# Phase space formulation of Classical Electrodynamics

Is there a Lorentz invariant formulation of classical electrodynamics whose source terms are the components of a phase space distribution of current densities? I assume that from one perspective it would provide a net current density equal to an integral over all current velocities (and corresponding current densities) at each point in configuration space. I've done a lot of searching online, but have only found articles that discuss plasma dynamics and quantum mechanics.

• By phase space you mean Hamiltonian phase space? – Qmechanic Dec 26 '19 at 5:30
• Yes, Hamiltonian phase space, where the coordinates are, e.g., $q_{\mu}$ and $p_{\mu}$ for a set of particles; but I'm asking about a formulation in which $q_{\mu}$ and $p_{\mu}$ are components of a field vector (or tensor) and corresponding momenta respectively. – S. McGrew Dec 26 '19 at 15:02
• So $\mu=0,1,2,3$? – Qmechanic Dec 26 '19 at 17:46
• That's correct. – S. McGrew Dec 26 '19 at 17:51

If we are working with potentials (developing fomulations for fields is not much more cumbersome) then the wave equation for the four-potential ($$A$$) is:

$$\square^2 A^\nu = \mu_0 J^\nu$$

With Lorentz gauge:

$$\partial_\nu A^\nu =0$$

Where $$J^\nu$$ is four-current. Let the fourier tranforms be:

$$\tilde{J}^\nu \left(\mathbf{k},\omega\right) = \iiint d^3r \int cdt \exp\left(i\left(\frac{\omega}{c}\cdot ct - \mathbf{k}.\mathbf{r}\right)\right) J^\mu\left(\mathbf{r}, t\right)$$

Where $$c$$ is the speed of light. It makes sense to switch to four-position $$x^\nu = \left(ct,\mathbf{r}\right)^\mu$$ and four-wavevector $$k^\nu=\left(\frac{\omega}{c},\mathbf{k}\right)^\mu$$. Then:

$$\tilde{J}^\nu \left(k\right) = \int d^4x \exp\left(i k_\sigma x^\sigma\right) J^\mu\left(x\right)$$

Same for the four-potential. Then the equations become:

$$-k_\sigma k^\sigma\tilde{A}^\mu = \mu_0 \tilde{J}^\nu$$

$$k_\nu \tilde{A}^\nu = 0$$

Is this what you were looking for? For four current we have $$\partial_\nu J^\nu=0$$, so $$k_\sigma \tilde{J}^\sigma = 0$$, thus one of the four components (e.g. charge density) can be determined from the other ones in most cases.

• I think it is probably part of what I'm looking for. The momentum of a field in configuration space is the Fourier transform of the field, per "en.wikipedia.org/wiki/Position_and_momentum_space". So, I suppose a phase space density representation of a field would be the combination of the field and its Fourier transform. – S. McGrew Dec 26 '19 at 21:48