Relationship between Dark Energy and Dark Matter

I have encountered two seemingly contradictory theories on this(I don't know which one of them is correct, or where I am wrong):-

1. From Wikipedia:

Measurements of cosmic microwave background (CMB) anisotropies indicate that the universe is close to flat. For the shape of the universe to be flat, the mass/energy density of the universe must be equal to the critical density. The total amount of matter in the universe (including baryons and dark matter), as measured from the CMB spectrum, accounts for only about 30% of the critical density. This implies the existence of an additional form of energy to account for the remaining 70%."

The above statement, if correct, I suppose indicates that there is a critical density for which mass/energy has to account for. This means that if suppose there was no Dark Matter, our calculations/assumptions regarding the amount of Dark Energy would have been much higher(because then Dark Energy would have to account for a greater share of critical density).

2. The universe is accelerating in expanding. Dark Energy is supposed to cause this acceleration in expansion. While ordinary matter plus Dark Matter, is supposed to reduce this acceleration/expansion through gravity. So, according to this theory, if there would have been no Dark Matter, then there would be less Dark Energy required to account for the observed acceleration in expansion, so our calculations/assumptions regarding the amount of Dark Energy would have been lower.

First of all, if there were no dark matter (DM), you wouldn't ask this question, since structures — including galaxies, stars, planets, and you — wouldn't have had the time to form in the early Universe before it had expanded too much for gravitational collapse to occur. But let's use magic and make the galaxies anyway:

1. The CMB (specifically the power spectrum of the CMB) shows that the total density $\rho_\mathrm{tot}$ of mass/energy in the Universe is extremely close to the critical density $\rho_\mathrm{c}$. That is, $$\frac{\rho_\mathrm{tot}}{\rho_\mathrm{c}} \equiv \Omega_\mathrm{tot} \simeq 1.$$ The "$\Omega$" is a common way to express densities; as a ratio to the critical density. The CMB also gives some constraints on the total amount of mass (DM + "normal" matter, i.e. baryons), but it is better at constraining the ratio of baryons-to-DM. Together with the matter density $\Omega_\mathrm{M}=\Omega_\mathrm{b}+\Omega_\mathrm{DM}$ obtained from observations of supernovae and, in particular, baryonic acoustic oscillations, we then obtain the amount of DM. This fraction is roughly $\Omega_\mathrm{DM}=0.26$ (Planck Collaboration et al. 2016). If there were no DM, then $\Omega_\mathrm{tot}$ just wouldn't be $1$, but rather $\Omega_\mathrm{tot} - \Omega_\mathrm{DM} \simeq 0.74$.

2. Similarly, if there were no DM, there would be less matter to counteract the expansion of the Universe. That means that we wouldn't observe the same relation between the brightness and the distance of supernovae, from which we infer the presence of dark energy (DE). Instead, the brightnesses would be somewhat lower, because the supernovae would be farther away due to the faster expansion.

In other words, you are right that if there were no DM, and if we observed the same thing as we do, then there would be a contradiction. But if there were no DM, we wouldn't see what we see. Therein lies the resolution.

• I am unable to fully understand your explanation. I am not a pro in this subject. Can you please try to explain in more lucid terms. One thing specifically I didn't got, was the mathematical sign with subscripts 'tot' and 'DM'. What does it represent? – amsquareb Jun 27 '17 at 8:27
• @amsquareb: Sorry, I was being a bit too fast. Generally, the $\Omega$'s are used as the ratio between some component of the Universe (e.g. dark matter (DM), baryons (b), dark energy ($\Lambda$), radiation (rad), …) and the critical density. Specifically, $\{\Omega_\mathrm{DM}, \Omega_\mathrm{b}, \Omega_\Lambda, \Omega_\mathrm{rad}\} \simeq \{0.263, 0.045, 0.692, 10^{-5}\}$. I tried to explain better. Let me know if something's still unclear. – pela Jun 27 '17 at 12:08
• @amsquareb I think the real point to stress is the last two sentences. The contradictions pointed out don't really exist because the universe wouldn't be the same if you made the changes you proposed. You artificially caused the contradiction by saying, let's assume some parts of the universe are different but other parts aren't. – zephyr Jun 27 '17 at 15:24
• @pela Thanks a lot, I really appreciate. However, by the below statement:- "If there were no DM, then Ωtot just wouldn't be 1, but rather Ωtot−ΩDM≃0.74." - Do you mean that Ωtot would be .74 if there were no DM? – amsquareb Jun 27 '17 at 18:06
• @pela Also, it seems from your answer that DM and DE are completely independent of each other. Is that really the case? Suppose we lower or increase one entity, will there be no effect on the total amount of the other? – amsquareb Jun 27 '17 at 18:15

One explanation I found helpful went as follows: consider a volume of space which expands with the universe by a factor of $$a$$. The radiation density in it varies by a factor of $$a^{-4}$$ ($$a^{-3}$$ because the photons get more spread out, and a further $$a^{-1}$$ due to the cosmological redshift (each photon becomes less energetic)). The matter (dark and baryonic both) density changes by $$a^{-3}$$ (particles become more spread out). When we plug this into models for the expansion of the universe we find that a third term is needed whose energy density is constant. With enough data we can fit the curve

$$\rho(a) = pa^{-4} + qa^{-3} + r$$

to the observed density and find $$p$$, $$q$$ and $$r$$. We can also try adding a few terms for other powers of $$a$$, but they don't seem to help. There's no simple tradeoff between them, although there may be more uncertainty about some combinations than others.

The division of $$p$$ between regular and dark matter can be determined from other measurements.

• While your math is correct, the fact that you use $z$ for the expansion factor is hurtful to the eyes! The term $z$ is used for the redshift, while the expansion factor is called $a = 1/(1+z)$. – pela Aug 26 at 10:43