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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?

Geert VS
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  • Why are you thinking that the photons would lose more energy if the mirrors are farther away? And how are you going to make mirrors for gamma rays? – naturallyInconsistent Jun 25 '23 at 04:40
  • Can you be clearer that you are talking about cosmological redshift. Can you also say how it differs from the lots of other questions about cosmological redshift and energy conservation. e.g., https://physics.stackexchange.com/questions/410392/energy-conservation-on-expanding-universe https://physics.stackexchange.com/questions/13577/photons-in-expanding-space-how-is-energy-conserved and links therein. – ProfRob Jun 25 '23 at 06:45
  • Thanks for the comments. @naturallyInconsistent: further distance means longer transit in expanding space and thus more redshift, therefore less momentum to impart on the mirrors. You could even choose the distance such that you get optimal circumstances so that the mirror would reflect visible light which once was a gamma ray, right? ProfRob: I've checked a lot of pages but I can't check/find them all. What is special about my question is that it has a frame of reference that coincides with the emission ánd absorption of a photon (two photons). – Geert VS Jun 25 '23 at 10:25
  • You cannot have cosmological redshift in an inertial frame. This question doesn’t make sense to me – Dale Jun 26 '23 at 23:27
  • @Dale, why isn't the frame I've described inertial? The experimenter (in the middle) is not accelerating – Geert VS Jun 26 '23 at 23:34
  • @GeertVS it is non inertial because the metric is not $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$ which is the metric in an inertial frame – Dale Jun 26 '23 at 23:55
  • @Dale, I would argue this is the metric for the experimenter. His/her reference frame is local and doesn't encompass the space between the mirrors on each sides, right? – Geert VS Jun 27 '23 at 00:27
  • @GeertVS the space between the mirrors is part of the experiment too. You don’t get your non-conservation without it. The experiment is not confined to the experimenter’s local frame, by design. You just can’t have it both ways. Either you keep your experiment local and inertial and there is no cosmological expansion or you include cosmological expansion and the frame is non local. As it stands the question is self contradictory – Dale Jun 27 '23 at 00:29
  • @Dale going along with your assumptions, that doesn't mean conservation of energy holds. Do you agree we have to let go of this conservation principle (time symmetry) in GR? One more thing I've been thinking about: if we include the kinetic energy the mirrors got from the expansion of space (from the point of view of the experimenter in the middle), the equations look different. But the experimenter is not an observer of the reflection, so it's weird to include it into his/her reference frame. – Geert VS Jun 27 '23 at 01:25
  • sequel of previous comment: If one is allowed to include receding objects into their frame, then you don't even need photons. The kinetic energy of these objects is a violation of energy conservation in and of itself. It definitively doesn't scale with the amount of vacuum energy that was added to the space in between since all masses get the same velocity relative to our observer. More mass: more kinetic energy, independent of the amount of vacuum energy. – Geert VS Jun 27 '23 at 01:27
  • @GeertVS asked “Do you agree we have to let go of this conservation principle (time symmetry) in GR?” Absolutely. In the FLRW spacetime there is no timelike Killing vector. So there is no conserved quantity associated with a timelike Killing vector (energy) in the FLRW spacetime. In contrast there is a timelike Killing vector in Minkowski spacetime so you do get energy conservation in an inertial frame. So you can have either energy non-conservation or inertial frame. Not both. That is the issue. The question is self contradictory as written – Dale Jun 27 '23 at 02:10
  • @Dale thanks. And yet people (experts?) still seem to disagree about the status of conservation of energy in GR. Things are clearer for me now. Usually you need to wrestle with ideas before you get a good grasp of them, so I don't know whether I should change the title or the question. This way others can see the journey and learn more, I think (I hope). – Geert VS Jun 27 '23 at 11:11
  • @GeertVS the lack of a conserved energy in spacetimes without a timelike Killing vector is well known. Certainly it should be known by anyone who merits the designation of “expert”. I don’t care which you change, but you should fix this question so it isn’t self contradictory – Dale Jun 27 '23 at 11:21
  • This edit might seem it makes the question not self-contradictory. It might seem to have fixed the question. It might seem that I think this is a good revision. – Dale Jun 27 '23 at 13:55
  • Haha, thanks, much appreciated – Geert VS Jun 27 '23 at 14:45
  • It should be noted that the cosmological redshift is no different from a kinematic Doppler shift. So it's worth thinking about how your manipulations differ from those you would use to study reflection off a moving mirror in static coordinates, and whether (and why) you would find energy to be conserved in that case. – Sten Jun 27 '23 at 15:52

1 Answers1

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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.

JEB
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  • thanks for your answer. But I think you're getting conservation of energy by construction: you assume the transfer to the mirror is equal to the difference of the initial and final states. No surprise everything checks out then. I've made an edit/update to my question to show the mirror(s) can't account for the energy change. Perhaps I'm wrong, I'm happy to see where I might have made a mistake. – Geert VS Jun 26 '23 at 23:27