We get taught in school that energy can neither be created, nor destroyed. The law of energy conservation is confirmed by many processes. It is an incredibly accurate assumption for every-day life.

When we study physics at university level, we get taught the "fine-print", the exceptions to the rule. For example in cosmology we learn that the universe as a whole is expanding. The expansion affects the wavelength of photons: the wavelength gets stretched (while c remains constant), which means photons lose energy. To nothing. The energy is just gone. Experimental evidence is the cosmic microwave background (CMB), which is red-shifted to a black-body spectrum of 2.7K.

If we estimate the number of CMB photons originally produced and their energy (1eV?), subtract photon energy as observed today, then how much photon energy has been lost during the universe's expansion (in Joules)? Is it a significant amount, say, more than 1% of total hadron mass?

  • 2
    $\begingroup$ I would love to give a detailed answer to this part: "photons lose energy. To nothing." - it is not true. Their wavelength is not stretched by the expansion of the universe, simply we are moving with a very high velocity relative to the emitter of that photon. When we interact with the photon, then unsurprisingly, we find it can give off less energy than if we didnt move relative to the emitter, and everything is in full accord with special relativity (which conserves energy). $\endgroup$
    – Kostas
    Oct 17, 2019 at 21:02
  • $\begingroup$ @Kostas But in GR, energy is not conserved, I've heard. Why would you want to focus solely on SR instead of say, GR? $\endgroup$ Oct 18, 2019 at 13:43
  • 1
    $\begingroup$ "that energy can neither be created, nor destroyed" this is true for energy which includes kinetic energy+potential energy in all its forms. When including special relativity where energy is a part of a four vector ,, defined here hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$
    – anna v
    Oct 18, 2019 at 13:55

4 Answers 4


To answer this question, a few assumptions must be made:

The Universe may or may not be infinite. It would therefore make sense to answer your question considering the observable Universe only. But the observable Universe increases in size, not only due to expansion (which doesn't add or remove photons), but also because light from more and more distant regions reaches us. In comoving coordinates — i.e. the coordinates that expand with space — the observable Universe has increased in linear size by a factor of 50 since then, so the comoving volume, and hence the total number of photons, has increased by a factor $\gt10^{5}$.

In other words, new photons enter our observable Universe all the time, and they do so at a faster rate than the individual photons lose energy.

Moreover, most of the photons were not created when the CMB was emitted — they had ben around since the end of inflation, scattering aorund on free electrons, until they were "released" at the decoupling/recombination era. I'll address this in the end of my answer.

I know this is not really what you had in mind, so to be specific, I'll compare the total amount of energy in the observable Universe today, to the same space when the CMB was emitted.

Each $\mathrm{cm}^3$ of space holds roughly $n_\mathrm{ph}$ = 411 CMB photons. There are also photons coming from various astrophysical processes (mostly star formation and dust emission), but those are smaller in number by more than two orders of magnitude, and smaller in energy by at least one order of magnitude, probably more (Hill et al 2018), so let's ignore those.

With a CMB temperature of $T_0 = 2.718\mathrm{K}$ (Planck Collaboration et al. 2016), the average energy of a CMB photon is $E_\mathrm{ph,now} = k_\mathrm{B}T_0 = 2.3\times10^{-4}\,\mathrm{eV}$. Since they're seen redshifted by $z\simeq1100$, each photon has lost energy by the same factor. With radius $R = 46.3\,\mathrm{Glyr}$ of the Universe, the total amount of energy lost is $$ \begin{array}{rcl} \Delta E & = & E_\mathrm{tot,then} - E_\mathrm{tot,now} \\ & = & (E_\mathrm{ph,then} - E_\mathrm{ph,now}) \,\times\, N_\mathrm{ph,tot}\\ & = & \left((1+z)E_\mathrm{ph,now} - E_\mathrm{ph,now}\right) \,\times\, n_\mathrm{ph} \,\times\, \frac{4\pi}{3}R^3\\ & \simeq & (1+z)E_\mathrm{ph,now} - E_\mathrm{ph,now} \,\times\, n_\mathrm{ph} \,\times\, \frac{4\pi}{3}R^3\\ & \simeq & 4\times10^{88}\,\mathrm{eV}\\ & \simeq & 6\times10^{76}\,\mathrm{erg}\\ & \simeq & 6\times10^{69}\,\mathrm{J}, \end{array} $$ where the first approximation acknowledges the fact that the energy density today can be neglected compared to the energy density when the photons were emitted.

If you want to know how much energy these photons have lost since they were initially created, at the end of inflation, you just use the redshift corresponding to this epoch — roughly $z\sim10^{26}$. In that case you get some $10^{93}\,\mathrm{J}$.

