About the standard derivation of the gravitational redshift

Question

The objective is to derive the gravitational redshift ONLY from the Einstein's equivalence principle (E.E.P.), without using the whole theory of Relativity.

This is the standard "informal" derivation of the gravitational redshift (For example Carroll in his book follows this way):

Consider an emitter, $E$, e.g. a vibrating atom, at rest at a point near the Earth's surface, say, of gravitational potential $\phi$. Let it send light, or any other electromagnetic, signals to a receiver $R$ at rest directly above $E$ and distance $h$ from it; the gravitational potential at $R$ is $\phi+\Delta\phi$, where $\Delta\phi = gh$, $g$ being the acceleration due to gravity. Let $\nu_E$ be the frequency of the signal as measured at $E$, and $\nu_R$ the frequency of the signal received, and measured, at $R$. Then it is used the relativistic Doppler effect, in the case where the receiver is moving with constant relative velocity $V$ respect to the emitter, to show that:

$$ \frac{\nu_R-\nu_E}{\nu_E}=-\frac{\Delta\phi}{c^2}=-\frac{gh}{c^2} $$

By the E.E.P. will follow easly the gravitational redshift.

And now my trouble:

In the derivation of the basic formula for the classical Doppler shift (which, it may be recalled, is the first approximation in $\frac{V}{c}$ of the corresponding special relativistic formula), on which the standard arguent is so decisively based, the emitter and the receiver move with constant velocities relative to an inertial frame and $V$ is the constant velocity of the receiver relative to the emitter and away from it. That is, the velocity of the emitter is the same at the instant of the emission and, likewise, the velocity of the receiver is the same at the instant of the reception. This is not the case when $E$ and $R$ are accelerating relative to an inertial frame.

So, should I conclude that the above argument is wrong?

Answer 1

I don't have Carroll's book, but I don't recognise the description you give of the derivation of the red shift, and in particular I don't see why the relativistic Doppler shift is relevant. The derivation I'm familiar with is to say that the change in potential energy is $mgd$, where $m$ is the effective mass given by $E = h\nu = mc^2$. So:

$$ h\nu_e - h\nu_r = \frac{h\nu_e}{c^2} gd $$

and a quick rearrangement gives:

$$ \frac{\nu_e - \nu_r}{\nu_e} = \frac{gd}{c^2} $$

No Doppler shift involved.

Answer 2

the emitter and the receiver move with constant velocities relative to an inertial frame and $V$ is the constant velocity of the receiver relative to the emitter and away from it.

No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $\Delta v$ between the time the photon was emitted and the time it was received.

In other words, consider an emitter and a receiver, both accelerating with a constant acceleration $g$, and suppose the emitter is a distance $h$ behind the receiver. Also, suppose that $g$ is low enough so that relativistic effects can be ignored.

If the (trailing) emitter sends out a photon with wavelength $\lambda_e$, it reaches the (leading) receiver after a time $\Delta t\approx h/c$ (ignoring the little extra distance that the photon has to travel because the receiver has accelerated).

During this time, the receiver has gained an extra velocity $\Delta v=g\Delta t \approx gh/c$. If $\Delta v$ is small, the standard Newtonian Doppler effect applies, and the wavelength of the received photon has changed as $$ \frac{\Delta\lambda}{\lambda_e} = \frac{\Delta v}{c} \approx \frac{gh}{c^2}, $$ or equivalently, the frequency has changed as $$ \frac{\Delta\nu}{\nu_e} = -\frac{\Delta v}{c} \approx -\frac{gh}{c^2}. $$ According to the EEP, the acceleration of the emitter and the receiver is equivalent to a gravitational field.