Electric field and magnetic field from dipole radiation The electric field and magnetic field from dipole radiation given in my book is:
$$\overrightarrow{E}=\frac{\overrightarrow{r}\times(\overrightarrow{r}\times \overrightarrow{\ddot{d})}}{4\pi \varepsilon_0 r^3 } $$
$$\overrightarrow{B}=\frac{ \overrightarrow{\ddot{d}}\times\overrightarrow{r}}{4\pi \varepsilon_0 r^2 } $$
I am trying to derive this. I have derived the sclar and vector electric potential :
$$\overrightarrow{A}=\frac{1}{4\pi\varepsilon_0r} \overrightarrow{\dot{d}}(t-\frac{r}{c})$$
$$\phi=\frac{\overrightarrow{r}\cdot\overrightarrow{\dot{d}}}{4\pi \varepsilon_0 r^2}$$
Note that $\overrightarrow{\dot{d}}$ is a function of both $t$ and $r$ in $(t-\frac{r}{c})$
The textbook has given no hint of how this was derived.
I tried using the equation  $\overrightarrow{E}=-\nabla \phi-\frac{\partial \overrightarrow{A}}{\partial t}$ and $ \overrightarrow{B}=\nabla \times \overrightarrow{A}$
However I am unsure how to proceed when
$\overrightarrow{\dot{d}}$ is function of two variables such as $t$ and $r$.
 A: First of all, there are several equations determining the problem. Important to note that for electric and vector potential you have:
\begin{equation}
\Box{\vec{A}=-\mu_0 \vec{j}}
\end{equation}
The latter is true when assuming a Lorenz gauge, namely:
\begin{equation}
div \vec{A}+\frac {1} {c^2}\partial_t \phi = 0
\end{equation}
The solution of this equation is given by a Green function formalism, which gives retarded potential:
$$ \vec{A} =\frac{ \mu_0 } {4 \pi} \int  \frac{\vec{j}(\vec{R},t- \frac{|\vec{r} - \vec{R}|} {c})} {{| \vec{r} - \vec{R}|}} \, d^3 \vec{R}$$
The current density is
$$ \vec{j} = e \dot{\vec{r}} = \dot{\vec{d}} $$
After integrating we have:
$$ \vec{A} = \frac{\mu_0} {4 \pi} \frac{  \dot{\vec{d}}(t-r/c) } {r}$$
I get a different result, so, maybe you are using other system of units. Next, calculating B field
\begin{equation}
 \vec{B} = \nabla \times \vec{A} = e_{ijk} \nabla_j (\frac{\mu_0} {4 \pi} \frac{  \dot{\vec{d}}_k (t-r/c) } {r}) = \frac{\mu_0} {4 \pi} e_{ijk} [\frac{\nabla_j \dot{\vec{d}}_k} {r} +\dot{\vec{d}}_k \nabla_j( \frac{1} {r}) ]
\end{equation}
what you need to know is that $\nabla_j (\frac{1} {r}) =  - \frac{r_i} {r^3}$. You can prove the formula in Cartesian coordinates by taking $ r = \sqrt{x^2+y^2+z^2}$. And
\begin{equation}
 \nabla_j \dot{\vec{d}}_k = \frac{\partial \dot{\vec{d}}_k}{ \partial (t-r/c)} \nabla_j r/c = -\ddot{\vec{d}}_k \frac{r_j} {rc} 
\end{equation}
From the equation on $\vec{B}$ there are two terms, you can show that the second one is smaller then the first for big r, so you have only
\begin{equation}
 \vec{B} = \frac{\mu_0} {4 \pi c} \frac{  \ddot{\vec{d}}(t-r/c) \times \vec{r}} {r^2}
\end{equation}
So, next, you can get an equation on $\vec{E}$ using the same formalism on the equation
\begin{equation}
 \vec{E} = -\nabla \phi- \frac{\partial \vec{A}} {\partial t}
\end{equation}
Taking into account
\begin{equation}
 \frac{\partial \dot{\vec{d}}(t-r/c)} {\partial t} = \ddot{\vec{d}}(t-r/c)
\end{equation}
and using del operator as previously shown
A: Hint: 1) these relations are deduced with a delay time: https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential
with : $\vec{B}=\vec{\nabla} \times \vec{A} = \vec{\nabla} (t-|\vec{r} -\vec{r}_{s}|/c)\times \vec{\dot{A}}=-\frac{\vec{r}}{r}\times\vec{\dot{A}}  $
(.) derivation with respect to $\; t_{r}$


*For a plane wave, we have  $\vec{E}=\vec{B}\times\frac{\vec{r}}{r}$
