Statistical Physics, Second Revised and Enlarged Edition

28 Two examples

T

2  3 

C

0.44 NkkkB

0 

Fig. 3. 3 The variation ofheat capacitywithtemperatureforaspin-^12 solid,showingthe peakofthe
Schottkyanomaly(see section 3.1.2).

atlow temperatures. It alsogoes to zero, albeit rather moregentlyas 1 /T^2 ,at
high temperatures. (Prove this last statement? Seen byexpandingthe exponentials
in(3.4).)

EntropyS Finallyweobtainthe entropyS. Againthisisathermalproperty only,
independent of the zero-point term. With section 2.5 before us we have threepossible
derivations. Method 1 can be used (together with Stirling’s approximation) to give the
answerfrom thedistribution numbers (3.2) without recourse to calculus.Alternatively
(method 2)one can obtainSfrom an integration ofCupfrom the absolute zero (since
S=0 there), i.e.S=

∫

0

∫∫

(C/T)dT. This is possible but not recommended. Instead let
us use method3. From (2.28) we writedown

F=−NkkkBTlnZ

=Nε 0 −NkkkBTln[ 1 +exp(−ε/kkkBT)] (3.5)

substituting (3.1)forZ. Again note the zero-point and thermal terms in (3.5).
The entropyis obtained byone differentiation, sinceS = −(∂F/∂T)V,N, the
differentiation being at constant energy levels and number. The result

S=NkkkB

{

ln[ 1 +exp(−θ/T)]+

(θ/T)exp(−θ/T)

[ 1 +exp(−θ/T)]

}

(3.6)

isillustratedinFig. 3.4.
Thehighandlow temperature values ofSare worthnoting.WhenTθ,Sisagain
frozen out exponentially towards the value 0 which it must attain eventually (the third
law!). AsT→0allthe particles enter the groundstate, andthe assemblybecomes
completelyordered(i.e.=1). Athightemperatures,however, the particles are
randomized between the two states,so that= 2 N,andSreaches the expected value
NkkkBln 2 (againlikethe penny-tossing problem). The approachto thehighTlimit
againgoes as 1 /T^2.