Relativistic Beaming of Light

Figure 3.11: Relativistic beaming of light.

Consider a point source that emits light isotropically in all directions in its rest frame, $S$ Let us observe this source in a frame of reference, $S'$, that moves with velocity ${\bf v}= - v\,{\bf e}_x$, and is in a standard configuration, with respect to frame $S$. Thus, the source appears to move with velocity $v\,{\bf e}_x$ in frame $S'$. See Figure 3.11. Now, half of the emitted radiation in $S$ emerges in the region $ABC$, bounded by the rays $A$ and $C$ shown in the figure. Likewise, half the emitted radiation in $S'$ emerges in the region $A'B'C'$, bounded by the rays $A'$ and $C'$ shown in the figure. Ray $A$ has the phase velocity $(0,\,c, \,0)$. Likewise, ray $A'$ has the phase velocity $(c\,\cos\theta,c\,\sin\theta,\,0)$, where the angle $\theta$ is shown in the figure. By symmetry, the angle subtended between $A'$ and $B'$ is the same as that subtended between $C'$ and $B'$. The transformation of velocity, (3.122)–(3.124), yields

$$c\,\cos\theta = v,$$ (3.134)

$$c\,\sin\theta = \frac{c}{\gamma},$$ (3.135)

or

$$\sin\theta = \frac{1}{\gamma}.$$ (3.136)

It follows, that in a frame of reference in which the source moves with velocity ${\bf v}$, half of the emitted radiation is beamed into a cone whose axis is ${\bf v}$, and whose half-angle is $\sin^{-1}(1/\gamma)$. If the source is moving very close to the velocity of light then $\gamma\gg 1$, and $\theta\simeq 1/\gamma \ll 1$. In other words, the emitted radiation is beamed very strongly in the direction of motion of the source.