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Do photons lose energy after radiation pressure is applied to a perfect reflector? I was readng Wikipedia article (English one) about radiation pressure because there is something I still cannot figure out. As I understand it, radiation pressure emerges from conservation of momentum. Photons or electromagnetic waves possess momentum and when they are absorbed, reflected or even emitted, the aborber/reflector/emitter experiences a pressure that is proportional to the irradiance (in watts per square meter).
OK. In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction. And I do not understand how something can be set in motion (the reflector), which amounts to do work, yet the source of this work does not lose energy. I mean, if the radiation, after a U turn, meets a second reflector, it would have pushed two reflectors a way, yet it would continue in its original direction as if nothing happened...? What am I missing here? Should not photons lose something?
| There will be a recoil momentum ($2p$ and energy ($2p^2/M$ of the mirror. The photon energy will therefore be $pc-2p^2/M$. For a mirror of one mole of glass (60g) and a photon of 500 nm (~2eV) that is a relative frequency change of $2p/Mc= 2/(6\cdot 10^{23}\cdot 10^9 \approx 3\cdot10^{-32}$. However this is not what will happen, as the mirror will not move as a rigid body. Instead a collective electronic excitation in a very thin layer of metal will absorb the momentum. As a consequence I guess that the momentum will be carried by some $10^9$ electrons and the effect will be a relative frequency change of somewhere around $10^{-10}$, give or take a few orders of magnitude.
| {
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Why do fluids not accelerate? A fluid flowing in a horizontal pipe must be flowing at a constant velocity because of the conservation of mass.
However, considering how there would be a pressure and hence force acting behind the fluid, for it to have a constant velocity, there must be an equal force slowing it down (depicted as $F?$).
I can't see a force that would be as big as the driving force. Can someone explain to me what this force is and how it's created?
| The thing missing from your picture is the steep pressure gradients (and associated acceleration) at the inlet of the pipe. Far from the pipe inlet at the bottom of the tank, the pressure will be approximately hydrostatic; near the pipe inlet, it is anything but. The same thing can be observed at the drain of an emptying bathtub. Once in the pipe, if the flow in the pipe is inviscid and the pipe is horizontal, then it will be at atmospheric pressure everywhere in the pipe (ignoring some subtleties like the finite pipe height in the vertical direction) after the flow becomes parallel, and the flow will not accelerate further.
If there is wall friction in the pipe (which there will be in real pipes), then there will be a force resisting the flow, the result of which is a pressure gradient along the length of the pipe (a force is required to overcome the wall friction). The qualitative picture isn't changed by wall friction however.
| {
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Solving a momentum problem without the assumption I encountered the problem in Physics HRK:
A stream of bullets whose mass m is each $3.8$ g is fired horizontally
with a speed v of $1100 \frac{m}{s}$ into a large wooden block of mass that is
initially at rest on a horizontal table. If the block
is free to slide without friction across the table, what speed will it
acquire after it has absorbed eight bullets?
The solution is already written in the book. It's pretty simple, however, it assumes that all $8$ bullets are in flight before the first bullet strikes the block. The next paragraph intrigues me:
For simplicity in solving this problem, we have assumed that the
bullets are fired so rapidly that all eight are in flight before the
first bullet strikes the block. Can you solve this problem without
making this assumption?
Wouldn't it be the same?
Edit: This is the book's solution simplified:
The initial (horizontal) momentum, measured while the
bullets are still in flight and the block is at rest, is
$P_{ix}=N(mv)$
in which $mv$ is the momentum of an individual bullet and $N=8$.
The final horizontal momentum, measured when all the bullets are
in the block and the block is sliding over the table with horizontal
velocity V, is
$$P_{fx} = (M+Nm)V$$
Conservation of momentum requires that
$$P_{ix} = P_{fx}$$
Solving for $V$ yields $V = 2.8 \frac{m}{s}$
| You will have to consider a series of incidents taking place.
When the first bullet strikes
*
*$$P_{f 1} = (M+m)V1=mv$$
After this as the mass adds to the block
*
*$$P_{f2} = (M+2m)V2=mv+(M+m)V1$$
.....`v is initial velocity of bullet
And so on you can calculate
`
| {
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Can rolling without slipping occur without friction? If a body is rolling without slipping is it necessary that there is friction acting on it ? I encountered a question in which there is a spherical body and a force is being applied on its top point ...so if there is only force then it should do translation motion only .. If there is friction also then it then only it can rotate with translation am I right?
| As outlined in my answer here this is definitely possible.
If you have an object with moment of inertia $I=\gamma mR^2$ on a horizontal surface and you apply a constant horizontal force $F$ some distance $\beta R$ above the center of the object, then for rolling without slipping to occur we need a friction force
$$f=\frac{\beta-\gamma}{1+\gamma}F$$
So whenever $\beta=\gamma$ we would have rolling without slipping with no friction. Intuitively, in this case the accelerations associated with the forces and torques balance exactly in such a way so that $v=R\omega$ without friction needing to alter translation and/or rotation to make this true.
Some examples (some might be hard to realize in reality, but work fine mathematically):
*
*The force applied to the top of a circular hoop
*The force applied a distance $R/2$ above the center of a cylindrical disk
*The force applied a distance $2R/5$ above the center of a solid sphere
In your case with the force being applied at the top of the sphere ($\beta=1$), there would need to be friction though.
| {
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Wave motion in a transverse wave I was learning about wave motion and how in transverse waves, each particle executes SHM up and down. If that is the case, how is it so that energy is still transferred onto the next particle? The logical answer should be that it disturbs the other particle, but if it moves up and down how does that happen?
| In the case of transverse waves, the oscillations are at right angles to the direction of propagation of the wave. The wave itself is the transfer of energy from one point to another. Say the wave moves to the right (e.g., x-direction). Then the particles in the medium of the wave are moving "up and down" (y-direction) only. There does not need to be particle motion from "left to right". There is however energy being transferred from left to right.
| {
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Probability of measuring state $|+\rangle$ and state $|-\rangle$ given a state and a basis I am given a basis $|+\rangle = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$ and i am given a three qubit state $|\phi\rangle = \frac{1}{\sqrt 3}|1\rangle |0\rangle |1\rangle + \frac{2}{\sqrt 3}|0\rangle |1\rangle |0\rangle $
What is the probability of measuring state $|+\rangle|+\rangle|+\rangle$ and what is the probability of measuring state $|-\rangle|-\rangle|-\rangle$?
I know how to express the state $|0\rangle $ and $|1\rangle $ in the basis above, that would be $|0\rangle = \frac{1}{\sqrt 2}(|+\rangle + |-\rangle)$ and $|1\rangle = \frac{1}{\sqrt 2}(|+\rangle - |-\rangle)$
but when attempting to do $|\left(\langle+|\langle+|\langle+|\right)|\phi\rangle|^2$ I can't really get anywhere.
How do I solve this?
| So I don't think your state is normalized and probably what was meant was $$|\phi\rangle = \sqrt{\frac13}|101\rangle + \sqrt{\frac23}|010\rangle.$$(But as written you can also normalize it by multiplying by $\sqrt{3/5}$ and converting the denominator to 5.)
There are several ways to do this problem. Possibly the easiest is to rewrite the state that you are dotting with, $$|{+}{+}{+}\rangle = \sqrt{\frac18} \big(|000\rangle + |001\rangle + \dots + |111\rangle\big),$$ to reveal that $$\langle{+}{+}{+}|\phi\rangle = \sqrt{\frac18}\sqrt{\frac13} + \sqrt{\frac18}\sqrt{\frac23},$$ while $|{-}{-}{-}\rangle,$ having negative signs in front of every term with an odd number of ones, instead generates
$$\langle{-}{-}{-}|\phi\rangle = \sqrt{\frac18}\sqrt{\frac13} - \sqrt{\frac18}\sqrt{\frac23},$$and the probabilities are of course the squares of these amplitudes. So, uh, $\frac18 \pm \frac1{12}\sqrt{2}$ if I am doing the quadratic equation right in my head?
The other way is indeed as the other answer says, to convert the above into the appropriate basis, so each $|1\rangle$ becomes a $|+\rangle - |-\rangle$ and each $|0\rangle$ becomes a $|+\rangle + |-\rangle$ and so we have$$\begin{align}
|\phi\rangle =& \sqrt{\frac1{24}}\big(|{+}{+}{+}\rangle - |{+}{+}{-}\rangle +
|{+}{-}{+}\rangle-|{+}{-}{-}\rangle \\
&\hphantom{\frac1{24}}~~-|{-}{+}{+}\rangle + |{-}{+}{-}\rangle - |{-}{-}{+}\rangle + |{-}{-}{-}\rangle\big)\\
&+\sqrt{\frac1{12}}\big(|{+}{+}{+}\rangle + |{+}{+}{-}\rangle -
|{+}{-}{+}\rangle-|{+}{-}{-}\rangle \\
&\hphantom{+\frac1{12}}~~~~+|{-}{+}{+}\rangle + |{-}{+}{-}\rangle - |{-}{-}{+}\rangle - |{-}{-}{-}\rangle\big),\\
\end{align}$$
from which you can read not just those amplitudes but diverse other ones that you might be interested in in the Hadamard basis. Note that there is really no “hard work” here in either case, it is all mostly just keeping track of signs of terms. This is one reason why the Hadamard basis is so nice to think about problems with.
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How is it possible that combustion of coal releases similar energy as TNT explosion while intuitively we would not expect that? According to Wikipedia, the energy released in a TNT explosion is 4 × 106 J/kg.
https://en.wikipedia.org/wiki/TNT
According to web, combusion of coal is around 24 × 106 J/kg.
https://www.world-nuclear.org/information-library/facts-and-figures/heat-values-of-various-fuels.aspx
This looks rather counter-intuitive: TNT is famous for the explosion, thus I would expect that it releases a lot of energy, but actually, it seems much smaller than coal combustion...
How is that possible that combustion of coal releases similar energy as a TNT explosion while intuitively we would not expect that?
| You’re basically talking about the difference between power and energy. A TNT explosion has vastly more power than the combustion of an equivalent amount of coal (in normal circumstances), because it happens over a vastly shorter timescale. And it’s the power that you notice, not the energy.
| {
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Do bodies radiate energy by emitting gravitons? As gravitons are formulated in current theories, are bodies supposed to radiate energy as they emit gravitons?
I thought that if bodies radiate energy, for big bodies this would be difficult to measure, as they emit a lot more energy in the form of heath, and smaller bodies in their orbits would get the energy back by feeling the gravity of the big ones. But what would happen to a small shard of ice projected by a collision between comets far from the gravitational pull of any star? Is it supposed to evaporate?
Furthermore let's suppose we have a sphere of a very dense material, it is attached to a probe, but by a long arm, so long that the gravity radiated by the probe reaching the sphere is far less than the gravity radiated by the sphere. Let's suppose that the probe is travelling along the path of the pioneer far away from the sun. At a certain point the sphere could radiate more gravity than it receives, should there be any detectable effect?
| Yes bodies radiating gravitons do radiate energy. That's why we were able to detect gravitational waves a few years ago, as you can see from the Wikipedia article on gravitational waves:
On 11 February 2016, the LIGO collaboration announced the first observation of gravitational waves, from a signal detected at 09:50:45 GMT on 14 September 2015 of two black holes with masses of 29 and 36 solar masses merging about 1.3 billion light-years away. During the final fraction of a second of the merger, it released more than 50 times the power of all the stars in the observable universe combined. The signal increased in frequency from 35 to 250 Hz over 10 cycles (5 orbits) as it rose in strength for a period of 0.2 second. The mass of the new merged black hole was 62 solar masses. Energy equivalent to three solar masses was emitted as gravitational waves.
However simply existing will not radiate gravitational waves; you need to accelerate, and in the situations you describe there is no acceleration.
| {
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Transformers: How does current in primary coil change? I was doing a question on transformers and found this really confusing question:
A 100% efficient transformer converts a 240V input voltage to a 12V
output voltage. The output power of the transformer can be a maximum
of 20W. The output is connected to two 0.30A bulbs in parallel. One of
the bulbs fails. How does the current in the primary coil change?
What I did:
Since I have been given power and voltage I thought it would be helpful to figure out the current, so I did that which was fairly easy
20W/12V = 1.67 A
the output is connected to 2 0.3A bulbs so if one fails that's only 1 0.3A bulb
What I am stuck on
Now, this is where I kinda fall apart since I don't know where to go, I thought I would need to figure out some sort of ratio due to the transformer rule of the ratio of coil turns is equal to the ratio of voltages, but I don't really see where I would get information of the number of coils turns, however, my intuition is telling me that the current would increase...
Am I missing something that I haven't calculated from the question?
| So the first thing we need to do is find the "step-down ratio". Due to how a transformer works, a small change in the secondary coil/output current will result in an even smaller change in the primary coil current.
*
*Ratio = $\frac{\text{Output Voltage}}{\text{Input Voltage}}$
= $\frac{12V}{240V}$
= $0.05A$
*Use the ratio to calculate the change in current in the primary coil. Because the bulbs are in parallel, the current will be shared equally to between each of them. So good news, we don't need to do anything fancy here - simply calculate the change of current if we lose one bulb.
= $\frac{12V}{240V} \times 0.3A$
= $0.015A$ decrease in the primary coil
| {
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Are distributions of position and momentum assumed to be independent in quantum mechanics? Given a wave-function of a single particle we can calculate probability density for positions. We can also calculate probability density for momenta. Are these probability densities assumed to be always independent?
Or, in other words, if we measure position and momentum of a particle (for example electron in hydrogen being in the ground state), should we expect that these two random quantities independent?
|
if we measure position and momentum of a particle (for example electron in hydrogen being in the ground state), should we expect that these two random quantities independent?
No. The momentum probability distribution and the position probability distribution have an inverse relationship with each other, which is a direct consequence of the Heisenberg's uncertainty relation.
Suppose you have a certain wave packet. What is the special thing about a wave packet? Well, it's confined to a certain region of space. Let's say that it is confined within a length $a$. So, you know that the particle can not exist outside of that region of space, and so the uncertainty principle puts you into a position where you can not measure the momentum arbitrarily accurately. This is why, its probability distribution is also dependent upon that of the position. Mathematically speaking, the momentum and position wavefunctions are related by a fourier transform-
$$\psi(x)=\dfrac{1}{\sqrt{2\pi \hbar}}\displaystyle \int_{-\infty}^{+\infty}\Psi(p)\text{e}^{ipx/\hbar}\ dp$$
From this formula, we can see that the position probability density, which is $|\Psi(x)|^2$ is obviously dependent on the functional form of $\Psi(p)$. I hope this answers your question.
| {
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Direction of flow of Electrons during an Electric Shock Question:
What is the direction of the flow of electrons during an electric shock?
I was studying electrostatic force, suddenly a question struck my mind “What will be the direction of flow of electrons when I touch a live bare wire standing barefoot on the ground?” (Obviously I am not going to do that). Assuming that the current in the wire is small, I thought that the electrons would move from wire to my body then to the earth as electrons move from low potential (Wire) to high potential (Earth (Earth is neutral). But on thinking more deeply, I thought I am touching the wire of an electric circuit, not the negative terminal, then how can I comment that the wire I touched, the potential is low. I am seriously confused now.
What is the correct explanation for all this?
| You are correct that if the wire has a lower voltage than the ground, then electrons would flow from the wire to the ground. If the wire has a higher voltage than the ground, then electrons would flow the other way. This completely depends on the voltage of the wire relative to the ground, which can be pretty much anything depending on the circuit.
It's probably worth noting, though, that the electrons themselves don't move very quickly. It is the collective effect of a lot of electrons moving that creates electric current. So when you are getting shocked, you are feeling the very slow movement of a lot of electrons.
| {
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Why does metal change its color under polarized light? I've taken two photos of a metal in an experimental setup.
The first image shows the metal illuminated by a halogen-lamp from above. The second image shows the same metal illuminated by the same lamp but there are two additions: There is a linear-polarizing filter in front of the lamp and one in front of the camera.
I was expecting all specular reflection to be eliminated and only the diffuse reflection to be visible. But as you can see, the metal seems to have changed its color as well. Why is that? Shouldn't the metal still appear to be yellowish?
I've made two more photos of a similar object; the setup is the same.
The object appears to be blue when the polarizers are added to the setup.
Both metals are anodized aluminum.
I've done the same "experiment" with wood, plastic, and fabric; they don't appear in a different color as the aluminum does. I've also tried out white paper to see if white-balance might be the cause: no difference, the white paper stays white.
| I’ll venture a guess at what’s going on. Take it with salt.
