The current density in an antenna lying along the $z$-axis takes the form

$$\mathbf{J}(t, \mathbf{x})=\left\{\begin{array}{ll} \hat{\mathbf{z}} I_{0} \sin (k d-k|z|) e^{-i \omega t} \delta(x) \delta(y) & |z| \leqslant d \\ \mathbf{0} & |z|>d \end{array},\right.$$

where $I_{0}$ is a constant and $\omega=c k$. Show that at distances $r=|\mathbf{x}|$ for which both $r \gg d$ and $r \gg k d^{2} /(2 \pi)$, the retarded vector potential in Lorenz gauge is

$$\mathbf{A}(t, \mathbf{x}) \approx \hat{\mathbf{z}} \frac{\mu_{0} I_{0}}{4 \pi r} e^{-i \omega(t-r / c)} \int_{-d}^{d} \sin \left(k d-k\left|z^{\prime}\right|\right) e^{-i k z^{\prime} \cos \theta} d z^{\prime}$$

where $\cos \theta=\hat{\mathbf{r}} \cdot \hat{\mathbf{z}}$ and $\hat{\mathbf{r}}=\mathbf{x} /|\mathbf{x}|$. Evaluate the integral to show that

$$\mathbf{A}(t, \mathbf{x}) \approx \hat{\mathbf{z}} \frac{\mu_{0} I_{0}}{2 \pi k r}\left(\frac{\cos (k d \cos \theta)-\cos (k d)}{\sin ^{2} \theta}\right) e^{-i \omega(t-r / c)}$$

In the far-field, where $k r \gg 1$, the electric and magnetic fields are given by

$$\begin{aligned} &\mathbf{E}(t, \mathbf{x}) \approx-i \omega \hat{\mathbf{r}} \times[\hat{\mathbf{r}} \times \mathbf{A}(t, \mathbf{x})] \\ &\mathbf{B}(t, \mathbf{x}) \approx i k \hat{\mathbf{r}} \times \mathbf{A}(t, \mathbf{x}) \end{aligned}$$

By calculating the Poynting vector, show that the time-averaged power radiated per unit solid angle is

$$\frac{d \mathcal{P}}{d \Omega}=\frac{c \mu_{0} I_{0}^{2}}{8 \pi^{2}}\left(\frac{\cos (k d \cos \theta)-\cos (k d)}{\sin \theta}\right)^{2}$$

[You may assume that in Lorenz gauge, the retarded potential due to a localised current distribution is

$$\mathbf{A}(t, \mathbf{x})=\frac{\mu_{0}}{4 \pi} \int \frac{\mathbf{J}\left(t_{\mathrm{ret}}, \mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} d^{3} \mathbf{x}^{\prime}$$

where the retarded time $\left.t_{\text {ret }}=t-\left|\mathbf{x}-\mathbf{x}^{\prime}\right| / c .\right]$