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What's the entropy of the universe today? How does one go about calculating this? I've heard the statement that black holes account for the bulk of the entropy in the universe today, but don't know why this would be true or the relationship between black holes and entropy.

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    $\begingroup$ Maybe this paper might interest you. See Table $I$ page $4$. $\endgroup$
    – Trimok
    Jan 30 '14 at 13:45
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The paper suggested by Trimok seems to answer your question.

The paper gives an entropy for the observable universe of

\begin{equation} S_{obs U}= 3.1×10^{104} k \approx 10^{104}\,\mathrm{bits}, \end{equation} where $k$ is the Boltzmann constant and $S_{obs U}$ is the entropy.

However, I would like to answer your two questions with a back-of-the-envelope calculation.

If we use the Bekenstein bound, then

\begin{equation} S \leq \frac{2 \pi c R M}{\hbar \ln 2} \approx 2.577\times 10^{43} M R, \end{equation}

where $R$ is the radius of a sphere that can enclose the given system, $M$ is the mass of the system, $\hbar$ is the reduced Planck constant, $c$ is the speed of light and $S$ is the entropy in bits.

Now, for the observable universe,

$R=8.8×10^{26}\,\mathrm{m};$

$M=10^{53}\,\mathrm{kg}.$

Then, we have for the observable universe

\begin{equation} S_{obs U}\le \frac{2 \pi c R M}{\hbar \ln 2} \approx 10^{123}\,\mathrm{bits}. \end{equation}

John Baez did a similar calculation assuming the universe is a black hole and found the saturated bound.

He obtained

$S_{obs U}\le 1.4 × 10^{124}\,\mathrm{bits}.$

There is also this question (and answer) that might help.

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