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).

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).

  • $\begingroup$ After claiming that the gauge-condition removes one degree of freedom, it is insufficient to simply state that equations of motion remove another degree of freedom. I would like to know how exactly the equations of motion in the Lorenz gauge removes the second degree of freedom. $\endgroup$
    – QuantumDot
    Commented Jun 18, 2014 at 9:12
  • $\begingroup$ For the same question in arbitrary dimensions, see my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Commented Jul 6, 2014 at 13:08

2 Answers 2


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



$$k \cdot \epsilon = 0 $$

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.


The sourceless non-homogeoneous Maxwell equations are given by: $$ 0 = \partial_\mu F^{\mu \nu} = \partial^2 A^\nu - \partial_\mu \partial^\nu A^\mu $$ After imposing the Lorentz gauge, this becomes: $$ 0 = \partial^2 A^\nu $$ Clearly, this equation has the solution (remember that $p^\mu p_\mu = E^2 - \mathbf{p}^2=0$): $$ A^\mu = N \epsilon^\mu_r e^{-i p \cdot x} $$ where $N$ is a normalization factor and $\epsilon_r^\mu$ denotes the polarization vector. The Lorenz gauge condition, $\partial_\mu A^\mu = 0$, can now be written as: $$ \epsilon^\mu_r p_\mu = 0 \tag{1} $$ Now, via a gauge transformation $A'^\mu=A^\mu + \partial^\mu \alpha$, we see that gauge fields in the Lorenz gauge are still related to each other by: $$ \partial^2 \tilde{\alpha} = 0 $$ which is the residual gauge freedom. It should be clear that $\tilde{\alpha}$ satisfies the above equation for: $$ \tilde{\alpha} = N_1 e^{-i p \cdot x} $$ if $p^\mu$ satisfies $p^\mu p_\mu = E^2 - \mathbf{p}^2=0$. Therefore, a residual gauge transformation corresponds to: $$ A^\mu \rightarrow N \epsilon^\mu_r e^{-i p \cdot x} - \partial^\mu \tilde{\alpha} = N\epsilon^\mu_r e^{-i p \cdot x} + i N_1 p^\mu e^{-i p \cdot x} $$ which implies: $$ \epsilon^\mu_r \rightarrow \epsilon'^\mu_r = \epsilon^\mu_r + i \frac{N_1}{N} p^\mu $$ Note that: $$ \epsilon'^\mu p_\mu = (\epsilon^\mu_r + i \frac{N_1}{N} p^\mu ) p_\mu = 0 $$ due to equation (1). This implies that we can find a residual gauge transformation such that: $$ \epsilon^0_r = 0 $$ and so equation (1) becomes: $$ \epsilon^i_r p_i = \epsilon^1_r p_1 + \epsilon^2_r p_2 + \epsilon^3_r p_3 = 0 $$ This means that the are only two independent polarization vectors.

  • $\begingroup$ This is a naive question, is there any difficult while chosing purely imaginary polarization vectors as thats what you are doing when you try to put $\epsilon^0_r=0$, what do complex polarization vectors mean for that matter? $\endgroup$
    – user7757
    Commented Jun 18, 2014 at 16:22
  • $\begingroup$ @ramanujan_dirac we need to satisfy $\epsilon^i_r p_i = 0$, and so, for instance, for a wave with $p^\mu = (p^0,0,0,p^0)$, we can choose $\epsilon_1 = (1,0,0)$ and $\epsilon_2 = (0,1,0)$, which is not complex. But, apparently, it is also possible/conventional to choose complex polarization vectors, but I'm not really familiar with this. $\endgroup$
    – Hunter
    Commented Jun 18, 2014 at 17:00
  • $\begingroup$ I am talking about your "residual gauge transformation" such that $\epsilon^{\mu}_r=0$, so from the equation above that you need to set $\epsilon^{\prime \mu}_{r}--i\frac{N_1}{N}p^{0}$, the minus sign flips if you make the prime epsilon zero which is just as good. This is imaginary and also depends on $p^0$. $\endgroup$
    – user7757
    Commented Jun 18, 2014 at 17:06

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