What's the physical reason that a massive vector field has only three linearly-independent physical polarizations?

Question

While a four-vector field $A_\mu$ has four components, for a massive field there are only three linearly independent combinations of these components that correspond to physical situations. This follows from Maxwell's equations which yield the condition \begin{equation} \epsilon_\mu p^\mu = 0 \, . \end{equation} What's the physical meaning of this condition? In other words, what's the physical reason that a massive vector field only possesses three polarization degrees of freedom?

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In a previous related question of mine, @ACuriousMind mentioned that

The other degree of freedom (the one also having to be eliminated for a massive vector field with three degrees of freedom) is eliminated by a constraint setting the timelike Hamiltonian canonical momentum to zero, which has nothing to do with Maxwell's equations (it's simply the almost trivial statement that $F^{00} =0 $).

However, I'm not really able to understand the physical content of this comments.

Answer 1

A generic four-vector field contains four degrees of freedom that upon quantization describe a spin-1 and a spin-0 particle. In the action for the Proca field, the spin-0 component is projected out by using an antisymmetric kinetic term. So the answer to your question really is: there are only three degrees of freedom in the Proca field since we want it to describe a spin-1 particle. Why spin one corresponds to three spin polarizations is ordinary quantum mechanics. Your equation $\epsilon_\mu p^\mu=0$ is just a Lorent-covariant form of the statement that in the rest frame, the polarization vector of a spin-1 particle is purely spatial.