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It was previously my understanding that the reason blackbodies only emit light was because light was the only massless particle, so there exist excitations of the electromagnetic field of arbitrarily low energy. And that if a blackbody somehow had a temperature near $500\text{ keV}$ it would also emit electrons and positrons.

However it occurs to me that in many quantum theories of gravity (string theory and perturbative quantum gravity, but not loop quantum gravity), there's a graviton which is also massless.

And if I assume that neutrino masses scale in a similar way to all the other standard model particles (successive generations have geometrically increasing masses), then the lightest neutrino is a few meV, or a few tens of kelvin times the Boltzmann constant. So anything with a few tens of kelvin of temperature might emit them by this argument. But there's no way to emit them without violating charge or lepton number conservation, so I think that's why neutrinos aren't blackbody radiated. Either that or the cross section for interaction with neutrinos is really small for everyday objects, so they cannot be said to be "in equilibrium with the neutrino field." (any comments on this?)

But I don't see any such argument applying to gravitons. Somehow this naiive argument is really, really wrong, because it implies that the blackbody emission of gravitons should roughly double the cooling rate of blackbody radiators (adding one more way they can radiate). This is obviously at odds with observations.

And before someone says "we don't have a theory of quantum gravity." I'm asking about a feature one would generically expect (or rather not expect) from any theory of quantum gravity. Or perhaps I'm asking why string theory and perturbative quantum gravity don't predict this.

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    $\begingroup$ It depends on what is meant by black body. The concept is usually introduced in respect to absorbing and emitting electromagnetic radiation, so it is not clear how it is meant to be applied for gravitational waves - I suggest clarifying the question. $\endgroup$
    – Roger V.
    Commented Apr 27, 2023 at 11:04
  • $\begingroup$ @RogerVadim I don't agree that this is a relevant distinction. Of course, the typical derivation of blackbody radiation is specifically for radiation of light. Obviously I mean the same exact concept, but applied to excitations of the gravitational field instead of excitations of the electromagnetic field. And there isn't a word for this because it isn't a thing (for a reason I'm asking for help understanding) so the best existing word for it is blackbody radiation. $\endgroup$
    – AXensen
    Commented Apr 27, 2023 at 11:18
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    $\begingroup$ Black body is usually defined as an object that absorbs all the radiation that is incident on it, so that it can come in equilibrium with this radiation. Note also, that this is a somewhat outdated concept - since nowadays we can derive black body radiation without resorting to the concept of black body at all. (See also this answer So it is important to clarify what you mean. $\endgroup$
    – Roger V.
    Commented Apr 27, 2023 at 11:35
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    $\begingroup$ As far as thermal coupling goes: heavy ion collisions emit thermal pions (with the appropriate generalized BB spectrum), and some large fraction of neutrinos emitted in neutron-star formation are thermal (and thus are equally distributed among the 3 flavors, unlike the inverse beta decay one). $\endgroup$
    – JEB
    Commented Apr 27, 2023 at 13:52
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    $\begingroup$ Are relic Big Bang GWs thermal? Maybe it's the only thing that can overcome the weak coupling problem. $\endgroup$
    – JEB
    Commented Apr 27, 2023 at 13:53

4 Answers 4

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There is thermal emission of gravitational waves for any body where the thermal fluctuations generate time-dependent quadrupoles, and that means almost any body in practice. However as Anders Sandberg notes, the emission rate is very small in practice owing to the weakness of the coupling constant.

As a result of the weak coupling (smallness of $G^3/c^5$) most bodies are not 'black' to gravitational radiation: they are very nearly fully transparent. One exception is a black hole (for wavelengths smaller that its radius).

The question asserted an intuition that for a true black body (i.e. fully absorbing for both types of radiation) in thermal equilibrium the contributions of electromagnetic and gravitational radiation to the emitted power would agree. I think this is plausible.

Finally, do black holes emit thermal gravitational radiation? The argument leading to Hawking radiation involves electromagnetic interactions in a curved space. It is not self-evident that the argument carries over to graviton emission, but it might. Hopefully someone else who knows may add an answer.

