How much photon energy has already been destroyed?

Question

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?

Answer 1

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.

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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 $10^{93}\,\mathrm{J}$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

EDIT

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!

EDIT@safesphere

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!