Aluminum is not a perfect conductor, but I do not see how how the variation of reflection coefficient with wavelength or the slight difference between HH and VV reflections could be responsible.
Your material is polycrystalline, so the surface is never truly smooth, and the surface will look rougher with respect to shorter (bluer) wavelengths. Specular reflections from tilted grain facets will have slightly altered polarization, exactly orthogonal to the direction of arrival at your eye or camera, so you cannot expect orthogonal polarization filters to achieve ideal cancellation. The uncancelled polarization should skew blue.
If reflections from grain boundaries dominate, the strength of reflections will depend on both polarization and wavelength. (For example, in knife-edge diffraction, polarizations parallel and perpendicular to the edge behave differently, and backscattered power is proportional to wavelength.) I would be surprised if reflections from grain boundaries favored blue instead of red.
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Time transfer from proper to coordinate: apparent Special / General Relativity mismatching in theory In SR we've learned that the time dilation for an observer moving clock w.r.t one fixed in a frame at rest is
$$\tau = \gamma \tau_0 = \frac{\tau_0}{\left(1-v^2/c^2\right)^{1/2}}$$
ref: "Special Relativity - A.P. French" and many others
In this case being gamma > 1, it implies delta t < delta tau
No moving to GR, the basic starting expression for calculating the elapsed coordinate time from proper time for a observer clock located in a mass gravitational field and moving with velocity v w.r.t a frame at rest in the body mass center is (approximating square root at first order for v << c)
$$\Delta t = \int_A^B \left(1+\frac 1 {c^2} U + \frac{1}{2c^2} v^2\right)d\tau$$
ref: "Relativistic time transfer - ITU-R TF.2118-0" and many other
To note that all terms in the integral are positive, also excluding the presence of gravity (U=0), meaning that it would always result delta t > delta tau
This is an opposite result w.r.t. SR expression!
Can anyone clarify this (apparent) contradiction? Thanks in advance.
| Using bionomial approximation, French's equation equals:
$$\tau = \gamma \tau_0 = \frac {\tau_0} {\left(1-v^2/c^2\right)^{1/2}}\approx{\left(1+\frac{v^2}{2c^2}\right)\tau_0}$$
The second equation introduced by you, substituting $U=0$, also implies:
$$\Delta t\approx{\left(1+\frac{v^2}{2c^2}\right)\Delta\tau}$$
There is no contradiction, I think!
@Gianni:
... the elapsed coordinate time from proper time for an observer clock located in a mass gravitational field and moving with velocity $v$ w.r.t a frame at rest in the body mass center ...
However, remember that an observer located at the center of a planet measures that the clock located on the surface of the planet runs faster. I am doubtful that the boldfaced sentence may not indicate the Schwarzschild observer but rather the observer at the center of the planet. This can justify the difference between the two equations if you think $Δτ$ does not match $τ_0$.
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Representation of homogeneous Lorentz transformation In Page 63, Section 2.5 of Weinberg's QFT Volume 1, on "One-particle states", he considers the representation of homogeneous Lorentz transformation, $U(\Lambda, 0) \equiv U(\Lambda)$
$$
U(\Lambda) \Psi_{p, \sigma}=\sum_{\sigma^{\prime}} C_{\sigma^{\prime} \sigma}(\Lambda, p) \Psi_{\Lambda p, \sigma^{\prime}}
$$
then he claims that,
In general, it may be possible by using suitable linear combinations of
the $\Psi_{p, \sigma}$to choose the $\sigma$ labels in such a way that the matrix $C_{\sigma^{\prime} \sigma}(\Lambda, p)$
is block-diagonal; in other words, so that the $\Psi_{p, \sigma}$ with $\sigma$ within any
one block by themselves furnish a representation of the inhomogeneous
Lorentz group.
Now my question: the effect of $U(\Lambda)$ is to bring the state $\Psi_{p, \sigma}$ to $\Psi_{\Lambda p, \sigma^{\prime}}$, so the space that the $C_{\sigma'\sigma}$ acts is different for different $\Lambda$. But this "conclusion" is weird to me since I think the representation space of a group should be the same for the group elements.
| This is an example of induced representation. Consider two groups, $K < G$. Let a representation $D(K)$ act in a vector (usually, Hilbert) space ${\mathbb{V}}$. Based on this, we now wish to construct a representation of $G$. In mathematics, this (so-called "induced") representation is denoted with $\operatorname{Ind}_K^GD$ or simply $D(K)\uparrow G$.
Such a representation will be a fiber bundle whose base is the quotient space $G/K$, with copies of ${\mathbb{V}}$ playing the role of fibers. Simply speaking, we take multiple copies of ${\mathbb{V}}$ and agree that the action of $G$ is two-fold. It permutes the unit vectors within each copy of ${\mathbb{V}}$, and it also permutes the copies of ${\mathbb{V}}$.
Does this answer your question?
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Why can we see light further than it shines? This is a question I’ve been thinking about for a while. If I’m standing in the middle of a straight road, during night, I can see a car coming towards me because of its lights even if it is kilometers away. Notwithstanding, the driver can not see me because the car will brighten the road only few hundred meters further. What physics properties of the light causes this phenomenon?
| Strictly speaking the following statement is wrong:
the car will brighten the road only few hundred meters further
The fact that you can see the light emitted by the car is evidence for the light illuminating all the way between you and the car.
Two main reasons can be named why you see the car but the driver does not see you:
*
*Scattering of light
*The sensitivity of the human eye (and brain) for scattered light
The first effect leads to the car's light being scattered, the second to the driver not seeing this light.
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Application of Noether Theorem I attempt to understand one of the examples of the application of Noether theorem given in Peskin & Schroeder's An Introduction to Quantum Field Theory (Page no. 18, Student Economy Edition). The relevant portion of the text is given below.
If I understand the derivation and the corresponding discussion here properly, then it was assumed that the Lagrangian density $\mathcal{L}$ satisfies the Euler-Lagrange equation:
$$\frac{\partial\mathcal{L}}{\partial\phi} = \partial_{\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right].$$
My Confusion: I don't see how $\mathcal{L} = \frac{1}{2} (\partial_{\mu} \phi)^2$ satisfies the Euler-Lagrange equation. Because on the left hand side, I get $\frac{\partial\mathcal{L}}{\partial\phi} = 0$, and on the right hand side, I get $\partial_{\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right] = \partial_{\mu} \partial^{\mu} \phi.$ If the given $\mathcal{L}$ doesn't satisfy the Euler-Lagrange equation, then how Peskin & Schroeder's formulation can be applied to this case? What am I missing here?
| The Euler-Lagrange equation is not automatically satisfied by $\mathcal{L}$. It's the other way around. Given $\mathcal{L}$, you can find the classical equation of motion satisfied by the field $\phi$. This is like giving you a formula for the force in Newtonian mechanics. Even if you know $F$, you still need to know Newton's second law $F=ma$ to find the motion. Here too: given $\mathcal{L}$, you still need the "law" (E-L equation) to find the motion.
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Why do electric field lines curve at the edges of a uniform electric field? I see a lot of images, including one in my textbook, like this one, where at the ends of a uniform field, field lines curve.
However, I know that field lines are perpendicular to the surface. The only case I see them curving is when drawing field lines to connect two points which aren't collinear (like with charged sphere or opposite charges) and each point of the rod is collinear to its opposite pair, so why are they curved here?
| Rather than thinking of the plates as solid line charges, think of them as lines of infinitely many point charges.
2 point charges will have a straight field-line directly between them, and weaker curved field-lines outside of that. If you place 2 pairs of point charges next to each other — positive with positive, and negative with negative — then their field lines can overlap. However, much like with waves, electric fields can interfere with each other, both constructively and destructively.
This means that the overlapping curved field lines will average out as a straight field line, through the middle of the point-charge pairs. The outermost edges of the electric field, on the other hand, will have nothing to interfere with, and remain curved.
| {
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Particle as wave, stable? I've started reading about the wave-particle duality but, after a few steps, reached a dead end:
*
*Schrodinger equation solutions for a free particle is a sum of terms of the form:
$$\psi(\mathbf{r}, t) = Ae^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$$
however, a single element of this form can not normalize, thus, can not exists alone. That is, a particle must be an addition of several terms, a wave packet. (TODO: verify a Gaussian wave packet normalizes :-).
*The restriction:
$$ \omega = \frac{\hbar k^2}{2m} $$
applies to previous wave function. That means that each component of the wave packet has a different propagation speed. As consequence, the particle spreads.
The question: particles tends to disperse (dissolve) ? If so, how to explain the stable existence of protons, fermions, ... ?
| It's right, you cannot normalize a plane wave.
But... remember that plane waves cannot exist in nature, because nothing is actually infinite. Plane waves do not exist, no matter what waves we're talking about.
But! They are a nice approximation for many situations. You can approximate many usual waves to be plane, and it works well, because it is a very good approximation.
In the same way, tehre are never 100% free particles, but if they are far enough from other objects, the free particle approximation is a good one. Also is its solution.
So yes, if we were extremely strict, we shouldn't admit plane waves, but we do because tehy work well under many situations.
Note, altough many states are not normalizable, mean values can be normalized, because the divergence vanishes. Some sort of...
$$\langle A\rangle = \dfrac{\langle\psi|A|\psi\rangle}{\langle\psi|\psi\rangle}$$
The terms that would cause the divergence just cancel out. SO, numerator and denominator do not work well separately, but put together, they give a finite value.
So that's why we can make an exception, admittin plane waves
| {
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A photon scatters an electron at an angle... Does it imply electron having an area greater then the photon's? Even we don't know much about scattering areas of photons and electrons does the fact that a photon scattering an electron at an angle mean that the photon cross-section area hits only a small lateral area of the electron causing it to move at an angle. If the photon cross-section area was greater than the electron's one would the electron be scattered only in the direction of the photon incoming motion because the whole its area would be uniformly objected to photon's pressure? So should the electron have a finite volume?
| Photons and electrons are elementary particles in the standard model of particle physics, SM , a quantum field theory model.
They are by axiomatic definition point quantum mechanical particles and their behavior is perfectly described by the SM. Their scattering can only be described within this quantum mechanical framework.
So, as the SM fits the great plethora of particle physics data and predict new observations, it is accepted that the electron and the photon do not have a crossection. Their scattering is due to the electric field of the electron and the coupling of the photon to the electromagnetic field as given by the Feynman diagrams that can compute the probability of the scattering, example:
The experiments have put an experimental limit on the "size"of the electron , see this, as < $10^{-18 } m$ , which gives a limit to the validity of the standard model.
| {
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Am I understanding the equivalence principle correctly? So Einstein stated that gravity and acceleration are the same things in a local reference frame (please correct me if I misstated that).
Here is what I think I understand and want you to verify if it is correct:
In Special Relativity, the faster you go, the more time dilation and length contraction an outside observer sees you experience.
In General Relativity, the closer you are to a mass, the more time dilation an outside observer sees you experience.
Because gravity and acceleration are the same thing, does that mean that the time dilation when near a gravitating mass is the same time dilation as when you are at a very high speed? Are these two things connected/the same?
And does this prove that gravity and acceleration are the same thing (at least locally)?
Extra question: when you are near a gravitating mass, does an outside observer see you as length contracted too?
|
Because gravity and acceleration are the same thing, does that mean that the time dilation when near a gravitating mass is the same time dilation as when you are at a very high speed? Are these two things connected/the same?
No. In fact concerning GPS satellites SR effects and gravitational effects can compete each other. During the fist launches of GPS satellites, Russians have chosen orbits in which the two effects have compensated.
From this document from the Perimeter Institute
Einstein's theories of special relativity and general relativity have opposing effects on time in the GPS. Einstein’s theory
of special relativity states that the clocks inside GPS satellites run slower than a stationary clock on Earth by 8.3 x 10-11 s per second. This is due to the speed of the satellites. Einstein’s theory of general relativity says the satellite clocks also run faster than those on Earth by 5.2 x 10-10 s per second because Earth’s gravity is weaker at the satellites’ altitude.
| {
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How to measure the force produced by a finger? If I push my finger onto a weighing balance and it reads 100g, then how do I calculate the force produced by my finger? Do I just use F = mg? Or is this even possible?
| Yes, You are correct. A balance actually measures force, but the readout is conveniently scaled to tell you how many grams of mass would feel that same force from gravity near the surface of the Earth.
All you have to do is reverse that scaling to get the force in Newtons.
| {
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Force exerted by blocks on an angled rail I'm trying to figure out the force exerted on blocks positioned on an angled rail due to gravity. This is the scenario I have (apologies for the poor graphic):
where each block (red square) is equipped with a wheel (black circle) which is fitted on a rail (black lines). The top section of the rail is angled at 50 degrees (or just $\theta$) and the bottom section is vertical. I'm trying to figure out on the force exerted on the bottom wheel (green arrow). Here is what I have so far:
We have 3 forces: 1) The force exerted by the top block parallel to the rail is $mg\sin(\theta)$; and 2) The force exerted by the middle and bottom blocks are $mg$. I end up with the following vector triangle:
$R$ is the force exerted on the wheel. Am I missing something?
Any advice appreciated.
| In this situation, the vertical component of the (normal) force of the track acting on the upper block supports part of the weight of the upper block. The problem is to find that normal force. To keep the upper block from rotating there must be a horizontal friction force from the block below acting on the bottom of the upper block. That counters the torque caused by the vertical force from the lower block. To find the normal force I'm going to take the torques about the center of where the blocks meet. I will assume that each block has a width, x, that the upper blocks overlap by a distance, d, that the two upper centers are separated by a distance of, 2L, (center to center), θ = $50^o$, and the wheels are very small. Then the torques about the center point are produced only by the normal and gravity. NL – mg(x/2 – d/2). But (x/2 – d/2) = Lcos(θ). Together these give N = mgcos(θ). (I'm not allowed to give a complete answer.)
| {
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Why do aluminium sheets wrinkle easily but are hard to un-wrinkle? I notice when I buy aluminium sheets (the kind used for wrapping food) they come in straight, smooth rolls. Once I use them though, they become wrinkled, and they are impossible to un-wrinkle. Image of a wrinkled sheet:
Why is this the case? Only thing I can think of is the Second Law of Thermodynamics, but I can't see why that would be applicable. It's still the same sheet of aluminium, after all, and it's not like it's getting mixed with something else (like when making coffee). Furthermore, the manufacturers are able to make it unwrinkled, and it doesn't become wrinkled until it's perturbed. I'm tagging the question with thermodynamics anyway, because I don't know what else it could be.
| When the Aluminium sheet has been deformed beyond the elastic limit the bonds between the Aluminium atoms have been broken/deformed irreversibly.
This is called plastic deformation.
So a small amount of deformation (elastic) will allow the Aluminium sheet to revert to its original condition when the deforming force is removed but large deformations (plastic) do not.
Aluminium sheet is produced by passing heated ingots of aluminium through rollers numerous times and then processing them as necessary to obtain the required properties.
The rolling process is irreversible but controlled to produce a smooth surface.
Aluminium is ductile ie has ability to be permanently deformed without fracturing and you will notice this property when you crease Aluminium sheet.
| {
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What does $\hat{\phi}$ mean in cylindrical coordinates? When talking about the unit vectors in cylindrical coordinates, $\hat{\phi}$ often comes up. However, I cannot find a straightforward meaning for it. However, I do know that it is perpendicular to $\hat{\rho}$. How is that significant?
| From the linked article it shows exactly what the vector $\hat{\theta}$ means
There it is, a unit vector pointing around the "hoop" direction, tangent to the surface and perpendicular to the axial direction $\hat{z}$.
| {
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Is it possible to bend light without changing its color? It seems to me that whenever you change the direction of a wave it also affects frequency. Would this not also be true of light waves bending from, for example, gravity?
| You are specifically asking about gravitational lensing, that is a distribution of matter between a distand lightsource and the observer, that is capable of bending the light from the source as the light travels towards the observer.
If the (light) source, the massive lensing object, and the observer lie in a straight line, the original light source will appear as a ring around the massive lensing object (provided the lens has circular symmetry).
https://en.wikipedia.org/wiki/Gravitational_lens
Now light in this case follows a geodesic, that is, the world line of a particle, free from all external (non-gravitational) influences follows a geodesic.
