Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose you had an isolated cloud of gas at a low temperature in a vacuum with no external sources of radiation (e.g. no CMB). The gas would clearly cool via the emission of low-energy photons. But where do these photons come from? The thermal energy of the gas is simply manifest as elastic collisions between atoms where no energy should be lost. Do we have to appeal to the vacuum state of the electromagnetic field? But aren’t these virtual photons which could not be transformed into thermal radiation?

share|cite|improve this question
This follows up a previous question, "Quantum mechanics of thermal radiation":… – Nigel Seel Jan 18 '11 at 19:36
Classical answer is that collisions are connected with acceleration and both electrons and protons are charged. – mbq Jan 18 '11 at 20:03
If that were the case there would surely be no elastic collisions at all between atoms - ever? – Nigel Seel Jan 19 '11 at 9:10
up vote 2 down vote accepted

The particles of the gas undergo a process called radiative collision. For example, two atoms collide, and you end up with the same two atoms plus one or more photons.

A Feynman diagram for this process would have two atoms (complicated bound states of nucleons and electrons) coming in, and the same two atoms plus some photons coming out. It should be easy to see that such Feynman diagrams are possible. (And just as for any other scattering process, the sum of the amplitudes of all possible diagrams is related to the cross section.)

share|cite|improve this answer
So you are saying that in the simplest case of two atoms colliding at arbitrarily low velocity there is always a non-zero amplitude for photon emission, right? Hence there are no elastic collisions in QFT? – Nigel Seel Jan 19 '11 at 9:32
Right. For photons of any definite energy, the cross section goes to zero as the initial kinetic energy goes to zero (as it must by conservation of energy), but as the photon energy goes to zero the cross section actually diverges. This is known as "soft bremsstrahlung" and must be included in QFT radiative correction calculations because it cancels out a divergence involving low-energy VIRTUAL photons. Basically if the photon energy is too low you can't tell if it's virtual or real because you can't observe it. Read about it here: – Keenan Pepper Jan 19 '11 at 18:12

You need to be a little clearer about your assumptions. You haven't said what holds the cloud of gas together. Is it contained in a zero temperature transparent box - e.g., a very deep non-rel. QM deep square well? What is the gas made of - atoms?

If it's atoms confined by a transparent box that only interacts with them by reflecting them, then the cooling would have to proceed by exciting internal electronic states of the atoms in occasional inelastic high-energy pair collisions, followed by energy loss via electromagnetic decay. No QFT required.

share|cite|improve this answer
"Exciting internal electronic states" doesn't have to happen. The intermediate states could be short-lived virtual states rather than real states, which is the same thing as saying there's radiative collision going on. – Keenan Pepper Jan 18 '11 at 23:40
It's not necessarily an equilibrium state. We could imagine a large cloud of, say, atoms of a noble gas - no molecules to worry about, and just consider the cooling of a deeply interior part in dt (e.g. the first microsecond). Then the question is: why is d(temperature)/dt not zero? – Nigel Seel Jan 19 '11 at 9:13
We can arrange matters that the collision energy is too low to kick atoms up to their first excited state with any appreciable probability (necessary so that they don't drop down again with spontaneous emission of a photon and so create thermal radiation in that way). Basically this is a question of whether elastic collisions are even possible in QM or whether there is always an amplitude to emit photons and lose KE. – Nigel Seel Jan 19 '11 at 9:47
I agree with Keenan below. QFT does get you extra inelastic modes. – Foster Boondoggle Jan 19 '11 at 15:02

I have been pondering this question some. I would answer yes, if I understand the question in one sense, but if I interpret the question another way I am tempted to say no. The vacuum is a set of quantum occupation sites $\{|0_k\rangle\}$. A hot body usually emits a blackbody spectrum of particles which fill these slots in a thermal distribution. So one can say the vacuum cools a body by permitting the emission of radiation This is pretty standard fair.

The one thing which I am pondering is whether this question is meant to ask whether the vacuum can absorb thermal energy in some way and remain a vacuum. In the above example one has a vacuum plus a hot body, and the vacuum transforms into a set of boson states plus vacuum plus a cool body. So in a box you end up with radiation and not a “pure vacuum.” A vacuum can change its energy vev. After all that is what all of this business of Higgs symmetry breaking and inflation and …, is all about. So I might interpret the question as asking whether the thermal energy of a body can be absorbed by the vacuum so it changes its vev. However, the vacuum is a zero temperature state. So it is hard to see how one can cool an object with the vacuum without radiation emitted. This also appears to be a violation of thermodynamics.

