What I know (and please correct me if I'm stating malarkey): The entropy of the Universe (its description) is contained in Weyl tensor. Einstein's field equations don't directly relate the entropy with the universe and its curvature/geometry. We can obtain Weyl tensor by contracting the Riemann tensor which in $4D$ has only $20$ independent components (and Ricci tensor has $10$).

In Cosmology, to describe the evolution of the Universe, Einstein equations are not enough (indeed we need Friedmann Equations too and else), hence to understand the evolution what we do is to look at the independent components of the Weyl tensor.

But when one comes to deal with Big Bang, then the Weyl tensor vanishes, whereas they become larger and larger the more the Universe expands. This could be the explanation of why the entropy does always increase (at least without starting to run exotic physics and so on).

Now, the Universe is not a closed system, and it cannot be described by usual Thermodynamics because it has no volume and no temperature, and we cannot run experiments in the thermodynamics sense to study it in that way. Hence to speak about Entropy in the Clausius sense we need to consider it as a sum of portions (read: closed systems) and look at the interactions in the neighborhood.

Question: What happens to the curvature of the Universe if the Entropy of the universe wouldn't conserve? Is the increasing / conservation / non-conservation of the Entropy, related to something like the the energy density of the Universe? Maybe it would be comparable to the critical energy density?

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    $\begingroup$ i'd suggest you pick one of those 3 questions. otherwise by definition its too broad and should be closed $\endgroup$ Mar 19, 2018 at 11:02
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    $\begingroup$ Friedman's equations are derived from Einstein equation with FRW metric plus perfect fluid stress-energy tensor $T^{\mu\nu}$. And the entropy density of the universe $s$ is related to the energy density $\rho$ and pressure $p$ as $s=(\rho+p)/T$. @VonNeumann $\endgroup$
    – SRS
    Mar 19, 2018 at 11:28
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    $\begingroup$ @SRS - but that is the entropy of the matter contents, not the Weyl spacetime entropy. The question is somewhat confusingly expressed since it seems to say that the spacetime entropy is all there is, but maybe we can ignore the matter contribution. $\endgroup$ Mar 19, 2018 at 12:09
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    $\begingroup$ the entropy of the universe (its description) is contained in Weyl tensor This seems a little garbled to me. A universe with maximal entropy would probably be something like the "mixmaster" spacetimes, with most of the energy and entropy locked up in gravitational waves. In that scenario, I suppose it's true that the Weyl tensor would contain all the information necessary to get the main contribution to the entropy. But our actual universe looks nothing like this. For reasons that AFAIK are not understood, the early universe did not have the gravitational d.f. activated thermodynamically. $\endgroup$
    – user4552
    Apr 2, 2019 at 15:31
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    $\begingroup$ "Now, the Universe is not a closed system, and it cannot be described by usual Thermodynamics because it has no volume and no temperature, and we cannot run experiments in the thermodynamics sense to study it in that way. " That is utterly wrong as it stands and needs rewording perhaps $\endgroup$
    – ohneVal
    Aug 11, 2021 at 9:02

2 Answers 2


I'm not sure why you are mentioning the Weyl tensor in the context of FRW-cosmology. The Weyl tensor describes the curvature for vacuum solutions of the Einstein field equations (where the energy-stress tensor vanishes), e.g. the Schwarzschild solution.

As to the entropy of the universe, it is dominated by the huge number of photons. There are way more photons than baryons. Nevertheless the energy density of the photon bath is negligible compared the the matter density and thus not "comparable to the critical density".


Universe is closed system (hence the entropy S has to increase or remain constant) and we believe that it has gone through adiabatic expansion (thus entropy has to constant).

But since the volume is changing, we define a quantity called specific entropy s = S/V = $\frac{\rho+p}{T}$which decreases as $a^{-3}$ where, $a$ is scale factor. Specific entropy turns out to be very important parameter to as it relates energy density and temperature and is used to fix the degrees of freedom of particles (number of neutrino flavors etc) from CMB data. (ref to any Cosmology lecture notes; example Ch-3)

Now, if we stay with S, in terms of number of configurations of microstates W, it is given by, S = $k_B$ln W. If we mix two different color paints (fluids), the highest entropy state would be a complete mixture of two colors. On the contrary, from 'first' light (CMB) the Universe we appears to be in thermal equilibrium (highest entropy state), hence it appears to be going in the opposite direction of second law of thermodynamics. But there are small perturbations in this equilibrium temperature (for order $10^{-4}$) which comes from gravitational collapse of these fluids. Hence we have to take into account not only thermal entropy but also gravitational entropy (highest in black holes).

Weyl curvature is the remaining tonsorial part (WT: Weyl tensor) when we take out the curvature information defined using Ricci tensor from the Reimann tensor - hence it is traceless (and has the property of conformal invariance and is no dependent on $T_{\mu\nu}$). For FRW metric, WT vanishes for both early and late universe as they describe homogeneous and isotropic universe. It is large for Schwarzschild metric. Hence, Weyl curvature behaves like gravitational entropy from zero entropy from the early universe (no gravitational degrees of freedom - hot plasma) gravitational entropy is zero and then it takes over as the structures form. Currently maximum gravitational entropy is in Black Holes.


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