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Why electron do not eject out even when there is photon of less threshold energy with increasing in time When photon having certain energy less than threshold energy strikes on the electron of metallic plate. Electron do not eject out. But my question is when photon are falling continuously then electron must gain the energy to eject out.as electron have the energy transferred from earlier photon and next photon will transfer his energy too.so combination of these energy must be greater than the required energy for moving out from his path.
I know it's a silly question but i have doubt so asking it...
| This is just a matter of statistics. The number of atoms is typically many orders of magnitude greater than the number of photons. Therefore the mean time between excitations of the same atom is very long -- much longer than the time required for the atom's excitation energy to be thermalized.
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Why does the opposing force differ in when falling on concrete vs on water in spite of Newton's third law? If a person jumps from the first floor of a building and lands on a concrete surface, they will suffer serious injury because of Newton's third law.
If the same person jumps the same distance and lands in swimming pool filled with water, however, then there will not be any serious injury.
The person in both cases lands with same amount of force. Why doesn't water offer the same amount of force in return as concrete?
|
But the person in both case lands with same amount of force. Then why doesn't water offer the same amount of force in return as concrete does?
This is not correct: the force that the person applies to concrete/water is the same as the force that the concrete/water applies to the person (Newton's third law). The force is different in the two cases, as the person is slowered down for different amount of time, i.e., their acceleration is different: when they fall on concrete, they are stopped almost immediately, the force is high (Newton's second law), whereas when falling in water they are slowered down gradually, while submerging to a noticeable length - the force is smaller.
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Why does this fan with one blade missing rotates counterclockwise while running? Video: Fan with one blade missing rotates while running.
The fan worked just fine until my friend tried to stop the spinning blades with her finger and knocked one off. Now it always rotates counterclockwise when running. Can someone explain in details why? Does this have something to do with the shape of the blades?
| From the video, it looks like that the axis of the fan rotor is not orthogonal to the vertical direction.
When a motor-driven rotor spins in a direction, a torque acts on it due to aerodynamic drag in the opposite direction, i.e. the torque can be represented by a vector orthogonal to the plane of the rotor, that can be approximately written as
$\mathbf{M}^{aero} = - \dfrac{1}{2} \, C_T \rho R^3 |\mathbf\Omega| \mathbf{\Omega} = - \dfrac{1}{2} \, C_T \rho R^3 |\Omega| \Omega \,\mathbf{\hat{n}}$ ,
being
*
*$C_T$ the torque coefficient depending on the parameters of the blades, of the rotor as a whole and non-dimensional coefficients describing the motion, like Reynolds number
*$\rho$ is the air density
*$R$ the radius of the rotor
*$\mathbf{\Omega} = \Omega \, \mathbf{\hat{n}}$ is the angular velocity of the rotor around its axis, following the right-hand rule for angular velocity.
If we choose $\mathbf{\hat{n}}$ so that $\Omega > 0$, from what I get from the video, $\mathbf{\hat{n}}$ points slightly downward, so that we can write $\mathbf{\hat{n}} \cdot \mathbf{\hat{z}} = \sin \alpha < 0$, defining the angle $\alpha$ as the angle between $\mathbf{\hat{n}}$ and the horizontal plane. Now,
*if the axis of the fan is orthogonal to the vertical direction $\mathbf{\hat{z}}$, the rotor doesn't introduce any torque around the $\mathbf{\hat{z}}$-axis, since $\mathbf{\hat{z}} \cdot \mathbf{\hat{n}} = 0$ and this aerodynamic action is balanced by normal reactions exchanged by the table and the base of the fan
*if the axis of the fan is not horizontal (as it looks like from the video), the aerodynamic torque has a vertical component
$ M_z^{aero} = \mathbf{\hat{z}} \cdot \mathbf{M}^{aero} = - \dfrac{1}{2} \, C_T \rho R^3 |\Omega| \Omega \,\mathbf{\hat{n}} \cdot \mathbf{\hat{z}} = - \dfrac{1}{2} \, C_T \rho R^3 |\Omega| \Omega \sin \alpha > 0$,
that makes the fan rotate around a vertical axis, in the counter-clockwise direction if we look at the fan from above, if the friction between the base and the table is not enough to balance the external action.
This process is independent from the missing blade, but only on the inclination of the rotor plane. You could test this explanation if you have a fan with all the blades, and play around with the direction of the axis of rotation. Even if you only have the broken one, you can change the orientation of the axis, so that it point upwards and see if the fan rotates around z-axis in the opposite direction.
Other contributions (I guess minor, but I currently have no time to do the calculations) could come from the unbalanced rotor.
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Why is it easier to raise AC current to high voltage than DC? In my country (and maybe all around the world I don't know) once electricity has been generated, it is then raised to 200k Volts for transportation.
I know this is to reduce the loss. Given $P=U.I$ and $P=I^2.R$, raising U will lower I and so limit the loss by joule effect.
From what I've read, one of the reason electricity is transported in AC is because this is easier/cheaper to raise AC to 200k Volts than if it was in DC.
Why?
| You are absolutely right, the higher voltage, the less energy loss due to Joule heating.
The main reason why electricity is generated and transported in AC is that the generated electricity from electromagnetic motor due to mechanisms of electric induction is in fact in AC. So it would require a rectifier for converting AC to DC and thanks to Joule heating, energy will be lost.
Moreover, many electrical appliances run on AC (e.g. hair dyer, electric vehicles...) due to the fact they convert AC into kinetic energy by electric motors. And if electricity supplied to our homes were DC, then we would need a MOSFET or IGBT to convert them back to AC which would entail additional loss of energy.
It is not necessarily easier to raise AC current to hight voltage, but because DC is not an option due to the need of several conversions.
To put it simple:
*
*Electricity generated is in AC
*Most electric appliances need to run on AC
Enjoy!
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The current in an $LR$ circuit I am trying to understand $LR$ circuits better. I am under the impression that inductors are resistant to change in current.
So, the premise here is that the circuit was switched to side 1 for a long time and switched on to side 2.
What I thought was the 4H inductor would be "okay" with the 5A current when the switch is in side 1. When the switch is switched on to side 2, i.e. when the current should change to 2.5A, the 4H inductor would behave as if 5A was passing through it, but the current is changing gradually to 2.5A. So, without that 1H inductor, the current should be
$$I=2.5(1+e^{-(4t/5)})$$
Because the 1H inductor is then applied to the circuit, then the current should be
$$I=2.5(1+e^{-(4t/4)})(1-e^{4t})$$
I know I took a bunch of wrong assumptions, but how should we find the answer?
| The problem is not with you. This is an example where the assumption of ideal circuit behavior leads to inconsistencies.
Assuming ideal inductors, immediately after switching the current in the 1 H inductor has to be zero, as it was before switching, while the current in the 4 H inductor in series with it has to be 5 A, as it was before switching. This gives the inconsistent result of initially having two different values of current for a series circuit.
Hope this helps.
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Nature of tangential friction force When a ball rolls down a ruff slope the frictional force acts tangent to the ball and causes the angular acceleration of the ball but at the same time the frictional force is acting to reduce the translational acceleration of the ball. How is this possible when the frictional force is acting only tangentially and not through the centre of mass of the ball?
| Forces don't need to act through the center of mass to produce translational acceleration. Consider a rod lying on a horizontal frictionless surface - applying a force at one end of the rod perpendicular to its axis will cause the rod to both spin and move translationally. Or, consider a rolling object placed on a rug - pulling the rug causes the object to roll and move toward you, the object doesn't just spin in place.
Forces acting through the center of mass don't induce rotation, but that doesn't imply that forces not acting through the center of mass don't induce translation.
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Question Regarding The Movement of Charges We know that two electrons repel each other since they have like charges, which means they move in opposite directions. But how can they move if they exert equal and opposite charges, aren't the forces balanced which means there is no movement?
| As a motivational example, consider your air-table puck experiment from highschool. If you have a pair of pucks at rest, and this one pushes that one, then the two of them will move off in opposite directions. If they are equal mass then they move with equal speeds. The total momentum is therefore zero, since this one is exactly opposite that one.
The thing that lets you easily calculate this is the center of momentum. This is the reference frame where the total momentum adds to zero. In this frame, when you have collisions and forces and so on, the total momentum remains zero.
So two for electrons. This one provides a force on that one. That one provides a force on this one.
The forces are equal and opposite. That is, the change in momentum of this one is the same size and opp. direction as that one.
The center of momentum does not change its motion. If you start in the center of momentum frame then the total momentum starts zero and stays zero. But each electron does move.
| {
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Illuminance Formula This page says illuminance is
$$E=\frac{I}{L^2} cos \space \alpha$$
This page does something similar, but it ignores the $cos \space \alpha$ factor. Which is the correct formula?
Note: I don't have a physics background. I was looking at optimization problems in Calculus (which is why I came across the first page).
| A physical formula always applies to a given situation, and it is important to first check whether your situation matches the one this formula is meant for.
The formula without the $\cos \alpha$ term is meant for a situation where the light hits the surface in a right angle. It's a special case of the more general formula, using $\alpha = 0°$ (because then $\cos \alpha$ is 1).
Both pages you linked describe the same situation, one where this isn't true, so the $\cos \alpha$ term is needed.
Nitpicking addition:
The $E=\frac{I}{L^2}\cos\alpha$ formula applies to angles between -90° and 90°, otherwise you'd get negative illuminations (physical nonsense) out of the formula.
Instead the physically correct result would be 0, as a surface pointing away from a light source gets no light at all.
You see, even the "general" formula still has its limitations, being applicable only under specific circumstances.
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Why does a rocket engine that produces a constant thrust over a set period of time have less energy if it has more mass? (Zero-$g$) A rocket engine with the thrust of 1N working for 10 seconds will add more kinetic energy to the rocket if it is attached to a 10kg rocket and less if it is attached to a 20kg rocket. The rocket should consume the same amount of fuel if producing the same thrust for the same time with the same engine and so convert the same energy. So why doesnt it have the same energy. (I know the formula but why)
| Even if you assume that the change of the mass of the rocket, due to fuel consumption is negligible, the answer is that the heavier the rocket the smaller the variation of the kinetic energy. Details follow.
| {
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What reason/evidence do we have to think that the Planck length is the smallest length possible? From what I've gathered, Planck length is the smallest measurable length, though we do not know whether it is the smallest length physically possible. The Planck temperature is called the theoretically highest temperature, meaning this theory assumes that the Planck length is the smallest possible length. Given that it is (presumably) a theory, that makes me wonder what reasons we have to think the Planck length is the smallest length. I get speculating that it is the smallest length, given that it is the smallest measurable length; but calling it a theory means there's actually some reasons, or even evidence. So, what are those reasons?
| I'ts only a reference that came from the universal constants, as far as I know there is no theoretical limit
| {
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Why do diffraction limits exist? My question does not deal with standard derivations of diffraction. I have no doubt that I can follow standard derivations for diffraction phenomena. Actually, I have probably been shown these before, it's just been so long that I forget them.
My question deals with thinking about diffraction phenomenon from the quantum/particle perspective. I was thinking about why processes such as photolithography, e-beam lithography, electron microscopes, or even optical microscopes should be limited wavelength.
In all of these processes we are firing a bunch of elementary particles at a target. We know that the wave-like properties of these particles disappear when an 'observation' takes place. My understanding is that 'observation' really means interaction. So once these particles interact with the target they should lose their wave-like qualities and behave like particles. Since they are then localized in space, this would give infinite resolution. Obviously there is something wrong with this way of thinking. So to remedy this I was thinking that maybe the loss in resolution arises when the particle reflects back to the detector. During this journey it would return to wave-form and in a sense lose the information of where it reflected from.
Hope that at least made enough sense for someone to explain where it is wrong.
| It’s all fields, forget particles. Particles is a misnomer. What are called particles in quantum field theory are actually quantized excitations in quantum fields. The quantized fields obey wave equations and diffractions limits follow from wave equations.
| {
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Friedmann equation I've seen in literature
$$\dot{H} + H^2=\ldots$$
Source: https://en.wikipedia.org/wiki/Friedmann_equations
Defining the LHS. Since
$$H = \frac{\dot{a}}{a}$$
And that
$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}(\rho + 3P)$$
Then replacing gives
$$H^2 = \frac{8\pi G}{3}(\rho + 3P)$$
So my question is how to you arrive at the additive Hubble term
$$\dot{H} + H^2 = \frac{8\pi G}{3}(\rho + 3P)?$$
| The forth equation you wrote, being
\begin{equation}
H^2=\frac{8\pi G}{3}(\rho+3P)
\end{equation}
is incorrect. The first Friedmann equation is given by
\begin{equation}
H^2=\frac{8\pi G}{3}\rho.
\end{equation}
Now, taking a time derivative in both sides, we find:
\begin{equation}
2H\dot{H}=\frac{8\pi G}{3}\dot{\rho}.
\end{equation}
Knowing that for a bariotropic fluid
\begin{equation}
\dot{\rho}=-3H(\rho+P),
\end{equation}
we have
\begin{equation}
\dot{H}=-4\pi G(\rho+P).
\end{equation}
Adding the first Friedmann equation to this one we obtain:
\begin{equation}
H^2+\dot{H}=-\frac{4 \pi G}{3}(\rho+3P),
\end{equation}
which is the correct form for the equation you are looking for (note that it agrees with the equations in the Wikipedia link you shared.
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What makes a photon a photon? As i understand photons are excitation of the electromagnetic field. Therefore charged particles are affected by this excitation. But what if we have (highly theoretically) a particle that has the exact same properties like a photon (spin 1, no electric charge, no color charge, no mass etc.) but is an excitation of an other field. Is this even (again, theoretically) possible? Is a particle with the properties of an photon always the excitation of EM fields? And would this photon-like particle interact with charged particles (because they are not belong to EM fields)?
| Simple answer: Yes. It's possible to have 2 different types of photon where only one type interacts with the electron and the other type doesn't. In our universe, however, there doesn't seem to be this other type of photon.
| {
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Why is current finite for point charges? If an electron passes through a flat plane, then there will only be a single point in its entire path which lies on the plane,i.e the entire charge of an electron passes through in an instant (as it is a point charge), then why isn’t the current infinite at that instant and zero at all the others?
| WARNING: there are many paradoxes that arise by treating an electron as a classical point particle. You are best off to avoid this concept entirely. Electrons are not classical point particles, they are particles as defined by quantum electrodynamics.
If you are still reading then you have ignored the warning, like many before you. If despite the warning you insist on speaking of a classical point charge, then you are talking about classical electromagnetism where the appropriate concept is current density. The current density of a point particle is $$\vec j(t,x,y,z)=q \ \vec v(t) \ \delta(x-r_x(t),y-r_y(t),z-r_z(t))$$ where $q$ is the charge of the point charge, $\vec v$ is its velocity, $\vec r=(r_x,r_y,r_z)$ is its position, and $\delta$ is the Dirac delta distribution.
The current density of a classical point charge is thus infinite at the location of the charge and 0 elsewhere. But as warned, this is not a good model of an electron, so it is best not to call it that.
It is far better, classically, to consider charge density to be a continuum and then current density to be simply $$\vec j = \rho \vec v$$ where $\rho$ is the charge density. This will avoid the various point charge paradoxes and is a better basis when moving to QED and how electrons are actually treated.
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What are the significant digits, when the standard deviation is larger than the value? Let's say I have some quantity $a = 1.234\,a.u.$ with a standard deviation of $\sigma = 123.4\,a.u.$. How do I express the uncertainty in this case?