In general relativity, a geodesic generalizes the notion of a "straight line" to curved spacetime. Importantly, the world line of a particle free from all external, non-gravitational forces is a particular type of geodesic. In other words, a freely moving or falling particle always moves along a geodesic.
https://en.wikipedia.org/wiki/Geodesics_in_general_relativity
As per GR, light in your case, during gravitational lensing, when gravity bends its path, will retain its frequency unchanged (if there are no other effects), and this means, that the answer to your question about gravity bending light is that in reality this effect does not change lights frequency.
| {
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How do astronomers detect the spin of the black holes? There are various theoretical models for accretion disks and relativistic jets from black holes that considers the black hole to be either rotating (Kerr geometry) or non-rotating (Schwarzschild geometry). But when astronomers detect a source, they detects whether the source is rotating or non-rotating. One such example is the GRS 1915+105 which is an X-ray binary star system featuring a regular star and a black hole is observed to rotate at close to 1,150 times per second with a spin parameter value between 0.82 and 1.00 (maximum possible value).
So my question is the following:
How do astronomers detect whether a black hole is rotating or
non-rotating? And if the black hole is rotating, how do they
calculate the spinning rate?
| Here is from the abstract of a paper about theory and observation of the spin of a black
..non-zero spin leaving an indelible imprint on the space-time closest to the black hole. As a consequence of relativistic frame-dragging, particle orbits are affected both in terms of stability and precession, which impacts on the emission characteristics of accreting black holes both stellar mass in black hole binaries (BHBs) and supermassive in active galactic nuclei (AGN). Over the last 30 years, techniques have been developed that take into account these changes to estimate the spin ....
It is not a simple formula, but the paper goes into the mathematical details.
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Why do waves, specifically light, diffract through a slit? I've been wondering this for a while now, and have thus far only come across answers that seem to use an equation as an explanation. I've also looked at Huygens' principle (albeit not in-depth), but this doesn't make much logical sense to me.
I'd sincerely appreciate if anyone could shed any light on the topic for me (pardon the pun :P), or try to explain Huygens' principle to me.
Thanks in advance!
| Light diffracts primarily because there is an interaction of the EM field of the aperture material with the EM field of the photon or wave. Diffraction is interesting in itself but it is also the diffraction pattern or "interference" pattern that results which causes a lot of discussion. The word
"interference" is historical but also somewhat misleading, 2 photons never "interfere" or cancel, that would be a violation of energy conservation, instead photons take paths that are of certain pathlengths that are integer multiples of its wavelength ... this can explain the "interference" pattern. Mathematically this coincides with the use of Huygens principle in double slit experiments.
For water or air you could have the movement of 1 molecule radiate outwards and effect many other molecules in a circular pattern. Light is different, a quanta will radiate in one direction only.
| {
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Why are the capacitors in this circuit in parallel but not in series?
In the circuit, the capacitors are said to be connected in parallel. Why is that so?
Edit: The switch will be closed and C2 is fully charged by C1 and no more current will flow between C1 and C2. The question asks for the voltages and charges hold by C1 and C2. In the solution, it is mentioned that C1 and C2 are connected in parallel (V1 = V2), which is the part I don't quite understand.
| They are in series, one end of the first capacitor contacts one end of the other. They are also parallel when the switch is on, because they both connect two ends.
Parallel connection means both ends of the two elements are connected together. This happens when you turn the switch on. Both ends of C1 becomes connected to both ends of C2. Before that, only one end of C1 is connected to one end of C2. The two remaining ends are open.
| {
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Electric Potential Energy - How a charge can be brought from infinity to a point without accelerating it? Question:
My Book says:
ELECTRIC POTENTIAL ENERGY: Electric potential energy of a charge($q_o$) at a point(A) in the electric field due to any charge is given by the work done by an external force to displace $q_o$ without acceleration from infinity to that point(A).
$\color{red}{\text{How it is possible to displace a charge from infinity to a point without accelerating it}?}$
Suppose a point test charge q is located at:
(1) Infinity
Then the moment an external force $F_{external}$, the charge gets accelerated.
(2)A point in Q's electric field
Then the charge Q would exert an electrostatic force($F_Q$) and point charge q would accelerate due to this force. My book says now an external force $F_{external}$ is exerted to move it without any acceleration. Book does not specifies which force is larger.
Now three cases arrive:
$$(1) F_{external}>F_Q$$$$(2)F_{external}=F_Q$$ $$(3) F_{external}<F_Q$$
In (1) and (2) cases there would be some net force, so there would be acceleration.
In (3) case the net force would be zero so the charge would be at rest.
| You can assume that in the beginning you have a very, very small acceleration resulting in v, from then on you have no force so your last equation ist true, at the end you still have m/2v^2 left for your energy balance.
| {
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What is the polarization of an EM wave after it suffers scattering? If a linearly polarized classical monochromatic electromagnetic radiation undergoes a scattering, does the scattered electric field have the same polarization as the incident electric field? I am looking for an answer (or deduce the conclusion mathematically) from classical electromagnetic theory of scattering.
| It depends on what the wave is scattered from. The simplest case to study (as an exercise to get intuition) is scattering of a monochromatic EM wave in vacuum from an infinite metal plane: in this case the boundary condition is that the component of the electric field along the surface should be zero. This may add a phase of $\pi$ to one or both components of the electric field, but no other transformations occur.
In other words:
*
*linear polarization remains linear polarization (taking account that it is polarized perpendicular to the direction of the wave propagation, which is generally not the same for the incident and the scattered waves)
*circular polarization might change its direction of rotation - I can't be sure without doing a calculation that I described above
*obviously, if the waves are not in vacuum, or if the scattering surface has complex geometry, or if it is not a metallic surface, the answer may become a lot more complicated.
| {
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How the matrix representation of a Hamiltonian affects the eigenvalues? Suppose we're given the following Hamiltonian: $$\hat{H}=\frac{\omega}{\hbar} \left(\hat{S}_+^2+\hat{S}_-^2\right)$$ Suppose also that we measure $\vec{S}^2$ and get $6\hbar^2$, i.e. reduced to the $s=2$ subspace, and want to find all the possible energies (aka the eigenvalues of the Hamiltonian operator in the relevant basis).
Obviously the relevant basis consists of the eigenstates $\{|2,m\rangle\}$ of the spin projection operator. I calculated the matrix elements of $\hat{H}$ in this basis and got the following $5\times 5$ matrix:
$$\hat{H}=\hbar \omega \begin{pmatrix}
~ & |2,2\rangle & |2,1\rangle & |2,0\rangle & |2,-1\rangle & |2,-2\rangle\\
|2,2\rangle & 0 & 0 & 2\sqrt{6} & 0 & 0\\
|2,1\rangle & 0 & 0 & 0 & 6 & 0\\
|2,0\rangle & 2\sqrt{6} & 0 & 0 & 0 & 2\sqrt{6}\\
|2,-1\rangle & 0 & 6 & 0 & 0 & 0\\
|2,-2\rangle & 0 & 0 & 2\sqrt{6} & 0 & 0
\end{pmatrix}$$
It can be shown that the eigenvalues are $E=\pm 4 \sqrt{3} \hbar \omega, \pm 6\hbar \omega , 0$, which is indeed correct. However calculating them was a little bit tedious.
Now, it turns out that there exists a simpler matrix representation of $\hat{H}$ in the same basis. It has to do with the special structure of the Hamiltonian which has both raising and lowering operators squared. This naturally splits the basis into two groups: $\{ |2,2\rangle, |2,0\rangle,|2,-2\rangle \}$ and $\{ |2,1\rangle, |2,-1\rangle \}$ which are closed under the actions of $\hat{S}^2_{\pm}$. We can thus re-order the basis and get the following block-diagonal form
$$\hat{H}=\hbar \omega \begin{pmatrix}
~ & |2,1\rangle & |2,-1\rangle & |2,2\rangle & |2,0\rangle & |2,-2\rangle\\
|2,1\rangle & 0 & 6 & 0 & 0 & 0\\
|2,-1\rangle & 6 & 0 & 0 & 0 & 0\\
|2,2\rangle & 0 & 0 & 0 & 2\sqrt{6} & 0\\
|2,0\rangle & 0 & 0 & 2\sqrt{6} & 0 & 2\sqrt{6}\\
|2,-2\rangle & 0 & 0 & 0 & 2\sqrt{6} & 0
\end{pmatrix}$$
which is very convenient because now in order to find the eigenvalues we can analyze two smaller matrices. Thankfully, the eigenvalues turn out to be the same.
Question: we know from linear algebra that, in general, swapping/changing the order of rows/columns (which is exactly what happened here) changes the eigenvalues. However in this case the eigenvalues remained the same. I understand the physical reason behind it, but how can it be justified mathematically? Suppose we knew nothing about the structure of the Hamiltonian (or, alternatively, weren't smart enough to recognize that the basis can be conveniently broken into two "special" subgroups). Is there a mathematical way of finding the "best" ordering of basis vectors such that the matrix representation of a given operator assumes a block-diagonal form? And is there a mathematical justification for why the eigenvalues remain the same after we change the order of rows/columns? Maybe it has to do with the fact that the matrix (operator) is symmetric (Hermitian)?
| Your 2 matrices only differ by the same rearrangement of rows and columns, which does not change the eigenvalues. To be precise, let $P$ be the permutation that takes $\{|2,2\rangle,|2,1\rangle , |2,0\rangle , |2,-1\rangle , |2,-2\rangle)\}$ to
$\{|2,1\rangle , |2,-1\rangle , |2,2\rangle, |2,0\rangle , |2,-2\rangle)\}$.
$P$ can be constructed by considering for instance
\begin{align}
|2,2\rangle\mapsto \left(\begin{array}{c}1\\ 0\\0 \\ 0 \\0 \end{array}\right)
\end{align}
etc, so that the first column of $P$ would be $(0,0,1,0,0)^\top$ and P would take
$(1,0,0,0,0)^\top \to (0,0,1,0,0)^\top$.
Then your matrices are related by the similarity transformation
$$
P\hat H_1 P^{-1}= \hat H_2
$$
and thus the eigenvalues of both are the same. Alternatively your matrices are related by a change of basis generated by $P$, so they both have the same eigenvalues.
(NB: I hope I have my $P^{-1}$ and $P$ in the right place, but the argument is sound.)
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Why is the mean density the same for all nuclei? Tell me if this is a correct theory? So the radius $R$ of the nucleus is directly proportional to $A^{1/3}$ (the nucleon number).
As $$V = \frac 43 \pi r^3,$$ this makes $V$ directly proportional to $R^2$. Also, as the nucleon number increases, the mass also increases and as the masses of protons and neutrons are similar you could say that the mass of the nucleon is directly proportional to the nucleon number.
If you put all of this together, you get the mass of the nucleon being directly proportional to the volume where the constant is the density.
Thus, that is why the density is constant for all nuclei?
| Density is $$\rho = \frac mV$$
Expressing for nuclei mass and volume, gives :
$$ \rho = \frac{A\,\mu}{4/3 \,\pi \left(r_o A^{1/3}\right)^3} $$
Simplifying gives :
$$ \rho = \frac{\mu}{4/3 \,\pi~ r_o^3 } $$
Where $r_o = 1.25 ~\text{fm}$ and $\mu$ is typical nucleon mass. Thus nuclei density is constant.
| {
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Eddy current confusion Every source tells that eddy currents are produced by change in magnetic flux in conductor, but according Gauss' law for magnetism net flux over a closed surface is 0, then how magnetic flux will change to produce eddy current?
| Gauss' law for magnetism net flux over a closed surface is 0.
The polarity of static fluxes balances out to 0.
Static magnetic fields do not create eddy current losses because of the lack of motion of charges. You either need a moving magnetic material or a dynamic/alternating current or both.
The external flux is reduced by the amount of eddy currents and a drag force and heat loss occurs. The net flux is still zero even though the external flux on either side is weaker from the circulating eddy currents on the edges.
In a water stream the eddy currents on both edges of a stream reduce the overall flow rate, however the net water stored =0 as output = input.
Eddy current losses increase with the square of charge velocity which can be excited by magnet strength or AC coil voltage or frequency or any combination of these and other factors such as laminate thickness.
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Ceiling fans with just one large winglet? I am from India and in India ceiling fans have generally three wings.
Today while laying on my bed a question came in my mind.
First of all, I know that if we use just a single wing at a time of same dimensions as each wings of a three winged fans have then it will reduce the amount of air being pushed down.
But what if we just weld the edges of the three wings to make a single wing ? Will we get the same amount of air as we get generally from a three winged fan ? Which factors determine the amount of air being pushed down ?
I think by this technique the amount of air being pushed downward due to the curvature of the three wings (as a whole) will not be affected. And we can get exactly the same amount of air.
| @trula comment is correct. The fewer the blades the more unbalanced the air load on the fan putting more stress on the bearings. It is similar to what happens to a top loading washing machine in the spin cycle if the clothes are concentrated on one side of the tub as opposed to being distributed around the center of rotation. The machine can shake violently.
what about the amount of air being pushed down ? You didn't mention
anything about that
Sure it will increase the amount of air being pushed down. But @Justjohan pointed out that increasing the number or size of the blades will result in more air drag as well as weight. Although air drag can be compensated for by using a bigger motor, the additional weight may exceed the max load capability of the ceiling structure.
Bottom line, you can add blades or use a bigger blade, and you can use a bigger motor, but the cost benefit ratio may be too high considering that the common 3 to 5 blade fans provide more than enough air flow. And then there are the safety concerns associated with excessive weight and an unbalanced load that also need to be considered.
Hope this helps.
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Find the emf in the loop by two methods In the Figure below assume the magnetics are shaped such that the magnetic field is in
the $z$ direction and varies as
$$
B_0 = B_m\left(1-\frac{x^2}{a^2}\right)\hat{z}
$$
Find the emf in the loop by two methods, a) the rate of change of flux b) the motional
emf method.
My try:
$$V_{emf} = -\int B \cdot \frac{\partial S}{dt}$$
$S=2a\ell\cos \theta \hat{z}$, $B=B_m\left(1-\frac{a^2\cos^2(\theta)}{a^2}\right)\hat{z}=B_m \sin^2(\theta)\hat{z}$
$$
V_{emf} = 2a\ell B_m\cos \theta \sin^2(\theta)
$$
I am not sure if my solution is correct.
| Ok, we want to calculate the emf induced in the rotating coil. The B magnitude in the z direction (horizontal) depends on x (vertical position) as shown below:
$$emf = \oint \mathbf{E} \cdot \mathrm{d}\boldsymbol{l} = - \frac{\mathrm{d}}{\mathrm{d}t} \int \mathbf{B} \cdot \mathrm{d}\mathbf{a}$$
In other words, the emf is generated by the change in total flux linking the coil. $$ emf = - \frac{\mathrm{d\Phi}}{\mathrm{d}t}$$
where,
$$\Phi = \int {\boldsymbol{B} \cdot d \boldsymbol{a}}$$
The total flux linking the coil is time dependent because of the constant rotational speed of the loop.
When the loop is horizontally oriented (θ = zero), no flux passes through the coil. When it is vertically oriented (θ = π/2) the maximum flux possible passes through the coil.
If the B field where constant (Bm) across the region then the problem would simplify to (at t=0 the coil is perfectly vertical),
$$\Phi = B_mA \int cos(\Omega t)$$
where, $$A = 2al$$
so,
$$ emf = - \frac{\mathrm{d\Phi}}{\mathrm{d}t} = -B_mA\Omega[sin(\Omega t)]$$
Back to this problem...
The rotational speed is Ω rad/s. The total flux linking the coil is thus time dependent. The x position of one coil side is positive while the other is negative (except at θ = zero where both are zero).
$$ x = a sin(\Omega t)$$
The angle between the flux density B field and the normal to the coil surface is,
$$ \Theta = \Omega t $$
So, write in the equation for the total flux linkage (time dependent) normal to the coil surface (dot product) using your equation for B and then differentiate.
Please do let us know what the solution actually is (i'd like to know if my result is correct - which i'm holding back for now as this is a homework problem).
| {
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Magnetic, geographic and geomagnetic poles
Can someone explain me in simple terms where and what are the magnetic, geographic and geomagnetic poles?
Some sites say that magnetic north pole is in the south and thus it attracts the south pole of the magnetic needle while some say the north pole of the needle points south. In each place, I seem to get a different angle between the magnetic and geographical axis (10°, 11°, 17° etc.). And I don't get clearly what the geomagnetic pole is.
I'm utterly confused, please help me out.
This question slightly addresses my doubt but the answers contradict each other.
