What would be the general form of Lagrange Equation when instead of a scalar field we have a vector potential? has anyone derived the klein gordon equation for a corresponding vector potential Lagrangian?


1 Answer 1


Yes, starting from a Lagrangian density of the form $$ \mathcal L = -1/4 F_{\mu\nu} F^{\mu\nu} + m^2 A_\mu A^\mu $$ one can find the so-called Proca equation from the Euler-Lagrange, $$ \partial_\mu F^{\mu\nu} + m^2 A^\nu = 0 $$ If the field is massless, one can choose the Lorenz gauge, $\partial_\mu A^\mu =0$. If the field is massive, $\partial_\mu A^\mu =0$ follows from applying $\partial_\nu$ from the left-hand side. In both cases, one finds that the Proca equation is equivalent to the Klein-Gordon equation, $$ (\partial^2 + m^2 )A^\mu =0 $$ If the field is massless, we find Maxwell's equations in covariant form.

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    $\begingroup$ With a massive gauge field, there is no need for a gauge fixing choice. The equation $\partial_\mu F^{\mu\nu} + m^2 A^\nu = 0$ implies $\partial_\mu A^\mu = 0$ whenever $m^2 \neq 0$ (act on the first equation with $\partial_\nu$). $\endgroup$
    – Prahar
    Commented Dec 1, 2013 at 15:17
  • $\begingroup$ Ah yes thanks. I've tried to change it to add/correct that detail $\endgroup$
    – innisfree
    Commented Dec 1, 2013 at 16:04

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