Statistical Mechanics - Homework Assignment 5
Alejandro G´omez Espinosa∗
April 28, 2013
Pathria 8.2 For a Fermi-Dirac gas, we may define a temperature T0 at which the chemical potential of
the gas is zero (z = 1). Express T0 in terms of the Fermi temperature TF of the gas. (Hint: Use
equation (E.16).)
Let us start with Equation (8.1.4): N V = g λ3f3/2(z) = (2πmkT )3/2 h3 gf3/2
where T = T0 as defined in the problem. Then, solving for T0:
T0=
N2/3h2 2πmk(gV f3/2(z))2/3
(1) Then, the Fermi temperature is defined by TF = εkF where the Fermi energy εF is given by the
equation (8.1.24): εF = 3N 4πgV 2/3 h2 2m (2)
Hence, comparing (2) and (1): T0 = N2/3h2 2πmk(gV f3/2(z))2/3 = 4π 3f3/2(z) 2/3 3N 4πgV 2/3 h2 2mπk = 4π 3f3/2(z) 2/3 F πk = 4 3√πf3/2(z) 2/3 TF (3)
Now, let us calculate the factor f3/2(z). Since µ = 0, therefore z = eµβ = 1, thus
f3/2(z = 1) = 1 Γ 32 Z ∞ 0 x1/2 ex+ 1dx (4)
That is an integral easy to calculate using equation (E.16): 1 Γ (j + 1) Z ∞ 0 ηj eη+ 1dη = 1 − 1 2j ζ(j + 1) ∗ [email protected]
Hence, equation (4) become: f3/2(z = 1) = 1 Γ 32 Z ∞ 0 x1/2 ex+ 1dx = 1 −√1 2 ζ 3 2 ≈ 0.77 (5)
Plugging this result into (3), we got the final result:
T0= 1.73 √ π 2/3 TF = 0.98TF
Finally, using this result, we combined with knowledge of the low and high temperature limits, to sketch µ as a function of temperature in Figure1.
Figure 1: Sketch of µ/F as function of kT /F using the relation µ
F =
kT F ln e
N/gkT − 1 where, in this
Pathria 8.13 Show that, quite generally, the low-temperatre behavior of the chemical potential, the spe-cific heat, and the entropy of an ideal Fermi gas is given by
µ ' εF " 1 −π 2 6 ∂ ln a(ε) ∂ ln ε ε=εF kT εF 2# (6) and Cv ' S ' π2 3 k 2T a(ε F) (7)
where a(ε) is the density of (the single-particle) states in the system. Examine these results for a gas with energy spectrum ε ≈ ps, confined to a space of n dimensions, and discuss the special cases: s=1 and 2, with n = 2 and 3. (Hint: Use equation E.18)
Let us start with equation (8.1.20):
N = Z F
0
a()d (8)
where the density of states a() is given by: a()d = g()f ()d where f () = (e(−µ)β+1)−1. Then,
N = Z F 0 g()d e(−µ)β+ 1 = 1 β Z F 0 g(βx)dx ex−µ/β+ 1 if x = β (9)
This integral is similar to equation E.18 given in the book: Z ∞ 0 ψ(x)dx ex−ε+ 1 = Z ε 0 ψ(x)dx +π 2 6 dψ dx x=ε + ... (10)
Using this relation into (9):
N = 1 β Z µ/β 0 g(βx) dx + π 2 6 dg(βx) dx x=µ/β + ... ! = 1 β µ β − ε0+ π2 6 dg(βx) dx x=µ/β + ... ! µ ' εF " 1 −π 2 6 ∂ ln a(ε) ∂ ln ε ε=εF kT εF 2#
Now, for the heat capacity at constant volume we need to calculate the internal energy first: U = Z εF 0 a()d = Z εF 0 g() e(−µ)β+ 1d
that is very similar of the result of the chemical potential. In a good approximation: U ' π 2 6 dg(ε) d(/β) (kT )2 F (11) and for the chemical potential:
Cv = ∂U ∂T V = ∂ ∂T π2 6 dg(ε) d(/β) (kT )2 F ' π 2 3 k 2T a( F) (12)
In the case that ≈ ps, the density of states is given by: a() = gV h34p 2f ()dp = gV 2π2 2m ~ 3/2 1/2 e(−µ)β+ 1d = gV 2π2 2m ~ 3/2 ps/2 e(ps−µ)β + 1sp s−1dp = gV s 2π2 2m ~ 3/2 p3s/2−1 e(ps−µ)β + 1dp
Then, we only need to calculate the term inside the parentesis for each case. For s = 1: a() = gV 2π2 2m ~ 3/2 p3/2−1 e(p−µ)β+ 1dp dg(βx dx = gV 2π2 2m ~ 3/2 d dβp 1/2 = gV 2π2 2m ~ 3/2 d dp βp1/2 = gV 4π2 2m ~ 3/2 βp−1/2
and the thermodinamic variables are: µ ' εF " 1 −π 2 6 gV 4π2 2m ~ 3/2 βp−1/2 kT εF 2# Cv ' S ' π2 3 k 2TgV 4π2 2m ~ 3/2 βp−1/2 For s = 2: a() = gV 2π2 2m ~ 3/2 p6/2−1 e(p2−µ)β + 1dp dg(βx dx = gV 2π2 2m ~ 3/2 d dβp 2 = gV 2π2 2m ~ 3/2 d dp β p1/2 2p ! = −gV 8π2 2m ~ 3/2 βp−3/2
and the thermodinamic variables are: µ ' εF " 1 +π 2 6 gV 4π2 2m ~ 3/2 βp−3/2 kT εF 2# Cv ' S ' π2 3 k 2TgV 4π2 2m ~ 3/2 βp−3/2
Pathria 10.5 Show that the first-order Joule-Thomson coefficient of a gas is given by the formula ∂T ∂P H = N Cp T∂(a2λ 3) ∂T − a2λ 3 (13) where a2(T ) is the second virial coefficient of the gas and H its enthalpy; see equation (10.2.1).
