Homework Homework 4
David David Sirajuddin Physics 715 - Statistical Mechanics Professor Michael Ramsey-Musolf April 12, 2010
1. Consider an ideal gas of diatomic molecules. Compute the specific heat associated with the molecular vibrations C C V vib (T ) T ). Express your results in terms of the zero point energy of vibration. The vibrational energy ε energy ε n of a diatomic molecule may be modeled as a quantum harmonic oscillator, where εn =
n +
1 ε = 2
n +
1 2
ω ,
n
∈N
where ε0 = 12 ε is the zero point energy. energy. The specific heat associated with vibrational vibrational motion is computed as per the procedure: (a) Compute the system the system canonical canonical partition function Q function Q N associated with vibration (b) Obtain Obtain the average average energy per molecule, E , from QN (c) The vibrational vibrational specific specific heat C heat C V vib
(T ) T ) = ∂ E /∂T
(a) Compute the system canonical partition function QN The canonical partition function in quantum statistical mechanics is represented by the trace of the density matrix ρmn , or equivalently, by a sum over all states of the system of the Boltzmann factor e factor e εn β : ∞
QN = Tr ρ =
e−εn β
n=0
where the inverse temperature β temperature β = (kT ( kT ))−1 is defined in terms of Boltzmann’s constant Expressin singg the energ energy y εn as prescribed above k and the absolute temperature T . T . Expres provides a straightforward calculation of Q Q N ,
1
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
∞
QN =
1
e−(n+
2
)εβ
n=0 ∞
=
1
e−
2
εβ −nεβ
e
n=0
∞
− 12 εβ
= e
e−nεβ
n=0 ∞
= e−
1 2
εβ
n
e−(εβ )
n=0 ∞
= e−
1 2
εβ
xn ,
x = e −εβ
n=0
The substitution x = exp( εβ ) allows the identification of the series
−
∞
xn =
n=0
1 1
−x
=
1
−
1 e−εβ
So that the infinite sum allows for the replacement above,
− −− 1
QN = e− = =
εβ
2
1
e− e
1 2
1 2
εβ
εβ
1 1 e−εβ 1 1 e−εβ e−
1 2
εβ
1 1 sinh QN = εβ 2 2
−1
−1
(1)
(b) Obtain the average energy per molecule, E , from QN
The definition of the canonical partition function Q N is the trace of the density matrix over all states for N particles in the system occupying a volume V . This implies a probability function may be defined: P n =
1 −βε n e QN
where P n denotes the probability the system occupies a state n. Thus, the partition function furnishes a normalization factor that enforces the probability of molecules occupying all possible states n sum to unity: 2
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
n
1 P n = QN
e−βε n =
n
1 QN = 1 QN
= QN
Thus, the average energy E may be obtained by weighting the probability function by the energy ε n
=
E
1 QN
εn e−βε n
n
−
=
∂Q N ∂β
− Q1
∂Q N N ∂β
which admits,
E = − ∂ ln∂β Q
N
(2)
Recalling Eqn. (1) and letting ξ = εβ/2, the operations on the partition function Q N in Eqn. (2) are performed successively,
− ln Q
N
1 1 2 sinh ξ 1 1 ln ln 2 sinh ξ ln 2 + ln sinh ξ ∂ [ln 2 + ln sinh ξ ] ∂β cosh ξ ∂ξ 0+ sinh ξ ∂β
=
− ln
=
⇒ − ln Q ⇒ − ∂ ln∂β Q
N
N
− ∂ ln∂β Q
N
−
= = =
−
= cothξ
Noting that ξ = εβ/2 implies
1 ∂ξ ∂ 1 = εβ = ε 2 ∂β ∂β 2 Such that the average energy is obtained, 1 E = εcoth 2 3
⇒
1 εβ 2
(3)
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
(c) C V vib (T ) = ∂ E /∂T
The vibrational specific heat at constant volume is given by C V vib = ∂ E /∂T , where E is provided by Eqn. (3). The derivative is computed more readily by translation ∂/∂T ∂/∂β :
→
⇒
∂ ∂ ∂β ∂β = ; = ∂T ∂β ∂T ∂T 1 1 ∂ ∂ = 2 kT 2 ∂β ∂T
− 12 kT 1
2
−
Where the relation β = (kT )−1 was used. Employing the above translation for differential operators, and invoking the average energy E (Eqn. (3)) allows for the computation of the vibrational specific heat
C V vib (T )
≡ ∂ E /∂T
= = = = =
E − 12 kT 1 ∂ ∂β 2
− − − −
1 1 ∂ 1 1 εcoth εβ 2 kT 2 ∂β 2 2 1 ε ∂ 1 coth εβ 4 kT 2 ∂β 2 1 ε 1 1 2 csch εβ ε 4 kT 2 2 2 1 ε2 1 + csch2 εβ 2 8 kT 2
