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Problems & Solutions

for

Statistical Physics of Particles

Updated July 2008 by Mehran Kardar Department of Physics Massachusetts Institute of Technology Cambridge, Massachusetts 02139, USA

Table of Contents

I.

Thermodynamics

II.

Probability

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

III.

Kinetic Theory of Gases

IV.

Classical Statistical Mechanics

V. Interacting Particles VI. VII.

1

. . . . . . . . . . . . . . . . . . . . . . . . 38 . . . . . . . . . . . . . . . . . . . . . 72

. . . . . . . . . . . . . . . . . . . . . . . . . . 93

Quantum Statistical Mechanics

. . . . . . . . . . . . . . . . . . . .

121

Ideal Quantum Gases . . . . . . . . . . . . . . . . . . . . . . . .

138

Problems for Chapter I - Thermodynamics 1. Surface tension:

Thermodynamic properties of the interface between two phases are

described by a state function called the surface tension S. It is defined in terms of the

work required to increase the surface area by an amount dA through d¯W = SdA.

(a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius R is larger than outside pressure by 2S/R. What is the air pressure inside a soap bubble of radius R? • The work done by a water droplet on the outside world, needed to increase the radius

from R to R + ∆R is

∆W = (P − Po ) · 4πR2 · ∆R, where P is the pressure inside the drop and Po is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy S∆A = S · 8πR · ∆R, where S is the surface tension, and

∆Wtotal = 0,

=⇒

∆Wpressure = −∆Wsurface ,

resulting in (P − Po ) · 4πR2 · ∆R = S · 8πR · ∆R,

=⇒

(P − Po ) =

2S . R

In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and Pfilm − Po = Pinterior − Pfilm = leading to Pinterior − Po =

2S , R

4S . R

Hence, the air pressure inside the bubble is larger than atmospheric pressure by 4S/R. (b) A water droplet condenses on a solid surface. There are three surface tensions involved S aw , S sw , and S sa , where a, s, and w refer to air, solid and water respectively. Calculate

the angle of contact, and find the condition for the appearance of a water film (complete wetting).

• When steam condenses on a solid surface, water either forms a droplet, or spreads on

the surface. There are two ways to consider this problem: Method 1: Energy associated with the interfaces 1

In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore dE = Saw dAaw + Sas dAas + Sws dAws = 0. Since the total surface area of the solid is constant, dAas + dAws = 0. From geometrical considerations (see proof below), we obtain dAws cos θ = dAaw . From these equations, we obtain dE = (Saw cos θ − Sas + Sws ) dAws = 0,

=⇒

cos θ =

Sas − Sws . Saw

Proof of dAws cos θ = dAaw : Consider a droplet which is part of a sphere of radius R, which is cut by the substrate at an angle θ. The areas of the involved surfaces are Aws = π(R sin θ)2 ,

and

Aaw = 2πR2 (1 − cos θ).

Let us consider a small change in shape, accompanied by changes in R and θ. These variations should preserve the volume of water, i.e. constrained by V =

πR3 cos3 θ − 3 cos θ + 2 . 3

Introducing x = cos θ, we can re-write the above results as 2 2 A = πR 1 − x , ws Aaw = 2πR2 (1 − x) , 3 V = πR x3 − 3x + 2 . 3

The variations of these quantities are then obtained from dR 2 (1 − x ) − Rx dx, dAws = 2πR dx dR (1 − x) − R dx, dAaw = 2πR 2 dx dR 2 3 2 dV = πR (x − 3x + 2) + R(x − x) dx = 0. dx 2

From the last equation, we conclude 1 dR x2 − 1 x+1 =− 3 =− . R dx x − 3x + 2 (x − 1)(x + 2) Substituting for dR/dx gives, dAws = 2πR2

dx , x+2

and

dAaw = 2πR2

x · dx , x+2

resulting in the required result of dAaw = x · dAws = dAws cos θ. Method 2: Balancing forces on the contact line Another way to interpret the result is to consider the force balance of the equilibrium surface tension on the contact line. There are four forces acting on the line: (1) the surface tension at the water–gas interface, (2) the surface tension at the solid–water interface, (3) the surface tension at the gas–solid interface, and (4) the force downward by solid–contact line interaction. The last force ensures that the contact line stays on the solid surface, and is downward since the contact line is allowed to move only horizontally without friction. These forces should cancel along both the y–direction x–directions. The latter gives the condition for the contact angle known as Young’s equation, S as = S aw · cos θ + S ws ,

=⇒

cos θ =

S as − S ws . S aw

The critical condition for the complete wetting occurs when θ = 0, or cos θ = 1, i.e. for cos θC =

S as − S ws = 1. S aw

Complete wetting of the substrate thus occurs whenever S aw ≤ S as − S ws .

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is S o ≈ 7 × 10−2 N m−1 . Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important. 3

• Typical length scales at which the surface tension effects become significant are given

by the condition that the forces exerted by surface tension and relevant pressures become

comparable, or by the condition that the surface energy is comparable to the other energy changes of interest. Example 1: Size of water drops not much deformed on a non-wetting surface. This is given by equalizing the surface energy and the gravitational energy, S · 4πR2 ≈ mgR = ρV gR =

4π 4 R g, 3

leading to R≈

s

3S ≈ ρg

s

3 · 7 × 10−2 N/m ≈ 1.5 × 10−3 m = 1.5mm. 103 kg/m3 × 10m/s2

Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel (about 1 atm) = surface tension of water, gives πgel ≈

2S N kB T ≈ , V R

where N is the number of counter ions within the gel. Thus, 2 × 7 × 10−2 N/m R≈ ≈ 10−6 m. 105 N/m2 ******** 2. Surfactants:

Surfactant molecules such as those in soap or shampoo prefer to spread

on the air-water surface rather than dissolve in water. To see this, float a hair on the surface of water and gently touch the water in its vicinity with a piece of soap. (This is also why a piece of soap can power a toy paper boat.) (a) The air-water surface tension S o (assumed to be temperature independent) is reduced

roughly by N kB T /A, where N is the number of surfactant particles, and A is the area. Explain this result qualitatively. • Typical surfactant molecules have a hydrophilic head and a hydrophobic tail, and prefer to go to the interface between water and air, or water and oil. Some examples are, CH3 − (CH2 )11 − SO3− · N a+ , 4

CH3 − (CH2 )11 − N + (CH3 )3 · Cl− , CH3 − (CH2 )11 − O − (CH2 − CH2 − O)12 − H. The surfactant molecules spread over the surface of water and behave as a two dimensional gas. The gas has a pressure proportional to the density and the absolute temperature, which comes from the two dimensional degrees of freedom of the molecules. Thus the surfactants lower the free energy of the surface when the surface area is increased. ∆Fsurfactant =

N kB T · ∆A = (S − So ) · ∆A, A

S = So −

=⇒

N kB T. A

(Note that surface tension is defined with a sign opposite to that of hydrostatic pressure.) (b) Place a drop of water on a clean surface. Observe what happens to the air-watersurface contact angle as you gently touch the droplet surface with a small piece of soap, and explain the observation. • As shown in the previous problem, the contact angle satisfies cos θ =

S as − S ws . S aw

Touching the surface of the droplet with a small piece of soap reduces S aw , hence cos θ

increases, or equivalently, the angle θ decreases.

(c) More careful observations show that at higher surfactant densities 2 ∂S 2a N N kB T − = ∂A T (A − N b)2 A A

,

∂T A − Nb ; =− ∂S A N kB

and

where a and b are constants. Obtain the expression for S(A, T ) and explain qualitatively the origin of the corrections described by a and b.

• When the surfactant molecules are dense their interaction becomes important, resulting

in

and

2 N kB T 2a N ∂S = , − ∂A T (A − N b)2 A A

Integrating the first equation, gives

∂T A − Nb . =− ∂S A N kB

N kB T +a S(A, t) = f (T ) − A − Nb 5

N A

2

,

where f (T ) is a function of only T , while integrating the second equation, yields S(A, T ) = g(A) −

N kB T , A − Nb

with g(A) a function of only A. By comparing these two equations we get N kB T +a S(A, T ) = S o − A − Nb

N A

2

,

where S o represents the surface tension in the absence of surfactants and is independent

of A and T . The equation resembles the van der Waals equation of state for gas-liquid systems. The factor N b in the second term represents the excluded volume effect due to the finite size of the surfactant molecules. The last term represents the binary interaction between two surfactant molecules. If surfactant molecules attract each other the coefficient a is positive the surface tension increases. (d) Find an expression for CS − CA in terms of ∂E . ∂T S

∂E ∂A T ,

S,

• Taking A and T as independent variables, we obtain δQ = dE − S · dA,

=⇒

and δQ =

∂S ∂A

T

, and

∂E ∂E δQ = dA + dT − S · dA, ∂A T ∂T A

∂E ∂E − S dA + dT. ∂A T ∂T A

From the above result, the heat capacities are obtained as

resulting in

∂E δQ CA ≡ δT = ∂T A A , ∂E ∂A ∂E δQ CS ≡ = −S + δT S ∂A T ∂T S ∂T S CS − CA =

Using the chain rule relation

∂T , ∂S A

∂A ∂E −S . ∂A T ∂T S

∂T ∂S ∂A · · = −1, ∂S A ∂A T ∂T S 6

for

∂E ∂T A

=

we obtain CS − CA =

∂E −S · ∂A T ********

3. Temperature scales:

−1

∂T ∂S A

·

.

∂S ∂A T

Prove the equivalence of the ideal gas temperature scale Θ, and

the thermodynamic scale T , by performing a Carnot cycle on an ideal gas. The ideal gas satisfies P V = N kB Θ, and its internal energy E is a function of Θ only. However, you may not assume that E ∝ Θ. You may wish to proceed as follows: (a) Calculate the heat exchanges QH and QC as a function of ΘH , ΘC , and the volume expansion factors. • The ideal gas temperature is defined through the equation of state θ=

PV . N kB

The thermodynamic temperature is defined for a reversible Carnot cycle by Qhot Thot = . Tcold Qcold For an ideal gas, the internal energy is a function only of θ, i.e. E = E(θ), and d¯Q = dE − d¯W =

dE · dθ + P dV. dθ

adiabatics (∆Q = 0)

pressure P

1 Q hot 2

θ hot isothermals

4 Q cold 3 volume V 7

θ cold

Consider the Carnot cycle indicated in the figure. For the segment 1 to 2, which undergoes an isothermal expansion, we have dθ = 0,

=⇒ d¯Qhot = P dV,

and P =

N kB θhot . V

Hence, the heat input of the cycle is related to the expansion factor by Z V2 dV V2 N kB θhot Qhot = . = N kB θhot ln V V1 V1 A similar calculation along the low temperature isotherm yields Z V3 V3 dV , = N kB θcold ln Qcold = N kB θcold V V4 V4 and thus

Qhot θhot ln (V2 /V1 ) = . Qcold θcold ln (V3 /V4 )

(b) Calculate the volume expansion factor in an adiabatic process as a function of Θ. • Next, we calculate the volume expansion/compression ratios in the adiabatic processes. Along an adiabatic segment d¯Q = 0,

=⇒

0=

dE N kB θ · dθ + · dV, dθ V

=⇒

dV 1 dE =− · dθ. V N kB θ dθ

Integrating the above between the two temperatures, we obtain Z θhot 1 dE 1 V3 =− · dθ, and ln V N kB θcold θ dθ 2 Z θhot 1 dE 1 V4 =− ln · dθ. V1 N kB θcold θ dθ

While we cannot explicitly evaluate the integral (since E(θ) is arbitrary), we can nonetheless conclude that

V2 V1 = . V4 V3

(c) Show that QH /QC = ΘH /ΘC . • Combining the results of parts (a) and (b), we observe that Qhot θhot = . Qcold θcold 8

Since the thermodynamic temperature scale is defined by Qhot Thot = , Qcold Tcold we conclude that θ and T are proportional. If we further define θ(triple pointH2 0 ) = T (triple pointH2 0 ) = 273.16, θ and T become identical. ******** 4. Equations of State:

The equation of state constrains the form of internal energy as

in the following examples. (a) Starting from dE = T dS − P dV , show that the equation of state P V = N kB T , in fact

implies that E can only depend on T .

• Since there is only one form of work, we can choose any two parameters as independent

variables. For example, selecting T and V , such that E = E(T, V ), and S = S(T, V ), we obtain

∂S ∂S dT + T dV − P dV, dE = T dS − P dV = T ∂T V ∂V T

resulting in

Using the Maxwell’s relation†

we obtain

Since T

∂P ∂T V

E = E(T ).

B = T Nk V

∂S ∂E =T − P. ∂V T ∂V T ∂S ∂P = , ∂V T ∂T V

∂P ∂E =T − P. ∂V T ∂T V ∂E = 0. Thus E depends only on T , i.e. = P , for an ideal gas, ∂V T

(b) What is the most general equation of state consistent with an internal energy that depends only on temperature? • If E = E(T ),

∂E = 0, ∂V T

=⇒

∂P T = P. ∂T V

The solution for this equation is P = f (V )T, where f (V ) is any function of only V . ∂Y ∂2L † dL = Xdx + Y dy + · · · , =⇒ ∂X ∂y = ∂x y = ∂x·∂y . x

9

(c) Show that for a van der Waals gas CV is a function of temperature alone. • The van der Waals equation of state is given by "

P −a

N V

2 #

· (V − N b) = N kB T,

or N kB T P = +a (V − N b)

N V

2

.

From these equations, we conclude that ∂E CV ≡ , ∂T V

=⇒

∂CV ∂ 2E ∂ 2 P ∂P ∂ = −P =T T = 0. = ∂V T ∂V ∂T ∂T ∂T V ∂T 2 V ********

5. Clausius–Clapeyron equation describes the variation of boiling point with pressure. It is usually derived from the condition that the chemical potentials of the gas and liquid phases are the same at coexistence. • From the equations

µliquid (P, T ) = µgas (P, T ),

and µliquid (P + dP, T + dT ) = µgas (P + dP, T + dT ), we conclude that along the coexistence line dP = dT coX

∂µg ∂T P ∂µl ∂P T

− −

∂µl ∂T P ∂µg ∂P T

.

The variations of the Gibbs free energy, G = N µ(P, T ) from the extensivity condition, are given by ∂G V = , ∂P T

In terms of intensive quantities

∂µ V , = v= N ∂P T

∂G S=− . ∂T P S ∂µ s= , =− N ∂T P

10

where s and v are molar entropy and volume, respectively. Thus, the coexistence line satisfies the condition dP Sg − Sl sg − sl = = . dT coX Vg − Vl vg − vl

For an alternative derivation, consider a Carnot engine using one mole of water. At the source (P, T ) the latent heat L is supplied converting water to steam. There is a volume increase V associated with this process. The pressure is adiabatically decreased to P − dP . At the sink (P − dP, T − dT ) steam is condensed back to water. (a) Show that the work output of the engine is W = V dP + O(dP 2 ). Hence obtain the Clausius–Clapeyron equation dP L = . dT boiling T V

(1)

• If we approximate the adiabatic processes as taking place at constant volume V (vertical lines in the P − V diagram), we find P dV = P V − (P − dP )V = V dP.

pressure

W =

I

P

liq.

1

Q hot

2 T

gas

V

P-dP 4 Q cold

T-dT

3

volume 11

Here, we have neglected the volume of liquid state, which is much smaller than that of the gas state. As the error is of the order of ∂V dP · dP = O(dP 2 ), ∂P S

we have

W = V dP + O(dP 2 ). The efficiency of any Carnot cycle is given by η=

W TC =1− , QH TH

and in the present case, QH = L,

W = V dP,

TH = T,

TC = T − dT.

Substituting these values in the universal formula for efficiency, we obtain the ClausiusClapeyron equation dT V dP = , L T

L dP = . dT coX T ·V

or

(b) What is wrong with the following argument: “The heat QH supplied at the source to convert one mole of water to steam is L(T ). At the sink L(T − dT ) is supplied to condense

one mole of steam to water. The difference dT dL/dT must equal the work W = V dP , equal to LdT /T from eq.(1). Hence dL/dT = L/T implying that L is proportional to T !” • The statement “At the sink L(T − dT ) is supplied to condense one mole of water” is

incorrect. In the P − V diagram shown, the state at “1” corresponds to pure water, “2”

corresponds to pure vapor, but the states “3” and “4” have two phases coexisting. In

going from the state 3 to 4 less than one mole of steam is converted to water. Part of the steam has already been converted into water during the adiabatic expansion 2 → 3, and

the remaining portion is converted in the adiabatic compression 4 → 1. Thus the actual

latent heat should be less than the contribution by one mole of water.

(c) Assume that L is approximately temperature independent, and that the volume change is dominated by the volume of steam treated as an ideal gas, i.e. V = N kB T /P . Integrate equation (1) to obtain P (T ). 12

• For an ideal gas N kB T V = , P

=⇒

LP dP = , dT coX N kB T 2

dP L = dT. P N kB T 2

or

Integrating this equation, the boiling temperature is obtained as a function of the pressure P , as P = C · exp −

L kB TBoiling

.

(d) A hurricane works somewhat like the engine described above. Water evaporates at the warm surface of the ocean, steam rises up in the atmosphere, and condenses to water at the higher and cooler altitudes. The Coriolis force converts the upwards suction of the air to spiral motion. (Using ice and boiling water, you can create a little storm in a tea cup.) Typical values of warm ocean surface and high altitude temperatures are 800 F and −1200 F respectively. The warm water surface layer must be at least 200 feet thick to provide sufficient water vapor, as the hurricane needs to condense about 90 million tons of water vapor per hour to maintain itself. Estimate the maximum possible efficiency, and power output, of such a hurricane. (The latent heat of vaporization of water is about 2.3 × 106 Jkg −1 .)

• For TC = −120o F = 189o K, and TH = 80o F = 300o K, the limiting efficiency, as that

of a Carnot engine, is

ηmax =

TH − TC = 0.37. TH

The output power, is equal to (input power) x (efficiency). The input in this case is the energy obtained from evaporation of warm ocean temperature; hence dQc TH − TC dW = × dt dt TC 6 1hr 1000kg 2.3 × 106 J 90 × 10 tons · · · × 0.67 ≈ 4 × 1013 watts. = hr 3600sec ton kg

Power output =

(e) Due to gravity, atmospheric pressure P (h) drops with the height h. By balancing the forces acting on a slab of air (behaving like a perfect gas) of thickness dh, show that P (h) = P0 exp(−mgh/kT ), where m is the average mass of a molecule in air. 13

• Consider a horizontal slab of area A between heights h and h + dh. The gravitational force due to mass of particles in the slab is dFgravity = mg

N P Adh = mg Adh, V kB T

where we have used the ideal gas law to relate the density (N/V ) to the pressure. The gravitational force is balanced in equilibrium with the force due to pressure ∂P Adh. dFpressure = A [P (h) − P (h + dh)] = − ∂h

Equating the two forces gives ∂P P = −mg , ∂h kB T

mgh , P (h) = p0 exp − kB T

=⇒

assuming that temperature does not change with height. (f) Use the above results to estimate the boiling temperature of water on top of Mount Everest (h ≈ 9km). The latent heat of vaporization of water is about 2.3 × 106 Jkg −1 .

• Using the results from parts (c) and (e), we conclude that PEverest 1 mg L 1 . ≈ exp − (hEverest − hsea ) ≈ exp − − Psea kB T kB TEverest (boil) Tsea (boil)

Using the numbers provided, we find TEverest (boil) ≈ 346o K (74o C≈ 163o F). ********

6. Glass:

Liquid quartz, if cooled slowly, crystallizes at a temperature Tm , and releases

latent heat L. Under more rapid cooling conditions, the liquid is supercooled and becomes glassy. (a) As both phases of quartz are almost incompressible, there is no work input, and changes in internal energy satisfy dE = T dS + µdN . Use the extensivity condition to obtain the expression for µ in terms of E, T , S, and N . • Since in the present context we are considering only chemical work, we can regard

entropy as a function of two independent variables, e.g. E, and N , which appear naturally from dS = dE/T − µdN/T . Since entropy is an extensive variable, λS = S(λE, λN ).

Differentiating this with respect to λ and evaluating the resulting expression at λ = 1, gives ∂S E ∂S Nµ E + N = S(E, N ) = − , ∂E N ∂N E T T 14

leading to µ=

E − TS . N

(b) The heat capacity of crystalline quartz is approximately CX = αT 3 , while that of glassy quartz is roughly CG = βT , where α and β are constants. Assuming that the third law of thermodynamics applies to both crystalline and glass phases, calculate the entropies of the two phases at temperatures T ≤ Tm .

• Finite temperature entropies can be obtained by integrating d¯Q/T , starting from S(T =

0) = 0. Using the heat capacities to obtain the heat inputs, we find T dScrystal Ccrystal = αT 3 = , N dT T dSglass Cglass = βT = , N dT

N αT 3 , 3

=⇒

Scrystal =

=⇒

Sglass = βN T.

(c) At zero temperature the local bonding structure is similar in glass and crystalline quartz, so that they have approximately the same internal energy E0 . Calculate the internal energies of both phases at temperatures T ≤ Tm .

• Since dE = T dS + µdN , for dN = 0, we have (

dE = T dS = αN T 3 dT dE = T dS = βN T dT

(crystal), (glass).

Integrating these expressions, starting with the same internal energy Eo at T = 0, yields αN 4 E = Eo + T 4 E = E + βN T 2 o 2

(crystal), (glass).

(d) Use the condition of thermal equilibrium between two phases to compute the equilibrium melting temperature Tm in terms of α and β. • From the condition of chemical equilibrium between the two phases, µcrystal = µglass , we obtain

1 1 − 3 4

1 · αT = 1 − · βT 2 , 2 4

15

=⇒

αT 4 βT 2 = , 12 2

resulting in a transition temperature Tmelt =

r

6β . α

(e) Compute the latent heat L in terms of α and β. • From the assumptions of the previous parts, we obtain the latent heats for the glass to

crystal transition as

3 αTmelt L = Tmelt (Sglass − Scrystal ) = N Tmelt βTmelt − 3 2 αTmelt 2 2 2 = N Tmelt (β − 2β) = −N βTmelt < 0. = N Tmelt β− 3 (f) Is the result in the previous part correct? If not, which of the steps leading to it is most likely to be incorrect? • The above result implies that the entropy of the crystal phase is larger than that of

the glass phase. This is clearly unphysical, and one of the assumptions must be wrong.

