If a photon (wave package) redshifts (stretches) travelling in our expanding universe, is its energy reduced?

If so, where does that energy go?

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    $\begingroup$ The same question has been asked several times, in different guises, on this website. Please see: physics.stackexchange.com/q/1327/829 physics.stackexchange.com/q/2597/829 physics.stackexchange.com/q/10309/829 physics.stackexchange.com/q/296/829 The answer basically is: you need to be careful what you mean by "energy conservation" in the context of general relativity. $\endgroup$ Aug 15, 2011 at 15:34
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    $\begingroup$ This should be closed. It's an exact duplicate of the questions Willie Wong has linked to. $\endgroup$
    – user4552
    Aug 16, 2011 at 1:09
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    $\begingroup$ Since Marek's post in 1327 covers at least 95% of this question, second the move to close. Short addendum: there is no local coordinate-independent measure of gravitational energy, nor can there be one, because its existence would violate the equivalence principle. So the answer is just "it isn't!" $\endgroup$
    – Stan Liou
    Aug 16, 2011 at 5:06
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    $\begingroup$ Ted's answer below seems to address the matter of local definitions more squarely that the answers I read on 1327, and I am tempted to leave this open just for that reason. However, it is more the statement of the answer rather than the content which is different so I could still be talked into using my diamond power close. Thought? $\endgroup$ Aug 16, 2011 at 16:58
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    $\begingroup$ In some other stackexchanges, I believe the moderators will close a duplicate question and move a good answer for this question to be an answer to the original question. This gathers all the good answers together, and avoids the confusion of duplicate questions. Is that possible here? $\endgroup$ Aug 17, 2011 at 11:18

3 Answers 3


Since you say you're talking about what happens locally (in a small volume), I'll answer from that point of view. The usual formulation of energy conservation in such a volume is that energy is conserved in an inertial reference frame. In general relativity, there are no truly inertial frames, but in a sufficiently small volume, there are reference frames that are approximately inertial to any desired level of precision. If you restrict your attention to such a frame, there is no cosmological redshift. The photon's energy when it enters one side of the frame is the same as the energy when it exits the other side. So there's no problem with energy conservation.

The (apparent) failure of energy conservation arises only when you consider volumes that are too large to be encompassed by a single inertial reference frame.

To be slightly more precise, in some small volume $V=L^3$ of a generic expanding Universe, imagine constructing the best possible approximation to an inertial reference frame. In that frame, observers near one edge will be moving with respect to observers near the other edge, at a speed given by Hubble's Law (to leading order in $L$). That is, in such a frame, the observed redshift is an ordinary Doppler shift, which causes no problems with energy conservation.

If you want more detail, David Hogg and I wrote about this at considerable (perhaps even excessive!) length in an AJP paper.

  • $\begingroup$ I see, so photon did not actually lost any energy, but it is measured in a different reference frame. $\endgroup$
    – IljaBek
    Aug 16, 2011 at 16:01
  • $\begingroup$ @troyaner: Saying that the photon "did not actually lost any energy" is not really right. It assumes there is some way to talk about conservation of energy nonlocally, and there just isn't. $\endgroup$
    – user4552
    Aug 17, 2011 at 20:31
  • $\begingroup$ If measured in comoving coordinates the photon actually lost its energy. $\endgroup$
    – Anixx
    Aug 19, 2011 at 7:35
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    $\begingroup$ That's true. The quantity energy-as-measured-in-comoving-coordinates is not a conserved quantity. Personally, I'd rather not interpret that sentence as meaning that energy is not conserved, but rather as meaning that energy-as-measured-in-comoving-coordinates is not a concept that rightfully deserves to be considered as "energy." $\endgroup$
    – Ted Bunn
    Aug 19, 2011 at 13:48
  • $\begingroup$ Why do people have to make this all so complicated. Why not just accept the fact that photons in the past were less energetic than their modern day counterparts. I.e that dark energy is converted to normal energy during expansion. $\endgroup$
    – Alan Gee
    Dec 3, 2018 at 7:58

It goes to make work to expand the universe against the forces of gravity and inertia. This is like adiabatically expanding volume of gas: the gas becomes cooler as the volume increases. Where the energy goes?

  • $\begingroup$ wow, so you are suggesting that (0°K) shielding of some volume and reducing the photons inside can affect the expansion of space-time in this volume? I'm not sure. $\endgroup$
    – IljaBek
    Aug 15, 2011 at 14:08
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    $\begingroup$ I cannot prove the opposite, but expansion of space fueled by redshifting of photons seems improbable and has at least two issues - cause-and-effect and universality of physical laws. (is the expansion of space accelerated in photon-rich environment e.g. stars? I take a 0°K black-body, does the space ceases to expand inside it?) $\endgroup$
    – IljaBek
    Aug 15, 2011 at 14:18
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    $\begingroup$ Regarding cause and effect, take a syringe and quickly expand its internal volume. The gas inside will cool. Its energy went to expansion of the syringe but was not its cause. $\endgroup$
    – Anixx
    Aug 15, 2011 at 14:33
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    $\begingroup$ This answer is incorrect. There is no known conserved, scalar measure of mass-energy that applies to cosmological spacetimes. $\endgroup$
    – user4552
    Aug 16, 2011 at 1:08
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    $\begingroup$ This answer is correct. Even if you think there is something wrong with energy conservation the effect that radiation has on the expansion rate of the universe if determined by the Friedmann equations which are not controversial en.wikipedia.org/wiki/Friedmann_equations .Of course the effect is too small to be detectable which is perhaps why some people are confused. In any case energy conservation works perfectly well in this case see vixra.org/abs/1305.0034 $\endgroup$ May 6, 2013 at 19:01

This answer was intended to stay in this question.