1 = 2
2
1 ε 2
1 csch2 2 kT
1 ε 2 kT
Recalling the zero-point energy ε 0 = 12 ε, the final form of the specific heat is obtained,
C V vib (T )
k ε 0 = 2 kT
4
2
csch2
ε 0 kT
(4)
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
2. A container is divided into two parts by a partition containing a small hole of diameter D (Figure 1). Helium gas in the two parts is held at temperatures T 1 = 150◦ K and T 2 = 300◦ K, respectively, through heating of the walls. (a) How does the diameter D determine the physical process by which the gases come into steady state? The appropriate insights are described by Huang, pgs. 95-96 [1], in the discussion of effusion. (i) Small diameter hole ( 1,2 D) If the hole diameter is small, then the mean free paths of particles from either side of the chamber, 1,2 , is large compared to the hole diameter D (1,2 D). This implies the gas particles will not interact strongly as they traverse the hole (i.e. collisionless, or effusion regime). A two-way “leakage” will occur between both sides of the chamber until steady state is obtained, marked by the condition of the fluxes from both sides of the chamber are equal ( I 1 = I 2 ).
(ii) Large diameter hole 1,2 D) If the hole diameter D is large, then the mean free paths of particles from either side of the chamber, 1,2 , is small by comparison (1,2 D). These conditions allow for many collisions between gas particles while traveling through the hole (i.e. collisional, or hydrodynamic , regime). Steady state is then achieved by a measure of the interactions between the particles being balanced. Specifically, this condition is reached upon the pressures in both chambers being matched (P 1 = P 2 ).
(b) What is the ratio of the mean free paths 1 /2 between teh two parts when D 1,2 and the system has reached a steady state? The diameter D of the hole being much smaller than the mean free paths ell 1,2 of gas particles in either part of the chamber corresponds to the effusion regime (i). As per Figure 1, the small hole of diameter D is normal to the x-direction. In the absence of collisions (effusion), the normal flux I incident on the hole from one side of the chamber is proportional to the velocity of the molecules v as prescribed by the distribution function f ( p):
dI = d3 p(v xˆ)f ( p) = d3 pv[cos(θ)]f ( p) 3 dI = d pvx f ( p)
·
⇒
(5)
for the normal velocity v x . Physically, the velocity v x must be positive in order to impinge on the small hole. Given that the scenario of interest involves steady state conditions, the distribution function f ( p) is taken to be the Maxwell-Boltzmann distribution, f 0 ( p): 5
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
f 0 ( p) =
n e− p 3/2 (2πmkT )
2
/2mkT
where the quanties k, n, and m denote Boltzmann’s constant and the gas particle density and mass, respectively. All other quantities are as previously defined. This definition facilitates integration of Eqn. (5) such that total flux I on the small hole from one side of the chamber may be obtained. The result is quoted from Huang [1], I =
d3 pvx f 0 ( p)
vx >0
kT ⇒ I = n 2πm √ ⇒ I ∝ T As described in part (a), steady state implies the fluxes are equal, I 1 = I 2
k n1 T 1 2πm
k = n2 T 2 2πm
n1
T 1 = n2 T 2 n2 T 1 = n1 T 2
(6)
The above relation may be put in terms of the mean free paths 1,2 by noting 1,2 = (n1,2 σ)−1 for an interaction cross-section σ. This implies, 1 = 2
T 1 = T 2
150◦ 300◦
K = K
1 2
(7)
(c) What is 1 /2 when D 1,2 ? The condition D 1,2 is indicative of the hydrodynamic regime (ii) as in part (a). Accordingly, equilibrium is established upon the pressures P 1,2 are balanced. For temperatures on this scale, helium gas is approximately ideal, such that it is permissible to express P = nkT as per the ideal gas law.
