# Gravitational entropy

Science Advisor
In ordinary QM and QFT entropy is defined using a density operator for a generalized state:

$$S = -\text{tr}\left(\rho\,\ln\rho\right)$$
b/c for the gravitational field we do neither know the fundamental degrees of freedom nor the Hilbert space states, a definition like

$$\rho = \sum_np_n\,|n\rangle\langle n|$$
is not available.

Questions:
1) are there attempts to formulate entropy for a QFT on a gravitational background for a "finite volume"?
2) are there attempts to formulate entropy for the gravitational field "within this finite volume"?
3) has this been done in string theory and / or spin networks for several different spacetimes (black holes, some other finite volume, expanding spacetime with e.g. co-moving dust, ...)?
4) how does the holographic principle show up?

(I know some special cases like the state counting for black holes in LQG, but I have never seen a general construction)

## Answers and Replies

In ordinary QM and QFT entropy is defined using a density operator for a generalized state:

$$S = -\text{tr}\left(\rho\,\ln\rho\right)$$
b/c for the gravitational field we do neither know the fundamental degrees of freedom nor the Hilbert space states, a definition like

$$\rho = \sum_np_n\,|n\rangle\langle n|$$
is not available.

(I know some special cases like the state counting for black holes in LQG, but I have never seen a general construction)

Carlo Rovelli gave a talk related to this today, maybe a new paper will come out soon.

Science Advisor
great!

bcrowell
Staff Emeritus
Science Advisor
Gold Member
Is it even well understood how to define gravitational entropy in the classical case? I was under the impression that there were nontrivial issues. Do we have a general definition that applies to cases like inflationary models, spacetimes with CTCs, spacetimes that aren't time-orientable, ...? Is this the reason for Tom's "finite volume" with quotation marks? Is it hopeless to try to cover the classical case before moving on to quantum gravity because accretion of hot matter onto a black hole then violates the second law?

Science Advisor
Is it even well understood how to define gravitational entropy in the classical case?
You can't define a "gas of gravitons" like in classical statistical mechanics. And it wouldn't make sense b/c you don't want to split spacetime in background + gravitons. But in statistical mechanics there is no entropy if there are no states, so I would say that in GR the entropy of spacetime is zero.

Do we have a general definition that applies to cases like inflationary models, spacetimes with CTCs, spacetimes that aren't time-orientable, ...?
Are you talking about the gravitational entropy or about the entropy of quantum fields on a background?

Is this the reason for Tom's "finite volume" with quotation marks?
The problem is that for finite volume the volume (the region of space) may change, i.e. it may expand, and that particles may cross the boundary of that region. So I don't know if it makes sense to consider entropy of a finite volume. But I also don't think that one can integrate over inifinite volume. In addition I do not see (but this may be due to a lack of understanding) how to write down the relevant thermodynamic expressions in a covariant way and for general spacetome.

@tom: I don't think there needs to be a problem. As usual, the starting point is to pick an ensemble, which matches a known set of constraints (and which experimentally is known to be sufficient to predict some other set of observables). All that is required is a known set of physically distinct states (which in LQG is given by the boundary data?). Then one proceeds and can calculate all the relevant quantities. Now, the problem is really the "sufficient" bit of all that --- without experiments or some deep reasoning, it is hard to see what are the relevant constraints (e.g. with ideal gasses you have volume and internal energy). For blackholes (non-rotating, equilibriated, etc.), it seems like area is sufficient --- but there we could actually be wrong. In the general case, it seems like one would struggle to find general observables which are always sufficient, but maybe Rovelli has done exactly that!