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Statistical Mechanics - Homework Assignment 2 Alejandro G´omez Espinosa∗ February 17, 2013

Pathria 2.7 Derive (i) an asymptotic expression for the number of ways in which a given energy E can be distributed among a set
of N one-dimensional harmonic oscillators, the energy eigenvalues of the 1 oscillators being n + 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 two results, 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:    X N 1 ~ω = R + ~ω E= nr + 2 2 r subsequently: E N − (1) ~ω 2 Then, to determine the number of distinct ways in which this energy can be distributed, let us use eq. 3.8.25: (R + N − 1)! (R + N )! RN Ω= ≈ ≈ R!(N − 1)! N 1 R!N ! RN N ! R=

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: N X p2r mω 2 qr2 H= + 2m 2 r=1

√p 2m

Making the following substitution x = tion:

and y = q

Z Z



PN

H≤E

4m mω 2

N/2

=

r=1

mω 2 2

for convenience in the volume calcula-

dN p dN q

V (E) = 0≤

q

Z Z P 2 2 0≤ N r=1 xr +xr ≤E

∗

[email protected]

1

dN x dN y

where the last integral has the same form as the volume of an n-dimentional sphere compute in the Appendix C. Therefore, using eq. C.7a from such Appendix for a two dimensional case:  N N N   2 π E 1 2πE N V (E) = = (3) ω N! N! ω Finally, dividing (3) by (2), we find the correspondence between the two results:
1 2πE N V N! ω = (E/~ω)N = (2π~)N = hN Ω N!

Pathria 2.8 Following the method of Appendix C, replacing equation (C.4) by the integral Z ∞ e−r r2 dr = 2

(4)

0

show that Z V3N =

Z ...

0≤

PN

i=1 ri ≤R

N Y i=1


(8πR3 )N 4πri2 dri = (3N )!

(5)

Using this result, compute the volume of the relevant region of the phase space of an extreme relativistic gas (ε = pc) of N particles moving in three dimensions. Hence, derive expressions for the various thermodynamic propierties of this system and compare your results with those of Problem 1.7. Using (4) and the procedure of Appendix C: 2

n

2N (4π)N 3N (3N − 1)! (8π)N (3N )!

P Zri =∞Z
exp (− ni=1 ri ) Y 2 = ... 4πr dr i i (4π)N ri =−∞ Z ∞
1 2 e−R 3N CN R3N −1 dR = N (4π) Z0 3N CN ∞ −R2 3N −1 = e R dR (4π)N 0 3N CN (3N − 1)! = (4π)N = CN = CN

Replacing in eq. C.2: V3N = CN R3N =

(8πR3 )N (3N )!

(6)

Then, to derive thermodynamic propierties of this system, let us calculate the multiplicity Γ of the microstates accessible to the system:   ω (8πR3 )N V 3N V R 3N (8π)N Γ= = = (7) ω0 (3N )! h3N h (3N )! 2

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

3N X

3N

X Xq E ⇒ p2xi + p2yi + p2zi = pi = c

pi c

i

i

i

We find a relation for the entropy:  S = k ln Γ = k ln

VR 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 ∂S = T ∂V N,E !!   ∂ EV 3N (8π)N = k ln ∂V hc (3N )! = = = = PV

=



∂ k ln(EV )3N (8π)N − ln(hc(3N )! ∂V

∂ k N ln(8π(EV )3 ) − ln(hc(3N )! ∂V
kN 24πE 3 V 2 3 8π(EV ) 3N k V 3N kT

Pathria 3.15 Show that the partition funtion QN (V, T ) of an extreme relativistic gas consisting of N monatomic molecules with energy-momentum relationship ε = pc, c being the speed of light, is given by (  3 )N 1 kT QN (V, T ) = 8πV (9) N! hc Study the thermodynamics of this system, checking in particular that 1 P V = U, 3

U = 3kT, N

γ=

4 3

(10)

Next, using the inversion formula (3.4.7), derive an expression for the density of states g(E) of this system. To calculate the partition function, we can use the result from 3.5.8. But in this case H(q, p) =