Thanks in advance.
| The North geographic pole is where the effective axis of rotation meets the surface at the North end of the earth. The North (seeking) pole of a magnet is the end which swings toward the North (if its mount allows it to turn). The magnetic field of a magnet emerges from the North (seeking) pole, loops around, and reenters at the “South pole”. The magnetic field of the earth emerges from a North magnetic pole, which is near (but not at) the South geographic pole, loops around and reenters at a South magnetic pole which is near (but not at) the North geographic pole. I have just learned that there are two types of magnetic pole. A compass locates a magnetic pole on the surface of the earth,but the field in space directs the charged particles which cause the aurora toward a geomagnetic pole. The two do not coincide.
| {
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Newtonian approximation of the metric tensor I was reading Dirac's General Theory of Relativity. In chapter 16, the Newtonian approximation, we start with
Let us consider a static gravitational field and refer it to a static coordinate system. The $g_{\mu\nu}$ are then constant in time, $\frac{\partial g_{\mu\nu}}{\partial x^0}=0$. Further, we must have
\begin{equation}
g_{m0}=0, \text{ }(m=1,2,3)
\end{equation}
This leads to
\begin{equation}
g^{m0}=0, \text{ }g^{00}=(g_{00})^{-1}
\end{equation}
How does having a static gravitational field lead to $g_{m0}=0$?
| I think the key phrase is "and refer it to a static coordinate system."
It isn't so much that a static spacetime implies the metric takes a particular form, but that you can choose a coordinate system such that $g_{m0}=0$.
A good exercise would be to start with with a metric where every component was time independent, and $g_{m0}$ was nonzero, and explicitly construct a coordinate transformation which set the $g_{m0}$ to zero. Essentially this should amount to a kind of space-time dependent boost.
| {
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Will energy in an LC circuit become 0 if its is disconnected when the capacitor is fully discharged? In a recent test I had a question in which there was an LC circuit with an inductor a capacitor and a switch. According to the answer key
If switch is opened when capacitor is fully charged energy of LC system remains same.
If switch is opened when capacitor is fully discharged energy of LC system becomes 0.
I can understand the first one but not the second one. The answer keys to this particular exam do tend to be wrong once in a while so I thought I'd get a second opinion.!
| When the capacitor has no charge, the energy is in the magnetic field of the inductor, which is associated with a current flow. If the switch is open, the current cannot flow. The magnetic field collapses, leaving no energy. (The collapse of the field will cause a large voltage spike, and probably an arc across the opening switch. The energy is dissipated there.)
| {
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Will the sky be full of rainbows? I saw this article and something really wierd came in my mind.
What would the sky look like on those planets of our solar system where rainfall of diamonds occur? Will it be full of rainbows? Has it been ever captured?
I don't know whether it will be full of rainbows or not but there can be surely more than two rainbows which we rarely observe here on the Earth because of multiple reflection of light inside the diamonds.
| Unlikely. The strong primary rainbow on earth happens because the sun location and the water locations are limited. If you tried to make a sky full of rainbows, it would just smear together into white.
Ice crystals commonly form in earths atmosphere. Like diamonds, and unlike water droplets, these disperse light in different directions depending on the orientation of the crystal. Because in most situations there is no orientation to the crystals, all colors go in all directions, and the overall impression is white. This is the same thing that happens in foams and snow. Further, we see a rainbow because there is little rain between us and the edge of the rain area. If the rain were everywhere in the sky, the light would rebound around and you'd get the white of clouds.
There is a type of ice rainbow that can happen when a type of hexagonal ice forms, and then orients itself face up as it falls. This directionality allows them to disperse in a particular direction and such ice rainbows are possible. Sky diamonds would need some sort of orientation mechanism to do the same thing.
All told, you could possibly get diamond rainbows in the same way you get ice rainbows on earth. But a sky full of diamonds would probably look white (assuming white light), not spectacularly colorful.
| {
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Ordering ambiguity in the Feynman propagators obtained using Wick's theorem Applying Wick's theorem to a string of four field operators, $\phi_a\equiv\phi(x_a)$:
$$T(\phi_1\phi_2\phi_3\phi_4)=\{...\}, \tag{1}$$
we obtain several terms, three of which are fully contracted fields: $$\phi_1^{\bullet}\phi_2^{\bullet}\phi_3^{\bullet\bullet}\phi_4^{\bullet\bullet},\quad \phi_1^{\bullet}\phi_2^{\bullet\bullet}\phi_3^{\bullet}\phi_4^{\bullet\bullet},\quad \phi_1^{\bullet}\phi_2^{\bullet\bullet}\phi_3^{\bullet\bullet}\phi_4^{\bullet}. \tag{2}$$
Where I have given contracted fields the same number of dots. Each contracted field gives the associated Feynman propagator: $\phi_1^{\bullet}\phi_2^{\bullet}\equiv D_F(x_1-x_2)$.
My question is, when we have terms with more than one contraction, which propagator goes first? Based on what I am reading in Peskin and Schroeder we order them according to the ordering of the left-most contraction arm, however the book only demonstrates this for terms with four operators (so far), and I am unsure if this relation holds for terms with more fields.
| In Wick's theorem, it is usually assumed that contractions supercommute with all pertinent operators, cf. e.g. my Phys.SE answer here. With this assumption there are no ambiguities in operator-ordering as long as one observes the sign rule for manipulating Grassmann-graded objects.
| {
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Increase in internal energy and temperature of a thermodynamics system A banquet hall ($150 m^3$) is to be used for a formal dinner for 20 persons. Each person occupies $0.075 m^3$ of space and has an average heat transfer rate of $500 \frac{kJ}{hr}$. If the AC system fails, determine the increase in the internal energy and temperature of the air in the hall during the first 15 minutes of failure. Take the hall to be well insulated from outside. If the hall and all its contents are considered as system how much will be the increase in the internal energy of the system.
I am trying to use $Q= U + W$, don't have an idea about the data related to each person, why it is given. Thanks, please help.
| If the walls are rigid then the system is isolated the total change in internal energy of the system (humans +room air) would be zero.
But when we consider the air in the room as the system the humans are then the surroundings. Any work done by or on the air in the room is due to the expansion and contraction of the human chests during breathing and would cancel out for a net work of zero on the room air. Therefore for the air in the room $\Delta U=Q$ where $Q$ is the heat transfer from the humans to the air in the room.
If we can assume the humans inhale and exhale the same volume of gas then the number of moles of gas is in the room is constant and can be calculated based on the initial condition information given (T,P,V) and ideal gas law. Finally, $\Delta U=nc_{v}\Delta T$ for an ideal gas, which we can assume for the air even though its composition is changing slightly (increase in carbon dioxide and decrease in oxygen).
Hope this helps
| {
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Solutions of the Harmonic Oscillator are $not$ always a Combination of Separable Solutions? Are there solutions of the Schrödinger equation that are not a linear combination of separable solutions and how do we find them?
In Griffiths, Quantum, Prob. 2.49, there is a solution of the (time-dependent) Schrödinger equation, which reads
$$
\Psi(x,t)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left[-\frac{m\omega}{2\hbar}\left(x^2+\frac{a^2}{2}(1+e^{-2i\omega t})+\frac{i\hbar t}{m}-2axe^{-i\omega t} \right)\right].
$$
It seems that this is not a linear combination of the stationary states that he found previously in the chapter.
If it is the caes, does that mean that solving the time-dependent Schrödinger equation by separation of variables does not yield the general solution as the author claimed? if so, how do we find the other solutions?
| Sometimes the expansions are not obvious. For example The harmonic oscillator time-dependent Schr"odinger equation
$$
i\partial_t \psi = -\frac 12 \partial^2_x \psi +\frac 12 \omega^2 x^2 \psi
$$
has a ``breathing'' solution
$$
\psi(x,t)= \left(\frac{\omega}{\pi}\right)^{1/4}\frac 1{\sqrt{e^{i \omega t} +R e^{-i\omega t}}}\exp\left\{ - \frac \omega 2 \left(\frac{1-R\,e^{-2i\omega t}}{1+R\,e^{-2i\omega t}}\right)x^2\right\},
$$
where the parameter $|R|<1$.
Mehler's formula gives expansion in terms of the states as
$$\psi(x,t) {=}\pi^{1/4}\sum_{n=0}^\infty e^{-i(n+1/2) \omega t} \varphi_n(0)(i\sqrt R)^n \frac{\varphi_n(\sqrt{\omega} x)}{(\omega)^{1/4}}.
$$
Here
$$
\varphi_n(x)\equiv \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} H_n(x) e^{-x^2/2}
$$
is the normalized $\omega=1$ harmonic oscillator wavefunction. Now
$\varphi_n(0)$ vanishes if $n$ is odd, and
$$
\pi^{1/4}\varphi_{2n}(0)= \frac{1}{\sqrt{4^n (2n)! } } \frac{(2n)!}{n!}(-1)^{n}.
$$
so one has found as set of quite "non obvious" expansion coefficients.
| {
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How does light, which is an electromagnetic wave, carry information? We see an object when light from a source strikes the object and then reaches our eyes. How does light, which is an electromagnetic wave, gets encoded with the information about the object? Do the individual photons get encoded with this information or is it the wave nature of light that gets modified to carry information about the object?
Also, if light hits an object and then another on the way to our eyes, does it only carry information from the very last interaction it had? How is the information due to all previous interactions erased (if it is indeed)?
| Assume Light is a being.
Light doesn't carry information. All it does is, just pass through, if the object it hits, allows it to pass through or it gets reflected back.
Example 1 :
Infrared Laser in Scanning barcodes.
Barcodes are nothing but alphanumerics shrinked in size. When infrared laser is allowed to hit it. Each letter in that alphanumeric information lets the laser pass through the area, only surrounding its shape but not the light that hit on the shape/surface ( gets blocked and reflects back ) paving way for the information detection.
Example 2:
Kepler-1649c - Transit method - Dips in brightness
The Method used to collect data over years to discover an earth like planet. For more info, check this link
| {
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Doubt on net acceleration during non-uniform circular motion During non-uniform circular motion, the direction of net acceleration is not in the direction of the centripetal acceleration, then why does a particle still move in a circular path, please explain.
| Think of a car going round a circular track. At an instant when it has speed $v$ it has an acceleration of magnitude $\frac{v^2}{r}$ towards the centre of the circle. The car is gaining velocity towards the circle centre. But suppose that, at this instant, the driver is making the car go faster. The car will also be gaining velocity in a direction tangential to the circle. That doesn't interfere with its gaining of velocity towards the circle centre.
Looking at it in terms of forces, the road is exerting a frictional force on the powered wheels of the car that has both a forward component, giving the car its increase in speed, and a sideways component towards the circle centre, allowing the car to go at speed $v$ in a circle of radius $r$.
| {
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If an electron-hole pair is formed, where does the electron "go"? I've seen the explanation before that holes are basically electron deficiencies in an atom and that the hole "moves around" by electrons from surrounding atoms shuffling to fill that spot which is a very spatial way of viewing what a hole is. However, when I think about the band structure picture, the excitation of an electron from the valence band to the conduction band leaves behind an empty spot in the valence band hence being the "hole" but that's just a momentum space picture.
What's the connection and which one is more correct? How can I reconcile the bandstructure picture and the emergence of holes in a semiconductor. Is the "spatial picture" just a useful but incorrect tool for explaining the origin of holes?
| It is also not very clear to me. I try below an explanation.
For a single piece of doped semiconductor, I agree that there is no meaning in talking about holes "moving" from place to place in the k-space.
But in a junction, the band structure is not constant due to the diffusion of the dopants atoms. It can be thought as several thin slices of different materials held together.
When a hole is created in one of that slices, it is in this case localized. Because it is related to the local band structure.
As the slices are in contact, holes, as available energy states, can diffuse to the neighbours.
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Understanding the Functional Poisson Bracket In classical field theory (for a single field $\psi$) the dynamical variables are defined to be functions of the fields $\psi$, $\pi$, $\partial_{x_{i}}\psi$ and maybe $\mathbf{r}$, where $\pi$ is the conjugated field to $\psi$.
For $F=\int\mathcal{F}\,d\mathbf{r}$ and $G=\int\mathcal{G}\,d\mathbf{r}$, where F and G are dynamical variables, the functional Poisson bracket can be defined according to (José and Saletan, “Classical Dynamics: A Contemporary Approach”, cap 9))
$$\left\{ F,G\right\} ^{f}=\intop\left(\frac{\delta F}{\delta\psi}\frac{\delta G}{\delta\pi}-\frac{\delta F}{\delta\pi}\frac{\delta G}{\delta\psi}\right)d\mathbf{x},$$ where the derivatives are functional derivatives. The fields themselves have the canonical property
$$\left\{ \psi(\mathbf{y}),\pi(\mathbf{z})\right\} =\delta(\mathbf{y}-\mathbf{z}),$$
$$\left\{ \pi(\mathbf{y}),\pi(\mathbf{z})\right\} =\left\{ \psi(\mathbf{y}),\psi(\mathbf{z})\right\} =0.$$
So far so good, but I'm not sure how to handle the functional derivatives. I'm interested, for example, in the following bracket
$$\left\{ F(\mathbf{x}),\pi(\mathbf{z})\right\} ^{f}$$ $(F(\mathbf{x})\equiv F(\psi(\mathbf{x}),\pi(\mathbf{x}),\partial_{x_{i}}\psi(\mathbf{x}))$
Using $\frac{\delta\pi(\mathbf{z})}{\delta\psi}=0$ and $\frac{\delta\pi(\mathbf{z})}{\delta\pi}=\delta(\mathbf{y}-\mathbf{x})$, I think the answer is
$$\left\{ F(\mathbf{x}),\pi(\mathbf{z})\right\} ^{f}=\frac{\delta F(\mathbf{z})}{\delta\psi}.$$
Is this result correct?
| You are on the right track, but some clarification is needed. First as you mentioned, the Poisson bracket is defined over functionals of the local fields, i.e. an object of the form $F=\int d^n x {\cal F}[\psi(x),\pi(x)]$. Note that here we integrate over $x$, so $F$ is not a function of $x$. Accordingly, your last equation should read
\begin{align}
\{F,\pi(z)\}&=\int d^nx \{{\cal F}[\psi(x),\pi(x)],\pi(z)\}=\int d^nx \frac{\delta {\cal F}}{\delta \psi(x)}\frac{\delta \pi(z)}{\delta \pi(x)}=\int d^nx \frac{\delta {\cal F}}{\delta \psi(x)}\delta^n(z-x)=\frac{\delta {\cal F}[\psi(z),\pi(z)]}{\delta \psi(z)}
\end{align}
So in your last equation, there is two typos: on the lhs, there is no argument $x$ for $F$, and on the rhs, it is the density $\cal{F}$ instead of the functional $F$.
If you insist on having the left hand side of your last equation as it is, you can start by smearing the density $\cal{F}$ with a delta function, i.e. you take your functional to be $F(x)=\int d^nx' \delta^n(x'-x){\cal F}[\psi(x'),\pi(x')]$. Now repeating the procedure you find
\begin{align}
\{F(x),\pi(z)\}&=\frac{\delta {\cal F}[\psi(z),\pi(z)]}{\delta \psi(z)}\,\delta^n(x-z).
\end{align}
| {
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In turbulence what to we mean that an eddy has a wavenumber k? In turbulence what to we mean that an eddy has a wavenumber k?
|
In turbulence what to we mean that an eddy has a wavenumber k?
Not all eddies are created equal, if you will. That is, they need not be all the same spatial size. Different wavenumbers correspond to different length scale fluctuations. One of the basic ideas of turbulence is that energy cascades from large to small scales, thus, a spectrum of different wavenumbers play a role.
A specific type of eddy is generated by the Kelvin-Helmholtz instability, which can be described using the typical wave phase parameters, i.e., $\phi = \mathbf{k} \cdot \mathbf{x} - \omega \ t$, where $\phi$ is the phase, $\mathbf{k}$ is the wave vector, $\mathbf{x}$ is the vector position of the phase, $\omega$ is the angular frequency, and $t$ is the time. If the system is turbulent, an initial, large eddy can generate within itself smaller and smaller eddies, each with different $k$ and $\omega$.
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Are there Interpretations of Classical Mechanics, Classical Electrodynamics, and/or Relativity (either Special or General) Just like Quantum Mechanics has many interpretations (Copenhagen, Many-Worlds etc.), do any other theories in Physics have multiple interpretations? For example, does Classical Mechanics, Classical Electrodynamics, Special Relativity, or General Relativity have multiple interpretations?
| Classical theories do not have multiple interpretations in the same way quantum Mechanics does. All theories require some sort of interpretation - i.e. a way to tie the mathematics back to the "real" world, however classical theories use concepts such as mass, position, time, EM fields about which there is little real argument (has not always been the case).
The concept of space-time introduced by Relativity is sufficiently similar to classical geometry that once again there is little real argument.