The questionable step is the assumption that the glass phase is subject to the third law of thermodynamics, and has zero entropy at T = 0. In fact, glass is a non-ergodic state of matter which does not have a unique ground state, and violates the third law. ******** 7. Filament: For an elastic filament it is found that, at a finite range in temperature, a displacement x requires a force J = ax − bT + cT x, where a, b, and c are constants. Furthermore, its heat capacity at constant displacement is proportional to temperature, i.e. Cx = A(x)T . (a) Use an appropriate Maxwell relation to calculate ∂S/∂x|T . • From dF = −SdT + Jdx, we obtain ∂S ∂J =− = b − cx. ∂x T ∂T x (b) Show that A has to in fact be independent of x, i.e. dA/dx = 0. 16

∂S = A(x)T , where S = S(T, x). Thus • We have Cx = T ∂T

∂ ∂S ∂ ∂S ∂A = = =0 ∂x ∂x ∂T ∂T ∂x from part (a), implying that A is independent of x. (c) Give the expression for S(T, x) assuming S(0, 0) = S0 . • S(x, T ) can be calculated as T ′ =T

∂S(T ′ , x = 0) ′ S(x, T ) = S(0, 0) + dT + ∂T ′ T ′ =0 Z x Z T ′ (b − cx′ )dx′ AdT + = S0 + Z

Z

x′ =x x′ =0

∂S(T, x′ ) dx ∂x′

0

0

c = S0 + AT + (b − x)x. 2

(d) Calculate the heat capacity at constant tension, i.e. CJ = T ∂S/∂T |J as a function of

T and J.

• Writing the entropy as S(T, x) = S(T, x(T, J)), leads to ∂S ∂S ∂x ∂S = + . ∂T J ∂T x ∂x T ∂T J = b − cx and ∂S x = A. Furthermore, From parts (a) and (b), ∂S ∂x T ∂T ∂x ∂x − b + cx + cT ∂T = 0, i.e. a ∂T

Thus

Since x =

b − cx ∂x = . ∂T a + cT

∂x ∂T J

is given by

(b − cx)2 . CJ = T A + (a + cT ) J+bT a+cT

, we can rewrite the heat capacity as a function of T and J, as "

CJ = T A +

2 (b − c J+bT a+cT )

(a + cT ) (ab − cJ)2 . =T A+ (a + cT )3

#

******** 8.

Hard core gas:

A gas obeys the equation of state P (V − N b) = N kB T , and has a

heat capacity CV independent of temperature. (N is kept fixed in the following.) 17

(a) Find the Maxwell relation involving ∂S/∂V |T,N . • For dN = 0,

d(E − T S) = −SdT − P dV,

=⇒

∂P ∂S = . ∂V T,N ∂T V,N

(b) By calculating dE(T, V ), show that E is a function of T (and N ) only. • Writing dS in terms of dT and dV , dE = T dS − P dV = T

! ∂S ∂S dT + dV − P dV. ∂T V,N ∂V T,N

Using the Maxwell relation from part (a), we find ∂S dT + dE(T, V ) = T ∂T V,N

But from the equation of state, we get N kB T P = , (V − N b)

=⇒

P ∂P = , ∂T V,N T

! ∂P T − P dV. ∂T V,N =⇒

i.e. E(T, N, V ) = E(T, N ) does not depend on V .

∂S dT, dE(T, V ) = T ∂T V,N

(c) Show that γ ≡ CP /CV = 1 + N kB /CV (independent of T and V ). • The hear capacity is

∂Q ∂E + P V = CP = ∂T P ∂T

But, since E = E(T ) only,

= ∂E + P ∂V . ∂T P ∂T P P

∂E ∂E = = CV , ∂T P ∂T V

and from the equation of state we get N kB ∂V = , ∂T P P

=⇒

CP = CV + N kB ,

=⇒

γ =1+

N kB , CV

which is independent of T , since CV is independent of temperature. The independence of CV from V also follows from part (a). 18

(d) By writing an expression for E(P, V ), or otherwise, show that an adiabatic change satisfies the equation P (V − N b)γ =constant. • Using the equation of state, we have dE = CV dT = CV d

P (V − N b) N kB

=

CV (P dV + (V − N b)dP ) . N kB

The adiabatic condition, dQ = dE + P dV = 0, can now be written as 0 = dQ =

CV 1+ N kB

P d(V − N b) +

CV (V − N b)dP. N kB

Dividing by CV P (V − N b)/(N kB ) yields d(V − N b) dP +γ = 0, P (V − N b)

ln [P (V − N b)γ ] = constant.

=⇒

******** 9. Superconducting transition: Many metals become superconductors at low temperatures T , and magnetic fields B. The heat capacities of the two phases at zero magnetic field are approximately given by (

Cs (T ) = V αT 3 Cn (T ) = V βT 3 + γT

in the superconducting phase in the normal phase

,

where V is the volume, and {α, β, γ} are constants. (There is no appreciable change in volume at this transition, and mechanical work can be ignored throughout this problem.)

(a) Calculate the entropies Ss (T ) and Sn (T ) of the two phases at zero field, using the third law of thermodynamics. • Finite temperature entropies are obtained by integrating dS = d¯Q/T , starting from S(T = 0) = 0. Using the heat capacities to obtain the heat inputs, we find

dSs , dT dS Cn = V βT 3 + γT = T n , dT Cs = V αT 3 = T

19

=⇒ =⇒

αT 3 , 3 3 . βT Sn = V + γT 3 Ss = V

(b) Experiments indicate that there is no latent heat (L = 0) for the transition between the normal and superconducting phases at zero field. Use this information to obtain the transition temperature Tc , as a function of α, β, and γ. • The Latent hear for the transition is related to the difference in entropies, and thus L = Tc (Sn (Tc ) − Ss (Tc )) = 0. Using the entropies calculated in the previous part, we obtain r αTc3 3γ βTc3 = + γTc , =⇒ Tc = . 3 3 α−β (c) At zero temperature, the electrons in the superconductor form bound Cooper pairs. As a result, the internal energy of the superconductor is reduced by an amount V ∆, i.e. En (T = 0) = E0 and Es (T = 0) = E0 −V ∆ for the metal and superconductor, respectively. Calculate the internal energies of both phases at finite temperatures.

• Since dE = T dS + BdM + µdN , for dN = 0, and B = 0, we have dE = T dS = CdT .

Integrating the given expressions for heat capacity, and starting with the internal energies E0 and E0 − V ∆ at T = 0, yields h α 4i Es (T ) = E0 + V −∆ + 4 T β 4 γ 2 . En (T ) = E0 + V T + T 4 2

(d) By comparing the Gibbs free energies (or chemical potentials) in the two phases, obtain an expression for the energy gap ∆ in terms of α, β, and γ. • The Gibbs free energy G = E − T S − BM = µN can be calculated for B = 0 in each

phase, using the results obtained before, as h h α 3 α 4i α 4i T − T V T = E − V ∆ + T G (T ) = E + V −∆ + 0 0 s 4 3 12 β 4 γ 2 β 3 β 4 γ 2 . Gn (T ) = E0 + V T + T − TV T + γT = E0 − V T + T 4 2 3 12 2

At the transition point, the chemical potentials (and hence the Gibbs free energies) must be equal, leading to ∆+

α 4 β 4 γ 2 Tc = T + Tc , 12 12 c 2

=⇒ 20

∆=

γ 2 α−β 4 T − T . 2 c 12 c

Using the value of Tc =

p

3γ/(α − β), we obtain ∆=

3 γ2 . 4α−β

(e) In the presence of a magnetic field B, inclusion of magnetic work results in dE = T dS +BdM +µdN , where M is the magnetization. The superconducting phase is a perfect diamagnet, expelling the magnetic field from its interior, such that Ms = −V B/(4π) in

appropriate units. The normal metal can be regarded as approximately non-magnetic,

with Mn = 0. Use this information, in conjunction with previous results, to show that the superconducting phase becomes normal for magnetic fields larger than T2 Bc (T ) = B0 1 − 2 , Tc giving an expression for B0 . • Since dG = −SdT − M dB + µdN , we have to add the integral of −M dB to the Gibbs

free energies calculated in the previous section for B = 0. There is no change in the metallic phase since Mn = 0, while in the superconducting phase there is an additional R R contribution of − Ms dB = (V /4π) BdB = (V /8π)B 2 . Hence the Gibbs free energies

at finite field are

h B2 α 4i + V T G (T, B) = E − V ∆ + s 0 12 8π . β 4 γ 2 Gn (T, B) = E0 − V T + T 12 2

Equating the Gibbs free energies gives a critical magnetic field

γ α−β 4 3 γ2 γ α−β 4 Bc2 = ∆ − T2 + T = − T2 + T 8π 2 12 4α−β 2 12 " # 2 2 α−β 3γ 6γT 2 α−β = − Tc2 − T 2 , + T4 = 12 α−β α−β 12 where we have used the values of ∆ and Tc obtained before. Taking the square root of the above expression gives T2 Bc = B0 1 − 2 , Tc

where

B0 =

r

21

2π(α − β) 2 Tc = 3

s

p 6πγ 2 = Tc 2πγ. α−β

******** 10. Photon gas Carnot cycle:

The aim of this problem is to obtain the blackbody 4

radiation relation, E(T, V ) ∝ V T , starting from the equation of state, by performing an

infinitesimal Carnot cycle on the photon gas.

P P+dP

T+dT

P V

T

V+dV

V (a) Express the work done, W , in the above cycle, in terms of dV and dP . • Ignoring higher order terms, net work is the area of the cycle, given by W = dP dV . (b) Express the heat absorbed, Q, in expanding the gas along an isotherm, in terms of P , dV , and an appropriate derivative of E(T, V ). • Applying the first law, the heat absorbed is Q = dE + P dV =

∂E ∂T

dT +

V

∂E ∂V

dV + P dV

T

= isotherm

∂E ∂V

T

+ P dV.

(c) Using the efficiency of the Carnot cycle, relate the above expressions for W and Q to T and dT . • The efficiency of the Carnot cycle (η = dT /T ) is here calculated as η=

dP dT W = = . Q [(∂E/∂V )T + P ] T

(d) Observations indicate that the pressure of the photon gas is given by P = AT 4 , 3

4 where A = π 2 kB /45 (¯ hc) is a constant. Use this information to obtain E(T, V ), assuming

E(T, 0) = 0. • From the result of part (c) and the relation P = AT 4 , 4

4AT =

∂E ∂V

4

+ AT ,

T

22

or

∂E ∂V

= 3AT 4 , T

so that E = 3AV T 4 .

(e) Find the relation describing the adiabatic paths in the above cycle. • Adiabatic curves are given by dQ = 0, or 0=

∂E ∂T

dT +

V

∂E ∂V

dV + P dV = 3V dP + 4P dV, T

i.e. P V 4/3 = constant.

******** 11. Irreversible Processes: (a) Consider two substances, initially at temperatures T10 and T20 , coming to equilibrium at a final temperature Tf through heat exchange. By relating the direction of heat flow to the temperature difference, show that the change in the total entropy, which can be written as Z

∆S = ∆S1 + ∆S2 ≥

Tf

T10

d¯Q1 + T1

Z

Tf

T20

d¯Q1 = T2

Z

T1 − T2 d¯Q, T1 T2

must be positive. This is an example of the more general condition that “in a closed system, equilibrium is characterized by the maximum value of entropy S.” • Defining the heat flow from substance 1 to 2 as, d¯Q1→2 , we get, ∆S = ∆S1 + ∆S2 ≥

Z

Tf

T10

d¯Q1 + T1

Z

Tf

T20

d¯Q2 = T2

Z

T1 − T2 d¯Q1→2 . T1 T2

But according to Clausius’ statement of the second law d¯Q1→2 > 0, if T1 > T2 and d¯Q1→2 < 0, if T1 < T2 . Hence, (T1 − T2 )d¯Q1→2 ≥ 0, resulting in ∆S ≥

Z

T1 − T2 d¯Q1→2 ≥ 0. T1 T2

23

(b) Now consider a gas with adjustable volume V , and diathermal walls, embedded in a heat bath of constant temperature T , and fixed pressure P . The change in the entropy of the bath is given by ∆Sbath =

∆Qbath ∆Qgas 1 =− = − (∆Egas + P ∆Vgas ) . T T T

By considering the change in entropy of the combined system establish that “the equilibrium of a gas at fixed T and P is characterized by the minimum of the Gibbs free energy G = E + P V − T S.”

• The total change in entropy of the whole system is, 1 1 ∆S = ∆Sbath + ∆Sgas = − (∆Egas + P ∆Vgas − T ∆Sgas ) = − ∆Ggas . T T By the second law of thermodynamics, all processes must satisfy, 1 − ∆Ggas ≥ 0 ⇔ ∆Ggas ≤ 0, T that is, all processes that occur can only lower the Gibbs free energy of the gas. Therefore the equillibrium of a gas in contact with a heat bath of constant T and P is established at the point of minimum Gibbs free energy, i.e. when the Gibbs free energy cannot be lowered any more. ******** 12. The solar system originated from a dilute gas of particles, sufficiently separated from other such clouds to be regarded as an isolated system. Under the action of gravity the particles coalesced to form the sun and planets. (a) The motion and organization of planets is much more ordered than the original dust cloud. Why does this not violate the second law of thermodynamics? • The formation of planets is due to the gravitational interaction. Because of the attrac-

tive nature of this interaction, the original dust cloud with uniform density has in fact

lower entropy. Clumping of the uniform density leads to higher entropy. Of course the gravitational potential energy is converted into kinetic energy in the process. Ultimately the kinetic energy of falling particles is released in the form of photons which carry away a lot of entropy. (b) The nuclear processes of the sun convert protons to heavier elements such as carbon. Does this further organization lead to a reduction in entropy? 24

• Again the process of formation of heavier elements is accompanied by the release of large amounts of energy which are carry away by photons. The entropy carry away by these

photons is more than enough to compensate any ordering associated with the packing of nucleons into heavier nuclei. (c) The evolution of life and intelligence requires even further levels of organization. How is this achieved on earth without violating the second law? • Once more there is usage of energy by the organisms that converts more ordered forms of energy to less ordered ones.

********

25

Problems for Chapter II - Probability 1. Characteristic functions:

Calculate the characteristic function, the mean, and the

variance of the following probability density functions: (a) Uniform

1 2a

p(x) =

−a < x < a , and

for

• A uniform probability distribution, 1 p(x) = 2a 0

p(x) = 0

for − a < x < a

otherwise;

,

otherwise

for which there exist many examples, gives 1 f (k) = 2a = Therefore,

a 1 1 exp(−ikx)dx = exp(−ikx) 2a −ik −a −a

Z

a

∞ X 1 (ak)2m sin(ka) = (−1)m . ak (2m + 1)! m=0

m1 = hxi = 0, (b) Laplace

p(x) =

• The Laplace PDF,

1 2a

exp

− |x| a

and

m2 = hx2 i =

1 2 a . 3

;

1 |x| p(x) = , exp − 2a a

for example describing light absorption through a turbid medium, gives Z ∞ |x| 1 dx exp −ikx − f (k) = 2a −∞ a Z ∞ Z 0 1 1 dx exp(−ikx − x/a) + dx exp(−ikx + x/a) = 2a 0 2a −∞ 1 1 1 1 = = − 2a −ik + 1/a −ik − 1/a 1 + (ak)2 = 1 − (ak)2 + (ak)4 − · · · .

Therefore, m1 = hxi = 0,

and

26

m2 = hx2 i = 2a2 .

(c) Cauchy

p(x) =

a π(x2 +a2 )

.

• The Cauchy, or Lorentz PDF describes the spectrum of light scattered by diffusive

modes, and is given by

p(x) =

π(x2

a . + a2 )

For this distribution, ∞

a exp(−ikx) dx 2 π(x + a2 ) −∞ Z ∞ 1 1 1 exp(−ikx) dx. − = 2πi −∞ x − ia x + ia

f (k) =

Z

The easiest method for evaluating the above integrals is to close the integration contours in the complex plane, and evaluate the residue. The vanishing of the integrand at infinity determines whether the contour has to be closed in the upper, or lower half of the complex plane, and leads to Z 1 exp(−ikx) dx = exp(−ka) − 2πi x + ia C f (k) = Z exp(−ikx) 1 dx = exp(ka) 2πi B x − ia

for k ≥ 0

= exp(−|ka|).

for k < 0

Note that f (k) is not an analytic function in this case, and hence does not have a Taylor expansion. The moments have to be determined by another method, e.g. by direct evaluation, as m1 = hxi = 0,

and

2

m2 = hx i =

Z

dx

π x2 · 2 → ∞. a x + a2

The first moment vanishes by symmetry, while the second (and higher) moments diverge, explaining the non-analytic nature of f (k). The following two probability density functions are defined for x ≥ 0. Compute only

the mean and variance for each. (d) Rayleigh

p(x) =

x a2

2

x exp(− 2a , 2)

• The Rayleigh distribution, x2 x p(x) = 2 exp − 2 , a 2a 27

for

x ≥ 0,

can be used for the length of a random walk in two dimensions. Its characteristic function is Z ∞ x2 x f (k) = exp(−ikx) 2 exp − 2 dx a 2a 0 Z ∞ x x2 = [cos(kx) − i sin(kx)] 2 exp − 2 dx. a 2a 0 The integrals are not simple, but can be evaluated as Z ∞ ∞ X n (−1)n n! x2 x 2a2 k 2 , cos(kx) 2 exp − 2 dx = a 2a (2n)! 0 n=0 and

Z

0

resulting in

∞

Z x x x2 1 ∞ x2 sin(kx) 2 exp − 2 dx = sin(kx) 2 exp − 2 dx a 2a 2 −∞ a 2a r 2 2 π k a , ka exp − = 2 2 r 2 2 ∞ X (−1)n n! k a π 2 2 n 2a k −i . f (k) = ka exp − (2n)! 2 2 n=0

The moments can also be calculated directly, from r Z ∞ 2 Z ∞ 2 x π x x2 x2 m1 = hxi = dx = dx = exp − exp − a, 2 a2 2a2 2a2 2 0 −∞ 2a 2 Z ∞ 2 Z ∞ 3 x x2 x2 x x 2 2 exp − 2 dx = 2a exp − 2 d m2 = hx i = 2 2 a 2a 2a 2a 2a2 0 0 Z ∞ y exp(−y)dy = 2a2 . = 2a2 0

q

2

2

x (e) Maxwell p(x) = π2 xa3 exp(− 2a . 2) • It is difficult to calculate the characteristic function for the Maxwell distribution r 2 x2 x2 exp − 2 , p(x) = π a3 2a

say describing the speed of a gas particle. However, we can directly evaluate the mean and variance, as r Z ∞ x3 2 x2 exp − 2 dx m1 = hxi = π 0 a3 2a r 2 Z ∞ 2 x x2 x 2 =2 a exp − 2 d 2 π 0 2a 2a 2a2 r r Z ∞ 2 2 a y exp(−y)dy = 2 a, =2 π 0 π 28

and 2

m2 = hx i =

r

2 π

Z

∞

o

x4 x2 exp − 2 dx = 3a2 . a3 2a

******** 2. Directed random walk:

The motion of a particle in three dimensions is a series

of independent steps of length ℓ. Each step makes an angle θ with the z axis, with a probability density p(θ) = 2 cos2 (θ/2)/π; while the angle φ is uniformly distributed between 0 and 2π. (Note that the solid angle factor of sin θ is already included in the definition of p(θ), which is correctly normalized to unity.) The particle (walker) starts at the origin and makes a large number of steps N .

(a) Calculate the expectation values hzi, hxi, hyi, z 2 , x2 , and y 2 , and the covariances hxyi, hxzi, and hyzi.

• From symmetry arguments,

hxi = hyi = 0,

while along the z-axis, hzi =

X i

hzi i = N hzi i = N a hcos θi i =

Na . 2

The last equality follows from Z Z π 1 hcos θi i = p(θ) cos θdθ = cos θ · (cos θ + 1)dθ 0 π Z π 1 1 (cos 2θ + 1)dθ = . = 2 0 2π The second moment of z is given by X XX

2 X zi2 hzi zj i = hzi zj i + z = i

i,j

=

XX i

Noting that

i6=j

i

i6=j

X hzi i hzj i + zi2 i

2

= N (N − 1) hzi i + N zi2 .

2 Z π Z π zi 1 1 1 2 = cos θ(cos θ + 1)dθ = (cos 2θ + 1)dθ = , 2 a 2 0 π 0 2π 29

we find

a 2

2 a2 a2 +N z = N (N − 1) = N (N + 1) . 2 2 4

The second moments in the x and y directions are equal, and given by XX X

2 X x2i = N x2i . hxi xj i = hxi xj i + x = i,j

i

i

i6=j

Using the result

2

xi = sin2 θ cos2 φ 2 a Z π Z 2π 1 1 2 dθ sin2 θ(cos θ + 1) = , dφ cos φ = 2 2π 0 4 0 we obtain

2 2 N a2 x = y = . 4

While the variables x, y, and z are not independent because of the constraint of unit length, simple symmetry considerations suffice to show that the three covariances are in fact zero, i.e. hxyi = hxzi = hyzi = 0. (b) Use the central limit theorem to estimate the probability density p(x, y, z) for the particle to end up at the point (x, y, z). • From the Central limit theorem, the probability density should be Gaussian. However,

for correlated random variable we may expect cross terms that describe their covariance. However, we showed above that the covarainces between x, y, and z are all zero. Hence we can treat them as three independent Gaussian variables, and write (x − hxi)2 (y − hyi)2 (z − hzi)2 p(x, y, z) ∝ exp − − − . 2σx2 2σy2 2σz2 (There will be correlations between x, y, and z appearing in higher cumulants, but all such cumulants become irrelevant in the N → ∞ limit.) Using the moments hxi = hyi = 0,

and

a hzi = N , 2

a2 2 = σy2 , σx2 = x2 − hxi = N 4 30

σz2

and we obtain

a2 2 = z 2 − hzi = N (N + 1) − 4

p(x, y, z) =

2 πN a2

3/2

"

Na 2

2

=N 2

x2 + y 2 + (z − N a/2) exp − N a2 /2

a2 , 4

#

.

