The cosmological constant $\Lambda$ is roughly the inverse square of the size $R$ of the observable universe, or horizon radius (with proper constants to get the right units). $\Lambda$ has a solid basis in observations. Observations also suggest that $\Lambda$ is the same for distant galaxies. Various research proposals even explain the value of $\Lambda$ with the radius of the observable universe. (If my reading of the literature is correct, no explanation of the value of $\Lambda$ unrelated to $R$ has ever been proposed, but I might be wrong.)

But if $\Lambda$ is related to the size of the observable universe $R$, its value should have been higher in the past, when the observable universe was smaller. This would imply that $\Lambda$ is not a constant, and that its value should be higher for distant galaxies. This last conclusion contradicts observations.

How can one reconcile these two contrasting arguments?

Edit: Note that I might not have distinguished $\Lambda$ and $\Omega_\Lambda$ clearly in this question. My question is about energy density: Measurements show that the energy density due to $\Lambda$ is constant over time. But then, why is $\Lambda \approx 1/R^2$ ($R$ being the present radius)? This coincidence that two numbers have the same value, one constant, the other time dependent, is really astonishing.

(Even Milgrom's https://arxiv.org/abs/2001.09729 does not attempt an answer.)

Edit: Many recent papers speculate that dark energy density (assuming this is Lambda) is related to the temperature and entropy of the horizon (like the Hawing radiation for a black hole). If one adds numbers, this come out about right. However, this would also mean that the temperature and dark energy density must have been higher in the past. And again, this last conclusion is not backed up by data.

Edit: As Koshi writes, the question is one of the great unsolved problems of physics. So there is no reason to expect a simple answer.

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    $\begingroup$ "The cosmological constant $\Lambda$ is roughly the inverse square of the size $R$ of the universe." Uh... who told you that? $\endgroup$
    – knzhou
    Aug 30, 2021 at 6:35
  • $\begingroup$ Measurements. The value of $\Lambda$ is 1/(10^26 m)^2. $\endgroup$
    – user85598
    Aug 30, 2021 at 6:45
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    $\begingroup$ @Christian all that says is the value at a single data point, that doesn't prove a functional relationship that isn't constant. Where specifically did someone mention $1/R^2$, ie a paragraph from a publication, article, textbook..? $\endgroup$
    – Triatticus
    Aug 30, 2021 at 8:45
  • $\begingroup$ @Triatticus Milgrom is already cited. Milgrom has thought a lot about the issue, a he writes about it in most of his papers. In the question, I added a summary about what is happening in research by others. $\endgroup$
    – user85598
    Aug 30, 2021 at 11:52
  • $\begingroup$ Related question on cosmological coincidences: physics.stackexchange.com/questions/412998/… $\endgroup$
    – MadMax
    Sep 7, 2021 at 15:23

3 Answers 3


If I understand correctly, I think what you are referring is the following, and I will go trough all of the steps to get there: The cosmological constant pops up as a possible constant in the Einstein equations, as $\Lambda$, and this why it pops up in the Friedman equations: $$ H^2+\frac{k}{a^2}=\frac{8\pi G}{3}(\rho_m+\rho_r)+\Lambda.$$

