electromagnetism - Counting the number of propagating degrees of freedom in Lorenz Gauge Electrodynamics

How do I definitively show that there are only two propagating degrees of freedom in the Lorenz Gauge $\partial_\mu A^\mu=0$ in classical electrodynamics. I need an clear argument that

1. 
   

   involves the equations of motion for just the potentials $A^0$ and $\mathbf{A}$, and not the electric and magnetic fields.

   
   
   


2. 
   

   includes sources $\rho$ and $\mathbf{J}$ in the equations of motion. This is to justify the assertion that there are degrees of freedom that decouple from the rest of system.

   
   


3. 
   

   does not critically rely on quantum field theoretic arguments (although any supplementary remarks are welcome).

   
   

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To illustrate the level of clarity I expect, I provide an argument in the Coulomb gauge $\nabla\cdot \mathbf{A} = 0$:

Of the four field degrees of freedom, the gauge condition $\nabla\cdot \mathbf{A} = 0$ removes one degree of freedom (the longitudinally polarized EM waves).

To show that among the remaining three degrees of freedom only two are propagating, consider the field equations of motion in the Coulomb gauge:

\begin{align} \nabla^2 A^0 &= -\rho/\epsilon_0,\\ \big[\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\big]\mathbf{A}&= \mu_0 \mathbf{J}-\nabla \frac{1}{c^2}\frac{\partial}{\partial t}A^0. \end{align}

The first equation is NOT a wave equation for $A^0$, and thus does not propagate. The final equation IS a wave equation, and describes the propagation of two degrees of freedom (Gauss' law in first eq. can be solved, and then inserted into the second equation to show that $\mathbf{A}$ only couples to the Solenoidal part of the current).

Answer

I would like to know how exactly the equations of motion in the Lorenz gauge removes the second degree of freedom.

In the Lorenz 'gauge', we have

$$\Box A^{\mu} = \mu_0j^{\mu}$$

If $A^{\mu}$ is a solution, then so is $A^{\mu} + N\epsilon^{\mu}e^{-ik\cdot x}$ if

$$\Box (N\epsilon^{\mu}e^{-ik\cdot x}) = 0$$

Consistency with the Lorenz condition

$$\partial_{\mu}A^{\mu}=0$$

requires

$$k \cdot \epsilon = 0 $$

and consistency with the equation of motion requires that

$$k^2 = k \cdot k = 0$$

But this means that if a polarization four-vector $\epsilon$ satisfies the condition

$$k \cdot \epsilon = 0$$

then $\epsilon' = \epsilon + \alpha k$ also satisfies this condition

$$k \cdot \epsilon' = k \cdot (\epsilon + \alpha k) = k \cdot \epsilon + \alpha k^2 = 0$$

This means we can choose an $\epsilon^{\mu}$ such that $\epsilon^0 = 0$ and then the Lorenz condition implies that the wave and polarization 3-vectors are orthogonal

$$\vec k \cdot \vec\epsilon = 0$$

so there are only two independent polarization vectors (for freely propagating wave solutions).

To summarize, the Lorenz condition implies that the wave and polarization four-vectors are (Minkowski) orthogonal leaving three polarization degrees of freedom.

The equation of motion implies that the wave four-vector is null. Since null-vectors are self-orthogonal ($k^2 = 0$), we are left with two physical polarization degrees of freedom.