******** Consider any probability density p(x) for (−∞ < x < ∞),

3. Tchebycheff inequality:

with mean λ, and variance σ 2 . Show that the total probability of outcomes that are more than nσ away from λ is less than 1/n2 , i.e. Z |x−λ|≥nσ

dxp(x) ≤

1 . n2

Hint: Start with the integral defining σ 2 , and break it up into parts corresponding to |x − λ| > nσ, and |x − λ| < nσ.

• By definition, for a system with a PDF p(x), and average λ, the variance is Z 2 σ = (x − λ)2 p(x)dx. Let us break the integral into two parts as Z Z 2 2 (x − λ) p(x)dx + σ = |x−λ|≥nσ

|x−λ|0 py 0, i.e. f (px , x > 0) = A (px ) e−ax/px + feq. (px ) . The constant A (px ) can be determined by matching to solution for x < 0 at x = 0, and is related to the incoming flux. The penetration depth d is the inverse of the decay parameter, and given by d=

px , a

with

a = σst c (n0 − n1 ) > 0. 81

******** 11. Equilibrium density: Consider a gas of N particles of mass m, in an external potential U (~q ). Assume that the one body density ρ1 (~p, ~q, t), satisfies the Boltzmann equation. For a stationary solution, ∂ρ1 /∂t = 0, it is sufficient from Liouville’s theorem for ρ1 to satisfy ρ1 ∝ exp −β p2 /2m + U (~q ) . Prove that this condition is also necessary by using the

H-theorem as follows.

(a) Find ρ1 (~ p, ~q ) that minimizes H = N

R

d3 p~d3 ~qρ1 (~ p, ~q ) ln ρ1 (~ p, ~q), subject to the con-

straint that the total energy E = hHi is constant. (Hint: Use the method of Lagrange

multipliers to impose the constraint.)

• Using Lagrange multipliers to impose the constraints, hHi = E and

minimizing H with the given constraints reduces to minimizing, N

Z

R

d3 p~d3 ~q ρ1 = 1,

d3 ~pd3 ~q(ρ1 ln ρ1 + βρ1 H + αρ1 ) − βE − αN.

Differentiating with respect to α, β, and the function ρ1 we get, N Z

Z

3

3

d p~d ~q ρ1 = N 3

3

d ~pd ~q ρ1 H = E

→ →

Z

Z

d3 p~d3 ~q ρ1 = 1, d3 p~d3 ~q ρ1 H = E/N,

ln ρ1 + βH + α = 0 → ρ1 = exp(−βH − α),

respectively. Hence we conclude, ρ1 = R

where β is determined by, R

exp(−βH) d3 p~d3 ~q exp(−βH)

,

d3 ~pd3 ~q H exp(−βH) E R = . 3 3 N d ~pd ~q exp(−βH) (a)

(b) For a mixture of two gases (particles of masses ma and mb ) find the distributions ρ1 (b)

and ρ1 that minimize H = H(a) + H(b) subject to the constraint of constant total energy. Hence show that the kinetic energy per particle can serve as an empirical temperature. 82

• If we have Na and Nb of each particle type with total energy E, then H is minimized with the total energy constraint by extremizing, Z (a) (a) (b) (b) (a) (b) d3 p~d3 ~q(Na ρ1 ln ρ1 + Nb ρ1 ln ρ1 + β(Na Ha ρ1 + Nb Hb ρ1 ) . (a) ′ (b) ′ +Na αρ1 + Nb α ρ1 ) − βE − αNa − α Nb (a)

(b)

Differentiating this expression with respect to α, α′ , β, ρ1 , and ρ1 , we get, Z (a) d3 p~d3 ~q ρ1 = 1, Z (b) d3 p~d3 ~q ρ1 = 1, Z (a) (b) 3 3 d p~d ~q Na Ha ρ1 + Nb Hb ρ1 = E, (a)

ln ρ1 + βHa + α = 0, (b)

ln ρ1 + βHb + α′ = 0. So we get, (a) ρ1 = R (a) ρ2 = R

exp(−βHa ) 3 d ~pd3 ~q exp(−βH

a)

,

exp(−βHb ) . d3 ~pd3 ~q exp(−βHb )

where β is obtained by, R 3 3 R 3 3 d p~d ~qHa exp(−βHa ) d p~d ~qHb exp(−βHb ) Na R 3 3 + Nb R 3 3 = E. d p~d ~q exp(−βHa ) d p~d ~q exp(−βHb )

Note that β is a value defined for both gases a and b, and hence can serve as an empirical temperature. For the specific case of Ha =

p2 + Ua (~q), 2ma

Hb =

p2 + Ub (~q), 2ma

the kinetic energy per particle in a distribution with equal β is also equal, since R 3 3 p2 R 3 R p2 d p~d ~q 2ma exp −β(p2 /2ma + Ua (~q)) exp(−βp2 /2ma ) d ~q exp(−βUa ) d3 p~ 2m a R R R = d3 p~d3 ~q exp(−βHa ) d3 ~q exp(−βUa ) d3 p~ exp(−βp2 /2ma ) R∞ p4 4π 0 dp 2m exp(−βp2 /2ma ) a . R = ∞ 4π 0 dp p2 exp(−βp2 /2ma ) R∞ 2 3 1 0 dt t4 e−t = = R∞ 2 β 0 dt t2 e−t 2β 83

So we see that the kinetic energy per particle for the gas can also serve as an empirical temperature in this case. ******** 12. Moments of momentum: Consider a gas of N classical particles of mass m in thermal equilibrium at a temperature T , in a box of volume V . (a) Write down the equilibrium one particle density feq. (~p, ~q ), for coordinate ~q, and momentum p~. • The equilibrium Maxwell-Boltzmann distribution reads f (~ p, ~q ) =

n (2πmkB T )

3/2

exp −

D

p2 2mkB T

(b) Calculate the joint characteristic function, exp −i~k · p~

• Performing the Gaussian average yields

E

.

, for momentum.

D E mkB T 2 −i~ k·~ p ~ k . p˜ k = e = exp − 2

n (c) Find all the joint cumulants pℓx pm y pz c .

• The cumulants are calculated from the characteristic function, as

ℓ m n px py pz c =

∂ ∂ (−ikx )

ℓ

∂ ∂ (−iky )

m

∂ ∂ (−ikz )

n

ln p˜ ~k ~

= mkB T (δℓ2 δm0 δn0 + δℓ0 δm2 δn0 + δℓ0 δm0 δn2 ) ,

i.e., there are only second cumulants; all other cumulants are zero. (d) Calculate the joint moment hpα pβ (~ p · p~ )i. • Using Wick’s theorem

hpα pβ (~p · p~ )i = hpα pβ pγ pγ i = hpα pβ i hpγ pγ i + 2 hpα pγ i hpβ pγ i 2

2

= (mkB T ) δαβ δγγ + 2 (mkB T ) δαγ δβγ 2

= 5 (mkB T ) δαβ . 84

k=0

Alternatively, directly from the characteristic function, ∂ ∂ ∂ ∂ hpα pβ (~ p · p~ )i = p˜ ~k ∂ (−ikα ) ∂ (−ikβ ) ∂ (−ikγ ) ∂ (−ikγ ) ~ k=0 i h mk T 2 ∂ ∂ 2 ~2 − 2B ~ k 3mkB − (mkB T ) k e = ~ ∂ (−ikα ) ∂ (−ikβ ) k=0 2

= 5 (mkB T ) δαβ .

******** 13. Generalized ideal gas: Consider a gas of N particles confined to a box of volume V in d-dimensions. The energy, ǫ, and momentum, p, of each particle are related by ǫ = ps , where p = |p|. (For classical particles s = 2, while for highly relativistic ones s = 1.) Let

f (v)dv denote the probability of finding particles with speeds between v and v + dv, and n = N/V .

(a) Calculate the number of impacts of gas molecules per unit area of the wall of the box, and per unit time as follows: (i) Show that the number of particles hitting area A in a time dt arriving from a ~ with a speed v, is proportional to A · vdt cos θ · nf (v)dv , specific direction Ω, ~ and the normal to the wall. where θ is the angle between the direction Ω ~ with the same polar angle θ, demonstrate that (ii) Summing over all directions Ω dN (θ, v) = A · vdt cos θ · nf (v)dv ·

Sd−1 sind−2 θ dθ Sd

,

where Sd = 2π d/2 /(d/2 − 1)! is the total solid angle in d dimensions. (iii) By averaging over v and θ show that Sd−1 N · nv , = A dt (d − 1)Sd

where

v=

Z

vf (v)dv

is the average speed.

(b) Each (elastic) collision transfers a momentum 2p cos θ to the wall. By examining the net force on an element of area prove that the pressure P equals

s d

.

E V ,

where E is the

average (kinetic) energy. (Note that the velocity v is not p/m but ∂ǫ/∂p.) Hint: Clearly

upon averaging over all directions cos2 θ = 1/d.

(c) Using thermodynamics and the result in (b) show that along an adiabatic curve P V γ is constant, and calculate γ. 85

(d) According to the equipartition theorem, each degree of freedom which appears quadratically in the energy has an energy kB T /2. Calculate the value of γ if each gas particle has ℓ such quadratic degrees of freedom in addition to its translational motion. What values of γ are expected for helium and hydrogen at room temperature? (e) Consider the following experiment to test whether the motion of ants is random. 250 ants are placed inside a 10cm × 10cm box. They cannot climb the wall, but can escape

through an opening of size 5mm in the wall. If the motion of ants is indeed random, and they move with an average speed of 2mms−1 , how may are expected to escape the box in the first 30 seconds? ******** 14. Effusion:

A box contains a perfect gas at temperature T and density n.

(a) What is the one-particle density, ρ1 (~v), for particles with velocity ~v ? A small hole is opened in the wall of the box for a short time to allow some particles to escape into a previously empty container. (b) During the time that the hole is open what is the flux (number of particles per unit time and per unit area) of particles into the container? (Ignore the possibility of any particles returning to the box.) (c) Show that the average kinetic energy of escaping particles is 2kB T . (Hint: calculate contributions to kinetic energy of velocity components parallel and perpendicular to the wall separately.) (d) The hole is closed and the container (now thermally insulated) is allowed to reach equilibrium. What is the final temperature of the gas in the container? (e) A vessel partially filled with mercury (atomic weight 201), and closed except for a hole of area 0.1mm2 above the liquid level, is kept at 00 C in a continuously evacuated enclosure. After 30 days it is found that 24mg of mercury has been lost. What is the vapor pressure of mercury at 00 C? ******** 15. Adsorbed particles: Consider a gas of classical particles of mass m in thermal equilibrium at a temperature T , and with a density n. A clean metal surface is introduced into the gas. Particles hitting this surface with normal velocity less than vt are reflected back into the gas, while particles with normal velocity greater than vt are absorbed by it. 86

(a) Find the average number of particles hitting one side of the surface per unit area and per unit time. (b) Find the average number of particles absorbed by one side of the surface per unit area and per unit time. ******** 16. Electron emission:

When a metal is heated in vacuum, electrons are emitted from

its surface. The metal is modeled as a classical gas of noninteracting electrons held in the solid by an abrupt potential well of depth φ (the work function) relative to the vacuum. (a) What is the relationship between the initial and final velocities of an escaping electron? (b) In thermal equilibrium at temperature T , what is the probability density function for the velocity of electrons? (c) If the number density of electrons is n, calculate the current density of thermally emitted electrons. ********

87

Problems for Chapter IV - Classical Statistical Mechanics 1. Classical harmonic oscillators:

Consider N harmonic oscillators with coordinates and

momenta {qi , pi }, and subject to a Hamiltonian N 2 X pi mω 2 qi2 . + H({qi , pi }) = 2m 2 i=1 (a) Calculate the entropy S, as a function of the total energy E. (Hint: By appropriate change of scale, the surface of constant energy can be deformed into a sphere. You may then ignore the difference between the surface area and volume for N ≫ 1. A more elegant method is to implement this deformation through a canonical

transformation.)

• The volume of accessible phase space for a given total energy is proportional to Z 1 dq1 dq2 · · · dqN dp1 dp2 · · · dpN , Ω= N h H=E where the integration is carried out under the condition of constant energy, N 2 X pi mω 2 qi2 . E = H ({qi , pi }) = + 2m 2 i=1 Note that Planck’s constant h, is included as a measure of phase space volume, so as to make the final result dimensionless. The surface of constant energy is an ellipsoid in 2N dimensions, whose area is difficult to calculate. However, for N → ∞ the difference between volume and area is subleading in N , and we shall instead calculate the volume of the ellipsoid, replacing the constraint H = E by H ≤ E. The ellipsoid can be distorted into a sphere by a canonical transformation, changing coordinates to

qi′ ≡

√

mωqi ,

and

pi . p′i ≡ √ mω

The Hamiltonian in this coordinate system is E=

H ({qi′ , p′i })

N ω X ′2 pi + qi′2 . = 2 i=1

88

Since the canonical transformation preserves volume in phase space (the Jacobian is unity), we have Z 1 ′ dq1′ · · · dqN dp′1 · · · dp′N , Ω≈ N h H≤E where the integral is now over the 2N -dimensional (hyper-) sphere of radius R = As the volume of a d-dimensional sphere of radius R is Sd Rd /d, we obtain N

=

S ≡ kB ln Ω ≈ N kB ln

2πeE N hω

2π N 1 Ω≈ · (N − 1)! 2N

2E hω

2πE hω

N

p

2E/ω.

1 . N!

The entropy is now given by

.

(b) Calculate the energy E, and heat capacity C, as functions of temperature T , and N . • From the expression for temperature, 1 N kB ∂S ≈ ≡ , T ∂E N E

we obtain the energy

E = N kB T, and the heat capacity C = N kB .

(c) Find the joint probability density P (p, q) for a single oscillator. Hence calculate the mean kinetic energy, and mean potential energy for each oscillator. • The single particle distribution function is calculated by summing over the undesired

coordinates and momenta of the other N − 1 particles. Keeping track of the units of h used to make phase space dimensionless, gives p(p1 , q1 )dp1 dq1 =

1 dq2 · · · dqN dp2 · · · dpN (H≤EN −1 ) hN −1 R 1 dq1 dq2 · · · dqN dp1 dp2 · · · dpN (H≤E) hN

R

89

×

dp1 dq1 , h

where EN−1 = E − p21 /2m − mω 2 q12 /2. Using the results from part (a), p(p1 , q1 ) =

=

=

Ω (N − 1, EN−1 ) hΩ(N, E) 2 N−1 π N −1 hω

ω N 2π E

p2

2

1 E − 2m − mω q12 (N−1)! 2 2 N πN N h hω N! E !N−1 p21 mω 2 2 + q 1 2 . 1 − 2m E

N−1

Using the approximation (N − 1) ∼ N , for N ≫ 1, and setting E = N kB T , we have ω N p(p1 , q1 ) = 2π N kB T

1−

!N 2 2 + mω q 1 2 N kB T ! 2 2 + mω q 1 2 . kB T

p21 2m

p2

1 ω ≈ exp − 2m 2πkB T

Let us denote (p1 , q1 ) by (p, q), then ω p2 mω 2 q 2 p(p, q) = , exp − − 2πkB T 2mkB T 2kB T is a properly normalized product of two Gaussians. The mean kinetic energy is

p2 2m

=

Z

p(p, q)

p2 kB T dqdp = , 2m 2

p(p, q)

mω 2 q 2 kB T dqdp = . 2 2

while the mean potential energy is also

mω 2 q 2 2

=

Z

******** 2. Quantum harmonic oscillators: Consider N independent quantum oscillators subject to a Hamiltonian

N X

1 ¯hω ni + H ({ni }) = , 2 i=1 where ni = 0, 1, 2, · · ·, is the quantum occupation number for the ith oscillator. 90

(a) Calculate the entropy S, as a function of the total energy E. (Hint: Ω(E) can be regarded as the number of ways of rearranging M = N − 1 partitions along a line.)

P

i

ni balls, and

• The total energy of the set of oscillators is E=h ¯ω

N X

N ni + 2 i=1

!

.

Let us set the sum over the individual quantum numbers to M≡

N X

ni =

i=1

E N − . ¯hω 2

The number of configurations {ni } for a given energy (thus for a given value of M ) is

equal to the possible number of ways of distributing M energy units into N slots, or of

partitioning M particles using N − 1 walls. This argument gives to the number of states as

Ω=

(M + N − 1)! , M ! (N − 1)!

and a corresponding entropy M M +N −1 N −1 − M ln . S = kB ln Ω ≈ kB (M + N − 1) ln − (N − 1) ln e e e (b) Calculate the energy E, and heat capacity C, as functions of temperature T , and N . • The temperature is calculated by 1 kB kB ∂S M +N −1 ≈ = ≡ ln ln T ∂E N ¯ω h M ¯hω

E h ¯ω

+

E h ¯ω

N 2

−

−1

N 2

!

kB ln ≈ ¯hω

E+ E−

By inverting this equation, we get the energy

N exp (¯ hω/kB T ) + 1 1 1 E = ¯hω , = N ¯hω + 2 exp (¯ hω/kB T ) − 1 2 exp (¯ hω/kB T ) − 1 and a corresponding heat capacity 2 ¯hω exp (¯ hω/kB T ) ∂E = N kB C≡ 2. ∂T N kB T [exp (¯ hω/kB T ) − 1] 91

N ¯hω 2 N ¯hω 2

!

.

(c) Find the probability p(n) that a particular oscillator is in its nth quantum level. • The probability that a particular oscillator is in its nth quantum level is given by summing the joint probability over states for all the other oscillators, i.e. P X Ω N − 1, E − (n + 12 )¯ hω (E−(n+1/2)¯ hω) 1 P = p(n) = p(ni ) = Ω(N, E) (E) 1 {ni6=1 }

[(M − n) + N − 2]! M ! · (N − 1)! · (M − n)! · (N − 2)! (M + N − 1)! M (M − 1) · · · (M − n + 1) · N ≈ (M + N − 1)(M + N − 2) · · · (M + N − n − 1) =

≈ N (M + N )−n−1 M n ,

where the approximations used are of the from (I − 1) ≈ I, for I ≫ 1. Hence, p(n) = N

E N − +N ¯hω 2

N = E h ¯ω +

N 2

E h ¯ω E h ¯ω

−n−1

N 2 N 2

− +

!n

N E − ¯hω 2

n

,

which using 1 kB = ln T ¯ω h

E+ E−

N¯ hω 2 N¯ hω 2

!

,

E h ¯ω E h ¯ω

=⇒

+ −

N 2 N 2

!n

n¯hω , = exp − kB T

leads to the probability ¯ω h ¯hω p(n) = exp −n 1 − exp − . kB T kB T (d) Comment on the difference between heat capacities for classical and quantum oscillators. • As found in part (b), Cquantum = N kB

¯ω h kB T

2

h

h ¯ω kB T

exp i2 . h ¯ω exp kB T − 1

In the high temperature limit, h ¯ ω/(kB T ) ≪ 1, using the approximation ex ≈ 1 + x for

x ≪ 1, gives

Cquantum = N kB = Cclassical . 92

At low temperatures, the quantized nature of the energy levels of the quantum oscillators becomes noticeable. In the limit T → 0, there is an energy gap between the ground state

and the first excited state. This results in a heat capacity that goes to zero exponentially, as can be seen from the limit ¯hω/(kB T ) ≫ 1, Cquantum = N kB

¯ω h kB T

2

¯ω h . exp − kB T

******** 3. Relativistic particles: N indistinguishable relativistic particles move in one dimension subject to a Hamiltonian H ({pi , qi }) =

N X i=1

[c|pi | + U (qi )] ,

with U (qi ) = 0 for 0 ≤ qi ≤ L, and U (qi ) = ∞ otherwise. Consider a microcanonical ensemble of total energy E.

(a) Compute the contribution of the coordinates qi to the available volume in phase space Ω(E, L, N ). •

Each of N coordinates explores a length L, for an overall contribution of LN /N !.

Division by N ! ensures no over-counting of phase space for indistinguishable particles. (b) Compute the contribution of the momenta pi to Ω(E, L, N ). Pd (Hint: The volume of the hyper-pyramid defined by i=1 xi ≤ R, and xi ≥ 0, in d dimensions is Rd /d! .)

• The N momenta satisfy the constraint

PN

i=1

|pi | = E/c. For a particular choice of the

signs of {pi }, this constraint describes the surface of a hyper-pyramid in N dimensions. If

we ignore the difference between the surface area and volume in the large N limit, we can calculate the volume in momentum space from the expression given in the hint as 1 · Ωp = 2 · N! N

E c

N

.

The factor of 2N takes into account the two possible signs for each pi . The surface area √ √ of the pyramid is given by dRd−1 /(d − 1)!; the additional factor of d with respect to 93

dvolume/dR is the ratio of the normal to the base to the side of the pyramid. Thus, the volume of a shell of energy uncertainly ∆E , is Ω′p

√

N · =2 · (N − 1)! N

E c

N−1

·

∆E . c

We can use the two expressions interchangeably, as their difference is subleading in N . (c) Compute the entropy S(E, L, N ). • Taking into account quantum modifications due to anisotropy, and phase space measure, we have

√ N−1 1 LN N N E ∆E Ω(E, L, N ) = N · ·2 · · · . h N! (N − 1)! c c

Ignoring subleading terms in the large N limit, the entropy is given by S(E, L, N ) = N kB ln

2e2 L E · · hc N N

.

