Given the current density ${\bf j}({\bf r},t) = \mathbf{v}_{0}\,\omega\, \sin(\omega\,t)\,\delta({\bf r}-{\bf r}_0),$ what is the vector potential?

From a previous question I noticed the density is similar to the one of a Hertzian dipole. Now the vector potential of any current density is found to be $$\mathbf{A}(r,t) = \dfrac{\mu_0}{4\pi}\,{\large\int}\dfrac{{\bf j}\left({\bf r}',t-\frac{|{\bf r}-{\bf r}'|}{c}\right)}{|{\bf r}-{\bf r}'|}\,\mathrm{d^3}{\bf r}'$$ in $\underline{\text{Lorenz gauge}}$ by solving the tricky PDE $$\nabla^{2}\mathbf{A}- \dfrac{1}{c^2}\,\partial_t^2\mathbf{A} = -\mu_0{\bf j}.$$

For the current density above, the vector potential should integrate to $$\mathbf{A} = \dfrac{\mu_0}{4\pi} {\large\int} \dfrac{{\bf v}_{0}\omega\sin\left[\omega\left(t-\frac{|{\bf r}-{\bf r}'|}{c}\right)\right]\delta({\bf r}'-{\bf r}_0)}{|{\bf r}-{\bf r}'|}\,\mathrm{d^3}{\bf r}' = \dfrac{\mu_0}{4\pi}\dfrac{{\bf v}_{0}\omega\sin\left[\omega\left(t-\frac{|{\bf r}-{\bf r}_{0}|}{c}\right)\right]}{|{\bf r}-{\bf r}_{0}|},$$ thanks to the properties of the $\delta$-distribution. Is this solution deployed correctly at all?

Now heading to the scalar potential: Here in Lorenz gauge the determining formula looks quite similar: $$\phi({\bf r},t) = \dfrac{1}{4\pi\varepsilon_0}{\large\int}\dfrac{\rho\left({\bf r}',t-\frac{|{\bf r}-{\bf r}'|}{c}\right)}{|{\bf r}-{\bf r}'|}\,\mathrm{d^3}{\bf r}',$$ solving a similar but still tricky PDE, $$\nabla^2\phi-\dfrac{1}{c^2}\,\partial_t^{2}\phi = -\dfrac{\rho}{\varepsilon_0}$$ From the previous question I had posted I determined the charge density $\rho$ to be $$\rho = -\int \nabla\cdot{\bf j}\,\mathrm{d}t = \mathbf{v}_0[\cos(\omega\,t)-1]\cdot\nabla \delta({\bf r}-{\bf r}_0) \quad \text{with} \quad \rho({\bf r},0) = 0.$$

The main problem I'm facing: How to solve the integral yielding the scalar potential, $$\phi({\bf r},t) = \dfrac{1}{4\pi\varepsilon_0} {\large\int} \dfrac{{\bf v}_{0}\left\{\cos\left[\omega\left(t-\frac{|{\bf r}-{\bf r}'|}{c}\right)\right]-1\right\}\cdot\nabla\delta({\bf r}'-{\bf r}_0)}{|{\bf r}-{\bf r}'|}\,\mathrm{d^3}{\bf r}'?$$

Alright so trying to put the partial integration into praxis:

Here's what I try to decompose:

$$\phi({\bf r},t) = \dfrac{1}{4\pi\varepsilon_0} {\large\int} \dfrac{{\bf v}_{0}\left\{\cos\left[\omega\left(t-\frac{|{\bf r}-{\bf r}'|}{c}\right)\right]\right\}\cdot\nabla\delta({\bf r}'-{\bf r}_0)}{|{\bf r}-{\bf r}'|} - \dfrac{\mathbf{v_0} \,\{\nabla\delta({\bf r}'-{\bf r}_0)\}}{|{\bf r}-{\bf r}'|} \,\mathrm{d^3}{\bf r}'$$