The Unruh effect is related to this. The field theory for gravitons is not quite the same as for photons, because gravitons have spin 2. But I would hazard a guess that one would find a gravitational Unruh effect.

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    $\begingroup$ Thanks for the excellent answer (and for previous great answers to my questions I believe?). I too was wondering about gravitational hawking radiation of black holes. Small note: $G^3/c^5$ is not yet unitless (although I see that its $10^{-73}$ in MKS). Any idea what relevant parameters of the "radiator" get added to make a dimensionless "small parameter"? $\endgroup$
    – AXensen
    Commented Apr 27, 2023 at 12:57
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    $\begingroup$ Don Page derived graviton emissions from Hawking radiation, finding that 2% of the mass would turn into gravitons: journals.aps.org/prd/abstract/10.1103/PhysRevD.13.198 $\endgroup$ Commented Apr 27, 2023 at 13:01
  • $\begingroup$ @AndersSandberg That number changes significantly if the black hole is rotating see Page's follow up: journals.aps.org/prd/abstract/10.1103/PhysRevD.14.3260 $\endgroup$
    – TimRias
    Commented Apr 28, 2023 at 13:20
  • $\begingroup$ No, this answer is not correct. A normal object is not internally at thermal equilibrium with the gravitational-wave degrees of freedom, so the radiation is not thermal and you cannot apply equilibrium thermodynamics to it. There are good reasons why you cannot do so. $\endgroup$
    – Void
    Commented Apr 28, 2023 at 15:36
  • $\begingroup$ @Void The term "thermal" does not necessarily imply equilibrium. Oscillations in ordinary bodies owing to their non-zero temperature produce grav waves. That, and that alone, is what is being asserted in the first paragraph. $\endgroup$ Commented Apr 28, 2023 at 17:31
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Yes, there should be gravitational blackbody radiation. Randomly oscillating particles in molecules have quadrupole momentum and should emit and receive gravitons... but the cross section is tiny: it has a prefactor on the order of $G^3/c^5$. This is why the cooling rate is not doubled. This also means that under current conditions equilibrium is not reached in normal systems, and usually it cannot be confined to a box.

This may have been different during the very early eras, producing primordial gravitational radiation with a blackbody spectrum (most primordial gravitational waves are expected from inflation, though, and has a power spectrum).

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  • $\begingroup$ Thanks for the answer and for the links to highly relevant papers! Any idea why, for electromagnetic blackbody radiation, the thing doesn't need to be confined to a box in any way for the equations to be valid? $\endgroup$
    – AXensen
    Commented Apr 27, 2023 at 11:34
  • $\begingroup$ The box is mostly a conceptual tool for deriving the equations for the properties of the field, and in the last step dropped. That is why it works on the outside. $\endgroup$ Commented Apr 27, 2023 at 12:57
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Good question

But I don't see any such argument applying to gravitons. Somehow this naiive argument is really, really wrong, because it implies that the blackbody emission of gravitons should roughly double the cooling rate of blackbody radiators (adding one more way they can radiate). This is obviously at odds with observations.

Electromagnetic waves (ie light) is emitted proportional to dipole fluctuations of electromagnetic charges (ie an electron moving back en forth); gravitational waves are emitted proportional to quadrupole fluctuations of a mass... as the gravitational force is much weaker than electromagnetism and microscopic particles are comparatively much stronger in charge than in mass, any realistic black body couples much stronger to the electromagnetic field than to the gravitational field (even without assuming any particular quantum gravity theory).

Actually, in this sense, it is indeed a somewhat similar situations to neutrino's (at least to my understanding) as neutrino's don't interact electromagnetically either; however neutrino's can still interact with matter through the weak force. For gravitons, there really would only be gravity left as an interaction channel, which effect is completely negligible in any regular black body setup.

