromagnetic spectrum. However, there is an older convention which distinguishes them by energy. This convention is still common in astronomy and astrophysics. From Wikipedia : In astrophysics, gamma rays are conventionally defined as having photon energies above 100 keV and are the subject of gamma ray astronomy, while radiation below 100 keV is classified as X-rays and is the subject of X-ray astronomy. This convention stems from the early man-made X-rays, which had energies only up to 100 keV, whereas many gamma rays could go to higher energies. Here's a diagram from that article. In practice, gamma ray energies overlap with the range of X-rays, especially in the higher-frequency region referred to as "hard" X-rays. This depiction follows the older convention of distinguishing by wavelength. In both conventions, x-rays are electromagnetic radiation with wavelength shorter than ~10 nm (and hence energy ~125 eV) and EM radiation just below that energy is considered to be ultraviolet light. Here's another relevant passage from that Wikipedia article: Due to this broad overlap in energy ranges, in physics the two types of electromagnetic radiation are now often defined by their origin: X-rays are emitted by electrons (either in orbitals outside of the nucleus, or while being accelerated to produce bremsstrahlung -type radiation), while gamma rays are emitted by the nucleus or by means of other particle decays or annihilation events. There is no lower limit to the energy of photons produced by nuclear reactions, and thus ultraviolet or lower energy photons produced by these processes would also be defined as "gamma rays". The only naming-convention that is still universally respected is the rule that electromagnetic radiation that is known to be of atomic nuclear origin is always referred to as "gamma rays", and never as X-rays. However, in physics and astronomy, the converse convention (that all gamma rays are considered to be of nuclear origin) is frequently violated. Bremsstrahlung is braking radiation. It is any radiation produced due to the acceleration of a charged particle. A table of gamma emitters from Professor Peter Siegel's page on the California State Polytechnic University site lists the energies of a wide range of radioisotopes. The lowest energy in the table is that of Erbium-169, which decays by beta emission (with a half-life of 9.4 days) but some decays also release a gamma ray. The energy of that gamma photon is a mere 8 keV, well below the astronomical threshold of 100 keV. It has a wavelength of
0.155 nm, so a bit shorter than high energy UV, and at the low end of x-ray energies. | {
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562,101 | Recently I chewed the fat with a physics student and got intrigued by him mentioning "the Devil's problem," which he described as a simply worded mechanics problem that is extremely difficult to solve and has an answer of exactly 13 despite the formulation having no numbers and being very natural. That's kinda crazy, so I laughed and said he was kidding, but he responded that no, he wasn't kidding. He then explained me the formulation of the problem. You've got a plastic tube, like a tube used to carry posters to conferences, but open on both ends. You firmly attach a thin but heavy rod to the inner surface of the tube, parallel to the tube's axis. The tube is then laid on a floor so that the rod is in the uppermost position, and then is released to roll away from that unstable equilibrium position. The floor isn't slippery, so there's no sliding. How many times heavier than the original tube must the rod be for the tube to jump? I'm a student studying something unrelated to physics, and although I liked physics at school, this problem is too tough for me to crack, so I can't tell whether he was fooling me or whether what he said is true. I tried to find the problem on the Internet, but to no avail, so I'm posting it here. It's mystical and a bit scary if the Devil's dozen really pops out of nowhere in such a simply stated problem, but I guess the student was bluffing, counting on my inability to solve such problems. I don't even understand why the tube would jump. I just made an illustration to help understand the description of the problem: Can the rod actually jump? If so, how can one approach this problem? Can you help me