Partially integrated this becomes:

$$\begin{align} \phi(r,t) = & \\[12pt] &\dfrac{1}{4\,\pi\varepsilon_0}\left( \,\left[ \dfrac{{\bf v}_{0}\left\{\cos\left[\omega\left(t-\frac{|{\bf r}-{\bf r}'|}{c}\right)\right]\right\} \,\delta(\mathbf{r'}-\mathbf{r_0})}{|\mathbf{r}- \mathbf{r'}|} \right]^{+\textstyle \infty}_{-\textstyle \infty} - \\[12pt] {\bf v}_{0}\,{\large \int} {\large \nabla} \dfrac{\left\{\cos\left[\omega\left(t-\frac{|{\bf r}-{\bf r}'|}{c}\right)\right]\right\}}{|\mathbf{r}- \mathbf{r'}|} \cdot \delta(\mathbf{r'}-\mathbf{r_0})\,\mathrm{d^3r}\right) \\[12pt] &+ \dfrac{1}{4\,\pi\varepsilon_0}\left(\left[- \dfrac{\mathbf{v_0}}{|\mathbf{r}- \mathbf{r'}|}\,\delta(\mathbf{r}-\mathbf{r_0})\right]^{+\textstyle \infty}_{-\textstyle \infty}\right) + \mathbf{v_0}\,\int \nabla\dfrac{1}{|\mathbf{r} - \mathbf{r'}|}\cdot \delta(\mathbf{r}- \mathbf{r'}) \end{align}$$

Now assuming the field are vanishing in infinite space (as commonly expected) the scalar potential eventually should read:

$$\phi(r,t) = \dfrac{1}{4\,\pi\varepsilon_0}{\bf v}_{0}\,\left(-{\large \nabla} \dfrac{\left\{\cos\left[\omega\left(t-\frac{|{\bf r}-{\bf r_0}|}{c}\right)\right]\right\}}{|\mathbf{r}- \mathbf{r_0}|} +\nabla\dfrac{1}{|\mathbf{r} - \mathbf{r_0}|} \right)$$

Could it be?

  • 1
    $\begingroup$ When derivatives act on delta functions, use integration by parts to get them off the delta function. $\endgroup$
    – Ghoster
    Commented Nov 8, 2022 at 21:37
  • $\begingroup$ Alright, thanks for the hint, I'll try my best. The potential for A is correct tho? $\endgroup$
    – Leon
    Commented Nov 8, 2022 at 21:38
  • 1
    $\begingroup$ Other than the fact that you wrote $A$ so that it doesn’t look like it’s a vector, I think it’s right. I think one reason you are having some difficulties is that you aren’t being careful with vector notation. The fact that you wrote $v_0\nabla\delta()$ without a dot is an indication of this. You just jammed two vectors up against each other without realizing it. ($\nabla$ counts as a vector because it is a vector operator.) $\endgroup$
    – Ghoster
    Commented Nov 8, 2022 at 21:44
  • $\begingroup$ yea, I'm apologetic for being this reckless at throwing together vectors and scalers like no other. I hope the intermediate result above obeys all rules. $\endgroup$
    – Leon
    Commented Nov 9, 2022 at 16:58
  • 1
    $\begingroup$ Your scalar potential has the gradient of a vector, which is a tensor term , so immediately I can say that it is incorrect $\endgroup$ Commented Nov 9, 2022 at 17:06

1 Answer 1


Not sure what your question is, if $r_{0}$ is just a time independant vector, then the field is the same as the regular hertzian dipole, just translated, so replace $\vec{r}$ with $\vec{r} - \vec{r_{0}}$ and you are done, [obviously depending on the form of your solution you'd have to find the relevant changes to the spherical variables]

If $r_{0}$ is a function of time

$$\vec{J}(r',t) =\vec{V_{0}} \sin(\omega t) \delta^3(\vec{r} - \vec{r}_{0}(t))$$

$$\vec{J}(r',t_{r}) =\vec{V_{0}} \sin(\omega t_{r}) \delta^3(\vec{r} - \vec{r}_{0}(t_{r}))$$

Simply replacing r' with $r_{0}$ and calling it quits doesn't work, since $r_{0}$ is a function of the retarded time, and thus you cannot get anywhere when actually trying to calculate the field.

You may want to look at:https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential in order to know how to proceed from here.


Use the gauge fixing condition to solve for V,

$$\nabla \cdot \vec{A} + \mu_0 \epsilon_0 \frac{\partial V}{\partial t } = 0$$


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