<p>One explanation I found helpful went as follows: consider a volume of space which expands with the universe by a factor of <span class="math-container">$a$</span>.  The radiation density in it varies by a factor of <span class="math-container">$a^{-4}$</span> (<span class="math-container">$a^{-3}$</span> because the photons get more spread out, and a further <span class="math-container">$a^{-1}$</span> due to the cosmological redshift (each photon becomes less energetic)). The matter (dark and baryonic both) density changes by <span class="math-container">$a^{-3}$</span> (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</p>
<p><span class="math-container">$$\rho(a) = pa^{-4} + qa^{-3} + r$$</span></p>
<p>to the observed density and find <span class="math-container">$p$</span>, <span class="math-container">$q$</span> and <span class="math-container">$r$</span>. We can also try adding a few terms for other powers of <span class="math-container">$a$</span>, 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.</p>
<p>The division of <span class="math-container">$p$</span> between regular and dark matter can be determined from other measurements.</p>

CC BY-SA 4.0
2020-08-26T10:43:42.330Z
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
CC BY-SA 4.0
2020-08-25T19:05:17.027Z
2020-08-26T13:24:24.363Z
2020-08-26T13:24:24.363Z
Steve Linton
<p>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:</p>

<ol>
<li><p>The CMB (specifically the <em>power spectrum</em> 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 <em>total</em> amount of mass (DM + "normal" matter, i.e. baryons), but it is better at constraining the <em>ratio</em> 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, <a href="https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations" rel="nofollow noreferrer">baryonic acoustic oscillations</a>, we then obtain the amount of DM. This fraction is roughly $\Omega_\mathrm{DM}=0.26$ (<a href="https://arxiv.org/abs/1502.01589" rel="nofollow noreferrer">Planck Collaboration et al. 2016</a>).
If there were no DM, then $\Omega_\mathrm{tot}$ just wouldn't be $1$, but rather $\Omega_\mathrm{tot} - \Omega_\mathrm{DM} \simeq 0.74$.</p></li>
<li><p>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 <em>lower</em>, because the supernovae would be farther away due to the faster expansion. </p></li>
</ol>

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

CC BY-SA 3.0
2017-06-27T08:27:55.677Z
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
CC BY-SA 3.0
2017-06-27T12:08:38.897Z
@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
CC BY-SA 3.0
2017-06-27T15:24:47.447Z
@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
CC BY-SA 3.0
2017-06-27T18:06:46.307Z
@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
CC BY-SA 3.0
2017-06-27T18:15:08.360Z
@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
CC BY-SA 3.0
2017-06-27T22:57:02.130Z
@amsquareb: The _existence_ of DM and DE are completely independent of each other. However, it's not really correct to say that Ωtot would be 0.74 (but you're right, that's what I said). The value of Ωtot would depend on your assumed hypothetical universe. If you just remove the DM, then Ωtot is no longer unity, since you have removed part of it. Such a universe would be "open" and have a different geometry. If you conjecture a "flat" universe, like ours, but _replace_ all DM with DE, expansion would be even faster. If you replace it with baryons, galaxies would look very different.
pela
CC BY-SA 3.0
2017-06-27T22:59:27.017Z
But @zephyr is right, the important point is that if you change the values of the Ω's, the Universe would simply look different.
pela
CC BY-SA 3.0
2017-06-28T07:36:08.210Z
@pela From the wikipedia quote, I mentioned in my problem statement, it seems that they have first concluded that universe is close to flat thence Ωtot must be 1, and then they are trying to account for that amount of mass-energy density. So, according to you that is not the case?
amsquareb
CC BY-SA 3.0
2017-06-28T07:52:51.553Z
@amsquareb: No, that wasn't what I intended to say. The CMB shows us that the Universe is (close to) flat, so Ωtot must be ~1. It also gives constraints on the fraction of energy in mass (DM+baryons), but not very good constraints, actually; we get better constraints on this from [baryonic acoustic oscillations](https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations). This fraction is 0.31. Hence, we can conclude that "something else" must account for the 0.69.
pela
CC BY-SA 3.0
2017-06-28T07:59:16.760Z
It _does_ give some constraints on DE, through the intergrated Sachs-Wolfe effect, as far as I remember. I edited a bit more to clarify where we get the ΩM from.
pela
CC BY-SA 3.0
2017-06-26T20:28:59.493Z
2017-06-28T07:58:45.953Z
2017-06-28T07:58:45.953Z
pela
<p>I have encountered two seemingly contradictory theories on this(I don't know which one of them is correct, or where I am wrong):-</p>

<ol>
<li><p>From Wikipedia:</p>

<blockquote>
  <p>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%."</p>
</blockquote>

<p>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 <strong>if suppose there was no Dark Matter,</strong> <strong>our calculations/assumptions regarding the amount of Dark Energy would have been much higher</strong>(because then Dark Energy would have to account for a greater share of critical density).</p></li>
<li><p>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, <strong>if there would have been no Dark Matter</strong>, then there would be less Dark Energy required to account for the observed acceleration in expansion, so <strong>our calculations/assumptions regarding the amount of Dark Energy would have been lower.</strong></p></li>
</ol>

<dark-matter><dark-energy>
Relationship between Dark Energy and Dark Matter
amsquareb
# Relationship between Dark Energy and Dark Matter

(Asked by: username_0 on 2017-06-26)

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.**

**Tags:** dark-matter, dark-energy

---

## 2 answers

### Answer by username_2 on 2017-06-26. Score: 6 | Accepted answer

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](https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations), we then obtain the amount of DM. This fraction is roughly $\Omega\_\mathrm{DM}=0.26$ ([Planck Collaboration et al. 2016](https://arxiv.org/abs/1502.01589)).
   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.

> **Comments:**
> **username_0**: 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?  
> **username_2**: @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.  
> **username_3**: @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.  
> **username_0**: @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?  
> **username_0**: @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?  
> **username_2**: @amsquareb: The _existence_ of DM and DE are completely independent of each other. However, it's not really correct to say that Ωtot would be 0.74 (but you're right, that's what I said). The value of Ωtot would depend on your assumed hypothetical universe. If you just remove the DM, then Ωtot is no longer unity, since you have removed part of it. Such a universe would be "open" and have a different geometry. If you conjecture a "flat" universe, like ours, but _replace_ all DM with DE, expansion would be even faster. If you replace it with baryons, galaxies would look very different.  
> **username_2**: But @zephyr is right, the important point is that if you change the values of the Ω's, the Universe would simply look different.  
> **username_0**: @pela From the wikipedia quote, I mentioned in my problem statement, it seems that they have first concluded that universe is close to flat thence Ωtot must be 1, and then they are trying to account for that amount of mass-energy density. So, according to you that is not the case?  
> **username_2**: @amsquareb: No, that wasn't what I intended to say. The CMB shows us that the Universe is (close to) flat, so Ωtot must be ~1. It also gives constraints on the fraction of energy in mass (DM+baryons), but not very good constraints, actually; we get better constraints on this from [baryonic acoustic oscillations](https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations). This fraction is 0.31. Hence, we can conclude that "something else" must account for the 0.69.  
> **username_2**: It _does_ give some constraints on DE, through the intergrated Sachs-Wolfe effect, as far as I remember. I edited a bit more to clarify where we get the ΩM from.  

---

### Answer by username_1 on 2020-08-25. Score: 1

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.

> **Comments:**
> **username_2**: 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 $a = 1/(1+z)$.  

---

