1 SPA Homework solutions – week V
Marta Wac lawczyk & Maciej Lisicki
1. Based on the Boltzmann distribution P (E) = e−βE/Z(T, V, N ) calculate the internal energy U (T, V, N ) = hEi and its dispersion σ(U )2 = hE2i − hEi2 = kBT2Cv. Recall the definition of the heat capacity Cx= (δQ/dT )x at constant x and from the first law of thermodynamics, find the formula for Cv= (∂U/∂T )V,N.
Solution By definition, the average (internal) energy in the canonical ensemble is hEi =
Z
E(ΓN)ρ(ΓN)dΓN = 1 Z(T, V, N )
Z
Ee−βEdΓN, which can be written as
hEi = − ∂ log Z
∂β N,V For the fluctuations, we find
σ2E=E2 − hEi2=X
n
pnEn2− X
n
pnEn
!2
= P
nEn2e−βEn
Z − 1
Z2 X
n
Ene−βEn
!2
(1)
= −1 Z
∂
∂β X
n
Ene−βEn
!
−∂Z−1
∂β X
n
Ene−βEn
!
= ∂
∂β 1 Z
X
n
(−En)e−βEn
!
(2)
= ∂
∂β
∂ log Z
∂β = ∂2
∂β2(log Z) = − ∂
∂β hEi = − ∂
∂T hEi∂T
∂β = kT2CV (3)
Since CV ∼ N and hEi ∼ N , we have σE/ hEi ∼ N−1/2, so fluctuations decrease with increasing the system size and vanish in the thermodynamic limit of an infinite system.
5. Consider N magnetic moments, which have two allowed orientations ±µ in an external magnetic field B (the energy of each dipole can take values ±µB). Within the canonical ensemble, find the relative dispersion
of the magnetisation
σM
M =
q
hM2i − hM i2 hM i .
Solution Each moment can be oriented along the field or against it. The probabilities of these orientations are
p+ =eβµB Z1 , p− =e−βµB Z1
,
where Z1= 2 cosh βµB (= p++ p−). The moments are independent, so the average magnetisation is simply hM i =
N
X
i=1
hµii = N hµ1i = N µ(P+− p−) = N µ tanh βµB,
while the variance (using the independence of moments) is
σM =
* X
i
µi− hµii
!2+
=X
i
(µi− hµii)2 =X
i
(µ2i − hµii2) = N µ2 cosh2βµB.
