# Is cosmological constant really constant?

As the Universe expands, the dark energy in it also increases. I heard that the cosmological constant $\Lambda$ represents dark energy, so that constant must change as time passes, right? Correct me if I'm wrong...

• The cosmological constant is an energy density, so it could be constant, even if the total amount of dark energy and the total size of the universe were increasing. Is it constant? Probably not. What causes it/dark energy and how does it/dark energy change in time? We don't know. Jun 18, 2016 at 6:30

Whether the dark energy is constant or not will ultimately be determined by experiment. At the moment there is no evidence that the dark energy is changing, but the experimental errors are still quite large so a change is not ruled out. There are lots of papers on this subject, but as yet no firm conclusions.

It is important to be clear that dark energy does not necessarily respond to the expansion in the way matter does. We describe the expansion using a scale factor $a(t)$, where we take the value of $a$ to be one right now. So $a=2$ means the universe has doubled in size. The average density of matter is given by:

$$\rho_m(t) = \frac{\rho_0}{a^3(t)}$$

where $\rho_0$ is the current average density. This should make sense - if the universe expands by a factor of $2$ then the volume of any region within it increases by a factor of $2^3 = 8$ and the density of matter falls by a factor of eight.

However if dark energy is a cosmological constant then it is a property of spacetime itself and it does not change as the universe expands. This seems counter-intuitive, but it is why dark energy behaves as it does. It is precisely because a cosmological constant does not change as the universe expands that it causes the expansion to accelerate.

The default expectation is that the density of dark energy is constant, for theoretical reasons. That's why it was historically called a cosmological constant. If you take the Einstein field equations, $$G_{ab}=8\pi T_{ab}+\Lambda g_{ab}$$, and take the divergence of both sides, you get $$8\pi \nabla^a T_{ab}=-g_{ab}\nabla^a\Lambda$$. A nonvanishing left-hand side would show up experimentally as a violation of conservation of energy, which we don't observe and don't have in any of our theories. That means the gradient of $$\Lambda$$ on the right has to be zero. It's possible, but not easy, to find ways to write down a nonvanishing dark energy that works around this problem.

I heard that the cosmological constant Λ represents dark energy, so that constant must change as time passes, right?

Yeah, this is a perfectly reasonable thing to expect, if you're imagining dark energy as some stuff that gets diluted by cosmological expansion. But that isn't how the math works out, as shown above. For some motivation, consider the fact that dilution works out differently for ordinary matter than for photons, because photons not only get farther apart, they also suffer redshifts that reduce their energy. That means that photon energy is diluted faster than matter energy. Different forms of stuff have different values of a parameter called $$w$$, which measures this. The value of $$w$$ for dark energy is 0. There is a more recent

Empirically, cosmological observations do not seem to support any variation in the density of dark energy over time. Example: Carnero 2011, http://arxiv.org/abs/1104.5426 They measure $$w$$ to be statistically consistent with zero and to show no trend over time. I believe the SH0ES project may have something better and more recent, but I haven't seen it. There are serious discrepancies between their value of the Hubble constant and the value from the Planck collaboration, so until that is clarified, it may not be possible to understand what is going on with more speculative stuff about dark energy.

• Correct, and I would like to add that the measurements remain a bit rough. The cited paper quotes $w = -1 \pm 0.16$ in the abstract, which is impressive accuracy for such a parameter but it is not possible to say much about trends over time. Hence the assertion that the measurements "show no trend over time", while true, perhaps gives a misleading impression. Measurements to date can rule out large variations but beyond that we don't know. Also I think the quoted accuracy of $0.16$ on $w$ does not allow for inhomogeneity effects. ... Sep 8, 2022 at 14:07
• ... After allowing for that I think the precision is probably no better than $\pm 0.5$, and therefore one cannot even rule out $w=0$. Sep 8, 2022 at 14:09

Let it be $$G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2}Rg_{\mu \nu}$$. Any equation that is directly proportional to $$G_{\mu \nu}$$ will always admit a cosmological constant $$\beta \in \Bbb{R}$$ as a generalization. This is why Einstein introduced it “ad-hoc” in first place, a decision he seemed to regret as he’s biggest blunder. The reason is that the gradient of both $$G_{\mu \nu}$$ and $$g_{\mu \nu}$$ vanish, and therefore so does $$\alpha G_{\mu \nu}$$ and $$\beta g_{\mu \nu}$$ with $$\alpha, \beta \in \Bbb{R}$$.

