Consider a particle of charge $q$ moving with 3 -velocity $\mathbf{v}$. If the particle is moving slowly then Larmor's formula asserts that the instantaneous radiated power is

$$\mathcal{P}=\frac{\mu_{0}}{6 \pi} q^{2}\left|\frac{d \mathbf{v}}{d t}\right|^{2}$$

Suppose, however, that the particle is moving relativistically. Give reasons why one should conclude that $\mathcal{P}$ is a Lorentz invariant. Writing the 4-velocity as $U^{a}=(\gamma, \gamma \mathbf{v})$ where $\gamma=1 / \sqrt{1-|\mathbf{v}|^{2}}$ and $c=1$, show that

$$\dot{U}^{a}=\left(\gamma^{3} \alpha, \gamma^{3} \alpha \mathbf{v}+\gamma \dot{\mathbf{v}}\right)$$

where $\alpha=\mathbf{v} \cdot \dot{\mathbf{v}}$ and $\dot{f}=d f / d s$ where $s$ is the particle's proper time. Show also that

$$\dot{U}^{a} \dot{U}_{a}=-\gamma^{4} \alpha^{2}-\gamma^{2}|\dot{\mathbf{v}}|^{2} .$$

Deduce the relativistic version of Larmor's formula.

Suppose the particle moves in a circular orbit perpendicular to a uniform magnetic field $\mathbf{B}$. Show that

$$\mathcal{P}=\frac{\mu_{0}}{6 \pi} \frac{q^{4}}{m^{2}}\left(\gamma^{2}-1\right)|\mathbf{B}|^{2}$$

where $m$ is the mass of the particle, and comment briefly on the slow motion limit.