*
*$a = (1.2\pm 123.4)\,a.u.$
*$a = (0.0\pm1.2)\times10^2\,a.u.$
*$a = (0\pm120)\,a.u.$
*$a = (1.2\pm120)\,a.u.$
*Something else
| The official recommendations can be found in section 7 of the BIPM's "Guide to the expression of uncertainty in measurement"
https://www.bipm.org/documents/20126/2071204/JCGM_100_2008_E.pdf/cb0ef43f-baa5-11cf-3f85-4dcd86f77bd6
It recommends to avoid the use of the ± symbol entirely unless you are reporting a confidence interval or what they call an expanded uncertainty. Their recommendation would be something like:
"$a = 1.2 \mathrm{\ a.u.}$ with a combined standard uncertainty $u_c = 120 \mathrm{\ a.u.}$"
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Shouldn't the velocity of the wave associated with a particle be equal to the velocity of the particle? If a free particle of mass $m$ is moving with a velocity $v$, then it's kinetic energy is $\frac{mv^2}{2}$, therefore its frequency is $\nu = \frac{E}{h} = \frac{mv^2}{2h}$ where $h$ is Planck's constant, and it's wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$, then the velocity of the wave associated with this particle is $V = \nu \lambda = \frac{v}{2}$, hence we see that the Velocity of the wave $\neq$ Velocity of the particle, but shouldn't they be equal?
Where is the mistake here? We've just started learning Quantum Mechanics in this semester so please help me understand this. Thanks a lot!
| The velocity you're talking about is the phase velocity $v_p$
$v_p$ is defined as $v_p = \frac{\omega}{k}$ or more commonly as $v_p=\lambda \nu$
The problem with the phase velocity is that it is only defined for purely sinusoïdal waves.
When we deal with wavepackets (made of a continuous sum of purely sinusoïdal waves) we use instead the group velocity $v_g$ defined as $v_g = \frac{\partial \omega}{\partial k}_{|k_0}$ (the derivation can be found here, you will probably see it in class with your teacher).
There is a special case however, where the phase velocity can be pertinent for a wavepacket: it is when the medium is non-dispersive. In that case, the dispersion relation $\omega = f(k)$ is linear (note that for $k=0$, $\omega$ should always be $0$, so it can't be affine)
and the phase velocity is equal to the group velocity ($v_p= v_g$)
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Does the earth’s rotational angular velocity change? This is what is written in The Feynman Lectures on Physics, Vol. 1 (ch.5)
We now believe that, for various reasons, some days are longer than others, some days are shorter, and on the average the period of the earth becomes a little longer as the centuries pass.
Why should some days be longer than the others? There is no “gravitational” source of external torque acting on the earth, so why does its rotational angular velocity change?
| To add to @gandalf61's answer: You can also look up solar time.
Due to the orbit around the Sun, the Earth has to rotate a bit more than 360° for the sun to get back to the same apparent position in the sky. Then, since the orbit around the sun is elliptical, the Earth moves around the sun at different speeds depending on it position along the orbit. Therefore the effect of the orbit on the solar day varies throughout the year.
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If reference frames are equally valid, then why do teachers say the geocentric view is wrong? If all reference frames are valid, then why is the geocentric model taught as "wrong" in schools?
I've checked many websites but none of them clear the issue. Wiki says that in relativity, any object could be regarded as the centre with equal validity. Other websites and answers make a point on the utility of the heliocentric model (simplicity, Occam's razor...) but just because something is not so easy to deal with doesn't mean it is wrong.
Note: I am not asking for evidence that geocentrism is wrong; I am asking for a way to resolve the contradiction (from what I see) between relativity and this "geocentricism is wrong" idea.
| Sean Carroll wrote a blog post about this exact queston years ago: https://www.preposterousuniverse.com/blog/2005/10/03/does-the-earth-move-around-the-sun/.
As Sean E. Lake says, it's perfectly true that the Sun goes around the Earth ... in some reference frame. And that reference frame is actually a quite convenient one to use to describe many terrestrial physics situations. But it's very inconvenient to use that reference frame for solar-system-scale calculations, because it is not anywhere close to an inertial global frame even in the Newtonian limit.
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Why Is Capacitance Not Measured in Coulombs? I understand that the simplest equation used to describe capacitance is $C = \frac{Q}{V}$. While I understand this doesn't provide a very intuitive explanation, and a more apt equation would be one that relates charge to area of the plates and distance between them, I'm having trouble understanding it in general. Capacitance seems to be describing, well, the capacity of two plates to store charge (I understand that the electric field produced between them is generally the focus more so than the actual charge). Shouldn't it just be measured in units of charge such as coulombs? I'm sure this is due to a lack of more fundamental understanding of electric potential and potential difference but I'm really not getting it.
| An analogy here would be to a pressure vessel and asking what mass of air will fit inside.
While the tank has a fixed volume, the amount of air that will go inside depends on the pressure you that you use to force it in. For quite a while the relationship is linear. At double the pressure, you have double the mass of air.
Similarly, the capacitor doesn't have a fixed amount of charge that will fit. The amount depends on the electrical "pressure" (voltage) that is used.
Actually your initial equation is the useful one. Unless we're constructing one, we usually do not care about the physical particulars of a capacitor. Instead we want to know how much charge will move if we change the voltage. For a "larger" capacitor (higher capacitance), more charge will fit at a given voltage.
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Brewster's angle for Metal? As metal has a high refraction index, it is possible to have Brewster's Angle for Metal when the light incident from the air? Is it possible to derive from any formula?
| So there is pseudo-Brewster angle for metals (Citation: Journal of Applied Physics 78, 4799 (1995); doi: 10.1063/1.359763):
It is a subject of textbooks that the reflectivity of metals is a
function of the angle of incidence and that it has a minimum for a
given wavelength at the pseudo-Brewster angle. Consequently, the
absorption is maximal at this angle and can be much higher than the
usual low absorption of metals for normal incidence.
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Is the kind of physics proposed for "warp drive" related to the way that space really is expanding? As I understand, some parts of Universe really are moving faster than light -- is this expansion something we think we can create artificially?
And if we can do this artificially, could it be done a on very small scale in the near future? Or is it possible that space is already being affected in this way in, say, particle accelerators or during nuclear explosions?
| At the current moment, we don't even know what dark energy is which is the reason the universe expands faster than the speed of light. There is not really a way we could recreate it per se but there are talks about how we can create an effect similar to dark energy which has some kind of "negative energy effect".
There has been talks about how Harold White with his Eagle Works Laboratory has been working on like a potential precursor to the warp drive and might use the Casmir effect to cause some kind of dark energy like effect which could maybe used on warp drives. However, there has not really been any real success and for the moment it is pretty much impossible to harness such a faster than light expansion.
In the end, such a thing has a really low chance of existing but there might be a tiny chance that something might be discovered. Even if it is discovered, harnessing it is very difficult thing to do.
There is not really any other known way to create a faster than speed of light flow of space(maybe aside from black holes).
| {
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Does temperature depend on the frame of reference? The way I understand it, temperature is the average kinetic energy of the particles of a system. However, kinetic energy depends on the velocity of the particles, which is relative. If a cup of coffee were traveling at 99% the speed of light relative to an observer, its particles would have much higher kinetic energy compared to a cup of coffee at rest. Would it be hotter to this observer then? Would this observer disagree on the temperature with another observer traveling along side the cup of coffee?
| Well let's take the case of you in your spaceship with a cup of hot water in your hand, and a thermometer in the cup registering "90 Celsius". You speed up to 99% of the speed of light and go zooming right past the rest of us who are watching nearby. As this occurs you hold the cup up to a porthole so we can read the thermometer.
It reads "90 Celsius". But one of us spectators has a magic telescope that allows us to watch the water molecules jiggling around in the cup of water and measure their average speed. "Gadzooks!" she exclaims. "Those water molecules are HARDLY jiggling!"
How can that be? How can those water molecules be almost frozen in time, but the thermometer tells us the cup of water containing them is at 90 C?
This suggests that we have to know something about the frames of reference in which the hot object resides and in which we observe thermometers- and that if we do not shift our experiences between those frames according to the rules of relativity, we'll get the wrong answers and draw the wrong conclusions from what we are observing.
| {
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Explain Heisenberg's uncertainty principle There was one homework question that asks what Heisenberg uncertainty tell us about the energy of an electron in an infinite square well when the length of the well decreases. The correct answer is that the energy decreases when length increases. I know that the energy should decrease by the formula for energy eigenstate, but I feel like this has nothing to do with Heisenberg's uncertainty principle. Uncertainty principle only tells us how accurate is the measurement. Can someone explain how is the uncertainty in energy related to the actual energy of the electron?
| At first, uncertainty principle tells us not about how accurate measurement can be, but how accurate a state of particle can be regardless of any measurement devices used for measuring anti-commutating operators. Second, when you decrease length of well, then particle position becomes more accurate, hence by Heisenberg uncertainty, momentum uncertainty increases, and so does expected value spread of it as per : $$\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle ^2}.$$
Bigger variance of momentum means that you'll notice far more often higher electron momentum, which was not expected before in a wider setting of well size. And momentum uncertainty is related to energy uncertainty by :
$$ \Delta E=\Delta pc. $$
Btw, as a side-note, using this relation is quite easy to jump from momentum-position uncertainty to energy-time uncertainty :
$$ \begin{align}
\Delta p \cdot \Delta x &> \hbar/2 \to\\
\Delta pc \cdot \Delta xc^{-1} &> \hbar/2 \to\\
\Delta E \cdot \Delta t &> \hbar/2.
\end{align}
$$
| {
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Peskin and Schroeder's QFT book page 289 On Peskin and Schroeder's QFT book page 289, the book is trying to derive the functional formalism of $\phi^4$ theory in first three paragraphs. But the book omits many details (I thought), so I have some troubles here.
For the free Klein-Gordon theory to $\phi^4$ theory:
$$ \mathcal{L}=\mathcal{L}_0-\frac{\lambda}{4 !} \phi^4. $$
Assuming $\lambda$ is small, we expand
$$\exp \left[i \int d^4 x \mathcal{L}\right]=\exp \left[i \int d^4 x \mathcal{L}_0\right]\left(1-i \int d^4 x \frac{\lambda}{4 !} \phi^4+\cdots\right). $$
Here I thought the book use one approximation, since $\phi^4$ don't commute with $\mathcal{L_0}$, their have $\pi$ inside $\mathcal{L_0}$. And according to Baker-Campbell-Hausdorff (BCH) formula, the book omit order to $\lambda$. Is this right?
The book further says on p. 289:
"Making this expression in both the numerator and denominator of (9.18), we see that each is expressed entirely in terms of free-field correlation functions. Moreover, since $$ i \int d^3 x \mathcal{L}_{\mathrm{int}}=-i H_{\mathrm{int}},$$ we obtain exactly the same expression as in (4.31)."
I am really troubled for this, can anyone explain for me?
Here eq. (9.18) is
$$\left\langle\Omega\left|T \phi_H\left(x_1\right) \phi_H\left(x_2\right)\right| \Omega\right\rangle=$$
$$\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\int \mathcal{D} \phi~\phi\left(x_1\right) \phi\left(x_2\right) \exp \left[i \int_{-T}^T d^4 x \mathcal{L}\right]}{\int \mathcal{D} \phi \exp \left[i \int_{-T}^T d^4 x \mathcal{L}\right]} . \tag{9.18} $$
And eq. (4.31) is
$$\langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} . \tag{4.31} $$
| In the path integral formalism, $\phi$ is not an operator, it is an integration variable. In other words, inside the integral $\int D\phi$, $\phi$ is just an ordinary classical field. So there’s no need to worry about Baker-Campbell-Hausdorff and such.
| {
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Isentropic fluid: cross product of gradients is zero, why? In the vorticity equation we have the baroclinic term of the form: $$\frac{ {\nabla}\rho}{\rho}\times\frac{ {\nabla}{P} }{\rho}.$$
Why does it go to zero for isentropic flow?
I understand that, if the flow is barotropic, the above term vanishes. However, an isentropic (reversible $dS=Q$ + adiabatic $Q=0$, i.e. $dS=0$) flow is more general, in the sense that the pressure depends on both the density and temperature.
| Write pressure as a function of density and specific entropy, $P(\rho, s)$, so that its differential reads
$dP = \left( \dfrac{\partial P}{\partial \rho}\right)_s d \rho + \left( \dfrac{\partial P}{\partial s}\right)_{\rho} d s$
and its gradient
$\nabla P = \left( \dfrac{\partial P}{\partial \rho}\right)_s \nabla \rho + \left( \dfrac{\partial P}{\partial s}\right)_{\rho} \nabla s$.
For isentropic flow, with uniform entropy, $s = \overline{s}$, $ds = 0$, and thus
$\nabla P = \left( \dfrac{\partial P}{\partial \rho}\right)_s \nabla \rho = c^2(\rho, \overline{s}) \nabla \rho$,
the gradient of pressure field is proportional to (aligned with) the gradient of the density field in every point of the space. Taking the vector product of two proportional vectors, you get zero.
| {
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Low temperature behavior for ferromagnets: theoretical and experimental discrepancies This is in reference to page 326, 327 of introduction to solid state physics, 8th edition by Charles Kittel
The mean field theory does not give a good description of the variation of $M$ at low temperature. For $T<<T_{c}$ the argument of $\tanh$ in (9) is large and $\tanh(\xi) = 1 - 2e^{-2\xi}$ + ..
This appears to be a Taylor expansion for $\tanh$ but the fact that there is an exponential term is foreign to me.
Question:
*
*Can anyone shed some insight on how the expression for $\tanh(\xi)$ provided by the author came to be?
*Am I correct to understand that $M(0)$ is the temperature at $T =0$ for the magnetisation given in (8) on page 362? Even better is someone could show the steps behind which the magnetic deviation $\delta M$ in (10) came about.
| Start from the definition of $\tanh x$ in terms of exponentials
\begin{align}
\tanh x &= \frac{e^x -e^{-x}}{e^x +e^{-x}}\\
&= \frac{1 - e^{-2x}}{1 + e^{-2x}}\\
&=(1 - e^{-2x})\left(1 - e^{-2x} + O(e^{-4x})\right)\\
&=1 - 2e^{-2x} + O(e^{-4x})\;,
\end{align}
where in the second last line we have Taylor expanded in powers of $e^{-2x}$.
| {
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I'm having trouble understanding the intuition behind why $a(x) = v\frac{\mathrm{d}v}{\mathrm{d}x}$ I was shown
\begin{align}
a(x) &= \frac{\mathrm{d}v}{\mathrm{d}t}\\
&= \frac{\mathrm{d}v}{\mathrm{d}x}\underbrace{\frac{\mathrm{d}x}{\mathrm{d}t}}_{v}\\
&= v\frac{\mathrm{d}v}{\mathrm{d}x}
\end{align}
However, this feels somewhat unintuitive, and somewhat questionable mathematics-wise. Perhaps it's the best way to explain it, but I was hoping for a more intuitive understanding of this formula.
| This intuition is for one dimensional motion: Consider a particle moving in a line. At any time $t$, its velocity is $v(t)$, its acceleration is $a(t)$. The change in velocity after a small time $dt$ will be $a(t)dt$. The change in position will be $dx=v(t)dt$
The identity you wrote is expressing the fact that $vdv=adx$. This can be verified: $vdv=v(adt)=a(vdt)=adx$
Keep in mind that you do not need $v$ to be a function of $x$ for this identity to work. It's only saying that, at any time $t$, $a(t)$ multiplied by $dx$ will be equal to $v(t)$ multiplied by $dv$. $dx$ and $dv$ are the changes in position and velocity after an infinitesimally small time $dt$.
For three dimensional motion, the identity is $\vec{a}\cdot \vec{dx}=\vec{v}\cdot \vec{dv}$.
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Thermal Equilibrium and adiabatic walls - Zemansky In Zemansky's "Heat and Thermodynamics" it is stated that:
A thermodynamic system is in thermal equilibrium with its sorroundings iff it is in mechanical and chemical equilibria with its sorroundings, it is delimited by diathermic walls and its macroscopic coordinates do not change with time (hence they may be called thermodynamic coordinates).
A thermodynamic system is in thermodynamic equilibrium with its sorroundings iff it is in mechanical, chemical and thermal equilibria with its sorroundings.
Now, I have two questions concerning these definitions:
*
*Is there any difference between thermal and thermodynamic equilibria? (it seems like there should be, at least that is what I have read in the dedicated wikipedia page: https://en.wikipedia.org/wiki/Thermal_equilibrium);
*If diathermic walls are needed in the definition of thermal equilibrium, how is it possible to use the same concept with systems that are delimited by adiabatic walls? (indeed, Zemansky speaks of such systema as if they can be in an equilibrium state).
