User talk:Xinxtang

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Hamiltonian Mechanics - charged particles in EM field

Hey Xinxtang. You added a note in https://wiki.riteme.site/w/index.php?title=Hamiltonian_mechanics that the Lagrangian of a non-relativistic classical particle in an electromagnetic field is of this form only in Lorenz gauge. I believe it is of this general form for all gauges, but would like you to confirm. What sources did you use? Thnx — Preceding unsigned comment added by Ponor (talk • contribs) 06:28, 8 May 2020 (UTC)[reply]

@Ponor: The Lagrangian itself is gauge invariant; However, the values of vector potential and scaler potential would differ from gauge to gauge. Therefore, the expression of Lagrangian in terms of vector potential and scalar potential has to be different for different gauges. See Doughty's Lagrangian Interaction and Srednicki's Quantum Field Theory. - Xinxtang (talk) 17:47, 8 May 2020 (UTC)[reply]

@Xinxtang: I understand that Ai's and

\varphi

will be different from one to another gauge, and that even the Lagrangian 'value' will be different from gauge to gauge. However, the Lagrangian will always be

{\mathcal {L}}=\sum {\tfrac {1}{2}}m{\dot {x}}_{i}^{2}+\sum e{\dot {x}}_{i}A_{i}-e\varphi

, and will always produce physically relevant equations like

m{\ddot {\mathbf {x} }}=e\mathbf {E} +e{\dot {\mathbf {x} }}\times \mathbf {B}

no matter what the gauge is. The thing is, when you say this Lagrangian is valid for the Lorenz gauge only, in invalidates the minimal substitution formula for all other gauges, and I know that one is used in gauges other than Lorenz. Unfortunately, I don't have access to those two books. What would the Lagrangian and the minimal coupling look like in, say, Coulomb or in

\varphi =0

gauge? Thnx Ponor (talk) 18:18, 8 May 2020 (UTC)[reply]

@Ponor: I noticed that you have updated the wikis for Hamiltonian mechanics and Lagrangian mechanics. Thank you for your work. I do agree with most of your changes, but there is a caveat. Under a gauge transformation, the Lagrangian technically picks up extra terms, but they add up to a total time derivative of a function, and therefore won't change the Euler-Lagrange Equation. In short, I'm sorry for being sloppy and saying that Lagrangian itself is gauge invariant; Instead, it picks up a total time derivative during gauge transformation. By the way, can you provide more details about how minimal coupling applies to, say, Coulomb guage?--Xinxtang (talk) 06:15, 13 May 2020 (UTC)[reply]

@Xinxtang: It's been a while since I last dealt with all this, so I also consulted my students who are now experts. The thing is, first, you have to fix the gauge and continue working in it. I don't see the need for switching from one to another. So if you chose your A and fi in one, you put those into the Lagrangian as given. If you chose another gauge, you'll again have an A and a fi, and those will go into the Lagrangian. Invariance doesn't mean those two Lagrangians will evaluate to the same number. It means there will be no mixing of fi and x', or As alone, or fis and As squared... that would give different physics. Even adding a constant to the potentials is a gauge transformation, but you don't see us writing different Lagrangians for every possible constant added. I did a little exercise since these things are not easy to find in the literature. If you start with this Lagrangian, and put it into Euler-Lagrange eqs, you will need no gauge relations to get mx"=eE+evxB (just remember that A=A(x,y,z,t) and be mindful of total and partial derivatives). This is to the best of my knowledge, and I'm open to discussion. Pleasure talking to you. PS I don't know where the Coulomb gauge is used, but fi=0 is popular in condensed matter when the interaction of photons with Bloch electrons in the solid are considered (transitions between electron's states): one can use the Fermi's golden rule by just substituting p with p+eA in the kinetic part of the original Hamiltonian, which gives a relatively simple perturbation term. Ponor (talk) 12:07, 13 May 2020 (UTC)[reply]

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September 2020

Hi Xinxtang! I noticed that you recently marked an edit as minor at Lambert W-function that may not have been. "Minor edit" has a very specific definition on Wikipedia – it refers only to superficial edits that could never be the subject of a dispute, such as typo corrections or reverting obvious vandalism. Any edit that changes the meaning of an article is not a minor edit, even if it only concerns a single word. Please see Help:Minor edit for more information. Thank you. JBL (talk) 11:52, 23 September 2020 (UTC)[reply]

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