What is the maximum entropy a system can have based on its energy and size?

[PeterDonis]

So, if I have a system with very small mass, very large radius, and very large entropy, then it seems that I can violate the second law of thermodynamics by dumping the system into a small black hole.

First, you can't specify the mass, radius, and entropy of the system independently. Second, the bound is derived on the assumption that the black hole the system falls into is much larger than the system itself.

 

[jartsa]

First, you can't specify the mass, radius, and entropy of the system independently. Second, the bound is derived on the assumption that the black hole the system falls into is much larger than the system itself.

I did not violate the Bekenstein bound when I assumed a system with low mass, large entropy and sufficiently large size.

When this system is put into a container with a small black hole, the second law of thermodynamics will not be violated, because the black hole will evaporate before the system has been sucked in.

 

[PeterDonis]

I did not violate the Bekenstein bound when I assumed a system with low mass, large entropy and sufficiently large size.

When this system is put into a container with a small black hole, the second law of thermodynamics will not be violated, because the black hole will evaporate before the system has been sucked in.

Then I'm confused; I thought you were claiming that your scenario could violate the second law. What exactly are you trying to say?

 

[PeterDonis]

I think this is because expanding the system changes the details of the process of it falling into the black hole, in such a way as to make the lower bound on the area increase larger.

I've read through Appendix A a number of times now, and I think I have a basic understanding of what it's saying.

(1) First, consider the idealized limiting case of R = 0; in other words, we drop a point particle of initial rest mass M into a black hole. The way to do this that minimizes the area increase is to release the particle exactly at the horizon into an orbit with zero energy at infinity and zero angular momentum. In other words, we drop it from rest just at the horizon. (We ignore technicalities such as the fact that a particle of nonzero rest mass can't be at rest at the horizon, even instantaneously; this is just heuristic, after all.) In this idealized limiting case, the area increase of the hole is zero, because the hole's mass and angular momentum are unchanged by this process.

(We also ignore the fact that we had to lower the particle to the horizon by some process from which we extracted work equal to the particle's initial rest mass M, so that its energy at infinity when we release it at the horizon is zero. That process is considered to be already completed while the particle is fully outside the hole, so it doesn't figure into the calculation of what happens when the particle falls into the hole.)

(2) Now consider what happens if the particle has a nonzero radius R. This means that we can't lower it all the way to the horizon; we can only lower it to a proper distance R outside the horizon. That means we can't release it with zero energy at infinity (although we *can* release it with zero angular momentum); so it will deliver some nonzero energy to the hole, increasing the hole's mass by some minimum amount. For very small R (i.e., R much less than the hole's horizon radius), we expect the hole's mass increase to be linear in R. We also expect it to be linear in M, the initial rest mass of the particle.

(3) Now, the key point: the energy at infinity of the particle with nonzero radius R (and zero angular momentum) when it is released is also proportional to the *temperature* of the black hole that the particle is falling into! This is the key fact that makes the hole's parameters cancel out of the area increase: heuristically, entropy is energy divided by temperature, so if the hole's mass increase is proportional to its temperature, then its entropy increase (i.e., the area increase) will be independent of the hole's temperature. Since the hole's temperature is the only way in which the hole's parameters enter into the calculation, this means the hole's parameters cancel out.

Why is the energy at infinity of the particle at release proportional to the temperature? The temperature formula given by Bekenstein is

$$\Theta = \frac{1}{4} \frac{r_{+} - r_{-}}{\alpha}$$

where $\alpha = r_{+}^2 + a^2 = A / 4 \pi$ is the "normalized area" of the horizon. The $r_{+} - r_{-}$ in the numerator is a measure of how "extreme" the hole is in terms of angular momentum vs. mass; but since the infalling particle is adding zero angular momentum in the case of minimal area increase, this quantity does not change in the process, so we can heuristically consider the simplest case of a Schwarzschild hole, where $r_{+} = 2M$ and $r_{-} = 0$. This makes the temperature inversely proportional to the horizon radius. And *that* means the energy at infinity of a particle with a given initial rest mass M released at a given proper distance R above the hole's horizon is proportional to the particle's initial rest mass M, times the ratio of R to the hole's horizon radius. This makes sense: the energy at infinity is really a dimensionless fraction times the initial rest mass M, and so we would expect it to be determined by a dimensionless ratio derived from the particle radius R, rather than just by R itself.

So the overall calculation looks like this (where "=" here means "proportional to"):

(area increase) = (entropy increase) = (energy increase) / (temperature) = (M * R / horizon radius) / (temperature) = (M * R * temperature) / (temperature) = (M * R)

The hole parameters cancel out and we are left with just the particle parameters M and R.

 

[michael879]

This article I find unconvincing but understandable:

http://www.scholarpedia.org/article/Bekenstein_bound

First we derive a bound that has the radius of the black hole as one parameter.

Then we replace "the radius of the black hole" with "couple of times the radius of the dropped object".

Then we find the exact value of "couple of times".

Yes, I've seen this article. It doesn't answer any of my questions though..