P 1 n1 kT 1 n2 n1 1 2
= P 2 = n2 kT 2 T 1 = T 2 T 1 = (1,2 = (n1,2 σ)−1 ) T 2 6
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
Thus, 150◦ K 1 1 T 1 = = = 300◦ K 2 2 T 2
(8)
3. (Huang 7.2) Consider a classical system of N noninteracting diatomic molecules enclosed in a box of volume V and temperature T . The Hamiltonian for a single molecule is taken to be ˆ ( p 1 , p 2 , r1 , r 2 ) = 1 ( p2 + p2 ) + 1 K r1 H 2 2m 1 2
| − r | 2
2
where p 1 , p 2 , r1 , r2 , are the momenta and coordinates of the two atoms in a molecule. Find The solution presented proceeds by the extraction of all desired parameters from both the canonical partition function QN and the grand partition function (z , V , T ) , defined in terms of the fugacity z = e µβ , volume V , and temperature T , where the inverse temperature β = (kT )−1 and µ is the chemical potential. This formalism allows for interchanges of particle number N between systems. Given that the system of interest is singular, the use of the grand partition function is, in some sense, overly robust for the intended purpose since the canonical partition function Q N of the system may be used to obtain compatible results. The distinction between the two formalisms will be evident in that the grand partition admits factors proportional to the fugacity z in the free energy calculation. It should be noted that the use of the parameter z is only valid due to the system of interest consisting of indistinguishable particles. Further use of the grand partition function proceeds with the presumptuous enforcement that the chemical potential µ is known. The grand partition function (µ,V,T ) is used for illustrative purposes, to show that consistent results may be obtained in this fashion, while the problem statement intends for the canonical partition function QN is to be used.
Z
Z
(a) the Helmholtz free energy of the system; The free energy F is found by aid of Huang Eqn. (7.64):
− β 1 ln( Z /z ) where the definition of the grand partition function Z (z , V , T ) = F (N, V ) =
− β 1 ln Q
N (V )
=
N
(9)
∞ N =0
z N QN (V, T ) has been used. Thus, calculation of the grand partition function (z , V , T ) allows for the computation of the free energy F . the grand partition function is defined in terms of the canonical partition function Q N (V, T ) of the system, QN (V, T ) =
Z
d3N p 1 d3N p 2 d3N r1 d3N r2 −β H ˆ ( p , p , r e N 6N (2!) N !h 1
7
2
1
, r2 )
(10)
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
The integration is extended over both atoms making up each diatomic molecule of the gas consisting of N molecules. A factor of 1/N ! has been introduced in suit with proper Boltzmann counting given the molecules are indistinguishable. The division by a factor (2!) N is included to prevent overcounting of configurations given two atoms per diatomic molecule. Further, a factor of h6N has been included 3N
3N
N N
3 i=1
3N 3N
3 j=1
for the normalization over (d p 1 d r1 )( p2 d d r2 ) = dp1i dr1i ) dp2 j dr2 j for vector components (i, j) = 1, 2, 3 . The cardinality of the integration then admits 6N factors of dimensions dpdr h. This normalization is substantiated in the discussion prior to Eqn. (7.64) of Huang. For convenience, QN is rewritten in terms of the single molecule partition function Q1 :