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  • $\begingroup$ Thanks for the answer. Can you reconcile this with the fact that the blackbody emission rate isn't proportional in any way to the electric polarizability of the material or even $e$ the coupling constant between light and charged particles? Somehow the resolution has something to do with assuming the EM field is in equilibrium with your radiator, and therefore coupling is no longer needed to bring it into equilibrium. But I'm missing some details of the argument. $\endgroup$
    – AXensen
    Commented Apr 27, 2023 at 10:47
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    $\begingroup$ @AXensen I am not an expert, so I may be wrong; but from the top of my head I would think that in real life, getting to equilibrium with the EM field is something that happens fast compared to experimental timescale, so it is a useful approximation to study the equilibrium case only. For the gravitational case, such relaxation would take an extremely long time so that the equilibrium case never applies to any practical situation... $\endgroup$
    – Wouter
    Commented Apr 27, 2023 at 10:58
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Recall the essential steps that allow statistical mechanics to derive things such as black-body radiation and other thermodynamic phenomena:

  • There is an idealized system known as the microcanonical ensemble. This ensemble keeps its total energy fixed and its components (possibly amounting to an infinite number of degrees of freedom) interact only with itself and do not leave the system. However, it is tacitly assumed that the components do interact with itself enough to reach properties such as ergodicity or similar. In practice, this requires some sort of ability to impose "reflection" or restricting the spatial movement of the components trying to leave the system, and none of the components being essentially inert to the other.
  • Another step is the canonical ensemble, which is an idealized system that does not exchange particles (components) with its surroundings, while it is allowed to transfer energy to its surroundings. The canonical ensemble is assumed to be a part of a much larger microcanonical ensemble, and from this all its properties are derived. This gives rise to the notion of temperature, for instance.
  • Then there is the grandcanonical ensemble, whis can exchange both particles (in both directions) and energy while being a subsystem of a much larger canonical or microcanonical ensemble. This gives rise to the notion of a chemical potential.

Equilibrium thermodynamics assumes that we are able to somehow achieve at least some of these ensembles to some degree of approximation.


For example, electromagnetic black-body radiation is achievable because almost every molecule has an electric dipole moment, and radiation will react with it. Similarly, the free ions in plasma will generally scatter, absorb and re-radiate radiation very well. How much radiation interacts with a chunk of material is characterized by its optical thickness $\tau \in (0,1)$. Only bodies that retain a significant amount of a burst of radiation passing through them, $\tau \sim 1$, can be considered as reaching grandcanonical ensembles with EM radiation in their interiors. This is because energy gets exchanged at many points, and in many ways between the matter and radiation degrees of freedom. And only these bodies will give off a spectrum of photons that is close to the black-body radiation computed by the statistical-mechanics idealizations mentioned above. That is, you will get (roughly) the Planck-law distribution of radiation with temperature matching the Maxwell-Boltzmann distribution of the matter degrees of freedom near the surfaces of only optically thick media.

However, there are also optically thin media. One of them is air. Air would have to be closed in a cavity for a long time before it would reach equilibrium. In the vast majority of practical situations, a probe placed in the middle of air will not measure thermal radiation of the same spectral temperature as the surrounding air. We are thankful for this, since it is this property that also allows us to, like, see things in our everyday lives on Earth.

Many other interesting examples of optically thin media can be found in astrophysics. For example, the plasma in the accretion disk around the black hole in the center of our galaxy is assumed to have such a low density of particles that Coulombic collisions happen very little, radiation is produced rarely by the usual mechanisms, and the particles have a hard time staying in thermal equilibrium amongst themselves, let alone radiation. However, free electrons have a three orders of magnitude higher higher specific charge (charge per mass) than protons, and thus cool down separately by getting accelerated by random magnetic fields and produce synchrotron radiation. The result of this is a "two-temperature" plasma of electrons and ions of very different effective temperatures that produces EM radiation with a spectrum very far from the usual Planck's law. The radiation is random, has a broad distribution, but is not thermal in any sense of the word.