Note that interpretation of a theory is different from arguments about whether the theory is "correct".
Re interpretations: Quantum Mechanics is a completely different ball game; there are fundamental arguments over even what "elements of reality" the theory should be tied back to.
| {
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Does the oscillating electric and magnetic field of a photon generate gravitational waves? From my understanding, little as it may be, because photons have energy they warp spacetime. The energy is expressed as an oscillating electric and magnetic field. Would this mean that the energy is also oscillating and would generate a gravitational wave?
| A “photon” is a quantum entity while “gravitational wave” is an entirely classical (non-quantum) concept. Generally, a good classical description of a quantum physical system could only be achieved when the number of quanta is large. Consequently, question about a single photon generating a (classical) gravitational wave does not have a consistent answer. One should either stick to fully classical description of both electromagnetic and gravitational field or instead allow for gravitons, quanta of gravitational field. Note, that while full theory of quantum gravity is still lacking, there is universal consensus that perturbative quantum gravity should provide a good description of phenomena involving gravitons at energies below the Planck scale.
So here are a few facts about interaction of electromagnetic and gravitational fields:
*
*A wave packet of EM radiation could produce gravitational waves. Standing EM wave inside a resonator could also produce gravitational waves. Because in nearly-flat spacetime gravitational waves couple to spatial stresses, oscillating EM field is a more “effective” source of gravitational waves than moving nonrelativistic bodies per same mass–energy.
*A single photon traveling in flat spacetime cannot produce a graviton. Such process is prohibited by conservation laws (of energy, momentum, angular momentum), just like a single electron alone cannot produce a photon.
*A photon traveling in background (e.g. magnetic) field can undergo conversion to graviton.
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Perpetual motion: Conditions for rolling a wheel sliding on a bar This is a basic question about energy conservation and classical mechanics:
Question: Under what situations can this motion be perpetual?
*
*Without gravity and without frictions.
*Without gravity and with frictions.
*With gravity and without frictions.
*With gravity and with frictions.
*Others setup (please state the setup)
*Impossible to be perpetual
Gravity (say) is along the vertical $y$ direction, with a constant gravitational force and a linear gravitational potential $V(y)=mgy$.
| Yup, perpetual motion. The playground for physics hobbiests. Inside a proton, perhaps..But nowhere else. So your question, to spite the conditions you established, is moot.
There is no perpetual motion.
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How can a 2-sphere exist in Euclidean 3-space? I don't know if this is a simple question to answer however, I have trouble understanding how a spherical object (such as a planet) with positive curvature can exist in Euclidean 3-space with no curvature. From my understanding Euclidean geometry seems to be the most likely description of space as we know it. Is our spherical planet resting atop a flat piece of spacetime or is it surrounded by a flat plane?
| I think you are confusing the positive-curvature of the object's surface with the curvature of the ambient space.
We can draw a 2D sphere in 3D Euclidean space. the curvature of the surrounding space would be 0, but the induced curvature on the sphere would be positive.
Just as a comment, a physical massive object does influence the spacetime curvature and makes it non-Euclidean.
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Similarity of particle in a box and free particle It can be shown that particle in a box and free particle have the same energy at certain wavenumbers (at an integer multiple of $\pi/L$ , where $L$ is the length of the box)
I am aware that the general wavefunctions of the two particles spoken of are different, but I can't get over the fact that both particles have a $k^2$ dependence on energy.
If we can do the correlation, it seems as if the free particle becomes a particle in a box at these wavenumbers. Can I visualize the free particle becoming a standing wave in such situations? But a length is not defined for the free particle so that wouldn't make sense for such a particle.
Is there any similarity between them?
| The particle in a box is simply a free particle which lives in a compact interval $[0,L]$ rather than $\mathbb R$, and whose energy eigenstates are chosen to vanish at the endpoints. In other words, the free-particle energy eigenfunctions$^\dagger$ satisfy $-\frac{\hbar^2}{2m}\psi''(x) = E\psi(x)$, while the particle-in-a-box energy eigenstates satisfy $-\frac{\hbar^2}{2m}\psi''(x) = E\psi(x)$ and the boundary conditions $\psi(0)=\psi(L)=0$.
Any function on $[0,L]$ can be extended to a function on $\mathbb R$ by copying and pasting it along the whole real line. From the above argument, it follows that (the extensions of) the particle-in-a-box energy eigenstates are also free particle eigenstates which additionally obey the prescribed conditions at $x=0$ and $x=L$.
$^\dagger$The free particle energy "eigenstates" are non-normalizable, but this is somewhat beside the point of this disccusion.
| {
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If objects in motion experience time differently, how does my body stay synced when I move my legs or arms? If I move my swing my arm really fast, the matter in my arm should experience time slower than the matter in my body.
So how does my body still sync with each other?
And a more general question that derives from this: A lot of matter move at different speeds inside our body, how does anything ever stay synced?
|
If I move my arm really fast ...
We humans think we can move quickly. We can't. When compared with the micro-machines that make up our bodies, we are actually huge giants moving incredibly slowly on a low-gravity planet. We only think we move fast because our brains are slow. Luckily they are fast enough to keep our bodies balanced.
Of course "slow" is comparative. But, compared to the finest divisions of time, we live for an eternity.
physicists have successfully recorded an internal atomic event with an
accuracy of a zeptosecond (a trillionth of a billionth of a second).
Their measurement is the smallest division of time ever to be observed
and recorded by humans.
https://futurism.com/physicists-have-measured-the-smallest-division-of-time-ever-observed
The relativistic effects of moving our body are negligible (almost zero) compared to the speed of light (the fastest thing possible). We are kept in balance by nerve signals.
Approximate speeds in metres per second
Light 299 792 458 metres per second
Nerve impulses 120 metres per second (from brain to muscles)
Humans (Usain Bolt) 12 metres per second (approximately)
Answer
The brain has constant feedback from the inner ear for balance, and proprioception tells us where our arms, legs, and other parts are at any given instant. Because the brain works faster than our muscles, it can keep recalculating all the time to prevent us falling over. In fact the brain has numerous ways of synchronising movement. I could write an essay on this, however maybe those questions would be better answered under Biology.
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Many-world interpretation of double slit experiment Different interpretations of young's double slit experiment is available, I read Copenhagen and Feynman's path integral interpretation of double slit experiment; former uses the idea of wave function and later uses the idea of infinite paths and sum over the weight factors of each path obeys causality.
I Googled lot of time to see how double slit experiment is explained using Many-world interpretation of quantum mechanics, but unfortunately I don't find anything.
It would be great helpful if someone give a nice explanation that, how famous double slit experiment is explained using Many world interpretation of quantum mechanics?
|
It would be great helpful if someone give a nice explanation that, how famous double slit experiment is explained using Many world interpretation of quantum mechanics?
In the "one photon double slit experiment" an infinite number of photons leave the light source, go through an infinite number of screens, and hit an infinite number of photographic plates.
A physicist, which is really an infinite number of physicists, adjusts the light source so that on the average it takes, let's say one minute, for some plate to become hit by some photon.
Some areas of the plate cluster absorb more photons than other areas, therefore there exists an "interference pattern". The infinite number of plates interact with each other somehow, that must be the reason for the pattern.
Some areas of the plate cluster are hit by incoherent stream of photons, those areas are the darker ones. Here "incoherent" means "two equal streams of photons whose phases are not the same, but rather opposite ones to each other".
(I used the word "incoherent" because I wanted to put forward the idea that the brighter area receives more coherent light than the darker area)
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Why is the isotope Nitrogen-14 formed preferentially to Nitrogen-15 in the CNO-cycle in stars? Most of the universe's nitrogen is formed in larger, main sequence stars using the CNO Cycle, right?
But I cannot find a good, specific explanation as to why $^{14}$N, with both an odd number of neutrons and protons, is formed preferentially to $^{15}$N?
| Both $^{14}$N and $^{15}$N are produced as part of the CNO cycle during the hydrogen-burning main sequence phase of stars more massive than the Sun. However $^{15}$N reacts rapidly with protons to (re)form $^{12}$C and an alpha particle, whereas the much slower $^{14}$N$(p,\gamma){}^{15}$O reaction allows $^{14}$N abundances to build up and dominate when the CNO cycle reaches an equilibrium.
Details:
The addition of protons to $^{14}$N
$$ p + {}^{14}{\rm N} \rightarrow {}^{15}{\rm O} + \gamma$$
is the slowest reaction in the CNO cycle and hence at equilibrium there is a build up of $^{14}$N.
As to why this is the slowest reaction in the cycle; it is likely because:
*
*(i) Of the other proton addition reactions in the cycle, adding a proton to a carbon nucleus has a lower Coulomb barrier so is faster.
*(ii) The beta decay reactions, although governed by the weak interaction, are not subject to the high Coulomb barriers of the proton addition reactions and so they are faster.
*(iii) That leaves
$$ p + {}^{15}{\rm N} \rightarrow {}^{12}{\rm C} + \alpha$$
which is a faster reaction than adding a proton to $^{14}$N because whilst the latter is a "radiative capture" reaction involving an electromagnetic transition resulting in a gamma ray (see Brune & Davids 2015), the former is a more rapid (by 4 orders of magnitude) strong force interaction. This will also be the reason why $^{15}{\rm N}(p,\alpha){}^{12}$C is totally dominant over $^{15}{\rm N}(p,\gamma){}^{16}$O and as a result allows the $^{12}$C to be regenerated; and means that there is a CNO cycle at all!
| {
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How the classical linear correlation of spin is derived?
I found this image on Wikipedia under Bell's theorem. I understand the blue curve generated by quantum mechanics, but couldn't understand how the classical curve (red curve) is generated (I don't understand how correlation changes with angle, classically). This has been asked here before but I didn't find any relevant answer. Please help me.
| The classical correlation is obtained by averaging the measurement results depending on the direction of measurement and the incoming photon polarization (physically, but the lambda could be any data, provided all of them are averaged) :
$$C(a,b)=-\int A(a,\lambda)A(b,\lambda)d\lambda$$
The case of linear correlation is when the result is 1 when the projection of the incoming photon polarization is positive on the direction of measurement :
$$A(a,\lambda)=sign(\cos(a-\lambda))$$
The correlation is simply obtained by graphically drawing the above integral.
But classical correlation could be other, for example
Suppose we take a nonrelative separable function for :
$$A(a,\lambda)=sign(\cos a\cos\lambda)$$
Then it is easy to see that
$$C(a,b)=-sign(\cos a\cos b)$$
This is very sensitive at $\frac{\pi}{2}$ because of the sign function and any small variation could make the CHSH jump.
Suppose then wlog setting the zero of both angles with : $-a$ then we obtain a step function for the relative angle :
$$C(0,b-a)=\left\{\begin{array}{c} -1, 0\leq b-a\leq \frac{\pi}{2}\\+1, \frac{\pi}{2}\leq b-a\leq \pi\end{array}\right.$$
| {
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Why do the Dirac-Maxwell Lagrangian and the QED Lagrangian look the same? I know that QED is some kind of second quantized version of the Maxwell-Dirac theory. But why is it that this modification to a second quantized version is just to replace the scalar function $\Psi$ by a field operator $\hat{\Psi}$?
| Your observation is that the field operator $\hat\Psi$ is supposed to satisfy the field equation of the classical field $\Psi$, i.e. the Dirac-Maxwell equation derived from the classical Lagrangian $$\mathcal{L}=\Psi (i\gamma^\mu D_\mu - m) \Psi \quad+\quad ...$$ where $D_\mu = \partial_\mu -i e A_\mu$ is the covariant derivative. Now let's take a step back and look at "simple" quantum mechanics. Take a harmonic oscillator, classically discribed by the Lagrangian function $$L = \frac{1}{2} m \dot x^2 - \frac{1}{2} m\omega^2 x^2$$ leading to the equation of motion $m\ddot x = - \omega^2 x$. Now we proceed to the quantum harmonic oscilator in the Heisenberg picture, i.e. the observable $\hat x$ obeys the time evolution $\dot{\hat{x}} = i\hbar[H,\hat x]$. This yields the time-dependent observable $$\hat x(t) = \hat x_0 \cos(\omega t) + \frac{\hat p_0}{m\omega}\sin(\omega t),$$ see this SE post. It is immediate that this suffices the classical equation of motion. The same is true for other canonically quantized systems in the Heisenberg picture and so is the case in QED. However, replacing $\Psi$ by an operator-valued field $\hat\Psi$ in the equation of motion does not at all grasp the full procedure of second quantisation.
So to draw a conclution, the operator field satisfies the classical equations of motion by construction of canonical quantization (in the Heisenberg picture). However, second quantization needs more than just replacing a classical field by an operator-valued field! There is a lot of standard liturature on this topic, so I would recomment to check it out if you want to understand the full extend of quantum field theory. Hope this could help to clearify the matter. Cheers!
P.S. Be carefull when calling the spinor field $\Psi$ a scalar function, it is way more then that! ;)
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Why Bose-Einstein condensate is superconducting Im looking into Quantum Computing, where the BCS Theory is used to build Qubits with a BEC.
Why does the Bose-Einstein Condensate not interact with other particles and hence has no dissipation?
In other words how does the BEC form a superconducting material. Why is there no dissipation for BEC's?
| So I assume you know in the BCS theory, fermions form Cooper Pairs (bound electrons) at very low temperatures. Since their paired state has a lower energy than the Fermi energy, they are bound. So the pair is now a composite boson with spin 0 or 1 instead of the fermions spin 1/2.
This allows them to "condense" into the same quantum state which is what we see in BECs. So really the fact that the pair essentially behaves like a boson allows for the superconducting effect to happen. Since electrons follow the Pauli Exclusion Principle, you cannot just break up one Cooper Pair without changing energies of all other pairs. This fundamentally causes an "energy gap" at very low temperatures, meaning that phenomena such as electron scattering and other single-particle excitations (small excitations) are forbidden. Hence why you would not see dissipation effects or other low energy interactions. You can then describe the system with a macroscopic wavefunction, as is done in BEC.
There are special cases (experimental conditions) that have been experimentally observed where Superconductivity and dissipation coexist but I'm not familiar with the theory or experimental results so I cannot really comment on that.
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How to convince myself that capacitance cannot be a function of voltage? My textbook states that:
...The capacitance $C$ depends only on the geometrical configuration (shape, size, separation)of the system of two conductors. [As we shall see, later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]...
Now I wanted to know if these were the only two factors on which the capacitance of a conductor depends. So I tried thinking about it from the basic definition, which is:
Capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. (Source: Wikipedia)
i.e.,
$$C= \frac {dQ}{dV}$$
where, $V$ is the potential of the conductor with respect to zero potential at infinity.
Now, I can not think of why $C = f(V)$ isn't a possible scenario, assuming that shape size, etc remains constant.
So:
*
*Why cannot $C$ be a function of $V$?
| Capacitance does depend on the applied voltage.
I think what the author means is that for many substances, capacitance will not change regardless of the potential differences between the plates. He could also mean that for the same voltage different substances will cause different capacitance.
Consider two plates with a certain voltage in between (we won’t change anything about the plates - shape, size and separation). Now consider keeping this voltage constant whilst we insert then remove different dielectric substances. Every time we insert one we measure the capacitance $Q/V$. We will keep getting different values of $C$ for different substances. This maybe what he means by capacitance not being a function of voltage.
But changing the voltage will change capacitance (but once again this is not true for many substances due to internal properties of these substances) and therefore $C = f(V)$.
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Why doesn't planet Earth expand if I accelerate upwards when standing on its surface? According to General Relativity I am being accelerated upwards by planet earth while writing this question. But a curious person on the the other side of the planet relative to me would have the same experience. That means we are accelerated in opposite directions, although earths diameter do not seem to increase. How can this be?
| That is because we are just accelerating radially outwards and not "moving" radially outwards. This case is analogous to circular motion where there is radial acceleration but no radial movement.
You can refer to my article in the link below for detailed explanation:
https://paribeshregmi.medium.com/a-soft-intro-to-general-relativity-aa46da221747
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What is the difference between Luminous intensity and intensity of illumination? What is the difference between the Luminous intensity and intensity of illumination? Please explain with units and dimensions as well! I googled but I partially understood it!
| From this Wikipedia article:
The luminous intensity for monochromatic light of a particular wavelength λ is given by:
$$I_{\nu}=683\cdot \bar y(\lambda)\cdot I_e,$$
where
$I_{\nu}$ is the luminous intensity in candelas (cd),
$I_e$ is the radiant intensity in watts per steradian (W/sr),
$\bar y(\lambda)$ is the standard luminosity function.
