a  ^?? Представьте, что вы физик XIX века, который не знает теорию относительности, но знает о показателе преломления и законе сложения скоростей Галилея. Какое время $T$ займет пусть света от A до B согласно такой теории? В ответе оставьте только ведущий член $v/c$.

Naively, the light moves with speed $c$ in free space, and speed $v_\mathrm{in} = c/n + v$ inside the slab, by Galilean velocity addition. So when the light is in the slab, the relative speed of the light and slab is exactly $v_\mathrm{rel} = c n$ .

Therefore, by routine kinematics, the time spent in the slab is $t_\mathrm{in} = D v_\mathrm{rel}$ during which the light moves forward by $D + vt_\mathrm{in}$. The rest of the time is $t_\mathrm{out} = L − D − vt_\mathrm{in} c$ . Adding these together gives a total time of

\[T = \frac{L}{c} + D \left( \frac{1}{v_\mathrm{rel}} - \frac{1}{c} - \frac{v}{cv_\mathrm{rel}} \right) = \frac{L}{c} + \frac{D}{c} \left( n - 1 - \frac{vn}{c} \right) \]

b  ^?? Какое время $T$ на самом деле займет путь света от A до B? В ответе также оставьте только ведущий член $v/c$.

The slab length contracts, but this is second order in $v/c$, while we’re just interested in the first order effect. The key difference is that because of relativistic velocity addition, the light in the slab moves with speed

\[ v_\mathrm{in} = \frac{v + \frac{c}{n}}{1 + \frac{v}{nc}} \simeq \frac{c}{n}  + \left( 1 - \frac{1}{n^2} \right) v\]

Thus, to leading order in $v/c$, when the light is in the slab, the relative speed of the light and slab is, in the lab frame, $v_\mathrm{rel} \simeq c/n − v/n^2 $. The rest of the above derivation goes through unchanged, giving 

\[ T = \frac{L}{c} + D \left( \frac{1}{v_\mathrm{rel}} - \frac{1}{c} - \frac{v}{cv_\mathrm{rel}} \right) \simeq \frac{L}{c} + \frac{D}{c} \left( n - 1 - \frac{v(n-1)}{c} \right)  \]

again to first order in v/c. Before the advent of relativity, this result was explained by an “ether drag” coefficient of $1 − 1/n^2$.