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?

  • $\begingroup$ Remember entropy is k times the log of the number of possible states. $\endgroup$ – Bob Bee Mar 30 '17 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 Mar 31 '17 at 5:27
  • $\begingroup$ @doetoe I have updated my answer. $\endgroup$ – lemon Mar 31 '17 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 Mar 31 '17 at 19:30

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.

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  • $\begingroup$ 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 Apr 1 '17 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 Apr 1 '17 at 15:13

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