If more than one wavelength is present (as is usually the case), one must sum or integrate over the spectrum of wavelengths present to get the luminous intensity:
$$I_{\nu}=683\int_0^\infty\bar y (\lambda)\frac{dI_e(\lambda)}{d\lambda}d\lambda$$
Here one can read:
Intensity of illumination. Optics: the intensity of light falling at a given place on a lighted surface; the luminous flux incident per unit area, expressed in lumens per unit of area.
The luminous intensity corresponds with radiation coming from an object while the intensity of illumination corresponds with the radiation falling upon an object.
What more can I write? You can compare the two easily and discover the difference.
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How to determine the power of a magnifier lens empirically when no values are given? I have several magnifiers/magnifying-glasses of varying quality with either nothing inscribed on them, or power ratings that I think are wrong.
How can the power of a magnifier be determined when no values are given?
Kyle Downs at quora.com states:
Magnification $M$ of a single lens is $M=\frac{f}{f-d}$
I don't know for sure if my magnifiers are "single-lenses", so that may complicate this. According to edmundoptics.com:
"Most high quality magnifiers use achromats [a positive simple lens cemented to a negative simple lens] to eliminate color fringing at the edge of objects."
Thus I imagine my higher-quality "jeweler's loupes" may in fact have multiple lenses.
As I understand, focal length can be found by moving a shining a light source and moving a lens so that the projected image of the source displays sharpest on the surface. The distance from surface to lens is the focal length.
(Image from wikipedia.org)
To experiment, I measured the focal length of one my loupes to be about 9mm, and the diameter of the lens about 16mm. Using the formula mentioned above, that would give me a magnification power of 9/(9-16) = -1.29x ...Which I feel cannot be right. It is much stronger and should be more around 20x or 30x. Plus, the image is not inverted, so the magnification should be a positive value, correct?
Clearly I am going about this wrong.
| The power of a magnifying glass is relative to the near-limit distance of the standard human eye (25 cm). The lens makes it possible to bring the object closer to the eye, so that it subtends a larger angle than at the near limit. The ratio of the (tangent of the) angles is proportional to the ratio of these distances.
So if the focal length of a loupe is 10 mm, its M = 25.
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Rotating current carrying loop Consider a circular loop of wire fixed on the rim of a wheel. This wire carries a current 'i' in it. When the wheel is at rest, which basically means that the current carrying loop is at rest, the magnitude of magnetic field at the center is, say B1. If I set the wheel in motion with a constant angular velocity with the center of the wheel at rest and without changing the current in the loop, which implies that the current carrying loop rotates about its center point, will the magnitude of magnetic field change at the center point?
I think that it will change because number of charges passing through a unit cross section which is at relative rest with respect to the center point changes and hence current effectively changes and hence magnetic field changes.
Is this correct?
| The situation resembles alot to a rotating loop. So I'm using this as a hint .The formula for calculating magnetic field at the centre of a current carrying loop is directly proportional to i/R. We see that when the ring rotates the current in a small area remains constant.
This is because when the loop rotates the part of loop which crosses a region and the part which enters that region remains same . Considering that the current is constant the magnetic field has to remain constant
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When a sphere is in pure rotation, will all particles in its surface have same linear speed? Imagine this sphere to rotate about its diameter, from the centre to the point of surface if we take all of them have equal distance that is 'r(radius of the sphere)'.So same linear speed right? I looked up many sites but they all say "The linear speed v = ɷr. That means the particles at different r will have different linear speed".I am not able to digest it. Can anyone please make me understand in a better way?
| The distance to be considered is axial. You have various slices wich give different circumferences (varying radius) with a maximum at the equator.
Take for instante a cylinder: then you have only one axial radius. But the sphere has different level curves
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Is spin necessary for electromagnetism? I know that spin is needed for defining the magnetic moment of any particle, and I have also read that the spin actually is the reason why some materials are magnetic. What I want to know is whether spin is necessary for the some interactions in the electromagnetic field.
Let me expound a bit: in classical electromagnetic field theory, the electric and the magnetic fields could be considered as some combinations of partial derivatives of the vector potential ($A_\mu$). Any particle couples with the field and interacts with other particles through it.
Moving on, if we consider the quantum field theory version, we have two particles coupled with the electromagnetic field, which then interact with the exchange of bosons (photons). My question is: how big of a role does spin play in the interactions which happen through the electromagnetic field? Are there some interactions which spinless particles cannot have, but those with spin can?
| Classical electrodynamics is formulated in terms of macroscopic (i.e. averaged over many atoms/particles) fields and sources (currents and charges). It is fully condensed in Maxwell equations and supporting material equations (describing how the sources respond to the fields). As such it does not need spin, simply because it doesn't care about the origin of the magnetic moments involved. Indeed, interpreting spin in macroscopic terms, as a current due to the particle rotation, is known to be incorrect: quantitatively for the charged particles, and qualitatively for the neutral ones (such as neutron).
From the quantum electrodynamics point of view, the spin is a distinction between the cariers of interaction (bosons, which have integer spin, such as photons) and the fermions that couple to the carriers of interaction via their charge and spin. It is not quite clear how one can throw away the spin in this picture without destroying it completely.
Finally, spin affects the interactions inexplicitly via the particle statistics, i.e. via the exclusion principle.
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Are wormholes evidence for traversal of a higher dimension? Warning, pop science coming.. please correct what I’m getting wrong. Einstein’s equations of relativity showed the potential for existence of wormholes that can connect different points in space time. I understand the mechanisms for their practical implementation are nothing near feasible. However, based on the equations of gravitational “tunneling”, I can traverse back and forth between times and locations. Wouldn’t this require a higher dimension than 4d space time?
That is, we’re moving from a point that we would think of as the present to another point we would think of as the present. If this were feasible, Would These “presents” need to be on a traversable continuum?
To my lay brain, This seems as though there are points along a higher dimension where what we would consider the future is currently present, and what we consider the past is also present. That the world we see is determined and laid out as slices in a higher dimension that would be traversed with a wormhole, and that we normally traverse in a single direction.
| Agree to Rd Basha. Embedding spaces are only necessary for the mathematical constructions. They don't necesserily have physical reality.
Like the mathematics of a 2-sphere is easier if it's embedded into a 3-dimensional Euclidean space.
But the 2-sphere happily exists without a third physical dimension.
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Why does light hitting a charged particle cause it to oscillate? I am trying to study the classical Physics interpretation of light hitting a charged particle.
Why does light hitting a charge particle, such as an electron, cause it to oscillate?
Is it because the light hitting it produce an electromagnetic wave, which causes the particle to vibrate? Are photons moving in space creating electric fields for charges to move, which create a magnetic field, which then repeats the cycle?
| In classical electromagnetism, light is an electromagnetic (EM) wave. Photons are quantum objects that are absent in classical physics.
EM-waves are composed of oscillating electric and magnetic fields. Charged particles are sensitive to EM fields, which make them move in a definite direction. Then if the EM fields oscillate, then the particle oscillates as well.
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Skyrmion number The skyrmion number is defined as
$$n=\frac{1}{4\pi}\int\mathbf{M}\cdot\left(\frac{\partial\mathbf{M}}{\partial x}\times\frac{\partial\mathbf{M}}{\partial y}\right)dxdy$$
where $n$ is the topological index, $\mathbf {M}$ is the unit vector in the direction of the local magnetization within the magnetic thin, ultra-thin or bulk film, and the integral is taken over a two dimensional space.
It is known that $\mathbf{r}=\left(r\cos\alpha,r\sin\alpha\right)$ and $\mathbf{m}=\left(\cos\phi \sin\theta,\sin\phi \sin\theta,\cos\theta\right)$.
In skyrmion configurations the spatial dependence of the magnetisation can be simplified by setting the perpendicular magnetic variable independent of the in-plane angle ($ \theta \left(r\right)$) and the in-plane magnetic variable independent of the radius ($ \phi \left(\alpha\right)$). Then the skyrmion number reads:
$$n=\frac{1}{4\pi}\int_0^\infty dr\int_0^{2\pi}d\alpha\ \frac{d\theta\left(r\right)}{dr}\frac{d\phi\left(\alpha\right)}{d\alpha}\sin\theta\left(r\right)=\frac{1}{4\pi}\ [\cos\theta\left(r\right)]_{\theta\left(r=0\right)}^{\theta\left(r=\infty\right)}[\phi\left(\alpha\right)]_{\theta\left(\alpha=0\right)}^{\theta\left(\alpha=2\pi\right)}$$
My question is: is $\frac{\partial\mathbf{M}}{\partial x}\times \frac{\partial\mathbf{M}}{\partial y}$ a curl product and what is the output of this term? How to reach to the final equation then?
| It is not a curl. This can be seen by expressing the curl in vector components.
$$\nabla \times \mathbf M=\begin{pmatrix}
\partial_yM_z-\partial_z M_y\\
\partial_zM_x-\partial_x M_z\\
\partial_xM_y-\partial_y M_x
\end{pmatrix}$$
Here $\partial_x$ denotes the partial derivative with respect to $x$. The quantity $\partial_x\mathbf M$ is a vector just like $\mathbf M$. It has components
$$\partial_x \mathbf M=\begin{pmatrix}
\partial_xM_x\\
\partial_xM_y\\
\partial_xM_z
\end{pmatrix}$$
Calculating the quantity $\partial_x\mathbf M\times\partial_y\mathbf M$ is then just a matter of applying the cross product.
$$\partial_x\mathbf M\times\partial_y\mathbf M=\begin{pmatrix}
\partial_xM_y\partial_yM_z-\partial_xM_z\partial_yM_y\\
\partial_xM_z\partial_yM_x-\partial_xM_x\partial_yM_z\\
\partial_xM_x\partial_yM_y-\partial_xM_y\partial_yM_x
\end{pmatrix}$$
This is a daunting expression and you probably won't get a lot of intuition from looking at the components. What you can say about it is that $\mathbf A\cdot(\mathbf B\times \mathbf C)$ forms the vector triple product. This gives the volume spanned by (the parallelepiped of) $\mathbf A,\mathbf B$ and $\mathbf C$. So the quantity you're integrating is the volume spanned by $\mathbf M,\partial_x \mathbf M$ and $\partial_y \mathbf M$.
To calculate the integral in your last equation is just a matter of plugging everything in in my last expression for $\partial_x\mathbf M\times\partial_y\mathbf M$. This is tedious but should be doable.
EDIT I'll add some more info to make the calculation less tedious. The partial derivatives can be expanded using the chain rule $\partial_x=\frac{\partial r}{\partial x}\partial_r+\frac{\partial \alpha}{\partial x}\partial_\alpha$. These can be calculated to be
$$\partial_x=\cos\alpha\partial_r-\frac{\sin\alpha}r\partial_\alpha\\
\partial_y=\sin\alpha\partial_r+\frac{\cos\alpha}r\partial_\alpha$$
Next note that $\partial_r\mathbf M=\frac{d\theta}{dr}\partial_\theta\mathbf M$ and $\partial_\alpha\mathbf M=\frac{d\phi}{d\alpha}\partial_\phi\mathbf M$. If we name these partial derivative vectors $\mathbf e_\theta=\partial_\theta\mathbf M$ and $\mathbf e_\phi=\partial_\phi\mathbf M$ then the cross product becomes
$$\partial_x\mathbf M\times\partial_y\mathbf M=\left(\cos\alpha\frac{d\theta}{dr}\mathbf e_\theta-\frac{\sin\alpha}r\frac{d\phi}{d\alpha}\mathbf e_\phi\right)\times\left(\sin\alpha\frac{d\theta}{dr}\mathbf e_\theta + \frac{\cos\alpha}r\frac{d\phi}{d\alpha}\mathbf e_\phi\right)$$
Finally you can calculate that $\mathbf e_\theta\times \mathbf e_\phi=\sin\theta \,\mathbf M$ and you should be able to do this calculation without explicitly calculating all the components.
And yes you should add the factor $r$ when you switch to polar coordinates like you mentioned in your comment.
| {
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Is $U^\dagger(R)\hat{H}U(R)=\hat{H}$ always true? Consider a Rotation transformation on momentum state,
$$U^\dagger(R)\hat{\mathbf{p}}U(R)=R\hat{\mathbf{p}}$$
Now the question is whether,
$$U^\dagger(R)\hat{H}U(R)=\hat{H}\,?$$
Here, $\hat{H}$ is the Hamiltonian of a free particle. Is it always true? Is there any counter examples?
My attempt:
\begin{align}
U^\dagger(R)\hat{H}U(R)&=\frac{1}{2m}U^\dagger(R)\hat{\mathbf{p}}^2U(R)\\
&=\frac{1}{2m}U^\dagger(R)\hat{\mathbf{p}}U(R)U^\dagger(R)\hat{\mathbf{p}}U(R)\\
&=\frac{1}{2m}(R\hat{\mathbf{p}})(R\hat{\mathbf{p}})
\end{align}
Is this always true that $$\frac{1}{2m}(R\hat{\mathbf{p}})(R\hat{\mathbf{p}})=\frac{1}{2m}\hat{\mathbf{p}}^2\, ?$$
If it is why? If not when it is not?
Note: This is an exercise from Coleman's course 253a (https://arxiv.org/abs/1110.5013). See equation (1.8) there. It would be better if the answer is provided using his notations.
| Base transformations of a scalar operator (H) and the vector operator's spatial rotation are confused here, I think. Try to rewrite it in bra-ket notation.
Coordinate -> momentum space transformations also done in $ UHU^{-1} $ way and it really changes H.
| {
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Why is the potential not infinite? One way to calculate potential (using infinity as our reference point) is to sum all the contributions of charges that are around.
Let's say I want to calculate the potential at some point on charged surface.
At that point, there is some charge (can be infinitesimal) and that charge should contribute something divided by zero (since the distance is zero) to the potential at that point. Using that logic, every point on charge distribution should have infinite potential.
What is flawed with this argument?
| No, it would not necessarily mean infinite. This is a classic mathematical misunderstanding regarding limits.
If you have a fraction,
$$\frac ab,$$
and you let the numerator tend to zero, $a\to 0$, then the fraction might tend towards zero:
$$\frac ab\to\frac 0b=0\; \text{ for } \;a\to 0.$$
If you instead let the denominator tend to zero, $b\to 0$, then the fraction might tend towards infinity:
$$\frac ab\to\infty\; \text{ for } \;b\to 0$$
But what if both happen simultaneously? Which one will then win? Is the numerator or the denominator winning? Will the fraction as a whole tend towards zero or infinity, or something in between (converging)?
This is the situation you have. And you have entirely dismissed the numerator which is infinitesimal and are then claiming that the whole fraction is only based on the denominator tending towards zero. This is a mathematical misunderstanding. Firstly we can't directly know the answer, and secondly the answer depends on "how much" or "how fast" the number in the numerator and in the denominator tend towards their limits.
| {
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Curl of electric field is not zero in the case of a steady current in a loop Say we got conducting circular loop connected to a battery . The electric field inside the loop obeys equation $\vec{J}=\sigma \vec{E}$.
Since the current flows in a circumferential way around the loop the electric field will be circumferential as well which implies that the curl of the electric field will be non zero.
Which is a contradiction ,what's wrong in the above reasoning.
Thank you
| You can easily see that for a stationnary circuit the curl of the electric field should be $0$ (as you mentionned, this comes from the Maxwell-Faraday equation).
However, this is not contradictory with the fact that $\overrightarrow{E}$ seems to be circumferential. $\overrightarrow{J}$ is indeed circumferential but $\overrightarrow{E}$ is not. That's because $\overrightarrow{J} = \sigma \overrightarrow{E}$ only holds in a ohmic conductor, and you must have a generator in your circuit (otherwise $\overrightarrow{J} = \overrightarrow{E} = \overrightarrow{0}$), and the electric field in your generator is actually opposite to the current, in such a way that $\int_C \overrightarrow{E} \cdot \overrightarrow{dl} = 0$ (so no contradiction with $\overrightarrow{\nabla} \times \overrightarrow{E} = \overrightarrow{0}$).
| {
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Can we apply Stefan's law to find how much energy a body absorbs? Stefan's law tells gives an expression for thermal radiation emitted per unit time by a body of surface area A and temperature $T$
$$ u = \sigma A e T^4$$
In my book, it is written that in thermal equilibrium the energy of a body radiated out by stefan's law is equal to the energy radiated out. So, if a body is initially at temperature $T_o$ then it's absorbed heat is given by :
$$ u = \sigma A e (T_o)^4$$
And now suppose the body has it's temperature raised but room temperature is constant, then the energy radiated out per unit time is $ u = e \sigma A T^4$ and the absorbed is said to be $ u_o = e \sigma (T_o)^4$
Now, this is where I'm confused, why is the heat absorbed out at thermal equilibrium the same amount which is absorbed out when the temperature of the body is raised? Is there a proof for this?
