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I wanted to see if the divergence of the lienard wiechert field follows Maxwell's equations (gauss' law):

$$\nabla \cdot \vec{E} = 0$$

for

$$ E(r,t)=\frac{e}{\gamma^2 R^2} \frac{n-\beta}{(1-n\cdot\beta)^3} + \frac{e}{cR} \frac{n \times ((n-\beta)\times \beta')}{(1-n\cdot\beta)^3}$$

But I get a non-zero answer! I calculated it for a very simple test case of a stationary observation point at $(x,y,z)$ and a source charge at some retarded time, t', is located at the origin $(0,0,0)$ with a velocity $\vec\beta=\beta<1,0,0>$ and acceleration $\beta'=\frac{\beta^2 c}{\rho} <0,-1,0>$.

This is simple circular motion. After computing the field, I compute the divergence ($\nabla = \frac{\partial}{\partial x} \hat{x}+ \frac{\partial}{\partial y} \hat{y}+ \frac{\partial}{\partial z} \hat{z}$), but again I do not get a zero answer!

Am I doing this correctly? Do I have to take the divergence of evaluated at some retarded time? I thought that the divergence of the field should be respective of the observation point and therefore be in the current time-space coordinates, not the retarded.

Any help would be greatly appreciated!

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  • $\begingroup$ Do you get non-zero divergence even for the case where $\beta'$ is zero all the time (rectilinear motion)? $\endgroup$ – Ján Lalinský Oct 9 '18 at 9:50
  • $\begingroup$ shouldn't it satisfy $\nabla \cdot E = \delta(x-x_0)$, with $x_0$ the location of the source? $\endgroup$ – Lorenz Mayer Oct 12 '18 at 16:40
  • $\begingroup$ Thats exactly what I thought! Afterall, this is how the LW field is solved for in the first place. But, is that only for the point $x_0$ where the charge is located? Kind of like the static case ($\nabla \cdot E =0$ except at $x_0$)? Im just kind of confused what to expect! $\endgroup$ – Donkey Kong Oct 13 '18 at 1:03
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The Lienard-Wiechert solution describes the electromagnetic fields emitted from a single point charge, ignoring backreaction of the emitted fields on the charge. That is, take a trajectory

$$\mathbb{R} \mapsto \mathbb{R}^4 $$ $$ \tau \mapsto x^\nu(\tau) $$

Then the current corresponding to a particle of charge $e$ moving along the trajectory is

$$ j^\nu(x) = e \int_{-\infty}^\infty \frac{ d x^\nu(\tau)}{ d \tau} \delta^{(4)}(x - x(\tau)) d \tau $$

Then the Lienard-Wiechert field is the solution to

$$ \frac{\partial}{\partial x^\mu} F^{\mu \nu}(x) = j^\nu(x) \ .$$

Note that a reasonable assumption on the trajectory of the particle is that it should move forward in time, i.e.

$$ \frac{ d x^0(\tau)}{ d \tau} > 0 $$

Then $x^0(\tau)$ is a strictly monotonous function of $\tau$, and thus, if $\tau_0$ is the unique solution to $x^0 = x^0(\tau_0)$ for fixed $x_0$,

$$ \delta(x^0 - x^0(\tau)) = \frac{\delta(\tau - \tau_0)}{\left| \frac{ d x^0}{ d \tau}(\tau_0) \right|} \ .$$

We may thus evaluate the current to

$$ j^0(x_0,\mathbf{x}) = e \delta(\mathbf{x} - \mathbf{x}(\tau_0)) \ , \ \ \ \mathbf{j}(x^0,\mathbf{x}) = e \frac{\frac{ d \mathbf{x}}{ d \tau}(\tau_0)}{\left| \frac{ d x^0}{ d \tau}(\tau_0) \right|} \delta(\mathbf{x} - \mathbf{x}(\tau_0)) \ . $$

Note that this depends on $x^0$, since $\tau_0$ is a function of $x^0$. For simplicity one could choose $\tau_0 = x^0$, but this is not necessary.

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