2. A point charge q with mass m and initial relativistic velocity $v_0$ is slowed down in
linear motion with constant acceleration a. The acceleration will result in radiation,
which is called Bremsstrahlung.
(a) Show that the instantaneous intensity of the radiation is given by
$$\frac{dP}{d\Omega} = \frac{\mu_0 q^2 a^2}{16\pi^2 c} \frac{\sin^2 \theta}{(1 - \beta \cos \theta)^5},$$
where $\theta$ is the angle between the line of motion and the direction to the observer,
$\beta = v/c$ the velocity in units of c, and a the acceleration of the particle. Sketch
the angular distribution for a relativistic velocity.
(b) For which angle $\theta_{\text{max}}$ is the intensity of the radiation maximal?
(c) Calculate the total radiated power from the result of part (a) and confirm that
it is consistent with Lienard's generalization of Larmor's formula,
$$P = \frac{\mu_0 q^2 \gamma^6}{6\pi c} (a^2 - |\beta \times a|^2).$$
Hint: $\int_0^1 (1 - x^2)(1 - ax)^{-5} dx = \frac{1}{3}(1 - a^2)^{-3}$
(d) What is the fraction of the electron's initial energy that is radiated off? The
remainder would be absorbed by whatever mechanism slows the electron down.
