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Statistical Mechanics - Physics at Oregon State University

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240 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.<br />

and the average dipole moment is<br />

which simplifies to<br />

or<br />

< P >= <br />

s1,···,sN<br />

< P >= N<br />

<br />

P (sj)P rob(s1, · · · , sN)<br />

j<br />

1<br />

eaE<br />

( 2ea)e+ eaE<br />

2ea)e− 2kT<br />

eaE<br />

eaE<br />

e + 2kT + e− 2kT<br />

2kT + (− 1<br />

< P >= 1<br />

Nae tanh(eaE<br />

2 2kT )<br />

which reduces for small fields, or eaE ≪ 2kT to<br />

Problem 7.<br />

< P >≈ NE a2 e 2<br />

The probability of finding<br />

<br />

a system in a st<strong>at</strong>e s is Ps. In this case the entropy<br />

of the system is S = −kB Ps log(Ps). Assume th<strong>at</strong> the system consists of two<br />

s<br />

independent subsystems and th<strong>at</strong> the st<strong>at</strong>e s is the combin<strong>at</strong>ion of subsystem 1<br />

being in st<strong>at</strong>e s1 and system 2 being in st<strong>at</strong>e s2. Use the formula for the entropy<br />

given above to show th<strong>at</strong> S = S1 + S2.<br />

The st<strong>at</strong>e of the system is given by the st<strong>at</strong>es of the subsystems, s = (s1, s2).<br />

Because the systems are independent we have Ps = Ps1Ps2 and hence:<br />

4kT<br />

<br />

<br />

S = −kB Ps log(Ps) = −kB Ps1Ps2 log(Ps1Ps2)<br />

With Psi = 1 this gives<br />

Problem 8.<br />

s<br />

s1,s2<br />

<br />

S = −kB Ps1Ps2 (log(Ps1) + log(Ps1))<br />

s1<br />

s2<br />

<br />

S = −kB Ps1 log(Ps1) + <br />

s1<br />

s2<br />

Ps2 log(Ps1) = S1 + S2<br />

The energy eigenvalues for a single particle in a given potential are ɛn. These<br />

energy values are independent of how many other particles are present. Show<br />

th<strong>at</strong> the partition function Z(T, N) obeys the rel<strong>at</strong>ion Z(T, N) = (Z(T, 1)) N .

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