Statistical Mechanics - Homework Assignment 2

Alejandro Gomez Espinosa∗

February 17, 2013

Pathria 2.7 Derive (i) an asymptotic expression for the number of ways in which a given energy E canbe distributed among a set of N one-dimensional harmonic oscillators, the energy eigenvalues of theoscillators being

(n+ 1

2

)~ω, n = 0, 1, 2, 3.., and (ii) the corresponding expression for the volume of

the relevant region of the phase space of this system. Establish the correspondence between the tworesults, showing that the conversion factor ω0 is precisely hN .

To answer the first question, let us first define the total energy of N harmonic oscillators:

E =∑r

(nr +

1

2

)~ω =

(R+

N

2

)~ω

subsequently:

R =E

~ω− N

2(1)

Then, to determine the number of distinct ways in which this energy can be distributed, let us useeq. 3.8.25:

Ω =(R+N − 1)!

R!(N − 1)!≈

N1

(R+N)!

R!N !≈

RN

RN

N !

Replacing (1) in the last result and, considering that R N , we found:

Ω ≈ (E/~ω)N

N !(2)

which is the asymptotic expression for the number of ways we can distribute the given energy.

Next, to calculate the volume of the phase space, let us recall the Hamiltonian of our system:

H =

N∑r=1

p2r2m

+mω2q2r

2

Making the following substitution x = p√2m

and y = q√

mω2

2 for convenience in the volume calcula-tion:

V (E) =

∫ ∫0≤

∑Nr=1H≤E

dNp dNq

=

(4m

mω2

)N/2 ∫ ∫0≤

∑Nr=1 x

2r+x

2r≤E

dNx dNy

∗gomez@physics.rutgers.edu

1

where the last integral has the same form as the volume of an n-dimentional sphere compute in theAppendix C. Therefore, using eq. C.7a from such Appendix for a two dimensional case:

V (E) =

(2

ω

)N πNEN

N !=

1

N !

(2πE

ω

)N(3)

Finally, dividing (3) by (2), we find the correspondence between the two results:

V

Ω=

1N !

(2πEω

)N(E/~ω)N

N !

= (2π~)N = hN

Pathria 2.8 Following the method of Appendix C, replacing equation (C.4) by the integral∫ ∞0

e−rr2 dr = 2 (4)

show that

V3N =

∫...

∫0≤

∑Ni=1 ri≤R

N∏i=1

(4πr2i dri

)=

(8πR3)N

(3N)!(5)

Using this result, compute the volume of the relevant region of the phase space of an extreme rela-tivistic gas (ε = pc) of N particles moving in three dimensions. Hence, derive expressions for thevarious thermodynamic propierties of this system and compare your results with those of Problem1.7.

Using (4) and the procedure of Appendix C:

2n =

ri=∞∫...

∫ri=−∞

exp (−∑n

i=1 ri)

(4π)N

∏(4πr2i dri

)=

1

(4π)N

∫ ∞0

e−R2 (

3NCNR3N−1) dR

=3NCN(4π)N

∫ ∞0

e−R2R3N−1 dR

=3NCN(4π)N

(3N − 1)!

2N (4π)N

3N(3N − 1)!= CN

(8π)N

(3N)!= CN

Replacing in eq. C.2:

V3N = CNR3N =

(8πR3)N

(3N)!(6)

Then, to derive thermodynamic propierties of this system, let us calculate the multiplicity Γ of themicrostates accessible to the system:

Γ =ω

ω0=

(8πR3)N

(3N)!

V 3N

h3N=

(V R

h

)3N (8π)N

(3N)!(7)

2

Using the energy in the case of the extreme relativistic gas:

E =3N∑i

pic ⇒ E

c=

3N∑i

pi =∑i

√p2xi + p2yi + p2zi

We find a relation for the entropy:

S = k ln Γ = k ln

((V R

h

)3N (8π)N

(3N)!

)≈ k ln

((EV

hc

)3N (8π)N

(3N)!

)(8)

Finally, let us compare our results with the ones found in Pathria 1.7.

P

T=

(∂S

∂V

)N,E

=∂

∂V

(k ln

((EV

hc

)3N (8π)N

(3N)!

))

=∂

∂V

(k(ln(EV )3N (8π)N − ln(hc(3N)!

))=

∂

∂V

(k(N ln(8π(EV )3)− ln(hc(3N)!

))=

kN

8π(EV )3(24πE3V 2

)=

3Nk

VPV = 3NkT

Pathria 3.15 Show that the partition funtion QN (V, T ) of an extreme relativistic gas consisting of Nmonatomic molecules with energy-momentum relationship ε = pc, c being the speed of light, is givenby

QN (V, T ) =1

N !

