Does spin affect the expression for a Bose-Einstein Condensate's critical temperature?
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In an assignment, we were asked to find the critical temperature of a collection of Rubidium-87 atoms. The answer used an expression derived for spin-zero bosons in Schroeder's Thermal Physics (which I have also found elsewhere online):
kBTc=0.527(h22πm)(NV)2/3kBTc=0.527(h22πm)(NV)2/3
Schroeder uses the expression g(ϵ)g(ϵ) to denote the density of states, and in derivation of the above expression, explicitly assumes the spin of the bosons to be zero. However, the spin of a Rubidium-87 nucleus appears to be 3/23/2, which combined with its valence electron, can either form a boson with spin 11 or 22. For spin SS, this would introduce a factor of (2S+1)(2S+1) into g(ϵ)g(ϵ), as the number of available states would increase by this multiple. Since N∝g(ϵ)N∝g(ϵ), this would decrease the value for TcTc by a factor of (2S+1)2/3(2S+1)2/3.
kBTc=0.527(h22πm)(NV)2/3kBTc=0.527(h22πm)(NV)2/3
Schroeder uses the expression g(ϵ)g(ϵ) to denote the density of states, and in derivation of the above expression, explicitly assumes the spin of the bosons to be zero. However, the spin of a Rubidium-87 nucleus appears to be 3/23/2, which combined with its valence electron, can either form a boson with spin 11 or 22. For spin SS, this would introduce a factor of (2S+1)(2S+1) into g(ϵ)g(ϵ), as the number of available states would increase by this multiple. Since N∝g(ϵ)N∝g(ϵ), this would decrease the value for TcTc by a factor of (2S+1)2/3(2S+1)2/3.
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Explanation:
Superheated Bose-Einstein condensate exists above critical temperature Apr 10, 2013 First observation of spin Hall effect in a quantum gas is step.
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