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How do experimental physicists measure the temperature in cold atomic gases?

+ 6 like - 0 dislike
425 views

In cold-atom experiment, physicists can realize a Bose-Einstein Condensate by lowering the temperature via laser trapping and magneto-optical cooling. How do physicists measure the temperature of the cold atom system? 

asked Mar 19, 2015 in Experimental Physics by Timothy (30 points) [ revision history ]
edited Apr 1, 2015 by Arnold Neumaier

2 Answers

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There are a variety of ``thermometers'' for cold atomic gases. The simplest method is to fit a Bose-Einstein (or Fermi-Dirac distribution) to the dilute ``wings'' of the cloud. The reason we have to use the dilute part is to ensure that the distribution is not modified by interactions among the atoms. The temperature is then determined by the fit. This method can be used for strongly interacting gases (because even in that case the interaction is small in the dilute region), but it is more difficult because the signal/noise is poor. An example of a more sophisticated thermometer for strongly interacting gases is the compressibility thermometer developed by the MIT group, see http://arxiv.org/abs/1110.3309.

answered Mar 20, 2015 by Thomas [ revision history ]
edited Apr 1, 2015 by Jia Yiyang
+ 2 like - 0 dislike

I happened to talk to a cold atom faculty in my department, indeed the most common method is like what Thomas said, measuring the distribution(though I have no idea what a "dilute wing"is). But I'd like to elaborate what this means.

Suppose the cold atom gas is trapped in a small region, and suppose the gas is dilute(which is not the case for BEC), this means there's almost no interaction among the atoms to each other. Now you suddenly remove the trapping potential, which means the kinetic energy stays the same for every particle, then just let the gas thermally expand. The particles with larger momentum travels faster, so after a period of time, the particle with the largest momentum will be at the outermost part of the gas, and more generally, an atom's momentum should be proportional to the distance to the original trapping potential(assume the gas has expanded to a large enough volume so that the original trapping potential can be viewed as a single point.) Now measure the density of the expanded gas with respect to the distance to the trapping potential, you will get a density distribution of atoms' momenta and hence the energies $\frac{p^2}{2m}$. If we assume the shape of the trapping potential is not too weird so that Virial theorem applies, we can legitimately claim the energy distribution of the gas in the trap is the same as the free particle energy distribution. Now just do a fit to Boltzmann/Bose-Einstein/Fermi-Dirac distribution, you will get the temperature.

For BEC where the dilute assumption fails, we can still let the gas thermally expand but one typically needs to run a computer simulation to take the inter-atom interactions into account(a detailed profiling of the trapping potential might also be needed).

answered Apr 1, 2015 by Jia Yiyang (2,465 points) [ revision history ]
edited Apr 1, 2015 by Jia Yiyang

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