As always, any comment or answer is much appreciated and let me know if I can explain myself clearer!
| As I understand it:
Mechanical equilibrium: Equal pressures (forces).
Chemical equilibrium: Equal chemical potentials (fugacities)
Thermal equilibrium: Equal temperatures
Thermodynamic equilibrium: The state in which all three types of equilibria are satisfied.
| {
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How to derive the $vx/c^2$ term from first principles? In Lorentz transforms, the formula for time transformation is
$$t' = \gamma \left( t - \frac{v x}{c^2} \right)$$
I understand that the term $\frac{v x}{c^2}$ represents "time delay" seen by a stationary observer but I don't understand how to derive it from first principles. I understand $v/c$ as speed and $x/c$ as distance. Why multiply speed with distance? I thought time is distance divided by speed?
| One route (explained broadly) is as follows:
Once you've written down $x'(x,t)$ as a length contraction, (which itself follows from time dilation), then it follows from the first postulate there exists a mode of description for $x(x', t')$ in terms of $-v$. Writing both sides equal to $\gamma$ (which is with the square magnitude of $v$, irrespective of its direction) you can set equal and solve for $t'(x,t)$
| {
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Is there a known closed-form expression for the susceptibility of the 2-D Ising model at $B = 0$? The Onsager solution for the 2-D Ising model allows us to find (among other things) complicated expressions for the internal energy of the system (in the thermodynamic limit and in zero magnetic field):
$$
u \equiv \frac{U}{JN} = - \coth \frac{2}{t} \left\{ 1 + \frac{2}{\pi} \left[ 2 \tanh^2 \left( \frac{2}{t} \right) - 1 \right] K\!\left[4 \, \text{sech}^2 \left( \frac{2}{t} \right) \tanh^2 \left( \frac{2}{t} \right) \right] \right\}
$$
where $t \equiv kT/J$ is the dimensionless temperature and $K(x)$ is a complete elliptic integral of the first kind. We can then (in principle) find a closed-form expression $C = \partial U/\partial T$.
Further, the net mean magnetization is known to be
$$
m = \begin{cases} \left[ 1 - \text{csch}^4 (2/t) \right]^{1/8} & t < 2/\ln(1 + \sqrt{2}) \\ 0 & t > 2/\ln(1 + \sqrt{2})
\end{cases}
$$
The question is then:
Is there a known closed-form expression for the magnetic susceptibility $\chi$ of the 2-D Ising model at zero field?
My (limited) intuition tells me that there should be, because energy and heat capacity are related to the first and second derivatives of the partition function with respect to $\beta$, and we have closed-form expressions for both of those quantities. Similarly, the magnetization and susceptibility are related to the first and second derivatives of the partition function with respect to the external field—but I have not been able to find a source that discusses a closed-form expression for $\chi$, only for $m$. Am I just looking at the wrong sources, or is there not actually a known expression for $\chi$ at zero field?
| As far as I know, there is no closed form for the partition function in the presence of a field. Therefore there is no closed form for the susceptibility.
The closest I know of can be found here. Below are some screenshots of the relevant part of the paper. If you cannot get your hands on it, give me your email, I will send it to you.
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Could you feel your weight falling through the a tube drilled through the center of the earth? Suppose you drill a hole through the center of the earth (assume the earth is uniform and no air resistance) and you jump in. Would you be "weightless" throughout the entire fall?
The reason I ask is that if you jump off a cliff, throughout the entire fall you feel weightless (just like when astronauts train for the weightless feeling in orbit, they practice by going in an airplane and having the airplane fall to approximate the experience). Does this same weightless experience happen when you are falling through the center-of-the-earth tube?
I know that if you are stationary at the center of the earth, then you are weightless; but, I'm interested in falling throughout the entire hole.
The reason why I'm confused is that it's well-known that when you fall, you oscillate (simple harmonic motion) up and down the tube and this oscillation seems to imply that you will feel your weight.
| You would experience the feeling of weightlessness throughout the entire vibration as you move back and forth through the Earth. You hint at this in your question when you make reference to astronauts training for weightlessness in aircraft. The aircraft in this case is not simply moving down (and horizontally) but rather flying in a parabolic arc. Throughout the entire up and down motion the passengers experience the same sensation of weightlessness. You could think of this as half of an oscillation.
As explained in Cleonis's answer weightlessness will be experienced anytime the force of gravity is the only force acting on a person.
| {
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Correlation of the strength of the Balmer lines with the age of a galaxy I don't understand why a strong absorption line H$\delta$ indicates a young star population.
First of all, a strong H$\delta$ line just means that I have more A-type stars, doesn't it? Where is the connection to the age of a galaxy?
| A-stars have a main sequence lifetime of < a billion years and are much more numerous in a burst of star formation than the higher mass, and much more luminous O/B stars. The Balmer lines, particularly H$\delta$ are particularly prominent in A-stars compared with both hotter and cooler stars.
Lots of H$\delta$ tells you that the light is being dominated by main-sequence stars that are aged between about 200 Myr (after which the more luminous OB stars have died) and 1 Gyr, after which the A-stars will have died and H$\delta$ would be considerably weaker in cooler, longer-lived stars.
This simple argument applies to a single burst of star formation - a simple, single stellar population. Prolonged, or multiple epochs, of star formation will confuse the issue and require more detailed modelling.
| {
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Regarding Lenz's Law presented in hyperphycsics The following diagram is presented in hyperphysics as an introduction of Faraday's Law and Lenz's Law.
If the red arrows represent the direction of current, then what do the positive and negative poles across the resistor means? From my understanding, resistors do not produce a potential difference therefore poles shouldn't be present in the first place.
|
From my understanding, resistors do not produce a potential difference
Whenever there is a current $I$ through a resistor with resistance $R$, there is a potential drop $V$ across the resistor equal to $V=IR$. The potential gradient is always opposite the direction of the (conventional) current. This is why in the diagrams the current is moving from + to -. If you were to put the ends voltmeter at those locations in the diagram you would find the signs of your readings to match the diagram.
The diagram is not showing "permanent poles" that are inherent to the resistors themselves. The signs are dependent on the current itself.
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Does a random number generator have real entropy? In thermodynamics, entropy is defined for gases. Of course, my laptop is not a gas. However, it contains a random number generator and I have seen the word ‘entropy’ being used in this context. Is this the same entropy? How can this entropy be linked to the definitions from thermodynamics?
UPDATE
I think my question is different from this question. That question is about information content, for example a book. However, this question is about the entropy of a random number generator. Those seem to be different because the contents of a book are fixed while the output of a random number generator is not yet fixed.
| The answer is simply that a random number generator's entropy and Thermodynamic entropy are not the same things.
A couple of distinct features of thermodynamic entropy is the following.
*
*Every thermodynamic entropy depends on a variable corresponding to the thermodynamic energy of the system.
*Thermodynamic entropies are defined for equilibrium states of macroscopic systems.
In the absence of these two conditions, whatever concept we give the name of entropy will be useless as a thermodynamic quantity. Said in another way, non-thermodynamic entropies are entirely decoupled from thermodynamics.
Nothing prevents us from associating the name of entropy with one or more concepts connected to random number generators (RNG). For instance, we can associate an RNG with Shannon's or Kolmogoroff's entropies (conceptually different entropies). However, no established or reasonable concept of equilibrium can be associated with RNGs, and there is no energy they depend on. We could say that we have a composite system consisting of the material the computer is made of and the RNG, and that the total entropy is the sum of the thermodynamic entropy of the hardware plus Kolmogoroff's (or Shannon's) entropy of RNG. However, that's all. There is no way to speak about a redistribution of energy maximizing the total entropy.
Notice that the decoupling between different entropies does not mean that the common features of the two definitions can't help shed light on the concept itself.
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Laser safety of standard laser pointer diode I am thinking about designing vector display with standard pointer laser diodes. I would make display be diffuser of some sort like milky glass or paper, but I want to make it see through, rather than seeing dot on paper. Would that be okay or is it possible to mess up somehow?
| It is absolutely possible to mess up, but you can take measures to understand the consequences.
The most useful thing is to understand the classes of lasers. I'm a fan of this image from Laser Safety Facts.
I would recommend taking the second most conservative approach, and assuming that your diffuser fails completely. Laser safety isn't all about what happens when things go well, its about how things go when stuff breaks too! Your laser should be classified as one of these levels (although beware: cheap products may be more powerful than they are supposed to be. This table will show what happens if the laser diode's output directly hits someone's eyes at various distances.
I mentioned this was the second most conservative approach. When dealing with Class 2 lasers, its good to make sure that none of the failure mechanisms can create a lens. The safety of this cannot be easily captured in a table. There are professionals who do this (which is why the table has so many "Consult an LSO" entries).
I am not a lawyer nor a professional that deals with laser safety, but in general this table should provide you the information you need to make an educated decision about how safe your laser is.
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In an entangled system, what happens to Alice's wavefunction right after Bob makes a measurement? Suppose two entangled particles are far apart. One is with Alice and the other is with Bob. The relative velocity between Alice and Bob is zero (and spacetime is flat), so that we can define a notion of simultaneity that is agreed upon by both observers.
Before measurement, the joint is wavefunction of the particles is $\frac{1}{\sqrt{3}}|up, down\rangle +\frac{\sqrt{2}}{\sqrt{3}} |down,up\rangle$. The second spin label is of Bob's particle
Suppose, at time $t_0$, Bob measures his particle and observes $|down\rangle$.
Can we say that, after time $t_0$, Alice's description of the joint system should become : $|up, down \rangle$?
Or should it become:
$$\frac{1}{\sqrt{3}}|up, down, \text{Bob measured up}\rangle +\frac{\sqrt{2}}{\sqrt{3}} |down,up, \text{Bob measured down}\rangle$$
If the first option is correct, how does it not violate locality? I am thinking that the first option involves the information, that Bob has made a measurement, to instantaneously travel to Alice's end.
| If Bob observes "down", due to the wavefunction collapse the full system will be described by :
$$
\left|down,up\right\rangle
$$
There is indeed some sort of "spooky action at a distance", but this action cannot be used to transfer information.
When Bob observes "down", he will instantaneously know the state of the particle in Alice's possession, and yes, that state will instantaneously change for Alice. However, this cannot be used to transfer information. This is mainly because Bob cannot choose what he observes. He will observe $1/3$ of the time "down", and $2/3$ of the times "up".
Using entanglement, you can create correlations which are not possible to make without using the entangled pair (this is at the core of Bell's inequalities), but this does not violate "locality" in the sense that no information or signal can be transmitted faster than light. You are right that there is "something" non-local happening at the moment of the wavefunction collapse.
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How do we prove that the 4-acceleration transforms as a 4-vector in Special Relativity? In order to define the acceleration of a body in its own frame, we need to first prove that the acceleration is a four-vector so that its dot product with itself can then be labeled as acceleration squared in the rest frame. For velocity and displacement vectors, we can show that they have a constant dot product. But how do we prove that for acceleration?
| Since you accept that four-velocity is a four-vector, this is an argument that four-acceleration is a four-vector:
$$a^{\mu}=\lim _{h\rightarrow 0} \frac{v^{\mu}(\tau+h)-v^{\mu}(\tau)}{h}$$.
The path is parametrized by $\tau$, the proper time, which is a scalar because it's equal to the spacetime interval (upto maybe a sign)
You can imagine carrying out this limit calculation in two different frames. If you're doing a numerical calculation, you will take $h$ to a small finite number.
The numerator will be a difference of four-vectors, hence it is a four-vector. The denominator is a scalar. Hence, the fraction is a four-vector.
This isn't a proof of course. A proof will try to argue that the limit of the sequence four-vectors, as $h\rightarrow 0$, will also be a four-vector (which makes sense sort-of)
EDIT Ok so, let's say $v^{\mu}(h)$ is a sequence of four-vectors paramerized by a real parameter $h$. Let $v^{\nu} (h)=\Lambda v^{\mu}(h)$, $\Lambda$ is a Lorentz transform.
Then,
$$\lim_{h\rightarrow 0}v^{\nu} (h)= \lim_{h\rightarrow 0}\Lambda v^{\mu}(h)$$
$$=\Lambda \lim_{h\rightarrow 0} v^{\mu}(h)$$
So you see, the limit in one frame is the Lorentz transform of the limit in another frame (We could pull $\Lambda$ out of the limit because it's a constant matrix)
| {
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What is the entropy of the system, surroundings, and universe of a reversible adiabatic process? I had a homework question that confused me a little bit. We were told to find the change in entropy of the system and the universe of a process. This process is an adiabatic, reversible process. From what I understand, reversible indicates that change in entropy of the universe is zero. Adiabatic means the change in entropy of the surroundings is zero. This would mean that the entropy of the system is zero. I feel like I am missing something here.
| You are not missing anything.
The change in entropy of the system for a reversible adiabatic process is zero. There is no transfer of entropy to or from the system because there is no heat transfer to or from the system due to it being an adiabatic process. Likewise since only heat can transfer entropy to the surroundings the entropy of the surroundings is also zero.
Finally, since a change in entropy of the system is the sum of entropy transfer and entropy generation, there is no generation of entropy in the system because the adiabatic process is reversible (carried out quasi-statically and without friction).
Hope this helps.
| {
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Why does a sensitive thermometer absorb little heat? In an experiment to measure the specific heat capacity of water I'm trying to make it as accurate as possible. And somewhere I read that a sensitive thermometer absorbs little heat. By "sensitive" I am referring to the amount of change in thermometric property for a unit change in temperature.
| The sensitivity of a thermometer is defined as the smallest temperature change which can be measured, thus if you have two mercury in glass thermometers and the only difference between them is the size of the bulb, the one with the larger bulb will be more sensitive.
The thermometer with the smaller bulb will react faster to temperature changes and affects the temperature readings less as it has a smaller thermal capacity.
It appears that in the answer given (and hence the question) the word sensitive has been used incorrectly and the question should have been in terms of the thermal capacity of the thermometer.
| {
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Why does using images that are not really formed work in ray optics? It's all in the title. For instance, if I have two lenses , I have been taught to first find the position of the image formed by the first lens, and then use that image to find the final image formed by the 2nd lens, if the first image is formed beynd the 2nd lens. Why does this work?
edit:- image for reference
| Because the pattern of rays emerging from the first image is (more or less) the same as the pattern of rays that would emerge from a real object at that point.
| {
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Definition of momentum We say that momentum is the measure of how a body is moving or the quantity of movement inside a body
But what this definition really mean?
This terms are very vague
$p=mv$,why the movement inside the body depend on it's mass?
| The definition is simply $\mathbf{p} = m \mathbf{v}$.
In classical mechanics, this quantity is conserved in absence of an external net force, since the second principle of dynamics reads
$\dfrac{d \mathbf{p}}{dt} = \mathbf{F}^{ext}$,
and thus $\mathbf{p}(t) = \text{const.}$ if $\mathbf{F}^{ext} = \mathbf{0}$.
