Proof for photon energy loss due to cosmological redshift

Question

According to Doppler redshift, the frequency of the EM radiation decreases if the emitting object is receding from the observer - this produces the problem in the quantization of energy - the observed photon energy should be less than the energy emitted, so is the conservation of energy violated?

Unable to find an explanation here, I asked this question one day, and my physics teacher produced proof that total energy is conserved between the emission and the effect of redshift. He has given me permission to post the proof here in case anyone is interested in having a look.

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Here's the proof:

When a photon is received from a galaxy receding from us with a recession velocity $v$, its observed frequency is reduced by $\Delta f$ where: $\Delta f/f \approx v/c$.

This means its energy appears reduced by an amount:

$$h \Delta f ≈ hvf/c \tag{1}$$

However, the photon has momentum, so the object in the receding galaxy that emitted the photon will experience an equal and opposite change in momentum:

$$m \Delta v = h/\lambda \tag{2}$$

This will increase its kinetic energy in our frame of reference by an amount ΔKE, where:

\begin{align*} \Delta \textrm{KE} &= \frac{1}{2} m(v+\Delta v)^2 - \frac{1}{2} mv^2 \\ &= \frac{1}{2} mv^2+ \frac{1}{2} \times 2mv \Delta v + \frac{1}{2} m \Delta v^2 - \frac{1}{2} mv^2 \\ &\approx mv \Delta v \tag{3} \end{align*} (since the $\frac{1}{2} m \Delta v^2$ term is negligible in comparison)

Substituting equation (2) into (3), we have:

$$\Delta \textrm{KE} ≈ hv/\lambda ≈ hvf/c$$

We can see that the loss of energy of the redshifted photon in our frame of reference is exactly equal to the gain in kinetic energy of the object that emitted the photon as observed in our frame of reference, thus there is no violation of energy conservation.

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We'd appreciate it very much if anyone could kindly criticize this proof =)

Answer 1

Your proof is obviously correct in a local frame of reference for a flat non-accelerating universe (e.g. in the Milne model), because your example is equivalent to the galaxies simply flying apart in a non-expanding space. The proof is even simpler, because a galaxy emits in all directions. As a result, its extra momentum and energy are zero, but the opposite photon is equally blue-shifted. Simpler yet, in the frame of the emitting galaxy, the emitted photon is never redshifted no matter how far it has gone.

However, globally energy does not conserve, because all emitted photons (when detected) appear redshifted to all observers in the universe. This is easy to understand for a closed subluminal universe. The opposite photon that is blue-shifted in your frame would come to you deeply redshifted after a full circle. In other words, if you look at a galaxy directly or in the opposite direction (around the global circle), either way it would appear receding from you due to the space expansion.

Global energy does not have a single universally accepted definition in General Relativity, so the global energy conservation is a subject of a debate. One view is that energy conservation follows from the Noether theorem as a consequence of the uniformity of time. The flip side of this theorem is that, if time is not uniform, energy does not conserve. For reversible processes this can be viewed simply as "potential energy". For example, time moves slower closer to the Earth. For this reason, energy is released when things fall down. Alternatively this can be viewed as a transfer of the "potential gravitational energy" to the kinetic energy and back, because we can reverse the process and everything zeroes out over a full cycle. Well, the expansion of the universe is not reversible (at least in our lifetime), but if it were, any energy lost during the expansion would be gained back during the contraction.

Answer 2

First, photons' as they travel from far away receding galaxies, they eventually will travel through regions of space where there is no matter that would have gravitational effects to withstand the expansion of the universe. In these regions, between galaxy clusters, the photon is traveling in expanding space, and the photon's wavelength is getting stretched too. It's frequency decreases, and so it;s energy too.

Photons' wavelength gets longer proportional to the expansion of the universe. Their energy decreases by its reciprocal. So the energy just disappears.

Cosmology does not have energy conservation.

According to GR total energy conservation is not valid unless the universe is asymptotically flat.

As per GR, Noether's theorem is broken because time translational symmetry is not valid.

Answer 3

I don't believe the solution proposed by your teacher is correct. First of all, we don't have to prove that energy is conserved, because we're taking into consideration two different frames of reference, so energy shouldn't even be equal, although relatable, as we will show. Second, I think that the proof is conceptually wrong, because the photons have only one frequency when you stick to a frame. Nonetheless, you could think proving it by saying that the photon has a frequency $f$ at the beginning and a frequency $f^{'}$ after some time, so the change in momentum of the photon is $\Delta p = \frac{h}{c} \Delta f$, thus that on the galaxy must be equal and opposite i.e. $\ m \Delta v = - \Delta p$ . Using your result for a small change in kinetic energy we get $\Delta K = v m \Delta v = - v \Delta f \frac{h}{c}$, which doesn't add up to zero with the change in enegy for the photon, which is $\Delta E= h \Delta f$ Third, if you're not convinced of my second remark, this proof may seem to work because we're in the limit of low speeds, as experience tells us energy should be equal after all, but upon replacing the classical doppler shift and the kinetic energy with the relativistic equivalent , we see that the equations differ. Having said that conservation of energy is true if we stick to a single frame, and that in this frame the energy of the photon is alway the same, we can nonetheless show, using transformation of $E$ from one frame to another, that the doppler shift obeys this law.

The relativistic doppler shift is given by:

$f = \sqrt {\frac{1+\beta}{1-\beta}} f^{'}$

Where $\beta= v/c$. This equation can be obtained by considering dilatation of time and lenght contraction simultaneously. From this we get a relation for the energy in the two frames:

$E= \sqrt {\frac{1+\beta}{1-\beta}} E^{'} $

Now, the equation for the transformation of $E$ and $p$ take the same form for those of $x$ and $t$, and we may write:

$E = \gamma (E^{'}+ vp^{'})$

For a photon we have $E=pc$, as it's mass is zero; thus the previous equation becomes:

$E= \frac{1+ \beta}{\sqrt{1- {\beta}^2}} E^{'} $

$ \Rightarrow E= \sqrt {\frac{1+ \beta}{1-\beta}}$

As it should.