There appears to be a loop hole, but I think it can be closed. Suppose we have a hot body with a temperature $T$ and we accelerate it to $a$. On the frame of the accelerated body there is a temperature $T’~=~2\pi a$ from the horizon, called Unruh radiation, which bounds its Rindler wedge. On the accelerated frame the pure vacuum observed in an inertial frame becomes a vacuum plus a thermal distribution of boson states. So it is tempting to think one could accelerate a body with $a$ so that for $T’~<~T$ and the radiation from the body crosses the Rindler horizon with a rate of energy flow that is faster than the energy it emits. This state of affairs approximates the situation for a body accelerated near the event horizon of a black hole. So for this hot body suspended above a black hole with a temperature $T~>~2\pi a$ the black hole will absorb more energy than it emits, as observed on that frame.

So what is the problem? For one the limiting case with $a~\rightarrow~0$ recovers the case above. So the vacuum does not absorb this energy and change the vacuum vev. The other problem is that an inertial observer does not detect Unruh radiation. The body will absorb this radiation, but the inertial observer still detect nothing. The reason is the time dilation slows down or redshift any radiation the body absorbs so as to cancel out any apparent heating up. A body then will not be observed to actually heat up, or if it is hotter than the Unruh radiation it is not observed to be influenced by any exterior radiation source.

What has bothered me is the near horizon approximation for a particle near a black hole. For the accelerated observer the life time of the exterior world races by in a flash. For a stellar mass black hole it requires billions of g-forces to remain a few meters from the horizon, and to get within centimeters requires about a billion billion g's of acceleration. If by some means an observer could do this the outside world would be racing by, say for a small proper time with $t~=~g^{-1} cosh(gs)$. So the proper time element is $s~\simeq ~ g^{-1}ln(gt)$ for a time unit $t$ outside. As a result for t the lifetime of the black hole $\sim~ 10^{67}$year, $g$ in units of distance $\sim~ 1cm ~ 10^{-10} sec~\sim~ 10^{-17}$year the proper time the observer on the accelerated frame observes the BH to evaporate is $$ s~\sim~ 5\times 10^{18} years. $$ This is much shorter than the BH life time measured by the exterior world. Assume you get that acceleration up to $10^{33}cm^{-1}$ or $10^{43}sec^{-1}$ or $10^{50}year^{-1}$, then you are hovering practically on the stretched horizon. The BH evaporates in about $10^{-42}$ seconds, or close to the Planck unit of time! Bang!; which means all that ingoing and outgoing radiation which interacts with the black hole hits you at once is a colossal thunderclap. The event horizon appears for larger g close in to be more of a singularity, or a surface region of huge energy density that is radiating and absorbing energy at a ferocious rate.

So I think the result hold for a body suspended near a black hole as well. The above argument means the same time dilation for the absorption and emission of radiation will hole in the BH case, just as with the accelerated frame case. This question is interesting for this reason, and frankly this may still bother me for a while. Yet I feel safe in saying that the vacuum will not cool a body by adjusting its vev in some way and without the emission of radiation.

share|cite|improve this answer
Some fairly mind-blowing scenarios! For what it's worth my personal intent with this question was to understand how low-energy, apparently elastic collisions, between atoms can nevertheless lead to the emission of one or more photons and a decrease in the KE of the colliding atoms (= cooling). I couldn't understand the precise mechanism for photon emission. I'm awaiting clarification of "radiative collision" as the relevant Feynman process as suggested in some of the other answers. – Nigel Seel Jan 19 '11 at 9:42
The question you raised was odd in how it was phrased, and struck me as having some odd meanings. A body with excited states is a set of separable states. However, the huge vacuum of occupation states for photons results in spontaneous emission. If the excited atoms in the body could be characterized and in a perfectly reflecting box, then the photons emitted would be in entanglements with the atomic states. In principle things would be reversible. The reservoir of photon states is large so the system is not reversible FAPP. The physics is approximated by Fermi’s golden rule. – Lawrence B. Crowell Jan 19 '11 at 13:51

A basic part of the theory is to remember the Boltzmann distribution: N = exp (-E/kT) for each energy level E, where N is the occupation number and T is the Temperature (I am omitting subscripts). So for a given T (=/=0) not everything is at the ground state. These nonground state atoms can decay by spontaneous emission as stated in other Answers.

Spontaneous Emission does have a modern (post Einstein) explanation in terms of the Quantum Vacuum however, which might also interest you.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.