In order to understand the meaning of the mass in the momentum, you need to evaluate the influence a force (and the impulse of the force, i.e. its integral in time) in changing the momentum
$\displaystyle m (\mathbf{v}_1 - \mathbf{v}_0) = \mathbf{p}_1 - \mathbf{p}_0 = \int_{t_0}^{t_1} \dfrac{d \mathbf{p}}{dt} dt = \int_{t_0}^{t_1} \mathbf{F}^{ext}(t) dt = I$.
| {
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Optical theorem Peskin and Schroeder I'm trying to understand the optical theorem of Peskin and Schroeder
$$\tag{7.50} \text{Im} M(k_1,k_2\rightarrow k_1,k_2)=2E_{cm}p_{cm}\sigma_{tot}(k_1,k_2\rightarrow\text{anything})$$
which Peskin and Schroeder says follows from
$$\tag{7.49} -i[M(a\rightarrow b)-M^\ast(b\rightarrow a)]=\sum_f\int d\Pi_f M^\ast(b\rightarrow f)M(a\rightarrow f)$$
and $$\tag{4.79} d\sigma=\frac{1}{2E_A2E_B|v_A-v_B|}\times\prod_f\frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f}\times |M(p_A,p_B\rightarrow\{p_f\})|^2(2\pi)^4\delta^{(4)}(p_A+p_B-\sum p_f)$$
and $$\tag{4.80} \int d\Pi_n=\prod_f\frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f}\times(2\pi)^4\delta^{(4)}(P-\sum p_f)$$
But how did we integrate (4.79) to get (7.50)?
| I am also reading the book, I may not provide an answer, as far as I can understand:
$-i[M(k_1,k_2\rightarrow k_1,k_2)-M^*(k_1,k_2\rightarrow k_1,k_2)]=2 Im (M(k_1,k_2\rightarrow k_1,k_2))$
In eqn (4.79), if we consider $(E_A,\vec{p_A})=(E_1,\vec{p_{cm}})$ and $(E_B,\vec{p_B})=(E_2,-\vec{p_{cm}})$ as two incoming particles, and if we are considering two particles of same mass, ie $m_1=m_2$, we have $E_1=E_2$ and the particles are traveling in z-direction only, we have $2E_A=E_A+E_B=E_1+E_2=E_{cm}$ then in eqn (4.79)
$2E_12E_2|v_1-v_2|=(E_1+E_2)(2)|\vec{p}_{cm}-(-\vec{p}_{cm})|=E_{cm}(4)p_{cm}$
If we integrate over every $p_f$ in 4.79 and sum over them, and define
$\sigma_{tot}(k_1,k_2\rightarrow anything)$
$ =\frac{1}{4E_{cm}p_{cm}} \Sigma_n(\Pi_i^n\int \frac{d^3q_i}{(2\pi)^3}\frac{1}{2E_i})|M(k_1k_2\rightarrow {q_i})|^2 \times (2\pi)^4\delta^{(4)}(k_1+k_2-\Sigma q_i)$
... work from (4.79)
$\Sigma_n(\Pi_i^n\int \frac{d^3q_i}{(2\pi)^3}\frac{1}{2E_i})|M(k_1k_2\rightarrow {q_i})|^2 \times (2\pi)^4\delta^{(4)}(k_1+k_2-\Sigma q_i)$
is just the RHS of 7.49, so
$2 Im (M(k_1,k_2\rightarrow k_1,k_2))$
$=\Sigma_n(\Pi_i^n\int \frac{d^3q_i}{(2\pi)^3}\frac{1}{2E_i})|M(k_1k_2\rightarrow {q_i})|^2 \times (2\pi)^4\delta^{(4)}(k_1+k_2-\Sigma q_i)$
$=4E_{cm}p_{cm}\sigma_{tot}(k_1,k_2\rightarrow anything)$
| {
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What are the consequences of the tidal friction on the galaxy? In the Earth-Moon system tidal-friction slows down the rotation, so does it do the same for the galaxy?
*
*If not how come, and why it's different than on Earth?
*If so can this slow-down be the reason for collapse of the matter to quasars? If not why?
Sorry if this is a silly question but I was not able to find anything on Google.
| Tidal friction is caused by big objects that can "feel" gradient in the gravitational field of another object. It works for moon, because it's diameter is close to the distance from Earth to Moon. Moon deformation requires energy that goes from orbital rotation. But objects, orbiting galaxy has too small sizes to feel gradient of galaxy's gravity field.
If so can this slow-down be the reason for collapse of the matter to quasars? If not why?
Yes it can, the difference in matter's orbital velocity make friction that consumes orbital energy, but it's not the only reason of collapse. Near the black holes gravity field is so big, that another parts of Taylor expansion appears, field is not only $1/r^2$ anymore and orbits become unstable.
| {
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Relation between velocity and mobility of electrons and holes I have been studying band theory and semiconductors in condensed matter physics and I am confused about the relation
between mobility and velocity of electrons and holes in semiconductors.
My standard text book reference, Introduction to Solid State Physics, by Charles Kittel, says this:
i.e., the velocities of electrons and holes are the same in a semiconductor.
However, I was also reading about the dependence of the Hall coefficient on temperature and found this:
Now I can't understand how the mobilities of electrons and holes are different if their velocities are the same. What am I missing here?
Also, intuitively why should the mobilities be different for electrons and holes? Does it depend on doping too? Holes are just the gaps left behind by electrons and can practically be regarded as positive versions of electrons. Is it due to the mass factor coming into play due to electrons having some mass but holes being massless? Even then, holes should be more mobile than the electrons, right?
|
The velocity of a hole is equal to the velocity of the missing electron
The missing context: in the same band.
Typically we talk about "electrons" in the conduction band and "holes" in the valence band. But the quote is talking about electrons filling or leaving a vacancy in the valence band.
The velocity of the hole is the same as the velocity of the electron whose absence means a hole exists.
Also, intuitively why should the mobilities be different for electrons and holes (does it depend on doping too?) ? Holes are just the gaps left behind by electrons and can practically be regarded as a positive version of an electron.
Charge carriers in a semiconductor behave differently than their free space counterparts. An electron in the conduction band may have a different effective mass than a hole in the valence band.
Holes are the missing electron, but in a different band with a different effective mass. So they act differently.
| {
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How does thermal motion tend to change the direction of electron spin? Purcell(3rd edition, Chap-11, p-548) in context of paramagnetism writes
Thermal agitation tends always to create a random distribution of
spin axis directions.
I have trouble understanding how do molecular collisions change the direction of electron spin?
My thought process: Assume that each molecule of a gas filled in a container contriubutes one upaired electron.
Now what's the direction of it's spin ? Well random , we don't know .
Now if we turn ON external magnetic field, then spin would align parallel to B (taking uncertainty principle into account).
Now, how exactly does thermal motion changes the direction of spin ? I understand that collisions can change the direction of axis of rotation(AOR) of molecules. So is direction of spin somehow related to the direction of AOR ? (seems unlikely).
| A clue is already hidden in your question. You understand that the spin is influenced by the external magnetic field. If you replace spin with magnetic dipole, the phenomenon of thermal disturbance of the alignment of the spins becomes much clearer.
The disturbance occurs because an electron is not only an electric charge, but also a magnetic dipole. To operate here with the spin only complicates the matter. The valence electrons are (as long as we are not talking about an ionised gas) bound with their magnetic moment to the total moment of the atom / molecule. In an external magnetic field, this total moment is rotated in the direction of the external field as well as the valence electrons themselves. However, the thermal collisions keep disturbing this alignment. The phenomenon occurs up to temperatures before the Bose-Einstein condensate.
| {
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Does Proca's hypothesis make sense of giving mass to the photon in reference to special relativity? The Romanian physicist Proca formulated his famous Lagrangian to describe a hypothetical massive photon. From it we derive, as equations of motion, the relations that the electric and magnetic fields must obey (the analogous of Maxwell's equations for massive photons). We know that Einstein deduced the Lorentz transformations by raising the constancy of the speed of light to a principle. In this sense, giving a mass to the photon would mean going against this principle. However, we also know that transformations between inertial systems can be obtained based on the principles of relativity, isotropy and causality. So there are two types of groups (excluding space-time inversions)
which simultaneously satisfy this principles: Galileo's and Lorentz. In the last $c$ appears as limit speed that cannot be exceeded and which, a priori, is not related to the speed of light. Despite this, however, we choose to consider the Lorentz transformations and to discard those of Galileo due to the fact that, since Maxwell's equations must be invariant, the speed of light must remain the same in all inertial reference systems. So, even in this way, Proca's hypothesis seems to make no sense. Can you explain to me why the Proca hypothesis makes sense?
Furthermore, in this perspective, since the speed of light can no longer be the limiting speed in the Lorentz transformations, what meaning should we give to $c$?
| Calling $c$, the parameter in Lorentz transformations, the speed of light is something of a misnomer. It should more properly be called speed of massless particles -- or as robphy suggested in the comments, '“maximum signal speed” (putting more emphasis on the causal structure).' Whether or not the photon is actually a massless particle is not relevant for the logic of special relativity.
| {
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Velocity in power calculations in different inertial frames In calculating power using the formula $\underline{F}\cdot\underline{v}$, what is the correct velocity to use? Does one use the velocity of the body on which the force is acting, or the velocity of the body providing the force? I always thought it was the former (at least because in the case of force fields the field doesn't have a velocity, so the only velocity is that of the body the force is acting on).
However, when I use this understanding on an example problem I seem to end up with results about power calculations in different reference frames that I am struggling to make sense of. I have posted this question here for you to see the numbers.
Any clarity people can provide on this point (either in general or in specific relation to the example question I posted) would be much appreciated.
|
Does one use the velocity of the body on which the force is acting, or the velocity of the body providing the force?
The velocity of the body on which the force is acting. More specifically, the mechanical power delivered to a system is $\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material of the system at the point of application of the force $\vec F$ on the body.
If the velocity of the material of the system at the point of application of the force is the same as the velocity of the environment at the point of application of the force, then all of the mechanical energy that leaves the environment enters the system. If the velocities differ then mechanical energy is being destroyed at the point of contact, e.g. it is converted to heat with sliding friction.
I seem to end up with results about power calculations in different reference frames that I am struggling to make sense of
Mechanical power is frame dependent, but energy is conserved in all frames. Things like the mechanical power converted to heat from sliding friction is the same in all frames, even though the amount of mechanical power transferred varies.
| {
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Why is the mass of small elements taken as $∆m$ in center of mass of a continuous body? A continuous body has continuous distribution of mass. Doesn't $\Delta m$ mean $m_f - m_i$? But, is the mass Changing? If yes, how is the mass varying? Why is the mass of the small elements in a body taken as $\Delta m$? Why isn't it taken just as $m$ (mass of the small element)?
| It's a good question.
Normally speaking, we consider a quantity written as $\Delta x$ to conceptually mean a "change in a quantity". For example a "tiny change in time" or a "tiny change in position".
However, the usage in case of center of mass calculation is in a different conceptual sense. We can think of chopping up the body into some chunks with each chunk being of mass:
$$ dm = \rho dV$$
Here I am considering that we have a 3-D body but it could be that we have 2d or 1d as well.
In a way, we can think of the two approaches as actually being the same. Suppose you had a rod of length L lying along the x-axis from $x=0$ to $x=L$. You can imagine a situation where tiny rods of mass $dm$ are placed consecutively with tail of one at head of previous as time passes to reconstruct the original rod.
| {
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Connection between covariant derivative operators upon conformal compactification I'm having trouble determining the connection between two covariant derivative operators. These are: the one associated with the original space-time (and thus with the metric $ \tilde{g}_{ab}$) and the one associated with the non-physical one, which is acquired by a conformal compactification of the original space-time, i.e. $g_{ab}=\Omega^2\tilde{g}_{ab}$.
The two derivative operators are, using the notation as above, $\tilde{\nabla}_a$ and $\nabla_a$, the old and the new one, respectively. It can be shown that their effect is the same on scalar fields.
My main goal is to somehow determine the difference between their actions on a tensor, namely the metric $\tilde{g}_{ab}$.
It can also be shown that the difference between them can be completely characterized by a tensor field $C^{c}{}_{ab}$.
From this we have:
$\left(\nabla_a-\tilde{\nabla}_a\right)g_{bc}=-C^{e}{}_{ab}g_{ec}-C^{e}{}_{ac}g_{be}$
by definition $\nabla_ag_{bc}=0$ and I calculated that $\tilde{\nabla}_ag_{bc}=2\Omega\tilde{g}_{bc}\tilde{\nabla}_a\Omega=2\Omega\tilde{g}_{bc}\nabla_a\Omega$
From these is it true that $\left(\nabla_a-\tilde{\nabla}_a\right)g_{bc}=-2\Omega\tilde{g}_{bc}\nabla_a\Omega$ ?
(keep in mind that $\tilde{g}_{ab}$ is the original metric and $g_{ab}$ is the conformally rescaled one)
| I redid the calculations (and included them for future readers). The first $\Omega$ in your equations should be $\Omega^{-1}$, but despite that the last formula is correct (and I think the previous ones are as well).
The Christoffel symbol transforms as:
\begin{align*}
\widetilde\Gamma_{ab}^d
&=\frac{1}{2}\widetilde g^{cd}\left(\partial_a\widetilde g_{bc}+\partial_b\widetilde g_{ac}-\partial_c\widetilde g_{ab}\right) \\
&=\underbrace{\frac{1}{2}\Omega^{-2}g^{cd}\cdot\Omega^2\left(\partial_a g_{bc}+\partial_bg_{ac}-\partial_cg_{ab}\right)}_{=\Gamma_{ab}^d}
+\frac{1}{2}\Omega^{-2}g^{cd}\cdot 2\Omega\left(g_{bc}\partial_a\Omega+g_{ac}\partial_b\Omega-g_{ab}\partial_c\Omega\right) \\
&=\Gamma_{ab}^d
+\Omega^{-1}\left(\delta_b^d\partial_a\Omega+\delta_a^d\partial_b\Omega-g_{ab}g^{cd}\partial_c\Omega\right).
\end{align*}
The Weyl transformation $\widetilde{g}_{ab}=\Omega^2g_{ab}$ is often written as $\widetilde{g}_{ab}=e^{-2\omega}g_{ab}$ with $\Omega=e^{-\omega}$ to simplify calculations. You can see it here, the two appearances of $\Omega$ get shortened to one appearance of $\omega$:
$$\widetilde\Gamma_{ab}^d
=\Gamma_{ab}^d
-\left(\delta_b^d\partial_a\omega+\delta_a^d\partial_b\omega-g_{ab}g^{cd}\partial_c\omega\right).$$
Now considering the difference between covariant derivatives, the difference between partial derivatives vanishes and with the relation between Christoffel symbols, we get:
\begin{align*}
\left(\nabla_a-\widetilde\nabla_a\right)g_{bc}
&=\left(-\Gamma_{ab}^d+\overline\Gamma_{ab}^d\right)g_{cd}
+\left(-\Gamma_{ac}^d+\overline\Gamma_{ac}^d\right)g_{bd} \\
&=-\left(\delta_b^d\partial_a\omega+\delta_a^d\partial_b\omega-g_{ab}g^{de}\partial_e\omega\right)g_{cd}
-\left(\delta_c^d\partial_a\omega+\delta_a^d\partial_c\omega-g_{ac}g^{de}\partial_e\omega\right)g_{bd} \\
&=-g_{bc}\partial_a\omega-g_{ac}\partial_b\omega+g_{ab}\partial_c\omega
-g_{bc}\partial_a\omega-g_{ab}\partial_c\omega+g_{ac}\partial_b\omega \\
&=-2g_{bc}\partial_a\omega.
\end{align*}
Using $\nabla_a\Omega=\partial_a\Omega=-\Omega\partial_a\omega$, this is not your result, but it would be, if there was a $\Omega^{-1}$ instead of $\Omega$ at the front, which results from the inverse metric tensor $\widetilde{g}^{ab}=\Omega^{-2}g^{ab}$ at the front of the Christoffel symbol.
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Why does high frequency have high energy? The electromagnetic spectrum's wavelengths all travel at the same speed, $c$. Also, the wavelength $\lambda$ and frequency $\nu$ are related by $c = \lambda \cdot \nu$. Since all moving particles here would have the same speed, why would higher frequencies have more energy?
| The relation just means that the wave can be decomposed (very roughly said) into quanta that have certain energy. For higher frequency waves the individual quanta have bigger energy than for the lower frequency waves. That is an experimental fact - many phenomena can only be explained by this. But the wave itself may contain almost an arbitrary amount of energy (provided it is much more than the energy of the individual quantum). It would just be decomposed into fewer or more quanta.
Since the light is completely relativistic and massless, it is better to not think about the speed of the particle when arguing about the energy. The correct relativistic energy-momentum relation is
$$
E^2 = (pc)^2 + (mc^2)^2,
$$
and since the light quanta are massless, the energy is just $E=pc$. The momentum is universally related to the wavenumber ($k=2 \pi /\lambda$) by $p=\hbar k$. This also holds for massive particles according to the the de-Broglie hypothesis. This relation is central to the wave-partical duality in quantum mechanics. The energy of a massless particle is then $E = \hbar k c = h \nu$.