(d) Calculate the one dimensional pressure P . • From dE = T dS − P dV + µdN , the pressure is given by N kB T ∂S = . P =T ∂L E,N L (e) Obtain the heat capacities CL and CP . • Temperature and energy are related by N kB ∂S 1 = = , T ∂E L,N E

=⇒

E = N kB T,

=⇒

∂E = N kB . CL = ∂T L,N

Including the work done against external pressure, and using the equation of state, ∂E ∂L CP = +P = 2N kB . ∂T P,N ∂T P,N (f) What is the probability p(p1 ) of finding a particle with momentum p1 ? 94

• Having fixed p1 for the first particle, the remaining N − 1 particles are left to share

an energy of (E − c|p1 |). Since we are not interested in the coordinates, we can get the

probability from the ratio of phase spaces for the momenta, i.e.

Ωp (E − c|p1 |, N − 1) Ωp (E, N ) " N−1 # N ! c N E − c|p1 | 2N−1 × N · = · (N − 1)! c 2 E N cN c|p1 | cN |p1 | cN ≈ . · 1− · exp − ≈ 2E E 2E E

p(p1 ) =

Substituting E = N kB T , we obtain the (property normalized) Boltzmann weight c|p1 | c . · exp − p(p1 ) = 2kB T kB T ******** 4. Hard sphere gas: Consider a gas of N hard spheres in a box. A single sphere occupies volume ω, while its center of mass can explore a volume V (if the box is otherwise empty). There are no other interactions between the spheres, except for the constraints of hard-core exclusion. (a) Calculate the entropy S, as a function of the total energy E. 2 (Hint: V − aω V − (N − a)ω ≈ V − N ω/2 .)

• The available phase space for N identical particles is given by Z 1 Ω= d3 ~q1 · · · d3 ~qN d3 p~1 · · · d3 ~pN , N !h3N H=E where the integration is carried out under the condition, N X p2i E = H(~qi , ~pi ) = , 2m i=1

N X

or

p2i = 2mE.

i=1

The momentum integrals are now performed as in an ideal gas, yielding (2mE)3N/2−1 · Ω= N !h3N

2π 3N · 3N 2 −1 ! 95

Z

d3 ~q1 · · · d3 ~qN .

The joint integral over the spacial coordinates with excluded volume constraints is best performed by introducing particles one at a time. The first particle can explore a volume V , the second V − ω, the third V − 2ω, etc., resulting in Z

d3 ~q1 · · · d3 ~qN = V (V − ω)(V − 2ω) · · · (V − (N − 1)ω).

Using the approximation (V − aω)(V − (N − a)ω) ≈ (V − N ω/2)2 , we obtain Z

N Nω d ~q1 · · · d q~N ≈ V − . 2 3

3

Thus the entropy of the system is "

e S = kB ln Ω ≈ N kB ln N

3/2 # Nω 4πmEe V − . 2 3N h2

(b) Calculate the equation of state of this gas. • We can obtain the equation of state by calculating the expression for the pressure of the gas,

which is easily re-arranged to,

N kB ∂S P ≈ = , T ∂V E,N V − Nω 2 P

Nω V − 2

= N kB T.

Note that the joint effective excluded volume that appears in the above expressions is one half of the total volume excluded by N particles. (c) Show that the isothermal compressibility, κT = −V −1 ∂V /∂P |T , is always positive. • The isothermal compressibility is calculated from N kB T 1 ∂V = κT = − > 0, V ∂P T,N P 2V

and is explicitly positive, as required by stability constraints. ******** 96

5. Non-harmonic gas:

Let us reexamine the generalized ideal gas introduced in the

previous section, using statistical mechanics rather than kinetic theory. Consider a gas of N non-interacting atoms in a d-dimensional box of “volume” V , with a kinetic energy H=

N X i=1

s

A |~ pi | ,

where p~i is the momentum of the ith particle. (a) Calculate the classical partition function Z(N, T ) at a temperature T . (You don’t have to keep track of numerical constants in the integration.) • The partition function is given by 1 Z(N, T, V ) = N !hdN 1 = N !hdN

Z

···

Z Z

Z

"

dd ~q1 · · · dd ~qN dd p~1 · · · dd ~pN exp −β d

s

d

d ~qd p~ exp (−βA |~ p| )

N

N X i=1

A |~ pi |

s

#

.

Ignoring hard core exclusion, each atom contributes a d–dimensional volume V to the integral over the spatial degrees of freedom, and Z N VN s d d p~ exp (−βA |~p | ) . Z(N, T, V ) = N !hdN Observing that the integrand depends only on the magnitude |~ p | = p, we can evaluate the R d R d−1 integral in spherical coordinates using d p~ = Sd dpp , where Sd denotes the surface area of a unit sphere in d–dimensions, as N Z ∞ VN d−1 s dpp exp (−βAp ) . Z(N, T, V ) = Sd N !hdN 0

Introducing the variable x ≡ (βA)1/s p, we have −dN /s Z ∞ N A V N SdN d−1 s dxx exp(−x ) Z(N, T, V ) = N !hdN kB T 0 −dN /s N 1 V Sd A N = C (d, s) , N! hd kB T where C denotes the numerical value of the integral. (We assume that A and s are both real and positive. These conditions ensure that energy increases with increasing |~ p |.) The integral is in fact equal to

C(d, s) =

Z

∞ 0

dxx

d−1

1 exp(−x )dx = Γ s s

97

d , s

and the partition function is 1 Z= N!

V Sd hd s

N

A kB T

−dN /s N d Γ . s

(b) Calculate the pressure and the internal energy of this gas. (Note how the usual equipartition theorem is modified for non-quadratic degrees of freedom.) • To calculate the pressure and internal energy, note that the Helmholtz free energy is F = E − T S = −kB T ln Z, and that

∂F P =− , ∂V T

while

First calculating the pressure:

∂ ln Z E=− . ∂β V

∂ ln Z N kB T ∂F . = kB T = P =− ∂V T ∂V T V

Now calculating the internal energy:

∂ dN d ∂ ln Z A = − − = E=− ln N kB T. ∂β V ∂β s kB T s

Note that for each degree of freedom with energy A|~ pi |s , we have the average value,

hA|~ pi |s i = ds kB T. This evaluates to 32 kB T for the 3–dimensional ideal gas. (c) Now consider a diatomic gas of N molecules, each with energy t (1) s (2) s (1) (2) Hi = A p~i + p~i + K ~qi − ~qi ,

where the superscripts refer to the two particles in the molecule. (Note that this unrealistic potential allowsthe two atoms to occupy the same point.) Calculate the expectation value t (1) (2) ~qi − ~qi , at temperature T .

• Now consider N diatomic molecules, with H=

N X i=1

Hi ,

where

t (1) s (2) s (1) (2) Hi = A ~p i + p~ i + K ~q i − ~q i . 98

The expectation value t (1) (2) ~q i − ~q i =

1 N!

t P (1) (2) ~q i − ~q i exp [−β i Hi ] , R QN P (1) d (2) d (1) d (2) 1 d~ d q d ~ q d p ~ d p ~ exp [−β H ] i i i i i i=1 i N! d (1) d (2) d (1) d (2) q i d ~q i d p~ i d p~ i i=1 d ~

R QN

is easily calculated by changing variables to ~x ≡ ~q (1) − ~q (2) ,

and

~y ≡

q~ (1) + ~q (2) , 2

as (note that the Jacobian of the transformation is unity) t R dd ~xdd ~y · |~x |t · exp [−βK|~x |t ] (1) R ~q − ~q (2) = dd ~xdd ~y · exp [−βK|~x |t ] R d d ~x · |~x |t · exp [−βK|~x |t ] R . = dd ~x · exp [−βK|~x |t ]

Further simplifying the algebra by introducing the variable ~z ≡ (βK)1/t~x, leads to t (βK)−t/t R dd ~z · |~z |t · exp [−|~z |t ] d kB T (1) (2) R . = · − ~q = ~q d t t K d ~z · exp [−|~z | ]

Here we have assumed that the volume is large enough, so that the range of integration over the relative coordinate can be extended from 0 to ∞. Alternatively, note that for the degree of freedom ~x = ~q (1) − ~q (2) , the energy is K|~x |t .

Thus, from part (b) we know that

d kB T K|~x |t = · , t K

i.e.

t |~x | =

t d k T (1) B (2) . − ~q = · ~q t K

And yet another way of calculating the expectation value is from N t X 1 ∂ ln Z N d kB T (1) (2) = · , ~q i − ~q i = − β ∂K t K i=1

(note that the relevant part of Z is calculated in part (d) below).

(d) Calculate the heat capacity ratio ratio γ = CP /CV , for the above diatomic gas. 99

• For the ideal gas, the internal energy depends only on temperature T . The gas in part

(c) is ideal in the sense that there are no molecule–molecule interactions. Therefore, dE + P dV dE(T ) d¯Q = = , CV = dT V dT dT V and

dE + P dV d¯Q = CP = dT P dT

Since P V = N kB T,

CP =

dE(T ) ∂V (T ) = . + P dT ∂T P P

dE(T ) + N kB . dT

We now calculate the partition function Z= =

1 N !hdN

Z Y N

i=1

(1)

(2)

(1)

(2)

"

dd ~q i dd ~q i dd p~ i dd p~ i exp −β

1 z1N , dN N !h

X i

Hi

#

where z1 = =

Z

Z

d (1) d (2) d

d ~q

d ~q

d p~

d (1) d (2)

d ~q

d ~q

(1) d

d p~

(2)

t (1) s (2) s (1) (2) exp −β · A ~p + A p~ + K ~q − ~q

t Z 2 (1) (1) s (2) d (1) exp −βK ~q − ~q · d p~ exp −βA p~ .

Introducing the variables, ~x, ~y , and ~z, as in part (c), N Z ∞ 2N Z ∞ VN −d/t d−1 t d−1 s (βK) z exp(−z )dz · p exp(−βAp )dp Z∝ N! 0 0 N 2N 1 d d VN −d/t 1 −d/s (βK) · (βA) Γ Γ = N! t t s s ∝

VN −dN/t −2Nd/s (βK) (βA) . N!

Now we can calculate the internal energy as ∂ ln Z d 2d hEi = − = N kB T + N kB T = dN kB T ∂β t s

1 2 + t s

From this result, the heat capacities are obtained as ∂V 2d d ∂E +P = N kB + +1 , CP = ∂T P ∂T P s t 2 1 ∂E = dN k . CV = + B ∂T V s t 100

.

resulting in the ratio γ=

CP 2d/s + d/t + 1 st = =1+ . CV 2d/s + d/t d(2t + s)

******** 6. Surfactant adsorption:

A dilute solution of surfactants can be regarded as an ideal

three dimensional gas. As surfactant molecules can reduce their energy by contact with air, a fraction of them migrate to the surface where they can be treated as a two dimensional ideal gas. Surfactants are similarly adsorbed by other porous media such as polymers and gels with an affinity for them. (a) Consider an ideal gas of classical particles of mass m in d dimensions, moving in a uniform attractive potential of strength εd . By calculating the partition function, or otherwise, show that the chemical potential at a temperature T and particle density nd , is given by µd = −εd + kB T ln nd λ(T )d ,

where

λ(T ) = √

h . 2πmkB T

• The partition function of a d-dimensional ideal gas is given by ( " #) Z Z Y Nd Nd 2 X p ~ 1 i ··· dd ~qi dd ~p i exp −β Nd εd + Zd = Nd !hdNd 2m i=1 i=1 Nd 1 Vd = e−βNd εd , Nd ! λd where λ≡ √

h . 2πmkB T

The chemical potential is calculated from the Helmholtz free energy as ∂ ln Zd ∂F = −kB T µd = ∂N V,T ∂Nd V,T Vd . = −εd + kB T ln Nd λd 101

(b) If a surfactant lowers its energy by ε0 in moving from the solution to the surface, calculate the concentration of floating surfactants as a function of the solution concentration n (= n3 ), at a temperature T . •

The density of particles can also be calculated from the grand canonical partition

function, which for particles in a d–dimensional space is Ξ(µ, Vd , T ) = =

∞ X

Nd =0 ∞ X

Nd =0

Z(Nd , Vd , T )eβNd µ 1 Nd !

Vd λd

Nd

−βNd εd βNd µ

e

e

= exp

Vd λd

β(µ−εd )

·e

.

The average number of particles absorbed in the space is 1 ∂ 1 ∂ hNd i = ln Ξ = β ∂µ β ∂µ

Vd λd

β(µ−εd )

·e

=

Vd λd

· eβ(µ−εd ) .

We are interested in the coexistence of surfactants between a d = 3 dimensional solution, and its d = 2 dimensional surface. Dividing the expressions for hN3 i and hN2 i, and taking

into account ε0 = ε3 − ε2 , gives

Aλ βε0 hN2 i = e , hN3 i V

which implies that n2 =

hN2 i = nλeβε0 . A

(c) Gels are formed by cross-linking linear polymers. It has been suggested that the porous gel should be regarded as fractal, and the surfactants adsorbed on its surface treated as a gas in df dimensional space, with a non-integer df . Can this assertion be tested by comparing the relative adsorption of surfactants to a gel, and to the individual polymers (presumably one dimensional) before cross-linking, as a function of temperature? • Using the result found in part (b), but regarding the gel as a df –dimensional container,

the adsorbed particle density is

hngel i = nλ3−df exp [β(ε3 − εgel )] . Thus by studying the adsorption of particles as a function of temperature one can determine the fractal dimensionality, df , of the surface. The largest contribution comes from the 102

difference in energies. If this leading part is accurately determined, there is a subleading dependence via λ3−df which depends on df . ******** 7. Molecular adsorption:

N diatomic molecules are stuck on a metal surface of square

symmetry. Each molecule can either lie flat on the surface in which case it must be aligned to one of two directions, x and y, or it can stand up along the z direction. There is an energy cost of ε > 0 associated with a molecule standing up, and zero energy for molecules lying flat along x or y directions. (a) How many microstates have the smallest value of energy? What is the largest microstate energy? • The ground state energy of E = Emin = 0 is obtained for 2N configurations. The largest microstate energy is N ǫ is unique.

(b) For microcanonical macrostates of energy E, calculate the number of states Ω(E, N ), and the entropy S(E, N ). • Let Nz = E/ǫ. Expressing Ω as the number of ways to choose the Ng excited molecules,

multiplied by the number of possible configurations Ω(E, N ) =

N! · 2N−Nz , Nz !(N − Nz )!

and S(E, N ) = Stwo−level system + kb (N − Nz ) ln 2 E E E E E = −N kB ln + (1 − ) ln(1 − ) + kb (N − ) ln 2. Nǫ Nǫ Nǫ Nǫ ǫ (c) Calculate the heat capacity C(T ) and sketch it. • The temperature dependence of the energy is obtained from the relation kB ∂S E kB 1 =− = ln( )− ln 2, T ∂E N ǫ Nǫ − E ǫ

whence

E= and

exp( kǫB ( T1

Nǫ Nǫ = , kB 2 exp( kBǫ T ) + 1 + ǫ ln 2)) + 1

2 exp( kBǫ T ) ǫ 2 dE = N kB ( ) . C= dT kB T (1 + 2 exp( kBǫ T ))2 103

C/kBN

(kεT)2e- ε/k T B

1 ε 2 4 kB T

( )

B

T

(d) What is the probability that a specific molecule is standing up? • The probability that a specific molecule is standing up is Ω(E − ǫ, N − 1) Ω(E, N ) (N − 1)! Nz !(N − Nz )! 1 = 2(N−1)−(Nz −1) (Nz − 1)!((N − 1) − (Nz − 1))! N! 2N−Nz E Nz = = N Nǫ 1 . = 2 exp kBǫ T + 1

p(~r1 = zˆ) =

(e) What is the largest possible value of the internal energy at any positive temperature? • Since

dE dT

> 0 for all T > 0, the energy is largest for → ∞, i.e. Emax =

Nǫ . 3

******** 8. Curie susceptibility: Consider N non-interacting quantized spins in a magnetic field ~ = B zˆ, and at a temperature T . The work done by the field is given by BMz , with a B PN magnetization Mz = µ i=1 mi . For each spin, mi takes only the 2s + 1 values −s, −s + 1, · · · , s − 1, s.

(a) Calculate the Gibbs partition function Z(T, B). (Note that the ensemble corresponding

to the macrostate (T, B) includes magnetic work.) • The Gibbs partition function is Z=

X

{mi }

~ ·M ~ = exp β B

X

exp βBµ

N X i=1

{mi }

104

mi

!

=

"

m i =s X

mi =−s

exp(βµB · mi )

#N

.

Thus we obtain the series N

Z = [exp(−βBµs) + exp(−βBµ(s − 1)) + · · · + exp(βBµ(s − 1)) + exp(βBµs)] . In general, to evaluate a geometrical series of the form S = x−s + x−(s−1) + · · · + xs−1 + xs , increase the order of the series by one, Sx = x−s+1 + · · · + xs + xs+1 , and subtract from the original series: (1 − x)S = x−s − xs+1 ,

=⇒

S=

x−s − xs+1 . 1−x

(Note that the same result is obtained whether s is an integer or half–integer quantity.) Using this expression, we get Z=

=

exp(−βBµs) − exp(βBµ(s + 1)) 1 − exp(βBµ)

N

exp(−βBµ(s + 1/2)) − exp(−βBµ(s + 1/2)) exp(−βBµ/2) − exp(−βBµ/2)

N

.

Substituting in the proper trigonometric identity,

sinh (βµB(s + 1/2)) Z= sinh(βµB/2)

N

.

(b) Calculate the Gibbs free energy G(T, B), and show that for small B, G(B) = G(0) −

N µ2 s(s + 1)B 2 + O(B 4 ). 6kB T

• The Gibbs free energy is G = E − BM = −kB T ln Z = −N kB T ln[sinh(βµB(s + 1/2))] + N kB T ln[sinh(βµB/2)]. 105

Using an approximation of sinh θ for small θ, 1 1 θ e − e−θ ≈ sinh θ = 2 2

θ3 2θ + 2 + O(θ 5 ), 3!

for

θ ≪ 1,

we find (setting α = βµB), G ≈ −N kB T

(

" 2 !# 1 α2 1 α α2 4 ln α s + 1+ s+ − ln 1+ + O(α ) . 2 6 2 2 24

Using the expansion ln(1 + x) = x − x2 /2 + x3 /3 − · · ·, we find

1 1 2 2 G ≈ −N kB T ln(2s + 1) + (α(s + 1/2)) − (α/2) + O(α4 ) 6 6 2 (s + s) ≈ −N kB T ln(2s + 1) − N kB T α2 6 2 2 N µ B s(s + 1) = G0 − + O(B 4 ). 6kB T

(c) Calculate the zero field susceptibility χ = ∂Mz /∂B|B=0 , and show that is satisfies Curie’s law χ = c/T.

• The magnetic susceptibility, χ = ∂Mz /∂B, is obtained by noting that the average

magnetization is

hMz i = kB T Thus χ=

∂G ∂ ln Z =− . ∂B ∂B

∂ hMz i ∂ ∂G N µ2 ts(s + 1) =− = , ∂B ∂B ∂B 3kB T

which obeys Curie’s law, χ = c/T , with c = N µ2 s(s + 1)/3kB . (d) Show that CB − CM = cB 2 /T 2 where CB and CM are heat capacities at constant B and M respectively.

******** 9. Langmuir isotherms:

An ideal gas of particles is in contact with the surface of a

catalyst. 106

(a) Show that the chemical potential of the gas particles is related to their temperature and pressure via µ = kB T ln P/T 5/2 + A0 , where A0 is a constant. • For convenience, we begin by defining the characteristic length, λ= √

h , 2πmkB T

in terms of which the free energy of ideal gas is given by, F = −N kB T ln

Ve N λ3

.

Then using the equation of state of ideal gas we get, ∂F µ= = −kB T ln ∂N "

V N λ3

= kB T ln(P T −5/2 ) + ln

= −kB T ln h3

kB T P λ3

5/2

kB 23/2 π 3/2 m3/2

!#

.

(b) If there are N distinct adsorption sites on the surface, and each adsorbed particle gains an energy ǫ upon adsorption, calculate the grand partition function for the two dimensional

gas with a chemical potential µ. • We have, Q(T, µ) =

N X N

n=0

n

N enβµ e−nβǫ = 1 + eβ(µ−ǫ) .

(c) In equilibrium, the gas and surface particles are at the same temperature and chemical potential. Show that the fraction of occupied surface sites is then given by f (T, P ) = P/ P + P0 (T ) . Find P0 (T ). • Since the average number of absorption sites occupied is, hN i = kB T

eβ(µ−ǫ) ∂ ln Q =N , ∂µ 1 + eβ(µ−ǫ)

and the fraction of occupied sites is given by, f (T, P ) =

eβ(µ−ǫ) . 1 + eβ(µ−ǫ)

107

Since the gas and surface particles have the same temperature and chemical potential, the realtion in (a), namely, P λ3 kB T

eβµ =

holds. Plugging this into the formula for f we obtain, f (T, P ) =

P , P + P0 (T )

with P0 (T ) =

kB T βǫ e . λ3

(d) In the grand canonical ensemble, the particle number N is a random variable. Calculate its characteristic function hexp(−ikN )i in terms of Q(βµ), and hence show that m

hN ic = −(kB T )

m−1

where G is the grand potential.

