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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?

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  • $\begingroup$ So the retarded Greens function for the d'Alembertian in 3+1D? $\endgroup$
    – Qmechanic
    Jun 23, 2021 at 2:04
  • $\begingroup$ @Qmechanic Yes that is correct $\endgroup$ Jun 23, 2021 at 15:33

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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.

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  • $\begingroup$ Maybe I'm being difficult, but the expression for the potential you state, $\Phi(\vec{x},t)$, is actually not the retarded potential expression in the literature; there is no integration over $dt'$. The integration over $dt'$ serves the same purpose as the heuristic argument (see Griffiths/Zangwill) of the doppler effect of the moving charge's through an initesimal spherical shell. So, you are imposing the effects of motion post-facto. So, if you don't know/assume the Green's function is symmetric, would you need to include the $\nabla_{\phi,\psi}$ components in the wave equation? $\endgroup$ Jun 23, 2021 at 21:08
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    $\begingroup$ It is absolutely the retarded potential in the literature. Often in text books, the t' integral is done using the retarded delta function so the source is evaluated at the retarded time. The form I wrote demonstrates the relativistic covariance by integrating over space time. This is just the solution of the mathematical problem with the appropriate boundary conditions. There are no assumptions, heuristics, or impositions of any effects. It's just mathematics. $\endgroup$
    – user200143
    Jun 24, 2021 at 19:27

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