All this does not explain why nature works like that, that must be showed by experiment. It just connects the experimental facts using some theory.
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Why does a small thermocol ball fall slower than a metal ball of the same volume and surface area (air resistance equal)? Suppose a thermocol ball and a metal ball of same volume and surface area (but different masses, obviously) are dropped from the same height from rest. The acceleration due to gravity is 'g' and the air resistance is also same in both the cases, then why is it that the metal ball reaches the ground first?
| Let $R$ be the resistive force.
Applying Newton's second law $mg-R = ma$ where $a$ is the acceleration of the ball.
Thus $a= g-\frac Rm$.
Assuming that the density of the metal ball is greater than that of the thermocol (made of expanded polystyrene beads) ball then the mass of the metal ball is greater than that of the thermocol ball, which in turn means that the second term on the right-hand side of the equation, $\frac Rm$, is smaller for a metal ball, and so the acceleration of the metal ball is greater this its time of travel is smaller than the thermocol ball.
| {
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Lorentz contraction using odometers? In principle, would cars moving between a pair of points at different speeds show different odometer readings due to Length contraction? When we use odometers to measure length between two points, what can we say of length contraction?
| Yes, in theory the odometer will measure a contracted length. To show that rigorously is quite a complicated matter, because the rotating wheel of the car is no longer circular in the frame of the car or of the road.
Relative to the road, the bottom of the wheel is stationary, the centre of the wheel is moving at the speed of the car, and the top of the wheel is moving faster still (you have to use the relativistic velocity formula to work out the speed).
Relative to the car, the section of the bottom of the wheel that is in contact with the road is moving at the same speed as the road, and will therefore be length contracted by the same amount as the road.
Generally speaking, the circumference of the wheel is no longer 2pi times the radius in either the frame of the car or the frame of the road.
Of course, that is all hypothetical, since you could never accelerate the wheel of a car to produce a measurable relativistic effect.
If you want to get into the subject in more detail, google the Ehrenfest paradox.
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What is the correct gravitational potential energy of a single particle in an $N$-body system? I am aware that the total gravitational potential energy of a system of $N$ particles is given by pairwise interactions, i.e., you start with a single particle in the system, and then calculate the work done (negative for an attractive force) to bring in every other additional particle. Like this:
$$U_{total}=-G\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}\frac{m_im_j}{r_{ij}}\tag{1}$$
However, does it make sense to talk about the gravitational potential energy of a single particle? Something like this:
$$U_i=-Gm_i\sum_{j=1,j\neq i}^{N}\frac{m_j}{r_{ij}}\tag{2}$$
However, as can be seen from equation 1, summing over these "individual" gravitational potential energies would result in pairwise interactions being counted twice. Thus, would this:
$$U_i=-Gm_i\frac{1}{2}\sum_{j=1,j\neq i}^{N}\frac{m_j}{r_{ij}}\tag{3}$$
... be a correct equation for the gravitational potential energy of the $i^{th}$ particle in an N-body system? At the very least, using equation 3 to calculate the potential energy of each particle would result in the correct total potential energy for the system when summing the inividual energies of the particles.
Any insight would be much appreciated.
| The total potential energy of a system of $N$ particles interacting through Newton's gravitational force can always be written either as
$$U_{total}=-G\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}\frac{m_im_j}{r_{ij}}\tag{1}$$
or as
$$U_{total}=-\frac{G}{2}\sum_{i=1}^{N}\sum_{j=1;j \neq i}^{N}\frac{m_im_j}{r_{ij}}\tag{2}.$$
In the form $(2)$, each pair of bodies is counted twice, then the necessity of a factor $1/2$.
Notice that $U_{total}$ is a property of the whole system. However, it is possible to interpret formula $(2)$ as the sum over all the bodies of the potential energy of one body at the time, in the presence of the remaining $N-1$. Therefore, your equation $(3)$.
| {
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Intuitive understanding of the unit $kg/s^3$ — the unit of sound volume I know that decibels are used to measure volume of sound, and they are basically a logarithm of $kg/s^3$. The best explanation that I have for the unit of $kg/s^3$ is that it is an alternate way of saying $W/m^2$, which measures intensity of sound. What is an intuitive explanation for the unit $kg/s^3$? What do the kilograms and each factor of a second represent? Is it a way of measuring the variation in how many kilograms of air are hitting your ear every second?
| Not every combination of base units has a clear physical meaning. The best way to understand any quantity is to look at the equation that produces it. For instance in your case sound intensity $I$ is an expression of the power $P$ traveling through a perpendicular area $A$,
$$I=\frac P A$$
So you correctly stated that the most natural unit for this is $\rm{W/m^2}$. I could choose to write this equivalently as $\rm{kg/s^3}$ or $\rm{kg \cdot Hz^3}$ or $\rm{eV/(millenia \cdot acre)}$. The units do not provide the meaning. What matters is the physical relationship, expressed in the equation.
| {
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Second quantization: Hamiltonian in field operators vs Tight binding form Hamiltonian written in terms of field operators: The kinetic energy (KE) part of Hamiltonian is
$$
H=-\frac{\hbar^2}{2m}\int d\mathbf{r} \Psi^\dagger(\mathbf{r}) \nabla^2\Psi(\mathbf{r}) \tag{1}
$$
Hamiltonian written in tight-binding form:
One reaches to tight-binding form by transforming the field operators as
$$
\Psi(\mathbf{r}) = \sum_i \phi_i(\mathbf{r})c_i \quad ; \quad
\Psi(\mathbf{r})^\dagger = \sum_i \phi_i^*(\mathbf{r})c_i^\dagger \tag{2}
$$
here $i$ goes over all the lattice sites inside the system, and $\phi_i(\mathbf{r})$ is the wavefunction of $i$-the lattice site (assume only one state per lattice site). When we put this transformation in $(1)$, we get
$$
H=\sum_{ij}c_i^\dagger \left[-\frac{\hbar^2}{2m}\int d\mathbf{r} \phi_i^*(\mathbf{r}) \nabla^2 \phi_j(\mathbf{r})\right] c_j
$$
$$
H=\sum_{ij}c_i^\dagger t_{ij} c_j\tag{3}
$$
Question:
In Hamiltonian $(1)$, the field operators create and destroy particles at $\mathbf{r}$ position. This position vector $\mathbf{r}$ includes all the lattice sites (for a discrete system). In the tight-binding Hamiltonian, the same job is done by operators ($c_i^\dagger, c_i$). So, can we say that the discrete version of the field operator Hamiltonian $(1)$ is equal to the tight-binding Hamiltonian $(3)$? What I mean is to discretize the position vector and $\nabla$ in $(1)$
$$
H=-\frac{\hbar^2}{2m}\sum_i \Psi^\dagger(\mathbf{r}_i) \nabla^2\Psi(\mathbf{r}_i)\\
H=-\frac{\hbar^2}{2m}\sum_i \Psi^\dagger(\mathbf{r}_i)
\left(\frac{\Psi(\mathbf{r}_i+a) - \Psi(\mathbf{r}_i) + \Psi(\mathbf{r}_i-a)}{a^2} \right) \tag{4}
$$
here $a$ is lattice constant, and $\Psi^\dagger(\mathbf{r}_i)$ creates particle at site $i$. This is exactly what $c_i^\dagger$ does. So, can replace $\Psi^\dagger(\mathbf{r}_i)$ with $c_i^\dagger$:
$$
H=-\frac{\hbar^2}{2ma^2}\sum_i
\left(c_i^\dagger c_{i+a} - c_i^\dagger c_{i} +c_i^\dagger c_{i-a} \right) \tag{5}
$$
Is not $(5)$ similar to $(3)$? Why do we need tight-binding model when we can just discretize field operators?
| The operator $\Psi^\dagger(\mathbf{r}_i)$ creates a particle at $\mathbf{r}_i$ which is infinitely localized (i.e. it has a delta-function wave function.) In contrast, $c_i^\dagger$ creates a particle which is localized at position $\mathbf{r}_i$, but has wavefunction $\phi_i(\mathbf{r})$. You should think of this as essentially an atomic wavefunction (e.g. 1s, 2s, 2p, etc.), not a delta-function.
The choice of $\phi_i(\mathbf{r})$ depends on the problem, and the chemistry involved in your materials. Take the simplest case: Assume you are dealing with a chain of hydrogen atoms. Here, $\phi_i(\mathbf{r})$ should be taken as the 1s orbitals localized on each hydrogen atom. Of course, you could in principle include in your tight-binding model the 2s,2p,3s,3p,3d,etc. orbitals, and this should make your model more accurate. However, such complications are often unnecessary to describe the essential physics of the problem.
| {
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Does it matter when we measure the spin of the other entangled particle? Let's say, we have 2 entangled particles: A and B which are a light-year away from each other.
We know if we measure the spin of particle A, we can be certain the spin of particle B will be in the opposite direction.
My question is: does it matter "when" we measure the spin of particle B?
E.g., if we measure the spin of particle A on the 27th of November 2022 at 14:15, and let's say the spin is UP, can we always be certain that if we measure the spin of particle B, let's say millions of years in the future, its spin will always be DOWN? Or must the measurement of particle B also be at the exact same time as A for us to be certain of its spin to be DOWN?
| No, it doesn't matter.
Say you have the entangled state:
\begin{equation}
|\psi_0\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle_A|\downarrow\rangle_B + |\downarrow\rangle_A|\uparrow\rangle_B \right)
\end{equation}
If A makes a measurement and finds, say, spin up, then the state $|\psi_0\rangle$ collapses into:
\begin{equation}
|\psi_0\rangle\longrightarrow|\psi\rangle=|\uparrow\rangle_A|\downarrow\rangle_B
\end{equation}
So it doesn't matter when particle B is measured, because the state is what it is.
This is assuming there's nothing that acts externally to change the state $|\uparrow\rangle_A|\downarrow\rangle_B$ in time. If that was the case, then yes it would matter, but only because the state would be evolving in time.
| {
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How can the Joule be the unit of both work and energy?
Say a person applies 1 N to a box with a mass of 1 kg, displacing 1 m. This is one Joule of work
(1 N for 1 m).
Now say the person applies 1 N to a box with double the mass, displacing 1 m as well. This is still one Joule of work, despite the person having to push with the same amount of force for longer.
So the person expended more energy (pushing with the same effort for longer), but does the same amount of "work". How can the joule be used to measure energy then?
| Work is a transfer of energy from one system to another (by any means other than heat). So it must have the same units as energy.
Similarly, if I transfer 5 EUR to your account then your account balance is 5 EUR larger. The transfer is work, and the balance is energy. They must be in the same units. It wouldn't make sense to maintain a balance in EUR but do transfers in liters.
So the person expended more energy (pushing with the same effort for longer), but does the same amount of "work". How can the joule be used to measure energy then?
In your example never use human beings to figure out work and energy. We are marvelously inefficient machines. Instead use an ideal spring or something similar. Two springs in your example would use the same energy. Yes, it would take longer for the spring in B to move the block, but the energy expended would be identical. The spring would operate at a lower power for a longer time.
| {
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What is the difference between mechanics and analog? What is the difference and the relationship between mechanics and analog/analogue? I have noticed that mechanical things are often considered analog.
Note: The difference between digital and analog is clear to me.
| Originally the term "analog" came from computing, where mechanical computing machines implemented systems that were analogous to the physical systems they simulated. When electronic analog computers came about, a key component was the "operational amplifier". Because they were developed to support math, it turned out that they were easier to understand through abstract circuit theory than simpler devices like triodes. That's convenient.
These days, if you want an electronic amplifier, you usually implement it as a simple analog computer, computing the desired output from the input. Opamp-based integrators and differentiators are useful building blocks of electronic signal processors. Other computation circuits like logarithmic amplifiers and multipliers are also useful.
So, as electronics processing signals understood as continuous variables evolved to resemble analog computers, "analog" came to mean "continuous", as opposed to the discrete signals a digital system processes.
| {
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Linear transformation in $pV$ Say ideal gas is in a state $p_1$, $V_1$, $T_1$ (known). I compress this gas such as $\frac{dp}{dV}$ is a constant. I want to reach $T_2$>$T_1$ and I need the values for $\frac{dp}{dV}$. Would it be right to picture the problem like that: say I have a point $(p_1,V_1)$ in a $pV$ diagram, and I draw the isothermal corresponding to $T_2$. Then would it be right to consider any line connecting $(p_1,V_1)$ to this isothermal as correct, as long as it corresponds to a compression?
| It certainly is correct. You need to get to a point with $T_2$, so the endpoint lies somewhere on the $T_2$ isotherm. The condition $\frac{dp}{dV}=\text{const.}$ means that the curve connecting the startpoint with the endpoint is really just a line segment. And there you have it, this is exactly what you have said.
| {
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What is the state of an entangled photon after its twin is absorbed? Let's two photons are entangled in polarization after a laser beam passes through a Betha Barium Borate crystal. They take different paths and one of them (1) is absorbed in a black sheet. What is the state of the leftover photon (2)? Is it in superposition of polarization h/v or it must flip spontaneously in a certain polarization? What if the black sheet atoms absorb photons only with a certain polarization (say h)? Will the absorbed photon (1) take h polarization in the process of absorption and hence the second twin flip to v?
| Take the initial state of the pair and project onto the outcome of the measurement.
| {
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How to calculate radioactivity concentration of components in reactor? I understand the reactor parts will become radioactivity because there are neutron flux in reactor. How should we calculate the reactivity of reactor parts? For example lets say there is a reactor core with thermal neutron flux of 1n/cm^2. If there is a steel board been exposed directly to the neutron flux for this 10 years, how do we calculate its radioactivity concentration?
Can anyone give me some key words I could look into?
| There are laws of activation (and deactivation) of materials. In a simple way, the activity is proportional to the number of target nuclei, to the flux of neutrons, to the cross section of activation.
In practice, these basic notions become very complex:
You must know the exact composition of the materials, take into account the radioactive filiations (Batman's equations), have the irradiation diagram (running periods, power of the running stages, reactor shutdown time), neutrons reactor are not mono-energetic, the cross sections vary with the energy, which will lead to multigroup calculations at each point of the material.
But all this is satisfactorily resolved today.
There are powerful neutron calculation codes in 2d, 3d,
taking into account all these variables, some codes even using probabilistic Monte-Carlo methods.
The application of these codes is the work of a specialist in neutronics using these codes full time.
| {
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Why are there hemispheres in Huygens principle? In the formulation of Huygens principle, it is said that the secondary waves are spheres but on the plots for determination of the wavefront there suddenly become hemispheres? What is the reason for this mismatch? Is there any physical explanation based on the momentum of the incoming wave in the sense that it influences the atoms of the media? What media does one have in case of EM waves? Maybe Huygens does not have any physical meaning but is just an observational geometrical rule?
|
In the formulation of Huygens principle, it is said that the secondary
waves are spheres but on the plots for determination of the wavefront
there suddenly become hemispheres?
The whole sphere is often not shown as a convenience in creating the plot. Also, showing only half of the sphere avoids the issue of the backward wave.
See my https://www.nature.com/articles/s41598-021-99049-7 ("Huygens' Principle geometric derivation and elimination of the wake and backward wave") for a discussion. This paper also shows how the backward wave is eliminated when the secondary waves, the Huygens' wavelets, are whole spheres. No inclination factor is needed.
| {
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Paradox regarding Young-Laplace equation Recently I've been working with the Young-Laplace equation and came across the following physical paradox that I couldn't wrap my head around. It goes like this:
Imagine a spherical droplet (filled with water) in air. By the Young-Laplace equation, we know that the pressure in the spherical droplet is larger than the pressure in the air:
$$P_{\rm droplet}-P_{\rm air}=\frac{2\gamma}{R}$$
Where $\gamma$ is surface tension and R is the radius of the droplet.
