The Relativity of Velocities — Question 6

Question 6

A backward-hemisphere photon can enter the forward hemisphere

In $S\prime$ a photon propagates at $\theta\prime=120^\circ$ from +x. The frame moves at $v=0.80c$ relative to S. For $S\prime$ moving at +v along x, use $u_x=(u_x\prime+v)/(1+vu_x\prime/c^2)$ and $u_y=u_y\prime/[\gamma_v(1+vu_x\prime/c^2)]$, with $\gamma_v=(1-v^2/c^2)^{-1/2}$.

1. Find the laboratory propagation angle.

2. Derive the condition on $\theta\prime$ for the laboratory x component to be positive.

Question 6 - Solution

A backward-hemisphere photon can enter the forward hemisphere

(a) Method and calculation

$$\cos\theta=\frac{\cos\theta\prime+\beta}{1+\beta\cos\theta\prime}=\frac{-.5+.8}{1-.4}=\boxed{.5}.$$$$\boxed{\theta=60^\circ},\qquad u_y/c=\frac{\sqrt3/2}{(5/3)(.6)}=\sqrt3/2.$$

(b) Interpretation and further calculation

$$1+\beta\cos\theta\prime>0\quad(|\beta|<1).$$$$\boxed{u_x>0\ \Longleftrightarrow\ \cos\theta\prime>-\beta}.$$

Physical check

Backward and forward hemispheres are frame dependent; the photon still propagates at c in each frame.