  • $\begingroup$ That's very strange, I had read that the observable universe is actually shrinking due to the accelerated expansion of universe and that in a far future we shouldn't be able to see more than just a couple of stars (if the Earth would remain still, that is). Wikipedia (giving a source), claims that in about 100 to 150 billion years, "The Universe's expansion causes all galaxies beyond the former Milky Way's Local Group to disappear beyond the cosmic light horizon, removing them from the observable universe". So the universe might become much bigger, but we will be able to see less and less. $\endgroup$ Oct 18, 2019 at 13:41
  • 1
    $\begingroup$ @thermomagneticcondensedboson The Wiki article is mixing up two different horizons. It quotes Loeb 2011, but I can't (from a quick browse) find exactly where. Galaxies leave our event horizon, which is the farthest point (currently ~16.5 Glyr) from which a signal may be emitted and still reach us before infinity. But the observable Universe is bound by the particle horizon (currently 46.3 Glyr) which always increases in size, both because the Universe expands, and because light from more and more distant region will reach us. [cont'd below] $\endgroup$
    – pela
    Oct 18, 2019 at 14:15
  • 1
    $\begingroup$ Because of my stupidity pela. I want to remove my downvote, but the system is even stupidier and I cannot, unless you make an edit to your answer. If you could add a comma somewhere or something like that, I'd upvote you. $\endgroup$ Oct 18, 2019 at 16:41
  • 1
    $\begingroup$ @thermomagneticcondensedboson I edited to use average energy rather than peak energy, which is more correct (but doesn't change the result much). $\endgroup$
    – pela
    Oct 18, 2019 at 20:26
  • 1
    $\begingroup$ :) From a -1 to +1 from me :) Thank you pela. $\endgroup$ Oct 18, 2019 at 21:19

add Would you ask: "How much electron energy has already been destroyed?" Energy is not something that can be destroyed, it is apportioned around according to the kinematics and the interactions and the inertial frames.

The "length" of the energy-momentum four vector is given by

$$\sqrt{P\cdot P} = \sqrt{E^2 - (pc)^2} = m_0 c^2$$

The photon by having mass zero has energy $E=pc$. Since also $E=hν$ it is obvious that the changes in the momentum of the inertial frame the photon first appeared, will have to change $ν$ as $h$ is Planck's constant.This is consistent with the Doppler shift expected by the mathematics for light waves, as seen here:

$$ \nu_\mathrm{observed} = \left[\frac{\sqrt{1-\frac{v^2}{c^2}}}{1-\frac vc}\right] \nu_\mathrm{source} $$ which can be rearranged to the form $$ \nu_\mathrm{observed} = \nu_\mathrm{source}\sqrt{\frac{1+\frac vc}{1-\frac vc}} $$ or in common relativity notation: $$ \nu_\mathrm{observed} = \nu_\mathrm{source}\sqrt{\frac{1+\beta}{1-\beta}} $$

Here v is the relative velocity of source and observer and v is considered positive when the source is approaching.

and $ν$ the frequency of light.

It is the change in the known spectra of specific atoms that shows how the motion of the various inertial frames changes the frequency.

You say:

The expansion affects the wavelength of photons: the wavelength gets stretched (while c remains constant), which means photons lose energy

Lets be clear, the photon does not have a wavelength in spacetime. Only a quantum mechanical probability wavefunction. It is the light waves which emerge from the confluence of zillions of photons that has a wavelength in space and time, that changes.

Lets go to a simple example

If you roll a ball up hill (no friction,on ice), it loses energy, no? Where does its energy go? Actually in moving the earth backwards , but the earth is so large it would never be measured.

It is similar with photons loosing energy, except it shows in the frequency of the emergent wave , not in the photon velocity which is always c. This is seen with the spectra according to the motion of the star they come from, if is is moving away from us, the spectra go to infrared, and corrected mistake to violate when moving towards us. The energy balances go with the motion of the sources. In cosmic photons it is the expansion of space that can explain the microwave background, and that is what is accepted as valid in the current model of the cosmos, the Big Bang theory.

Photons build up a light wave whose frequency spectrum is connected to the energy of the photon . When the particle photon loses energy the light frequency goes to the infrared. Two different frames.

  • $\begingroup$ can you please elaborate on " the photon does not have a wavelength in spacetime." What do you mean? $\endgroup$ Oct 18, 2019 at 20:51
  • 2
    $\begingroup$ When you roll a ball uphill it gains potential energy, right? You give it kinetic energy, and that energy gets converted. It doesn't lose energy. It gains energy. $\endgroup$
    – J Thomas
    Oct 21, 2019 at 8:59
  • $\begingroup$ @JThomas Yes, you are correct, I put it the wrong way. It is expansion after all, away from every point. Actually I woke up thinking to correct it, and then I forgot ( age is catching up with me), thanks $\endgroup$
    – anna v
    Oct 21, 2019 at 9:04
  • $\begingroup$ @ÁrpádSzendrei What I am saying, that for quantum mechanical entities the wavelength is seen in the probability distribution, which can only be measured by detecting many photons of the same energy. The intividual photon is a point particle in (x,y,z,t) according to the standard model. $\endgroup$
    – anna v
    Oct 21, 2019 at 9:13

Imagine a classical light wave. It has a beginning and an end. The beginning and end are both moving at lightspeed.