Now, lets take a stationary fluid element that encompasses both side of the droplet. It is clear that the pressure on both sides of the fluid element are not equal, and by Newton's third law, all internal forces within the fluid element cancels out.
In this case, what keeps the fluid element stationary?
| The Young-Laplace equation you give is exactly the force balance you are looking for.
Because it is curved (with total curvature $2H=2/R$), the surface area element inside the fluid element exerts a certain portion of its surface tension, namely $2H\gamma = \frac{2\gamma}{R}$, in the radial direction (add up the vectors from a curved portion of surface where all elements are pulling on each other: components parallel to the surface cancel, but small inward radial components add up).
The Laplace pressure $\Delta P$ that arises is defined as exactly the pressure difference needed to balance out the force from surface tension and keep the element stationary.
| {
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Generalized conjugate momenta and the generalized Fourier transform Let the Lagrangian be a functional of $\hat{\phi}$ and $\partial_{\mu}\hat{\phi}$, i.e. $\hat{L} = L(\hat{\phi},\partial_{\mu}\hat{\phi})$, where $\hat{\phi}$ is an operator.
The conjugate momenta is defined as $$\hat{\pi} = \frac{\delta \hat{L}}{\delta \partial_0{\hat{\phi}}}. \tag{1}$$
Let the eigenfunctions and eigenvectors of $\hat{\phi}$ and $\hat{\pi}$ at $t=0$, be defined as:
$$\hat{\phi}(0,{\bf x})|\phi\rangle = \phi({\bf x})|\phi\rangle\qquad\text{and}
\qquad\hat{\pi}(0,{\bf x})|\pi\rangle = \pi({\bf x})|\pi\rangle.\tag{2}$$
Question:
Based on (1) and (2), is it possible to derive that:
$$\langle \phi|\pi\rangle = \exp\left ( \frac{i}{\hbar}\int \phi({\bf x})\pi({\bf x}) d^3x \right )~?\tag{3}$$
Is the relation (3) valid for any arbitrary Lagrangian $L$?
| *
*Quantization of a theory starting from a classical Lagrangian density ${\cal L}(\phi,\partial\phi)$ is a non-trivial process. Let us assume that it is possible for OP's theory.
*There might be operator ordering issues when performing a Legendre transformation from the Lagrangian to the Hamiltonian formulation at the quantum level. To avoid this, in what follows, it will be more relevant to use the momentum operator $\hat{\pi}$ from the Hamiltonian formulation.
*Granted the equal-time CCR
$$ [\hat{\phi}({\bf x}),\hat{\pi}({\bf y})]~=~i\hbar\delta^3({\bf x}\!-\!{\bf y})\hat{\bf 1},$$
and the definition (2), then one may derive the overlap
$$\langle \phi |\pi\rangle ~=~\exp\left\{ \frac{i}{\hbar}\int_{\mathbb{R}^3} \!d^3{\bf x}~\phi({\bf x})\pi({\bf x})\right\} \tag{3}$$
(up to a phase factor, which is conventionally put to 1), cf. e.g. this and this related Phys.SE posts.
--
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Is energy "equal" to the curvature of spacetime? When you are solving the Einstein field equations (EFE), you basically have to input a stress–energy tensor and solve for the metric.
$$
R_{\mu \nu} - \frac{1}{2}R g_{\mu \nu} = 8 \pi T_{\mu \nu}
$$
For a vacuum solution we have:
$$
T_{\mu \nu} = 0
$$
This yields:
$$
R_{\mu \nu} = 0
$$
This means that the local curvature of an inertial frame of reference is zero.
But, setting the stress–energy tensor to zero, could be given in multiple situations: In flat spacetime, around a non-rotating black hole, around a planet, etc.
When I read about this equations in divulgation books, they portrait the Einstein field equations as:
$$
\text {Space-Time Curvature} = \text{Energy}
$$
But with this interpretation, by setting $T_{\mu \nu} = 0$, you are saying that the energy is zero, hence no curvature, but you are able to get more solutions than the Minkowski metric (which is the only solution with truly no energy and with no curvature).
Are this books interpretations wrong or is there something I'm not getting from the true meaning of the equation? How would you distinguish, while solving the EFE, from a truly flat spacetime, from a locally flat spacetime?
| The curvature of spacetime is proportional to energy.
Spacetime curves around that energy.
If you imagine a sphere with energy at its centre then the curvature is the same at every point on the surface of the sphere.
| {
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Is Hubbles law due to Gravity? Hubble's law states that Distance is proportional to Velocity. A ScienceDirect article states that Classical Hubble expansion is characterized by a proportional increase in the rate of expansion groups based on the distance from the main center of gravity
So is it due to gravity?
| No, Hubble's law is due to the expansion of the universe.
I can't say what the ScienceDirect article meant because you didn't link it, but the description of what it wrote does not sound sensible either.
Edit: from the link, the relevant quote is:
A special place in astrophysics is the effect of Hubble—expanding groups of galaxies, accompanied by a proportional increase in the rate of expansion groups based on the distance from the main center of gravity.
What this means is that if you have a group of galaxies,* then when space expands, the galaxies appear to recede from one another at a speed given by the Hubble Law. The Hubble Law itself is still caused by expansion of space.
*Note "groups of galaxies" (i.e. galaxy clusters) are gravitationally bound - that is, they do not expand because their gravitational attraction is strong enough to overrule cosmic expansion.
| {
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Given a magnetic field how to find its vector potential? Is there an "inverse" curl operator? For a certain (divergenceless) $\vec{B}$ find $\vec{A} $ such that $\vec{B}= \nabla \times \vec{A} $.
Is there a general procedure to "invert" $\vec{B}= \nabla \times \vec{A} $? An inverse curl?
(I was thinking of taking the curl of the previous equation:
$$ \nabla \times \vec{B}= \nabla \times \nabla \times \vec{A} = 0. $$
Then using the triple cross product identity $ \nabla \times \nabla \times \vec{V} = \nabla (\nabla \cdot V) - \nabla^2 V$ but that does not quite simplify things... I was hoping to get some sort of Laplace equation for $\vec{A}$ involving terms of $\vec{B}$.)
| You could try using the Helmholtz decomposition.
If $F$ is a twice-differentiable vector field on a bounded volume $V$ with boundary $S$, then it can be decomposed into divergence-free and curl free components.
$$F=-\nabla\Phi+\nabla\times \mathbf{A}$$
where
$$\Phi(\mathbf{r})=\frac{1}{4\pi}\int_V\frac{\nabla'\cdot F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dV'-\frac{1}{4\pi}\oint_S\mathbf{\hat{n}}'\cdot\frac{F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dS'$$
$$\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_V\frac{\nabla'\times F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dV'-\frac{1}{4\pi}\oint_S\mathbf{\hat{n}}'\times\frac{F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dS'$$
$\mathbf{\hat{n}}'$ is the unit outward normal and $\nabla'$ is the gradient with respect to $\mathbf{r}'$ rather than $\mathbf{r}$.
The curl on its own does not have a uniquely-defined inverse. However, the curl and divergence can be combined into a single operator that does have a unique inverse up to boundary conditions.
If the fields are assumed to approach zero at infinity, the boundary integral in each of the above expressions becomes zero. If the field is required to be solenoidal, then the divergence in the first expression will be zero.
| {
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Absorption spectrum dependence on concentration/pressure I am having difficulties understanding the relation between absorption spectrum and gas concentration. The online resources I found, including many questions here, do not clarify my doubts.
Take for example this picture from wikipedia. It lacks any information about what concentration/partial pressure of each species results in the given absorption spectrum:
I don't understand how the absorption spectrum in the picture above relates to the concentration of the species in the atmosphere. For example, assuming that the concentration used for that spectrum is the average earth atmospheric concentration (i.e. 340ppm of CO2), I wonder what happens when the concentration halves. Do we get only 50% absorbance in the 4.3 microns band for CO2 or do we get a bit more than that because CO2 will still be fairly opaque at this concentration/wavelength?
| The Beer-Lambert Law says that the transmittance of a sample is given by $T = e^{-n \sigma \ell}$, where $\ell$ is the path length, $\sigma$ is the absorption (or scattering) cross-section of the scattering species, and $n$ is the concentration of the species (molecules per volume.) If you halve the concentration ($n' = n/2$), then we can see from the above formula that the new transmittance is $T' = e^{-n' \sigma \ell} = (e^{-n \sigma \ell})^{1/2} =\sqrt{T}$.
In the case of the graphs above, it's important to note that what you're calling "100%" absorption is really something like 99.999999% absorption and 0.000001% transmission. There's never going to be perfect absorption in this sort of situation, because the scattering/absorption process is probabilistic. In general, the above argument says that if a fraction of $1 - T$ of the photons are absorbed before you halve the concentration, then $1 - \sqrt{T}$ will be absorbed afterwards. Exactly how big of a difference this makes depends on exactly how close you are to 100%. If you were scattering 99% of the light before, then you'll be scattering only 90% after. If it was 99.999999% $(1 - 10^{-8})$ before, then it will be 99.99% after ($1 - 10^{-4}$).
The graph below shows the new absorbance $A'$ (vertical axis) as a function of the old absorbance $A$ (horizontal axis). The fact that the slope of the graph becomes infinite as $A \to 1$ reflects the idea that it really matters how close to 100% absorbance you are. On the other hand, at very low absorbance (near $A = 0$), you can see that halving the concentration simply halves the absorbance.
| {
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Is there any physical difference between the absence of a force and a force with a magnitude of 0 Is there some sort of experiment one could perform that would differentiate between the absence of a force and a force with a magnitude of 0? Or are they just different ways of describing the same thing?
| Kinematatticaly they are the same but dynamically, there will be stresses built up in the object experiencing a net force of zero. Consider a spring for example, when you pull on its ends, even though the force is zero, there are still stresses in it.
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Impact of distance from galactic centre on the value of energy in the cosmic ray spectrum where knee is observed? This question is based on the recommendation and great explanation by @Kyle_Kanos. Is it known what causes the "knee" in the observed Cosmic Ray spectrum?
Accepting the reason for the occurrence of knee around a few PeV energies of Ultra High Energy Cosmic Rays (UHECR) to be the shift in majority source from intergalactic to extra-galactic due to Larmor radius becoming comparable to the sscale height perpendicular to the galactic plane. Does that mean that observers farther from the centre would observe the value of knee energy to be lower due them getting more vulnerable to lower energy extra-galactic sources than the knee value for observer closer to the centre?
| For a highly relativistic proton $v\approx c$, moving with momentum $p\approx\gamma mc$, the Larmor radius is given by -
\begin{equation}
r_L = \frac{p}{eB} \approx \frac{\gamma m c}{eB} \approx \frac{E}{eBc}
\end{equation}
So, higher the energy, lesser will be the deflection due to linearly increasing Larmor radius with energy for a relativistic particle. Due to this, for energies larger than the value for which the Larmor radius becomes comparable to the scale height of the galaxy in the direction perpendicular to the disk plane, the extragalactic cosmic rays cease to deflect away significantly, hence start to dominate the cosmic ray spectrum after that value. So, this 'knee' of sudden change could be caused by shift in majority source from intergalactic to extragalactic.\
So, the location of this knee should depend majorly on the galactic scale height at that location and average strength of magnetic field within the galaxy. The scale height(for a spiral galaxy similar to milky way) is further dependent on size of the galaxy and distance from the centre. I will use a few approximations to estimate the proportionality expression for scale height in terms of these quantities.\
Denoting D as the size of galaxy, d as the distance from the centre and t as the scale height there. I will assume that the shape of galaxies of these categories are about the same, therefore for a galaxy of twice the size 2D, the scale height at twice the distance 2d, from the centre should be twice 2t. Also, scale height is assumed to be decreasing linearly with distance from centre, which gives-
\begin{equation}
\frac{t2}{t1}=\frac{D2}{D1}\Big(\frac{D_1(d-D_2)}{D_2d_1-D1D_2}\Big)=\frac{D_2-d}{D_1-d_1}
\end{equation}
The Larmor radius, which is a measure of deflection at knee energy value, should be of order of this scale height. So,
\begin{equation}
\frac{E_1}{eB_1c}\approx t1
\end{equation}
Therefore,
\begin{equation}
\frac{E_1}{E_2}\frac{B_2}{B_1} = \frac{t1}{t2}= \frac{D1-d1}{D_2-d}
\end{equation}
| {
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What would a standing wave of light look like? I want to know what a standing wave of light would like and what properties it might have that are interesting.
| If you could stand at a node in a perfectly round monochromatic light two optic resonant cavity, you would see a small circle of light. If you could stand at the anti-node you would see a large circle of light.
No monochromatic light source is prefectly round so inside a real resonant cavity the light will have a central beam and what we call junk light around the central beam that will bounce around in the cavity by not produce a standing wave.
As the light leaves the resonant cavity it's no longer a standing wave and its properties are changed by the optics the light goes through or bounces off of, so what you see outside of the resonant cavity isn't what you would see inside the cavity.
| {
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What happens to circularly polarized light when it hits a linear polarizer? I have seen a lot of examples of what happens when circularly polarized light passes through a circular polarizer composed of a quarter-wave plate and a linear polarizer, but what would happen to the circularly polarized light if it passed through only the linear polarizer without a quarter-wave plate?
The reason I’m asking is because I’ve heard that in photography linear polarizers can cut through smog but not through fog, which generates circularly polarized light. This seems strange, because one would assume that the linear polarizer would absorb all of the circularly polarized light and would thus cut through the fog. Why is this not so?
| Circularly polarized light is composed of 50% linearly horizontally polarized light and 50% linearly vertically polarized light. The two linearly polarized components are 90 degrees out of phase. If perfectly circularly polarized light passes through a perfect linear polarizer that transmits horizontal (vertical) polarized light then the output will be perfectly linearly horizontal (vertical) polarized light with 50% of the power of the incident circular polarized light. The answer is the same whether the incident light is right- or left-hand circularly polarized.
| {
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Star convection gradient dependencies In star convection you can work with radiative and adiabatic temperature gradients, that for example the radiative one can be defined as,
$$\nabla_{rad}=\frac{3\kappa L P}{16 \pi a c G m T^4}$$
This can be seen for example in this source. Here they put emphasis in the dependency of luminosity $L$ and mass $m$ with radius, i.e. $L=L(r)$ and $m=m(r)$. But otherwise for the other star variables is not specified,
*
*$\kappa$ the opacity
*$P$ the pressure
*$T$ the temperature
As is not specified, my doubt is if this variables are used as radius dependent ones or fixed values. In this case what variable should I use? The star-center values, or the surface values?
| The opacity, pressure and temperature all vary as a function of position in a star. For a spherically symmetric star you can assume they vary only as a function of the radial coordinate $r$.
The reason that luminosity and mass are written as $L(r)$ and $m(r)$ is to remind us that these do vary with radial coordinate and avoid confusion with the luminosity and mass of the entire star, which are meaningful and observable quantities. $L_* = L(R)$ and $M_* = m(R)$ respectively, where $R$ is the radius of the star.
| {
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Does the 1-loop correction for $g\bar\psi\psi$ term leads to the counterterms for additional terms in the Lagrangian? I have the Lagrangian in 4 dimensions:
$$
L = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\sum_{i =1,2}\bar\psi_i(i\not\partial-m)\psi_i-g\phi\bar\psi_i\psi_i.
$$
Assuming there are $n$ external scalar lines, and no external fermion lines, then the superficial degree of divergence could be found as $\omega = 4-n$.
Suppose $n= 3$, I can draw this divergent diagram, with the counterterms (a) and (b):
If I now add another term $g_3\phi^3$ to this Lagrangian, we can draw another divergent graph (c) with the same superficial degree of divergence. My question is does the counterterm stay the same if I add this term? I'm not quite sure how to make sense of this.
