Suppose v and a are instantaneously collinear (at time t_r ), as, for example, in straight-line motion. Find the angular distribution of the radiation (Eq. 11.72) and the total power emitted.
\frac{dP}{d\Omega }=\left(\frac{\eta .u}{\eta c} \right) \frac{1}{\mu_0c}E^2_{rad}\eta ^2=\frac{q^2}{16\pi^2\epsilon _0}\frac{\left|\hat{\eta }\times (u\times a)\right|^2 }{(\hat{\eta }.u)^5}, (11.72)
In this case (u × a) = c( \hat{\eta} × a), so
\frac{dP}{d\Omega }=\frac{q^2c^2}{16\pi^2\epsilon _0}\frac{\left|\hat{\eta }\times (\hat{\eta}\times a)\right|^2 }{(c-\hat{\eta }.v)^5},Now
\hat{\eta}\times (\hat{\eta}\times a)=(\hat{\eta}.a)\hat{\eta}-a, so \left|\hat{\eta}\times (\hat{\eta}\times a)\right|^2=a^2-(\hat{\eta}\times a)^2.In particular, if we let the z axis point along v, then
\frac{dP}{d\Omega }=\frac{\mu_0q^2c^2}{16\pi^2c}\frac{\sin^2\theta }{(1-\beta \cos\theta)^5}, (11.74)
where β ≡ v/c. This is consistent, of course, with Eq. 11.69, in the case v = 0. However, for very large v (β ≈ 1) the donut of radiation (Fig. 11.11) is stretched out and pushed forward by the factor (1 − β cos θ)^{−5}, as indicated in Fig. 11.13. Although there is still no radiation in precisely the forward direction, most of it is concentrated within an increasingly narrow cone about the forward direction (see Prob. 11.15).
S_{rad}=\frac{1}{\mu_0c}\left(\frac{\mu_0q}{4\pi\eta } \right)^2 \left[a^2-(\hat{\eta }.a)^2\right]\hat{\eta }=\frac{\mu_0q^2a^2}{16\pi^2c}\left(\frac{\sin^2\theta}{\eta ^2} \right)\hat{\eta} , (11.69)
The total power emitted is found by integrating Eq. 11.74 over all angles:
P=\int{\frac{dP}{d\Omega }d\Omega} =\frac{\mu_0q^2a^2}{16\pi^2c}\int{\frac{\sin^2\theta }{(1-\beta \cos\theta)^5}}\sin\theta d\theta d\phi ,The φ integral is 2π; the θ integral is simplified by the substitution x ≡ cos θ:
P =\frac{\mu_0q^2a^2}{8\pi c}\int_{-1}^{+1}{\frac{(1-x^2)}{(1-\beta x)^5}}dx ,Integration by parts yields \frac{4}{3}(1 − β^2)^{−3}, and I conclude that
P =\frac{\mu_0q^2a^2\gamma ^6}{6\pi c}\ . (11.75)
This result is consistent with the Liénard formula (Eq. 11.73), for the case of collinear v and a. Notice that the angular distribution of the radiation is the same whether the particle is accelerating or decelerating; it only depends on the square of a, and is concentrated in the forward direction (with respect to the velocity) in either case. When a high speed electron hits a metal target it rapidly decelerates, giving off what is called bremsstrahlung, or “braking radiation.” What I have described in this example is essentially the classical theory of bremsstrahlung.
P =\frac{\mu_0q^2\gamma ^6}{6\pi c}\left(a^2-\left|\frac{v\times a}{c} \right|^2 \right), (11.73)
