2

I have a question about the entropy of black holes. Well, some time ago I was able to demonstrate theoretically that the entropy of black holes should be relative, that is, depend on the observer.

But when I submitted an article showing this to a scientific journal, they didn't accept it as one reviewer said my ideas seemed "weird". So I decided to show the calculations here and see what you think.

I'll simplify things a little, but it starts this way:

The first law of classical thermodynamics says that the internal energy of a system depends on the temperature, entropy and the work being done by the system: $$dE=TdS-dW$$

If we assume that the system is infinitesimal, therefore, it would not suffer the action of tidal forces, and that it follows a geodesic, therefore, there are no forces being made or suffered by the system, we can reduce this equation to just: $$dE=TdS$$

Then: $$dS=\frac{dE}{T}$$

In the study of Special Relativity, we learned that the total energy is equal to the Gamma Factor times the object's mass, times the speed of light squared: $$E=\gamma mc^2$$

It is known that the Gamma Factor is defined as $$\gamma \equiv \frac{\text{d}t}{\text{d}\tau}$$ (the coordinate-time by the proper-time)

This makes it possible to find the "Gamma Factor" of any spacetime as long as we have the metric for that spacetime. So, if we assume that we are talking about a Schwarzschild black hole, and that therefore the temperature of the equation is the temperature as Stephen Hawking derived: $$T=\frac{\hbar c^3}{8\pi GkM}$$

We have the following result: $$\text{d}S=\frac{8\pi Gkm^2}{\hbar c}\text{d}\gamma$$

Here, we say $d\gamma$ because this is the only thing that is changing if the black hole is not swallowing matter. As mass is not changing, then: $$S=\frac{8\pi G k m^2}{\hbar c}\Delta \gamma$$

For an observer initially at rest at infinity, falling radially into the Schwarzschild black hole, it is possible to show that the Gamma Factor will be $$ \gamma=\frac{1}{1-\frac{2Gm}{rc^2}}$$

You can see that the entropy value of a black hole changes depending on the observer, depending on where the observer is and how it moves.

That would mean that entropy is relative, right? Is there something wrong with my math?

Vincent Thacker
  • 7,926
Vinicius Araujo Ritzmann
  • 151
  • 2
    The Hawking temperature can be used to derive either an observer dependent or an observer independent notion of entropy. A covariant version of the first law (above equation 1.18 in 1205.5040) is $TS^\mu = -T^{\mu\nu}u_\nu$. What you're doing is taking the zeroth component and integrating both sides over space. But you can also contract both sides with $u_\mu$ and then integrate over space-time. Something which compares both approaches might satisfy the reviewer. – Connor Behan Aug 25 '21 at 00:29
  • 1
    Here's an alternative to the approach in @ConnorBehan's comment: we can treat $dE=T,dS$ as an identity that follows from specifying $S$ as a function of $E$, using $1/T=\partial S/\partial E$ as the definition of the temperature $T$, as usual in statistical mechanics. Then the entropy is invariant and the temperature is not. That's consistent with the invariance of the area (arXiv:0708.0276). The behavior of temperature under Lorentz boosts is the subject of another question. – Chiral Anomaly Aug 25 '21 at 01:23
  • By the way, to make the equations look nicer, Physics.SE lets you enclose them in dollar-signs just like in TeX. For example, writing $E=mc^2/\sqrt{1-v^2/c^2}$ gives the result $E=mc^2/\sqrt{1-v^2/c^2}$. For more info, see MathJax basic tutorial and quick reference. – Chiral Anomaly Aug 25 '21 at 01:28
  • @ConnorBehan Thank you very much. Do you know where I can find out more about this covariant version of the law? – Vinicius Araujo Ritzmann Aug 25 '21 at 12:30
  • @ChiralAnomaly Thank you I'll study the article and question you mentioned. – Vinicius Araujo Ritzmann Aug 25 '21 at 12:31
  • Any other source I recommend would just be the Google hits for "entropy current", "entropy flux" or something like that. – Connor Behan Aug 26 '21 at 13:10

0 Answers0