Actually, with $n = 2,3,4$, we will have this same question. Is it right if I say the 1-loop correction for the $g\bar\psi\psi$ term leads to the need of counterterms for $\phi^2$, $\phi^3$, and $\phi^4$?
| Yes, one must for consistency as a minimum include all possible renormalizable terms that are not excluded by symmetry, cf. my related Phys.SE answer here. The $\phi^n$ vertex with $n=1,2,3,4,$ is e.g. generated from a 1-loop diagram of Yukawa $\phi\bar{\psi}\psi$ vertices with the fermion running in a loop.
| {
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Why $F = m(v_f - v_0)/2$? Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$ where $m$ is mass, $v$ is final velocity, and $v_0$ is initial velocity?
| One way to convince yourself of this is that force is measured in $\rm{kg\cdot m/s^2}$ which is a Newton. This can be formed by multiplying mass kg times velocity m/s divided by time s. You cannot get there by adding the units.
More generally, you cannot add any units (or the corresponding dimensioned quantities) unless they are the same. You can add two masses
1 kg + 2 kg = 3 kg
Or two velocities
2 m/s + 4 m/s = 6 m/s
Or two forces
1 $\rm{kg\cdot m/s^2}$ + 0.5 $\rm{kg\cdot m/s^2}$ = 1.5 $\rm{kg\cdot m/s^2}$
But you cannot add a mass and a length
1 kg + 2 meters =.....
It doesn't work.
| {
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NIF Ignition Achievement Although the National Ignition Facility achieving positive energy output is hailed as an achievement in fusion energy, I am curious whether their configuration could ever be adapted to produce continuous power. The fuel is at the center of a symmetric array of lasers, and it is hard to see how a "feed system" could be developed to pull in new fuel constantly. Moreover, their approach requires tritium, which is seriously lacking and would need an entire industry built up to produce enough of it to power any industrial fusion plant.
To be frank, the NIF seems like a curiosity more than a serious energy breakthrough. I don't wish to sound needlessly skeptical or dismissive, so my honest question is: are there legitimately feasible design concepts that adapt the NIF approach to generate continuous power?
| Yes, there is the Laser Inertial Fusion Energy (LIFE) concept. It includes a light-gas gun that would inject 15 targets per second into the target chamber, and a tritium breeding ratio of 1.05. Pursuit of this concept was canceled after ignition was not achieved as promised by the end of 2012. However, the repeated demonstrations of ignition in 2021/22 invite renewed conversations about LIFE and have also emboldened startups such as Focused Energy and Marvel Fusion, which are developing their own design concepts.
For LIFE to become commercially viable, the laser efficiency and repetition rate has to be increased, and the target fabrication further optimized. For example, the NIF lasers required $300$ MJ of input energy to deliver $2$ MJ laser energy to the target chamber, which in early Dec 2022 generated $3$ MJ of fusion energy. The LIFE concept uses laser diodes that are more efficient and allow a higher rep rate than the flashtubes currently in use at NIF. The repeated demonstration of ignition is an important milestone that has moved the promise of inertial fusion energy from 5-6 decades away to perhaps only a few decades (in the cautious words of the LLNL director at the 20 min mark here).
| {
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Multiplication of probability in quantum mechanics Consider a ket-space spanned by the eigenkets of an observable $A$ and let $B$ be an additional observable on the same ket-space. We can build a filter that only lets an eigenvalue $a$ of $A$ through and afterwards subject the outgoing beam to a second filter that selects an eigenvalue $b$ of $B$.
If $|\alpha\rangle$ is an arbitrary ket and we're interested in computing the probability of a particle whose state is described by $|\alpha\rangle$ to survive the ensemble described above, we can firstly compute the probability of measuring $a$ in $|\alpha\rangle$ which is simply given by:
$$
P_a = |\langle a | \alpha \rangle | ^2
$$
If the particle comes through the first filter, its state is now described by the ket $|a\rangle$. Just like before, we can calculate the probability of the particle to pass through the second filter:
$$
P_b = |\langle b | a \rangle | ^2
$$
I'm not sure if my question is completely trivial, but I'm having trouble understanding why the probability for the initial state $|\alpha\rangle$ to survive the whole ensemble is then given by the product:
$$
P_{tot} = P_bP_a = |\langle b | a \rangle \langle a | \alpha \rangle | ^2
$$
Do we have to postulate this? This result is well known for the computation of the probability of two independent events in mathematics, but I don't know if I can just directly translate such results to the probabilities of quantum mechanics.
| The issue is whether the probabilities are independent. I believe that this is an assumption of the notion of "projective measurements" which are approximations to what actually happens in an experiment. You really need a Hilbert space ${\mathcal H}_A\otimes {\mathcal H}_B \otimes {\mathcal H}_{\rm system}$ where $A$ and $B$ are the detectors measuring the system. You then ask what is the amplitude for the system to evolve from $|0\rangle\otimes |0\rangle\otimes |\alpha \rangle$ an end up in $|a\rangle\otimes |b\rangle\otimes |{\rm something}\rangle$. Whether the square of this is the product $|\langle a|b\rangle|^2| \langle b|\alpha\rangle|^2$ depends on the physics of the situation, but experiments are usually designed so that it works. Mott shows how the projective measuremnts are approximations to reality in hos famous paper on cloud chamber tracks.
| {
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Kinematics 1D-Problem In the below question, what does the phrase "the time taken by the particle to hit the ground" mean? Does it mean time taken by the ball to cover distance from the point of projection to the ground or does it mean the time taken by the ball to cover the distance from the point of projection when it is under free fall?
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is
a) 2gH = n²u²
b) gH = (n-2)²u²
c) 2gH = nu²(n-2)
d) gH = (n-2)²u²
| The time taken to reach highest point is,
$$t_1=\sqrt{\frac{2H}{g}}=\frac{u}{g} $$
And when the particle reach the bottom again, displacement in y is 0. So,
$$0=ut_2-\frac{gt_2^2}{2}⟹t_2=\frac{2u}{g}$$
So, $2t_1=t_2⟹n=2$. Now you can see according to your options.
| {
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Do water wheels slow down the water downstream in a river? Since the speed of the water may vary across the river, let us focus on the speed at the river mouth.
When a water wheel is placed in a river, part of the kinetic energy of the water is stored. Therefore, the water should flow slower than without the water wheel. This is the case for every point between the water wheel and the river mouth.
On the other hand, the total amount of water that enters the river annually (through glaciers, rain, etc) remains the same. Let us call this amount A. Now the total amount of water that exits the river anually is also A. Let $\sigma$ be the area of the cross-section of the river at the mouth. Now the speed of the water at the cross section is proportional to $A/\sigma$. This is independent on the water wheel. Therefore, the river does not flow slower with the water wheel.
How can these two be reconciled?
UPDATE
I would like to emphasize that my question is NOT about the flow rate of the water, but about the velocity. Here flow rate is in liters per second, while velocity is in meters per second.
Also, the question is about the water downstream of the water wheel, not upstream.
Finally, I would like to know what the difference is between the situation without the water wheel, and the situation with the water wheel when equilibrium is attained.
| The water wheel cannot change the flow rate of the water. If it reduced the flow rate then water upstream of the wheel would flow towards the wheel faster than the water below the wheel was flowing away. That means the depth of the water upstream would increase steadily with time and eventually overflow the river banks.
What the wheel does is increase the depth of the water immediately upstream from it. That is, it behaves like a small dam. Then the energy to rotate the wheel and grind your corn (or whatever) comes from the change in gravitational potential energy of the water as is flows down the height difference between the upstream and downstream sides of the wheel.
| {
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Does the formula for time period of a simple pendulum hold up for larger angles? When calculating the time period of a simple pendulum as an experiment in junior classes we used the formula
$$T = 2\pi\sqrt{\frac{l}{g}}$$
But recently seeing the derivation of the formula using Simple Harmonic Motion, I can't understand how it holds for such large angles since $\sin{\theta}$ gets approximated to $\theta$ which only holds true for very small values of $\theta$ (or as it would be written mathematically for $\theta \to 0$).
So, how, if at all, does this formula hold for the larger angles and why is it used for calculating time period of a pendulum during experiments?
| The exact solution is given by this equation:
$$T=4\,\sqrt{\frac lg}\int_0^{\pi/2}\frac{d\vartheta}{\sqrt{1-k^2\,\sin^2(\vartheta)}}\tag 1$$
where $~k=\sin(\theta_0/2)~$ and $~\theta_0~$ is the pendulum start amplitude.
Thus if $~\theta_0=0~$ you obtain that period $~T_L=2\pi\sqrt{\frac lg}$
This graph gives you period $~\frac{T}{T_L}~$ via the pendulum start amplitude:
Thus (for example) for $\theta_0=50~$ degrees the pendulum period is 5% greater than the approximation $~T_L~$.
| {
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Understanding this Lagrangian calculation I was trying to understand this section of a Wikipedia article:
$$0 = \delta \int \sqrt{2T} d\tau =
\int \frac{\delta T}{\sqrt{2T}} d\tau =
\frac{1}{c} \delta \int T d\tau$$
For the life of me, I can't figure out how does one get from $\displaystyle \delta \int \sqrt{2T} d\tau$ to $\displaystyle \int \frac{\delta T}{\sqrt{2T}} d\tau$. What is $\delta$? I assume it is some kind of differential operator but in respect to what variable?
| $T(\tau)$ is a function. $dT(\tau)$ is also a function that is "small" i.e. less than some $\epsilon$ for all $\tau$. $T(\tau) + \delta T(\tau)$ means you perturb the function by adding a small perturbation at all $\tau$. If you do this then $\sqrt{2(T + \delta T)} \approx \frac{\delta T}{\sqrt{2T}}$. Here you kind of Taylor expand the function at each $\tau$ (also the arguments of the functions are suppressed for brevity). $\delta(F(T))$ means the expression you get when substituting $T + \delta T$ for $T$ in $F$, so $\delta\sqrt{2T} \equiv \sqrt{2(T+\delta T)}$.
To better understand all of this I would suggest reading a good textbook on the calculus of variations. There are many around (this is a relatively old discipline of mathematics).
| {
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Second Law of Thermodynamics Restatement with usable energy instead of entropy Is it technically accurate to state the Second Law of Thermodynamics as:
"The total amount of usable energy only decreases in a closed system"
I ask because it doesn't evoke the term "entropy", which usually only confuses the average person.
| What you suggest is the so-called Exergy, a term introduced in the fifties for a concept (the available energy) that dates back to Gibbs. The decrease in exergy is the counterpart of the usual increase in entropy.
However, I notice that after more than sixty years, the concept of exergy has not substituted entropy. Entropy may be confusing for the average person, but exergy is by no means a simpler concept.
| {
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Why does the motor "flinches" when its motor starts spinning? I'm wondering why motor "flinches" when it starts spinning.
Is there any physics law causing this?
If so, if a satellite in space has a rotating parts in it then would the satellite also moves due to the same reason?
| Every action has an equal and opposite reaction. The torque that accelerates the rotor is associated with an opposite torque on the motor body. If the motor is on flexible mounts, that torque will rotate the body until the mounts can provide enough torque to balance it. Of course, the mounts then torque whatever they're connected to, but it may have sufficient inertia that you don't notice its response.
If the motor is delivering torque to something else, the effect is stronger: the mounts have to balance both the acceleration torque and the output torque.
Without the mounts, the body of the motor would continue to spin. That's the working principle of a reaction wheel, commonly used to orient spacecraft in space.
| {
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About vertical circular motion I have two cases here.
In the first case, we have a body moving at the speed shown. As we know the normal force on it due to the surface under it is just 9.8 * 10 = 98 Newtons.
Now, in the second case, the body is moving in a vertical circle. But at that instant at the top, the velocity vector is still pointing exactly to the right (just like in the first case). Why then would the normal force be different?
It seems to me that at that instant when the body is at the top of the circle, both cases are identical.
| Force and acceleration are vectors. I assume the magnitude of the velocity is constant for both of your cases in the question.
For the first case (purely horizontal motion) the velocity vector is in the horizontal direction and does not change direction or magnitude, so the horizontal acceleration is zero. That means the net horizontal force to the right (applied force minus friction) is zero. (Of course some net force to the right had to be applied initially to accelerate the body from rest to the constant velocity.) There is no motion in the vertical direction, since the vertical velocity is zero. The net force upwards (normal minus gravity) is zero. In this case the upward normal force merely counteracts gravity to keep the mass fixed in the vertical direction.
For the second case, uniform circular motion, the velocity vector has constant magnitude but changes direction as the mass moves around the circle. The motion is easier to visualize and evaluate using polar coordinates. A change in the direction of the velocity vector is acceleration. The (instantaneous) acceleration is radially inward to the center of the circle and has magnitude ${v^2 \over r}$ where $r$ is the radius of the circle. Therefore, the net force in the inward radial direction has magnitude${mv^2 \over r}$. In the absence of gravity, this force is supplied by a rope or the circular surface. (With gravity, the net external force is more complicated; since the force of gravity is always vertically downward.)
In the first case, the normal force is vertically upwards and there is no vertical acceleration. In the second case there is a net force radially inwards that provides acceleration.
| {
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liquid-vapor interface goldstone modes The free energy of a liquid-vapor interface (far from critical point) can be approximates as
$$
\mathcal{F}=γ A+\frac{γ}{2}\int dxdy|\nabla h(x,y)|^2
$$
where $h(x,y)$ is the height of the interface at a given point (x, y), $\gamma$ is the surface tension, and A is the area of a completely flat interface. What are the goldstone modes appearing in the system, and which symmetry breaking is responsible for them?
I imagine the way to solve such a question would be to decompose h into its fourier components, but I am not sure how to word the symmetry broken here.
| No symmetries are broken at the liquid-gas phase transition. Therefore no Goldstone modes are expected either.
| {
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Why does water feel hotter at larger volume? Why does a 104°F pool/tub feel boiling hot, whereas a pot/cup of water at the same temperature does not feel hot at all.
(Normally a pot/cup of water won't be hot enough to cause one to immediately remove ones finger from the water till it's around 165-175°F).
Probably same conductivity, specific heat, BTU's (since the amount of BTU's needed to raise 1°F is proportion to volume of water. Meaning, 1 BTU is needed per pound, so no matter what the volume is it will contain the same BTU's per pound).
Perhaps there's much more "heat" (BTU's/Joules) available in a tub/pool to "refill" the spot which transferred into ones body (perhaps through conductivity) not allowing the area of water touched by ones body in the tub/pool to cool off fast enough.
Another possible factor might be that perhaps there's an increase in convection; not sure if the higher the volume of water, the higher the convection.
| It depends on the surface area of your body that is exposed to the hot water. If you get into a hot tub, most of your body is exposed to the heat, whereas dipping your fingers into the water only exposes a small surface area.
The sensitivity of different parts of your body will also play a part. Your fingers and hands are less sensitive to heat than other parts of your body. This also applies to the soles of your feet, where the skin is thicker. Which is why a hot water bottle that feels pleasantly warm under your feet will feel much hotter against your legs. This is also why mothers with young babies are advised to test the temperature of their baby’s bath water with their elbow, which is more sensitive to heat than their hand.
