 Solution

In the instantaneous rest frame the total radiation has zero net three-momentum, so its emitted four-momentum is proportional to the four-velocity. A Lorentz boost multiplies its energy by $\gamma$, while the elapsed coordinate time for the source is also multiplied by $\gamma$. Their ratio, total instantaneous radiated power, is therefore invariant. This concerns total emitted power, not the directional flux or an observer's photon arrival rate.

Use the metric of signature $(+---)$ and dots for proper-time derivatives. From $\gamma =(1-v^2{)}^{-1/2}$ obtain $\dot{\gamma }={\gamma }^3\alpha$, where $\alpha =\mathbf{v}\cdot \dot{\mathbf{v}}$. Differentiating $U=(\gamma ,\gamma \mathbf{v})$ givesSquaring in the Minkowski metric, using $1-v^2={\gamma }^{-2}$, gives $\dot{U}\cdot \dot{U}={\gamma }^6{\alpha }^2(1-v^2)-2{\gamma }^4{\alpha }^2-{\gamma }^2|\dot{\mathbf{v}}{|}^2$, henceIn the rest frame, $-{\dot{U}}^2$ is the squared ordinary acceleration, so invariance of power and the low-speed Larmor formula give $P=-({\mu }_0{q}^2/6\pi ){\dot{U}}^2$. In terms of coordinate-time acceleration $\mathbf{a}=d\mathbf{v}/dt$, this is the Lienard formulaFor circular motion perpendicular to $\mathbf{B}$, $\mathbf{a}\perp \mathbf{v}$ and the Lorentz force gives $\gamma m|\mathbf{a}|=|q|v|\mathbf{B}|$. ThusAt low speed ${\gamma }^2-1\sim v^2$, reproducing ordinary Larmor radiation with magnetic acceleration $|q|vB/m$.