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?

  • 1
    $\begingroup$ This is not the same as: How many bytes can the observable universe store? or How many bits are needed to simulate the universe? $\endgroup$
    – Skyler
    Commented Apr 9, 2014 at 17:54
  • 1
    $\begingroup$ What do you mean "a single instant"? How would one define an unambiguous slice of time across the universe? $\endgroup$
    – Jim
    Commented Apr 9, 2014 at 17:59
  • $\begingroup$ don't we consider Planck time the smallest sliver of time that exists. $\endgroup$
    – Skyler
    Commented Apr 9, 2014 at 18:14
  • $\begingroup$ ok, so you can (if you want) define the duration of an instant as the Planck time (or any other duration, makes not much difference). The question is how do you define when all of the different points of space in the universe are within the same instant of time? This is the problem of relativity of simultaneity. $\endgroup$
    – Jim
    Commented Apr 9, 2014 at 18:24
  • $\begingroup$ lets say at a determined Planck time in the snap of my finger the universe froze. Are you saying that the instant I chose would be different times for other regions (i.e.: its almost like my freezing has to propogate and thus can be non-instaneous) Or are you just simply stating it cant be the same for bodies (e.g.: an observer who has been travelling impossibly close to the speed of light for a billion years our time would have a t=12.8 billion vs my 13.8 billion)? $\endgroup$
    – Skyler
    Commented Apr 9, 2014 at 19:28

2 Answers 2


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 $\approx 10^{123}\text{ nats} = \log_2(e)\times 10^{123}\text{ bits}$.

  • $\begingroup$ how would that last fractional bit work out in terms of a state? $\endgroup$
    – Skyler
    Commented Apr 13, 2014 at 18:49
  • $\begingroup$ This is based on the absolute limit of possible states in a volume and uses lots of rounding (e.g. radius of the universe to nearest light-year). So uncertainty in N is much larger than a fractional bit. More generally Classical entropy is continuous, but statistical mechanics actually counts microstates. For Black holes this means a quantum theory of gravity (which we don't have). Also you specify the problem macroscopically (e.g. via size of our universe) so we're left with classical quantities (macroscopic surface area) for our calculation which always has some imperfect precision. $\endgroup$ Commented Apr 13, 2014 at 19:29
  • $\begingroup$ It's just that the log number is irational and so no matter how accurate a number of states you create there is always going to be a fractional remainder. What you are saying is we don't yet have a satisfying answer to that remainder bit? Or are you saying since that is our upper threshold (a black hole of radius observable univers) that we would instead have a discrete number of states. $\endgroup$
    – Skyler
    Commented Apr 13, 2014 at 19:49
  • $\begingroup$ The information is the log of the number of states so the number of states $N$ is the exponent (not the log) of the entropy. And since you're computing $N$ by taking the exponent of an area which will always have some error in the estimate, then if you end up with a fractional state then you can just assume this is a rounding error. If it makes you feel better you can just round $A/4\ell_p^2\approx \log(n)$ where $n$ is an integer, then you'd have exactly $N=n$ states. $\endgroup$ Commented Apr 13, 2014 at 22:02
  • $\begingroup$ This doesn't seem quite right to me. Nearly all of the possible microstates for this spherical region of space are ones in which all its matter is in a single black hole. Some justification would be required for the assumption that this black hole had the same radius as the radius of the observable universe. $\endgroup$
    – user4552
    Commented Aug 24, 2014 at 18:42

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.