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Telegrapher's equation and complex wave equations I am not sure if this falls into engineering or physics, but since I am confused more about the underlining mathematics, I am posting it here. For the transmission of a TEM wave, telegraphers equations are derived for the propagation of current and voltage over an infinitely small region of a transmission wire:
$$-\frac{\partial v(z,t)}{ \partial z} = R i(z,t) + L \frac{\partial i(z,t)}{\partial t} --- (1)$$
$$-\frac{\partial i(z,t)}{\partial z} = G v(z,t) + C \frac{\partial v(z,t)}{ \partial t} -- (2)$$
Often what is done is "complexifying" the voltage and current signal such that:
$$v(z,t) = Re(V(z)e^{i \omega t}) \hspace{1cm} i(z,t) = Re(I(z) e^{i \omega t})$$
Now, the phasor part is used to solve the equation yielding an equation of the form:
$$\frac{d^2 V(z)}{dz^2} = (R + i \omega L)(G + i \omega C) V(z) = \gamma^2 V(z)$$
Which is claimed to be a wave equation since the solution will be of the form: $V_0^+ e^{-\gamma z} + V_0^- e^{\gamma z}$. To convince myself that this is indeed some sort of travelling wave take the rightward travelling component and convert it back to the time domain as follows ($\gamma = \beta + i\alpha$):
$$Re(V_0^+ e^{-(\beta + i\alpha)} e^{i \omega t}) = V_0^+ e^{-\beta} \cos(\alpha + \omega t)$$
However, when I look at the original equations (1) and (2), and try to create a wave function in the original sense ($\partial^2_{xx} f = \alpha \partial_{tt}^2 f$), I get:
$$\frac{\partial^2 v}{\partial z^2} = -LC \frac{\partial^2 v}{\partial t^2} - GR v + (-RC - GL) \frac{\partial v}{\partial t}$$
Which obviously does not have the traditional form that I know how to solve, nor do I recognize the solution I got using phasors. Could someone help me conceptually reconcile this?
| Your last displayed equation is correct. The solutions of the telegrapher's equation describe damped (decaying) waves whose phase velocity depends on frequency unless the Heaviside condition $LG=RC$ is satisfied. They do not satisfy the usual wave equation unless both $R$ and $G$ are zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can you add pressure and internal energy in relativistic enthalpy? In relativistic fluid dynamics the relativistic enthalpy in natural units is defined as:
\begin{equation}
h = \frac{e+p}{\rho},
\end{equation}
Where $e$ is the total energy density and $p$ is the thermodynamic pressure.
In contrast, the Newtonian enthalpy is defined as:
\begin{equation}
h_N = \epsilon+\frac{p}{\rho},
\end{equation}
Where $\epsilon$ is the specific internal energy and $\rho$ is the rest mass density.
The question:
What are the units of $e$ and $p$ so that I can add them in the definition of $h$ above?
| So, following @Ghoster's comments, $e$ and $p$ simply have the same dimensions in both systems of units:
\begin{equation}
e = \underbrace{(M\ L^2\ T^{-2})}_{energy} \cdot \underbrace{L^{-3}}_{per\ unit\ volume} = \left( \underbrace{(M \ L \ T^{-2})}_{force} \cdot \underbrace{L}_{times\ length} \right) \cdot \underbrace{L^{-3}}_{per\ unit\ volume} = M\ L^{-1}\ T^{-2} \\
p = \underbrace{(M \ L \ T^{-2})}_{force} \cdot \underbrace{L^{-2}}_{per\ unit\ area} = M\ L^{-1}\ T^{-2} = e
\end{equation}
| {
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How to find the corresponding energy given the wave function? So I was struggling doing the following question:
Given the wave function $\psi(x) = A\, \mathrm{e}^{-ax^2} $ with potential $V = \frac12 kx^2$, find the corresponding total energy in terms of $k$ and $m$.
I did the calculation for $\left<x^2\right>$ and $\left<p^2\right>$ but it turns out to be expressions including $a$. How can I express $a$ in terms of $k$ and $m$?
| Since $\hat H\psi(x)=E\psi(x)$ just act on $\psi(x)$ using $\hat H$ and presumably the result is a multiple of $\psi(x)$.
Alternatively, to build on your calculation, if $\psi(x)$ is suitably normalized (and your expression is not):
$$
\langle H\rangle=\frac{1}{2m}\langle p^2\rangle+ \frac{1}{2}k\langle x^2\rangle
$$
and express $a$ in terms of $m,k,\hbar$ using the normalization of $\psi(x)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Atmospheric pressure in non-nertial frame? Any object kept in an accelerating container of water feels different pressure than unaccelerated. Because if we go into the frame of water the g effective changes. Since air is also a fluid, a container of liquid accelerating upwards should experience more atmospheric pressure than it feels at rest, but intuitively it does not feel so. Am I correct in assuming that liquid feels more atmospheric pressure? (Quantitatively, $P×(g+a)/g$)
| Existing answer is correct but I will present it in less technical terms.
If the air is not moving relative to the ground, then the atmospheric pressure at the ground is whatever is enough to prevent the air from the next layer up from falling down towards the ground.
If the whole system (ground and air) is accelerating upwards then in order to prevent the layer above from moving towards the ground, the pressure at the ground must be higher. Therefore the pressure is higher under that condition. It goes up in proportion to the "effective gravity" ($g+a$) as you suspected.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the Lorentz force equations, can the magnetic field be assumed to be singular at one or more points? Consider the relativistic Lorentz force equation (in a simplified form) given by
\begin{equation}\
\left(\frac{u'}{\sqrt{1-|u'|^2}}\right)'= E(t,u)+u'\times B(t,u).
\end{equation}
Here, $E$ and $B$ denote respectively the electric and magnetic fields and are given by
\begin{equation}
E=-\nabla_u V-\frac{\partial W}{\partial t}, \qquad B=\mbox{curl}_u\, W,
\end{equation}
with $V:[0,T]\times (\mathbb R^3\setminus\{0\}) \to\mathbb R$, $W:[0,T]\times\mathbb R^3\to\mathbb R^3$ two $C^1$-functions.
The models I have seen so far deals with $V$ singular at one (or more) points ($0$ in this case, e.g.). My question is: there exists a relevant physical model in which also $W$ is singular at one or more point?
| Of course, in case the potentials are due to a moving point particle, both scalar and vector potential are functions of distance with a component proportional to $1/r$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How distance affect wind speed from a fan? I noticed that when you feel the wind force from a fan close up, it feels like more force than from far away. Can someone give me an equation, where given a base wind speed $v$ in mph of the fan, and a distance $d$ in miles, one can get the wind speed from distance d away from the fan.
| Check Landau & Lifshits, vol.6 ("Fluid Mechanics"), ch.23 ("Exact solutions of the equations of motion for a viscous fluid"), item (3) "submerged jet". I believe this is the closest thing for an analytical expression for your problem.
| {
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Compressed water Normally, water is incompressible, but what happens if I have a fixed volume and keep pumping water into it?
Let's say I have a compartment full of water. The compartment is completely sealed off, with only one entrance where a pump is connected. What will happen if the pump keeps running, pumping water into the compartment even though it is full? Will the pressure rise? and is there a formula for this?
| for a positive displacement pump with a small motor on it, pumping into an enclosed volume will stall the pump. in the case of a large motor, it will burst the container, split the pipes, or break a connecting rod inside the pump mechanism.
For a non-positive displacement pump (like an impeller pump) the flow will stop but the pump keeps running, agitating the water inside the pump scroll. with a big enough motor, the impeller blades will begin whipping the water into vapor, a condition called cavitation.
In no instance of an ordinary mechanical pump working against water will the bulk compressibility of the water enter into the model of the pump/water system. For that to be significant you need high explosives machined into shaped charges.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Distinction between "types of heat" in thermal efficiency The definition of thermal efficiency I see in several sources is "total work" divided by "heat input".
Wikipedia, for example, says: "For a heat engine, thermal efficiency is the ratio of the net work output to the heat input".
I don't understand this definition. Net work is a perfectly valid concept, always given by $\oint pdV$. For a Carnot cycle, heat presents no problem, because the adiabatic processes involve no heat, while the isothermic processes are easily identified as consuming heat or producing heat.
However, for a general cycle this distinction between "heat input" and "heat output" is not clear. Just imagine a generic cycle in a $p-V$ diagram. How I am supposed to know which bits of the cycle are "heat input" and which are "heat output"?
| The way to track this in complete generality, for a reversible process, is to track the system entropy. If the entropy went up, then heat came in. If the system entropy fell then heat went out.
If you want a method that does not involve any mention of the concept of entropy, then proceed as follows. For a given change first calculate the work done on the system $W$, and then find out (from an energy equation or from previously acquired data) the change $\Delta U$ in the internal energy of the system. If $\Delta U > W$ then heat flowed in. If $\Delta U < W$ then heat flowed out.
| {
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Is it possible to statically generate lift with the difference in pressure like wings? If I understood it correctly, the shape of the wings and/or propellers generates lift/thrust with the difference in pressure in both sides of the wings/propellers; where the lower side has higher pressure airflow and the uper side has low pressure airflow.
With this in mind, I was wondering if it is possible to generate an area of low pressure around the upper part of the an aircraft without the moving balloons, wings or propellers/rotors.
A "static lift" is the best way I could put it.
So, would such thing be possible? Or lift would only be achieved with the airflow that wings already work around?
| The purpose of wings is to generate this lift in a dynamic situation.
If there is a low pressure area and a high pressure area in a system and they are connected, then air will flow from the high pressure area to the low pressure area. Air will go around the wing from the bottom to the top.
Something would have to be done to prevent it. There is a real life example of this: air ride suspension. On large trucks, a volume of high pressure air is trapped inside a rubber shock absorber so that it is not connected to the outside air and cannot travel from high pressure to low. Thus we have high pressure below the top of the shock, and low pressure above it. This generates lift, and is used to keep the semi trucks from falling to the ground.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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What fraction of heat is exchanged by thermal radiation (by ordinary objects)? I understand that heat can be transferred by conduction, convection and thermal radiation.
So, lets say when I bring a cold glass of milk in a room at room temperature, then what fraction of heat is transferred to the milk by radiation and what fraction by conduction?
(I guess that the heat exchange by thermal radiation would be very low)
| Heat transfer via radiation scales like the fourth power of the temperature in degrees absolute. It is very small for objects near room temperature like your glass of milk but climbs rapidly as you go above room temperature. at a couple hundred degrees above room temperature it is significant (think: hot stove) and a few hundred above that, it is dominant (think: glowing furnace). The guys on the engineering stack exchange can give you some formulas if you want to do the math.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does touching something cause quantum entanglement? Disclosure: I'm pretty uneducated when it comes to quantum entanglement.
I was aware that there are a few ways for particles to become entangled. Is physical interaction one of them?
I read this question and answer: Quantum entanglement question
Then I thought when I touch something, does that mean that trillion trillions of particles in my fingers have become partially or maximally entangled with the particles in the thing?
Say I touch a wall, is my hand partially quantumly entangled with the wall?
|
Say I touch a wall, is my hand partially quantumly entangled with the wall?
In principle yes, though in practice there is a process called decoherence that removes the entanglement of massive objects so fast that it could never be observed. So your hand is not entangled with the wall in any physically meaningful sense.
The Schrodinger equation for any system (including your hand is):
$$ -\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi = -i\hbar\frac{d}{dt}\Psi \tag{1}$$
and to understand entanglement the key parameter is the potential term $V$. Suppose we have $n$ particles at positions $r_1$, $r_2$ etc up to $r_n$. If the particles do not interact with other we can write this potential as a sum of a separate potential for each particle:
$$ V = V_1(r_1) + V_2(r_2) + \ldots + V_n(r_n) \tag{2} $$
In this case the Schrodinger equation can be separated into sum of separate equations for each particle, and the total wavefunction can be written as a product:
$$ \Psi = \psi_1(r_1)\psi_2(r_2) \ldots \psi_n(r_n) \tag{3} $$
This is what we mean by an unentangled system. Each particle is described by a wavefunction that does not depend on the positions of any other particle.
But suppose the particles do interact. Then the potential will contain terms that depend on the relative positions of the particles i.e. it will contain terms that depend on $r_i - r_j$. In that case the potential cannot be written in a form like equation (2), and the wavefunction will not have a form like equation (3). This is an entangled system. It's entangled because the position of one particle affects all the other particles.
This makes it clear what we mean by the term interaction. It is anything that causes the potential to depend on the relative positions of the particles. In the case of physically touching a wall the interaction obviously depends on the relative position of your hand and the wall, so your hand and the wall will (in principle) become entangled.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ represent a rotation operator? It is well known that $\exp(-i \sigma_k \theta)$ where $\sigma_k$ $(k=x,y,z)$ is a Pauli matrix, represents the rotation operator about $k$-th axis. What physical interpretation does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ have, where $\otimes$ is the tensor product?
|
It is well known that $\exp(-i \sigma_k \theta)$ where $\sigma_k$ $(k=x,y,z)$ is a Pauli matrix, represents the rotation operator about $k$-th axis. What physical interpretation does $\exp(-i \theta \sigma_m \otimes \sigma_n)$ have, where $\otimes$ is the tensor product?
If we use the common physics terminology and say that $\vec \sigma/2$ is an angular momentum (generator of rotations), then we can also say that $\vec \sigma_n\otimes\vec \sigma_m$ is not an angular momentum. It does not generate rotations, and it does not obey the correct commutation relations for an angular momentum.
Rather, the direct product of the two "spin-1/2" angular momenta $\vec \sigma/2$ can be decomposed into a direct sum of a "spin-1" or "triplet" angular momentum and a "spin-0" or "singlet" angular momentum. The symbolic equation for this "addition of angular momenta" is:
$$
\frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0\;,
$$
where the symbols refer to the "spin" value. Note that the spin 1/2 representation here is a 2x2 matrix, the spin zero representation is 1x1 (trivial) matrix, and the spin 1 representation is a 3x3 matrix, so the matrix dimensions work out correctly.
If you would like to get a better feel for what $e^{-i\theta \sigma_m\otimes\sigma_n}$ "looks like" you can expand the exponential and use
$$
(\sigma_m\otimes\sigma_n)^2 = 1
$$
to see that:
\begin{align}
e^{-i\theta \sigma_m\otimes\sigma_n}
&=1 + -i\theta \sigma_m\otimes\sigma_n + \frac{1}{2} \left(-i\theta \sigma_m\otimes\sigma_n\right)^2 + \frac{1}{3!}\left(-i\theta \sigma_m\otimes\sigma_n\right)^3+\ldots\\
&=\left(1 - \frac{\theta^2}{2} + \ldots\right)
-i \sigma_m\otimes\sigma_n \left(\theta - \frac{\theta^3}{3!}+\ldots\right)\\
&=\cos(\theta) - i\sin(\theta)\sigma_m\otimes\sigma_n
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Does rate of acceleration change as object gets closer or further to center of a mass? I learnt that newton's law of universal gravitation F = G(m1m2)/R^2, and thought if the R is distance and determined gravitational strength, why do we use 9.81 as default acceleration of earth's gravity when it is not even constant at different heights?
| The local gravitational acceleration at a height $h$ above the surface (of water) is given by $$g = \frac{G\,M_{\rm earth}}{(R_{\rm earth}+h)^2}$$
with $g_0 = 9.80665\;{\rm m/s^2} \approx 9.81$ when $h=0$ on average.
Your question is why use a constant value when there is a dependency on $h$?
Well, check how much does $g$ varies given reasonable values for $h$.
Heght, $h$ [m]
Gravitational Acceleration, $g$ [m/s^2]
Change
0
9.80665
0
100
9.80634
-0.003%
1000
9.80357
-0.03%
2000
9.80050
-0.06%
5000
9.79130
-0.16%
10000
9.77597
-0.32%
As you can see, most high school lever experiments take place under 10,000 meters of height, where 0.3% error due to gravity is perfectly acceptable.
| {
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"url": "https://physics.stackexchange.com/questions/747083",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Why is the current the same after passing through a resistor even when the drift velocity goes down? A resistor converts some of the electrical energy into heat energy, implying that the energy goes down, implying that the force with which an electron moves, and consequently, the drift velocity goes down.
Now, I=naeV where V is the drift velocity, so shouldn't the current go down after an electron has passed through a resistor?
I am familiar with the pipe-water-flow analogy, but my issue with that is it just involves water flowing, not the loss of any energy from that water.
Where am I going wrong?
This is with reference to steady state.
| Electrons are accelerated by the electric field, but slowed down by scattering with impurities, phonons, and other electrons. The velocity that enters the equation $I=nAeV$ is so-called drift velocity, i.e., the average velocity when both the acceleration of electrons by the field and their slowing down by collisions are accounted for. An analogy here is falling of an object with the constant speed, because it is slowered down by the friction against the air. This analogy is rather imperfect, but it serves as a basis of a very well known model of electric resistance.
Remark: The question also seems to confuse the steady state current, where the current is the same in all the cross-section in the circuit, and a transient regime - as could take place, e.g., when the switch has been just closed and the current in different parts of the circuit is different. During such a transient regime there may be regions in the circuit where the charge accumulates or becomes depleted, until the additional electric field created by this charge assures that the current is the same everywhere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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