What is the pressure inside of a black hole? Is it infinite or is it finite? Does it vary?
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3 Answers
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There is no pressure that we know of. In fact we really don't know what is inside a black hole (BH).
The classical solutions for BHs have a horizon (or two for the Kerr rotating BH solution) where the internal region is disconnected causally with the external region. In the internal region the spacetime is empty, nothing there, except at the singularity where the spacetime curvature becomes infinite.
Moreover, a person (or particle) going into the horizon (and in the coordinate frame of one going in, or the the particle's frame of reference, it does this in a finite period of time) does not see anything strange going on at the horizon (possible exception later in this answer), and inevitably falls into the singularity, and does that pretty quickly. The gravitational effect that observer experiences inside the horizon gets stronger till it becomes, in the classical solutions, infinite.
The exceptions, or caveats to this story, is that it does not take into account quantum gravity. We don't have an accepted theory of quantum gravity yet (we have some hypothetical theories, like string theory and loop quantum gravity), so as we get closer to the singularity General Relativity becomes invalid and we don't know yet what to have it take over. In fact, there are propositions that there's something called a firewall at the horizon, and everything is destroyed there. There's issues with the conservation of physical information in BHs, and some hypothesis are that the information gets frozen at the horizon, and kept there. This whole issue is in active current research.
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1$\begingroup$ Perhaps tidal forces could be regarded as a kind of pressure. If you fall into a black hole feet first, the different gravitational force on the different ends of your body will cause you to be stretched out and pulled thin like spaghetti. Google "spaghettification." $\endgroup$– JudyJun 29, 2017 at 11:00
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$\begingroup$ Spaggettification is the tidal forces, a grav field that has a gradient will do that. Pressure is not like that, just pushes or pulls, and it has to do with a field or matter. $\endgroup$– Bob BeeJun 29, 2017 at 20:11
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2$\begingroup$ Pressure is usually defined by force per area of surface. Since 'inside' a BH there's no physical dimension and therefore no surface, there's no way to define pressure. Actually, the inside of a black hole is nothing we currently understand. $\endgroup$– Zac67Jun 30, 2017 at 11:37
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It can be shown that the pressure of Schwarzschild black holes actually arises as a consequence of the quantum corrections due to local and non-local terms in the effective quantum gravity action.
For a Schwarzschild black hole it was found in $[1]$ that
$$P_{q,S} = − \gamma \frac{1}{2 G^4_N M^4}$$
where $\gamma$ is a constant appearing in the non-local part of the action.
However, charged (Reissner-Nordström) black holes do possess a classical pressure $[2]$, which is proportional to the charge, and hence is zero for Schwarzschild:
$$ P_{cl,RN}=-\frac{Q^2}{64\pi G^4_NM^4} $$
Quantum gravity effects modify this classical expression too. The full expression for the pressure was found in $[3]$ (with some approximations) and is
$$ P_{RN}=P_{cl,RN}+P_{q,S}+P_{q,RN}Q^2$$
where $P_{q,RN}$ is the a slightly more complicated generalisation of $P_{q,S}$:
\begin{multline} P_{q,RN}=\frac{1}{32G^5_NM^6}\Big[c_2+4c_3+2\beta(\gamma_E-4)+8\gamma(\gamma_E-5)+2(\beta+4\gamma)\ln(2G_NM\mu)\Big] \\+\frac{\pi}{9G^6_NM^8}\bigg\{54(\beta+4\gamma)\Big[c_1+2\alpha\ln\left(2G_NM\mu\right)\Big] +72(3\gamma_E-7)\beta\Big[c_2+2\beta\ln\left(2G_NM\mu\right)\Big]\\ +768(3\gamma_E-7)\gamma\Big[c_3+2\gamma\ln\left(2G_NM\mu\right)\Big] +6\Big[c_2\gamma(120\gamma_E-287)+3c_3\beta(40\gamma_E-91)\\ +160\beta\gamma(3\gamma_E-7)\ln\left(2G_NM\mu\right)\Big] +54\Big[c^2_2+4c_2\beta\ln\left(2G_NM\mu\right)+4\beta^2\ln^2\left(2G_NM\mu\right)\Big]\\ +576\Big[c^2_3+4c_3\gamma\ln\left(2G_NM\mu\right)+4\gamma^2\ln^2\left(2G_NM\mu\right)\Big] +360\Big[c_2c_3+2c_2\gamma\ln\left(2G_NM\mu\right)\\+2c_3\beta\ln\left(2G_NM\mu\right) +4\beta\gamma\ln^2\left(2G_NM\mu\right)\Big]+(\beta+4\gamma)\Big[36\alpha(3\gamma_E-7) \\+9\beta\big(8\gamma_E(3\gamma_E-14)+95+4\pi^2\big) +\gamma\big(192\gamma_E(3\gamma_E-14)+2095+120\pi^2\big)\Big] \bigg\}. \end{multline}
Here, $\alpha,\beta,\gamma$ belong to the non-local part of the action, $c_1,c_2,c_3$ belong to the local part.
The quantum corrections do not only modify the pressure, but also the entropy, the temperature etc. The effective field theory method for quantum gravity is useful because you can keep track of the quantum effects easily.
References
$[1]$ X. Calmet and F. Kuipers. Quantum gravitational corrections to the entropy of a Schwarzschild black hole. Phys. Rev. D 104 , 066012 (2021).
$[2]$Y.-H. Wei. Understanding first law of thermodynamics of black holes. Phys. Lett. B 672 98–100 (2009).
$[3]$ Campos Delgado, R. Quantum gravitational corrections to the entropy of a Reissner–Nordström black hole. Eur. Phys. J. C 82, 272 (2022).
answered Dec 20, 2022 at 15:57
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Black holes have quantum pressure, which is one of their fundamental characteristics along with others like their Hawking temperature, which may be expressed mathematically as follows: [1],
$$P = \frac {hc^9}{G^4m^4}$$
where all constants hold its usual meanings and $m$, $R$ are only variable used.
It denotes the quantum pressure of a black hole and corresponds to the Hawking temperature. Via this expression one can estimate its numerical value.
[1] Chandra, Kapil, Hawking Temperature and The Quantum Pressure of The Schwarzschild Black Hole (2022) http://dx.doi.org/10.2139/ssrn.4288217