call the bluff of the student, or is 13 really the answer? UPDATE: To clarify in response to a comment below, the rod and the wall of the tube are much thinner than the tube diameter and thus can be assumed to be infinitesimally thin. Likewise, an infinitesimal initial perturbation due to a slight asymmetry or thermal fluctuations is assumed. The problem is clearly well-posed from the mathematical standpoint, so the only question is how to solve it and what is the answer. UPDATE 2. It seems I've figured out why the tube will jump if the mass of the rod is large enough, but I can't calculate the exact threshold. My proof of the jump is in my answer below. | I don't know if there's a beautiful solution for this. I'd love to see it, if it exists. What I can do is show you how I slogged my way through it. All praise to the mighty Mathematica. Part I: Obtaining the Equations of Motion First, we can dispense with the cylinder and rod and consider only a point mass $M$ on a ring of mass $m$ and radius $R$ . Define $\theta$ as the angle the mass makes with the vertical, as shown: All of the dynamics of this problem can be framed in terms of this angle. Assuming a no-slip condition and purely horizontal motion, the linear velocity of the center of the ring is $R\dot \theta$ , where $R$ is the ring's radius. The components of the velocity of the point mass are $$v_x = R\dot \theta + \frac{d}{dt}\big(R\sin(\theta)\big) = R\dot\theta + R\cos(\theta) \dot\theta$$ $$v_y = \frac{d}{dt}R\big(1+\cos(\theta)\big) = -R\sin(\theta)\dot\theta$$ The total kinetic energy can be expressed as (i) the translational kinetic energy of the center of mass of the ring, plus (ii) the rotational kinetic energy of the ring about its center, plus (iv) the kinetic energy of the point mass. Summing all of these contributions yields $$T = \frac{1}{2}m(R\dot \theta)^2 + \frac{1}{2}(mR^2)\dot \theta^2 + \frac{1}{2}M\left(\big(R\dot \theta + R\cos(\theta)\dot \theta\big)^2 + \big(-R\sin(\theta)\dot\theta\big)^2\right)$$ $$ = mR^2\dot\theta^2 + \frac{1}{2}MR^2\dot \theta^2\left(2+2\cos(\theta)\right)$$ $$= MR^2\dot \theta^2\left(1+\cos(\theta)+\mu\right)$$ where $\mu\equiv \frac{m}{M}$ . The potential energy is simply $U=MgR(1+\cos(\theta))$ , so the Lagrangian for this system is $$L = MR^2\dot\theta^2\left(1+\cos(\theta)+\mu\right) - MgR(1+\cos(\theta))$$ and the total energy is $$E = MR^2\dot\theta^2\left(1+\cos(\theta)+\mu\right) + MgR(1+\cos(\theta))$$ Because the kinetic part of the Lagrangian is quadratic in $\dot \theta$ and there is no explicit time dependence, $E$ is a conserved quantity. If we assume that the initial condition is an infinitesimal distance away from $\theta=0$ , the total energy is equal to $2MgR$ ; this allows us to write $$\dot \theta^2 = \left[\frac{1-\cos(\theta)}{1+\cos(\theta)+\mu}\right]\frac{g}{R}$$ and via differentiation, $$\ddot \theta = \left[\frac{(1+\frac{\mu}{2})\sin(\theta)}{(1+\cos(\theta)+\mu)^2}\right]\frac{g}{R}$$ Part II: The "No-Jump" Condition The sum of the vertical components of the forces on the point mass is $$\sum F_y = F_R - Mg = M\dot v_y = -MR\big(\sin(\theta)\ddot \theta +\cos(\theta)\dot\theta^2\big)$$ where $F_R$ is the vertical component of the constraint force due to the ring. The sum of the vertical components of the forces on the ring is then $$\sum F_y = -F_R - mg + F_N = 0$$ where $F_N$ is the normal force on the ring due to the floor. The condition that the ring will never jump is that $F_N \geq 0$ ; this corresponds (after some algebra) to the condition $$(1+\mu)\frac{g}{R} -\sin(\theta)\ddot \theta -\cos(\theta)\dot\theta^2 \geq 0$$ Part III: Putting Things Together We already have expressions for $\dot\theta^2$ and $\ddot \theta$ ; our no-jump condition becomes (dividing by $g/R$ ) $$1+\mu- \left[\frac{(1+\frac{\mu}{2})\sin^2(\theta)}{(1+\cos(\theta)+\mu)^2}\right]-\left[\frac{(1-\cos(\theta))\cos(\theta)}{1+\cos(\theta)+\mu}\right] \geq 0$$ At this point, one can clearly see that there are circumstances under which the ring would jump. For $\theta=\pi+\epsilon$ , the left hand side diverges to negative infinity like $-1/\mu^2$ , meaning that for sufficiently small $\mu$ we can violate our no-jump condition. From here, it's a matter of rather unpleasant algebra. If you minimize the left hand side with respect to $\theta$ and grind through more algebra, the condition takes the form $$(2+\mu)^2(13\mu-1)\geq 0$$ $$\implies \mu \geq \frac{1}{13}$$ Therefore, if $m < \frac{M}{13}$ , the normal force from the floor on the ring would have to become negative; it follows that in the absence of any adhesive effects, the ring would jump up into the air. | {
