2
$\begingroup$

Let $\mathbf{w}(t)$ be the trajectory of a moving charge. Let the observation event be $(\mathbf{r},t)$.

The scalar potential is:

$$\varphi = \frac{q}{4\pi\epsilon_0}\int \frac{\delta\left(\mathbf{r'} - \mathbf{w}\left(t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}\right)\right)}{|\mathbf{r'}-\mathbf{r}|} \mathrm d^3\mathbf{r'}$$

It can be shown that at most only ONE event on the trajectory of the charge produces the potential at the observation event. This is the event $(\mathbf{w}(t_r),t_r)$, where $t_r$ is such that $|\mathbf{r}-\mathbf{w}(t_r)| = c(t-t_r)$.

Because the delta function is 0 apart from at one point, it seems to make sense that $\mathbf{w}(t_r)$ must be the point that it picks out. Is it then legitimate to write the scalar potential as:

$$\varphi = \frac{q}{4\pi\epsilon_0|\mathbf{r} - \mathbf{w}(t_r)|}\int \delta\left(\mathbf{r'} - \mathbf{w}\left(t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}\right)\right) \mathrm d^3\mathbf{r'}\;?$$

If not, why not? And what is the best way to calculate the remaining delta function integral?

$\endgroup$
  • $\begingroup$ It is hard to see whether this is valid without calculation. You can calculate the original integral using substitution $\mathbf y = \mathbf r' - \mathbf w(t-\frac{|\mathbf r-\mathbf r'|}{c})$ (this involves calculating Jacobi matrix and its determinant). $\endgroup$ – Ján Lalinský Sep 10 '15 at 19:23
3
$\begingroup$

The general rule (see the "Composition with a function" section of the Wikipedia article on Dirac delta functions) is (for suitably well-defined functions):

\begin{equation*} \int_{-\infty}^\infty {\mathrm dx \, f(x)\, \delta(g(x))} = \sum_i {\frac{f(x_i)}{|g'(x_i)|} } \end{equation*}

where $x_i$ are the roots of $g(x)$, so your "extraction" is justified.

One way to apply this result to the 3-dimensional delta function is to choose the axes so that 1) the particle is moving along the $x$-axis with velocity $v$ at time $t_r$ (e.g. $\boldsymbol{w}(t)=(a+vt,0,0)$ near time $t_r$), and 2) the observation point is at $\boldsymbol{r}=(0,y,0)$. Then:

\begin{align*} \int {\mathrm dx' \,\mathrm dy' \, \mathrm dz' \, \delta^3} &= \int {\mathrm dx' \, \mathrm dy' \, \mathrm dz' \, \delta \left(x' - a - v(t-|\boldsymbol{r-r'}|/c)\right)\, \delta(y')\, \delta(z')} \\ &= \int {\mathrm dx' \, \mathrm dy' \, \mathrm dz' \, \delta \left(x' - a - vt + (v/c) \sqrt{x'^2 + (y-y')^2 + z'^2} \right) \,\, \delta(y') \, \delta(z')} \\ &= \int {\mathrm dx' \, \delta \left(x' - a - vt + (v/c) \sqrt{x'^2 + y^2} \right)} \\ &= \int {\mathrm dx' \, \frac{\delta \left(x'- (a+vt_r)\right)}{1+\frac{v}{c}\frac{x'}{\sqrt{x'^2+y^2}}}} \\ &= \frac{1}{1-\boldsymbol{\beta \cdot n}} \end{align*}

where $\boldsymbol{\beta}=\boldsymbol{v}/c$ and $\boldsymbol{n}$ is the unit vector $\boldsymbol{n}= \frac{\boldsymbol{r-w}}{|\boldsymbol{r-w}|}$, evaluated at time $t_r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.