Unformatted text preview: 2π (2m)3/2 E 1/2 dE Assuming isotropy, we can write the volume element as: d3 p = 4π p2 dp In the nonrelativistic case: p2
E=
⇒ p2 = 2mE
2m
Differentiating both sides: √
2pdp = 2 2mEdp = 2mdE m
⇒ dp = √ E −1/2 dE
2m
combining these gives the form above (b) [10 pts] Enforcing the normalization f (E )d3 pd3 x = N and using the conversion from (a), show that the constant for the distribution is equal to (3/4)nεF
3/2 (3/8π)(2mεF)
3/2n, where n = N/V is the particle number density. Don’t forget about degeneracy! Writing out this integral by substituting in C for f(E), including a factor of 2 for degeneracy (spin), and substituting the momentum differential element as above, we can solve this integral: pF 3 pF 3 2Cd pd x = 2CV
0 0 = 2CV 2π (2m) d3 p F 3 /2 E 1/2 dE 0 8π
3 /2
CV (2m F )
=N
=
3
3
−3/2
(2m F )
⇒C=n
8π
(c) [15 pts] Using this distribution, find the average energy <E> per particle...
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 Fall '08
 Norman,M
 Physics, Work, Classless InterDomain Routing, dτ, the00, Kap, lim ξ

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