To get everything to the same footing, let us introduce the constant (!) energy density in 'dark energy' by setting $\rho_{\Lambda}=3\Lambda/(8\pi G)$. We can then write the first equation as: $$ H^2+\frac{k}{a^2}=\frac{8\pi G}{3}(\rho_m+\rho_r+\rho_{\Lambda}).$$ We now introduce density parameters $\Omega_{i,0}$, where $i$ can relate to matter, radiation, the cosmological constant (i.e. dark energy) and curvature via $$\Omega_{i,0}\equiv\frac{\rho_{i}(t_0)}{\rho_{\mathrm{crit}}(t_0)},\quad i\in\{m,r,\Lambda\},\quad\Omega_{k,0}\equiv\frac{k}{a(t_0)^2H^2(t_0)},$$ where $\rho_{\mathrm{crit},0}\equiv 3/(8\pi G) H_0^2$ is the critical density of the universe that would lead to a flat universe TODAY and $H_0=H(t_0)$. Inserting the known scaling of the different energy densities with the scale factor for matter, radiation and dark energy, and the obvious scaling of $\Omega_k$ we can read of from the definition, and set $a_0=1$, we find the Friedman equation to be $$H^2(t)=H_0^2\left(\Omega_{r,0}a^{-4}+\Omega_{m,0}a^{-3}+\Omega_{k,0}a^{-2}+\Omega_{\Lambda,0}\right).$$ This equation tells us that to find $H$ at a certain time, we have to insert the density parameters we find today and scale them with their behavior with $a$. We do not have to scale $\Omega_{\Lambda}$, since $\rho_{\Lambda}$ is constant, but that does not mean that $\Omega_{\Lambda,t}$ is constant.

I think this what you are referring to is the following: You can find $\Lambda$ by measuring $\Omega_{\Lambda,0}$ and $H_0$ (where both measurements are far from trivial) and use our definition for the density parameter to get $\Lambda$ from that by $$\Lambda=\Omega_{\Lambda,0}H_0^2$$ This is where the $H_0^2=1/R_0^2$ dependence pops up that you mentioned. For some this raises the question, since $\Omega_{\Lambda}$ is roughly 1, why the value for dark energy is so related to today's horizon size i.e. to the miniscule value of the 'Hubble constant'. BUT if one looks back at the definition of $\Omega_{\Lambda,0}$ (and of $\rho_{\Lambda}$), one sees that the factors of $H_0^2$ cancel and there is no time dependence in $\Lambda$. (The interesting question 'why is $\Lambda\approx H_0^2$?' should therefore maybe be rephrased to 'How come we live in the short time where $\Omega_{\Lambda}$ and $\Omega_m$ are roughly equal??' or maybe just 'Why is the cosmological constant so small?') If you would evaluate the equation for $\Lambda$ at a different time, then also $\Omega_{\Lambda,t}$ would be different, since the critical density that was used to define this parameter is time dependent, with $H^2$! There can sometimes be confusion about the density parameters, since some people only use today's values and skip the 0 index for that matter.

  • $\begingroup$ Maybe your text has the answer to my question. I always thought that a constant Lambda also means a constant Omega_Lambda. Are you saying that Omega_Lambda was higher in the past, and that it changes as 1/R^2? More precisely: how does energy density change over time? $\endgroup$
    – user85598
    Aug 30, 2021 at 10:02
  • $\begingroup$ @Christian By definition the energy density corresponding to a cosmological constant does not change over time. (Note that this does not necessarily need to be true for any kind of 'dark energy', but this is another story). $\Omega_\Lambda$ is defined by dividing this constant by a time dependent term, containing the Hubble factor, so it is time dependent because of that. Today, $\Omega_\Lambda$ is of order one and seems to dominate the universe. In the early universe, its contribution to the total density was minuscule. $\endgroup$
    – Koschi
    Aug 30, 2021 at 11:22
  • $\begingroup$ Ok thank you. So the original question is still open. $\endgroup$
    – user85598
    Aug 30, 2021 at 11:41
  • $\begingroup$ @Christian I don't understand, why. :-) Maybe then you have to rephrase the initial question. I thought that you wondered how $\Lambda$ can be constant, if it is also $\sim H^2$. I thought I explained that it is $\sim H_0^2$ and constant as asummed. It does not change since we have actually $\Lambda=\Omega_{\Lambda}H^2$ and the seemingly $\sim H^2$ relation is actually at all times canceled by the $H^{-2}$ dependence of $\Omega_{\Lambda}$. $\endgroup$
    – Koschi
    Aug 30, 2021 at 12:59
  • $\begingroup$ i.e. I hoped it became clear from my answer, why $\Lambda$ and also the corresponding energy density $3\Lambda/(8\pi G)$ is constant and does not actually depend on $H$, althought the equation $\Lambda=\Omega_{\Lambda,0}H_0^2$ might imply this. $\endgroup$
    – Koschi
    Aug 30, 2021 at 13:07