∂ m G , ∂µm T

• Note that in calculating Q(βµ) the term for N particles is proportional to eβµN . In calculating the average of e−ikN , we just replace the initial factor with e(βµ−ik)N , and hence

−ikN Q(βµ − ik) . e = Q(βµ)

For the cumulant generating function, we get

Hence,

ln e−ikN = ln Q(βµ − ik) − ln Q(βµ) = −βG(βµ − ik) + βG(βµ). ∂m (−βG(βµ − ik))|T ∂(−ik)m m ∂ m G 1−m ∂ G = −β = −β ∂(βµ)m T ∂µm T ∂ m G . = −(kB T )m−1 ∂µm T

hN m ic =

(e) Using the characteristic function, show that

2 ∂hN i . N c = kB T ∂µ T 108

•

2 ∂ G ∂hN i . = kB T hN 2 ic = −kB T ∂µ2 T ∂µ T

(f) Show that fluctuations in the number of adsorbed particles satisfy

2 N c 2

hN ic

• By definition of f ,

and since,

=

1−f . Nf

∂G hN ic = − = N f, ∂µ T

−βµ −βǫ ∂f e e−βǫ ∂ = βe = 2 = βf (1 − f ), ∂µ T ∂µ e−βµ + e−βǫ T (e−βµ + e−βǫ )

we get,

hN 2 ic =

1 ∂f 1 ∂N = N = N f (1 − f ), β ∂µ β ∂µ

leading to the equality,

hN 2 ic 2 hN ic

=

1−f . Nf

******** ~ of unity, i.e. S z is quantized to -1, 0, 10. Molecular oxygen has a net magnetic spin, S, ~ k zˆ is or +1. The Hamiltonian for an ideal gas of N such molecules in a magnetic field B H=

N X p~i i=1

2

2m

−

µBSiz

,

where {~ pi } are the center of mass momenta of the molecules. The corresponding coordi-

nates {~qi } are confined to a volume V . (Ignore all other degrees of freedom.)

(a) Treating {~ pi , ~qi } classically, but the spin degrees of freedom as quantized, calculate the ˜ partition function, Z(T, N, V, B). • Z=

X µs

−βH(µs )

e

=

X µs

−β

e

P

(~ p2i /2m−µBSiz )

109

1 = N!

V βµB (e + 1 + e−βµB ) λ3

N

,

where, λ= √

h . 2πmkB T

(b) What are the probabilities for Siz of a specific molecule to take on values of -1, 0, +1 at a temperature T ? • The probabilities for Siz of a given molecule to take values -1, 0, +1, are, e−βµB , 2 cosh βµB + 1

eβµB , 2 cosh βµB + 1

1 , 2 cosh βµB + 1

respectively. (c) Find the average magnetic dipole moment, hM i /V , where M = µ

•

1 ∂ ln Z 1 ∂N ln(2 cosh βµB + 1) = β ∂B β ∂B 2 sinh βµB . = Nµ 2 cosh βµB + 1

PN

i=1

Siz .

hM i =

(d) Calculate the zero field susceptibility χ = ∂ < M > /∂B|B=0 . •

2 cosh βµB(2 cosh βµB + 1) − 4 sinh βµB sinh βµB 2 χ = N βµ = N βµ2 . 2 (2 cosh βµB + 1) 3 B=0 2

********

~ of unity, i.e. S z is quantized to -1, 0, 11. Molecular oxygen has a net magnetic spin, S, ~ k zˆ is or +1. The Hamiltonian for an ideal gas of N such molecules in a magnetic field B H=

N X p~i i=1

2

2m

−

µBSiz

,

where {~ pi } are the center of mass momenta of the molecules. The corresponding coordi-

nates {~qi } are confined to a volume V . (Ignore all other degrees of freedom.)

(a) Treating {~ pi , ~qi } classically, but the spin degrees of freedom as quantized, calculate the ˜ partition function, Z(T, N, V, B). • Z=

X µs

−βH(µs )

e

=

X µs

−β

e

P

(~ p2i /2m−µBSiz )

110

1 = N!

V βµB (e + 1 + e−βµB ) λ3

N

,

where, λ= √

h . 2πmkB T

(b) What are the probabilities for Siz of a specific molecule to take on values of -1, 0, +1 at a temperature T ? • The probabilities for Siz of a given molecule to take values -1, 0, +1, are, e−βµB , 2 cosh βµB + 1

1 , 2 cosh βµB + 1

eβµB , 2 cosh βµB + 1

respectively. (c) Find the average magnetic dipole moment, hM i /V , where M = µ

•

1 ∂N ln(2 cosh βµB + 1) 1 ∂ ln Z = β ∂B β ∂B 2 sinh βµB . = Nµ 2 cosh βµB + 1

PN

i=1

Siz .

hM i =

(d) Calculate the zero field susceptibility χ = ∂ < M > /∂B|B=0 . •

2 2 cosh βµB(2 cosh βµB + 1) − 4 sinh βµB sinh βµB = N βµ2 . χ = N βµ 2 (2 cosh βµB + 1) 3 B=0 2

******** 12. Polar rods:

Consider rod shaped molecules with moment of inertia I, and a dipole

moment µ. The contribution of the rotational degrees of freedom to the Hamiltonian is given by Hrot.

1 = 2I

p2θ +

p2φ sin2 θ

!

− µE cos θ

,

where E is an external electric field. (φ ∈ [0, 2π], θ ∈ [0, π] are the azimuthal and polar

angles, and pφ , pθ are their conjugate momenta.)

(a) Calculate the contribution of the rotational degrees of freedom of each dipole to the classical partition function. 111

• The classical partition function is obtained by integrating over the angles θ and φ, and

the corresponding momenta as " Z π Z 2π Z ∞ 1 β Zrot = 2 dθ dφ dpθ dpφ exp − h 0 2I 0 −∞

p2θ +

p2φ sin2 θ

!

#

+ βµ cos θ .

The Gaussian integrations over momenta are easily carried out, and after the change of variables to x = cos θ, we find 2 Z +1 2πI 8π I sinh(βµE) βµEx Zrot = dxe = (2π) . 2 βh βh2 βµE −1 (b) Obtain the mean polarization P = hµ cos θi, of each dipole.

• From the form of the partition function, we note that ∂ ln Zrot 1 P = hµ cos θi = . = µ coth(βµE) − ∂βE βµE (c) Find the zero–field polarizability

• The susceptibility is

∂P χT = ∂E E=0

.

∂P 1 1 µ2 1 2 χT = = + = βµ − ∂E 3 kB T sinh2 (βµE) (βµE)2

for E = 0 .

(d) Calculate the rotational energy per particle (at finite E), and comment on its high and low temperature limits. • The energy stored in this degree of freedom is hErot i = −

∂ ln Zrot = 2kB T − µE coth(βµE). ∂β

At high temperature hErot i ≈ kB T coming entirely from the kinetic energies. At temper-

atures kB T ≪ µE, the polarization is saturated, and hErot i ≈ −µE + 2kB T . (e) Sketch the rotational heat capacity per dipole.

• The classical heat capacity is 2kB at low temperatures, and decreases to 1kB at high temperatures, with a ‘knee’ at kB T ≈ µE.

******** 112

Problems for Chapter V - Interacting Particles 1. Debye–H¨ uckel theory and ring diagrams:

The virial expansion gives the gas pressure

as an analytic expansion in the density n = N/V . Long range interactions can result in non-analytic corrections to the ideal gas equation of state. A classic example is the Coulomb interaction in plasmas, whose treatment by Debye–H¨ uckel theory is equivalent to summing all the ring diagrams in a cumulant expansion. For simplicity consider a gas of N electrons moving in a uniform background of positive charge density N e/V to ensure overall charge neutrality. The Coulomb interaction takes the form UQ =

X i Tc , T = Tc , and T < Tc . For T < Tc there is a range of compositions x < |xsp (T )| where F (x) is not convex and hence the composition is locally unstable. Find xsp (T ).

• The function F (x) is concave if ∂ 2 F /∂x2 < 0, i.e. if 2

x

Tc (g) In the (T, x) plane sketch the phase separation boundary ±xeq (T ); and the so called

spinodal line ±xsp (T ). (The spinodal line indicates onset of metastability and hysteresis

effects.)

• The spinodal and equilibrium curves are indicated in the figure above. In the interval

between the two curves, the system is locally stable, but globally unstable. The formation of ordered regions in this regime requires nucleation, and is very slow. The dashed area is locally unstable, and the system easily phase separates to regions rich in A and B. ********

145

T Tc x eq(T) x sp(T)

unstable

metastable

metastable

1

1

x

Problems for Chapter VI - Quantum Statistical Mechanics 1. One dimensional chain: A chain of N +1 particles of mass m is connected by N massless springs of spring constant K and relaxed length a. The first and last particles are held fixed at the equilibrium separation of N a. Let us denote the longitudinal displacements of the particles from their equilibrium positions by {ui }, with u0 = uN = 0 since the end particles are fixed. The Hamiltonian governing {ui }, and the conjugate momenta {pi }, is H=

N−1 X i=1

" # N−2 X p2i K 2 u21 + (ui+1 − ui ) + u2N−1 . + 2m 2 i=1

(a) Using the appropriate (sine) Fourier transforms, find the normal modes {˜ uk }, and the corresponding frequencies {ωk }.

• From the Hamiltonian H=

N−1 X i=1

# " N−1 X K p2i 2 (ui − ui−1 ) + u2N−1 , u21 + + 2m 2 i=2

the classical equations of motion are obtained as m

d 2 uj = −K(uj − uj−1 ) − K(uj − uj+1 ) = K(uj−1 − 2uj + uj+1 ), dt2 146

for j = 1, 2, · · · , N − 1, and with u0 = uN = 0. In a normal mode, the particles oscillate in phase. The usual procedure is to obtain the modes, and corresponding frequencies,

by diagonalizing the matrix of coefficeints coupling the displacements on the right hand side of the equation of motion. For any linear system, we have md2 ui /dt2 = Kij uj , and

we must diagonalize Kij . In the above example, Kij is only a function of the difference

i − j. This is a consequence of translational symmetry, and allows us to diagonalize the

matrix using Fourier modes. Due to the boundary conditions in this case, the appropriate transformation involves the sine, and the motion of the j-th particle in a normal mode is given by r

2 ±iωn t e sin (k(n) · j) . N The origin of time is arbitrary, but to ensure that uN = 0, we must set u ˜k(n) (j) =

k(n) ≡

nπ , N

n = 1, 2, · · · , N − 1.

for

Larger values of n give wave-vectors that are simply shifted by a multiple of π, and hence coincide with one of the above normal modes. The number of normal modes thus equals the number of original displacement variables, as required. Furthermore, the amplitudes are chosen such that the normal modes are also orthonormal, i.e. N−1 X j=1

u ˜k(n) (j) · u ˜k(m) (j) = δn,m .

By substituting the normal modes into the equations of motion we obtain the dispersion relation

where ω0 ≡

nπ h nπ i = ω02 sin2 , ωn2 = 2ω02 1 − cos N 2N

p K/m.

The potential energy for each normal mode is given by N N h nπ io2 KX K X n nπ 2 Un = |ui − ui−1 | = sin i − sin (i − 1) 2 i=1 N i=1 N N N X 1 4K 2 nπ 2 nπ i− . cos sin = N 2N i=1 N 2

Noting that N X i=1

cos

2

nπ N

1 i− 2

N

h nπ io N 1 Xn 1 + cos = (2i − 1) = , 2 i=1 N 2 147

we have 2

Uk(n) = 2K sin

nπ 2N

.

(b) Express the Hamiltonian in terms of the amplitudes of normal modes {˜ uk }, and evaluate the classical partition function. (You may integrate the {ui } from −∞ to +∞). • Before evaluating the classical partition function, lets evaluate the potential energy by first expanding the displacement using the basis of normal modes, as uj =

N−1 X n=1

an · u ˜k(n) (j).

The expression for the total potential energy is (N−1 )2 N N X X K KX . (ui − ui−1 )2 = an u ˜k(n) (j) − u ˜k(n) (j − 1) U= 2 i=1 2 i=1 n=1

Since N−1 X j=1

u ˜k(n) (j) · u ˜k(m) (j − 1) =

N−1 X 1 δn,m {− cos [k(n)(2j − 1)] + cos k(n)} = δn,m cos k(n), N j=1

the total potential energy has the equivalent forms U=

=

N−1 N X KX 2 (ui − ui−1 ) = K a2n (1 − cos k(n)) , 2 i=1 n=1

N−1 X i=1

a2k(n) ε2k(n)

= 2K

N−1 X i=1

a2k(n)

2

sin

nπ 2N

.

The next step is to change the coordinates of phase space from uj to an . The Jacobian associated with this change of variables is unity, and the classical partition function is now obtained from # " Z ∞ Z ∞ N−1 X 1 2 nπ 2 , da1 · · · daN−1 exp −2βK an sin Z = N−1 λ 2N −∞ −∞ n=1 √ where λ = h/ 2πmkB T corresponds to the contribution to the partition function from each momentum coordinate. Performing the Gaussian integrals, we obtain N−1 Z ∞ i h 1 Y 2 nπ 2 , dan exp −2βKan sin Z = N−1 λ 2N −∞ n=1 N2−1 N−1 Y h nπ i−1 1 πkB T = N−1 . sin λ 2K 2N n=1 148

D

2

(c) First evaluate |˜ uk |

E

, and use the result to calculate u2i . Plot the resulting squared

displacement of each particle as a function of its equilibrium position. • The average squared amplitude of each normal mode is R∞

2 nπ 2 2 da (a ) exp −2βKa sin n n n 2N −∞ R∞ 2 sin2 nπ da exp −2βKa n n 2N −∞ h nπ i−1 kB T 1 = 4βK sin2 = 2 nπ . 2N 4K sin 2N

2 an =

The variation of the displacement is then given by

2 uj =

*"N−1 X

2 = N

n=1

an u ˜n (j)

#2 +

=

N−1 X

N−1 X n=1

2 2 an u ˜n (j)

N−1

2 2 nπ kB T X sin2 j = an sin N 2KN n=1 sin2 n=1

nπ j N nπ 2N

.

The evaluation of the above sum is considerably simplified by considering the combination N−1

2 2

2 kB T X 2 cos uj+1 + uj−1 − 2 uj = 2KN n=1

2nπ 2nπ 2nπ j − cos (j + 1) − cos (j − 1) N N N 1 − cos nπ N N−1 1 − cos nπ kB T X 2 cos 2nπ kB T N j N = , =− nπ 2KN n=1 KN 1 − cos N

PN−1

cos(πn/N ) = −1. It is easy to check that subject to the

n=1

boundary conditions of u20 = u2N = 0, the solution to the above recursion relation is

where we have used

2 kB T j(N − j) . uj = K N

(d) How are the results modified if only the first particle is fixed (u0 = 0), while the other end is free (uN 6= 0)? (Note that this is a much simpler problem as the partition function can be evaluated by changing variables to the N − 1 spring extensions.)

• When the last particle is free, the overall potential energy is the sum of the contributions PN−1 of each spring, i.e. U = K j=1 (uj − uj−1 )2 /2. Thus each extension can be treated

independently, and we introduce a new set of independent variables ∆uj ≡ uj − uj−1 . (In 149

Amplitude squared

NkT/K B

free end

fixed end 0

N/2 Position j

0

N

the previous case, where the two ends are fixed, these variables were not independent.) The partition function can be calculated separately for each spring as Z ∞ Z ∞ N−1 X K 1 (uj − uj−1 )2 du1 · · · duN−1 exp − Z = N−1 λ 2k T B −∞ −∞ j=1 (N−1)/2 Z ∞ Z ∞ N−1 X K 2πk T 1 B 2 d∆uN−1 exp − . d∆u1 · · · ∆uj = = N−1 λ 2kB T λ2 K −∞ −∞ j=1

For each spring extension, we have

2

kB T ∆uj = (uj − uj−1 )2 = . K The displacement

uj =

j X

∆ui ,

i=1

is a sum of independent random variables, leading to the variance * j !2 + j X X

2 kB T 2 uj = (∆ui ) = ∆ui = j. K i=1

i=1

The results for displacements of open and closed chains are compared in the above figure. ******** 2. Black hole thermodynamics: According to Bekenstein and Hawking, the entropy of a black hole is proportional to its area A, and given by S=

kB c3 A . 4G¯h 150

(a) Calculate the escape velocity at a radius R from a mass M using classical mechanics. Find the relationship between the radius and mass of a black hole by setting this escape velocity to the speed of light c. (Relativistic calculations do not modify this result which was originally obtained by Laplace.) • The classical escape velocity is obtained by equating the gravitational energy and the

kinetic energy on the surface as,

2 Mm mvE G = , R 2

leading to vE =

r

2GM . r

Setting the escape velocity to the speed of light, we find R=

2G M. c2

For a mass larger than given by this ratio (i.e. M > c2 R/2G), nothing will escape from distances closer than R. (b) Does entropy increase or decrease when two black holes collapse into one? What is the entropy change for the universe (in equivalent number of bits of information), when two solar mass black holes (M⊙ ≈ 2 × 1030 kg) coalesce?

• When two black holes of mass M collapse into one, the entropy change is kB c3 kB c3 ∆S = S2 − 2S1 = (A2 − 2A1 ) = 4π R22 − 2R12 4G¯h " 4G¯h 2 # 2 3 πkB c 2G 8πGkB M 2 2G = 2M M > 0. − 2 = G¯h c2 c2 c¯h Thus the merging of black holes increases the entropy of the universe. Consider the coalescence of two solar mass black holes. The entropy change is 2 8πGkB M⊙ c¯h 8π · 6.7 × 10−11 (N · m2 /kg 2 ) · 1.38 × 10−23 (J/K) · (2 × 1030 )2 kg 2 ≈ 3 × 108 (m/s) · 1.05 × 10−34 (J · s)

∆S =

≈ 3 × 1054 (J/K).

151

In units of bits, the information lost is NI =

∆S ln 2 = 1.5 × 1077 . kB

(c) The internal energy of the black hole is given by the Einstein relation, E = M c2 . Find the temperature of the black hole in terms of its mass. • Using the thermodynamic definition of temperature

1 T

=

∂S ∂E ,

and the Einstein relation

2

E = Mc ,

1 ∂ 1 = 2 T c ∂M

"

3

kB c 4π 4G¯h

2G M c2

2 #

8πkB G = M, ¯hc3

=⇒

¯ c3 1 h T = . 8πkB G M

(d) A “black hole” actually emits thermal radiation due to pair creation processes on its event horizon. Find the rate of energy loss due to such radiation. • The (quantum) vacuum undergoes fluctuations in which particle–antiparticle pairs are

constantly created and destroyed. Near the boundary of a black hole, sometimes one member of a pair falls into the black hole while the other escapes. This is a hand-waving explanation for the emission of radiation from black holes. The decrease in energy E of a black body of area A at temperature T is given by the Stefan-Boltzmann law, 1 ∂E = −σT 4 , A ∂t

where

4 π 2 kB σ= . 60¯ h3 c2

(e) Find the amount of time it takes an isolated black hole to evaporate. How long is this time for a black hole of solar mass? • Using the result in part (d) we can calculate the time it takes a black hole to evaporate.

For a black hole 2

A = 4πR = 4π Hence

which implies that

2G M c2

2

=

16πG2 2 M , c4

4 d π 2 kB M c2 = − dt 60¯ h3 c2

M2

E = M c2 ,

16πG2 2 M c4

¯ c3 1 h 8πkB G M

dM ¯hc4 =− ≡ −b. dt 15360G2 152

and T =

4

,

¯ c3 1 h . 8πkB G M

This can be solved to give M (t) = M03 − 3bt

1/3

.

The mass goes to zero, and the black hole evaporates after a time τ=

M03 5120G2 M ⊙3 = ≈ 2.2 × 1074 s, 3b ¯hc4

which is considerably longer than the current age of the universe (approximately ×1018 s). (f) What is the mass of a black hole that is in thermal equilibrium with the current cosmic background radiation at T = 2.7K? • The temperature and mass of a black hole are related by M = h ¯ c3 /(8πkB GT ). For a

black hole in thermal equilibrium with the current cosmic background radiation at T = 2.7◦ K, M≈

1.05 × 10−34 (J · s)(3 × 108 )3 (m/s)3 ≈ 4.5 × 1022 kg. 8π · 1.38 × 10−23 (J/K) · 6.7 × 10−11 (N · m2 /kg 2 ) · 2.7◦ K

(g) Consider a spherical volume of space of radius R. According to the recently formulated Holographic Principle there is a maximum to the amount of entropy that this volume of space can have, independent of its contents! What is this maximal entropy? • The mass inside the spherical volume of radius R must be less than the mass that would

make a black hole that fills this volume. Bring in additional mass (from infinity) inside

the volume, so as to make a volume-filling balck hole. Clearly the entropy of the system will increase in the process, and the final entropy, which is the entropy of the black hole is larger than the initial entropy in the volume, leading to the inequality S ≤ SBH =

kB c3 A, 4G¯h

where A = 4πR2 is the area enclosing the volume. The surprising observation is that the upper bound on the entropy is proportional to area, whereas for any system of particles we expect the entropy to be proportional to N . This should remain valid even at very high temperatures when interactions are unimportant. The ‘holographic principle’ is an allusion to the observation that it appears as if the degrees of freedom are living on the surface of the system, rather than its volume. It was formulated in the context of string theory which attempts to construct a consistent theory of quantum gravity, which replaces particles as degrees of freedom, with strings. 153

******** 3. Quantum harmonic oscillator: Consider a single harmonic oscillator with the Hamiltonian H=

p2 mω 2 q 2 + , 2m 2

with p =

¯ d h i dq

.

(a) Find the partition function Z, at a temperature T , and calculate the energy hHi. • The partition function Z, at a temperature T , is given by Z = tr ρ =

X

e−βEn .

n

As the energy levels for a harmonic oscillator are given by 1 , ǫn = h ¯ω n + 2 the partition function is Z=

X n

=

1 exp −β¯hω n + = e−β¯hω/2 + e−3β¯hω/2 + · · · 2

1 1 . = 2 sinh (β¯hω/2) eβ¯hω/2 − e−β¯hω/2

The expectation value of the energy is ∂ ln Z 1 ¯hω cosh(β¯hω/2) ¯hω hHi = − = = . ∂β 2 sinh(β¯hω/2) 2 tanh(β¯hω/2) (b) Write down the formal expression for the canonical density matrix ρ in terms of the eigenstates ({|ni}), and energy levels ({ǫn }) of H.

