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A relativistic electron radiates while executing a nearly circular cyclotron motion in a uniform magnetic field B. Find how the function $\gamma(t) \equiv E(t)/mc^2$ decreases with time t starting from some initial value $\gamma(0)$. Assume that $\gamma(t) \gg 1$ and that the energy radiated during one period of cyclotron motion is small compared to the electron energy E(t).

Hint: the power W radiated by a relativistic electron can be written as

(1)

\begin{align} W=-\frac{2}{3}\frac{e^2}{m^2c^3}\frac{dp^{\mu}}{d\tau}\frac{dp_{\mu}}{d\tau} \end{align}

where $p^{\mu}=(E/c,-\mathbf{P})$ and $p_{\mu}=(E/c,\mathbf{P})$ are contravariant and covariant 4-momenta, respectively, and $\tau$ is the proper time.

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Answer

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The relation between the proper and lab time is

(2)

\begin{align} d\tau=dt\sqrt{1-\frac{v^2}{c^2}}=\frac{dt}{\gamma(t)} \end{align}

Therefore,

(3)

\begin{align} \frac{dp^{\mu}}{d\tau}=\gamma(t)\left(\frac{1}{c}\frac{dE}{dt},\frac{d\mathbf{P}}{dt} \right), \;\; \frac{dp_{\mu}}{d\tau}=\gamma(t)\left(\frac{1}{c}\frac{dE}{dt},-\frac{d\mathbf{P}}{dt} \right) \end{align}

For motion in magnetic field

(4)

\begin{align} \frac{dE}{dt}=0, \;\; \frac{d\mathbf{P}}{dt} = \frac{e}{c}\left[\mathbf{v} \times \mathbf{B} \right]. \end{align}

Hence,

(5)

\begin{align} W = \frac{2}{3}\frac{e^2}{m^2c^3}\gamma^2(t)\frac{v^2}{c^2}e^2B^2 \simeq \frac{2}{3}\frac{e^4}{m^2c^3}\gamma^2(t) B^2. \end{align}

The energy balance equation becomes

(6)

\begin{align} \frac{d}{dt}\left( \gamma mc^2\right)=-W = -\frac{e^4}{m^2c^3}\gamma^2(t)B^2, \end{align}

which has the solution

(7)

\begin{align} \frac{1}{\gamma(t)}=\frac{1}{\gamma(0)}+\frac{2}{3}\frac{e^4B^2}{m^3c^5}t. \end{align}