Are you talking about the Hubble sphere/horizon rather than the observable universe (which are not the same)? If so, Hubble horizon is defined as $R=1/(aH)$ (Planck units), and for a cosmological constant driving the expansion (present era) we have $H\sim \sqrt{\Lambda}$, so yes, $\Lambda\sim 1/R^2$ (at the present time fixing $a=1$). But this relation doesn't hold if other sources contribute to the expansion, like in the earlier universe where radiation and matter were the main contributors, or even earlier when inflation took place.

Edit: Hubble horizon in general is defined as $1/(aH)$ with the scale factor $a$ and the Hubble parameter both time-dependent of course. So if $H$ is constant, Hubble radius is still time dependent due to $a(t)$. The inverse $H$ is called Hubble time.

  • $\begingroup$ What you say is true; but let us go back just to z=0.3 or some similarly small amount. Was Lambda smaller then? $\endgroup$
    – user85598
    Aug 30, 2021 at 10:34
  • $\begingroup$ @Christian No, $\Lambda$ (assuming it is a cosmological constant) is, well, constant. But the relation $H\sim \sqrt{\Lambda}$ just means that $H$ approaches this constant value. It is an approximation. See also the answers to this question physics.stackexchange.com/questions/136056/… . They are quite detailed. $\endgroup$
    – Kosm
    Aug 30, 2021 at 10:42
  • $\begingroup$ $\Lambda\sim1/R^2$ is not true in general, since then it could not be constant. See my answer how this relation arises if we use the density parameter $\Omega_\Lambda$. $\Lambda\sim1/R_0^2$ could still be confusing, but at least now we relate constants. $\endgroup$
    – Koschi
    Aug 30, 2021 at 11:30
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    $\begingroup$ It is also maybe a little confusing that you say this relation does not hold in the early universe when matter and radiation were important. In an universe with ONLY the cosmological constant, $H$ is constant, since we have $a\sim\exp(Ht)$ with $H=$const. and the universe expands exponentially. But it is not obvious from your answer if you're talking about this. In such a universe it is also a little harder to define what the 'observable universe' is, but you already mentioned the confusion about this notion and the Hubble sphere. :-) $\endgroup$
    – Koschi
    Aug 30, 2021 at 11:36
  • $\begingroup$ @Koschi i should correct the definition of Hubble radius as $1/(aH)$ which includes the scale factor $\endgroup$
    – Kosm
    Aug 30, 2021 at 11:43

The OP's question is a variation of this question, with a different start point. The answer, therefore, is very similar:

Its often conjectured that the observable universe will asymptote to a de Sitter state (dark energy only, no matter, flat universe).

It is well known that the de Sitter characteristic length $l_\Lambda$ (i.e. future cosmic event horizon radius) is related to the cosmological constant $\Lambda$ (dimensions $L^{-2}$):

\begin{equation} {\Lambda} = \frac{3}{l_\Lambda^2} \end{equation}
Also, in a dS state, the future Hubble Horizon equals the cosmic event horizon, so: \begin{equation} {l_\Lambda} = \frac{c}{H(\infty)} \end{equation} That is the OP's $1/R^2$ relationship (also pointed out via the Hubble parameter by the other answers); the $R$ is the future de Sitter cosmic event horizon radius, which is a fixed number. At the present time, the radius of the observable Universe is around 2.9 times the future dS CEH radius.

TLDR: $\Lambda$ is related to the future cosmic event horizon radius, which is fixed. As an energy density (i.e. vacuum energy) the cosmological constant remains also fixed over time. The radius of the observable universe, on the other hand, is not fixed.


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