• Using the formal representation of the energy eigenstates, the density matrix ρ is ! X 1 β¯hω < n| . |n > exp −β¯hω n + ρ = 2 sinh 2 2 n In the coordinate representation, the eigenfunctions are in fact given by 2 mω 1/4 H (ξ) ξ n √ hn|qi = exp − , n π¯h 2 2 n! 154

where ξ≡ with

r

mω q, ¯h

n

d dξ

exp(−ξ 2 ) Hn (ξ) = (−1) exp(ξ ) Z exp(ξ 2 ) ∞ (−2iu)n exp(−u2 + 2iξu)du. = π −∞ n

2

For example, H0 (ξ) = 1,

and

H1 (ξ) = − exp(ξ 2 )

d exp(−ξ 2 ) = 2ξ, dξ

result in the eigenstates h0|qi = and h1|qi =

mω 1/4 π¯h

mω exp − q2 , 2¯ h

mω 1/4 r 2mω

mω q · exp − q2 . ¯h 2¯ h

π¯ h

Using the above expressions, the matrix elements are obtained as

1 · hq ′ |ni hn|qi exp −β¯ h ω n + 2 hq ′ |ρ|qi = hq ′ |n′ i hn′ |ρ|ni hn|qi = n P 1 exp −β¯ h ω n + ′ n 2 n,n X β¯hω 1 = 2 sinh · exp −β¯hω n + · hq ′ |ni hn|qi . 2 2 n P

X

(c) Show that for a general operator A(x), ∂A ∂ exp [A(x)] 6= exp [A(x)] , ∂x ∂x while in all cases ∂ tr {exp [A(x)]} = tr ∂x • By definition

unless

155

∂A = 0, A, ∂x

∂A exp [A(x)] . ∂x

∞ X 1 n A , e = n! n=0 A

and

But for a product of n operators,

The

∂A ∂x

∞ X ∂eA 1 ∂An = . ∂x n! ∂x n=0

∂A ∂A ∂A ∂ (A · A · · · A) = · A···A + A · ···A + ··· + A · A··· . ∂x ∂x ∂x ∂x can be moved through the A′ s surrounding it only if A, ∂A ∂x = 0, in which case ∂A n−1 ∂A =n A , ∂x ∂x

and

∂eA ∂A A = e . ∂x ∂x

However, as we can always reorder operators inside a trace, i.e. tr(BC) = tr(CB), and

∂A ∂A n−1 , tr A · · · A · · · · · · A = tr ·A ∂x ∂x

and the identity ∂A A , ·e ∂x can always be satisfied, independent of any constraint on A, ∂A . ∂x ∂ tr eA = tr ∂x

(d) Note that the partition function calculated in part (a) does not depend on the mass m, i.e. ∂Z/∂m = 0. Use this information, along with the result in part (c), to show that 2 p mω 2 q 2 = . 2m 2 • The expectation values of the kinetic and potential energy are given by 2 2 p mω 2 q 2 mω 2 q 2 p = tr = tr ρ , and ρ . 2m 2m 2 2 Noting that the expression for the partition function derived in part (a) is independent of mass, we know that ∂Z/∂m = 0. Starting with Z = tr e−β H , and differentiating ∂Z ∂ ∂ −β H −β H = 0, = tr = tr e (−βH)e ∂m ∂m ∂m

where we have used the result in part (c). Differentiating the Hamiltonian, we find that mω 2 q 2 −β H p2 −β H + tr −β = 0. e e tr β 2m2 2 156

Equivalently, p2 −β H mω 2 q 2 −β H tr = tr , e e 2m 2

which shows that the expectation values of kinetic and potential energies are equal.

(e) Using the results in parts (d) and (a), or otherwise, calculate q 2 . How are the results in Problem #1 modified at low temperatures by inclusion of quantum mechanical effects. −1

• In part (a) it was found that hHi = (¯ hω/2) (tanh(β¯hω/2)) . Note that hHi =

2

p /2m + mω 2 q 2 /2 , and that in part (d) it was determined that the contribution from

the kinetic and potential energy terms are equal. Hence,

1 −1 hω/2) (tanh(β¯hω/2)) . mω 2 q 2 /2 = (¯ 2

Solving for q 2 ,

2 q =

¯ h ¯h −1 (tanh(β¯hω/2)) = coth(β¯hω/2). 2mω 2mω

While the classical result q 2 = kB T /mω 2 , vanishes as T → 0, the quantum result satu rates at T = 0 to a constant value of q 2 = h ¯ /(2mω). The amplitude of the displacement curves in Problem #1 are effected by exactly the same saturation factors.

(f) In a coordinate representation, calculate hq ′ |ρ|qi in the high temperature limit. One approach is to use the result exp(βA) exp(βB) = exp β(A + B) + β 2 [A, B]/2 + O(β 3 ) . • Using the general operator identity exp(βA) exp(βB) = exp β(A + B) + β 2 [A, B]/2 + O(β 3 ) , the Boltzmann operator can be decomposed in the high temperature limit into those for kinetic and potential energy; to the lowest order as mω 2 q 2 p2 ≈ exp(−βp2 /2m) · exp(−βmω 2 q 2 /2). −β exp −β 2m 2 157

The first term is the Boltzmann operator for an ideal gas. The second term contains an operator diagonalized by |q >. The density matrix element < q ′ |ρ|q > =< q ′ | exp(−βp2 /2m) exp(−βmω 2 q 2 /2)|q > Z = dp′ < q ′ | exp(−βp2 /2m)|p′ >< p′ | exp(−βmω 2 q 2 /2)|q > Z = dp′ < q ′ |p′ >< p′ |q > exp(−βp′2 /2m) exp(−βq 2 mω 2 /2). Using the free particle basis < q ′ |p′ >=

h √ 1 e−iq·p/¯ , 2π¯ h

Z ′ ′ ′2 2 2 1 dp′ eip (q−q )/¯h e−βp /2m e−βq mω /2 < q |ρ|q >= 2π¯h !2 r r Z β 2m i 1 2m −βq 2 mω 2 /2 1 ′ ′ ′ ′ 2 =e dp exp − p + (q − q ) exp − (q − q ) , 2π¯h 2m 2¯ h β 4 β¯h2 ′

where we completed the square. Hence

1 −βq 2 mω2 /2 p mkB T ′ 2 2πmkB T exp − < q |ρ|q >= e (q − q ) . 2π¯h 2¯ h2 ′

The proper normalization in the high temperature limit is Z 2 2 2 Z = dq < q|e−βp /2m · e−βmω q /2 |q > Z Z 2 2 2 = dq dp′ < q|e−βp /2m |p′ >< p′ |e−βmω q /2 |q > Z Z ′2 2 2 kB T 2 = dq dp |< q|p >| e−βp /2m e−βmω q /2 = . ¯hω Hence the properly normalized matrix element in the high temperature limit is s mω 2 mkB T mω 2 2 ′ ′ 2 < q |ρ|q >lim T →∞ = exp − q exp − (q − q ) . 2πkB T 2kB T 2¯ h2 (g) At low temperatures, ρ is dominated by low energy states. Use the ground state wave-function to evaluate the limiting behavior of hq ′ |ρ|qi as T → 0.

• In the low temperature limit, we retain only the first terms in the summation ρlim T →0 ≈

|0 > e−β¯hω/2 < 0| + |1 > e−3β¯hω/2 < 1| + · · · . e−β¯hω/2 + e−3β¯hω/2 158

Retaining only the term for the ground state in the numerator, but evaluating the geometric series in the denominator, < q ′ |ρ|q >lim T →0 ≈< q ′ |0 >< 0|q > e−β¯hω/2 · eβ¯hω/2 − e−β¯hω/2 . Using the expression for < q|0 > given in part (b), ′

< q |ρ|q >lim T →0 ≈

r

h mω i mω q 2 + q ′2 1 − e−β¯hω . exp − π¯h 2¯ h

(h) Calculate the exact expression for hq ′ |ρ|qi.

********

4. Relativistic Coulomb gas:

Consider a quantum system of N positive, and N negative

charged relativistic particles in box of volume V = L3 . The Hamiltonian is H=

2N X i=1

c|~ pi | +

2N X i −1, or d > s. Therefore, Bose-Einstein condensation occurs for d > s. For a two dimensional gas, d = s = 2, the integral diverges logarithmically, and hence Bose-Einstein condensation does not occur. ******** 3. Pauli paramagnetism: Calculate the contribution of electron spin to its magnetic susceptibility as follows. Consider non-interacting electrons, each subject to a Hamiltonian p~ 2 ~ , − µ0 ~σ · B 2m ~ are ±B. where µ0 = e¯h/2mc, and the eigenvalues of ~σ · B ~ has been ignored.) (The orbital effect, p~ → ~p − eA, H1 =

(a) Calculate the grand potential G − = −kB T ln Q− , at a chemical potential µ. • The energy of the electron gas is given by X − E≡ Ep (n+ p , np ), p

where n± p (= 0 or 1), denote the number of particles having ± spins and momentum p, and 2 2 p p + − + Ep (np , np ) ≡ − µ0 B np + + µ0 B n− p 2m 2m 2 − − p − (n+ = (n+ + n ) p − np )µ0 B. p p 2m The grand partition function of the system is P + − N= (n +n ) ∞ Xp p X − exp −βEp (n+ Q= exp(−βµN ) p , np ) − {n+ p ,np }

N=0

X

=

− {n+ p ,np }

=

− + − exp βµ n+ p + np − βEp np , np

X

Y

p2 p2 · 1 + exp β µ + µ0 B − 1 + exp β µ − µ0 B − 2m 2m

p {n+ ,n− } p p

=

p2 p2 + exp β µ − µ0 B − np + µ + µ0 B − n− p 2m 2m

Y p

= Q0 (µ + µ0 B) · Q0 (µ − µ0 B) , 176

where

Y p2 Q0 (µ) ≡ 1 + exp β µ − . 2m p

Thus ln Q = ln Q0 (µ + µ0 B) + ln Q0 (µ − µ0 B) .

Each contribution is given by

Z p2 V p2 3 −β 2m ) = d p ln 1 + ze ln Q0 (µ) = ln 1 + exp β(µ − 2m (2π¯h)3 p r Z √ V 4πm 2m dx x ln(1 + ze−x ), where z ≡ eβµ , = 3 h β β X

and integrating by parts yields ln Q0 (µ) = V

2πmkB T h2

3/2

2 2 √ π3

Z

dx

x3/2 V − = f (z). z −1 ex + 1 λ3 5/2

The total grand free energy is obtained from ln Q(µ) = as

i V h − − −βµ0 B βµ0 B , ze ze + f f 5/2 λ3 5/2

G = −kB T ln Q(µ) = −kB T

i V h − − −βµ0 B βµ0 B . ze ze + f f 5/2 λ3 5/2

(b) Calculate the densities n+ = N+ /V , and n− = N− /V , of electrons pointing parallel and antiparallel to the field. • The number densities of electrons with up or down spins is given by

where we used

N± ∂ V − ze±βµ0 B , =z ln Q± = 3 f3/2 V ∂z λ z

∂ − − f (z) = fn−1 (z). ∂z n

The total number of electrons is the sum of these, i.e. V − − −βµ0 B βµ0 B . + f3/2 ze N = N+ + N− = 3 f3/2 ze λ 177

(c) Obtain the expression for the magnetization M = µ0 (N+ − N− ), and expand the result for small B.

• The magnetization is related to the difference between numbers of spin up and down

electrons as

i V h − − βµ0 B −βµ0 B − f3/2 ze . M = µ0 (N+ − N− ) = µ0 3 f3/2 ze λ

Expanding the results for small B, gives

∂ − − − − (z) ± z · βµ0 B f3/2 [z (1 ± βµ0 B)] ≈ f3/2 ze±βµ0 B ≈ f3/2 f3/2 (z), ∂z

which results in

M = µ0

V 2µ20 V − − (z) = (z). (2βµ B) · f · B · f1/2 0 1/2 3 3 λ kB T λ

(d) Sketch the zero field susceptibility χ(T ) = ∂M/∂B|B=0 , and indicate its behavior at low and high temperatures. • The magnetic susceptibility is 2µ20 V ∂M · f − (z), = χ≡ ∂B B=0 kB T λ3 1/2

with z given by,

N =2

V − · f3/2 (z). 3 λ

In the low temperature limit, (ln z = βµ → ∞) Z ln(z) n 1 [ln(z)] n−1 dx x = , Γ(n) 0 nΓ(n) T →0 0 2/3 √ 3N π 3 V 4(ln z)3/2 √ , λ , =⇒ ln z = N =2 3 · λ 8V 3 π 1/3 1/3 1/3 √ 2µ2o V 4µ20 V 4πmµ20 V 3N 3N 3N π 3 = χ= √ = . λ · · · πkB T λ3 8V kB T λ2 πV h2 πV

fn− (z)

1 = Γ(n)

Z

∞

xn−1 dx 1 + ex−ln(z)

≈

Their ratio of the last two expressions gives − µ20 f1/2 3µ20 1 3µ20 1 χ 3µ20 = = = . = − N T →0 kB T f3/2 2kB T ln(z) 2kB T βεF 2kB TF 178

In the high temperature limit (z → 0), →

z fn (z) z→0 Γ(n)

Z

∞

dx xn−1 e−x = z,

0

and thus → 2V N · z, 3 β→0 λ

=⇒

N N z≈ · λ3 = · 2V 2V

h2 2πmkB T

3/2

→ 0,

which is consistent with β → 0. Using this result, χ≈ The result

2µ20 V N µ20 · z = . kB T λ3 kB T

χ N

T →∞

=

µ20 , kB T

is known as the Curie susceptibility. χ/Nµο2

χ/Nµο2 ∼ 1/k BT

3/2k BTF

3/2k BTF

1/k BT

(e) Estimate the magnitude of χ/N for a typical metal at room temperature. • Since TRoom ≪ TF ≈ 104 K, we can take the low T limit for χ (see(d)), and 3µ20 3 × (9.3 × 10−24 )2 χ = ≈ ≈ 9.4 × 10−24 J/T2 , N 2kB TF 2 × 1.38 × 10−23 179

where we used µ0 =

eh ≃ 9.3 × 10−24 J/T. 2mc ********

4. Freezing of He3 :

At low temperatures He3 can be converted from liquid to solid by

application of pressure. A peculiar feature of its phase boundary is that (dP/dT )melting is negative at temperatures below 0.3 K [(dP/dT )m ≈ −30atm K−1 at T ≈ 0.1 K]. We will

use a simple model of liquid and solid phases of He3 to account for this feature.

(a) In the solid phase, the He3 atoms form a crystal lattice. Each atom has nuclear spin of 1/2. Ignoring the interaction between spins, what is the entropy per particle ss , due to the spin degrees of freedom? • Entropy of solid He3 comes from the nuclear spin degeneracies, and is given by Ss kB ln(2N ) ss = = = kB ln 2. N N

(b) Liquid He3 is modelled as an ideal Fermi gas, with a volume of 46˚ A3 per atom. What is its Fermi temperature TF , in degrees Kelvin? • The Fermi temperature for liquid 3 He may be obtained from its density as εF h2 TF = = kB 2mkB

3N 8πV

2/3

(6.7 × 10−34 )2 ≈ 2 · (6.8 × 10−27 )(1.38 × 10−23 )

3 8π × 46 × 10−30

2/3

≈ 9.2 K.

(c) How does the heat capacity of liquid He3 behave at low temperatures? Write down an expression for CV in terms of N, T, kB , TF , up to a numerical constant, that is valid for T ≪ TF .

• The heat capacity comes from the excited states at the fermi surface, and is given by CV = kB

π2 2 3N π2 T π2 kB T D(εF ) = kB T = N kB . 6 6 2kB TF 4 TF

180

(d) Using the result in (c), calculate the entropy per particle sℓ , in the liquid at low temperatures. For T ≪ TF , which phase (solid or liquid) has the higher entropy? • The entropy can be obtained from the heat capacity as T dS , CV = dT

⇒

1 sℓ = N

Z

T 0

CV dT π2 T = kB . T 4 TF

As T → 0, sℓ → 0, while ss remains finite. This is an unusual situation in which the solid

has more entropy than the liquid! (The finite entropy is due to treating the nuclear spins

as independent. There is actually a weak coupling between spins which causes magnetic ordering at a much lower temperature, removing the finite entropy.) (e) By equating chemical potentials, or by any other technique, prove the Clausius– Clapeyron equation (dP/dT )melting = (sℓ − ss )/(vℓ − vs ), where vℓ and vs are the volumes

per particle in the liquid and solid phases respectively.

• The Clausius-Clapeyron equation can be obtained by equating the chemical potentials at the phase boundary,

µℓ (T, P ) = µs (T, P ),

and µℓ (T + ∆T, P + ∆P ) = µs (T + ∆T, P + ∆P ).

Expanding the second equation, and using the thermodynamic identities ∂µ S V ∂µ = − , and = , ∂T P N ∂P T N results in

∂P ∂T

=

melting

sℓ − ss . vℓ − vs

(f) It is found experimentally that vℓ − vs = 3˚ A3 per atom. Using this information, plus the results obtained in previous parts, estimate (dP/dT )melting at T ≪ TF .

• The negative slope of the phase boundary results from the solid having more entropy than the liquid, and can be calculated from the Clausius-Clapeyron relation π2 T − ln 2 4 TF ∂P sℓ − ss = ≈ kB . ∂T melting vℓ − vs vℓ − vs

Using the values, T = 0.1 K, TF = 9.2 J K, and vℓ − vs = 3 ˚ A3 , we estimate ∂P ≈ −2.7 × 106 Pa ◦ K−1 , ∂T melting 181

in reasonable agreement with the observations. ******** 5. Non-interacting fermions:

Consider a grand canonical ensemble of non-interacting

fermions with chemical potential µ. The one–particle states are labelled by a wavevector ~k, and have energies E(~k). (a) What is the joint probability P ( n~k ), of finding a set of occupation numbers n~k , of the one–particle states?

• In the grand canonical ensemble with chemical potential µ, the joint probability of finding a set of occupation numbers n~k , for one–particle states of energies E(~k) is given by the Fermi distribution

i h ~ exp β(µ − E( k))n Y ~ k h i, P ( n~k ) = ~ 1 + exp β(µ − E(k)) ~ k

where

n~k = 0 or 1,

for each ~k.

(b) Express your answer to part (a) in terms of the average occupation numbers • The average occupation numbers are given by h i ~k)) exp β(µ − E(

h i, n~k − = ~ 1 + exp β(µ − E(k))

from which we obtain

h i exp β(µ − E(~k)) =

n o n~k − .

n~k −

. 1 − n~k −

This enables us to write the joint probability as Y n~k

1−n~k . 1 − n~k − P ( n~k ) = n~k − ~ k

(c) A random variable has a set of ℓ discrete outcomes with probabilities pn , where n = 1, 2, · · · , ℓ. What is the entropy of this probability distribution? What is the maximum possible entropy?

• A random variable has a set of ℓ discrete outcomes with probabilities pn . The entropy of this probability distribution is calculated from S = −kB

ℓ X

pn ln pn

n=1

182

.

The maximum entropy is obtained if all probabilities are equal, pn = 1/ℓ, and given by Smax = kB ln ℓ. (d) Calculate the entropy of the probability distribution for fermion occupation numbers in part (b), and comment on its zero temperature limit. • Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by S = −kB

X h

i

n~k − ln n~k − + 1 − n~k − ln 1 − n~k − . ~ k

In the zero temperature limit all occupation numbers are either 0 or 1. In either case the contribution to entropy is zero, and the fermi system at T = 0 has zero entropy.

(e) Calculate the variance of the total number of particles N 2 c , and comment on its zero temperature behavior.

• The total number of particles is given by N =

are independent

P

~ k

n~k . Since the occupation numbers

X X D E XD E

2

2 2 2 n~k − 1 − n~k − , − n~k − = n~k n~k = N c= ~ k

since

D

n~2k

vanishes.

E

−

c

~ k

−

~ k

= n~k − . Again, since at T = 0, n~k − = 0 or 1, the variance N 2 c

(f) The number fluctuations of a gas is related to its compressibility κT , and number density n = N/V , by

2 N c = N nkB T κT

.

Give a numerical estimate of the compressibility of the fermi gas in a metal at T = 0 in units of ˚ A3 eV −1 .

• To obtain the compressibility from N 2 c = N nkB T κT , we need to examine the behavior

at small but finite temperatures. At small but finite T , a small fraction of states around the fermi energy have occupation numbers around 1/2. The number of such states is roughly N kB T /εF , and hence we can estimate the variance as

2 1 N kB T . N c = N nkB T κT ≈ × 4 εF 183

The compressibility is then approximates as κT ≈

1 , 4nεF

where n = N/V is the density. For electrons in a typical metal n ≈ 1029 m−3 ≈ 0.1˚ A3 , and

εF ≈ 5eV ≈ 5 × 104 ◦ K, resulting in

κT ≈ 0.5˚ A3 eV −1 .

******** 6. Stoner ferromagnetism:

The conduction electrons in a metal can be treated as a

gas of fermions of spin 1/2 (with up/down degeneracy), and density n = N/V . The Coulomb repulsion favors wave functions which are antisymmetric in position coordinates, thus keeping the electrons apart. Because of the full (position and spin) antisymmetry of fermionic wave functions, this interaction may be approximated by an effective spin-spin coupling which favors states with parallel spins. In this simple approximation, the net effect is described by an interaction energy U =α

N+ N− , V

where N+ and N− = N − N+ are the numbers of electrons with up and down spins, and V is the volume. (The parameter α is related to the scattering length a by α = 4π¯h2 a/m.)

(a) The ground state has two fermi seas filled by the spin up and spin down electrons. Express the corresponding fermi wavevectors kF± in terms of the densities n± = N± /V . • In the ground state, all available wavevectors are filled up in a sphere. Using the appropriate density of states, the corresponding radii of kF± are calculated as N± = V

Z

k αc =

2/3 h ¯ 2 −1/3 4 3π 2 n . 3 2m 185

(e) Explain qualitatively, and sketch the behavior of the spontaneous magnetization as a function of α. • For α > αc , the optimal value of δ is obtained by minimizing the energy density. Since the coefficient of the fourth order term is positive, and the optimal δ goes to zero continuously

as α → αc ; the minimum energy is obtained for a value of δ 2 ∝ (α−αc ). The magnetization √ is proportional to δ, and hence grows in the vicinity of αc as α − αc , as sketched below.