Starting from the EFE $$G_{\mu \nu}+\Lambda g_{\mu \nu}=kT_{\mu \nu} \, (\Lambda,k \in \Bbb{R}) \tag{1}$$ The change you propose transforms $$(1)$$ into $$G_{\mu \nu}+\Lambda(\mathbf{x})g_{\mu \nu}=G_{\mu \nu}+f_{\mu \nu} (\mathbf{x})=kT_{\mu \nu} \tag{2}$$ With $$f_{\mu \nu} (\mathbf{x})$$ a tensor function of the coordinates. The gradient of $$(2)$$ is $$\nabla f=k\nabla T \tag{3}$$ Which allows to interpret $$f$$ as an energy and momentum creation (annihilation) operator. However, the integral of $$(3)$$ includes an integration constant, so $$(2)$$ can be generalized into $$f_{\mu \nu}+G_{\mu \nu}+\beta g_{\mu \nu}=kT_{\mu \nu} \, (\beta, k \in \Bbb{R}) \tag{4}$$ Where $$G_{\mu \nu}+\beta g_{\mu \nu}$$ is the integration constant. Equation $$(4)$$ are just the EFE plus the energy creation operator. You started with a non-constant cosmological constant $$\Lambda$$ but ended up with another cosmological constant $$\beta$$ that is constant.

In conclusion questions of this kind are actually asking: “is energy and momentum conserved or not?”.

• I don't see the point of introducing $f_{\mu\nu}$ in Einstein's equations. Is $f_{\mu\nu}$ a purely geometric quantity ? Also, is $\beta g_{\mu\nu}$ only possible integration constant? My point is, how do you show that $\beta g_{\mu\nu}$ is the only possible non-trivial factor that one can add on top of EFE, and it would still satisfy the divergence free condition?
– KP99
Sep 9, 2022 at 10:37
• @KP99 Me too. I’m just showing to the OP what it means by replacing $\Lambda \in \Bbb{R}$ a constant, into $\Lambda=\Lambda (\mathbf{x})$ a function of the coordinates (time and space) in EFE. I only introduced $f_{\mu \nu}=\Lambda (\mathbf{x})g_{\mu \nu}$ to get into a more general form of that substitution, and no matter the solution $f$, you get from $(3)$, it is always possible to add $G_{\mu \nu}+\beta g_{\mu \nu}$ into it, making $(4)$ the most general form of the EFE. Sep 9, 2022 at 11:10
• @KP99 I edited the question to address your concerns. I guess the shortcut I made by not mentioning that $f_{\mu \nu}(\mathbf{x})=\Lambda (\mathbf{x}) g_{\mu \nu}$ was not very clear there. More importantly, a mentioned $\beta g_{\mu \nu}$ as the integration constant only before, when in fact I meant $G_{\mu \nu}+\beta g_{\mu \nu}$. My mistake. Thanks for your comment :) Sep 9, 2022 at 11:21

No it's not a constant. These days we prefer to call it a parameter.

EDIT

I've been asked to expand the post. There may be numerous ways to explain why the parameter changes over time, but a fool proof explanation says that the metric changes with expansion. Since vacuum energy is manifestly important to any definition of the vacuum, a variation in the metric implies a change in the vacuum energy as it expands.

You may think also that it is an explicit rule that the stress energy is conserved

$$T^{\mu \nu}_{:\nu} = 0$$

(As in a) must be conserved case, but cosmologist Lloyd Motz points out that conservation is an unfounded assumption.

Susskind is also vocal in his online lectures that the CC is NOT a constant, and refers to the requirement of calling it a parameter.