# I Unruh effect, temperature and particle density...

1. Dec 2, 2016

### asimov42

Hi all,

This is a followup to a question I asked a number of years ago about the Unruh effect. I understand that an accelerated observer will see warm gas of particles following a blackbody distribution with some temperature T, where as an inertial observer would see none.

My question is: how does one determine the density of particles that the accelerated observer would be expected to see? I know the relevant distributions are Fermi-Dirac (for fermions) and Bose-Einstein (for bosons). Given that the range of particle momentums is continuous, would this imply that the accelerated observer would see an infinite density of particles? (that seems wrong) Or is it possible to compute an average density of particles?

As an example, given an observer accelerating at 1 g, what would be the expected density of particles seen by the observer due to the Unruh effect (roughly)? (clearly the temperature here would be very low)

Thanks!

2. Dec 2, 2016

### asimov42

Actually, perhaps the above is not the right question - since the number of particles will vary, I guess my real question is how to compute the energy density per unit volume of space filled with blackbody radiation at a specific temperature?

3. Dec 2, 2016

### Staff: Mentor

It should, since the same argument, if it were correct, would show that there is an infinite density of particles in the air that is surrounding you right now.

The simplest way to get an energy density, or at least an energy flux, would be to use a Planck black-body distribution for either bosons or fermions, as applicable, at that temperature.

4. Dec 2, 2016

### asimov42

Thanks PeterDonis - got it, sorry my thinking about about infinite number of particles was rather "dense" in retrospect Appreciate the reply!

5. Dec 2, 2016

### asimov42

One more quick question: from a previous post, The_Duck kindly answered my query about the contents of blackbody radiation:

"Particles besides EM radiation are present in blackbody radiation but in vanishingly small quantities. All particles get emitted, with probability roughly e^(-k * energy / temperature) where k is the Boltzmann constant. Since the minimum energy of, say, an electron is its rest mass, 511 keV, we should expect that electrons are emitted by blackbodies in roughly the same quantities as 511 keV gamma rays. But unless the temperature is really huge, the e^(-kE/T) probability factor above is vanishingly tiny for E this large. Roughly speaking, to emit a decent quantity of electrons you need temperatures of order (511 keV) / k = 6 billion Kelvin."

So, given the above, if I only want a rough estimate of the energy density per unit volume of space filled with blackbody radiation, which would include fermions, can I just use the Planck radiation law / the Stefan–Boltzmann law? Or is it really necessary to use the Fermi-Dirac distribution?

Thanks!

6. Dec 2, 2016

### Staff: Mentor

Yes, that will work fine.

7. Dec 2, 2016

### asimov42

Ok, last question, out of curiosity - if one wanted to calculate the true energy density per unit volume of space filled with blackbody radiation, including both bosons and fermions, would one simply sum the Bose-Einstein distribution contribution and the Fermi-Dirac distribution contribution? This seems like double counting... or can one consider a transition from the Bose-Einstein distribution and Planck's radiation law to use of the Fermi-Dirac statistics?

8. Dec 2, 2016

### Staff: Mentor

Not really, because as far as we know there are no massless fermions. (Neutrinos used to be thought to be massless, but now we know they actually have very small masses.) The black-body radiation formula assumes massless radiation, because if you are trying to radiate particles with mass, there is a minimum energy per particle required for the particle to exist at all, so the dependence of energy density on temperature changes. For massless radiation, each particle (that term has limits but it will do for here) can have an arbitrarily small energy; that's what the ordinary black-body formula's dependence on temperature assumes.

(Also, the ordinary black-body formula assumes that there is only one kind of boson--photons--but as far as we know there are no other massless bosons that could be radiated on macroscopic scales. The weak interaction bosons are not massless, and the strong interaction bosons, gluons, are confined the same way quarks are and can't be radiated.)

9. Dec 2, 2016

### asimov42

Hmm, ok, so I'm a bit confused - if particles besides photons are present in black-body radiation (at least in vanishing small quantities for reasonable temperatures), as the The_Duck said, then how does one actually calculate the energy density per unit volume under thermodynamic equilibrium, incorporating contributions from both bosons and fermions? The ordinary black-body formula can't be used, so what would be used instead?

If it's ok just to use the standard black-body formula as an approximation (for relatively low temperatures again), then the contribution from fermions and massive bosons (which can't radiate on a macroscopic scale) can be safely ignored?

10. Dec 3, 2016

### asimov42

Also, specifically for the Unruh effect, the accelerated observer should see a thermal bath - but I'm not clear if the black-body em curve is the right thing to use (alone) to determine things like energy density?

Lastly, even for the black-body em curve, at higher energies, wouldn't one expect pair production? So in some sense, the black-body em curve also results in fermions...

11. Dec 4, 2016

### A. Neumaier

Since this is about quantum gravity, gravitons also count. The original articles also consider neutrinos:
Page, Don N. "Particle emission rates from a black hole: massless particles from an uncharged, nonrotating hole." Physical Review D 13.2 (1976): 198.

12. Dec 4, 2016

### asimov42

Does this have any significant effect at all on the energy density? Shouldn't the black-body just radiate gravitons in accordance with the mass of the body and the 'mass' (energy) of other particles produced (photons, electrons, etc.)?

13. Dec 5, 2016

### Demystifier

Since the number of particles is not a conserved quantity, you have a canonical ensemble at finite temperature and zero chemical potential. Any textbook of quantum statistical mechanics can tell you the density in that case.