******** 7. Boson magnetism: Consider a gas of non-interacting spin 1 bosons, each subject to a Hamiltonian H1 (~ p, sz ) =

p~ 2 − µ0 sz B 2m

,

where µ0 = e¯h/mc, and sz takes three possible values of (-1, 0, +1). (The orbital effect, ~ has been ignored.) p~ → p~ − eA, (a) In a grand canonical ensemble of chemical potential µ, what are the average occupation n o numbers hn+ (~k)i, hn0 (~k)i, hn−(~k)i , of one-particle states of wavenumber ~k = p~/¯h?

• Average occupation numbers of the one-particle states in the grand canonical ensemble of chemical potential µ, are given by the Bose-Einstein distribution ns (~k) = =

1 eβ [H(s)−µ] − 1

,

(for s = −1, 0, 1)

1 h 2 i p exp β 2m − µ0 sB − βµ − 1

(b) Calculate the average total numbers {N+ , N0 , N− }, of bosons with the three possible

+ values of sz in terms of the functions fm (z).

186

• The total numbers of particles with spin s are given by Z X 1 V 3 h i . d k Ns = ns (~k), =⇒ Ns = p2 (2π)3 exp β − µ sB − βµ − 1 0 ~ 2m {k} After a change of variables, k ≡ x1/2

√

Ns = where + fm (z)

1 ≡ Γ(m)

Z

0

∞

2mkB T /h, we get

V + βµ0 sB f ze , λ3 3/2

dx xm−1 , z −1 ex − 1

λ≡ √

h , 2πmkB T

z ≡ eβµ .

(c) Write down the expression for the magnetization M (T, µ) = µ0 (N+ − N− ), and by expanding the result for small B find the zero field susceptibility χ(T, µ) = ∂M/∂B|B=0 . • The magnetization is obtained from M (T, µ) = µ0 (N+ − N− ) i V h + + −βµ0 sB βµ0 B . − f3/2 ze = µ0 3 f3/2 ze λ

Expanding the result for small B gives

∂ + + + + (z) ± z · βµ0 B f3/2 (z[1 ± βµ0 B]) ≈ f3/2 ze±βµ0 B ≈ f3/2 f3/2 (z). ∂z

+ + Using zdfm (z)/dz = fm−1 (z), we obtain

M = µ0 and

V 2µ20 V + + (z) = (z), (2βµ B) · f · B · f1/2 0 1/2 λ3 kB T λ3 2µ20 V ∂M = · f + (z). χ≡ ∂B B=0 kB T λ3 1/2

To find the behavior of χ(T, n), where n = N/V is the total density, proceed as follows: (d) For B = 0, find the high temperature expansion for z(β, n) = eβµ , correct to second order in n. Hence obtain the first correction from quantum statistics to χ(T, n) at high temperatures. 187

+ • In the high temperature limit, z is small. Use the Taylor expansion for fm (z) to write

the total density n(B = 0), as

3 + N+ + N0 + N− (z) = 3 f3/2 n(B = 0) = V λ B=0 z3 3 z2 ≈ 3 z + 3/2 + 3/2 + · · · . λ 2 3 Inverting the above equation gives z=

nλ3 3

1

−

23/2

nλ3 3

2

+ ···.

The susceptibility is then calculated as 2µ20 V · f + (z), kB T λ3 1/2 2µ20 1 z2 χ/N = z + 1/2 + · · · kB T nλ3 2 3 2 1 nλ 2µ0 1 2 . 1 + − 3/2 + 1/2 +O n = 3kB T 3 2 2 χ=

(e) Find the temperature Tc (n, B = 0), of Bose–Einstein condensation. What happens to χ(T, n) on approaching Tc (n) from the high temperature side? • Bose-Einstein condensation occurs when z = 1, at a density n=

3 + f (1), λ3 3/2

or a temperature h2 Tc (n) = 2πmkB

n 3 ζ 3/2

2/3

,

+ + (z) = ∞, the susceptibility χ(T, n) diverges (1) ≈ 2.61. Since limz→1 f1/2 where ζ 3/2 ≡ f3/2

on approaching Tc (n) from the high temperature side.

(f) What is the chemical potential µ for T < Tc (n), at a small but finite value of B? Which one-particle state has a macroscopic occupation number? −1 ~ • Since ns (~k, B) = z −1 eβ E s (k,B) −1 is a positive number for all ~k and sz , µ is bounded

above by the minimum possible energy, i.e. for

T < Tc ,

and

B finite,

zeβµ0 B = 1, 188

=⇒

µ = −µ0 B.

Hence the macroscopically occupied one particle state has ~k = 0, and sz = +1. (g) Using the result in (f), find the spontaneous magnetization, M (T, n) = lim M (T, n, B). B→0

• Contribution of the excited states to the magnetization vanishes as B → 0. Therefore the

total magnetization for T < Tc is due to the macroscopic occupation of the (k = 0, sz = +1) state, and M (T, n) = µ0 V n+ (k = 0) = µ0 V n − nexcited

3V = µ0 N − 3 ζ 3/2 . λ

******** 8. Dirac fermions are non-interacting particles of spin 1/2. The one-particle states come in pairs of positive and negative energies, p E ± (~k) = ± m2 c4 + h ¯ 2 k 2 c2

,

independent of spin. (a) For any fermionic system of chemical potential µ, show that the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. (δ is any constant energy.)

• According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] =

eβ(µ−E )n , 1 + eβ(µ−E )

for

n = 0, 1.

For a state of energy µ + δ, p [n(µ + δ)] =

eβδn , 1 + eβδ

=⇒

p [n(µ + δ) = 1] =

eβδ 1 = . 1 + eβδ 1 + e−βδ

Similarly, for a state of energy µ − δ, p [n(µ − δ)] =

e−βδn , 1 + e−βδ

=⇒

p [n(µ − δ) = 0] = 189

1 = p [n(µ + δ) = 1] , 1 + e−βδ

i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. (b) At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. Using the result in (a) find the chemical potential at finite temperatures T . • The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for an tem-

perature; any particle leaving an occupied negative energy state goes to the corresponding

unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enfrces µ(T ) = 0. (c) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) − E(0) = 4V

Z

d3~k E + (~k) (2π)3 exp βE (~k) + 1

.

+

• Using the label +(-) for the positive (energy) states, the excitation energy is calculated

as

E(T ) − E(0) =

X

k,sz

=2

[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]

X k

2 hn+ (k)i E + (k) = 4V

Z

d3~k E + (~k) . (2π)3 exp βE (~k) + 1 +

(d) Evaluate the integral in part (c) for massless Dirac particles (i.e. for m = 0). • For m = 0, E + (k) = h ¯ c|k|, and ∞

4πk 2 dk ¯hck E(T ) − E(0) = 4V = (set β¯hck = x) 8π 3 eβ¯hck + 1 0 3 Z ∞ 2V x3 kB T = 2 kB T dx x π ¯hc e +1 0 3 kB T 7π 2 V kB T . = 60 ¯hc Z

For the final expression, we have noted that the needed integral is 3!f4− (1), and used the given value of f4− (1) = 7π 4 /720. (e) Calculate the heat capacity, CV , of such massless Dirac particles. 190

• The heat capacity can now be evaluated as 3 ∂E 7π 2 kB T CV = = . V kB ∂T V 15 ¯hc (f) Describe the qualitative dependence of the heat capacity at low temperature if the particles are massive. • When m 6= 0, there is an energy gap between occupied and empty states, and we thus

expect an exponentially activated energy, and hence heat capacity. For the low energy excitations, E + (k) ≈ mc2 + and thus

¯ 2 k2 h + ···, 2m

√ Z ∞ 2V 2 −βmc2 4π π dxx2 e−x E(T ) − E(0) ≈ 2 mc e π λ3 0 2 48 V = √ 3 mc2 e−βmc . πλ

The corresponding heat capacity, to leading order thus behaves as C(T ) ∝ kB

V 2 2 −βmc2 βmc e . λ3 ********

9. Numerical estimates: The following table provides typical values for the Fermi energy and Fermi temperature for (i) Electrons in a typical metal; (ii) Nucleons in a heavy nucleus; ˚3 per atom). and (iii) He3 atoms in liquid He3 (atomic volume = 46.2A n(1/m3 )

m(Kg)

εF (eV)

TF (K)

electron

1029

4.4

nucleons

1044

9 × 10−31

5 × 104

liquid He3

2.6 × 1028

1.6 × 10−27

4.6 × 10−27

1.0 × 108 ×10−3

1.1 × 1012 101

(a) Estimate the ratio of the electron and phonon heat capacities at room temperature for a typical metal. • For an electron gas, TF ≈ 5 × 104 K, TF ≫ Troom ,

=⇒

Celectron π2 T ≈ · ≈ 0.025. N kB 2 TF 191

For the phonon gas in iron, the Debye temperature is TD ≈ 470K, and hence # " 2 T 1 Cphonon + . . . ≈ 3, ≈3 1− N kB 20 TD resulting in

Celectron ≈ 8 × 10−3 . Cphonon

(b) Compare the thermal wavelength of a neutron at room temperature to the minimum wavelength of a phonon in a typical crystal. • Thermal wavelengths are given by λ≡ √

h . 2πmkB T

For a neutron at room temperature, using the values m = 1.67 × 10−27 kg,

T = 300 K,

kB = 1.38 × 10−23 JK−1 ,

h = 6.67 × 10−34 Js,

we obtain λ = 1˚ A. The typical wavelength of a phonon in a solid is λ = 0.01 m, which is much longer than the neutron wavelength. The minimum wavelength is, however, of the order of atomic spacing (3 − 5 ˚ A), which is comparable to the neutron thermal wavelength. (c) Estimate the degeneracy discriminant, nλ3 , for hydrogen, helium, and oxygen gases

at room temperature and pressure. At what temperatures do quantum mechanical effects become important for these gases? • Quantum mechanical effects become important if nλ3 ≥ 1. In the high temperature

limit the ideal gas law is valid, and the degeneracy criterion can be reexpressed in terms of pressure P = nkB T , as nλ3 =

nh3 h3 P = ≪ 1. (2πmkB T )3/2 (kB T )5/2 (2πm)3/2

It is convenient to express the answers starting with an imaginary gas of ‘protons’ at room temperature and pressure, for which mp = 1.7 × 10−34 Kg, and

(nλ3 )proton =

P = 1 atm. = 105 Nm−2 ,

(6.7 × 10−34 )3 10−5 = 2 × 10−5 . −21 5/2 −27 3/2 (4.1 × 10 ) (2π · 1.7 × 10 ) 192

The quantum effects appear below T = TQ , at which nλ3 becomes order of unity. Using 3

3

nλ = (nλ )proton

m 3/2 p

m

and TQ = Troom (nλ3 )3/2 ,

,

we obtain the following table: (d) Experiments on He4 indicate that at temperatures below 1K, the heat capacity is given by CV = 20.4T 3 JKg −1 K−1 . Find the low energy excitation spectrum, E(k), of He4 . (Hint: There is only one non-degenerate branch of such excitations.) • A spectrum of low energy excitations scaling as E(k) ∝ k s , in d-dimensional space, leads to a low temperature heat capacity that vanishes as C ∝ T d/s . Therefore, from CV = 20.4 T 3 JKg−1 K−1 in d = 3, we can conclude s = 1, i.e. a spectrum of the form E(k) = h ¯ cs |~k|, corresponding to sound waves of speed cs . Inserting all the numerical factors, we have 12π 4 N kB CV = 5

T Θ

3

,

where

¯ cs h Θ= kB

6π 2 N V

1/3

.

Hence, we obtain E =h ¯ cs k = kB

2π 2 kB V T 3 5 CV

1/3

k = (2 × 10−32 Jm) k,

corresponding to a sound speed of cs ≈ 2 × 102 ms−1 . ********

10. Solar interior: According to astrophysical data, the plasma at the center of the sun has the following properties: Temperature: Hydrogen density: Helium density:

T = 1.6 × 107 K

ρH = 6 × 104 kg m−3

ρHe = 1 × 105 kg m−3 .

(a) Obtain the thermal wavelengths for electrons, protons, and α-particles (nuclei of He). 193

• The thermal wavelengths of electrons, protons, and α-particles in the sun are obtained

from

λ= √

h , 2πmkB T

and T = 1.6 × 107 K, as λelectron ≈ p λproton ≈ p

2π × (9.1 ×

10−31

2π × (1.7 ×

10−27

6.7 × 10−34 J/s

Kg) · (1.4 ×

10−23

Kg) · (1.4 ×

10−23

6.7 × 10−34 J/s

and

J/K) · (1.6 ×

107 107

K)

≈ 1.9 × 10−11 m, ≈ 4.3 × 10−13 m,

J/K) · (1.6 × K) 1 λα−particle = λproton ≈ 2.2 × 10−13 m. 2

(b) Assuming that the gas is ideal, determine whether the electron, proton, or α-particle gases are degenerate in the quantum mechanical sense. • The corresponding number densities are given by 3

nH ≈ 3.5 × 1031 m−3 , ρHe 3 ≈ 1.5 × 1031 m−3 , ρHe ≈ 1.0 × 105 Kg/m =⇒ nHe = 4mH ne = 2nHe + nH ≈ 8.5 × 1031 m3 . ρH ≈ 6 × 104 kg/m

=⇒

The criterion for degeneracy is nλ3 ≥ 1, and nH · λ3H ≈2.8 × 10−6 ≪ 1,

nHe · λ3He ≈1.6 × 10−7 ≪ 1, ne · λ3e ≈0.58 ∼ 1. Thus the electrons are weakly degenerate, and the nuclei are not. (c) Estimate the total gas pressure due to these gas particles near the center of the sun. • Since the nuclei are non-degenerate, and even the electrons are only weakly degenerate,

their contributions to the overall pressure can be approximately calculated using the ideal gas law, as P ≈ (nH + nhe + ne ) · kB T ≈ 13.5 × 1031 (m−3 ) · 1.38 × 10−23 (J/K) · 1.6 × 107 (K) 2

≈ 3.0 × 1016 N/m .

194

(d) Estimate the total radiation pressure close to the center of the sun. Is it matter, or radiation pressure, that prevents the gravitational collapse of the sun? • The Radiation pressure at the center of the sun can be calculated using the black body formulas,

P = as P =

1U , 3V

1U π 2 k4 4 c= T = σT 4 , 4V 60¯ h3 c3

and

4 4 · 5.7 × 10−8 W/(m2 K4 ) · (1.6 × 107 K)4 2 σT 4 = ≈ 1.7 × 1013 N/m . 8 3c 3 · 3.0 × 10 m/s

Thus at the pressure in the solar interior is dominated by the particles. ******** 11. Bose condensation in d–dimensions:

Consider a gas of non-interacting (spinless)

bosons with an energy spectrum ǫ = p2 /2m, contained in a box of “volume” V = Ld in d dimensions. (a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical + potential µ. Express your answers in terms of d and fm (z), where z = eβµ , and

1 = Γ (m)

+ fm (z)

Z

∞ 0

xm−1 dx. z −1 ex − 1

(Hint: Use integration by parts on the expression for ln Q.)

• We have

Q= =

∞ X

P

eNβµ

N=0

n =N

i X

i

{ni }

YX

exp −β

eβ(µ−ǫi )ni =

i {ni }

Y i

X i

ni ǫi

!

,

1

1 − eβ(µ−ǫi )

P P whence ln Q = − i ln 1 − eβ(µ−ǫi ) . Replacing the summation i with a d dimensional h i R d d R d−1 integration V dd k/ (2π) = V Sd / (2π) k dk, where Sd = 2π d/2 / (d/2 − 1)!, leads

to

ln Q = −

V Sd

d

(2π)

Z

2 2 k d−1 dk ln 1 − ze−β¯h k /2m .

The change of variable x = β¯h2 k 2 /2m (⇒ k = results in V Sd 1 ln Q = − d (2π) 2

2m ¯h2 β

d/2 Z 195

p p 2mx/β/¯h and dk = dx 2m/βx/2¯ h)

xd/2−1 dx ln 1 − ze−x .

Finally, integration by parts yields V Sd 1 ln Q = d (2π) d

2m ¯h2 β

d/2 Z

x

d/2

ze−x Sd dx = V 1 − ze−x d

i.e. Sd G = −kB T ln Q = −V d

2m h2 β

d/2

kB T Γ

2m h2 β

d/2 Z

dx

d (z) , + 1 f+ d 2 +1 2

which can be simplified, using the property Γ (x + 1) = xΓ (x), to G=−

V (z) . kB T f + d 2 +1 λd

The average number of particles is calculated as d/2 Z ze−x Sd 2m ∂ d/2−1 x dx ln Q = V N= ∂ (βµ) d h2 β 1 − ze−x , d/2 Sd 2m V + d + =V f d (z) = d f d (z) Γ 2 2 h2 β 2 λ 2 i.e. n=

1 + f d (z) . λd 2

(b) Calculate the ratio P V /E, and compare it to the classical value. • We have P V = −G, while E=−

d ln Q d ∂ ln Q = + = − G. ∂β 2 β 2

Thus P V /E = 2/d, identical to the classical value. (c) Find the critical temperature, Tc (n), for Bose-Einstein condensation. • The critical temperature Tc (n) is given by n=

1 + 1 f d (1) = d ζ d , d λ 2 λ 2

for d > 2, i.e. h2 Tc = 2mkB

n ζd 2

!2/d

(d) Calculate the heat capacity C (T ) for T < Tc (n). 196

.

xd/2 , z −1 ex − 1

• At T < Tc , z = 1 and ∂E G d d V d ∂G d d C (T ) = +1 = +1 kB ζ d +1 . =− =− 2 ∂T z=1 2 ∂T z=1 2 2 T 2 2 λd (e) Sketch the heat capacity at all temperatures. •

.

(f) Find the ratio, Cmax /C (T → ∞), of the maximum heat capacity to its classical limit,

and evaluate it in d = 3.

• As the maximum of the heat capacity occurs at the transition, Cmax

d = C (Tc ) = 2

Thus

d +1 2

V

ζ d /n 2

kB f + d 2 +1

Cmax = C (T → ∞)

d +1 2

d (1) = N kB 2

ζ d +1 2

ζd

d +1 2

ζ d +1 2

ζd

.

2

,

2

which evaluates to 1.283 in d = 3. (g) How does the above calculated ratio behave as d → 2? In what dimensions are your

results valid? Explain.

+ • The maximum heat capacity, as it stands above, vanishes as d → 2! Since fm (x → 1) →

∞ if m ≤ 2, the fugacuty z is always smaller than 1. Hence, there is no macroscopic

occupation of the ground state, even at the lowest temperatures, i.e. no Bose-Einstein condensation in d ≤ 2. The above results are thus only valid for d ≥ 2. 197

******** 12. Exciton dissociation in a semiconductor:

Shining an intense laser beam on a semi-

conductor can create a metastable collection of electrons (charge −e, and effective mass

me ) and holes (charge +e, and effective mass mh ) in the bulk. The oppositely charged particles may pair up (as in a hydrogen atom) to form a gas of excitons, or they may dissociate into a plasma. We shall examine a much simplified model of this process. (a) Calculate the free energy of a gas composed of Ne electrons and Nh holes, at temperature T , treating them as classical non-interacting particles of masses me and mh .

• The canonical partition function of gas of non-interacting electrons and holes is the product of contributions from the electron gas, and from the hole gas, as Ze−h

1 = Ze Zh = Ne !

V λ3e

Ne

1 · Nh !

V λ3h

Nh

,

√ where λα = h/ 2πmα kB T (α =e, h). Evaluating the factorials in Stirling’s approximation, we obtain the free energy Fe−h = −kB T ln Ze−h = Ne kB T ln

Ne 3 λ eV e

+ Nh kB T ln

Nh 3 λ . eV h

(b) By pairing into an excition, the electron hole pair lowers its energy by ε. [The binding energy of a hydrogen-like exciton is ε ≈ me4 /(2¯ h2 ǫ2 ), where ǫ is the dielectric constant,

−1 and m−1 = m−1 e + mh .] Calculate the free energy of a gas of Np excitons, treating them

as classical non-interacting particles of mass m = me + mh . • Similarly, the partition function of the exciton gas is calculated as 1 Zp = Np !

V λ3p

Np

e−β(−Np ǫ) ,

leading to the free energy Fp = Np kB T ln where λp = h/

p

Np 3 λ eV p

− Np ǫ,

2π (me + mh ) kB T .

(c) Calculate the chemical potentials µe , µh , and µp of the electron, hole, and exciton states, respectively. 198

• The chemical potentials are derived from the free energies, through ∂Fe−h 3 = k T ln n λ µe = , B e e ∂Ne T,V ∂Fe−h = kB T ln nh λ3h , µh = ∂Nh T,V ∂Fp = kB T ln np λ3p − ǫ, µp = ∂Np T,V

where nα = Nα /V (α =e, h, p).

(d) Express the equilibrium condition between excitons and electron/holes in terms of their chemical potentials. • The equilibrium condition is obtained by equating the chemical potentials of the electron

and hole gas with that of the exciton gas, since the exciton results from the pairing of an electron and a hole, electron + hole ⇀ ↽ exciton. Thus, at equilibrium µe (ne , T ) + µh (nh , T ) = µp (np , T ) , which is equivalent, after exponentiation, to ne λ3e · nh λ3h = np λ3p e−βǫ .

(e) At a high temperature T , find the density np of excitons, as a function of the total density of excitations n ≈ ne + nh .

• The equilibrium condition yields np = ne nh

λ3e λ3h βǫ e . λ3p

At high temperature, np ≪ ne = nh ≈ n/2, and n 2 h3 λ3 λ3 np = ne nh e 3 h eβǫ = 3/2 λp 2 (2πkB T ) ******** 199

me + mh me mh

3/2

eβǫ .

