I Something about retarded potentials for oscillating electric dipole

[Salmone]

In a problem of an oscillating electric dipole, under appropriate conditions, one can find, for the potential vector calculated at the point $\vec{r}$, the expression $\vec{A}=\hat{k}\frac{\mu_0I_0d}{4\pi}\frac{cos(\omega(t-r/c))}{r}$ where: $\hat{k}$ is the direction of the $z-axis$ where the dipole is oscillating, $I_0$ is the current ($I(t)=I_0cos(\omega t)$), $d$ is the distance between the charges of the dipole and $r$ is the distance between the origin of the system and the point where I want to calculate the potential vector. Let $\vec{p}=\hat{k}qd=\frac{\hat{k}dI_0}{\omega}sin(\omega t)$ be the dipole moment, it is possible to rewrite the potential vector as $\vec{A}=\frac{\mu_0}{4\pi}\frac{\vec{\dot p(t-r/c)}}{r}$ where $\vec{\dot p}$ is the derivative with respect to time.

 

[vanhees71]

You can start with the source of a point charge,
$$\rho(t,\vec{x})=q \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{j}(t,\vec{x})= q \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)],$$
where $\vec{y}(t)$ is the trajectory of the charge.

Now we assume that
$$\vec{y}(t)=\vec{d} \sin(\omega t).$$
For $r=|\vec{x}|\gg |\vec{d}|$ we can expand the charge-current distribution up to first order in $\vec{d}$,
$$\rho(t,\vec{x})=q \delta^{(3)}(\vec{x}) - q \vec{y}(t) \cdot \vec{\nabla} \delta^{(3)}(\vec{x}) + \mathcal{O}(\vec{d}^2), \quad \vec{j}(t,\vec{x})=q \dot{\vec{y}}(t) \delta^{(3)}(\vec{x}) + \mathcal{O}(\vec{d}^2).$$
From the first term of $\rho$ (of order $\mathcal{O}(d^0)$) you get the electrostatic Coulomb field of a charge at rest in the origin (which you can easily verify using the retarded potential too).

For the terms of order $\mathcal{O}(d)$ you get for $\vec{A}$ (in SI units)
$$\vec{A}(t,\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(t-|\vec{x}-\vec{x}'|/c,\vec{x}') \frac{1}{|\vec{x}-\vec{x}'|}=\frac{\mu_0 q \omega \vec{d}}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \cos[\omega (t-|\vec{x}-\vec{x}'|/c] \frac{\delta^{(3)}(\vec{x}')}{|\vec{x}-\vec{x}'|} = \frac{\mu_0 q \omega \vec{d}}{4 \pi r} \cos[\omega (t-r/c)].$$
With $I_0=q \omega$ that's the solution you are looking for.

The final equation, of course, must read
$$\vec{A}(t),\vec{x})=\frac{\mu_0}{4 \pi} \frac{\dot{\vec{P}}(t-r/c)}{r}.$$