# Something about retarded potentials for oscillating electric dipole

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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.

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$$\rho(t,\vec{x})=q \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{j}(t,\vec{x})= q \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)],$$
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
$$\vec{A}(t),\vec{x})=\frac{\mu_0}{4 \pi} \frac{\dot{\vec{P}}(t-r/c)}{r}.$$

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