# Pound-Rebka-Snider experiment in the inertial frame

In Schutz's book (page 120), Schutz first derives the gravitational redshift in the PRS experiment in a previous paragraph.

$\frac{\nu^{\prime}}{\nu}=\frac{m}{m+mgh+O(v^4)}=1-gh+O(v^4)$.

Here $\nu^{\prime}$ is the frequency of the photon at the top of the tower after the experiment, and $\nu$ at the bottom. (The photon starts at the top.)

In a later paragraph, (page 123). He shows that in an inertial frame, which is a free falling frame wrt Earth, the gravitational redshift is 0. I quote the section from Schutz.

The redshift experiment again. Let us now take a different point of view on the Pound-Rebka-Snider experiment. Let us view it in a freely falling frame, which we have seen has at least some of the characteristics of an inertial frame. Let us take the particular frame that is at rest when the photon begins its upward journey and falls freely after that. Since the photon rises a distance h, it takes time $\Delta t = h$ to arrive at the top. In this time, the frame has acquired velocity $gh$ downward relative to the experimental apparatus. So the photon's frequency relative to the freely falling frame can be obtained by the redshift formula

$\frac{\nu(free falling)}{\nu (apparatus at top)}=\frac{1+gh}{\sqrt{ 1-g^2h^2}}=1+gh+$ higher order terms

From the first formula in my question, we see that if we neglect terms of higher order , then we get $\nu$(photon emitted at bottom) = $\nu$(in freely falling frame when photon arrives at top). So there is no redshift in a freely falling frame. This gives us a sound basis for postulating that the freely falling frame is an inertial frame

I didn't understand the second formula that I have quoted. How did they get that stuff in the denominator? The frequency in the numerotor, is the freq of the photon at the bottom of the tower in the inertial frame which would be the same as $\nu$ in the denominator of the first equation, and $\nu(apparatus at top)$ is the freq at the top of the tower in the free-falling frame/ Is this right?

The end of the equation, where it's written $\dots = 1+gh$, shouldn't be there, right? This just means that the denominator was erased. The last $=$ only holds within an approximation, $gh\ll 1$, up to the leading term.

At any rate, due to the difference-of-squares identity, $$\frac{1+gh}{\sqrt{1-g^2h^2}} = \sqrt{\frac{1+gh}{1-gh}} = \sqrt{\frac{1+v}{1-v}}$$ which is nothing else than the usual relativistic Doppler factor with $v=gh$ being the final speed. We have set $c=1$ everywhere.

Effectively, the calculation using the freely falling frame – i.e. using the equivalence principle – converts the gravitational redshift of GR to the Doppler shift of SR.

Einstein's own first derivation of the gravitational redshift used a rotating frame (and the centrifugal force) instead of a uniformly accelerating frame (and the inertial force).

• Thanks a lot Lubos! Extremely dumb of me to have not noticed this. :(
– user7757
Mar 29 '13 at 10:00
• You are right, the second equation should have higher order terms to make it into an equality. I missed it out.
– user7757
Mar 29 '13 at 10:08