# Is the cosmological constant the simplest candidate for dark energy?

I'm reading this article on Dark matter and dark energy. There is a statement in the document which is as follows

What is the best bet for the nature of dark energy?

...The simplest candidate for dark energy is Einstein’s cosmological constant, which denotes a perfectly uniform fluid with negative pressure that is associated with the lowest energy (vacuum) state of the Universe. However, the observationally required value of the cosmological constant is $$10^{120}$$ times smaller than the theoretical expectation..

I'm not sure I understand the final sentence here. $$120$$ orders of magnitude difference is huge and I fail to see what the connection is here between the CC and simplicity?

Any suggestions?

• Essentially, the 120 orders of magnitude calculation has a lot of assumptions built-in. The cosmological constant is simple because it's an easy and natural addition to the Einstein equations. This is a perspective I like on this: arxiv.org/abs/1002.3966 Commented Jul 16, 2017 at 3:01

The $10^{120}$ still worries physicists a lot. It came from assuming the CC is due to vacuum energy in the universe, and the most natural number when one does basic calculations of it gives you the $10^{120}$. See how to get the number, and other description of the CC at https://en.m.wikipedia.org/wiki/Cosmological_constant

So the problem is that we don't know how to calculate for sure the CC from quantum field theory, or any other theory that is completed and accepted as valid. That will remain an issue, with the corresponding uncertainty about what the CC is until we discover what are the particles or 'things' that are responsible for the dark energy, and get a way to estimate the number that would lead to for the CC, from some accepted quantum gravity theory.

• The 120 orders of magnitude come from calculating the zero point energy density of all the quantum fields as the source of dark energy. This means that every Planck volume provides the same energy. However, if the Holographic principle is applied this is way over counting the degrees of freedom of the vacuum. If you assume that there is a zero-point contribution for each Planck area on the Hubble sphere, instead of for each Planck volume, you only miss the empirical energy density by two orders of magnitude...which could adjusted by a small factor on the Planck scale. Commented Jan 18, 2020 at 20:41

What they mean is that a cosmological constant is simpler than models in which dark energy varies dynamically. It's actually quite difficult to make a varying dark energy fit into the structure of general relativity. Naive attempts to make Λ variable cause the stress-energy tensor to have a nonvanishing divergence, which makes GR not self-consistent.

I would first look at my Stack Exchange post How did the universe shift from "dark matter dominated" to "dark energy dominated"? on a related topic. I derive the accelerated expansion of the universe using Newtonian mechanics. The surprising thing is that a constant positive energy or mass equivalent in the vacuum can result in a repulsive force. This basic result carries over with general relativity.

The $120$ orders of magnitude problem has to do with the nature of this vacuum and finding the actual energy density. The cosmological constant $\Lambda~\simeq~10^{-52}cm^{-2}$ is extremely small, which is one reason it took so long to discover the accelerated expansion of the universe. It required observations of type I supernova sufficiently far removed to detect this variation. This problem is a part of the difficulty with renormalization of gravitation.

The difficulty with the renormalization of gravity can be seen in the following argument. The main difficulty lies with loops. We consider the following loop

In this diagram we have vertices $V~\sim~p^2$ and internal lines $\sim~1/p^2$ and the loop with $O~\sim~\int d^4p$. The degree of divergence (the exponent on divergence) gives the $D~=~2V~-~2L~+~4O$ The Euler characteristic for this graph is $1~=~V~-~L~+~O$ or that $2(V~-~L)~=~2~-~2O$ so that $D~=~2(1~+~O)$. This means the divergence increases without limit as the order of the diagram increases.

We might though suppose that the order of these loop diagrams cuts off at one. We then have the integration from $k~=~0$to $k~=~1/\ell_{p}$, for $\ell_p~=~\sqrt{g\hbar/c^3}$ $=~1.6\times 10^{-35}m$ the Planck length. The divergence is then order $4$ or $\simeq~10^{140}m^{-4}$. The expected cosmological constant would then be $\Lambda_p~\simeq~10^{70}m^{-2}$ this is $122$ orders of magnitude larger than the measured cosmological constant.