Lecture15

3 2 e 0 e e kt 1 3 2 e 12 de change

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Unformatted text preview: E(1,1,1) dominates the distribu/on i.e. for many parUcles, there is a huge preference for atoms to be in the ground state 5 Total number of bosons # in a small energy range dE dN dn = N ( E ) dE dE !N 1 1 # 2m & populaUon Density of states D( E ) = = 2% 2 ( !E V 4" $ ! ' per state at E V " 2m % N= 2$ ' 4! # ! & 3/ 2 ) *e 0 E ( E (µ )/ kT (1 3/ 2 E 1/2 dE Change integraUon variables from E to x=E/kT: N 1 " 2 mkT % n= = 2$ 2 ' V 4! # ! & 3/ 2 ) x dx * exp( x ( µ / kT ) ( 1 0 !µ / kT > 0 As T decreases, this factor decreases so the integral must increase. µ " 0 as T 6 0 " A problem with μ As µ ! 0 as T ! 0 we reach temperature TE where μ cannot decrease further. 1 " 2 mkTE % n= 2$ ' 4! # ! 2 & 1 " 2 mTE % = 2$ 2 ' 4! # ! & 3/ 2 3/ 2 ) y dy * exp[ y] ( 1 0 + (1...
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