Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Statistical Mechanics: Cooling to Bose-Einstein condensate

  1. Oct 21, 2018 #1
    Hello,

    I have a question regarding the derivation for Bose Einstein condensation. I understand that in a boson gas for high temperatures the expectation value of the total number of particles should equal something like: $$ \langle N \rangle \sim T * \eta(z)$$ With ## z = exp(\frac {\mu} {k_b T})##, ##T## the temperature and $$ \eta(z) = \int_0^\infty \frac{x^2}{z^{-1}*exp(x)-1} \, dx$$ This function ##\eta(z)## increases for decreasing ##z## until it reaches a maximum at ##z = 1##. The graph looks like this:

    syDUVJG.png
    Now if I understand correctly, what you want to do is start cooling the system whilst keeping the number of particles constant. My question is, how? If you decrease ##T## then ##z## will increase thus lowering the value of ##\eta(z)##. Because of this the number of particles should always decrease right? I feel like I am missing something here so maybe someone could explain?

    Thanks!

    Edit: So I feel like I should clarify a couple of things. The above expression for ## \langle N \rangle## gives the number of particles in the excited state. In other words, all particles except the ones in the ground state. Now my lecture notes explain that one can lower the temperature whilst keeping the number of particles (in excited states) constant, up until the ##\eta(z)## maxes out at ##z=1##. After this point particles will start to move to the ground state which defines the critical temperature. I understand this, but still don't see how the number of particles can be held constant at all. Lowering ##T## also lowers ##\eta(z)## so number of particles will never be held constant right? Lowering the temperature immediately lowers the number of particles in the excited states.
     
    Last edited: Oct 21, 2018
  2. jcsd
  3. Oct 26, 2018 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted