763620S STATISTICAL PHYSICS Solution Set 11 Autumn 2012

1. Critical exponents of the van der Waals system Express the reduced van der Waals equation interms of small expansion parameteters

Tr = 1 + τ, vr = 1 + ω, pr = 1 + π.

By using the expansion, calculate the critical exponents δ and γ of the van der Waals system. Interpretyour results.

Solution We need to find the scaling exponents δ and γ defined as

m|T=Tc∼ h1/δ,

(mh

)T,h→0

∼ (T − Tc)−γ . (1)

Here, the order parameter is the density of the liquid phase minus the critical density, and the fieldparameter is the applied pressure:

m = ρ` − ρc, h = p− pc.

In an earlier problem, we found the following expansion of the equation of state near a critical point:

(2π − 8τ) + (7π − 16τ)ω + 8(π − τ)ω2 + 3(1 + π)ω3 = 0.

First, the exponent δ. Let T → Tc (τ → 0):

2π + 7πω + 8πω2 + 3(1 + π)ω3 = 0,

⇒ π = − 3ω3

2 + 7ω + 8ω3 + 3ω3.

Near the coexistance curve ω � 1 and so π ∼ ω3 or ω ∼ π1/3. The order parameter then scales like

ρ` − ρc ∝1

1 + ω− 1 ' 1

ω∼ π−1/3,

⇒ δ = 3.

To find γ, we need to calculate(∂m

∂h

)τ,h→0

=

(∂

∂π

)τ,π→0

(1

1 + ω− 1

)= − 1

(1 + ω)2

(∂ω

∂π

)τ,π→0

.

Differentiate the equation of state, Eq. (1) w.r.t. ω holding τ constant:

2 + 7ω + (7π − 16τ)

(∂ω

∂π

)τ

+ 2nd and higher order terms = 0 ⇒(∂ω

∂π

)τ

=2 + 7ω

7π − 16τ

Using the equation of state near ω = 0, one sees that τ ∝ π, and so ∂ω∂π ∝ π

−1. Thus,(∂m

∂h

)τ,h→0

∼ π−1 ⇒ γ = −1.

2. 1-Dimensional Ising Model Show that the critical exponent β = 12 for the one-dimensional Ising

model in the mean field approximation.

Solution Let T = Tc(1 + τ) where Tc is such that νJ/(2kBTc) = 1. With these definitions, the Isingmodel magnetization m = 〈σ〉 becomes

m = tanh

(m

1 + τ

).

Suppose now that m� 1, and replace the hyperbolic tangent with its Taylor series, expanded to thirdorder, tanhx ' x− 1

3x3:

m =1

1 + τm− 1

3

(1

1 + τ

)3

m3.

Ignoring the m = 0 solution, we get the following:

m2 = 3 (1 + τ)3

(1

1 + τ− 1

)= 3

[(1 + τ)

2 − (1 + τ)3]' 3 [1 + 2τ − (1 + 3τ)] = −3τ,

⇒ m ∼ t1/2.

3. Landau Theory Consider the Landau’s theory of phase transition presented in the lectures.

(a) Calculate the low-field susceptibility:

χ0(T ) =

(∂m

∂h

)T

.

(b) What are the critical exponents γ and δ?

Solution

(a) Let φ(t,m) be the free energy, expanded in m and with coefficients from the Landau theory:

φ(t,m) = φ(t, 0) + r1tm2 + s0m

4. (2)

Legendre transforming to a free energy parametrized by the field h one has

φ(t, h) = φ(t,m)− hm = −hm+ φ(t, 0) + r1tm2 + s0m

4,

which is now minimized by m with h held constant,(∂φ

∂m

)h

= 0.

Straight-forward differentiation gives

−h+ 2r1tm+ 4s0m3 = 0 (3)

⇒ h ' −2r1tm ⇒ m(h) = − 1

2r1th.

And from that, one quite trivially gets the susceptibility(∂m

∂h

)t

=1

2r1t. (4)

(b) First, the exponent δ in m|t=0 ∼ h1/δ. From Eq. (3) with t→ 0:

h = 4s0m3 ⇒ m =

(h

4s0

)1/3

⇒ δ = 3.

Second, the exponent γ in (∂m/∂h) ∼ t−γ . This follows trivially from Eq. (4):

γ = 1.

4. Widom scaling Consider the widom scalinq

φs(λpt, λqh) = λφs(t, h). (5)

Assume that the order parameter is defined as

m(t, h) = −∂φs(t, h)

∂h

and show that the critical exponents obey equations

α+ 2β + γ = 2,

β(δ − 1) = γ.

Solution

m ∼ tβ Differentiate Eq. (5) w.r.t. h:

λq∂φs∂h

(λpt, λqh) = λ∂φs∂h

(t, h).

Let λp → t, t→ 1 (λpt→ t) and h→ 0:

tq/p∂φs∂h

(t, 0)︸ ︷︷ ︸−m(t,0)

= t1/p∂φs∂h

(1, 0) ⇒ m(t, 0) = t1/p−q/pm(1, 0) ⇒ β =1− qp

.

m ∼ h1/δ Again differentiating Eq. (5) w.r.t. h, but now let λq → h, h→ 1 (λqh→ h) and t→ 0:

h∂φs∂h

(t, 0) = h1/q∂φs∂h

(0, 1) ⇒ m(t, 0) = t1/q−1m(1, 0) ⇒ δ =q

1− q.

∂2φs

∂t2∼ t−α Differentiate Eq. (5) twice w.r.t. t:

λ2p∂2φs∂t2

(λpt, λqh) = λ∂2φs∂t2

(t, h).

Let λp → t, t→ 1 (λpt→ t) and h→ 0:

t2∂2φs∂t2

(t, 0) = t1/p∂2φs∂t2

(1, 0) ⇒ ∂2φs∂t2

(t, 0) ∼ t1/p−2 ⇒ α = 2− 1

p.

∂2φs

∂h2 ∼ t−γ Finally, differentiate Eq. (5) twice w.r.t. h:

λ2q∂2φs∂h2

(λpt, λqh) = λ∂2φs∂h2

(t, h).

Let λq → t, t→ 1 (λqt→ t) and h→ 0:

t2q/p∂2φs∂h2

(t, 0) = t1/p∂2φs∂h2

(1, 0) ⇒ ∂2φs∂h2

(t, 0) ∼ t1/p−2q/p ⇒ γ =2q − 1

p.

And now we can check the identities given in the problem:

α+ 2β + γ = 2− 1

p+ 2

1− qp

+2q − 1

p= 2,

β(δ − 1) =1− qp

(q

1− q− 1

)=q

p− 1− q

p=

2q − 1

p= γ.