# Neat expression for the dipole potential in Fourier space?

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).

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 $$\nabla^2 \mathbf{A} \propto \mathbf{J}$$.