When we say dark energy makes up ~68% of the universe and the rest is dark/ordinary matter (plus light etc.), what quantities are actually being compared?

Is it just comparison of total dark energy to mass through E=mc2? Or something more sophisticated?


3 Answers 3


So, first let's focus on what quantities are being evolved in this comparison: (According to Lambda-CDM Cosmological model) Dark Energy - 68% Dark Matter - 27% Ordinary/Baryonic Matter - 5% While Ordinary matter that is visible is made up of Baryons and Anti-Baryons, Dark matter is composed of some underlying particles most probably outside of the standard model such as WIMPS (Weakly Interacting Massive Particles), and GIMPS (Gravitationally Interacting Massive Particles).

But Dark energy is different, it isn't matter but it is a form of energy which forms the basis for the cosmological constant, it might be the zero point energy from the vacuum i.e vacuum energy. So as to answer the question: While being referred to composition, it refers to the mapping of the mass-energy content of the universe, which means that while comparing matter/anti-matter to forms of energy like dark energy, the mass-energy equivalence($E=mc^2$) principal is used!


The dynamics of the universe and a careful analysis of the cosmic microwave background suggest that the energy density of the universe is very close to the critical density of a flat universe.

When we express the density of the various cosmological components (dark energy, dark matter, baryonic matter, radiation), it is convenient to express these as a fraction of the critical density. Where we are dealing with slowly moving masses, then these are converted to an energy density simply by multiplying by $c^2$.

Thus when you see "68% of the universe is dark energy", it almost certainly means that the dark energy density is 68% of the critical (energy) density. But since the universe appears to have an energy density that is very close to the critical density, the two are interchangeable.


I will give a basic explanation of one of the ways in which the density ratios of the components that compose the universe have been calculated: ESA's Planck satellite measured the temperature fluctuations of the cosmic microwave background (CMB) between 2009 and 2013, and from that produced the power spectrum plot.

power spectrum plot Look at the top of the plot, the blue dots are the measurements and the red line is the fit to the concordant ΛCDM model. The figures below are those published by the Planck Collaboration-2015.

1. The abscissa position of the peak I have marked as "1" indicates the curvature of the universe $\Longleftrightarrow$ total density. If the peak (and in general the whole curve) were further to the right the universe would be spherical and if it were further to the left it would be hyperbolic. In the position where it is, it tells us that the Universe is Euclidean (its density is the critical density) with 3 significant figures:

$$\Omega_{K_0} = 0.000 \pm 0.005 \ \Longleftrightarrow \ \Omega_0 = 1.000 \pm 0.005$$

2. The ratio of amplitudes, (in ordinates) between peak 1 and peak 3 gives the ratio of dark energy Λ to total matter M, (dark+baryonic) which together with the condition (1) that both ratios have to sum to 1 for the curvature to be zero, gives us:

$$\Omega_{\Lambda_0} = 0.6911 \pm 0.0062$$

$$\Omega_{M_0} = 0.3089 \pm 0.0062$$


$\Omega_{M_0}=\dfrac{\rho_{M_0}}{\rho_{c_0}}\simeq 31\%$

Where $\rho_{c_0}$ is the critical density and $\rho_{M_0}$ is de matter density both in the same units

Similary $\Omega_{\Lambda_0}=\dfrac{\rho_{\Lambda_0}}{\rho_{c_0}} \simeq 69\%$

The units for mass density are $kg/m^3$. But we can multiply all the mass densities by $c^2$

$\Omega_{M_0}=\dfrac{\rho_{M_0} c^2}{\rho_{c_0} c^2}$

$\Omega_{\Lambda_0}=\dfrac{\rho_{\Lambda_0} c^2}{\rho_{c_0} c^2}$

Now the numerators and denominators are of the type $\rho c^2$ and therefore are energy densities $J/m^3$.

This answers the question of the thread: the Omegas density ratios are quotients between mass densities or energy densities, since the energy densities are obtained by multiplying the mass densities by the square of the speed of light.

3. The amplitude (ordinate) of peak 2 with respect to peaks 1 and 3 gives the baryonic matter ratio

$$\Omega_{b_0} = 0.049 \pm 0.001$$

That leaves the dark matter density ratio $\Omega_{dm_0} = 0.2599$

4. Finally, the damping levels of peaks 4, 5, 6, ... are the ones that have allowed us to reaffirm the model and to fit the values we have given for the Omegas somewhat more precisely. The physical reasons why the curvature and energy ratios of the Universe can be deduced from the power spectrum can be found, for example, at waynehu. Power Spectrum

And following this link you can play with a "CMB Simulator" Planck CMB Simulator

Click on "Toggle power spectrum plot".

Click on "Toggle options", click on "Normalise scale" and "x" to exit.

With the bars on the left you can simulate different energy compositions of the Universe.

This programme is from 2013 (before the Planck-2015 results, which are from 2015), so there are small differences in the ratios compared to the ones I have written before. For this 2013 programme, the values of the ratios that have to be entered in the programme to obtain what it calls "100% similarity to our universe" are:

$$\Omega_{b} = 0.05$$ $$\Omega_{DM} = 0.275$$ $$\Omega_{\Lambda} = 0.675$$

Playing with the values of the density ratios, you can see how the peaks are shifted on the graph.

Best regards.


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