Imagine that the space is stretched.

The whole wave has just as much energy as it did before, but now it is longer. It is stretched out over a longer distance. At any one spot there is less energy, but the whole wave hasn't lost any.

  • $\begingroup$ But the classical picture breaks down when the photon is absorbed, which is a point-like event in space-time. The energy measured is empirically less and was lost. $\endgroup$
    – Jens
    Oct 22, 2019 at 8:31
  • $\begingroup$ It's two incompatible ways to look at it. If quanta are absorbed by an atom whenever the wave condition at that atom provides the "key" to open that "lock", then the wave can go right on, minus whatever part of a quantum was absorbed by that atom. If you demand that the quantum that one atom absorbs is the same quantum that another atom emitted then it doesn't work. But why assume that? $\endgroup$
    – J Thomas
    Oct 22, 2019 at 11:28
  • $\begingroup$ If two descriptions make different statements, then one is incorrect. The world is quantum by nature, not classical. $\endgroup$
    – Jens
    Oct 22, 2019 at 20:24
  • $\begingroup$ If two descriptions make different predictions, then one of them is incorrect. Interactions of radiation with atoms is quantum -- much radiation is absorbed and immediately re-radiated, and quantum amounts are absorbed or radiated later. I can't speak for classical physics, but are you certain that quantum descriptions are the only possible way to describe the motion of radiation through space, and the absorption and emission of radiation by free electrons? $\endgroup$
    – J Thomas
    Oct 22, 2019 at 21:54

The expansion affects the wavelength of photons: the wavelength gets stretched (while c remains constant), which means photons lose energy.

No, they are NOT. Magnitude of redshift associated with universe expansion is defined as:

$$ z = a_{\space t} / a_{\space t_0} - 1 $$ where $a$ is cosmic scale factor. So to get cosmic expansion redshift, we need to compare universe scale factor from different epochs! This means comparing different photons arriving from galaxies at a different distance from us!

Even if you had in mind not cosmological redshift, but ordinary Doppler Effect - still photon emitted from source with energy $h\nu$ arrives with same energy to observer. Energy difference happens only when we compare photons reaching us from moving objects at different radial speeds in relation to us. So still this is a photons comparison from different sources or from different epochs. Single photon it is as it is - without any changes.


After a long discussion with @pela, I stay by my view. If it is different photons, it's not an energy loss, technically. Just one photon has energy $h(\nu_0 + \Delta \nu_{_1})$ and another has $h(\nu_0 + \Delta \nu_{_2})$, because of different setup (distance covered of expanding space) between those two!


Energies of emission and absorption are measured in different frames "resulting" in the "energy loss". If you measure both in the same frame, the energy is conserved. In the frame of the receiver, photons are emitted already redshifted. In the frame of the emitter, photons are never redshifted.

Good point!

  • 4
    $\begingroup$ Photons do lose energy when cosmologically redshifted. $\endgroup$
    – pela
    Oct 18, 2019 at 12:00
  • 3
    $\begingroup$ Single photons do experience cosmological redshift. They were emitted with some energy, and are later observed with a smaller energy. So energy has been lost. This is not controversial, it's just a consequence of the lack of time symmetry in the expansion of the Universe. $\endgroup$
    – pela
    Oct 18, 2019 at 12:12
  • 3
    $\begingroup$ Because we're pretty sure physical processes in a distant galaxy are the same as here on Earth. A hydrogen atom decaying from its first excited state emits a photon of λ=1216 Å in a lab, so we assume a hydrogen atom in a distant galaxy emits a photon with the same wavelength. But if we observe it at λ=2432 Å, we know that it has been redshifted by a factor of 2 (i.e. z=1). The scale factor-comparing comes from the propagation of a light ray traveling along a null geodesic in the FLRW metric describing the expansion (e.g. Watson 2000). $\endgroup$
    – pela
    Oct 18, 2019 at 12:37
  • 2
    $\begingroup$ No, cosmological redshift is different. It happens along the way. An observer halfway between the emitter and us would see a smaller redshift than we do (not exactly half the redshift, since this depends on the expansion history of the Universe). $\endgroup$
    – pela
    Oct 18, 2019 at 12:54
  • 2
    $\begingroup$ Sure, of course not the exact same photon, but if two photons are emitted by the same process, and one is observed by us, and the other is observed by a midway observer… $\endgroup$
    – pela
    Oct 18, 2019 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.