A point particle of charge $q$ has trajectory $y^{\mu}(\tau)$ in Minkowski space, where $\tau$ is its proper time. The resulting electromagnetic field is given by the Liénard-Wiechert 4-potential

$$A^{\mu}(x)=-\frac{q \mu_{0} c}{4 \pi} \frac{u^{\mu}\left(\tau_{*}\right)}{R^{\nu}\left(\tau_{*}\right) u_{\nu}\left(\tau_{*}\right)}, \quad \text { where } \quad R^{\nu}=x^{\nu}-y^{\nu}(\tau) \quad \text { and } \quad u^{\mu}=d y^{\mu} / d \tau$$

Write down the condition that determines the point $y^{\mu}\left(\tau_{*}\right)$ on the trajectory of the particle for a given value of $x^{\mu}$. Express this condition in terms of components, setting $x^{\mu}=(c t, \mathbf{x})$ and $y^{\mu}=\left(c t^{\prime}, \mathbf{y}\right)$, and define the retarded time $t_{r}$.

Suppose that the 3 -velocity of the particle $\mathbf{v}\left(t^{\prime}\right)=\dot{\mathbf{y}}\left(t^{\prime}\right)=d \mathbf{y} / d t^{\prime}$ is small in size compared to $c$, and suppose also that $r=|\mathbf{x}| \gg|\mathbf{y}|$. Working to leading order in $1 / r$ and to first order in $\mathbf{v}$, show that

$$\phi(x)=\frac{q \mu_{0} c}{4 \pi r}\left(c+\hat{\mathbf{r}} \cdot \mathbf{v}\left(t_{r}\right)\right), \quad \mathbf{A}(x)=\frac{q \mu_{0}}{4 \pi r} \mathbf{v}\left(t_{r}\right), \quad \text { where } \quad \hat{\mathbf{r}}=\mathbf{x} / r$$

Now assume that $t_{r}$ can be replaced by $t_{-}=t-(r / c)$ in the expressions for $\phi$ and $\mathbf{A}$ above. Calculate the electric and magnetic fields to leading order in $1 / r$ and hence show that the Poynting vector is (in this approximation)

$$\mathbf{N}(x)=\frac{q^{2} \mu_{0}}{(4 \pi)^{2} c} \frac{\hat{\mathbf{r}}}{r^{2}}\left|\hat{\mathbf{r}} \times \dot{\mathbf{v}}\left(t_{-}\right)\right|^{2}$$

If the charge $q$ is performing simple harmonic motion $\mathbf{y}\left(t^{\prime}\right)=A \mathbf{n} \cos \omega t^{\prime}$, where $\mathbf{n}$ is a unit vector and $A \omega \ll c$, find the total energy radiated during one period of oscillation.