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562,149 | I am reading a book titled "Relativity Demystified --- A self-teaching guide by David McMahon". He explains the derivation of electromagnetic wave equation. $$
\nabla^2 \, \begin{cases}\vec{E}\\\vec{B}\end{cases} =\mu_0\epsilon_0\,\frac{\partial^2}{\partial t^2}\,\begin{cases}\vec{E}\\\vec{B}\end{cases}
$$ He then compares it with $$
\nabla^2 \, f =\frac{1}{v^2}\,\frac{\partial^2 f}{\partial t^2}
$$ and finally find $$
v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=c
$$ where $c$ is nothing more than the speed of light. The key insight to gain from this derivation is that electromagnetic waves (light) always travel at one and the same speed in vacuum. It does not matter who you are or what your state of motion is, this is the speed you are going to find. Now it is my confusion. The nabla operator $\nabla$ is defined with respect to a certain coordinate system, for example, $(x,y,z)$ . So the result $v=c$ must be the speed with respect to $(x,y,z)$ coordinate system. If another observer attached to $(x',y',z')$ moving uniformly with respect to $(x,y,z)$ then there must be a transformation that relates both coordinate systems. As a result, they must observe different speed of light. Questions Let's put aside the null result of Michelson and Morley experiments because they came several decades after Maxwell discovered his electromagnetic wave derivation. I don't know the history of whether Maxwell also concluded that the speed of light is invariant under inertial frame of reference. If yes, then which part of his derivation was used to base this conclusion? | Your question is an excellent one and you are right about the $\nabla$ operator. And you are also right about the insufficiency of the argument you report in the book you are reading. To make the argument more carefully, there are two options. The first would be to work out how the Maxwell equations themselves change as you go to another inertial frame. That would take a lot of calculating if you start from first principles. (And by the way, they don't change---you get back the same equations but now in terms of ${\bf E}', {\bf B}', \rho', {\bf j}', {\bf \nabla}', \partial/\partial t'$ ). A second option, mathematically easier but still requiring some work if you are not familiar with it, is to show that the $\nabla$ operator and the $\partial/\partial t$ operator have a special property: when you combine them in the combination $$
\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}
$$ then their effect is the same as $$
\nabla'^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t'^2}
$$ All the changes when moving from unprimed to primed coordinates cancel out. If you are familiar with partial differentiation then you could try checking this. When you learn the subject more fully, it becomes an example that can be handled more easily using the language of 4-vectors. I think that McMahon might possibly have not thought carefully enough about what he was deriving and what he was assuming in his argument. He might for example have been taking it for granted that the Maxwell equations themselves take the same form in all inertial frames. But if he did not first prove that in his book then he ought not to claim that the derivation of waves of given speed from them proves that the wave speed will be independent of the motion of the source. | {
