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From what I've read

  • The temperature of a black hole is measured by the amount of Hawking Radiation emitted.
  • For a photon to escape, it must travel perfectly perpendicular to the event horizon.
  • The larger the area of the event horizon, the more mass the black hole has.

What I do not understand is why "the larger the event horizon, the less paths there are that a photon could take."

https://phys.org/news/2016-09-cold-black-holes.html#:~:text=The%20larger%20the%20event%20horizon,energy%20to%20release%20these%20photons. (Link to where I read this)

Qmechanic
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Raul Bijy
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  • Related: https://physics.stackexchange.com/q/505044/2451 , https://physics.stackexchange.com/q/23099/2451 – Qmechanic Apr 26 '23 at 10:39

1 Answers1

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There are two things going on: photons (and other particles) forming due to the curvature of the spacetime, and how many of them get away to observers at infinity.

The first part is often popularly explained using particle-antiparticle pairs (an explanation mentioned in Hawking's original paper) but a more correct one is based on the Unruh effect: quantum fields subjected to strong accelerations (here by gravity) show a temperature, and this leads to particle production. This temperature is proportional to acceleration, and for black holes will be proportional to the "surface" gravity, making it scale as $T\propto 1/M$.

However, not all particles (or, strictly, their wave packets) get out since they get reabsorbed. This is determined by "greybody factors" that turn out to mainly depend on the wavelength of the particles (but also weakly on the radius), and this produces the overall power $\propto 1/M^2$. In a way these factors take the "lack of paths" into account by summing up the ways particles and waves can move to escape. But they are not calculated as actual trajectories but as solutions to the wave equations.

Anders Sandberg
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  • Thank you for your reply. Just to make sure I understand. The fewer paths a photon has to escape for a larger black hole results from fewer solutions to the wave equations? Would there happen to be a conceptual answer to this? (As my mathematical abilities are not that advanced) – Raul Bijy Apr 26 '23 at 15:43
  • @RaulBijy - The number of solutions of the wave equation has nothing to do with how difficult it is for the photon to escape. What the equation does is calculating the amount of incoming to outgoing radiation, giving a measure of how much gets trapped. You could say it is measuring all possible paths at the same time. – Anders Sandberg Apr 26 '23 at 17:00
  • Oh, so with larger black holes, the wave equations show there is more radiation getting trapped compared to smaller black holes? – Raul Bijy Apr 27 '23 at 14:35
  • @RaulBijy - The wave equation doesn't care about the absolute size of the hole, just the ratio between its size and the wavelength. – Anders Sandberg Apr 27 '23 at 22:30
  • Okay, hopefully, I can get it right this time. Greybody factors determine how many particles escape the black hole. But Greybody factors depend on the wavelength of the particles emitted. So what about the wavelength affects how many particles can escape? Does longer the wavelength mean less particles emitted and vice versa? Sorry if I'm not understanding, but thank you for responding. – Raul Bijy Apr 27 '23 at 23:55