Why Lagrangian of electromagnetism with Lorenz Gauge evolve Klein Gordon equation?

Question

Simply Lagrangian without a source for Maxwell equation is $$ L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} $$ Also Lorenz Gauge condition is $$ \partial_{\mu}A^{\mu}=0 $$ and if so I can briefly add this into Lagrangian like $$ L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}(\partial_{\mu}A^{\mu})^2 $$ and such a Lagrangian have equations of motions like below: $$ \partial_\nu\partial^{\nu}A^{\mu}=0 $$ Actually this is Klein-Gordon for massless particles, right? So it means I have a massless and spin-0 particle. What does it mean? Am I getting something different?

Answer 1

The equation of motion is

$$\partial^2 A_\mu = 0$$

which in fact is a set of four KG equations, one for each field $A_\mu$, $(\mu = 0,\dots,3)$.

This gives us four degrees of freedom for the $A_\mu$ field. But this cannot describe the electromagnetism, since the light has only two degrees of freedom (polarization).

We must not forget that we have set a gauge condition, $\partial_\mu A^\mu=0$, which removes a degree of freedom. The other degree of freedom is removed using another contraints.

Answer 2

yes, we have massless particle but not spin zero because A has 4 index.so it is basically four fields. but for EM we have only two states of polarization so it is cut down to two only not four. for refrence see peskin book.