Reference: In Concepts of Physics part-2 by H.C.-Verma
| If the question is why the energy absorbed per unit time is the same irrespective of the body's temperature, then the answer is: this is not strictly true, because emissivity $\epsilon$ and area of the surface of the body $A$ depend on temperature. But if the temperature changes are small, this effect may be too small to make a difference and the simple model with $\epsilon$ and $A$ independent of temperature is used.
If the temperature changes by a lot, then emissivity $\epsilon$ may change substantially enough so that the simple model may not be accurate enough.
| {
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Angular momentum operator for Dirac equation The orbital angular momentum operator is given by
$$L_i=\epsilon_{ijk}x_j p_k$$
where $x$ and $p$ are the position and momentum operators.
In the Dirac equation, the hamiltonian operator is a 4x4 matrix. Will $L_i$ then also be a 4x4 matrix, which is given by $$L_i=\epsilon_{ijk}x_j p_k I$$
where $I$ is the identitiy matrix? Or is it just still $L_i=\epsilon_{ijk}x_j p_k$ without the identity matrix?
| Formally all linear operators are $4\times 4$ matrices,
because they need to transform Dirac-spinors to Dirac-spinors.
Therefore all those operators, which don't mix the components of Dirac-spinors
(like $x_i$, $p_i$, $L_i$), contain $I$ (the $4\times 4$ identity matrix)
as a factor.
| {
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If you keep the center of gravity of two objects on each other would you never be able to separate them? To find the attractions between planets and stuff like that, you use the center of gravity/mass to apply to Newton's equation. So even if those planets collided into each other, you could separate them if you give enough force, because $r$ (distance between the center of gravity/mass of each planet) in the gravitation equation is not $0$ therefore $r^2$ is not $0$. But the problem comes when you put the centers of gravity/mass of two objects on each other. Then $r$ is $0$, $r^2$ is $0$ and when you divide by $r^2$ (in the gravitation equation), you’re dividing by $0$ which means the gravity is infinite; i.e you'll never be able to separate them. Now you might say that there will never be such an instance where the two centers of gravity/mass will never be on each other, but consider this-
Two hoops, one 1/2 in radius of the other, placed on a table such that the circumference of those 2 hoops are parallel (like a train track that goes in perfect circles). The center of mass of the bigger hoop will be at the very center of the area (circle) enclosed by the bigger hoop. The same goes for the second, smaller hoop. The center of mass of each hoop will lie on the same point. So does that mean no matter how much you tried, you'll never be able to separate them? This question has been puzzling me for ages so help would be great.
| No. When centered, your two concentric hoops don’t exert any net force at all on each other. There is a net force in other positions, but never an infinite force. You can separate them.
| {
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How to use Ampère's law in this problem? I want to prove that the magnetic field inside a cylindrical cavity in a long, cylindrical conductor carrying uniformly distributed current i and having radius R is uniform. The radius of cavity is 'a' and it's axis is parallel to the main conductor axis. However the axis of cavity is not coaxial with main conductor.
My attempt- I was able to think that the ampèrian loop should be rectangular with two long sides of length L and two short sides of length less than the radius of cavity. Now if one of the longer side is along the axis of cavity, the other longer side should be parallel to this and to ensure that the line integral along shorter side does not contribute significantly, we may take the longer side very long. After this I am having confusion in proving that the magnetic field at center of cavity is same as that at a off axis point. I am having confusion with directions and magnitudes of the field along sides of rectangle.
| This is a standard "trick" question. You compute the field due inside the fat conductor carrying a uniform current I amps per unit area and ignoring the cavity. This is a standard Ampere law calculation and gives you circumferential field proportional to the distance from center of the fat conductor. You then compute the field due a thin conductor inside the fat one and carrying uniform current density -I amps per square meter. Again this is an easy calculation. Superpose the two currents and you have no current in the overlap (cavity). The total field is the sum of the two fields, and is easy to compute in the cavity.
| {
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Is the Green function of electromagnetism a scalar or a tensor? When I check classical electromagnetism books Maxwell equations
\begin{equation}
\Box A^\nu (x)=\frac{4\pi}{c}j^\nu (x)
\end{equation}
can be solved using a scalar Green function $G(x,x')$
\begin{equation}
A^\nu (x)=\int G(x,x')j^\nu (x')d^4x'
\end{equation}
where the Green function satisfies
\begin{equation}
\Box G(x,x')=\frac{4\pi}{c}\delta^4(x-x')
\end{equation}
Examples of this are Jackson, eq. 6.48 on sec. 6.5. Also, on "The classical theory of fields" by Landau, on eqs. 62.9 and 62.10 he uses the scalar green function as well.
This immediately feels strange, since the 4-potential $A^\nu(x)$ could, in theory, have different boundary conditions for each component and a scalar Green function simply doesn't have enough degrees of freedom to accommodate that. Evenmore, in the context of quantum field theory, the photon propagator (which is essentially the Green function) is a tensor $\Pi_{\mu\nu}$ so I'm confused about the nature of the Green function in classical electromagnetism: Is the scalar Green function $G(x,x')$ the most general Green function or in a general case we need a tensorial Green function $G_{\mu\nu}(x,x')$?
Note: This question is explicitly about classical electromagnetism, I'm using the quantum field theory propagator as an example to show my confusion but the question applies to the classical theory.
| It is a scalar or a second rank unit tensor, which amounts to the same. The boundary condition argument does not hold because physically a boundary condition can only be realised by additional charge-current distributions, that is, the boundaries are sources of field themselves.
| {
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Why don't we use the concept of axis of mass in place of center of mass? Being a high school student, I read the concept of center of mass and it was written in my book that
When a spinning ball is projected with some velocity , then all the points on the ball have complicated paths except the center of that ball which follows the well known parabolic trajectory. And hence we define that point as center of mass.
However, I think that all the points on the any axis about which the ball is spinning follow the parabolic trajectory and are not influenced by spin .
Edit : Most of the answers argued that the rotation axis may change because of torque but the main point to note here is that we can't differentiate between two axis in case of a sphere since it is symmetric from all sides and also that a sphere can't rotate about more than one axis at a time . So saying that it will rotate about different axis is I think meaningless.
So is it okay to define axis of mass in place of center of mass for sphere or other symmetric bodies or am I wrong somewhere ? If not, give a proper reason.
| Take a globe. Spin it around its normal axis, the polar axis the way the Earth spins. Then pick the globe up and while it is still spinning the first way, flip it north to south the way you would a coin. There you have a sphere spinning over two axes at the same time and your axis of mass makes no sense. The center of mass and only the center of mass is traveling in the parabolic trajectory under your conditions.
| {
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Can quantum measurements be the origin of thermodynamic arrow of time? We can practically consider that the microscopic interactions are symmetric with respect to time(as we can neglect weak force for many cases which is the only interaction that can violate $T$ symmetry). So I thought that the asymmetry might be due to the irreversibility of quantum measurements. But this is only applicable for interpretations where wave function collapses like Copenhagen etc. What is the answer to this question in Many-worlds interpretation, Consistent histories, etc? Also in this page, they gave that the initial conditions of the universe are the reason for $T$ asymmetry in the 2nd law of thermodynamics. But I am not sure what they mean. Do they mean that the universe had a very low entropy at the beginning?
|
Do they mean that the universe had a very low entropy at the beginning?
Yes, the universe had a very low entropy immediately after the Big Bang. It was filled with a very uniform distribution of very energetic (very "hot") fundamental particles. Due to the effect of gravity, this uniform distribution is actually a highly improbable state, and so has an extremely low entropy. So the universe started in a low entropy state, and its entropy has been increasing ever since, following the second law of thermodynamics.
The natural follow-up question, which is very interesting, is why did the early universe have such a low entropy to start with ? Was this inevitable, or is it an unusual and unlikely feature of our particular universe ? Since we have no other universes to compare ours with, this is a very difficult question to answer !
| {
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How is a free theory defined? In field theory, I've seen a free theory described as
*
*A field with the specific Lagrangian density ${\cal L}=|\partial\phi|^2+m^2\phi^2$
*A field whose equation of motion yields a linear set of solutions
*A field with non-interacting i.e. free normal modes
The first seems too specific, the second seems too general, and the third seems ill-defined. I was hoping that these three could be extended to solve any of those problems or if there is some way to unify, say, the first and the second then maybe that final description would strike right.
| I think these might all be equivalent up to change-of-basis. The definition I have heard is "Lagrangian is bilinear in the fields", which I think is also the same.
In the basis in which the bilinear operator is diagonal, the equations of motion are linear, so (2) and (3) are the same up to change-of-basis.
If the field is scalar and all the terms in your bilinear operator involve 0 or 2 derivatives, you can use integration by parts to put the operator in the form of (1) (up to scaling). I think all Lorentz-invariant self-adjoint bilinear operators on scalar fields have to be of this form. However, there might be non-Lorentz-invariant operators which satisfy (2) and (3) but not (1). I'm not sure.
| {
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How is torque transmitted between inclined surfaces? In the picture below, in a), a body K1 is pivotably attached to a bearing. My question is about the torque that results from a force exerted onto a surface of the body K1.
A first force F1 applied orthogonally onto the surface should result in a torque M1 in clockwise direction.
Is it correct that a second force, F2, applied almost parallel to the surface will result in a torque M2 in counterclockwise direction?
My thoughts are, F2 is split into F2t and F2o (transversal and orthogonal components) by the surface of the body K1. To get a torque, F2o is multiplied by the lever b and F2t is multiplied by the lever a (M2 = F2t * a - F2o * b > 0). As a>b and F2t>F2o, the torque from the force F2 results in counterclockwise direction.
Applying these thought to the two bodies K1, K2 in b), a torque of M3 applied to the body K2 will result in a torque M4 in the body K1. (The bodies won't move because they are in each others movement path)
Is this correct or am I forgetting something? What is the job of friction in this case? From looking at b), K2 should push K1 away by applying a clockwise torque, but that is wrong then, right?
Suppose there is enough friction so that no slippage occurs.
| I think friction is required for any torque to be applied CCW, which be definition works against torque being applied CW. So with no friction it would be net torque CW, but with "infinite" friction (i.e. no slipping) it would be net torque CCW (and also locked up and not spinning). I don't think this question can be solved without some assumption or knowledge about the friction involved.
| {
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Does the Bell test preclude localism, realism, both, or just one of either (indeterminate)? I saw this excerpt from the wikipedia article on EPR paradox
They postulate that these elements of reality are, in modern terminology, local, in the sense that each belongs to a certain point in spacetime. Each element may, again in modern terminology, only be influenced by events which are located in the backward light cone of its point in spacetime (i.e., the past). These claims are founded on assumptions about nature that constitute what is now known as local realism.
Local realism is made up of the notions of locality and realism. Realism being the idea that nature exists independently of the observer, while the principle of locality states that an object is directly influenced only by its immediate surrounding.
Does the test preclude both locality and realism, or all that we can say is "local realism"? That is, that both things cannot simultaneously be true.
| The violation of Bell inequalities proves that quantum mechanics is incompatible with the assumption of local realism (rather than with the assumptions of locality and realism), in the sense that either the one or the other or even both turn out to be false.
| {
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Reduced density matrix: Derive or postulate? Let us consider a quantum mechanical system of interest S that interacts with the environment E. Then, the reduced density matrix
$$
\hat \rho_\mathrm{S} = \mathrm{Tr}_\mathrm{E} \{ \hat \rho \}
$$
is the partial trace over the environment, where $\hat \rho$ denotes the density matrix of the complete system (S + E).
Is this expression postulated or can it be derived (from basic postulates, such as the set given by Nielsen and Chuang for example)?
| As indicated in the answer by @Vadim the expression you give is a definition of the reduced density matrix. From this definition, the Born rule and some probability you can prove that the reduced density matrix carries information about the marginal probability distribution for the subsystem which it describes.
| {
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Why is the electrostatic force felt in straight lines? When two positive charges are kept close, they get repelled in the direction of a line joining both the charges. Why is it so?
Also, why is the repulsion in a straight path?
In both the cases, the potential energy of the charge which gets repelled decreases. What makes it repel in a straight line such that the line passes through both charges?
| Great question!
Perhaps you're familiar with the idea that the force on an object is the negative gradient of the potential energy created by that force*:
$-\vec{\nabla} PE(x,y,z) = \vec{F}(x,y,z)$
Now, imagine any slope. The gradient vector at any point on that slope points in the direction that you would step to increase you're altitude most quickly. The negative gradient vector does the opposite: it points in the direction that you would step to decrease you're altitude most quickly.
With this in mind, think about what this physics equation is really saying! The force on an object always points in the direction of travel that would most efficiently decrease it's potential energy.
So it's not JUST that the forces on the object want to get rid of it's potential energy, it's that (in this sense) they want to do it as efficiently as possible. This is why the positive charge wants to get away from the positive charge in a straight line: it wants to decrease it's potential energy as efficiently as possible.
*if the force is conservative, and here it is.
| {
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I Need the Z-Machine's Circuit Diagram Where can I find a copy of the circuit diagram for Sandia Lab's Z-Machine, or indeed for any pulse forming network that outperforms the ordinary Marx bank (if such a thing exists)? Google has let me down in my search for this information. TIA.
| A recent Physics of Plasmas on the Z Machine has a high level overview with copious references to the fundamental technology. However, it is not clear just how detailed of a 'circuit diagram' you want. Simulations using standard electrical engineering analysis tools, while useful as rapidly calculated analogs of the machine, do not capture the full physics of the power flow.
| {
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Describing small, NRQM systems purely in terms of photons Is there a canonical way to describe an open, non-relativistic quantum system with density matrix $\rho(t)$ entirely in terms of the light that it emits and absorbs (and vice versa?) Or is it possible in general for a density matrix trajectory $\rho(t)$ to be induced by several (e.g. possibly contrived and time dependent) photon baths?
| For Markovian systems, this is possible in a certain sense. If the quantum system is linearly coupled to the bath (which is usually the case, e.g. in light-matter interaction, in cavity-bath interaction etc.), one can write an input-output relation
$$\hat{b}_\mathrm{out}(t) = \hat{b}_\mathrm{in}(t) + i\kappa\hat{a}(t)$$
where $b$ are the bath operators, $a$ is the system operator and $\kappa$ the coupling between them.
This equation implies that if you are able to measure/specify the correlation function of the input and output operators, you are able to reconstruct the corresponding correlation function of the system operator.
| {
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Confusion about the dimension of a Hilbert Space in Quantum Mechanics In Quantum Mechanics, the quantum state of the physical system lives in an infinite-dimensional Hilbert space and can be written in terms of two different bases, the position basis (uncountably infinite) and the energy basis (countably infinite). Apparently, the two bases are of different cardinalities, which violates a theorem in Linear Algebra that all bases of a vector space must be of the same cardinality. How to explain this confusion?
| Note that the "position basis" is not a basis, because its elements $|x\rangle$ are not elements of the Hilbert space. It is a useful structure to work with, but making mathematically rigorous statements about it can be quite tricky.
| {
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Higgs Lagrangian and massive/massless fields In QFT yesterday we were talking about Higgs/SSB and mass terms in Lagrangian. Our professor wrote down some lagrangians
and asked us to explain certain things and also if it is for massive or massless field. One of them was this
Lagrangian (not a realistic one I think)
$\mathcal{L} = \frac{1}{2} (\partial_\mu \phi )^2 - e^{m \phi} + \lambda \phi^4 $
and we said this cannot represent massive field since there is no $\phi^2$ term therefore $m = 0$. He said we are wrong and the point is
that mass terms are not always available by inspection of $\phi^2$ term in some complex Lagrangians. The mass term could be hidden.
This confused us. Can anyone explain
why this is Lagrangian describing massive field?
| In this case you can
demonstrate that this Lagrangian represents a massive field. You do this by looking at the exponential term $e^{m\phi}$ in
$\mathcal{L} = \frac{1}{2} (\partial_\mu \phi )^2 - e^{m \phi} + \lambda \phi^4$
and performing a Taylor expansion. That is
$e^{m\phi} = 1 + m \phi + m^2 \large \frac{\phi^2}{2} + ....$
This means the Lagrangian can now be written
$\mathcal{L} = \frac{1}{2} (\partial_\mu \phi )^2 + ... - m^2 \large \frac{\phi^2}{2} + $ $....+ \lambda \phi^4$
where the only term in the expansion involving $m^2$ is explicit. So this Lagrangian does indeed represent a massive field.
| {
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Is Penrose's CCC consistent with Penrose's singularity theorem? According to Penrose's Conformal Cyclic Cosmology (CCC), there were universes prior to ours, prior to the singularity of our universe.
