How can Hawking radiation with a finite (greather than zero) temperature come from the event horizon of a black hole? A redshifted thermal radiation still has Planck spectrum but with the lower temperature (remember CMB with temperature redshifted by expansion of the universe). Now, redshift at the event horizon is infinite (time is frozen for a distant observer) so the temperature of the radiation would be zero for him/her, that is no radiation is detected

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    $\begingroup$ The radiation comes from the region just outside the horizon. If you search the site I'm sure something like this has been asked before. $\endgroup$ May 15, 2013 at 17:53
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    $\begingroup$ related: physics.stackexchange.com/q/22498/4552 $\endgroup$
    – user4552
    May 15, 2013 at 19:01
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    $\begingroup$ @JohnRennie I see - so the radiation comes from the just outside the horizon and the redshift is quite large but finite, reducing the temperature to low values observed but still greater than zero. $\endgroup$
    – Leos Ondra
    May 16, 2013 at 6:53
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    $\begingroup$ It's a bit more complicated than that because the radiation isn't really emitted in the sense that a hot filament emits EM. A distant observer calculates the QFT vacuum to be different to an observer nearer the event horizon, and as a result the distant observer finds the "vacuum" near the horizon to contain a finite particle density - this constitites the Hawking radiation. I'm pretty certain (but wouldn't swear to it) that the Hawking calculation calculates the temperature at infinity. An observer falling freely into the black hole would not see any Hawking radiation even at the horizon. $\endgroup$ May 16, 2013 at 7:46
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    $\begingroup$ physics.stackexchange.com/q/30597 is related, though not a duplicate. I'm not sure I understand this area well enough to hazard an answer rather than just this comment :-) $\endgroup$ May 16, 2013 at 7:49

1 Answer 1


Oh, one of my favourite questions. Let me try to explain why black holes radiate:

To understand the Hawking radiation, you need to know about the Bogoliubov-Valatin transformation, which is often use to diagonalise Hamiltonians, and which was actually developed in regards of superconductivity and superfluidity. If you have the creation and annihilation operators $$a^\dagger$$ and $$a,$$ you can define new operators

\begin{align} b&=ua+va^\dagger\\ b^\dagger&=u^*a^\dagger+v^*a. \end{align}

Just defining new operators is fun, but only really helpful if certain conditions are met. The Bogoliubov-Valatin transformation is the canonical mapping from the set of $a$-operators to the set of $b$-operators:

\begin{align} \left[b,b^\dagger\right]&=\left[ua+va^\dagger,u^*a^\dagger+v^*a\right]\\ &=\left[ua,u^*a^\dagger+v^*a\right]+\left[va^\dagger,u^*a^\dagger+v^*a\right]\\ &=\left[ua,u^*a^\dagger\right]+\left[ua,v^*a\right]+\left[va^\dagger,u^*a^\dagger\right]+\left[va^\dagger,v^*a\right]\\ &=u^2\left[a,a^\dagger\right]+uv\left[a,a\right]+uv\left[a^\dagger,a^\dagger\right]+v^2\left[a^\dagger,a\right]\\ &=u^2\left[a,a^\dagger\right]+uv\cdot0+uv\cdot0-v^2\left[a,a^\dagger\right]\\ &=\left(u^2-v^2\right)\left[a,a^\dagger\right]\\ &=\left(u^2-v^2\right), \end{align} where we have to set $u$ and $v$ in a way that our transformation is indeed canonical, i.e. that $\left(u^2-v^2\right)=1$. This is a transformation of the phase space. This transformation can be used to transform one coordinate system to another one, coordinate systems which are accelerated compared to each other! If we have one observer in the past and another one in the future, both observing an area where in between them a black hole came into existence, their coordinate systems are accelerated (compared to each other), because the black hole curves space-time. The past-observer sees a vacuum, maybe a star that will become a black hole, but otherwise a vacuum. The future-observer sees a strongly curved space-time full of radiation. But why?

The reason is the uncertainty principle! We don't know the energy state of the vacuum, it depends on our coordinate system! I am not talking here about the position-momentum uncertainty, I am talking about the energy-time uncertainty! The higher the precision of your energy measurement, the higher the uncertainty of your time measurement. And because the vacuum can have different energy states, it can also spontaneously create particles - the Hawking radiation.

Summary: A black holes radiates, because different observers can observe different energy state of the vacuum around a black hole. This has to do with the coordinate system and the uncertainty principle. The vacuum energy depends on both.

So, the redshift precisely at the event horizon might be infinite, but not a bit away from it.

  • $\begingroup$ This comment comes late. I understand the first part of your answer but do not understand how the energy-time uncertainty relation comes in here. Thus how does it imply that `the redshift precisely at the event horizon might be infinite, but not a bit away from it'? Can you elaborate a bit? $\endgroup$
    – damaihati
    Jun 23, 2021 at 5:15

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