{
} ∼
1 QN (V, T ) = N !
⇒Q
N (V,
T ) =
d3 p 1 d3 p 2 d3r1 d3r2 −β H ˆ ( p , p , r e 6 (2!)h 1
2
1
N
, r2 )
= Q1
1 N Q N ! 1
(11)
Which is justified by noting each partition function for each molecule gives identical contributions. ˆ in the problem statement, the single partition function Given the Hamiltonian H Q1 may be computed:
Q1
1 1 β 2 3 3 3 3 2 = exp ( + p ) + d p d p d r d r p K r1 1 2 1 2 1 2 (2!)h6 2m 2 1 3 − βm p 3 − βm p 3 3 − βK |r −r | = d p e d p e d r d r e 1 2 1 2 2h6 2
(2πmkT )3/2
⇒Q
1
=
1 (2πmkT )6/2 6 h
2 1
2 2
2
| − r
2
1
2
2
| 2
2
(2πmkT )3/2
d3r1 d3r2 e−
βK 2
| r1 −r2 |2
The integration over space is furnished by the transformation to the relative and 12 = 1 (r1 + r2 ), respectively. The center-of-mass coordinates, r12 = r1 r2 and R 2 choice of relative coordinates was chosen given the convenient transformation of the exponential argument of the integrand. Any linear combination of the coordinates (r1 , r2 ) may be used for the remaining choice in the coordinate pair transformation however, some selections may be more convenient and intuitive. Here, the remaining coordinates selected are center-of-mass coordinates, motivated by both the conventional pairing of relative to CM coordinates, and the ease of translation:
−
d3r1 d3r2 8
3
3 12 d R12
→ d r
N
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
where it may be shown that the Jacobian determinant is identical to unity. Thus, the above equation is equivalent to 1 = (2πmkT )3 6 2h
Q1
12 d3 R
d3r12 e−
βK 2
2 r12
12 over all space occupying the system Integration the center-of-mass coordinates R 12 = V ), and evaluation of the remaining admits a factor of the volume V ( d3 R Gaussian integral is facilitated by the change of variables, r12,i ui for r12 = 3 ˆi , i=1 r12,i x
ui =
βK r12,i 2
→
2 dui βK
dr12,i =
⇒
Which provides the following replacements in the integrand above,
2 = u 2 r12
3
and d r12 =
2 βK
3/2 3
d u =
2kT K
3/2
d3 u
Thus, V = (2πmkT )3 6 2h
Q1
3/2
2kT K
2
d3 ue−u = π 3/2
V = (2πmkT )3 6 2h
⇒Q
1
1 V = (2πmkT )3 6 N ! 2h
⇒Q
N
3/2
2πkT K
2πkT K
3/2 N
(12)
The free energy F may be calculated directly from Eqn. (9) using this canonical partition function Q N , where the definition β = (kT )−1 is invoked, F = =
−kT ln Q −kT ln N 1 ! N
− −
= +kT ln N ! F = kT ln N !
V (2πmkT )3 6 2h
2πkT K
V (2πmkT )3 NkT 6 2h
V NkT ln 2h6 9
3/2 N
2πkT K
3/2
3 + 3ln (2πmkT ) + ln 2
2πkT K
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
A more tractable form of the free energy may be obtained by approximating the term ln N ! N ln N N as per Stirling’s approximation, admitting
−
3 2πkT V (N kT ln N N kT ) NkT ln + 3ln (2πmkT ) + ln F 2h6 2 K 3 2πkT V = ln N + ln(e) + ln + 3ln (2πmkT ) + ln N kT 2h6 2 K
−
−
−
−
Thus, the form of the free energy F calculated as per the canonical partition function Q N is given by,
F (N , V , T )
eV NkT ln 2N h6
−
3 + 3ln (2πmkT ) + ln 2
2πkT K
(13)
where ln e = 1, i.e. e exp(1). Alternatively, an expression for the free energy (in terms of the fugacity z ) may obtained from the grand partition function . Rewriting Eqn. (12) according to,
≡
Z
Q1 =
V 1/3 2h2
8π3 K
3
1/2
m β −9/2
α
3 −9/2
Q1 = α β 1 N QN = Q N ! 1 1 3N −9N/2 QN = α β N ! where the dependence on the parameter β has been left in tact to ease later computation. The grand partition function (z , V , T ) is then calculated directly,
⇒ ⇒
Z
∞
Z (z , V , T )
=
N =0
=
N
=
N
=
N
⇒ Z (z , V , T )
z N QN (V, T ) 1 N 3N −9N/2 z α β N ! 1 (zα 3 β −9/2 )N N ! 1 N λ N ! 3 −9/2
= exp(zα β 10
),
N
λN = e λ N !
