This is a bit quirky: For a very long time I've found Stephen Hawking's evaporating small black holes a lot more reasonable and intuitive than large black holes.

The main reason is that gravity is relative only if your gravity vectors are all parallel. When that is true, you can simply accelerate along with the field and have a perfectly relativistic frame going for you.

Not so if your gravity vectors angle in towards each other, as is particularly true for very small black holes. In that case the energy inherent in the space around the hole becomes quite real and quite hot, and that's regardless of whether you have matter in the mix or not. (Hopefully that's fairly intuitive to everyone in this group?)

So, how can the space immediately surrounding a tiny black hole not be incredibly hot? By its very geometry it must be absolutely bursting with energy due to the non-parallel intersection of extremely intense gravity vectors. So, the idea of that energy evidencing itself in the creation of quite real particles outside of the event horizon seem almost like a necessity, a direct consequence of the energetic structure of space itself.

So, that's really the basis for my question: Isn't the curvature of space a better way to understand its entropy adding up the surface area of a black hole?

By focusing on curvature, all space has entropy, not just the peculiar variety of space found on event horizons. Flat space maximizes entropy, while the insanely curved space near a microscopic black hole maximizes it. I also like this because if you get right down to it, entropy is really all about smoothness, in multiple forms.

So: Is the inverse of space curvature considered an entropy metric? If not, why not? What am I missing?

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    $\begingroup$ I really like this intuition. I can't fault it, but unfortunately I can't answer your question either! One problem is that uniformly spread matter - begetting flat space - tends to clump together in things like black holes, so the gathered, clumped and thus curvy space state would correspond to the higher entropy if we believe the second law to hold in such systems. Another problem might be working these ideas into an entropy formula -I'm thinking aloud here - the generalised Gauss Bonnett theorem means that you could have difficulty defining an integrated-over-space curvature sum that ..... $\endgroup$ Nov 29 '13 at 1:07
  • $\begingroup$ ..... turns out other than something proportional to the Euler characteristic of space, so that it could only change in integer jumps (I'm not sure how you would cope with a singularity - but this could be a clue). $\endgroup$ Nov 29 '13 at 1:09

We may associate to some degree the curvature of a solution to its entropy. The Euclidean partition function in general relativity can be approximated by,

$$Z_{E} \approx \exp(-I_E)$$

where $I_E$ is the Euclidean action evaluated at all classical solutions which have a periodic $\tau$ coordinate with period $\beta=1/k_BT$. The entropy of the system is given approximately by,

$$S = k_B \frac{\partial }{\partial T}\left(T\log Z\right)$$

The Einstein-Hilbert action contains the Ricci scalar, i.e.

$$I_{EH}=\int_M \mathrm{d}^d x \, \sqrt{-g} \, R$$

Therefore, technically, the entropy is dependent on the curvature. In addition, even for non-trivial solutions such as the Schwarzschild metric, there may be cases when $R=0$. However, recall that to define a fully rigorous action principle, it should be supplemented by a boundary term,

$$I_{GH} = \int_{\partial M} \mathrm{d}^{d-1}x \, \sqrt{-h} \, K$$

where $K$ is the extrinsic curvature, and $h_{ab}$ the induced metric on the boundary $\partial M$.

Sample Calculation (Schwarzschild Solution)

After a Wick rotation, keeping in mind $\tau$ is periodic with period $\beta$, the metric is given by,

$$\mathrm{d}s^2 = \left( 1- \frac{2GM}{r}\right)\mathrm{d}\tau^2 + \left( 1-\frac{2GM}{r} \right)^{-1}\mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \mathrm{d}\phi^2$$

If we choose an appropriate inward pointing normal,

$$n^{a} = - \sqrt{1-\frac{2GM}{r}}\delta^a_r$$

the extrinsic curvature is given by $\nabla_{a} n^a$, i.e. the divergence of the normal. Intermediate steps require the introduction of a radial cut-off $R$, and the subtraction of the boundary term of flat space with the same boundary. Eventually, one finds, $$I_{GH} = \frac{\beta M}{2G} = \frac{A}{4G}$$

where $A$ is the area of the black hole. The entropy obtained via the formalism is in agreement with the Bekenstein-Hawking entropy. (In fact, the boundary term was partially due to Hawking.)