But how is this claim compatible with his famous singularity theorem, according to which spacetime geodesics cannot be extended beyond a singularity?
I believe Penrose doesn't deny the big bang singularity. Then how does he make sense of 'spacetime prior to the big bang singularity' in CCC?
| Long story short, the Big Bang is a singularly unique singularity which is mathematically no different than the massively expanded universe in the far, far future. Because they are the same, one infinitely expanded universe becomes the infinitely small start of the next.
The mathematics he uses to demonstrate this comparison is called conformal geometry, a math that remains consistent despite working with the cosmically infinite be it infinitely huge or small. Conformal geometry has some advantages, apparently, in that it "squashes" the infinities at the beginning and end of the universe into quantised concepts. It also has some advantages because it allows the Big Bang to occur without the need for Inflation in the very early moments of the universe.
In fact, one of the main theoretical arguments in favour of CCC is that it overcomes some issues with Inflation which the cosmic background radiation map presents, namely some ripples which should not exist with Inflation. CCC explains those ripples as the gravity outcomes of collisions between super massive blackholes or the final 'pops' as blackholes eventually evaporate in the previous universe. Gravity, the mysterious non-force force, traverses from one universe to the next and makes itself apparent in the ripples of the microwave background. He makes predictions about what those ripples would looks like and has a few people supporting him with claims that they see the suspected ripples.
| {
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Why does power increase as a constant force accelerates a body? If a constant force is being applied to a body, without any other external forces, F = ma says that that body will accelerate at a constant rate. This acceleration will continuously increase the body's velocity.
According to P = Fv, since the force is constant and the velocity is continuously increasing, the power required by the force will continuously increase.
I understand all the maths, but am trying to get a better intuitive understanding of this. I cannot seem to come to terms with the fact that a constant force will need to supply an increasing power. What is this power being represented by, if the force is constant? What typical inner-workings of such a force would require its power to increase, even though its ultimate "output" is the same? What actually constitutes "power" and "force" at the "force-side" of things?
| Suppose the body starts from rest at time $0$ and accelerates at a constant rate $a$. At time $t$ it has speed $v=at$ and kinetic energy $E=\frac 1 2 ma^2t^2$.
At time $t+\delta t$ it has speed $a(t+\delta t)$ and energy $E + \delta E = \frac 1 2 ma^2(t+\delta t)^2$. So
$\delta E = m a^2 t \delta t + \frac 1 2 ma^2 (\delta t)^2$
So the power applied to the body is
$\frac {\delta E}{\delta t} \rightarrow ma^2t$ as $\delta t \rightarrow 0$
So power increases linearly with time because kinetic energy increases as the square of speed.
| {
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Orbit with crash Let's assume I shoot an object from a high tower horizontal to the earth's surface. As far as I understand, depending on the velocity I will get different types of orbits. With decreasing velocity I will go from
*
*hyperbolic orbit where the focal point is the earth's center
*parabolic orbit where the focal point is the earth's center
*elliptic orbit where the earth's center is that focal point that is nearest to the tower
*circular orbit where the earth's center is the center of the circle
*crash orbit
My question is about the last orbit. If earth was transparent for the thrown object,...
Would orbit 5 be an ellipse with the center of earth being at the focal point furthest from the tower?
| The answer is yes, even though it doesn't make sense to talk about "orbit" in your case as the object crashes on the surface of the planet.
As a thought experiment, though, you can think of the Earth as a point particle, and your object being shot in space far from the surface. Then, as its trajectory starts "bending" downwards, it doesn't hit the surface of the Earth and can propagate.
This is best exemplified in the following image from here, where the blue planet would be my "point-like" Earth:
Out of completeness, it might be worth mentioning that for spherically symmetric vacuum solutions where GR effects are important, there is a correction to the Newton's potential that actually results in an innermost stable circular orbit ($R_{\text{ISCO}}$), below which there are plummeting orbits.
| {
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What is the equivalent to $\Box A^\alpha =- \mu_0 J^\alpha$ using differential forms? The set of equations $$\Box A^\alpha = -\mu_0 J^\alpha$$
can be found in section 12.3.5 of Griffiths's book. From what I understand, the real-valued functions on both side of the equations are the coefficients of some $1$-forms with respect to a chart. Thus, I am wondering how the equivalent index-free equation involving differential forms looks like.
| If $$\delta \equiv *d*$$
then the Maxwell equations amount to
$$\delta d A = J$$ with $\mu_0 =1$.
Starring the above equation, you get conservation of current. Taking the dual twice will get you to the form you started with, that is modulo some signs. Taking the dual of the above equation you get conservation of current
$$
*\delta d A \sim d*dA = *J
$$
now apply $d$ and get as a consistency on the equations of motion current conservation:
$$
d*J = 0.
$$
It is interesting exercise to consider the action from which those equations of motion come from, and discover how the variational derivative would be expressed in terms of differential forms. My suggestion, without the current, would be
$$
S_{EM} = \int dA \wedge * dA
$$
| {
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Understanding of electric potential's integration form I already known that the potential difference when a charge moves from A to B is
But I still have confusions about what does the infinitesimal of vector $s$ refers. I mean when you change the movement of the charge from B to A, the $\Delta V$ should be opposite number of it. But if the E and S's direction is opposite, the dot product of E and S should be negative. Since the range of integration is reversed, the out come of the delta Vab is same to delta Vba.
Help is really appreciated.
| If I understand your question, I think you are assuming that when you integrate from B to A two things change: the direction of ds (and therefore the dot product of E and ds) and the integral (because you changed the limits) and therefore you end up with the same as when you integrated from A to B. But integrating from B to A just changes ds to the negative of what it was when integrating from A to B. That's what changing the limits does. So the dot product is the negative of what it was when integrating from A to B and the end result is the negative of the original potential difference.
| {
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Precise zero energy bound for supersymmetry Usually we can shift the energy $E$ by any amount $\delta$ to redefine the lowest energy as
$$
E + \delta.
$$
However, in supersymmetry, there is a precise $E=0$ must be true, so that the supercharge $Q$ annihilates some state $|\psi_{min}\rangle$ to give the minimal energy
$$
Q |\psi_{min}\rangle =0
$$
and also
$$
H|\psi_{min}\rangle =Q^2|\psi_{min}\rangle=0.
$$
Question
This raises the question that we have a precise zero energy bound for supersymmetry theory.
Does it mean that we cannot shift the energy $E$ to $E + \delta$ in supersymmetry theory? What is the deep reason behind it?
| I will offer two arguments to try to exhibit what the problem with ground states with non zero energy in a supersymmetric theory is.
The case of gauged supersymmetry: Energy backreacts on the geometry.
If the energy of the ground states of a supersymmetric theory were non exactly zero the underlying geometry of the background would be allowed to be curved in arbitrary ways. The latter is imposible in a supersymmetric theory because supersymmetry enforce at least an spin structure of the underlying background geometry or possibly a trivial canonical bundle or a special holonomy (as in the Calabi-Yau, $G_{2}$ or $Spin(7)$ cases) on the target manifold.
Supersymmetry in quantum mechanics: The hamiltonian in a $(0+1)$-supersymmetric theory can be schematically written (see Supersymmetry and Morse theory) as $$H=\frac{1}{2}(Q_{1}^{2}+Q_{2}^{2}).$$
If for some ground state $\psi$ $$H\psi \neq 0,$$ it would follow that $\psi$ wouldn't be annihilated by the (positive definite) squares of the supercharges; then $\psi$ wouldn't preserve some supersymmetry.
It is illustrative to notice that in the case of a Riemannian manifold $\mathcal{M}$ the hamiltonian coincides with the laplacian of $\mathcal{M}$ (see chapter 2 in Supersymmetry and Morse theory); if the laplacian was not zero, then you can make an analogy with electrodynamics, you can't be possibly describing the vaccum of the theory, sources must be present.
| {
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What exactly means "local"? How local (and memoryless) is "local"? Local and memoryless are easily defined in quantized space and time:
Local: What happens from one time step to the next in one "cell" of quantized space is only influenced by the state of neighboring cells.
Memoryless: The state in the next time step is only influenced by the state in the previous time step (not earlier ones).
My understanding is when we talk about the laws of physics being local, what is actually meant is local and memoryless.
In a continuous spacetime, I guess one could define locality (and I assume that's how it is defined) roughly as "the shorter the time difference, the smaller the neighborhood that can influence what happens during this time". And that is indeed how it works since the light speed imposes exactly such a limit.
But how "local" is physics when we get to really small time differences and distances, since we also get things like uncertainty of position into the picture. That seems to impose a limit after which (I guess) physics seems to behave non-local? Does it still behave memoryless? If I reduce time difference further, does the distance also still shrink to whatever can influence what happens? Is "distance" even still well-defined when uncertainty of position becomes significant?
| Local may have different meaning depending on the context. One common use of this term is in describing the equations of physics as local, because a value of a function/field is determined only by the values of this and other functions/fields at the the same point (or infinitesimally close points), as reflected by the fact that the equations are differential equations of finite order.
| {
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Shouldn't We modify the field in force equation $\mathbf{F}=q\mathbf{E}$? Consider charge particle $q$ in electric field $\mathbf{E}$. The force on the charge is given by
$$\mathbf{F}=q\mathbf{E}$$
Now we know that charge $q$ will also produce an electric field. Due to this field, the field already present in the space should be modify. And thus we should use the modified version of the field. But we don't? (atleast I didn't see).
So the question is if the above reasoning is correct what should be the correct expression? If it's wrong why?
| The electric field that appears in this expression (and in Lorentz force equation more generally) is the total electric field, meaning the field as contributed from all sources. The reason the field due to the point particle in question (which would make it interact with itself) is usually ignored because the field of a point particle diverges at the location of the point particle and it becomes impossible to get anything approaching a reasonable result.
The exact calculation including the interaction of a particle with its own field can be performed but is extremely intensive from a technical standpoint (the calculation appears near the back of Classical Electrodynamics by Jackson). This issue exact issue with point particles can in some respects be pinpointed as a signal that point particles are probably only useful as an approximation than that a better description might involve fields that change in time instead of particles that move around (fields do not run into the same issues with divergent quantities).
| {
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Why not consider rate of change of acceleration? Why do we not consider rate of change of acceleration in the study of linear motion?
| Sometimes we do need to consider the rate of change of acceleration - sufficiently often for it to have been given a name. The rate of change of acceleration is called jerk. Limiting the magnitude of jerk is an important consideration in the real world design of railway tracks, elevators and roller coasters.
The initial kinematics problems that you encounter in a physics course tend to focus on motion with constant acceleration for the sake of simplicity, so as not to confuse the student with too many new concepts at once.
| {
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Differential operators in QM whose domain is a subspace of $L^2$ act on equivalence classes - How is that even defined? As far as I know, differential operators in Quantum Mechanics (for example the momentum operator) are defined on a subspace of $L^2$ (if $L^2$ is the Hilbert space). This means that they act on equivalence classes - for what subspace and how is the derivative defined?
| Each equivalence class in the domain must contain a differentiable element. The differential operator acts on those representatives. Notice that each such representative is necessarily unique, since two continuous functions which are different on a zero measure set are actually everywhere equal. So the action is well defined.
| {
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A Loop Quantum Gravity question Can someone answer this question, I have chosen this from Bodendorfer's article on 'An Elementary Introduction to Loop Quantum Gravity' from section 3 General relativity in the connection formulation and quantum kinematics exercise 3.6.10
Ashtekar-Lewandowski Vaccum
A Cylindrical Function Ψ = 1, this state is called Ashtekar-Lewandowski Vacuum. Show that it corresponds to a maximally degenerate Spatial Geometry by evaluating the Vacuum expectation value of the Flux Operator
| Well, the answer is in the question – you need to evaluate the flux operator!
It's not that hard to see actually. Draw an arbitrary auxiliary graph, and note that the Ashtekar-Lewandowski vacuum state corresponds to a spin network on this graph with all spins equal to zero (because the spin-0 representation of $SU(2)$ is the trivial representation that assigns $1$ to each of its elements, which leads exactly to the state $\Psi[A] = 1$).
Now remember that area in LQG is proportional to
$$
A \sim \sum \sqrt{j(j+1)}.
$$
Substitute $j = 0$ and you will get $A = 0$ – all areas vanish for such a state. I believe that's what "maximally degenerate spatial geometry" means.
| {
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Matsubara sum with log term How do I compute the Matsubara sum
$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right)?$$
If I have sums like
$\sum_n \frac{1}{i\omega_n -m}$, I can sum it up by calculating the sum of residues of the function $\frac{1}{z-m}g(z)$ at the poles where $g(z)=\begin{cases}
\frac{\beta}{\exp (\beta z)+1} \text{ for Fermions}\\
\frac{\beta}{\exp (\beta z)-1} \text{ for Bosons}
\end{cases}$
But how do I do I compute in this case where there is a $\log$ term and there are no poles.
| For this kind of stuff you usually integrate by parts.
First change your sum to an integral:
$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right) \Rightarrow \int \mathrm{d}\omega \, \log(f - \mathrm{i}\omega), $$
where $f$ here is $k^2/2m+\mu$ which I am assuming are not functions of $\omega$.
Then integrate by parts:
$$\int \mathrm{d}\omega \, \underbrace{1}_{u'} \cdot\underbrace{\log(f - \mathrm{i}\omega)}_v = \underbrace{\omega}_{u}\underbrace{\log(f - \mathrm{i}\omega)}_{v}\bigg\vert_0^{\omega_{\text{max}}} - \int\mathrm{d}\omega\, \underbrace{\omega}_u\cdot\underbrace{\frac{\mathrm{-i}}{f-\mathrm{i}\omega}}_{v'}$$
$$\Rightarrow \omega_{\text{max}}\log(f - \mathrm{i}\omega_{\text{max}})+ \int_0^{\omega_{\text{max}}} \mathrm{d}\omega\,\omega\cdot\frac{\mathrm{i}}{f-\mathrm{i}\omega}.$$
The first term is a constant energy offset. Usually it cancels out when you consider differences in energy.
| {
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Problem with the proof that for every timelike vector there exists an inertial coordinate system in which its spatial coordinates are zero I am reading lecture notes on special relativity and I have a problem with the proof of the following proposition.
Proposition. If $X$ is timelike, then there exists an inertial coordinate system in which $X^1 = X^2 = X^3 = 0$.
The proof states that as $X$ is timelike, it has components of the form $(a, p\,\mathbf{e})$, where $\mathbf{e}$ is a unit spatial vector and $\lvert a \rvert > \lvert p \rvert$. Then one considers the following four four-vectors:
\begin{align*}
\frac{1}{\sqrt{a^2 - p^2}}(a, p\,\mathbf{e}) & & \frac{1}{\sqrt{a^2 - p^2}}(p, a\,\mathbf{e}) & & (0, \mathbf{q}) & & (0, \mathbf{r})\,,
\end{align*}
where $\mathbf{q}$ and $\mathbf{r}$ are chosen so that $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ form an orthonormal triad in Euclidean space. Then the proof concludes that these four-vectors define an explicit Lorentz transformation and stops there.
For me this explicit Lorentz transformation is represented by the following matrix.
\begin{bmatrix}
\frac{1}{\sqrt{a^2 - p^2}} a & \frac{p}{\sqrt{a^2 - p^2}} & 0 & 0 \\
\frac{p}{\sqrt{a^2 - p^2}} e^1 & \frac{a}{\sqrt{a^2 - p^2}} e^1 & q^1 & r^1 \\
\frac{p}{\sqrt{a^2 - p^2}} e^2 & \frac{a}{\sqrt{a^2 - p^2}} e^2 & q^2 & r^2 \\
\frac{p}{\sqrt{a^2 - p^2}} e^3 & \frac{a}{\sqrt{a^2 - p^2}} e^3 & q^3 & r^3 \\
\end{bmatrix}
However, multiplying the column vector $(X^0, X^1, X^2, X^3)$ by the matrix above does not seem to yield a column vector whose spatial components are zero.
What did I miss?
| The matrix you wrote will take the standard basis $\{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)\}$ into the basis constructed from $\mathbf{X}$. Therefore, to take $\mathbf{X}$ into something proportional to $(1,0,0,0)$, you need to use the inverse matrix.
| {
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