Do the Lienard Wiechert Potentials satisfy the wave equation?

Question

The Lienard Wiechert potential (leaving out the vector potential for simplicity),

$$\left.\phi(\vec{r},t)=\frac{e}{4\pi\epsilon_0R\,(1-\hat{n}\cdot\vec{\beta})} \right|_{t'=t_{\rm ret}},$$

where the quantities should be evaluated in the retarded time given by $t_{\rm ret}=t-R/c$, should satisfy the wave equation. The wave equation for the potential is derived from Maxwell's equations in the Lorenz gauge and given by

$$\left(\nabla^2-\frac{1}{c^2}\partial_t^2\right)\phi(\vec{r},t)=-\frac{\rho}{\epsilon_0}, $$

where $\rho$ is the charge density and should be $\rho=e\,\delta(\vec{r}-\vec{r}')$ for the point charge. Plugging $\phi(\vec{r},t)$ into the wave equation we need to compute the second-order spatial and time derivatives with respect to the observer. To assist with the derivation, I have computed the following spatial derivatives in cartesian coordinates ($i=x,y,$ or $z$),

$$\partial_i R=n_i, $$ $$ \partial_i \frac{1}{R} = -\frac{n_i}{R^2},$$ $$\partial_i n_i = \frac{1-n_i^2}{R},$$ $$\partial_i (1-\hat{n}\cdot\vec{\beta}) = \frac{n_i(\hat{n}\cdot\vec{\beta})-\beta_i}{R},$$

and the following observer time derivatives (taking mind to switch from $\partial_t$ to $\partial_{t'}$ using $dt=dt'(1-\hat{n}\cdot\vec{\beta})$ when necessary),

$$\partial_t R=-c\frac{\hat{n}\cdot\vec{\beta}}{(1-\hat{n}\cdot\vec{\beta})}, $$ $$ \partial_t \frac{1}{R} = c\frac{\hat{n}\cdot\vec{\beta}}{R^2(1-\hat{n}\cdot\vec{\beta})},$$ $$\partial_t n_i = \frac{c}{R(1-\hat{n}\cdot\vec{\beta})}\left({n_i(\hat{n}\cdot\vec{\beta})-\beta_i }\right),$$ $$\partial_t (1-\hat{n}\cdot\vec{\beta}) = \frac{c\left({ \vec{\beta}\cdot\vec{\beta}-(\hat{n}\cdot\vec{\beta})^2 }\right)}{R\left({1-\hat{n}\cdot\vec{\beta}}\right)}.$$

Now, when I compute the second-order derivatives I get:

$$\nabla^2\phi(\vec{r},t) = -\frac{e}{2\pi\epsilon_oR^3(1-\hat{n}\cdot\vec{\beta})^3}\left({ 1-(\hat{n}-\vec{\beta})\cdot(\hat{n}-\vec{\beta}) }\right)$$

$$\frac{1}{c^2}\partial_t^2 = \frac{e}{4\pi\epsilon_oR^3(1-\hat{n}\cdot\vec\beta)^7} \left({(1-\hat{n}\cdot\vec\beta)^3 \left({(\hat{n}\cdot\vec\beta)^2-\vec\beta\cdot\vec\beta +2(\hat{n}\cdot \vec\beta)(\hat{n}\cdot\vec\beta-\vec\beta\cdot\vec\beta)}\right) + (\hat{n}\cdot\vec\beta-\vec\beta\cdot\vec\beta)((\hat{n}\cdot\vec\beta)^2-\vec\beta\cdot\vec\beta) }\right) $$

which I dont see any way that their difference will be equal to $-\rho/\epsilon_0$ for a point charge as dictated by the wave equation. Am I doing something wrong here?

Answer 1

Yes, the Lienard-Wiechert potential is a solution to the wave equation with a point charge source. That's how it is derived. However, if you want to take derivatives in order to show this explicitly from the differential equation, you must be very careful to not neglect any terms in the derivatives, and also be careful to be able to extract the delta function.

For the second point, the delta function occurs from derivatives like those in the simple Coulomb's law static case $\nabla^2 \frac{1}{r} = -4\pi \delta^3(\vec r)$. However, if you were to write $\nabla^2$ in spherical coordinates it would be $\frac{1}{r}\partial_r^2 r$ plus angular terms. Taking the derivatives above gives $\frac{1}{r}\partial_r^2 r \frac{1}{r} = 0$. The delta function is hidden at the coordinate boundary at $r=0$. Exactly this case will also occur for the the Lienard-Wiechert potential, so it is best to isolate a $\nabla^2 \frac{1}{R}$ term and realize this will give the delta function, by either using the divergence theorem, smearing out the singularlity, or changing to other coordinates.

For the derivatives, you just have to be meticulous. If you think of a derivative as being $\lim_{\delta\rightarrow 0} \frac{F(x+\delta)-F(x)}{\delta}$, then when you change the observation point $x$, $y$, $z$, or $t$, the retarded time changes. You have to take derivatives with respect to $t_{\rm ret}$ times derivatives of $x$, $y$, $z$, or $t$ using the chain rule.

So realizing that $R$ is a function of $t_{ret}$, $t_{\rm ret} = t-R(t_{\rm ret})/c$ \begin{equation} \partial_t t_{\rm ret} = \partial_t [t-R(t_{\rm ret})/c] = [1-\vec \beta(t_{\rm ret}) \cdot \hat R(t_{\rm ret})\partial_t t_{\rm ret}] \end{equation} gives \begin{equation} \partial_t t_{\rm ret} = \left . \frac{1}{1-\vec \beta(t')\cdot \hat R(t')} \right |_{t'= t_{\rm ret}} \end{equation} similarly \begin{equation} \partial_x t_{\rm ret} = -\left . \frac{1}{1-\vec \beta(t') \cdot \hat R(t')} \frac{X(t')}{cR(t')} \right |_{t'= t_{\rm ret}} \end{equation} with analogous expressions for $y$ and $z$. $X/R$ is the $x$ component of your $\hat n$.

Using the chain rule, your need to evaluate your derivatives as \begin{eqnarray} \partial_ t \left [ F \right ]_{\rm ret} &=& \left [ \frac{1}{1-\vec \beta \cdot \hat R} \dot F \right ]_{\rm ret} \nonumber\\ \partial_ x \left [ F \right ]_{\rm ret} &=& \left [ \partial_x F -\frac{X}{cR} \frac{1}{1-\vec \beta \cdot \vec r} \dot F \right ]_{\rm ret} \end{eqnarray} where the dot means take the partial derivative with respect to the time argument, and I have just written the subscript ret to indicate evaulating at the retarded tim. If this isn't clear, just carefully take the derivatives with the chain rule. The second terms are from the change in the retarded time.

If you do this and don't make an algebra mistake, you will find your desired result.