(14)
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
Then, by Eqn. (9), the free energy F is determined,
F = = = = =
− β 1 ln(Z /z ) − β 1 ln(exp(zα β )/z ) −kT lnexp(zα (kT ) ) − ln z −kT zα (kT ) − N ln z −zα (kT ) + NkT ln e , z = e kT + N µ −z 2hV (2π) m (kT ) K kT N
3 −9/2
3
3
3
9/2
9/2
11/2
6
N
µ/kT
9/2
=
N
3
µβ
= e µ/kT
11/2
3/2
α3
Which provides the final solution,
F (µ , N , V , T ) = N µ
−
V (2π)9/2 3 µ/kT (kT )11/2 m e 6 2h K
(15)
Notice that despite the manifestly distinct apperance of this result obtained via the grand partition function , this formulation is completely compatible with the free energy result (13) obtained by way of the canonical ensemble. In particular, it appears that the right-hand term is not proportional to the number of gas particles N (!) The fact that N does not appear here is an artifact, in that the free energy does indeed contain this dependence, yet it is not as readily evident in this expression. Thus, this particular form is not as transparent as result (13). The suppression of the particle number N traces back to the replacement of the infinite series with the exponential function in Eqn. (14). In passing, it is noted that the free energy may be written generally as F = Nµ P V , enabling the pressure of the system to be readily determined,
Z
−
(2π)9/2 m3 µ/kT (kT )11/2 P = e 6 2h K (b) the specific heat at constant volume; The remaining two parts of the problem may be calculated directly from either the canonical or grand partition functions. The grand partition function is used here. The specific heat at constant volume, C V ,
Z
C V =
∂E ∂T
where the energy E is extracted from the (7.37): 11
Z (or Q
N )
according to Huang Eqn.
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
−
⇒
∂ ln (z , V , T ) , or E = ∂β ∂ ln (z , V , T ) E = +kT 2 ∂T
Z
∂ ln QN ∂β
−
Z
Translating differential operators, ∂ ∂ = ∂β ∂T
−
1 = kβ 2
∂T ∂ = ∂β ∂T
∂ −kT ∂T 2
Thus, ∂ ln (z , V , T ) ∂T ∂ / ln ezα (kT ) kT 2 ∂T ∂ kT 2 zα 3 (kT )9/2 ∂T 9 2 3 kT zα (kT )7/2 2 9 3 zα (kT )11/2 2
E = kT 2
Z
3
=
9 2
= =
(16) ⇒ E = Seeking the specific heat C ∼ N kT , the expectation value of N is computed V
by Eqn. (7.36) of Huang,
N
⇒ N
∂ = z ln (z , V , T ) ∂z ∂ / = z ln ezα (kT ) ∂z ∂ = z zα 3 (kT )9/2 ∂z = zα 3 (kT )9/2
Z
3
9 2
(17)
Combining Eqns. (16) and (17), the energy E may be expressed in terms of particle number N :
9 3 zα (kT )11/2 2 9 = zα3 (kT )9/2 (kT ) 2
E =
= N
⇒ E
9 = N kT 2 12
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
Then, the specific heat C V = ∂E/∂T ,
C V =
9 N k 2
(18)
As calculated using the grand partition function . The identical expression with replacement N N is obtained when the canonical partition function QN is employed.