Derive an explicit expression for the Joule-Thomson coefficient in the case of a gas with interparticle interaction u(r) = +∞ for 0 < r < D, −u0 for D < r < r1, 0 for r1 < r < ∞, (14) and discuss the temperature dependence of this coefficient.
Let us calculate the JT coefficient: ∂T ∂P H = − ∂T ∂H P ∂P ∂H T = − ∂H ∂P T ∂H ∂T P = − 1 Cp ∂H ∂P (15) Then, we know that:
dH = CpdT + V 1 − T V ∂V ∂T P dP (16) therefore: ∂H ∂P T = V 1 − T V ∂V ∂T P (17) Plugging (17) into (15): ∂T ∂P H = − 1 Cp V 1 − T V ∂V ∂T P = 1 Cp T ∂V ∂T P − V (18) Now, let us calculate the partial derivative in the RHS using equation (10.2.1):
P v kT = ∞ X l=1 al(T ) λ3 v l−1 P = kT v ∞ X l=1 al(T ) λ3 v l−1 P = N kT V + N kT V2 a2(T )λ 3+ .... dP = 0 = N k V dT − N kT V2 dV + N k V2 a2λ3− T ∂(a2λ3) ∂T dT − 2N kTa2λ 3 V3 dV + ... N k V 1 + 1 V a2λ3− T ∂(a2λ3) ∂T dT = N kT V2 1 + 2a2λ 3 V dV ∂V ∂T P = V T 1 + 1 V a2λ3− T ∂(a2λ3) ∂T 1 + 2a2λ 3 V −1 ≈ V T 1 + 1 V a2λ3− T ∂(a2λ3) ∂T 1 − 2a2λ 3 V + ... ≈ V T 1 + 1 V a2λ3− T ∂(a2λ3) ∂T − 2a2λ 3 V = V T 1 −a2λ 3 V + T V ∂(a2λ3) ∂T
Plugging this result into (18), we found: ∂T ∂P H = 1 Cp T ∂V ∂T P − V = 1 Cp T V T 1 −a2λ 3 V + T V ∂(a2λ3) ∂T − V = 1 Cp V − a2λ3+ T ∂(a2λ3) ∂T − V = 1 Cp T∂(a2λ 3) ∂T − a2λ 3
Now, let us discuss this expression for a gas with interparticle interaction. The term a2 is given by
equation (10.3.1): a2= 2π λ3 Z ∞ 0 1 − e−u(r)β r2 dr (19)
For the case when 0 < r < D:
a2 = 2π λ3 Z D 0 1 − e−∞β r2 dr = 2π λ3 Z D 0 r2 dr = 2π λ3 r3 3 D 0 = 2πD 3 3λ3
and the JT coefficient: ∂T ∂P H = N Cp T∂ 2πD 3/3 ∂T − 2πD3 3 ! = −2πD 3N 3Cp
For the case when D < r < r1:
a2 = 2π λ3 Z r1 D 1 − eu0βr2 dr = 2π λ3 1 − eu0β Z r1 D r2 dr = 2π λ3 1 − eu0β r3 3 r1 D = 2π 3λ3 1 − eu0β(r3 1− D3)
and the JT coefficient: ∂T ∂P H = N Cp T∂ 2π 3 1 − e u0β (r3 1− D3) ∂T − 2π 3 1 − eu0β(r3 1 − D3) ! = N Cp −2πT 3 (r 3 1− D3) ∂ eu0/kT ∂T − 2π 3 1 − eu0β (r13− D3) ! = 2πN 3Cp (r31− D3)u0 kTe u0β−1 − eu0β
For the case when r1 < r < ∞: a2 = 2π λ3 Z ∞ r1 1 − e0 r2 dr = 0
and the JT coefficient:
∂T ∂P
H
= 0
From the expressions above, the only temperature dependence of the JT coefficient is when D < r < r1. Since temperature appears in the exponential and the denominator, when T → 0 this term
tends to 2πN3C
p(r
3