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562,305 | Setup: an official ping pong ball is floating inside a party plastic cup filled with clean water, which is then dropped from a certain height onto a soft mat. Observation: the ping pong ball shoots up to a height which is much higher than its initial position. Question: why does the ping pong ball do that? Why didn't the water and soft mat absorb the kinetic energy? Is this an inelastic collision? PS: the first time it was an accident, the second time the soft mat and I were thrown out XD | I've confirmed the experiment, using a McD_n_lds paper drinks cup and a beer can hollow plastic ball of about $5\mathrm{g}$ , of about the same diameter as a ping pong ball (PPB): The observed effect depends largely on the cup being soft and permanently deformable (like an object made of blutack or playdough), so its collision with Earth is inelastic. A stiff, hard cup (made of steel e.g.) would not work the same way here. The inelastic collision of the ensemble causes kinetic energy of cup and water, post-collision, to be small. The PPB bounces back quite high (from a quarter-filled cup) and the cup of water loses quite little water and doesn't really bounce at all. It's quite a sight to behold! A simple model can be set up a follows. We can write with Conservation of Energy (the collision is clearly not elastic - as evidenced by the permanent deformation of the bottom of the cup ): $$(M+m)gH=mgh+W+\Delta Q+K_{M+m}$$ where: $M$ is the mass of water plus cup and $m$ is the mass of the PPB $H$ is the height from which the cup, water and PPB are dropped and $h$ is the rebound height of the PPB, after the ensemble hits the Earth $W$ the work done on the cup's bottom $\Delta Q$ heat energy dissipated by various non-conservative forces $K_{M+m}$ the kinetic energy of water and cup, post-collision with Earth. Trouble is, we don't know the value of $W+\Delta Q+K_{M+m}$ . Direct observation suggests it is small, so we can write: $$(M+m)gH\geq mgh$$ Or: $$\boxed{h \leq H\Big(\frac{M+m}{m}\Big)}$$ If $M\gg m$ we can further approximate: $$h \leq \frac{M}{m}H$$ I wanted to confirm experimentally the effect of $M$ on $h$ . Using a nearly empty cup, one half-filled and one filled completely I can confirm increased $M$ increases $h$ . Some further experiments are planned. | {
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563,072 | I read about Kepler's Laws and in one of them he mentions that the path of a planet is an ellipse, with the sun as one of its foci (I'm narrowing down this to only our solar system). However though I'm not experienced in this subject, I had a doubt. I read in places that the Sun is not stationary. Please correct me if it's not the case. But if it is the case, then the path of the planets is an ellipse only with respect to the Sun. So the actual path of a planet observed from, let's say, a point in space, would differ from an ellipse? Or is this already factored into the law? Please help me because I'm new to this concept. | This is an interesting question, since it raises the problem of the reference frame where Kepler's laws are true, which is often neglected. As a consequence of Newton's laws, in the inertial reference frames where the center of mass (c.m.) is fixed (there is a triple infinity of them, differing only with respect to the position of the c.m.) both planet and Sun describe an elliptic motion having the center of mass as one focus of the ellipse. The two ellipses are similar, with a rescaling factor equal to the planet/Sun mass ratio. In every other inertial frame, the elliptic motion is combined with a uniform translation, therefore, in such systems, no closed orbit exist anymore. There are two additional reference frames where the orbit is an ellipse. Both are non-inertial. One is the non-inertial reference frame where the Sun is fixed. You correctly noticed that the Sun is non-stationary. But this is true in any inertial frame. If one picks precisely the non-rotating, non-inertial system where the Sun is fixed, it stays forever at the position of one focus of the elliptic orbit of the planet. Similarly, one could sit on the planet without rotations, and in that system the orbit of the Sun would be again an ellipse like the one of the planet, with the planet at one focus position. In conclusion, there is not the actual path . Shapes and properties of the orbits are not invariant with respect to changes of reference. | {
