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Recently I heard a talk by Bill Phillips, who talked about the coldest temperatures in the universe.

Among others, he sayed that the coldest temperatures created at the moment are BECs, which can reach roughly 300 pK. Furthermore he mentioned, that in the future it might be possible to create BECs in space at a satellite, where one could reach up to 1 pK.

I don't understand why a BEC in space would be so much colder. The low pressure and temperature in space cannot be the reason I think, because this is not the case in a satellite (electronics etc. would not work otherwise). The only other thing that is different is that such BECs are in a free-falling environment. But I don's see how one could get a factor of 300 in temperature, given that the BEC on earth is anyway trapped, and gravity is very weak.

Does anybody know an explanation or a reference for such a statement?

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  • $\begingroup$ The temperature in space in vacuum in absence of radiation etc is $2.7 \text{K}$ due to the CMB, while on Earth it is $300 \text{K}$ on average. This makes it easier to reach lower temperatures in space, but I cannot argue about specifics in this case. $\endgroup$ – auxsvr Jul 12 '14 at 12:59
  • $\begingroup$ Thought about this aswell, but I dont think this can be the reason. You are talking about temperature-differences that are 12 orders of magnitude bigger than what is can be reached with BECs. But of course, I can also not disprove you. $\endgroup$ – NiceDean Jul 12 '14 at 13:19
  • $\begingroup$ auxsvr: Also, the temperature and pressure in a satellite is much higher than in space, which is another hint that this cannot solve the question unfortunatly. $\endgroup$ – NiceDean Jul 12 '14 at 13:21
  • $\begingroup$ @NiceDean: What you would be probably most interested would be the heat flux into your system which is classically of orders $\lambda \Delta T$ with $\lambda$ some heat transfer coefficient. Note, that we are in a quantum regime around $T\sim 0$, so there might be severe corrections to such a linear law. When you put a satellite say to L2 where the Planck mission took place, you indeed reach a temperature around a few $K$ "effortlessly"... $\endgroup$ – Void Jul 13 '14 at 8:11
  • $\begingroup$ ...So $\Delta T$ and thus the flux to be handled is an order smaller than on earth, so orders of $10\, pK$ would seem as an appropriate estimate. It might be that Bill Phillips is also extrapolating that technology will give us an extra order in the future. But I don't really know the technology and setup involved, so don't take my word for it. $\endgroup$ – Void Jul 13 '14 at 8:13
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The temperature of a BEC formed from a dilute atomic gas (e.g. Rb87) isn't determined by the ambient radiation field, as the vast majority of the photons don't interact with the atoms. BECs are also produced inside ultra high vacuum vessels, which have a vacuum much better than near-Earth orbit, so the ambient pressure isn't the reason either. A satellite-borne BEC experiment would also use a vacuum chamber similar to those on Earth.

Rather, on Earth, the main heating sources come from noise. In a BEC formed in an optical dipole trap, some of this is from intensity and position fluctuations in the laser beam, and in a magnetic trap, it can come from magnetic fluctuations due to e.g. a.c. power or vibrations in a spatially non-uniform magnetic field.

While in the best labs, these can be shielded to one degree or another, going to a satellite would pretty much eliminate them as heating sources, as you're no coupled to sources of vibration, and are far from sources of changing magnetic fields.

An additional consideration is that the best cooling of BECs has been achieved by weakening and expanding the confining trap both to adiabatically decompress the BEC and to allow for further evaporation. In gravity, there's a limit to how weak you can make the trap, because you still need to provide a force that overcomes gravity. In free-fall, this limitation is removed, and opens the possibility of producing very shallow traps.

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  • $\begingroup$ The reasons you mention make immediate sense. Thank you for this really good answer, after 2.5years :-) $\endgroup$ – NiceDean Jan 15 '17 at 12:28

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