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Nov
25
comment Do bad clocks measure proper time?
Well, quite honestly, I had never even heard the term "bad clock" until I read this question. I think it's safe to say that, unless otherwise stated, if someone says "clock" they mean "good clock". With this assumption, then all three quotes are correct and they would not retract their claims.
Nov
24
comment Do bad clocks measure proper time?
In particular, in general, no single observer can measure the entire space-time metric everywhere. You can only measure the space-time metric locally.
Nov
24
comment Do bad clocks measure proper time?
I'm afraid I will not be able to explain how one can measure the metric in an understandable way without being able to draw a diagram, but the basic idea is one that comes up in a derivation of length-contraction/time-dilation. Briefly, you send a light beam to a point you want to determine the distance to (after you've placed a perfect mirror there), and by measuring the time (proper time, according to you) it takes to get back to you, you can determine the distance to that point. It's not quite this simple in curved space-time, but that's the basic idea.
Nov
24
comment Do bad clocks measure proper time?
By definition, bad clocks do not measure proper time. The argument I gave should show that this is equivalent to 'making the world-lines of free particles locally look curved'.
Nov
22
comment Work-Energy theorem vs conservation of mechanical energy?
Sure, but this is not the integral of $-kx$, it is the integral of $-kx(s)$ with respect to s.
Nov
21
comment Work-Energy theorem vs conservation of mechanical energy?
How do you get that result for the integral? It should not be the case that $x(s)=s$. Remember, the spring itself is moving as well, so it should be the case that $s\geq x(s)$.
Nov
14
comment A question from Schwinger's particles, sources and fields monograph
I am confused. In some places you write $\phi ^\mu$ and in others you write simply $\phi$. Is there a scalar field $\phi$ and a vector field $\phi ^\mu$?
Nov
14
comment Calculating the Sun's emitted power in a wavelength range?
See wikiwand.com/en/Stefan%E2%80%93Boltzmann_law#/… . I believe this calculation is very similar to what you would like to do.
Nov
14
comment Which renormalisation techniques are available for 3+1 QED?
You should probably distinguish between re-normalization and regularization. Pauli-Villars is a method of regularization. Two other ones that come to mind are dimensional regularization and the usual cut-off regularization.
Nov
12
comment Momentum conservation in an electromagnetic system?
Well, I honestly don't know why you would need to do this. Perhaps I misunderstand your question. I thought that you took note of the fact that momentum was not conserved, and was confused as to why that was not the case. And the reason of course is because we are applying a non-zero net force to the system. As a 'sanity check', I suppose you could actually calculate the change in momentum of those three things to verify that it does indeed come out to $f\Delta t$, but given that it's so tedious to do so, I don't know why you would want to do that.
Nov
12
comment Momentum conservation in an electromagnetic system?
So in particular, it shouldn't be balanced. They should differ by $f\delta t$. Also, I admittedly don't understand why you think that the momentum of the E&M field won't change unless the top charge accelerates to a non-negligible degree. The Liénard–Wiechert depend on the velocity, which, for the bottom particle, is changing to a non-negligible degree (though I guess it's possible that you might find this dependence 'cancels' when you do the actual computation).
Nov
12
comment Momentum conservation in an electromagnetic system?
. . . Then the momentum change of the E&M field would just have to be so that, after you add these three together, you get $f\Delta t$.
Nov
12
comment Momentum conservation in an electromagnetic system?
So I am admittedly trying to avoid details because I believe the actual computation will be at least slightly tedious. Nevertheless, I think it's easy to see that this is possible. The system has three things contributing to its momentum: the mechanical momentum of the top particle, the mechanical momentum of the bottom particle, and the momentum of the E&M field. If we declare that the positive direction is to the right, then the momentum of the bottom particle is increasing over time and the momentum of the top particle is decreasing over time . . .
Nov
12
comment Momentum conservation in an electromagnetic system?
@JohnEastmond That's right. The momentum of your entire system should change (in magnitude) by $f\Delta t$ over a time period of $\Delta t$.
Nov
11
comment Why do we have to use an integral in this scenario to figure out $v_{max}$?
@user42141 The kinetic energy does vary . . . $v$ depends on $r$.
Nov
11
comment Why do we have to use an integral in this scenario to figure out $v_{max}$?
@NeuroFuzzy Thanks for catching that.
Nov
10
comment Derivation of the irreducible representations of SO(3)
The same principle applies to both $SO(3)$ and $SU(2)$. All this essentially comes down to the trivial fact that, if $x=y$, then we must have that $f(x)=f(y)$. In $SO(3)$, the identity element and the element that rotates by $\pi$ are the same element. Hence, they must do the same thing in a representation. Obviously, this principle is still true for $SU(2)$; however, in this case, we have distinct $g,h\in SU(2)$ such that $\rho (g)=\rho (h)=1\in SO(3)$ ($\rho :SU(2)\rightarrow SO(3)$ is the map I mentioned in my answer).
Nov
9
comment Derivation of the irreducible representations of SO(3)
Also, it cannot be related to $L_i$ at all. $L_i$ is an element of the Lie algebra, and we know that $\mathfrak{su}(2)=\mathfrak{so}(3)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
This is exactly what I am suggesting. This has nothing to do with spin vs. orbital. If somebody has told you that spin in $SO(3)$ can change a sign after a rotation by $2\pi$, they were wrong. This can only happen for spin with $SU(2)$. The point is is that, in quantum mechanics, we don't really care about representations, but rather projective representations (basically representations up to phase). It turns out that projective representations of $SO(3)$ are in canonical one-to-one correspondence with actual representations of $SU(2)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
Suppose that the representation has spin $n/2$ with $n\in \mathbb{Z}^+$. Then, in particular, this representation will contain an element $v$ such that $S_zv=\frac{n}{2}v$. This is at the level of the Lie algebra. At the level of the Lie group, $R_z(\theta )=\exp (-i\theta S_z)$ is a rotation about the $z$-axis by an angle $\theta$. Hence, $R_z(\theta )v=\exp (-i\theta \frac{n}{2})$. For $\theta =2\pi$, we had better have that $R_z(2\pi )=R_z(0)$; however, using the formula above, $R_z(2\pi )=\exp (-i\pi)=-1$. Thus, this does not give us a representation of $SO(3)$ for half-integer spin.