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563,765 | When we pluck a string, it vibrates in all possible modes of vibrations. The lowest frequency possible is the fundamental frequency and it is the most significant part of sound. But why do the amplitude of higher harmonics decrease? Which formula is responsible? Also, how is the energy of wave distributed among different modes? A Google search didn't give any explained answer. | Why not calculate it? Consider a string of length $L$ , with its ends fixed at $x=\pm\frac{L}{2}$ . Let's assume for convenience that at time $t=0$ the string is "plucked" at $x = 0$ , so that the string displacement relative to its equilibrium position is given by $$f(x)=A\left|1-\frac{2x}{L}\right|.$$ The standing wave solutions to the wave equation obeying the boundary conditions are $$\psi_n(x)=\cos\left(\frac{(n-\frac{1}{2})2\pi x}{L}\right) $$ with $n\ge1$ , $n=1$ corresponding to the fundamental, $n=2$ to the third harmonic, $n=3$ to the fifth harmonic and so on. Note that I haven't included the odd solutions (even harmonics) here, because these modes won't be excited since $f(x)$ is even. It is a straightforward exercise to show that $\psi_n$ are orthogonal: $$\int\limits_{-L/2}^{L/2}\psi_m(x)\psi_n(x)dx=\frac{L}{2}\delta_{mn}$$ where $\delta_{mn}$ is the Kronecker delta . If $$f(x)=\sum\limits_{m=1}^\infty a_m\psi_m(x),$$ multiplying by $\psi_n$ , integrating and using the orthogonality relation yields $$a_n = \frac{2}{L}\int\limits_{-L/2}^{L/2}f(x)\psi_n(x)dx=\frac{4A}{L}\int\limits_{0}^{L/2}\left(1-\frac{2x}{L}\right)\cos\left(\frac{(n-\frac{1}{2})2\pi x}{L}\right)dx.$$ Evaluating the integral gives $$a_n=\frac{2A}{\pi^2\left(n-\frac{1}{2}\right)^2}.\tag{1}$$ So the amplitude of the harmonics decreases roughly as $1/n^2$ . You find that if you pluck the string closer to the ends, the amplitude of the harmonics goes down slower, i.e. there are more "overtones". Specifically, if the string is plucked a distance $\ell$ from one of the ends, the amplitudes are $$ b_n = \frac{2AL^2}{\pi^2\ell(L-\ell)n^2}\sin\left(\frac{n\pi\ell}{L}\right)\tag{2}$$ where the sine factor accounts for the slower decay of $b_n$ when $\ell$ is small. $(2)$ is more general than $(1)$ as it is also valid when the string is not plucked in the middle, and is also consistent with how a guitar string is normally picked. Note: the meaning of $n$ in $b_n$ is different from before: here, $n=1$ is the fundamental, $n=2$ is the second harmonic, $n=3$ is the third harmonic and so on. The difference is because when the string is plucked in the middle, the even harmonics are not excited. As for the energy distribution, the energy in the $n$ 'th harmonic is $$ E_n = \frac{1}{4}M\omega_n^2b_n^2 = \frac{1}{4}M\omega_1^2n^2b_n^2$$ where $M$ is the total mass of the string and $\omega_n=n\omega_1$ is the angular frequency of the $n$ 'th harmonic. | {
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563,889 | My text book ( Fundamentals of Physics by Halliday, Resnick, and Walker) mentions the following about the work done in internal energy transfers: An initially stationary ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F on her from the rail. However, that force does not transfer energy from the rail to her. Thus, the force does no work on her. Rather, her kinetic energy increases as a result of internal transfers from the biochemical energy in her muscles. This is confusing me a lot. The energy transfer is clearly internal but work must be done by the force as work done is defined as the (dot) product of force and displacement and the definition makes no reference to any transfer of energy. I thought work done by a force just means that the force is causing a transfer of energy to (or from) an object, and gives no information about whether the energy is coming from the object exerting the force. My confusion is not over whether work is being done or not but which force is doing the work which ends up causing the change in kinetic energy. | Let's make a simple example. A block with a compressed spring attached to it is on a frictionless horizontal surface against a stationary, immovable wall. The spring is released, and the block is then pushed away from the wall, thus gaining kinetic energy. The relevant forces here are 1) the force between the spring and the block and 2) the force between the spring and the wall. Which force does work here? Force 1 did, because it is applied over a distance. The energy is transferred from the potential energy stored in the spring to the kinetic energy of the block. In your example, the skater is the block, and the arms/muscles are the spring. | {
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563,971 | The following question is taken from 10th class science NCERT book chapter 14th. Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Sun’s energy?