See http://perimeterscholars.org/448.html; lecture 10 provides the calculation for the Schwarzschild black hole, and the previous lectures discuss the required mathematics of sub-manifolds.


Some remarks :

If we consider a Schwarzschild black hole, a local measure of the curvature is the square root of the Kretschmann scalar $K = R_{abcd}R^{abcd}$. If you consider the inverse of the curvature, you have :

$K(r)^{-\frac{1}{2}} \sim \dfrac{r^3}{GM} \tag{1}$

On the other way, the total entropy of the black hole is :

$S \sim GM^2 \tag{2}$


You may, of corse, think about $K^{-\frac{1}{2}}$, as an entropy volume density, but it is false, first we know that we have to do some integral of some quantity on the black hole horizon, which is a surface, but not a volume, second the dimensionality of $K^{-\frac{1}{2}}$ is not correct, it is $[K^{-\frac{1}{2}}] = M^{-2}$ ( and $[S] = M^0) $, while it should be $M^2$.

Moreover, flat space does not "maximize" entropy. If one imagine some empty region of space, that is without any information, there will be no entropy either. If we begin by fill this region of space with energy and momentum, there will be information, and entropy (seen as "uniform" information). The maximum of entropy arises when the energy is so important in the choosen space region that you have a black hole.

  • $\begingroup$ "You may, of corse, think about $K^{-\frac{1}{2}}$, as an entropy volume density, but it is false, first we know that we have to do some integral of some quantity on the black hole horizon, which is a surface, but not a volume" -- yes, but the integral of the inverse of the square root of the Kretschmann scalar over the interior of the black hole converges and is proportional to the surface area, no? $\endgroup$
    – Johannes
    Nov 29 '13 at 15:26
  • $\begingroup$ @Johannes: Why would you think that the integral converges? $\endgroup$
    – user23660
    Nov 29 '13 at 17:31
  • $\begingroup$ @Johannes : You have a problem of mass dimensionality (see below the EDIT part). the integral would have a mass dimensionality $M^{-2} M^{-3} = M^{-5}$, while entropy has mass dimensionality $M^0$ $\endgroup$
    – Trimok
    Nov 29 '13 at 19:29
  • $\begingroup$ @user23660 - in the interior away from the horizon the flatness measure (taken as the inverse square root of the Kretschmann scalar) quickly drops to zero. $\endgroup$
    – Johannes
    Nov 30 '13 at 4:32
  • $\begingroup$ @Johannes: But the 3-volume of any hypersurface $r=\mathrm{const}$ is infinite. $\endgroup$
    – user23660
    Nov 30 '13 at 4:37

If curvature/flatness of space is a measure of gravity, then this flatness can certainly be considered a way of indicating entropy. This is so for a very simple reason; and you do not need to go the extreme and consider black holes and temperature in their vicinity to see this.

Notice one thing - gravity and entropy are simply two opposing phenomena (forces). Entropy is what makes particles move away from each other. If unrestricted by any boundaries, entropy - changing naturally from low to high - will make particles keep spreading more and more (loosing energy and velocity at the same time, but this is not so important here) giving them more freedom. On the other hand, gravity makes particles to come together, to concentrate. Again, if unrestricted, gravity makes particles keep coming closer to each other, it restricts the freedom of particles.

Therefore, the flatter the space (lower gravity), the greater the freedom of particles (higher entropy). Obviously, the more particles there are, the more they will restrict each other's freedom, as the increasing number of particles not only blocks movement, but also increases the gravity they exert on each other. Still, given two areas of space - other things being equal - you can certainly say that the one with greater gravity (which you can measure by flatness if you want) will produce lower entropy.

EDIT: I realize that such an answer might seem all too simple when considering such lofty and mysterious entities as black holes, but ... well ...


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