I read an article about evaprotive cooling and there are two (or three) things that I do not understand.

1.) In the article it says that the cut energy $E_{cut}$ (the energy a particle needs to leave the trap) scales with the temperature, namely

$E_{cut}= \eta k_B T$.

This means that by lowering the temperature in the trap, the threshold that a particle has to overcome lowers itself. Does this mean that in "runaway Evaporation" one does not have to lower $E_{cut}$ (since it already lowers itself), but only has to accelerate the decrease of $E_{cut}$?

If I want to describe physical phenomena in daily life with evaporative cooling, I can't see why $E_{cut}$ should decrease with the temperature. If I want to describe the cooling of my tea in a Cup, I can't see any reason why $E_{cut}$ should decrease. Btw: What exactly is $E_{cut}$ in the latter example ?

2.) In the article it says as well, that the time for the rethermalization in the trap is essential for the evaporative cooling to work. That's why a huge particle density is desireable (--> the bigger the density, the higher the numer of collisions--> the shrter the rethermalization time). Rethermalization is process of shifting of the maxwell-boltzmann distribution to a lower velocity by collisions between the atoms in the trap.

There are so called good and bad collisions. Bad collisions are inelastic collisions and good collisions are elastic collisions. Bad or good collisions are called bad or good collisions since normally the atoms involved in the collisions leave or do not leave the trap, respectively.

I don't get why elastic collisions are good collisions or inelastic collisions are called bad collisions.

Per Definition in a elastic collision, there is no loss of energy. In contrast, in inelastic collisions there is an energy loss. Wouldn't it be better if there were a lot of inelastic collisions, but the atoms involved in the collisions do not leave the trap? If that was possible, the atoms would have less energy ( leads to temperature decrease) after the collision and the number of atoms in the trap would still be the same. If I'm not wrong, this should be the most desireable kind of collisions. So why don't they try to increase the number of these collisions?

thanks in advance for your help!


1 Answer 1


The $\eta$ parameter described in $E_{cut}$ is completely arbitrary and set by the method of evaporation. In practice, it's usually an RF field (sometimes called the RF "knife") tuned to eject all atoms above $E_{cut}$, and the experimentalist sets some sort of profile for how quickly $E_{cut}$ decreases by changing the RF frequency. It seems to work best to scale $E_{cut}$ with the temperature (with a constant $\eta\approx 5$) but that's fabricated, not natural. This has to be tuned depending on your trap geometry, evaporation coil, starting conditions, and so on. There is no analogous concept to $E_{cut}$ for evaporation in a coffee cup because that's just free evaporation, not forced by some energy-specific means.

You have the right definition of elastic collisions don't decrease the total energy of the atoms involved, and therefore they don't lower the temperature. Their function here, however, is to re-establish the Maxwell-Boltzmann distribution by rebuilding the high-energy tail, so that the forced evaporation has something to cut off. All of your cooling comes from throwing the highest energies away, but this process couldn't continue if the elastic collisions didn't happen often enough to rethermalize your cloud. People often compress their traps for evaporation to increase the density and elastic collision rate for better evaporation.

In inelastic collisions, you just lose atoms outright--usually this starts when two atoms form a molecule, and the released binding energy throws the molecule and another bystander atom out of the trap. (Yes, the molecule has lower overall energy, but usually the molecule either isn't trappable and falls out, or is not the right species for the experiment and is therefore useless.) That's called a three-body loss event, and it's bad because you often lose the low-energy atoms in this process; i.e., the cloud heats up. This particular loss mechanism gets worse at extreme densities, and sets a practical upper limit on how dense you can get the cloud to be.

So in summary, yes, evaporation is a lossy process, but you want to make sure you're losing atoms on the right side of the energy distribution. Good elastic collisions feed high energy atoms to the RF knife, while bad inelastic collisions throw cold ones out of your trap.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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