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A stationary observer near the black hole horizon will detect Hawking radiation proportional to their local acceleration.

If we call $\alpha=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-\frac{2GM}{r}}}$ the proper acceleration then the temperature measured is: $T=\alpha/(2\pi)$. In order to find the temperature measured at infinity we need to account for a redshift factor $\sqrt{1-\frac{2M}{r}}$ and set $r=2M$.

I have two questions:

  1. Why is the temperature zero if we take a stationary observer at infinity $\alpha(r\rightarrow\infty)=0$? Shouldn't we be able to obtain the result also this way?

  2. When including the redshift factor we then set $r=2M$. Why is this necessary? In other words, I don't understand why we must propagate the temperature observed near the horizon and not in any arbitrary distance from it?

Níckolas Alves
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George Fanaras
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    I think you might be confused. At infinity, you set $r \rightarrow \infty$, and the redshift factor becomes 1. The temperature is just the one you indicate. When you set $r=2M$ you ask what the temperature at the horizon is, in this case the redshift factor goes to zero and the temperature becomes arbitrarily high. – cesaruliana Jun 20 '22 at 01:17
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    Your expression is wrong. The actual temperature should have $\alpha = \frac{GM}{(2GM)^2} \frac{1}{\sqrt{1 - \frac{2 G M}{r}}}$ (units with $c=1$). I think the problem is somewhere in trying to just apply the Unruh effect in flat spacetime directly to a curved spacetime, but I can't think of a quick and intuitive explanation of why that isn't working. – Níckolas Alves Jun 20 '22 at 11:25
  • Related: Deriving Hawking temperature "from" Unruh effect and their relation – Níckolas Alves Jun 20 '22 at 11:27
  • Yes, I understand that. What I don't understand is why we have to first assume a stationary observer at the horizon and then redshift what he observes at infinity... – George Fanaras Jun 20 '22 at 16:50

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