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Hawking radiation is frequently described as a form of black-body radiation, the same as non-black hole objects emit, except with the black hole's temperature defined solely by its mass.

But is this true? Thermal radiation is entirely electromagnetic radiation; just photons. But Hawking radiation, if I understand correctly, can be any type of particle. So is there actually a difference between the two, such that if you only had access to a measurement of the radiation itself, you could tell which type of object produced it?

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In everyday life thermal radiation means electromagnetic radiation simply because in everyday life thermal energies are too low to produce massive particles. The lightest massive particle (apart from neutrinos - more on this below) are electrons and their mass is 511keV. Using a naive calculation that equates $mc^2$ to $kT$ we find we'd need a temperature of over a gigakelvin before the thermal energy could start creating electrons. Not even the centre of Sun gets even remotely close to that hot.

Still, if we could generate temperatures that high we would start seeing radiated electrons as well as photons, so strictly speaking thermal radiation does include massive particles. The mechanism is simply that at such high temperatures the thermal velocities of charged particles are so high that their collisions create new electron positron pairs just like collisions in particle colliders.

We don't know the mass of neutrinos, though we expect them to be at most about $\frac{1}{10}$ eV. This isn't so far from everyday temperatures since room temperature corresponds to about $\frac{1}{40}$ eV, so in principle hot objects are radiating neutrinos as well as photons. The problem is that the coupling constant for neutrinos is so small that the probability of creating a neutrino is effectively zero.

The bottom line is that while it is indeed true that the Hawking mechanism produces thermal radiation, this doesn't just include photons because thermal radiation can include any particle that is in principle (if not in practice) produced by thermal motion. There is nothing unique about the radiation from a black hole that distinguishes it from any other kind of sufficiently hot system.

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    $\begingroup$ Does the smallness of the neutrino coupling constant(s) imply that the probability of creating a neutrino is effectively zero, or that the probability of observing one is? $\endgroup$
    – Martin C.
    Commented Aug 30 at 8:32
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    $\begingroup$ @MartinC. Both! Though I meant the probability of creating a neutrino is effectively zero. Then the probability that you will observe a thermal neutrino is 0² :-) $\endgroup$
    – John Rennie
    Commented Aug 30 at 8:54
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    $\begingroup$ @AgniusVasiliauskas the probability of creating a particle is of the order of $\exp(-E/kT)$ where $E$ is the total energy of the particle given by $E^2 = p^2 c^2 + m^2 c^4$. For photons $m=0$ so $E$ can be arbitrarily small. That's why black holes can create photons even though their temperature is very low. For electrons $E$ is always greater than $511$ keV and that's why black holes cannot create electrons unless their temperature is greater than a GK. The temperature is inversely related to the black hole mass, so black holes can have arbitrarily high temperatures by making them small. $\endgroup$
    – John Rennie
    Commented Aug 30 at 8:58
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    $\begingroup$ @IsaacKing Because the term Hawking radiation describes the mechanism by which black holes of any size radiate. It is not restricted to the subset of black holes astronomers observe while doing their day job. $\endgroup$
    – John Rennie
    Commented Aug 31 at 4:31
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    $\begingroup$ FYI your fraction for the upper limit of neutrino eV just displays as boxes, at least on my browser. $\endgroup$
    – RC_23
    Commented Aug 31 at 16:51
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There are a few instances where EM loses to other thermal radiation:

  1. Core collapse supernova (100 GK): 99% of the BB radiation is neutrinos. When you have $10^{57}$ nucleons in the volume of the size of Los Angeles$^\frac 3 2$, neutrino can escape.

  2. Heavy Ion Collision (1 TK?): These radiate thermal pions. Afaik, the lifetimes of the collisions are on the order of $10^{-21}\,$s, so electromagnetism is still warming up before it's all over (I think: has anybody seen BB $\gamma$'s from them?)

2.b) QGP? I suppose BB gluons don't go anywhere?

  1. As an afterthought: any system with quasi-particles radiating? Phonons perhaps?
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    $\begingroup$ It would only be possible to radiate phonons into a medium that can support phonons, which vacuum cannot. "radiating phonons" is essentially just a weird way of talking about conducted heat transfer. $\endgroup$
    – Hearth
    Commented Aug 30 at 20:31
  • $\begingroup$ @AXensen why is that relevant? The former is EM, the later is weak. 26 ns is a long time. $\endgroup$
    – JEB
    Commented Aug 31 at 1:27
  • $\begingroup$ @JEB sorry I misinterpreted. I thought you were claiming the lifetimes of the pions were $10^{-21}\,\text{s}$. I deleted my comment $\endgroup$
    – AXensen
    Commented Aug 31 at 6:35
  • $\begingroup$ @AXensen no, that was not well written on my part. I've been trying to find the lifetime of RHIC collisions as a side quest, w/o no institutional access--and yeah--I forgot about grammar, I just assumed "afaik" ruled out the sentence being about pions. $\endgroup$
    – JEB
    Commented Aug 31 at 18:19
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Roughly speaking, a system is said to be thermal if the probability density of a particle at energy $E$ is $p(E) = {\cal N} e^{-E/T}$ (in natural units), where ${\cal N}$ is some constant that we fix so that the total probability is 1. That's it. That's the definition of thermal. The idea comes from the Maxwell-Boltzmann distribution

As @JohnRennie mentioned in his answer, in everyday life, the particles involved in thermal radiation are typically photons, but there is no restriction that this be the case for all thermal systems.

For Hawking radiation, the particles whose distribution is thermal are gravitons.

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  • $\begingroup$ A thermal EM spectrum is Planckian correct? Which is $\sim E^3/(e^{E/T}-1)$. How does that relate to your expression? $\endgroup$
    – RC_23
    Commented Aug 31 at 21:32

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