Problem with Bose-Einstein Condensation

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  • Thread starter ShayanJ
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  • #1
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In section 7.1 of his statistical mechanics, Pathria derives the formula ## N_e=V\frac{(2\pi m k T)^{\frac 3 2}}{h^3}g_{\frac 3 2}(z) ## where ## \displaystyle g_{\frac 3 2}(z)=\sum_{l=1}^\infty \frac{z^l}{l^{\frac 3 2}} ## and ## z=e^{\frac \mu {kT}} ##. This formula gives the number of particles that are not in the ground state w.r.t. the temperature.
The maximum of ## g_{\frac 3 2}(z) ## happens at ## z=1 ## and is equal to ## \zeta(\frac 3 2) ##. So whenever ## z=1 ##, ## N_e ## reaches its maximum and any other particle has to go to the ground state and ## z=1 ## happens at any ## T<T_c ##.
My problem is with ## T>T_c ##. I can calculate ## N_e ## for any temperature which gives me the capacity of the excited states at the given temperature. Now I put ## N>N_e ## particles in the energy levels and so the excited states become full and the rest of the particles have to go to the ground state and I get Bose-Einstein condensation again, this time for ## T>T_c ## which can't be right because we're supposed to have condensation only for ## T<T_c ##.
What's wrong here?
Thanks
 

Answers and Replies

  • #2
954
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The critical temperature is dependent on the particle density, so if you keep the volume constant and add more particles, ##T_{c}## is going to change as well.
 
  • #3
2,791
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Yeah, good point. Thanks.
 

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