Talk:Proca action

When mass is zero
What happens when m is set to zero? Does it describe a photon then? Headbomb {{{sup|ταλκ}}κοντριβς – WP Physics} 02:27, 17 February 2009 (UTC)


 * Photon has mass in Proca theory. Proca electrodynamics become Maxwell's electrodynamics in the case m=0. 18:57, 29 April 2010

Units?
Is action in Gauss (cgs) units? The permittivity is missing. 19:43, 29 April 2010


 * The action is the same in any units system. Units don't matter in the slightest. And what exactly do you need permittivity for? MuDavid (talk) 11:08, 21 May 2010 (UTC)

What does "conjunction" mean?
The phrase "is equivalent to the conjunction of" confused me. The sentence would be clearer if it said "is equivalent to [equation] with the gauge condition (in the massive case) [gauge condition]". I also don't quite see how the equation right after the Proca equation is equivalent to the Proca equation using the gauge condition. A little elaboration would be helpful at this step.

The article is good for its simplicity. I would consider this topic to be VERY important for modern particle physics, especially given the new work at the LHC.

75.164.86.127 (talk) 16:53, 7 May 2015 (UTC)K. Pomeroy


 * "Conjunction" means "and". The conjunction of two equations is a situation where both equations are true simultaneously. The article is correct. One can derive both the second and third equations from the first, and one can derive the first from the second and third working together. It is really easy, just try it. JRSpriggs (talk) 20:12, 7 May 2015 (UTC)

I think the splitting of the proca equation into Klein-Gordon equation and Lorentz gauge is still questionable, because the full proca equation might have solutions, which are not solutions of the pair of equations. Though the other direction is pretty obvious, i.e. solutions of the pair of equations are obviously solutions of the proca equation, the splitting could be a restriction of the original equation. --B wik (talk) 21:58, 30 September 2015 (UTC)


 * To B wik: Here we assume that m≠0. Let us start with
 * $$\partial_\mu(\partial^\mu A^\nu - \partial^\nu A^\mu)+\left(\frac{mc}{\hbar}\right)^2 A^\nu=0$$
 * Applying the partial derivative to this gives the gauge equation
 * $$0 = \partial_\nu 0 = \partial_\nu \partial_\mu(\partial^\mu A^\nu - \partial^\nu A^\mu) + \left(\frac{mc}{\hbar}\right)^2 \partial_\nu A^\nu = \partial_\nu \partial_\mu \partial^\mu A^\nu - \partial_\nu \partial_\mu \partial^\nu A^\mu + \left(\frac{mc}{\hbar}\right)^2 \partial_\nu A^\nu = \partial_\mu \partial^\mu \partial_\nu A^\nu - \partial_\nu \partial^\nu \partial_\mu A^\mu + \left(\frac{mc}{\hbar}\right)^2 \partial_\nu A^\nu = \left(\frac{mc}{\hbar}\right)^2 \partial_\nu A^\nu $$
 * using the commutativity of partial derivatives and the fact that one can change contracted indices. Dividing out a non-zero constant factor yields
 * $$\partial_\mu A^\mu = 0 \,.$$
 * It follows that
 * $$ 0 = \partial^\nu \partial_\mu A^\mu = \partial_\mu \partial^\nu A^\mu \,.$$
 * If we add this to the original equation, we get
 * $$\left[\partial_\mu \partial^\mu + \left(\frac{mc}{\hbar}\right)^2\right]A^\nu = 0 \,.$$
 * OK? JRSpriggs (talk) 10:47, 1 October 2015 (UTC)

OK, :-) . Thanks. B wik (talk) 17:46, 1 October 2015 (UTC)