Z
→
(c) the mean square molecule diameter r1 r2 2 . The grand partition function is used to calculate the mean square molecular diameter r1 r2 2 . Similar to the canonical partition function, = N N z QN (V, T ) presents a normalization factor such that a probability function P N may be defined,
| − |
Z
| − |
Z
P N =
z N QN (V, T ) N N z QN (V, T )
which describes the probability of a system in a state with N particles. The mean square diameter is then determined by the definition of an ensemble average: 2
|r − r | = 1
2
| − | − | − | N
r1
r2 2 z N QN (V, T ) N N z QN (V, T )
(19)
Noting that QN d3r1 d3r2 exp β 2 K r1 r2 2 , the square diameter may be extracted by differentiation with respect to the parameter K , which is contained in the constant α3 , defined in part (a) in the calculation of Q 1 .
∼
∂ = ∂K
Z − β |r − r | Z (z , V , T ) 2 1
2
2
Implying Eqn. (19) may be rewritten in order to find the
13
2
|r − r | , 1
2
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
2
|r − r | 1
2
= = = =
(z , V , T )/∂K − β 2 ∂ ZZ (z , V , T )
∂ −2kT ∂K ln Z (z , V , T ) ∂ ln e −2kT ∂K ∂ −2kT ∂K zα (kT ) zα 3 (kT )9/2
3
9/2
−
∂ V (2π)9/2 3 = 2kT z m (kT )9/2 6 3/2 ∂K 2h K α3
= 2kT z
3 2
1 V (2π)9/2 3 m (kT )9/2 6 3/2 K 2h K α3
2
|r − r | 1
2
= +
3kT 3 zα (kT )9/2 K
which implies,
|r − r | = 3N K kT 1
2
2
(20)
where the mean particle number N = zα3 (kT )9/2 . Again, the identical result is obtained using the canonical partition function with the replacement N N .
→
4. (Huang 7.3) Repeat problem (7.2), using the Hamiltonian ˆ ( p 1 , p 2 , r1 , r 2 ) = 1 ( p21 + p22 ) + r12 H 2m
| −r | ≡ |r − r |.
where and r0 are given positive constants and r12
1
0
2
5. An ideal classical (Maxwell-Boltzmann) gas of N particles at temperature T is in the spherical (3-dimensional) potential well V (r) =
V 0 ln(r/a) , ,
∞
r
≥
(a) Write down the normalized distribution of particles in 6-dimensional phase space (r, p). Given a Maxwell-Boltzmann gas, the distribution function f (r, p) may be written as per Huang (4.27): 14
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
f (r, p) = f 0 ( p)e−V (r)/kT where f 0 ( p) is the Maxwell-Boltzmann distriution function such that n, the total particle density, f 0 ( p) =
n e− p 3/2 (2πmkT )
2
d3 pf 0 ( p) =
/2mkT
The isotropic potential V (r) restricts the volume V of the system and allows for the definition of the particle density n in terms of the particle number N and the length parameter a, i.e. V = 43 πa3 n = N/V = 3N/4πa 3 . To express a ˆ r, p), ˆ normalized distribution function f ( that, seek a normalization factor A such
⇒
ˆ r, p) ˆ r, p) f ( = Af ( with the condition
ˆ r, p)d f ( 3rd3 p = 1 Aˆ
Aˆ
d3 pf 0 ( p)
f (r, p) = 1
d3re−V (r)/kT = 1
=n
ˆ An
d3re−V (r)/kT = 1
Using the expression for the potential V (r), integral
15
(21)
may
be evaluated,
