Homework 4 Solutions

These are all just moments of various forms of energy

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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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This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.

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