2
$\begingroup$

I would like to know how many degrees of freedom an atomic gas has. What about the Bose-Einstein condensation from that gas?

$\endgroup$
7
  • 1
    $\begingroup$ presumably 6 per molecule $\endgroup$ Commented Apr 24, 2017 at 1:02
  • 1
    $\begingroup$ @ZeroTheHero Ok, but all of them are in the same quantum state. It suggests the whole system has only 6 degrees of freedom. $\endgroup$
    – peterh
    Commented Apr 24, 2017 at 1:14
  • 1
    $\begingroup$ @peterh Not in a gas... Put another way: there is a reason why statistical physics is statistical. Or maybe I just don't understand the first part of the question (which could very well be...) $\endgroup$ Commented Apr 24, 2017 at 1:20
  • $\begingroup$ @ZeroTheHero I agree, I think this question is not so clear as for the first spot. $\endgroup$
    – peterh
    Commented Apr 24, 2017 at 1:30
  • $\begingroup$ @ZeroTheHero Hydrogen is a diatomic gas. Therefore it has 6 degrees of freedom. What about his Bose- Einstein condensate? $\endgroup$ Commented Apr 24, 2017 at 1:52

1 Answer 1

1
$\begingroup$

It all depends on the type of gas and on its temperature.

If the gas is made of single atoms like $^{87}Rb$ or $^{39}K$ (typical atoms that are Bose-Condensed, since you are asking), then it as $3$ degrees of freedom - motion along $x, y,$ and $z$.

If the constituents are diatomic molecules, like $H_2$, then the two atoms could also rotate about the centre of mass, which introduces $2$ new degrees of freedom - clockwise, counterclockwise.

They can also vibrate, as if they were connected by a spring - this introduces $2$ new modes, for in-phase and out-of-phase vibration.

Evidence for the $3$ translational, $2$ rotational and $2$ vibrational degrees of freedom is in the plot of heat capacities (below), where it can also be seen that, when present, the latter two modes only get unlocked (unfrozen) at higher temperatures. This is because the energy associated with that motion is on the order of several $\gg k_B T$:

enter image description here

So even it you made a BEC of diatomic molecules, you will not see the rotational and vibrational modes, as you are too cold.
This is why a very active field is the one of cold molecules, where you bind two (cold) atoms together with, e.g., a Feshbach resonance, so as for the energy of the mode to be much lower and hence accessible.

$\endgroup$

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.