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1d
comment Does the induced charge on a conductor stay at the surface?
This is true in the static case, not in general. In the general case you need to use the continuity equation and Ohm's law, which yield $\rho(t) = \rho(0)e^{-\frac{\sigma}{\epsilon}t}$; since $\vec{J}$ vanishes asymptotically (static case), so does $\vec{E}$.
1d
comment 2D Gauss law vs residue theorem
How do you find the Coulomb potential in dimension 2?
1d
comment 2D Gauss law vs residue theorem
Not a point charge but a homogeneous linear distribution, which is perpendicular to the complex plane.
2d
comment Feynman Green Function in D=2 for D'Alembertian
When we attempt to find the Green's function, we observe that the integral diverges (the PV part), therefore we must take the retarded or advanced solutions instead, which are on the left of the Sokhotski-Plemelj formula. If you start from the definition of the Green's function, you'll see that you did things backwards. Also, constants in the definition of Green's function are irrelevant.
2d
comment Feynman Green Function in D=2 for D'Alembertian
Apparently you used $\lim_{\epsilon \to 0}\frac{1}{x\pm i \epsilon} = \text{PV} \frac{1}{x} \mp i\pi \delta(x)$, right?
2d
comment Feynman Green Function in D=2 for D'Alembertian
Well, $\int \frac{d^4 p}{(2\pi)^4} \frac{e^{ip_\mu x^\mu}}{-p_\mu p^\mu -i \epsilon}$ is the retarded function and, since the (non-converging) Green's function must be real (the dalembertian is hermitian), your integral does yield the retarded Green.
2d
comment Feynman Green Function in D=2 for D'Alembertian
Sorry, I deleted the comment before I noticed your answer.
2d
comment Feynman Green Function in D=2 for D'Alembertian
By the way, the PV is meant to be a symbol accompanying the integral, it is meaningless next to a function except as notation implying the integral.
2d
comment Feynman Green Function in D=2 for D'Alembertian
It should be $\Theta(T-r/c)$, otherwise the square root becomes imaginary, which is not allowed by the definition of the roots of the argument of the delta function.
2d
comment The correspondence between Grassmann number and 4-spinor
I recall that the fact that a Dirac spinor consists of Grassmann numbers is important when one tries to find that a bilinear covariant is invariant under charge conjugation, otherwise you get spurious minus signs.
2d
comment Feynman Green Function in D=2 for D'Alembertian
What is $P$? I have a solution that looks similar to yours but without the integral.
Apr
13
comment What experiment would disprove string theory?
@DarenW Ancient Greek philosophers did mention atoms, but their atoms have nothing to do with what we know as atoms today, because άτομο means something that cannot be divided in constituents and atoms certainly can. One could say that quarks are the atoms of the philosophers, but nobody is certain. Also, their theories were not just constrained by deficiencies in terms of experimental methods, but they mostly were too vague to be of any practical use for anything at all. Not to mention that no matter what one philosopher would say, another would claim the opposite. That said, I'm Greek!
Apr
13
comment Relation between shm and circular motion
They are $-a\cos t\hat{x}$, $-a\sin t\hat{y}$.
Apr
13
comment Moment of inertia of a sphere
@DylanDang Yes, what David Z is saying is that $dm$ in the notes is not infitesimal mass in dimension 3, the notation is confusing. If you write this as $dI = 2\pi \rho dz \int_0^r r^{\prime 3} d r'$, it should be clear.
Apr
13
comment Relation between shm and circular motion
Yes, I tried to make things simple, so that the idea is easier to convey. This is the reason the equations above have $\omega =1$, so that we don't need to set initial conditions that might seem arbitrary at first sight.
Apr
12
comment Reason why $F^{\mu\nu}F_{\mu\nu}$ and $\tilde{F}_{\mu\nu}F^{\mu\nu}$ are Lorentz invariant
@RobinEkman Yes, that's why I'm used to combining the two, depending on which leads to simpler expressions. For example, I've never done contractions using the Hodge dual and I prefer to replace it with the expression in terms of tensors.
Apr
11
comment Force on a point charge q inside a cavity in an uncharged conductor
@Tobias Thanks. If you change the distribution of induced charges, this will affect the charge in the cavity before the induced charges settle down in their new distribution, because the former propagates at the speed of light and the latter at lower speed. You don't need to freeze the system arbitrarily. Also, Earnshaw's theorem implies that the configuration with the charge in the cavity is always unstable, I'm not sure where I implied that this is a stable system.
Apr
5
comment Are signal fronts in a beam not at rest to each other?
You still don't understand what I'm telling you. Synge writes about measuring the elapsed time between sending and receiving a photon, how do you measure that on a photon?
Apr
5
comment Are signal fronts in a beam not at rest to each other?
You're confused. The point is that there must exist an observer to measure something at rest, it is not that the wave fronts do not exchange photons that makes what you ask not make sense at all. The measurement process includes sending and receiving photons, but what use would that be even if photons could do that, since a photon cannot measure the distance? To make a measurement you'll need at least a clock and a device to send and receive photons, how can a photon on its own do that? Even Synge says so, the curves must be time-like, why do you keep ignoring this?
Apr
4
comment Equivalence between Hamiltonian and Lagrangian Mechanics
$\dot{q} = \partial H / \partial p$ is valid regardless of whether $\dot{q}$ is a derivative or not. It could be any function, but if it isn't a derivative of $q$, then this is not the equation of Hamilton.