13. Freezing of He4 :

At low temperatures He4 can be converted from liquid to solid by

application of pressure. An interesting feature of the phase boundary is that the melting pressure is reduced slightly from its T = 0K value, by approximately 20Nm−2 at its minimum at T = 0.8K. We will use a simple model of liquid and solid phases of 4 He to account for this feature. (a) The important excitations in liquid 4 He at T < 1K are phonons of velocity c. Calculate the contribution of these modes to the heat capacity per particle CVℓ /N , of the liquid. • The dominant excitations in liquid 4 He at T < 1K are phonons of velocity c. The corresponding dispersion relation is ε(k) = h ¯ ck. From the average number of phonons in D E −1 mode ~k, given by n(~k) = [exp(β¯hck) − 1] , we obtain the net excitation energy as ¯ ck h exp(β¯hck) − 1 ~ k Z 4πk 2 dk ¯hck =V × (change variables to x = β¯hck) 3 (2π) exp(β¯hck) − 1 4 Z ∞ 4 x3 6 kB T π2 kB T V dx x , ¯hc = V ¯hc = 2π 2 ¯hc 3! 0 e −1 30 ¯hc

Ephonons =

X

where we have used 1 ζ4 ≡ 3!

Z

0

∞

dx

x3 π4 = . ex − 1 90

The corresponding heat capacity is now obtained as dE 2π 2 CV = = V kB dT 15

kB T ¯hc

3

,

resulting in a heat capacity per particle for the liquid of 2π 2 CVℓ = kB vℓ N 15

kB T ¯hc

3

.

(b) Calculate the low temperature heat capacity per particle CVs /N , of solid 4 He in terms of longitudinal and transverse sound velocities cL , and cT . • The elementary excitations of the solid are also phonons, but there are now two transverse sound modes of velocity cT , and one longitudinal sound mode of velocity cL . The 200

contributions of these modes are additive, each similar inform to the liquid result calculated above, resulting in the final expression for solid heat capacity of 2π 2 CVs = kB vs N 15

kB T ¯h

3

×

2 1 + 3 3 cT cL

.

(c) Using the above results calculate the entropy difference (sℓ − ss ), assuming a single

sound velocity c ≈ cL ≈ cT , and approximately equal volumes per particle vℓ ≈ vs ≈ v. Which phase (solid or liquid) has the higher entropy?

• The entropies can be calculated from the heat capacities as sℓ (T ) = ss (T ) =

Z

Z

T 0 T 0

CVℓ (T ′ )dT ′ 2π 2 = kB vℓ T′ 45 2π 2 CVs (T ′ )dT ′ = kB vs T′ 45

kB T ¯hc

kB T ¯h

3

,

3

×

2 1 + 3 3 cT cL

.

Assuming approximately equal sound speeds c ≈ cL ≈ cT ≈ 300ms−1 , and specific volumes vℓ ≈ vs ≈ v = 46˚ A3 , we obtain the entropy difference 4π 2 kB v sℓ − ss ≈ − 45

kB T ¯hc

3

.

The solid phase has more entropy than the liquid because it has two more phonon excitation bands. (d) Assuming a small (temperature independent) volume difference δv = vℓ − vs , calculate the form of the melting curve. To explain the anomaly described at the beginning, which phase (solid or liquid) must have the higher density? • Using the Clausius-Clapeyron equation, and the above calculation of the entropy difference, we get

∂P ∂T

melting

sℓ − ss 4π 2 v = =− kB vℓ − vs 45 δv

kB T ¯hc

3

.

Integrating the above equation gives the melting curve π2 v Pmelt (T ) = P (0) − kB 45 δv

kB T ¯hc

3

T.

To explain the reduction in pressure, we need δv = vℓ − vs > 0, i.e. the solid phase has

the higher density, which is normal.

201

******** 14. Neutron star core:

Professor Rajagopal’s group at MIT has proposed that a new

phase of QCD matter may exist in the core of neutron stars. This phase can be viewed as a condensate of quarks in which the low energy excitations are approximately E(~k)± =± h ¯2

2 ~ | k | − kF 2M

.

The excitations are fermionic, with a degeneracy of g = 2 from spin. (a) At zero temperature all negative energy states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. By relating occupation numbers of states of energies µ + δ and µ − δ, or otherwise, find the chemical potential at finite temperatures T .

• According to Fermi statistics, the probability of occupation of a state of of energy E is eβ(µ−E )n p [n(E)] = , 1 + eβ(µ−E )

for

n = 0, 1.

For a state of energy µ + δ, eβδn , p [n(µ + δ)] = 1 + eβδ

=⇒

eβδ 1 p [n(µ + δ) = 1] = = . βδ 1+e 1 + e−βδ

Similarly, for a state of energy µ − δ, p [n(µ − δ)] =

e−βδn , 1 + e−βδ

=⇒

p [n(µ − δ) = 0] =

1 = p [n(µ + δ) = 1] , 1 + e−βδ

i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. This implies that for µ = 0, hn(E)i + hn(−E)i is unchanged for an temperature; for every particle leaving an occupied negative energy

state a particle goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Assuming a constant density of states near k = kF , i.e. setting d3 k ≈ 4πkF2 dq with q = |~k | − kF , show that the mean excitation energy of this system at finite temperature is k2 E(T ) − E(0) ≈ 2gV F2 π

Z

∞

0

202

dq

E + (q) exp (βE + (q)) + 1

.

• Using the label +(-) for the positive (energy) states, the excitation energy is calculated as

E(T ) − E(0) =

X k,s

=g

[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]

X k

2 hn+ (k)i E + (k) = 2gV

Z

E + (~k) d3~k . (2π)3 exp βE (~k) + 1 +

The largest contribution to the integral comes for |~k | ≈ kF . and setting q = (|~k | − kF ) and using d3 k ≈ 4πkF2 dq, we obtain

4πkF2 E(T ) − E(0) ≈ 2gV 2 8π 3

Z

∞

0

E + (q) k2 dq = 2gV F2 exp (βE + (q)) + 1 π

Z

0

∞

dq

E + (q) exp (βE + (q)) + 1

.

(c) Give a closed form answer for the excitation energy by evaluating the above integral. • For E + (q) = h ¯ 2 q 2 /(2M ), we have ∞

¯ 2 q 2 /2M h = (set β¯h2 q 2 /2M = x) β¯ h2 q 2 /2M + 1 e 0 1/2 Z ∞ 2 gV kF x1/2 2M kB T = dx k T B π2 ex + 1 ¯h2 0 1/2 √ ζ3/2 V kF2 1 gV kF2 π 1 2M kB T √ √ ζ = 1 − = 2 kB T 1 − kB T. 3/2 π 2 π λ ¯h2 2 2

k2 E(T ) − E(0) = 2gV F2 π

Z

dq

− For the final expression, we have used the value of fm (1), and introduced the thermal √ wavelength λ = h/ 2πM kB T .

(d) Calculate the heat capacity, CV , of this system, and comment on its behavior at low temperature. • Since E ∝ T 3/2 , √ 3ζ3/2 1 3E V kF2 ∂E √ 1 − = k ∝ = CV = T. B ∂T V 2T 2π λ 2

This is similar to the behavior of a one dimensional system of bosons (since the density of states is constant in q as in d = 1). Of course, for any fermionic system the density of states close to the Fermi surface has this character. The difference with the usual Fermi systems is the quadratic nature of the excitations above the Fermi surface. ******** 203

15.

Non-interacting bosons:

Consider a grand canonical ensemble of non-interacting

bosons with chemical potential µ. The one–particle states are labelled by a wavevector ~ q, and have energies E(~q). (a) What is the joint probability P ({nq~ }), of finding a set of occupation numbers {nq~}, of

the one–particle states, in terms of the fugacities zq~ ≡ exp [β(µ − E(~q))]?

• In the grand canonical ensemble with chemical potential µ, the joint probability of finding a set of occupation numbers {nq~}, for one–particle states of energies E(~q) is given by the normalized bose distribution P ({nq~ }) = =

Y q ~

Y q ~

{1 − exp [β(µ − E(~q))]} exp [β(µ − E(~q))nq~] n

(1 − zq~ ) zq~ q~ ,

with nq~ = 0, 1, 2, · · · ,

for each ~q.

(b) For a particular ~q, calculate the characteristic function hexp [iknq~ ]i. n • Summing the geometric series with terms growing as zq~ eik q~ , gives hexp [iknq~ ]i =

1 − zq~ 1 − exp [β(µ − E(~q))] = . 1 − exp [β(µ − E(~q)) + ik] 1 − zq~ eik

(c) Using the result of part (b), or otherwise, give expressions for the mean and variance of nq~ . occupation number hnq~ i.

• Cumulnats can be generated by expanding the logarithm of the characteristic function in powers of k. Using the expansion formula for ln(1 + x), we obtain

ln hexp [iknq~ ]i = ln (1 − zq~ ) − ln 1 − zq~ 1 + ik − k 2 /2 + · · · zq~ k 2 zq~ = − ln 1 − ik + +··· 1 − zq~ 2 1 − zq~ " 2 # zq~ zq~ zq~ k2 = ik − + +··· 1 − zq~ 2 1 − zq~ 1 − zq~ = ik

zq~ zq~ k2 − + ···. 1 − zq~ 2 (1 − zq~ )2

From the coefficients in the expansion, we can read off the mean and variance hnq~ i =

zq~ , 1 − zq~

and 204

2 nq~ c =

zq~

2.

(1 − zq~ )

(d) Express the variance in part (c) in terms of the mean occupation number hnq~ i. • Inverting the relation relating nq~ to zq~ , we obtain zq~ =

hnq~ i . 1 + hnq~i

Substituting this value in the expression for the variance gives

2 nq~ c =

zq~ (1 − zq~ )

2

= hnq~ i (1 + hnq~ i) .

(e) Express your answer to part (a) in terms of the occupation numbers {hnq~i}.

• Using the relation between zq~ and nq~ , the joint probability can be reexpressed as i Yh nq~ −1−nq~ . P ({nq~}) = (hnq~ i) (1 + hnq~ i) q ~

(f) Calculate the entropy of the probability distribution for bosons, in terms of {hnq~ i}, and comment on its zero temperature limit.

• Quite generally, the entropy of a probability distribution P is given by S = −kB hln P i. Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by S = −kB

X q ~

[hnq~ i ln hnq~ i − (1 + hnq~ i) ln (1 + hnq~i)] .

In the zero temperature limit all occupation numbers are either 0 (for excited states) or infinity (for the ground states). In either case the contribution to entropy is zero, and the system at T = 0 has zero entropy. ******** 16. Relativistic Bose gas in d dimensions:

Consider a gas of non-interacting (spinless)

bosons with energy ǫ = c |~ p |, contained in a box of “volume” V = Ld in d dimensions. (a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical

+ (z), where z = eβµ , and potential µ. Express your answers in terms of d and fm + (z) fm

1 = (m − 1)!

Z

0

205

∞

xm−1 dx. z −1 ex − 1

(Hint: Use integration by parts on the expression for ln Q.)

• We have

Q= =

∞ X

P

eNβµ

N=0

n =N

i X

i

exp −β

{ni }

YX

eβ(µ−ǫi )ni =

i {ni }

Y i

X

ni ǫi

i

!

,

1

1−

eβ(µ−ǫi )

P P i) whence ln Q = − i ln 1 − eβ(µ−ǫ . Replacing the summation i with a d dimensional iR h R∞ ∞ d d k d−1 dk, where Sd = 2π d/2 / (d/2 − 1)!, integration 0 V dd k/ (2π) = V Sd / (2π) 0

leads to

ln Q = −

V Sd

d

(2π)

Z

∞

0

k d−1 dk ln 1 − ze−β¯hck .

The change of variable x = β¯hck results in ln Q = −

V Sd d

(2π)

kB T ¯hc

d Z

∞ 0

xd−1 dx ln 1 − ze−x .

Finally, integration by parts yields V Sd 1 ln Q = d (2π) d

kB T ¯hc

d Z

∞ 0

ze−x Sd x dx = V 1 − ze−x d d

kB T hc

d Z

∞

dx

0

xd , z −1 ex − 1

leading to Sd G = −kB T ln Q = −V d

kB T hc

d

+ kB T d!fd+1 (z) ,

which can be somewhat simplified to G = −kB T

V π d/2 d! + f (z) , λdc (d/2)! d+1

where λc ≡ hc/(kB T ). The average number of particles is calculated as N =−

∂G ∂G V π d/2 d! + = −βz = d f (z) , ∂µ ∂z λc (d/2)! d

where we have used z∂fd+1 (z)/∂z = fd (z). Dividing by volume, the density is obtained as n=

1 π d/2 d! + f (z) . λdc (d/2)! d

206

(b) Calculate the gas pressure P , its energy E, and compare the ratio E/(P V ) to the classical value. • We have P V = −G, while ∂ ln Q ln Q E=− = +d = −dG. ∂β z β

Thus E/(P V ) = d, identical to the classical value for a relativistic gas. (c) Find the critical temperature, Tc (n), for Bose-Einstein condensation, indicating the dimensions where there is a transition. • The critical temperature Tc (n) is given by n=

1 π d/2 d! 1 π d/2 d! + f (z = 1) = ζd . λdc (d/2)! d λdc (d/2)!

This leads to hc Tc = kB

n(d/2)! π d/2 d!ζd

1/d

.

However, ζd is finite only for d > 1, and thus a transition exists for all d > 1. (d) What is the temperature dependence of the heat capacity C (T ) for T < Tc (n)? • At T < Tc , z = 1 and E = −dG ∝ T d+1 , resulting in π d/2 d! G V E ∂E k = −d(d + 1) = d(d + 1) ζd+1 ∝ T d . = (d + 1) C (T ) = B ∂T z=1 T T λdc (d/2)! (e) Evaluate the dimensionless heat capacity C(T )/(N kB ) at the critical temperature T = Tc , and compare its value to the classical (high temperature) limit. • We can divide the above formula of C(T ≤ T c), and the one obtained earlier for N (T ≥ T c), and evaluate the result at T = Tc (z = 1) to obtain d(d + 1)ζd+1 C(Tc ) = . N kB ζd In the absence of quantum effects, the heat capacity of a relativistic gas is C/(N kB ) = d; this is the limiting value for the quantum gas at infinite temperature. ******** 207

17. Graphene is a single sheet of carbon atoms bonded into a two dimensional hexagonal lattice. It can be obtained by exfoliation (repeated peeling) of graphite. The band structure of graphene is such that the single particles excitations behave as relativistic Dirac fermions, with a spectrum that at low energies can be approximated by E ± (~k) = ±¯hv ~k .

There is spin degeneracy of g = 2, and v ≈ 106 ms−1 . Recent experiments on unusual

transport properties of graphene were reported in Nature 438, 197 (2005). In this problem, you shall calculate the heat capacity of this material. (a) If at zero temperature all negative energy states are occupied and all positive energy ones are empty, find the chemical potential µ(T ). • According to Fermi statistics, the probability of occupation of a state of of energy E is eβ(µ−E )n p [n(E)] = , 1 + eβ(µ−E )

for

n = 0, 1.

For a state of energy µ + δ, eβδn , p [n(µ + δ)] = 1 + eβδ

eβδ 1 p [n(µ + δ) = 1] = = . βδ 1+e 1 + e−βδ

=⇒

Similarly, for a state of energy µ − δ, p [n(µ − δ)] =

e−βδn , 1 + e−βδ

=⇒

p [n(µ − δ) = 0] =

1 = p [n(µ + δ) = 1] , 1 + e−βδ

i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for all temperatures; any particle leaving an occupied

negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at

zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) − E(0) = 4A

Z

d2~k E + (~k) (2π)2 exp βE (~k) + 1 +

208

.

• Using the label +(-) for the positive (energy) states, the excitation energy is calculated as E(T ) − E(0) =

X

k,sz

=2

[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]

X k

2 hn+ (k)i E + (k) = 4A

Z

E + (~k) d2~k . (2π)2 exp βE (~k) + 1 +

(c) Give a closed form answer for the excitation energy by evaluating the above integral. • For E + (k) = h ¯ v|k|, and ∞

2πkdk ¯hvk = (set β¯hck = x) 4π 2 eβ¯hvk + 1 0 2 Z ∞ 2A x2 kB T = dx x kB T π ¯hv e +1 0 2 kB T 3ζ3 . AkB T = π ¯hv

E(T ) − E(0) = 4A

Z

For the final expression, we have noted that the needed integral is 2!f3− (1), and used f3− (1) = 3ζ3 /4. E(T ) − E(0) = A

Z

E + (~k) d2~k (2π)2 exp βE (~k) − 1

.

+

(d) Calculate the heat capacity, CV , of such massless Dirac particles. • The heat capacity can now be evaluated as 2 9ζ3 ∂E kB T = CV = . AkB ∂T V π ¯hv (e) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capacity of graphene. The typical sound velocity in graphite is of the order of 2×104 ms−1 . Is the low temperature heat capacity of graphene controlled by phonon or electron contributions? • The single particle excitations for phonons also have a linear spectrum, with E p = h ¯ vp |k| and correspond to µ = 0. Thus qualitatively they give the same type of contribution to 209

energy and heat capacity. The difference is only in numerical pre-factors. The precise contribution from a single phonon branch is given by Z ∞ 2πkdk ¯hvp k Ep (T ) − Ep (0) = A = (set β¯hck = x) 4π 2 eβ¯hvp k − 1 0 2 Z ∞ A x2 kB T = dx x kB T 2π ¯hvp e −1 0 2 2 ζ3 3ζ3 kB T kB T = AkB T , CV,p = . AkB π ¯hvp π ¯hvp We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since the speed of Dirac fermions is considerably larger than that of phonons, their contribution to heat capacity of graphene is negligible. ******** 18. Graphene bilayer:

The layers of graphite can be peeled apart through different

exfoliation processes. Many such processes generate single sheets of carbon atoms, as well as bilayers in which the two sheets are weakly coupled. The hexagonal lattice of the single layer graphene, leads to a band structure that at low energies can be approximated by E 1 layer (~k) = ±tk (ak), as in relativistic Dirac fermions. (Here k = ~k , a is a lattice ±

spacing, and tk is a typical in-plane hopping energy.) A weak hopping energy t⊥ between the two sheets of the bilayer modifies the low energy excitations drastically, to E bilayer (~k) ±

=±

t2k

2t⊥

(ka)2

,

i.e. resembling massive Dirac fermions. In addition to the spin degeneracy, there are two branches of such excitations per unit cell, for an overall degeneracy of g = 4. (a) For the undoped material with one electron per site, at zero temperature all negative energy states are occupied and all positive energy ones are empty. Find the chemical potential µ(T ). • According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] =

eβ(µ−E )n , 1 + eβ(µ−E )

for

n = 0, 1.

For a state of energy µ + δ, p [n(µ + δ)] =

eβδn , 1 + eβδ

=⇒

p [n(µ + δ) = 1] = 210

eβδ 1 = . βδ 1+e 1 + e−βδ

Similarly, for a state of energy µ − δ, e−βδn , p [n(µ − δ)] = 1 + e−βδ

p [n(µ − δ) = 0] =

=⇒

1 = p [n(µ + δ) = 1] , 1 + e−βδ

i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ − δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for all temperatures; any particle leaving an occupied

negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at

zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) − E(0) = 2gA

Z

d2~k E + (~k) (2π)2 exp βE (~k) + 1

.

+

• Using the label +(-) for the positive (energy) states, the excitation energy is calculated

as

E(T ) − E(0) =

X

k,sz ,α

=g

X k

[hn+ (k)i E + (k) + (1 − hn− (k)i) E − (k)] 2 hn+ (k)i E + (k) = 2gA

Z

d2~k E + (~k) . (2π)2 exp βE (~k) + 1 +

(c) Give a closed form answer for the excitation energy of the bilayer by evaluating the above integral. • Let E + (k) = αk 2 , with α = (tk a)2 /(2t⊥ ), to get ∞

2πkdk αk 2 = (set βαk 2 = x) 2 eβαk2 + 1 4π 0 Z ∞ x kB T gA dx x kB T = 2π α e +1 0 2 gπ π A kB T kB T = = AkB T t⊥ . 24 α 3 a2 tk

E(T ) − E(0) = 2gA

Z

For the final expression, we have noted that the needed integral is f2− (1), and used f2− (1) = ζ2 /2 = π 2 /12. 211

(d) Calculate the heat capacity, CA , of such massive Dirac particles. • The heat capacity can now be evaluated as 2π A ∂E = kB CA = ∂T A 3 a2

k B T t⊥ t2k

!

.

(e) A sample contains an equal proportion of single and bilayers. Estimate (in terms of the hopping energies) the temperature below which the electronic heat capacity is dominated by the bilayers. • As stated earlier, the monolayer excitations for phonons have a linear spectrum, with

E1

layer

= ±tk (ka). Their contribution to energy and heat capacity can be calculated as

before. Including the various prefactors (which are not required for the solution), we have Z ∞ 2πkdk tk ak 1 layer E (T ) = 2gA = (set βtk ak = x) 4π 2 eβtk ak + 1 0 2 Z ∞ x2 gA kB T dx x = kB T π tk a e +1 0 2 2 kB T kB T A A 1 layer ∝ 2 kB , CA . ∝ 2 kB T a tk a tk The bilayer heat capacity, which is proportional to T is more important at lower temper-

atures. By comparing the two expressions, it is apparent that the electronic heat capacity per particle is larger in the bilayer for temperatures smaller than T ∗ ≈ t⊥ /kB . (f) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capacity of graphene. The typical sound velocity in graphite is of the order of 2 × 104 ms−1 .

Is the low temperature heat capacity of (monolayer) graphene controlled by phonon or electron contributions? • The single particle excitations for phonons also have a linear spectrum, with E p = h ¯ vp |k| and correspond to µ = 0. Thus qualitatively they give the same type of contribution to

energy and heat capacity. The difference is only in numerical pre-factors. The precise contribution from a single phonon branch is given by Z ∞ 2πkdk ¯hvp k = (set β¯hck = x) Ep (T ) − Ep (0) = A 4π 2 eβ¯hvp k − 1 0 2 Z ∞ A x2 kB T = dx x kB T 2π ¯hvp e −1 0 2 2 3ζ3 kB T kB T ζ3 , CV,p = . AkB = AkB T π ¯hvp π ¯hvp 212

We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since the speed of Dirac fermions is considerably larger than that of phonons, their contribution to heat capacity of graphene is negligible. ********

213