GW radiation should be likened to the second type of situations. The "GW thickness" (analogue of $\tau$) of essentially any known matter object is close to zero. GW radiation passes through matter almost freely, and the only way an object retains any of it is by having some sort of internal friction that captures the contractions or dilation due to the passage of the wave. (As in the famous sticky bead argument.) Let us compare the strength of interaction of a generic simple molecule (such as $\rm H_2 O$) with EM radiation and GW radiation. The dipole EM radiation power is $$ P = \frac{\mu_0 p^2 \omega^4}{6 \pi c}$$ where $p$ is the charge dipole moment and $\omega$ is the molecule rotation frequency. The size of a molecule is $L\sim$ Angstroms $ = 10^{-10}$ meters. The dipole moment is typically a $\sim 0,1$ fraction of the elementary charge times the size of the molecule. The typical rotation frequency will be roughly given by the fact that equipartition energy $k_{\rm B}T/2$ is equal to rotation energy $\omega^2 I/2$, where the moment of inertia is roughly the mass of the molecule times size squared and a geometric prefactor, or $\omega^4 \sim k_{\rm B}^2 T^2 m^{-2} L^{-4}$. Mass is roughly few times the proton mass. Omitting geometrical prefactors we get $$P \sim \frac{\mu_0 e^2 k_{\rm B}^2 T^2}{c \,m^2 L^2} \sim 10^{-20} \left(\frac{T}{100 \,\rm Kelvin} \right)^2 \,\rm W $$ You can see that this dipole will be able to radiate its rotational energy at the order $k_{\rm B} T/2$ in the order of $10^{-1}$ seconds at Earth-like temperatures ($T\sim 100$ K). To keep equilibrium, we would maybe require even stronger coupling, but it is admissible that this gas can be kept in equilibrium with EM radiation.

As for the GW radiation, the leading-order radiation is given by the quadrupole formula $$P \sim \frac{G}{c^2} \left(\frac{d^3 Q}{d t^3}\right)^2$$ where $Q$ is the molecule mass quadrupole. The quadrupole itself scales as $Q\sim m L^2$, and its third derivative will scale as $\dddot{Q} \sim m L^2 \omega^3$ with frequency being again $\omega^2 \sim k_{\rm B} T m^{-1} L^{-2}$. Now we have the GW power $$P \sim \frac{G k_{\rm B}^3 T^3 }{c^2 m L^2} \sim 10^{-69} \left(\frac{T}{100 \, \rm Kelvin}\right)^3\,\rm W$$ Again, we can estimate that at Earth-like temperatures this would take $10^{47}$ seconds or $\sim 10^{30}$ times the age of the Universe to radiate energy of order $\sim k_{\rm B}T$. You would get similar rates for transfers of a Planck-type spectrum of GWs back into the gas. (And notice also that the $\sim \omega^3$ power will also make the spectral shape different from that of a Planck black-body spectrum in any situation of interest.) So yes, reaching GW thermodynamic equilibrium with normal Earth-like systems is just not going to happen.


As for black holes and thermal GW radiation. If we trust tree-level quantization of gravitational perturbations, then yes, a black hole will emit a thermal-like spectrum of gravitons (with the appropriate corrections due to BH greybody factors, of course). However, depending on who you ask, they may not trust even that about quantum gravity.

Note also that one is not in thermal equilibrium with the surrounding vacuum, the leakage of Hawking radiation into the surrounding is a non-equilibrium process that should lead to a true global thermal equilibrium at late times. There is much discussion about this. There is one gravitational system that is in thermal equilibrium by construction, and that is a black hole in an asymptotically Anti-deSitter (AdS) space-time. The AdS space-time has its temperature and a reflective boundary that does reflect massless particles back and forth (a property set by hand, other boundaries are admissible), so that one can talk about the entire universe being a microcanonical ensemble. The black hole can then be the small sub-part, a canonical ensemble. But you can have more fun interpreting these kinds of space-times, see black hole chemistry as proposed by my colleague David Kubiznak and Robert B. Mann.

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  • $\begingroup$ You were late to the game but thanks for a really complete answer $\endgroup$
    – AXensen
    Commented May 3, 2023 at 14:17

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