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I have read this question (in the comments):

Late response, but the energy really comes from the gravitational field, which exists outside the event horiz0n and in space. This field is what lets us measure the "mass" of the black hole. Loosely speaking, the field, not the mass, "is" the gravity of the black hole. Since there's energy in the field permeating a region near the horizon, pair-creation is possible by "borrowing" that energy to form particles. But if some of that energy escapes (the photons of the Hawking radiation), that energy cannot be returned to the field, so the field loses energy.

How does the Hawking Radiation mechanism cause a black hole to lose its mass?

And this one:

This paper discusses these issues in a fairly comprehensible way. Faraway observers (like your observer A) see thermal Hawking radiation with an effective temperature given by the Hawking temperature TH:=ℏc38πGMkB, where M is the black hole's mass. If an observer on a string is very slowly lowered toward the black hole (so that her dr/dτ is very small), then the effective temperature increases without bound and diverges at the horizon, so your observer B inevitably gets burned up.

Intensity of Hawking radiation for different observers relative to a black hole

So naively saying, the black hole loses mass (energy) by Hawking radiation in the form of photons that originates from the gravitational field of the black hole, and this radiation is observer dependent, ergo, the mass lost is observer dependent? So basically, different observers see the mass (energy) of the black hole decrease by different amounts (the amount of Hawking radiation they observe) so they calculate different amounts for the mass (energy) of the same black hole.

Question:

  1. If Hawking radiation is observer dependent then is the mass of a black hole too?
Qmechanic
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Árpád Szendrei
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  • Disregarding Hawking radiation, mass in GR can be tricky to covariantly define because mass=energy, and it's a very subtle question about how to disentangles the energy of matter/radiation from the energy within the gravitational field itself. See the wiki on the Landau–Lifshitz psuedo-tensor for more. But about black hole mass in particular, one usually needs an asymptotic boundary like those in Minksowki or AdS to rigorously define it. See ADM mass and Bondi mass. Coincidentally, the same is needed for Hawking temperature. – Michael C. Jan 25 '24 at 05:56

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Your question is simply about the gravitational red/blue-shift of the photons, for that alone you don't need a black hole or Hawking radiation. You could ask the same question about the sun whose photons are received redder far away and bluer close by, even if the red/blue-shift is smaller there the principle is the same.

So the answer is: that doesn't lead to a mass paradox at the sun, and it doesn't so at a black hole. Since the proper time τ of a close oberserver runs slower than the time t of the far away coordinate bookkeeper, the dM/dτ goes faster than the dM/dt, but that's just time dilation and doesn't lead to contradictions since you can always transform from one frame to the other and back.

Yukterez
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  • Thank you so much! – Árpád Szendrei Jan 25 '24 at 05:31
  • I think the question is not about the redshift, but about the fact that that a free falling and hovering observers see different amounts of the Hawking radiation, same as an inertial and accelerated observers see different amounts of the Unruh radiation. – safesphere Jan 25 '24 at 16:24
  • @safesphere - There are different interpretations about that, I can only talk about the common denominator from the relativistic perspective since the Hawking radiation still leaves a taste of 1+2+3+...=-1/12 in my mouth. Nevertheless, since the OP thanked me for my answer he still might have learned some basics which are also important. – Yukterez Jan 25 '24 at 19:56
  • @safesphere I have a new question on black holes: https://physics.stackexchange.com/questions/801287/can-a-single-black-hole-emit-gravitational-waves-and-evaporate – Árpád Szendrei Feb 12 '24 at 23:41
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Hawking radiation is not observer-dependent; it's a theoretical prediction that describes the gradual loss of mass and energy by black holes through quantum effects near the event horizon. However, the mass of a black hole is an intrinsic property and remains the same regardless of the observer's perspective.

If Hawking radiation is observer dependent then is the mass of a black hole too?

Faisal Ahmmed
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