You probably noticed that the radius of the (observable) Universe in Planck lengths $R$ (about $10^{60}$) and the cosmological constant $\Lambda$ (about $10^{-120}$) in natural units obey approximately:

$$\Lambda R^2 \approx1$$

Suggesting that the amount of dark energy is proportional to the surface area of the cosmological horizon.

In Hawking's black hole theory the entropy and temperature of a black hole is proportional to its surface area.

This suggests that dark energy should be inversely proportional to the information content of the Universe or proportional to the temperature of the cosmological horizon by the Unruh effect (or seen from the "outside"?)

(Also, it is known that a black hole at a certain size will increase exponentially, as if it has it's own "dark energy" because it is colder than the thermal equilibrium of outer space. Could there be a connection here between how black holes expand and the Universe expands?)

And yet.... the cosmological constant is supposed to be constant.

So is the relationship merely a coincidence? Are there any current theories that an account for this relationship?

  • $\begingroup$ Your value for $R$ is the radius of the observable universe, at the current time. The universe may be much bigger than the observable universe; in fact, current curvature measurements point to it probably being infinite. The radius of the observable universe does not necessarily correspond to the cosmological horizon, especially not anything like an event horizon of a black hole. $\endgroup$ Jun 21, 2018 at 16:08
  • $\begingroup$ Well depends what you mean by "current time". If I take the temporal slice equivalent to the past light cone it has a finite size. If I take the slice at current "cosmological" time it is infinite but mostly unobservable by ourselves until the future when the light reaches us. $\endgroup$
    – zooby
    Jun 21, 2018 at 16:44
  • $\begingroup$ You are right the observable horizon, essentially looking back towards the big bang is not like a black hole. More like a black hole turned inside out! $\endgroup$
    – zooby
    Jun 21, 2018 at 16:46
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    $\begingroup$ It is not clear what reason except pure numerology you have to believe that the cosmological constant and the radius of the observable universe are connected, i.e. as written, the reasoning in this question about the cosmological constant being not constant is rather reminiscient of England drifting out to sea. $\endgroup$
    – ACuriousMind
    Jun 21, 2018 at 16:47
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    $\begingroup$ One could compare the hawking radiation of a black hole with the cosmic background radiation. Both decrease as the surface area (of black hole or cosmic horizon) gets bigger. $\endgroup$
    – zooby
    Jun 21, 2018 at 16:48

1 Answer 1


UPDATE: The current theory that accounts for the OP’s relationship is holographic dark energy (HDE). See this paper by Lee et al. In General Relativity you can consider Λ as as related to the intrinsic torsion or curvature of vacuum. In a quantum universe, dark energy is the energy of the Hawking radiation from a cosmic horizon – the future cosmic event horizon.

To an observer in our universe, a de Sitter universe is in their infinite future. In a de Sitter Universe, the Hubble horizon and Event horizon are coincident. Now, the de Sitter length is equivalent to the event horizon vector $R_h$ to an observer in a dS universe. We assign the de Sitter length $l_Λ=R_h=2= 1/H_0$ and thus de Sitter $Λ=3/4$ in natural units (a bivector), with the energy of the event horizon then being one Planck-energy.

The cosmological constant problem (CCP) can be expressed by observing that the number of degrees of freedom of dark energy in QFT is much too large to explain the observational data. The CCP can be written as: $$ρ_P/ρ_Λ = 8πc^3/ℏGΛ=(4/3)πR_h^3=32π/3$$ This approach equates one degree of freedom to one unit volume. The CCP can be overcome by considering the holographic principle which equates the actual number of degrees of freedom of a region to its area not its volume. This is equivalent to stating that the quantum vacuum energy in a region cannot be larger than a black-hole mass of the same size.

The resultant acceleration $a_H$ of the dS event horizon is given by the Unruh relationship: $a_H=1/2=c.H_0.$ as $T= (ℏa_H)/(2πck_B )$ This result also aligns with the principle of maximum force in GR i.e. $F_m=(E_H/c^2).a_H=1/2.$ Also, the 'entropic force' at the event horizon is $F_e=-2TS.a_H=-c^4/G$ i.e. Planck force.

From the Freidman equation, the vacuum energy density: $$p_Λ= 3H_0^2/8πG=Λc^2/8πG$$ Then, extending this answer, the degrees of freedom $N_s$ (dimensionless) on the dS event horizon surface area $A_s$ is: $$E=2TS=(1/2)N_s k_B T=2$$ $$A_s=4π.R_h^2$$ $$N_S=16π=A_s/L_Planck^2$$
We can thus associate one degree of freedom with one Planck area at the surface of the event horizon, which defines a theory of emergent space. i.e.

‘dark energy is proportional to the surface area of the cosmological horizon’.

Fine. Now what about the OP’s ‘numerological relationship’? Well, in our universe the Freidman equation is:

enter image description here

As dark energy will be the same density in our infinite future as now: $$(Λ/3).R_h^2=1$$

You can also think of $Λ=1/l_Λ^2 [1/L^2 ]$ as some kind of cosmic length scale, which, from 2018 Planck data is around $l_Λ=3.08$ GPc. The current particle horizon $R_P$ is $\approx$3 (2.72) times $R_h$ in GPc’s, so if you sub $R_P=2.72\times R_h=5.45$GPc you find $(Λ/3)R_P^2\approx$1

If you feel this is coincidence, well, HDE solves (Page 5) the cosmic coincidence problem too, as the holographic principle gives an upper bound of N~65 e-folds from the original dS universe.

  • $\begingroup$ Are you using "natural" units where ther radius of the universe is 2? $\endgroup$
    – zooby
    Jan 26, 2020 at 13:39
  • $\begingroup$ @zooby yes, that is the future event horizon radius corresponding to one planck energy. $\endgroup$ Jun 28, 2020 at 4:05
  • $\begingroup$ Thanks for the update! I will check out the paper. :) $\endgroup$
    – zooby
    Jun 29, 2020 at 2:36

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