Tuesday, March 01, 2011

Ricci flow and the entropy of space

The entropy of the universe appears to be increasing, in accordance with the Second Law of thermodynamics, and this is commonly taken to imply that the early universe resided in a state of extremely low entropy. However, given that the geometry of space in the early universe was highly uniform, and given that matter at this time resided in thermal equilibrium, there is a problem. If the universe began in what appeared to be a state of equilibrium (i.e., a maximum entropy), then why did it not remain in such a state?

The common answer proffered is to propose that there is some form of gravitational entropy, and this type of entropy is maximised by clustering, not by homogeneity. There are two primary observations which seem to support this argument: (i) a homogeneous distribution of matter is clearly unstable under gravitational attraction, and over time matter clusters into stars and galaxies; (ii) the entropy of a black hole, the most concentrated form of gravitational collapse conceivable, is huge.

Philosopher of physics David Wallace has already dispelled the complacent notion that the matter in the early universe was in a state of equilibrium. As David explains, the rate of expansion of the universe displaced the matter in the early universe from equilibrium, leaving most of it stranded in the form of hydrogen and helium. The formation of stars increases entropy because it permits such lighter elements to undergo nuclear fusion. Black holes, argues Wallace, are a special case. In addition, let us not forget that all gravitationally bound systems eventually evaporate, which is hardly consistent with clustering itself being the driving force behind the Second Law.

But what about the other half of that complacent assumption concerning the early universe? Does a homogeneous geometry of constant curvature really possess low gravitational entropy? Well, here's a possible reason for thinking that it doesn't: the Ricci flow:



The expression on the left here is the rate-of-change of the metric tensor, whilst the expression on the right is a negative multiple of the Ricci tensor. The Ricci flow is an equation used by mathematicians for, (amongst other things), evolving the geometry on a 3-dimensional manifold. There are two crucial facts about this equation: (i) it evolves a geometry of varying curvature into one of constant curvature; and (ii) it's a kind of diffusion equation.

Now, the point about diffusion equations is that they increase entropy. In the field of radiation transport, for example, whilst the radiative transfer equation is the relevant equation from non-equilibrium statistical mechanics, it is fiendishly difficult to solve, and a diffusion equation is used as a macroscopic approximation instead.

So, if the evolution of an initially random 3-dimensional geometry towards constant curvature, is represented by a diffusion equation, and if diffusion equations represent physical processes which increase entropy, doesn't this suggest that a geometry of constant curvature possesses greater geometrical entropy than one of varying curvature? Could the Ricci flow ultimately be cast as a macroscopic approximation to a random walk in the space of underlying quantum geometries?

Which is all very well, but how do we reconcile this with the fact that black holes possess a huge entropy? Well, recall Roger Penrose's long-held proposal that gravitational entropy will be associated in some way with Weyl curvature. Black holes possess zero Ricci curvature and non-zero Weyl curvature, whilst the big bang possesses the opposite combination of zero Weyl curvature and non-zero Ricci curvature. If we restrict attention to 3-dimensional geometries of zero Weyl curvature, we may be able to argue that it is actually those with the highest degree of symmetry, those of constant curvature, which possess the highest entropy. Nevertheless, it may be that gravitational entropy is increased by clustering, not because it reduces geometrical homogeneity or symmetry, but purely because it increases the Weyl curvature.

No comments: