4
$\begingroup$

In textbook electromagnetism we are used to seeing neat, coordinate-free, expressions for the scalar potential from an electric dipole (using Gaussian units)

$$\phi(\mathbf{r}) = \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}$$

and vector potential from a magnetic dipole

$$\mathbf{A}(\mathbf{r}) = \frac{\mathbf{\mu} \times \mathbf{r}}{r^3}$$

My question is: are there neat expressions for the spatial Fourier transforms of these dipole potentials?

Doing a quick search, and spending more time than I'd like to admit, I haven't found an equally nice expression for these potentials in the Fourier domain although I would expect that there should be one. I vaguely recall there is an identity such that $\mathcal{F}(\frac{\mathbf{r}}{r^3}) \sim -\frac{\mathbf{k}}{k^2}$ which would easily solve this, but was not able to confirm that after doing some quick math. Moreover, I would naively expect that all multipole potentials should have very straightforward Fourier transforms (maybe in terms of spherical harmonics).

$\endgroup$
1
$\begingroup$

The dipole potential is given by $$ \phi_\text{dip} (\mathbf{r}) = (\mathbf{p} \cdot \mathbf{\nabla}) \phi_\text{mono}(\mathbf{r}) $$ where $\phi_\text{mono}$ is the familiar monopole potential $\phi_\text{mono}(\mathbf{r}) = -1/r$. This follows from the fact that the idealized monopole & dipole charge distributions satisfy the same relation: $$ \rho_\text{dip}(\mathbf{r}) = (\mathbf{p} \cdot \nabla) \rho_\text{mono} (\mathbf{r}), $$ where $\rho_\text{mono} = \delta^{(3)}(\mathbf{r})$. To show the latter relation, write down the distribution for two point charges $\pm q$ separated by a finite distance $d$ along a given direction; then take the limit as $d \to 0$ while holding the quantity $p = qd$ constant. In this limit, the difference of the two delta-functions turns into a directional gradient of a delta-function.

Once you accept this, you can easily transfer my first equation into Fourier space. The gradient turns into a factor of $i \mathbf{k}$ and we have $$ \tilde{\phi}_\text{dip} (\mathbf{r}) = i (\mathbf{p} \cdot \mathbf{k}) \tilde{\phi}_\text{mono}(\mathbf{k}), $$ where $\tilde{\phi}_\text{mono}(\mathbf{k})$ is the Fourier transform of the monopole potential.

Similarly, the vector potential's Fourier transform can be found by noting that $\mathbf{A}_\text{dip} = (\mathbf{\mu} \times \nabla) \phi_\text{mono}(\mathbf{r})$, with the result that $$ \tilde{\mathbf{A}}_\text{dip} = i (\pmb{\mu} \times \mathbf{k}) \tilde{\phi}_\text{mono}(\mathbf{k}) $$ I don't think that there's an intuitive physical reason that this is true like there is for the electric dipole. Mathematically, it stems from the fact that $\phi_\text{mono}(\mathbf{r})$ is (proportional to) the Green's function for the Laplacian operator, and you have both $\nabla^2 \phi \propto \rho$ and $\nabla^2 \mathbf{A} \propto \mathbf{J}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I guess your first equation checks out, but how did you arrive at it? $\endgroup$ – KF Gauss Jun 30 at 20:20
  • $\begingroup$ @KFGauss: See my edit. $\endgroup$ – Michael Seifert Jun 30 at 20:45
0
$\begingroup$

Remember that polynomials become differential operators and viceversa when switching between spatial and Fourier coordinates, while products become convolutions and viceversa. Hence it is expected that the Fourier of that potential are sort of convolutions of third-order Green operators.

On the case of inverse-square potentials, the second-order Green operator is essentially the solution of the Laplace equation with a Delta-Dirac source. In your case, I would expect you would have to consider solutions of the Laplace equation with a directional derivative of the Delta-Dirac as source, or alternatively some directional derivative of the Laplacian operator with a Delta-Dirac as source after applying integration by parts

| cite | improve this answer | |
$\endgroup$

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.