Einstein and Gravitation — Question 6

Question 6

An accelerating cabin makes a locally straight light ray appear to bend

At the instant a cabin is at rest in an inertial frame, light enters horizontally across width $L=30\,\mathrm m$. The cabin accelerates upward at $a=10\,\mathrm{m/s^2}$. Use light travel time $L/c$, with $c=3\times10^8\,\mathrm{m/s}$, and neglect higher-order relativistic corrections over this short transit.

1. Find the downward displacement of the ray relative to the cabin at the far wall.

2. Explain what this thought experiment does and does not establish about light passing a star.

Question 6 - Solution

An accelerating cabin makes a locally straight light ray appear to bend

(a) Derivation and calculation

The leading crossing time is $$t_{\mathrm{cross}}=L/c=30/(3\times10^8)=10^{-7}\ \mathrm s.$$ While the inertial light path is horizontal, the cabin moves upward by $$\boxed{d=\frac12at_{\mathrm{cross}}^2=\frac{aL^2}{2c^2} =5.0\times10^{-14}\ \mathrm m}.$$ The ray therefore arrives that distance below its entry height relative to the cabin.

(b) Result and interpretation

Using $x\simeq ct$, the leading cabin-frame trajectory is $$\boxed{y_{\mathrm{cabin}}(x)\simeq-\frac{ax^2}{2c^2}},\qquad 0\le x\le L.$$ This local accelerated-frame result illustrates the equivalence argument. It does not determine the accumulated deflection along a stellar trajectory, which needs the gravitational geometry over the full path.

Physical check

The small parameter $aL/c^2\simeq3.33\times10^{-15}$ supports neglect of higher-order corrections; local light speed remains $c$.