How many states are there in the observable universe?

Question

If we took a single instant and considered all possible states of all energy and matter do we have any bounds on how much that would be? Would that number be related to information?

Answer 1

If you knew the maximum entropy $S_{\text{max}}$ possible for a system then you know how many possible states there are because $$S_{\text{max}}=\sup_{p_n}\left\{-\sum_nkp_n\log p_n\right\}=k\log N,$$ where $k$ is Bolzmann's constant, and $N$ is the number of states. There is a limit to the amount of entropy a volume of space can hold, and such a maximally entropic state in a region of space is given by a black hole. Now the entropy of a black hole is proportional to the surface area $A$ (surprisingly not the volume of the space) and is given by $$S=\frac{kA}{4\hbar G/c^3}=\frac{kA}{4\ell_p^2},$$ where $G$ is Newton's gravitational constant, $\hbar$ is Plank's constant, $c$ is the speed of light, and $\ell_p$ is the Plank length. So the number of possible states in the universe is given by $$N=\exp(A/4\ell_p^2)=\exp(4\pi R^2/\ell_p^2),$$ where $R$ is the radius of the observable universe.

Now the radius of the observable universe is about $$R=47\text{ billion light-years}\approx10^{26}\text{ meters},$$ and $\ell_p\approx 10^{-35}\text{ meters}$, so $$N\approx\exp\left(10^{123}\right).$$

Would that number be related to information?

The entropy is the (Shannon) information (if we set $k=1$) so the information is $\approx 10^{123}\text{ nats} = \log_2(e)\times 10^{123}\text{ bits}$.

Answer 2

The Bekenstein bound,

$$ S \le \frac{2\pi rE}{\hbar c},$$

is a limit on the natural log of the number of possible states (i.e., the information content) of a spherical region of space of radius $r$, containing mass-energy $E$. The mass of the hydrogen atoms in the observable universe is $\sim 10^{54}$ kg, and nonbaryonic dark matter is probably about 5 times that, or call it $10^{55}$ kg. The radius of the observable universe is about $4\times10^{26}$ m. I don't know whether dark energy should be counted in here or not, so I won't count it. This gives

$$ S \lesssim 10^{125}.$$

So the number of possible microstates might be something like $\exp(10^{125})$. Nearly all of these correspond to macrostates in which all the mass of the observable universe had hypothetically been concentrated into a single black hole. The actual entropy of the observable universe has been estimated to be about $10^{102}$ [Frampton 2008]. The fact that this number is much smaller tells us that the universe hasn't experienced heat death.

In general, I'm not sure how seriously to take the estimate using the Bekenstein bound. In relativity, unlike nonrelativistic physics, the total volume of the universe isn't fixed. That helps to make the cosmological notion of entropy fuzzy, and I think this is probably not a settled question, since we don't have a theory of quantum gravity. I don't know if there's any meaningful way to answer the question, "How different would the volume of this region of spacetime be if it were in a different state?"

Frampton et al., "What is the entropy of the universe?," 2008, https://arxiv.org/abs/0801.1847