8πV

(kT

hc

)3N

(9)

Study the thermodynamics of this system, checking in particular that

PV =1

3U,

U

N= 3kT, γ =

4

3(10)

Next, using the inversion formula (3.4.7), derive an expression for the density of states g(E) of thissystem.

To calculate the partition function, we can use the result from 3.5.8. But in this case

H(q, p) =

N∑i=1

εi =N∑i=1

pic

3

then:

QN (V, T ) =1

N !h3N

[V

∫ ∞0

epc/kT(4πp2 dp

)]N=

1

N !

[4πV

h3

∫ ∞0

epc/kT p2 dp

]N=

1

N !

[4πV

h32

(kT

c

)3]N

=1

N !

[8πV

(kT

hc

)3]N

(11)

Then, using (11), let us calculate the internal energy U :

U = − ∂

∂βlnQ

= − ∂

∂βln

1

N !

[8πV

(1

βhc

)3]N

= −N ∂

∂βln

(8πV

(N !)1/N

(1

βhc

)3)

= N∂

∂βln

(h3c3(N !)1/N

8πVβ3

)

= N∂

∂β

(lnh3c3(N !)1/N

8πV+ lnβ3

)

= N∂

∂β

(lnh3c3(N !)1/N

8πV+ 3 lnβ

)

=3N

βU

N= 3kT (12)

To compute the pressure, we must first define Helmholtz free energy:

A(N,V, T ) = −kT lnQN (V, T ) = NkT

(lnh3c3β3(N !)1/N

8π− lnV

)thus,

P = −(∂A

∂V

)N,T

=NkT

V=

N

βV

and replacing (12):

P =N

V β=

U

3V(13)

Finally, to calculate γ:

CV =

(∂U

∂T

)V

= 3Nk (14)

4

CP =

(∂

∂T(U + PV )

)N,P

= 3Nk +Nk = 4Nk (15)

using (14) and (15) :

γ =CPCV

=4NkT

3NkT=

4

3(16)

Pathria 3.30 The energy levels of a quantum-mechanical, one-dimensional, anharmonic oscillator maybe approximated as

εn =

(n+

1

2

)~ω − x

(n+

1

2

)2

~ω; n = 0, 1, 2, ... (17)

The parameter x, usually 1, represents the degree of anharmonicity. Show that, to the first orderin x and the fourth order in u(= ~ω/kT ), the specific heat of a system of N such oscillators is givenby

C = Nk

[(1− 1

12u2 +

1

240u4)

+ 4x

(1

u+

1

80u3)]

(18)

Note that the correction term here increases with temperature.

Let us calculate the partition function for a single harmonic oscillator:

Q1(β) =∞∑n=0

exp

(−β~ω

[(n+

1

2

)− x

(n+

1

2

)2])

=

∞∑n=0

exp

(−u

[(n+

1

2

)− x

(n+

1

2

)2])

=∞∑n=0

exp

(−u(n+

1

2

))(1 + ux

(n+

1

2

)2

+ux2

2

(n+

1

2

)2

+ ...

)Keeping only the first order in x:

Q1(β) =∞∑n=0

exp

(−u(n+

1

2

))(1 + ux

(n+

1

2

)2)

=

∞∑n=0

exp

(−u(n+

1

2

))+ ux

(n+

1

2

)2

exp

(−u(n+

1

2

))

=eu/2

1− e−u+∞∑n=0

ux

(n+

1

2

)2

exp

(−u(n+

1

2

))

=

(1

u− u

24+

7u3

5760+ ...

)+ ux

d2

du2

∞∑n=0

exp

(−u(n+

1

2

))=

(1

u− u

24+

7u3

5760+ ...

)+ ux

(2

u3+

7u

960− 31u3

48384+ ...

)

where the last calculations were done using Mathematica. Therefore, the N-oscillator partitionfunction, keeping only until the fourth order in u, is given by:

QN (β) = (Q1)N =

[(1

u− u

24+

7u3

5760

)+ ux

(2

u3+

7u

960

)]N(19)

5

Let us compute now the internal energy:

U = − ∂

∂β(lnQN ) = −~ω ∂

∂u(lnQN )

= −kTu ∂∂u

ln

([(1

u− u

24+

7u3

5760

)+ ux

(2

u3+

7u

960

)]N)

= −kTNu ∂∂u

ln

((1

u− u

24+

7u3

5760

)+ ux

(2

u3+

7u

960

))= kTNu

(1

u− u

12+

u3

240

)+ 4ux

(1

u2+u2

80

)= kTN

(1− u2

12+

u4

240

)+ 4ux

(1

u+u3

80

)Finally, let us calculate the specific heat of the system:

C =∂U

∂T= kN

(1− u2

12+

u4

240

)+ 4ux

(1

u+u3

80

)(20)

6