What percentage of the mass/energy of the universe is in the form of electromagnetic waves? [duplicate]

Is there any theoretical framework or model that would lead to a prediction, either precise or approximate, about how much or what proportion of the universe's total mass is in the form of photons, or electromagnetic waves? In lieu, is there any way to guesstimate or place a boundary on that proportion? Could that proportion vary over time, universally?

An answer to this question Where is radiation density in the Planck 2013 results? gives a calculated estimate of the photon density $$\rho_\nu = 3.21334\times 10^{-31}\;\text{kg}\,\text{m}^{-3},$$ but neither the question nor the answers explicitly show the total energy density, or express the photon density as a proportion of the total energy density.

In addition, I am interested as much in the theoretical framework that would predict the photon to total energy density as I am in the current measurements that lead to the estimate.

• Related Physics SE post, with quantitative answers: Where is radiation density in the Planck 2013 results? Sep 2, 2019 at 4:04
• Sep 2, 2019 at 7:13
• @BenCrowell is ΩR,0=9.23640×10−5 kg/cubic meter in the answer, the total energy density? and ρν=3.21334×10−31kgm−3 the photon density? That would be a much smaller ratio than the answer here of about 5 x 10-4 proportion of photon energy. Sep 2, 2019 at 22:17
• $\Omega_{R,0}$ is dimensionless; it doesn’t have units of kilograms per cubic meter. Kilograms per cubic meter are units for mass density, not energy density. $\rho_\nu$ is the neutrino mass density. Sep 3, 2019 at 0:19
• The photon ratio 0.005% that I mentioned is $\Omega_\gamma$. Sep 3, 2019 at 0:24

Cosmologists tend to talk in terms of the mass density or energy density of the universe, rather than its total mass or energy (which may well be infinite) or the total mass or energy of the observable part of the universe.

Photons are believed to currently make up only about 0.005% of the energy density of the universe. Their energy density is about $$4\times 10^{-14}\,\text{J/m}^3$$, while the total energy density is about $$8\times 10^{-10}\,\text{J/m}^3$$. So they are a small contribution now.

As you go back in time, the percentage contribution from photons gets larger and larger; the photon energy density varies as the inverse fourth power of the Friedmann scale factor, while the matter energy density varies as the inverse third power, and the dark energy density stays constant. For this reason, the early universe was actually radiation-dominated. The universe then evolved through an era where matter dominated, and now dark energy dominates.

The current theoretical model for understanding cosmology is the Lambda-CDM model, which is a spatially-flat, homogeneous, and isotropic Friedmann universe with three different kinds of contents: dark energy, matter, and radiation. As you can find in the Lambda-CDM Wikipedia article, the percentage of the current energy density due to dark energy is 69.11%, and the percentage due to matter (both normal and dark) is 30.89%. These add up to 100.00% because the tiny radiation contribution from neutrinos and photons is now so small that it is lost in the third decimal place.

The dominant contribution to the photon energy density is from the cosmic microwave background (CMB), not starlight. The temperature $$T$$ of the CMB is about $$2.7\,\text{K}$$ and the energy density of this photon gas is $$4\sigma T^4/c$$ where $$\sigma$$ is the Stefan-Boltzmann constant. This is where the $$4\times 10^{-14}\,\text{J/m}^3$$ comes from, as you can easily calculate.

• "the photon energy density varies as the inverse fourth power of the Friedmann scale factor, while the matter energy density varies as the inverse third power" The third power comes from the simple fact that things spread out as space expands. The additional power for photons comes from the fact that the expansion of space redshifts photons, but doesn't really do anything to each individual piece of mass. Sep 3, 2019 at 9:21
• Shouldn't your answer apply to the entire universe, not just the observable part? Sep 3, 2019 at 19:04
• @D.Halsey It does apply to the entire universe. We assume that the part that we can observe is typical of the whole, and that the entire universe, not just the observable part, is spatially-flat, homogeneous, and isotropic. The energy density of the observable universe is the energy density of the entire universe, and the same for each component, such as the photon energy density. Sep 3, 2019 at 19:26
• Sorry, I didn't read the 1st sentence carefully enough & jumped to the wrong conclusion when I read the word "observable". Sep 3, 2019 at 19:48