(a) geothermal energy (b) wind energy
(c) nuclear energy (d) bio-mass. The answer is given as (c) nuclear energy. I understand that the wind moves because of the uneven heating of the earth by the sun. And biomass uses solar energy for photosynthesis. How is geothermal energy ultimately derived from the sun? | It is not a correct statement: Geothermal energy comes from the heat within the earth. The word "geothermal" comes from the Greek words geo, meaning earth," and therme, meaning "heat." People around the world use geothermal energy to produce electricity, to heat buildings and greenhouses, and for other purposes. The earth's core lies almost 4,000 miles beneath the earth's surface. The double-layered core is made up of very hot molten iron surrounding a solid iron center. Estimates of the temperature of the core range from 5,000 to 11,000 degrees Fahrenheit (F). Heat is continuously produced within the earth by the slow decay of radioactive particles that is natural in all rock italics mine. Geothermal energy comes from the original energy of the matter solidifying into the sun-planetary system, ultimately from the Big Bang, and from continuous nuclear decays and reactions . | {
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564,096 | In General Relativity we see spacetime as a manifold; in this context vectors can't be defined on the manifold but need to be defined on the tangent space of the manifold. So each point of the manifold has its own tangent space and different vectors in different tangent spaces cannot be easily compared. At last each tangent space has its own metric tensor $g_{\mu \nu}=\partial _\mu \cdot \partial _\nu$ , where $\partial _\mu,\partial _\nu$ are the base of the tangent space. Problem is: my geometrical intuition makes me think about the tangent space as a flat space; if you have any 2D or 3D object in everyday experience the tangent space at one point is always a flat one. But it's not only intuition: spacetime locally looks like $\mathbb{M}^4$ , so locally it looks flat or to say it better: locally can be approximated with a flat spacetime; but seems to me that the tangent space at one point is simply the space that better approximate the area around that point; this also push me to say that the tangent space should be always flat. So is the tangent space of a manifold always flat? Or equivalently is the tangent space always $\mathbb{M}^4$ ? Based on the upper reasoning seems to me that the answer should be yes, but this seems to create a problem: in GR we use the Christoffel connection so the curvature can be calculated using only the metric tensor $g_{\mu\nu}$ , but if the tangent space is always flat then the metric tensor is always "a flat one", in the sense that it generates always flat curvature. This is obviously absurd. How can we get out of this apparent contradiction? Edit: Based on the answer of Javier tangent space is indeed always flat. Does it mean that I can take any tangent space (with metric tensor $g_{\mu\nu}$ ), apply a change of coordinates and get the metric tensor of $\mathbb{M}^4$ ( $\eta _{\mu\nu}$ )? This is important because this is what flatness means ; am I right? And also: we state that the metric is calculated by looking at the rate of change of the metric $g_{\mu\nu}=\partial _\mu \cdot \partial _\nu$ , but the metric is always flat! So the rate of change is always zero because every tangent space is flat! How can we deal with this? | Short answer: yes each tangent space is flat. In principle the tangent space is a vector space, not a Riemannian manifold, and so the concept of curvature technically doesn't apply if that's all you have. To define curvature you need to define parallel transport; for that, you need to think of the tangent space as a manifold, and that implies looking at the tangent spaces of the tangent space! And if you do that, the vector space structure gives you a canonical way to define parallel transport, and this parallel transport ends up being flat. Finally, the solution to your paradox is simple: the curvature depends on derivatives of the metric, that is, on how it changes from tangent space to tangent space. The metric at a point is irrelevant because all vector spaces with an inner product are isometric; what matters is how it varies in space. Edit in response to your edit: You're playing too fast and loose with the words. Differential geometry is complicated, and we need to be precise in how we speak. You can always make any tangent space into Minkowski space by choosing an appropriate basis, not because it is flat, but because it is a vector space. The difference is subtle but important. A vector space has a single metric tensor: it takes pairs of vectors and returns a number. A manifold has a metric tensor field : a metric tensor at each point, which takes pairs of tangent vectors. The fact that the tangent space is flat is a red herring. One definition of flatness (of a manifold) is that you can use a single coordinate system in which the metric tensor is everywhere Minkowski. This statement is different from the statement about the tangent space; here you're choosing a different basis at each point, which are related by coming from a single coordinate system. In a single tangent space, you only have one basis. And you can always have $g_{\mu\nu} = \eta_{\mu\nu}$ at a single point (and hence at a single tangent space), but making it so at every point with the same coordinate system may or may not be possible, and that's what flatness means. | {
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