Do the laws of thermodynamics hold true everywhere in universe ?

What about black holes and worm holes ?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/204099 $\endgroup$
    – Kyle Kanos
    Oct 5, 2015 at 20:35
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    $\begingroup$ As far as I know, there's a good chance wormholes don't and can't exist, which makes that part of the question rather Moot. It's a good question for black holes though. $\endgroup$
    – userLTK
    Oct 5, 2015 at 23:43
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    $\begingroup$ Black holes do not violate second law of thermodynamics because black holes eating more matter or energy get more surface area. This also hints to the idea that all the information in a black hole is stored on its surface area. $\endgroup$ May 20, 2020 at 20:27

1 Answer 1


No, black holes do not violate the 2nd Law of Thermodynamics.

Imagine that we want to violate the 2nd Law of Thermodynamics by throwing some volume of ideal gas into a black hole. This would seem to violate the 2nd law because when it is outside the black hole the ideal gas contributes some calculable amount of entropy to the total entropy of the universe. Once the gas crosses the event horizon of the black hole, however, it is hidden from all observers who are outside the black hole. We conclude then that in this process \begin{equation} dS_{universe} < 0. \end{equation} That is, we have decreased the entropy of the universe. This is a big problem and we need a way to resolve it. The answer is to assume that black holes carry entropy.

An important mathematical theorem in General Relativity is the Area Theorem, which states that the surface area of a black hole's event horizon can never decrease (given reasonable physical assumptions). So \begin{equation} dA \geq 0. \end{equation} In the early 1970s, Jacob Bekenstein had the insight that he could relate the area of the event horizon of a black hole to another quantity that we know can never decrease: entropy. By dimension analysis, Bekenstein wrote down the equation for black hole entropy: \begin{equation} S_{BH}=\alpha\frac{k_BAc^3}{G\hbar}, \end{equation} where $\alpha$ is some dimensionless constant (which Hawking proved is equal to exactly $1/4$).

Now, let's return to the problem of throwing an ideal gas into a black hole. We know that the mass of the black hole will increase, hence the area of the event horizon will increase. If, for example, our ideal gas also carries charge and angular momentum, then the charge and angular momentum of the black hole will change as well. We can relate the change in the area of the event horizon $dA$, the change in the mass of the black hole $dM$, the change in the charge of the black hole $dQ$, and the change in the angular momentum of the black hole $dJ$ via \begin{equation} \frac{\kappa}{8\pi}dA=dM + \Phi dQ - \Omega_H dJ, \end{equation} where $\kappa$ is the surface gravity of the black hole, $\Phi$ is the electrostatic potential of the black hole horizon, and $\Omega_H$ is the angular velocity of the event horizon. This equation should look familiar. It is the 1st Law of Black Hole Thermodynamics, and it is directly analogous to the ordinary 1st law.

We can write down a generalized 2nd Law of Thermodynamics:

\begin{equation} S_{tot} = S_{universe} + S_{BH}, \end{equation} so that $dS_{tot} \geq 0$, even when we consider processes involving black holes.

This is far from the end of the story, and many interesting questions abound. For example, the entropy of a thermodynamic system is usually explained as characterizing the number of accessible microstates of the system. If black holes have a well defined entropy, then it should be possible to express this entropy as the logarithm of the number of accessible black hole microstates. In 1995, Andrew Strominger and Cumrun Vafa showed that by counting certain states of black holes in string theory, they could correctly reproduce the Bekenstein-Hawking entropy of those special types of black holes.

  • $\begingroup$ Doesn’t Hawking radiation imply that black holes do decrease surface area? $\endgroup$
    – profPlum
    Jun 29, 2022 at 13:03
  • $\begingroup$ My understanding was that it meant the Mass of a black hole would decrease over time? Wouldn’t that cause surface area to go too? $\endgroup$
    – profPlum
    Jun 29, 2022 at 13:04

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