HW3_solution

An Introduction to Thermal Physics

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PHY306 HW#3 Solutions February 23, 2009 Problem 2.5 To represent each microstate, a sequence of digits is used, for the number of energy units in the first, second, and third oscillators, respectively. (a) N=3, q=4 : (400) (310) (013) (220) (211) (040) (301) (103) (202) (121) (004) (130) (013) (022) (112) 15 microstates exist. This number can be calculated from ± 4 + 3 - 1 4 ² = 15 . (b) N=3, q=5 : 500 410 041 320 032 311 221 050 401 104 302 203 131 212 005 140 014 230 023 113 122 21 microstes exist. ± 5 + 3 - 1 5 ² = 21 . (c) N=3, q=6 : 600 501 015 042 141 033 132 060 150 420 204 114 321 213 006 051 402 024 330 312 123 1
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510 105 240 411 303 231 222 28 microstates. ± 6 + 3 - 1 6 ² = 28 . (d) N=4, q=2 : 2000 0020 1100 1001 0101 0200 0002 1010 0110 0011 10 microstes. ± 2 + 4 - 1 2 ² = 10 . (e) N=4, q=3 : 3000 2100 0210 0021 1110 0300 2010 0201 1002 1101 0030 2001 1020 0102 1011 0003 1200 0120 0012 0111 20 microstates. ± 3 + 4 - 1 3 ² = 20 . (f) If N = 1, then all the energy must belong to the one and only oscillator, so there’s only one microstate, which we would denote simple “q”. And according to the formula, the multiplicity should be ± q + 1 - 1 q ² = 1 . (g) If q = 1, then there’s only one unit of energy to distribute among the N oscillators, so the allowed state would be 1000 ··· ,0100 ··· ,0010
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HW3_solution - PHY306 HW#3 Solutions February 23, 2009...

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