N X i=1

3

εi =

N X i=1

pi c

then:   Z ∞
N 1 pc/kT 2 e 4πp dp V N !h3N 0  N Z 1 4πV ∞ pc/kT 2 e p dp N ! h3 0 "  3 #N 1 4πV kT 2 3 N! h c "  3 #N 1 kT 8πV N! hc

QN (V, T ) = = =

=

(11)

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

∂ ln Q ∂β   "  3 #N ∂ 1 1  = − ln  8πV ∂β N! βhc   ! 1 3 ∂ 8πV = −N ln ∂β (N !)1/N βhc ! h3 c3 (N !)1/N 3 ∂ ln β = N ∂β 8πV ! ∂ h3 c3 (N !)1/N = N ln + ln β 3 ∂β 8πV ! ∂ h3 c3 (N !)1/N = N ln + 3 ln β ∂β 8πV = −

= U N

3N β

= 3kT

(12)

To compute the pressure, we must first define Helmholtz free energy: A(N, V, T ) = −kT ln QN (V, T ) = N kT

h3 c3 β 3 (N !)1/N ln − ln V 8π

!

thus,  P =−

∂A ∂V

 = N,T

N kT N = V βV

and replacing (12): P =

N U = Vβ 3V

(13)

Finally, to calculate γ:  CV =

∂U ∂T

4

 = 3N k V

(14)

 CP =

∂ (U + P V ) ∂T

 = 3N k + N k = 4N k

(15)

N,P

using (14) and (15) : γ=

4N kT CP 4 = = CV 3N kT 3

(16)

Pathria 3.30 The energy levels of a quantum-mechanical, one-dimensional, anharmonic oscillator may be approximated as     1 2 1 ~ω − x n + ~ω; n = 0, 1, 2, ... (17) εn = n + 2 2 The parameter x, usually  1, represents the degree of anharmonicity. Show that, to the first order in x and the fourth order in u(= ~ω/kT ), the specific heat of a system of N such oscillators is given by     1 2 1 1 4 1 3 C = Nk 1 − u + (18) u + 4x + u 12 240 u 80 Note that the correction term here increases with temperature. Let us calculate the partition function for a single harmonic oscillator: "    #! ∞ X 1 1 2 Q1 (β) = exp −β~ω n+ −x n+ 2 2 n=0 "     #! ∞ X 1 1 2 = exp −u n + −x n+ 2 2 n=0 !        ∞ X 1 2 ux2 1 2 1 1 + ux n + + n+ + ... = exp −u n + 2 2 2 2 n=0

Keeping only the first order in x:    !   ∞ X 1 2 1 1 + ux n + Q1 (β) = exp −u n + 2 2 n=0         ∞ X 1 1 2 1 = exp −u n + + ux n + exp −u n + 2 2 2 n=0      ∞ X eu/2 1 2 1 exp −u n + = + ux n + −u 1−e 2 2 n=0      ∞ 1 u 7u3 d2 X 1 = − + + ... + ux 2 exp −u n + u 24 5760 du 2 n=0     1 u 7u3 2 7u 31u3 = − + + ... + ux + − + ... u 24 5760 u3 960 48384 where the last calculations were done using Mathematica. Therefore, the N-oscillator partition function, keeping only until the fourth order in u, is given by:    N u 7u3 2 7u 1 N QN (β) = (Q1 ) = − + + ux + (19) u 24 5760 u3 960 5

Let us compute now the internal energy: U

∂ ∂ (ln QN ) = −~ω (ln QN ) ∂β ∂u   N !  2 7u 1 u 7u3 ∂ + ux + ln − + −kT u ∂u u 24 5760 u3 960     ∂ 1 u 7u3 2 7u −kT N u ln − + + ux + ∂u u 24 5760 u3 960     u u3 1 u2 1 − + + 4ux + kT N u u 12 240 u2 80     1 u3 u2 u4 + 4ux kT N 1 − + + 12 240 u 80

= − = = = =

Finally, let us calculate the specific heat of the system:     ∂U u2 u4 1 u3 C= = kN 1 − + + 4ux + ∂T 12 240 u 80

6

(20)