Conservation of the energy is (it used to be) a cornerstone in the physics framework. Without that anything can happen.

Let's see how energy can be conserved.

Galaxies are moving dragged by the space expansion. When atoms are in motion the doppler effect will shift the spectra of the emitted photons, as @anna answer showed in the link above.

The proton-to-electron mass ratio, $\frac{m_e}{m_p}$ has been measured constant along the history of the universe, but nothing can be said about the constancy of the electron's mass (to the downvoters: a reference is welcome).

The photon's energy obey the Sommerfeld relation, $E_{jn}=-m_e*f(j,n,\alpha,c)$, as seen here, and it is evident that a redshifted spectrum is obtained with a larger $m_e$.

The spectra lines are not only due to the Hydrogen atom; there are other spectral lines due to molecular interactions, due to electric/magnetic dipoles, etc, and so the electromagnetic interaction,the Coulomb's law, $F_{}=\frac{1}{4\cdot \pi\cdot \varepsilon_0}\cdot \frac{q1\cdot q2}{d^2}$ must be analyzed.

If we scale the mass $m_e$ by the relation $\alpha(t)$ (not related with the above fine structure constant), where $t$ is time (past), we should also scale the charge and the distance by the same factor, giving exactly the same value $F_{}=\frac{1}{4\cdot \pi\cdot \varepsilon_0}\cdot \frac{q_1\cdot q_2\cdot \alpha^ 2(t)}{d^2\cdot \alpha^2(t)}$. Thus the system with and without the transformation behaves in the same manner. The same procedure shows that the universal gravitational law is also insensitive to the scaling of the atom. This should not be a complete surprise because the scaling of masses, charges, time units and distances is routinely used on computer simulations that mimic the universe in a consistent way.

The conclusion is that there is no way to distinguish between the spectrum of an atom in motion and the one of a scaled atom.

The photons that were emitted by a larger atom in the past are received now without any change in its wavelength and, thus, with energy conservation.

The mainstream viewpoint not being aware that scaling the atom gave the same observational results, adopted the receding interpretation long time ago. As a consequence the models derived from that interpretation (BB, Inflation, DE, DM, ) do not obey the general laws of the universe, namely the energy conservation principle.

My viewpoint offers a cause for the space expansion. You can think about that, unless you are comfortable with: 'space expands', period, without a known cause.

Physics is about causes and whys, backed by proper references. I used the most basic laws to show that another viewpoint is inscribed in the laws of nature. I've only used Basic laws that do not need to be peer-reviewed as they are mainstream physics.

When I graduated as electronic engineer, long time ago, I accepted naively that the fields (electrostatic and gravitational) are sourced by the particles, and expand at $c$ speed, without being drained. But now, older but not senile, I assume without exception, that in the universe there are no 'free lunches' and thus the energy must be transferred from the particles (shrinking) to the fields (growing).

This new viewpoint is formalized and compared to the $\Lambda CDM$ model in a rigourous document, with the derivation of the scale relation $\alpha(t)$ that corresponds to the universe's evolution, at:
A self-similar model of the Universe unveils the nature of dark energy
preceded by older documents at arxiv:
Cosmological Principle and Relativity - Part I
A relativistic time variation of matter/space fits both local and cosmic data

Ps: Can someone provide a way to distinguish between the spectrum of an atom in motion and the one of a scaled atom ? (maybe probing the atom's nucleus and find the isotope ratio's abundance (D/H evolution and others) as Mr Webb has done)

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    $\begingroup$ I have a problem with this sentence: " As a consequence the models derived from that interpretation (BB, Inflation, DE, DM, ) do not obey the general laws of the universe, namely the energy conservation principle." The models do obey the laws of the universe. However, in GR it is generally understood that there is no "conservation of energy" law. There is only conservation of the energy-momentum tensor. In a non-expanding universe, this leads to energy conserved as well. But energy-momentum is still conserved in expanding universes; energy itself is not, but that is not an issue $\endgroup$
    – Jim
    Jun 10, 2014 at 20:51
  • $\begingroup$ the problem is "a way to distinguish between the spectrum of an atom in motion and the one of a scaled atom". If you like I'can remove that sentence, because I do not wish to make comparisons with SM (broad questions are outside of the spirit of the site). $\endgroup$ Jun 10, 2014 at 21:33
  • $\begingroup$ " is evident that a redshifted spectrum is obtained with a larger $m_e$. " How would you distinguish this change in the spectrum from e.g. lowering of $c$, which is also present in the formula? in general you need robust justification for assuming change of natural constants, not other way round. $\endgroup$
    – IljaBek
    Aug 27, 2014 at 14:43
  • $\begingroup$ The limits on by how much $\alpha$ can change with time are now very small and come from observations of absorption along the lines of sight to distant quasars. These techniques do indeed distinguish between a doppler shift and a "scaled atom" because fine structure splitting depends on $\alpha^4$, whilst overall energies of transitions depend on $\alpha^2$. $\endgroup$
    – ProfRob
    Nov 24, 2016 at 16:48
  • $\begingroup$ @RobJeffries The constants are indeed constants. I scale the atom and keep constants the same, because the books on physics are word by word, value by value the same for a scaled atom. Someone asserted elsewere with conviction: No change in the atoms were detected. Sure it is so. $\endgroup$ Nov 29, 2016 at 12:54

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