Violation of energy conservation: cosmological Doppler-effect in what might seem an inertial frame of reference

Question

I've seen disagreement about whether the cosmological Doppler-effect violates conservation of energy. It can be complicated to analyze since there is no reference frame of a photon. My question is: would the following (thought-)experiment prove it does violate energy conservation?

You are in the center of mass-frame where a stationary electron annihilates with a stationary positron. Two photons of 511 keV go out in opposite directions, both heading towards equidistant mirrors through nothing but expanding space. Perhaps your buddies at the mirrors notice a small momentum change when a photon bounces back. No absorption, although the photons have to lose some energy at the reflection. Anyway, both photons head back to center of mass-frame and at the point of their creation you measure their energy.

The further away the mirrors are, the bigger the energy-discrepancy between their creation and your final measurement, right? While your buddies at the mirrors would measure a smaller (not bigger!) kinetic energy of a mirror if the mirrors are further away, right? So: can we conclude the (cosmological) Doppler-shift violates conservation of energy?

Edit/update:

Without (cosmological) redshift the photon imparts the following momentum and kinetic energy to the mirror:

$p_m= h(\frac{1}{\lambda_0}+\frac{1}{\lambda_1})$

$\Delta E_f+\Delta E_k =0$ which means that $hc(\frac{1}{\lambda_0}-\frac{1}{\lambda_1})=\frac{p²_m}{2m}$

Defining $a=\frac{1}{\lambda_0}$ and $b=\frac{1}{\lambda_1}$, using Wolfram Alpha, and a Taylor-expansion for the sum under the square root I get:

$\Delta p_f=h(\frac{1}{\lambda_1}-\frac{1}{\lambda_0})=\frac{-2h²}{mc \lambda_0}$

Since the photons are 511 keV their wavelength is roughly 2.4 picometer. Filling in the numerical values for Planck's constant and the speed of light, for a 1 kg mirror we get a percentage change in momentum of:

$\frac{\Delta p_f}{p_{f,0}}=\frac{-2h}{mc \lambda_0}=-1,84\cdot10^{-28}$ %

Pretty small! While with cosmological redshift the photons will return back to the place they started with way less energy and momentum if the distance to the mirror is big. Also, the photons will impart less momentum and enery to the mirror if they have redshifted (where the accent indicates a redshifted wavelength):

$p_m= h(\frac{1}{\lambda_0'}+\frac{1}{\lambda_1'})$

Redshifted wavelengths are larger so their reciprocal is smaller. The kinetic energy that the mirror takes away depends on a difference:

$hc(\frac{1}{\lambda_0'}-\frac{1}{\lambda_1'})=\frac{p²_m}{2m}$

But the difference between the positive reciprocal wavelenghts is always smaller than the first term of this difference, which can become arbitrary small when the distance becomes large enough.

So with cosmological redshift the experimenter in the center of mass-frame will detect photons with energy way lower than 511 keV, while the mirrors can't account for this energy difference. Therefore: violation of energy conservation. My question remains: is my conclusion correct?

Answer 1

The 4-momentum transfer of a photon to the mirror needs to be orthogonal to the mirror's 4-momentum, to keep the mirror "on shell". The photon initial/final 4-momenta are ($c=1$):

$$ k_{\mu} = (k,k) $$ $$ k_{\mu}' = (k',-k')$$

so the 4-momentum transfer to the mirror is:

$$ q_{\mu} = k_{\mu}'-k_{\mu}= -(k-k', k+k') $$

Meanwhile, the mirror is moving away, so the four-velocity is:

$$ u_{\mu} = \gamma(1, v) $$

The condition that:

$$ q_{\mu}u^{\mu} = 0$$

means

$$ (k'-k) - v(k+k') = 0 $$

or

$$ \frac{k'}k = \frac{1-v}{1+v} = f_D^2$$

That is, the two-way Doppler shift is the square of the one-way Doppler shift.

The missing energy is transferred to the mirror.

Fixing the coordinates to co-moving coordinates means they're moving away.