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
=
d3re−V (r)/kT ∞
=
e−V (r)/kT r2 dr
dΩ
4π
0
= 4π
a
= 4π
0
a
r a
= 4π
0
−V 0 /kT
a
= 4π
−V (r>0)/kT r dr + 4π e r2 dr
rV 0 /kT
0
a
=0for V (r)→∞
r 2 dr
r2
dr
= 4πa
∞
−V 0 ln(r/a)/kT 2
e
aV /kT 0
a
V 0 /kT
drr 2−V /kT 0
0
V 0 /kT
4πa a3−V /kT 3 V 0 /kT 4πa3 = 3 V 0 /kT =
⇒
0
− −
Inserting result into Eqn. (21) provides the value of the normalization constant ˆ A,
ˆ An
d3re−V (r)/kT = 1
ˆ An
4πa3 = 1 3 V 0 /kT
− ˆ 3 − V /kT = 3 − V /kT ; ⇒ A = 0 4πa3 n
0
n =
3N
N (4/3)πa3
ˆ r, p) Then, the normalized distribution f ( may be written down, ˆ r, p) ˆ (r, p) f ( = Af ˆ 0 ( = Af p)e−V (r)/kT
ˆ r, p) f ( =
1 3 V 0 /kT e 4πa3 (2πmkT )3/2
−
16
1 − kT
»
–
p 2 +V (r) 2m
(22)
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
(b) Find the mean energy of a particle, ¯, where p2 + V (r) 2m The average energy is calculated routinely, =
ε¯ =
ˆ r, p) dΠεf ( , where dΠ = d 3rd 3 p ˆ r , p) dΠf (
Noting the definition of ε, result (22) may be recased: ˆ r, p) f (
ˆ = 1 3 − V /kT e → f (ε) 4πa (2πmkT ) 0
−ε/kT
3
3/2
ˆ into the relation for the average energy ε, the multiplicative prefInserting f (ε) actors cancel,
ˆ r, p) dΠεf ( ∂ = kT 2 ln ˆ r, p) ∂T dΠf (
ε¯ =
−
−
p 2 + V (r) 2m
dΠexp( ε/kT )
Computing the integral,
d3 pd 3r exp
dΠexp( ε/kT ) =
−
d3 p exp( p2 /2mkT )
=
Where the right-hand integral
And, integral Thus,
is
=
1 kT
−
was
d3r exp( V (r)/kT )
−
computed in part (a):
4πa3 d r exp( V (r)/kT ) = 3 V 0 /kT 3
−
−
a Gaussian integral that introduces a factor of (2πmkT )3/2 .
4πa3 ∂ 3/2 ln (2πmkT ) ε¯ = kT 3 V 0 /kT ∂T ∂ 3 = kT 2 ln(2πmkT ) + ln(4πa3 ) ln(3 ∂T 2 3 ∂ 3 = kT 2 ln(2πmk) + ln T + ln(4πa 3 ) 2 ∂T 2 2
−
−
− V /kT ) 0
− ln(3 − V /kT )
−
0
∂ 3 3 ∂ ∂ ∂ 3 + ln T + ln(4πa ) ln(3 ε¯ = kT 2 ln(2πmk) 2 ∂T ∂T 2 ∂T ∂T =0
= 1/T
17
=0
− V /kT ) 0
Sirajuddin, David
Homework 4 – Physics 715, Spring 2010
Evaluating derivative
:
∂ ln(3 V 0 /kT ) ∂T 1 V 0 = + 2 3 V 0 /kT kT 1 V 0 = T 3kT V 0 Combining these results in the relation for ¯ε admits the final result,
=
−
−
⇒
−
2
ε¯ = kT implying,
3 1 1 V 0 2 T T 3kT V 0
−
−
3 ε¯ = kT 2
−
V 0 3kT V 0
−
(23)
(c) The net energy of a gas is E = N ¯ and the heat capacity is C = dE/dT . Compute C (T ). Using result (23) above, the heat capacity calculation proceeds directly,
−
3 dE d(N ¯ ε) d V 0 = = N C = kT 2 3kT V 0 dT dT dT 3 V 0 d V 0 = Nk + NkT 2 3kT V 0 dT 3kT V 0 = Nk C =
− − 3 2
Nk 3kT V 0
−
which yields,
− − − − − − − − − −(3kT −V 0 )−2 3kV 0
3kV 0 V 0 3kT V 0 (3kT V 0 )2 3 3kV 0 (3kT V 0 ) V 0 2 3kT V 0
−
9 Nk C (T ) = kT 3kT V 0 2
−
−
1 V 0 2
−
3kV 0 3kT V 0
−
(24)
References [1] Huang, Kerson Statistical Mechanics 1987. John Wiley & Sons, Inc. Canada. [2] Ramsey-Musolf, Michael Physics 715 - Statistical Mechanics Class Notes, University of Wisconsin - Madison. Spring 2010. 18
