1
$\begingroup$

I heard the following way to estimate the entropy of the universe:

using that the entropy is dominated by photons, in particular the cosmic microwave background radiation, which has a wavelength of the order of millimeters, the entropy is estimated as the number of cubic millimeters in the observable universe.

(If I remember and understood correctly). Could​ someone explain how this works?

Improve this question
$\endgroup$
4
  • $\begingroup$ Remember entropy is k times the log of the number of possible states. $\endgroup$
    – Bob Bee
    Commented Mar 30, 2017 at 20:01
  • $\begingroup$ Yes, without a doubt k has to be set to 1 for the statement to hold. Are you suggesting it is obvious that the number of states is exp(N), where N is the number of cubic mm in the universe? $\endgroup$
    – doetoe
    Commented Mar 31, 2017 at 5:27
  • $\begingroup$ @doetoe I have updated my answer. $\endgroup$
    – lemon
    Commented Mar 31, 2017 at 12:00
  • $\begingroup$ Doetoe @Lemon has the right answer, from what we know now. Thing is we still don't know what happens at the Plank scale, i.e., whether M might be a small or large number. The current practice is to figure it's 0 or 1 or a relatively small number. There's another issue, and that is the holographic principle that says that the physics in a volume can be described on its boundary, and the entropy is proportional to the area and not the volume. That happens in a black hole. But that's beyond the more popular statement you highlighted, and undecided. So N is used as a current measure. $\endgroup$
    – Bob Bee
    Commented Mar 31, 2017 at 19:30

1 Answer 1

Reset to default
2
$\begingroup$

You can imagine that each EM wave spans, in space, an integer multiple of ~1 mm. In which case, if you divide the universe into a $1\times 1\times 1$ mm$^3$ grid, each cell will 'contain' an integer number of EM waves. So in order to fully describe the state of this simplified universe (containing only microwaves) you need the number $N$ of 1 mm$^3$ cubes that fill it (specifically, you would have $N$ integers, e.g. 0,0,1,9,0,2,..., that counts the number of EM waves/energy density in each cell).

Each cell therefore has an integer value that will obey some probability distribution $P:\mathbb{N}_0\to\mathbb{R}$ and will be capped at some maximum value $M$, i.e. $P(M > 0)=0$. So the number of microstates can be approximated by $M^N$, and thus the entropy will be in the region of $k_B\log(M^N)=k_B\log(M)N$.

The value of $N$ is $\sim 10^{90}$, which will dwarf $\log(M)$. So, in atomic units ($k_B=1$), $N$ is a good estimate for the entropy of the universe.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks, sounds good. Maybe a slight variation could be that in order to uniquely specify any wave containing only frequency components with wavelength at least one mm, you have to sample over a grid with spacing 1/2 mm (by the sampling theorem). This means that the state space for a single photon of frequency not exceeding microwaves has dimension $N = 10^{90}$. The state space of $M$ such photons has dimension $M^N$. $\endgroup$
    – doetoe
    Commented Apr 1, 2017 at 15:11
  • $\begingroup$ Assuming that the present macroscopic equilibrium state of the microwave radiation in the universe corresponds with a subspace that is overwhelmingly larger than that of any other state, in particular itself also of the order $M^N$, we obtain the result. $\endgroup$
    – doetoe
    Commented Apr 1, 2017 at 15:13

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.