# Coulomb gauge and two degrees of freedom of EM field

The EM field has two possible polarizations, which is caused by spin-one nature of field (leads to the Lorenz gauge) and massless of the field. Really, the Klein-Gordon equations for the EM field $$\square A^{\mu} = 0$$ does not leave a $A^{\mu}$ uniquely determined: $$A^{\mu} \to A^{\mu}{'} = A^{\mu} + \partial^{\mu}\varphi , \quad \square \varphi = 0 \Rightarrow \varphi^{0}{'} = 0, \quad \partial_{\mu}A^{\mu} = (\nabla \mathbf A ) = 0,$$ which is called Coulomb gauge. So EM field has two components. But after using Lorentz-transformations the new $A^{0}$ component isn't equal to zero. So I'm not imagine, what does it mean. It looks like that EM field begin to have three independently components in the new frame. Of course, this is not true, but I want to know, can the Coulomb gauge be an argument of existence only two independently components.

Can you help me?

• – Qmechanic Sep 21 '13 at 21:31
• @Qmechanic . My question isn't about the reason of why prohon has two helicity states. It is about Coulomb gauge and it's connection with that. – user8817 Sep 21 '13 at 21:45

## 1 Answer

The Coulomb gauge $\partial_i A^i=0$ is not Lorentz invariant. Therefore, you should not expect this condition to be true in another reference frame (even though the gauge is still fixed, but by a more complicated relation between the $A^\mu$).

• Of course, as I wrote. But it's strange that it's non-invariance gives seemingly independent third component. – user8817 Sep 21 '13 at 22:46
• It won't be independent, it would be a linear combination of the old potentials. – user7757 Jun 18 '14 at 16:07