The wave equation Green's function in the relativistic limit

Question

The wave equation with a given source, $f(\vec{x},t)$, is given by

$$\left(\nabla^2-\frac{1}{c^2}\partial_t^2\right)\psi(\vec{x},t) = -f(\vec{x},t),$$

and can be solved using the concept of a Green's function, $G(\vec{x},\vec{x}',t,t')$. The Green's function represents the response of the system to a unit source and given by,

$$\left(\nabla^2-\frac{2}{c^2}\partial_t^2\right)G(\vec{x},\vec{x}',t,t') = -\delta(\vec{x}-\vec{x}')\delta(t-t').$$

The $\psi(\vec{x},t)$ function can be constructed via a convolution integral with the Green's function.

My questions lies in the derivation of the Green's function. In many sources I've looked into, a key assumption is that the system is spherically symmetric (due to the delta function only depending on, ($\vec{x}-\vec{x}'$)). Now, this may be true in the non-relativistic limit but is this also true in the ultra-relativistic limit?

In the ultra-relativistic limit, the field of the charge is contracted heavily in the direction of motion (in a $1/\gamma$ cone), so can we assume the Green's function in the ultra-relativistic regime is spherically symmetric? Can we use the non-relativistic Green's function, $G=\delta(t-R/c)/4\pi R$, in ultra-relativistic applications?

Answer 1

First, I assume the $\frac{2}{c^2}$ is a typo and should be $\frac{1}{c^2}$, and the Green's function you write should be $\delta(T-R/c)/4\pi R$. This Green's function is a solution to the Green's function equation with the usual retarded boundary conditions. You do not need any additional assumptions like spherical symmetry to derive it. Since it is spherically symmetric, realizing that fact can simplify the derivation. It is fully relativistic since it is a solution to the wave equation. It is not a solution for $\psi(\vec x,t)$ for a particle moving at a velocity $\vec v$. So for example if this wave equation was for the scalar potential in Lorentz gauge, $\Phi(\vec x,t)$, for a charge $q$ moving at velocity $v$ along $x$, the source would be \begin{equation} f(\vec x,t) = q\delta(x-vt)\delta(y)\delta(z) \end{equation} and the retarded Lorentz gauge scalar potential would be \begin{equation} \Phi(\vec x,t) = \int d^3x' dt' \frac{\delta(t-t'+|\vec x -\vec x'|/c)}{4\pi|\vec x-\vec x'|} q\delta(x'-vt')\delta(y')\delta(z') \end{equation} Doing the integrals will give you the well known Lienard-Wiechert potential which has the nonspherical symmetry you are talking about.