I have seen data showing that the estimated mix of dark energy in the universe is 68.3%, the mix of dark matter is 26.8% and the mix of ordinary matter is 4.9%1. However, within "ordinary matter," what is the mix of stable atomic matter (solids, liquids, gases, plasma) within stars and planets vs. radiated EM energy (light, gamma rays, neutrinos, etc.)?

1 See Wikipedia here and here.


Photons (radio waves, "light", gamma-rays, etc.) and neutrinos contribute negligibly to the total energy density of the Universe. By far, most of the photons that exists today are the cosmic microwave background, with 450 photons per cm$^{3}$ (e.g. Hobson et al. 2006). The number density of neutrinos is similar, 330 per cm$^{3}$. In total, the energy density parameter (the fraction of energy in a given constituent compared to the total energy in the Universe) for photons is $\Omega_\mathrm{\gamma,0} = 3.777\times10^{-5}$, and that of neutrinos is $\Omega_\mathrm{\nu,0} = 5.460\times10^{-5}$ (where the subscripts "0" refer to today's value).

You can compare these numbers to the ones you give for dark energy ($\Lambda$), dark matter (DM), and ordinary matter ("baryons", b): $\{\Omega_{\Lambda,0},\Omega_\mathrm{DM,0},\Omega_\mathrm{b,0}\} = \{0.683, 0.268, 0.049\}$.

Evolution with time

As the Universe expands, the density of ordinary ("baryonic") and dark matter falls off as $a^{-3}$ (where $a$ is the scale factor), since the volume is proportional to $a^3$. The number density of photons scales the same way, but since additionally photons are redshifted proportionally to $1/a$, its energy density goes as $a^{-4}$. For neutrinos, the dependency is a bit more complex, since initially they relativistic (scaling as $a^{-4}$) but at some point became non-relativistic (scaling as $a^{-3}$). That means that in earlier times, radiation contributed more to the total energy density, and it is rather easy to calculate that until the Universe was $\sim50,000$ years old, it was dominated by radiation.

On the other hand, if you go forward in time (which I suppose you do), radiation will become even more negligible. So will baryonic and dark matter, but since dark energy is a property of the vacuum itself, its density does not decrease, and eventually, it will completely dominate the Universe.

DIY guide

The figure below shows the various components as a function of time (and redshift). Subsequently I describe how it was made.


In general, if you want to calculate the distribution of energies at a given time in the history of the Universe, you use the relationships given above. The scale factor $a$ is defined to be one today, so that when the Universe was, say, half the (linear) size of today, $a$ was equal to $0.5$ (when it was half the volume of today, $a$ was $0.5^3 = 0.125$). Thus, the density $\rho_i$ of the $i$'th constituent (where $i = \{\Lambda,\mathrm{DM},\mathrm{b},\gamma,\nu\}$) as a function of $a$ is $$ \rho_i(a) = a^{x_i} \rho_{i,0} = a^{x_i} \rho_\mathrm{tot,0} \,\Omega_{i,0}, $$ where $x_i = \{0,-3,-3,-4,-4\}$ are the exponents of the corresponding constituents (I think it's a fair approximation to use $x_\nu = -4$ at all times, since after all the density of neutrinos is negligibly at late times when $x_\nu=-3$), and $$ \rho_\mathrm{tot,0} = \frac{3H_0^2}{8\pi G} $$ is the total density of all constituents, which depends on the Hubble constant $H_0 = 67.8\,\mathrm{km}\,\mathrm{s}^{-1}$ (and on G).

The density parameters (the $\Omega$'s) are then $$ \Omega_i(a) \equiv \frac{\rho_i(a)}{\rho_\mathrm{tot}(a)} = a^{x_i} \frac{\rho_\mathrm{tot,0}}{\rho_\mathrm{tot}(a)} \Omega_{i,0} = a^{x_i} \frac{3H_0^2/8\pi G}{3H^2(a)/8\pi G} \Omega_{i,0}\\ = a^{x_i} \frac{H_0^2}{H^2(a)} \Omega_{i,0}. $$

The (squared) ratio between the Hubble constant now ($a=1$) and then ($a<1$) happens to be easy to calculate (from the Friedmann equation, note that this is it's inverse): $$ \frac{H^2(a)}{H_0^2} = \Omega_\Lambda + \Omega_\mathrm{DM} a^{-3} + \Omega_\mathrm{b} a^{-3} + \Omega_\gamma a^{-4} + \Omega_\nu a^{-4}. $$ (Here, I have omitted a term $\Omega_k$ corresponding to the global curvature of the Universe. Observationally, this term is found to be very close, and possibly exactly equal, to zero, but if included, it scales as $a^{-2}$).

Now if you want the distribution not at a given size of the Universe, but at a given time $t$, the only thing missing is the relationship between $a$ and $t$. If turns out that this relation only depends on all the $\Omega$'s and Hubble constant. Using instead of $a$ the redshift $z = 1/a - 1$, the result is $$ t = \int_z^\infty \frac{H(z')}{(1+z')}\,dz', $$ where $H(z)$ is given by the equation above.

However, it also turns out that in general it cannot be calculated analytically, but must be calculated numerically. This can be done using an online cosmological calculator such as Ned Wright's, or in Python using the package CosmoloPy, which is what I used to create the plot above.

  • $\begingroup$ thank you. I wonder if this mix has been plotted over time, starting just after the Big Bang? Based on what you've said, at the time, radiation must have dominated? $\endgroup$ – Brad Cooper - Purpose Nation Jul 25 '15 at 20:31
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    $\begingroup$ @PurposeNation: There are plenty of pie charts that show the energy distribution today and at the time of the CMB, e.g. this one. I only found one with the times in between, but that doesn't contain radiation. But you can calculate it yourself, using $\rho_i(t) = \rho_{i,0}a^{x_i}$, where $x_i$ is the exponent of the $i$'th component, i.e. $x_i = \{0,3,3,4\}$ for $\{\Lambda,\mathrm{DM},\mathrm{M},\mathrm{rad}\}$. $\endgroup$ – pela Jul 26 '15 at 9:23
  • $\begingroup$ Oops, the "M" above should be "bar". Usually, M is used for the combined mass of DM + baryons. $\endgroup$ – pela Jul 26 '15 at 9:36
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    $\begingroup$ Okay, I edited my answer to include a better description of how to do it. $\endgroup$ – pela Jul 26 '15 at 10:50
  • $\begingroup$ @PurposeNation: …and now I felt like making the plot you seem to request, so I edited to include the parameters as a function of redshift and time. $\endgroup$ – pela Jul 27 '15 at 22:53

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