The electric pickle is often used as an example of a non-ohmic resistor. In the experiment, electric current excites the sodium ions inside pickle, producing very bright and intense light effect. What I am wondering here is that why in many experiments only one end of the pickle glows?
@DanielSank Er ... the delta function is non-zero in a circle, but you integrate along the x-axis, so you encounter it twice.
But I'm not sure exactly what the properties of $\int dx\,\delta(x^2-x_0^2)$ are. I'd hesitate to assume they are the same as $\int dx\,\delta(x-x_0)$.
I'm sure I knew that stuff in grad school. Or at least the week I passed the comps.
I want your opinions on how useful it is to throw second quantization at students when they have not understood first quantization. I tend to add a simpler answer when I see this.
Also the emphasis on second quantization and statements of the type "all space is filled with electron fields /parti...
@Qmechanic: I am confused. I'm trying to help @EmilioPisanty with his zilch conservation, and I agree with your repeated statement in your answers that an on-shell symmetry is a vacuous notion.
However, the derivations in one of the papers seems to use that if I get $\delta S = \int \mathrm{d} C = 0$ for any on-shell transformation (and I will get that for every on-shell transformation, as you say), then $C$ is a conserved current form. Is there something wrong with it, i.e. is such a conserved quantity somehow inferior to a "true" conservation law derived by Noether's theorem, or is this reasoning flawed altogether?
If there is nothing wrong with it, what's so brilliant about Noether's theorem? It would seem such a reasoning somehow renders it moot to look for symmetries if I can get on-shell conserved quantities by any old transformation
I'm currently thinking that since Noether's theorem uses off-shell identities that just become conservation laws on-shell, those contribute "additional information" about the system, while if I use the equations of motion to get the identity in the first place, then I've gained no new information
This, however, would render this whole business about "conserved quantities of Maxwell's equation" pretty boring from a theoretical point of view.
At least when compared with British food, so take that with a grain of sand.
The statement, not the leberkäse
@ACuriousMind Yeah, if you're confused then I've got very little chance, it's way more technical on those Noether fronts than I'm used to at the moment so I'm slogging through this stuff.
@Danu Look in the other room for a bit more, I'm talking about this. I agree that they are conserved, I disagree that that particular paper uses a "symmetry" and "Noether's theorem" to derive that.
Oh btw if you want to have something nice yet basic to eat here you could just buy some nice Brezen, those are pretty good.
@0celo7 Many types of Asian food, Latin-American food when done well, Medditerranean etc
I tend to get a bit bored when eating typical West-European food because of the lack of interesting flavors (of course, this completely depends on the price level you're looking at, but that goes without saying).
This is exercise 7.8 b) of Nakahara's GTaP: Let $\omega\in\Omega^1(M)$ be a 1-form on a Riemannian manifold with Levi-Civita connection $\nabla$. Prove that
$$
\mathrm{d}\omega=(\nabla_\mu\omega)_\nu\, \mathrm dx^\mu\wedge\mathrm dx^\nu
$$
I proved it using the fact that $\mathrm dx^\mu\wedge\m...
@Bass 1. Just use that the contraction of a symmetric object (that $\Gamma$) and an antisymmetric object is always zero (although this just hides your explicit argument there, everyone just proves that once generally and then just uses it precisely to not quit the summation convention) 2. The r.h.s. has picked coordinates, how do you think you can prove that in a coordinate-free way?
@ACuriousMind with 1., I get rid of the second line, but what about the 3rd one?
2. I just thought maybe there is a coordinate-free way to express the r.h.s., then there might be a way to prove it coordinate-free.. apparently there isn't?