Frequency shift for accelerating source due to equivalence principle? Do cosmologists consider it alongside the Doppler shift? It's a question that's been bugging me since I first read Einstein's paper on gravitational frequency shift. He derives it using the equivalence principle, and considers an accelerating source first.
Now, for an accelerating source the shift is approximately proportional to acceleration and the distance to observer.
$$
\Delta \nu \propto a L
$$
But I never heard that cosmologists consider this kind of shift. I only hear about Doppler shift all the time, proportional to velocity.
$$
\Delta \nu \propto v
$$
Even Hubble's law is explained by Doppler shift. But wouldn't it make more sense to explain it by shift due to acceleration of all the stars? It is proportional to the distance after all. And every star or galaxy is always accelerating.
Or am I completely wrong, and $L$ is not distance to observer in this case?
 A: In the case of a constant gravitational field and a non-relativistically moving observer, the redshift $z$, i.e. the relative change in frequency $z=\frac{\nu_e}{\nu_o}-1$, is (there's a nice short explanation here under 'Gravitational Redshift'):
$$z=\frac{gL}{c^2}$$
The analog in cosmology would be Hubble's law:
$$v = H_0D$$
with $H_0$ being Hubble's constant (it has dimensions of frequency, by the way). Since $v=cz$ (implicitly assuming reasonably small $D$ here):
$$z = \frac{H_0D}{c}$$
$D$ and $L$ are both the distance to the source, so equating the two expressions for redshift above and re-arranging:
$$g = H_0c$$
So the expansion rate of the Universe (described by the FLRW metric), which here we've approximated as being $\sim$constant, causes the same redshift as a constant gravitational field of magnitude $H_0c$. Hopefully you can see the analogy with your first expression (the proportionality to distance is obvious, anyway).
This is in contrast to the other component of redshift often mentioned in cosmology, that caused by the peculiar velocity of an object, which is simply explained as a Doppler shift, analogous to your second expression.
