Introduction to Statistical Physics
Solution Manual
Chapter 1
1.1
Mass of water =106g, temperature raised by 20◦C. Heat needed Q = 2 × 107cal = 8.37×107J.=23.2 kwh.
Work needed = mgh = 14×150×29000 = 6.09×107 ft-lb =22.9 kwh.
1.2
Work done along various paths are as follows ab: Z b a P dV = N kBT1 Z b a dV V = N kBT1ln Vb Va cd: Pd(Vd− Vb) = N kBT3 µ 1 −VVb d ¶ de: N kBT3 Z e d dV V = N kBT3ln Va Vd
No work is done along bc and ea. The total work done is the sum of the above. Heat absorbed equals total work done, since internal energy is unchanged in a closed cycle. 1.3 (a) α = 1 V ∂V ∂T = bV0Tb−1 Tb 0V (b) ∆V = bV0T b−1 Tb 0 ∆T P = N kBT V = N kBT0b V0 T1−b Work done = P ∆V = bN kB∆T 1
1.4
Consider an element of the column of gas, of unit cross section, and height between z and z+dz. The weight of the element is −gdM, where dM is the mass of the element: dM = mndz, where m is the molecular mass, and n = P/kBT
is the local density, with P the pressure. For equilibrium, the weight must equal the pressure differential: dP = −gdM .Thus, dP/P = −(mg/kBT )dz. At
constant T , we have dp/P = dn/n.Therefore n(z) = n(0)e−mgz/kBT
1.5
No change in internal energy, and no work is done. Therefore total heat absorbed ∆Q = ∆Q1+ ∆Q2= 0. That is, heat just pass from one body to the
other. Suppose the final temperature is T . Then ∆Q1= C1(T − T1), ∆Q2= C2(T − T2). Therefore
T =C1T1+ C2T2 C1+ C2
1.6
Work done by the system is −RHdM . Thus the work on the system is Z HdM = κ T Z HdH =κH 2 2T 1.7
Consider the hysteresis cycle in the sense indicated in Fig.1.6. Solve for the magnetic field:
H = ±H0+ tanh−1(M/M0)
( + for lower branch, − for upper branch.). Using W = −RHdM , we obtain
W = − Z M0 −M0 dM [H0+ tanh−1(M/M0)] − Z −M0 M0 dM [−H0+ tanh−1(M/M0)] = −4M0H0 1.8
3 A log log plot of mass vs. A is shown in the following graph. The dashed line is a straightline for reference.
10 100 1000 10000
A
log M
1
Chapter 2
2.1
Use the dQ equation with P, T as independent variables: dQ = CPdT + [(∂U/∂P )T + P (∂V /∂P )T]dP
For an ideal gas (∂U/∂P )T = 0, P (∂V /∂P )T = −V. Thus
dQ = CPdT − V dP.
The heat capacity is given by C = CP− V (∂P/∂T )path.
The path is P = aVb, or equivalently Pb+1= a(N kBT )b by the equation of
state. Hence
V (∂P/∂T )path= [ab/(b + 1)]V (N kBT )bT−1 = bN kB/(b + 1). Therefore
C = CP −
b b + 1N kB This correctly reduces to CP for b = 0.
2.2
Use a Carnot engine to extracted energy from 1 gram of water between 300 K and 290 K.
Max efficiency η = 1 − (290/300) = 1/30. W = ηC∆T = 1
30(4.164 J g
−1K−1× 1 g × 10 K) = 1.39 J
Gravitational potential energy = 1 g × 9.8 kg s−2× 110 m = 1.08 J
2.3
The highest and lowest available temperatures are, 600 F = 588.7 K and 70 F = 294.3 K.
The efficiency of the power plant is W/Q1= 0.6[1 − (294.3/588.7)] = 0.3.
In one second: W = 106 J.
So Q2= 2.33 × 106J = CV∆T . Use CV = 4.184 J g−1K−1,
Flow rate = 6000 ×(0.305m)3 ∆T = 2.33 × 10 6 J (4.184 J g−1K−2)6000(0.305 m)3 106cm2/m3 = 3.27 × 10 −3K 2.4 (a)
Since water is incompressible, a unit mass input gives a unit mass output. The net heat supplied per unit mass is ∆Q = C(T1− T ) − C(T − T2),
where C is the specific heat of water (per unit mass.) In steady state v2/2 =
∆Q. This gives
v =p2∆Q =p2C(T1+ T2− 2T )
(b)
The entropy depends on the temperature like ln T . A unit volume of water from each of the input streams has total entropy ln T1+ ln T2 This makes two
unit volumes in the output stream, with entropy 2 ln T . Therefore the change in entropy is ln¡T2/T 1T2¢≥ 0. Thus T ≥√T1T2, and vmax= √ 2C¯¯¯pT1− p T2 ¯ ¯ ¯ 2.5 (a) P V1γ = 2P0V0γ, P V γ 2 = 2\P0V0γ (V1/V2)γ = 2. £ ( ¯L + a)/(L − a)¤γ = 2. a L = 21/γ− 1 21/γ+ 1 (b) ∆U = ∆Q − W , ∆Q = 0. CV ∆T = −W , ∆T = −W/CV. T1= 2T0+ ∆T = 2T0− (W/CV), T2= T0− ∆T = T0+ (W/CV). P =RT1 V1 = R [2T0− (W/CV)] A (L + a) (c) W = AR0adx(P1− P2) P1= 2P0V0γ/ [A(L + x)] γ , P2= P0V0γ/ [A(L − x)] γ . W = P0V0 γ − 1 ³ 1 − La−γ´ µ1 − 2aL ¶
7 2.6 (a) P V = U/3, U = σV T4. P = σT4/3. dS = dQ/T = (dU + P dV )/T.
Integrate along paths with T =const, V =const. S = 4 3σV T 3. (b) S =Constant. ∴ T3∼ V−1. Thus T ∼ R−1 2.7
The heat absorbed by an ideal gas in an isothermal process is ∆Q = N kT ln(Vf/Vi)
where Vf and Vi are respectively the final and initial volume.The temperature
T in this formula is the ideal-gas temperature.
Draw a Carnot cycle on the P V diagram, and label the corners 1234 clock-wise from the upper left.
The heat absorbed at the upper temperature T2, and the heat rejected at
the lower temperature T1, are
Q2= N kT2ln(V2/V1)
Q1= N kT1ln(V3/V4)
Because 23 and 12 lie on adiabatic lines, we have V2T2γ−1= V3T1γ−1
V1T2γ−1= V4T1γ−1
Dividing one equation by the other yields V2/V1= V3/V4.
The efficiency of the cycle is therefore η = 1 −QQ1 2 = 1 − T1 T2 2.8 Diesel cycle: Q2= CP(T3− T2) Q1= CV(T4− T1) η = 1 − (Q1/Q2) = 1 − γ−1[(T4− T1)/(T3− T2)] We have P3= P2, hence T3/T2= V3/V2= rc
The processes 12 and 34 are adiabatic, with T Vγ−1= constant. V 4= V1.
T3V3γ−1= T4V1γ−1
T2V2γ−1= T1V1γ−1
Using the three relations derived, we obtain η = 1 −γ1 r γ c − 1 rγ−1(rc− 1) 2.9 Otto cycle: Q2= CV(T3− T2) Q1= CV(T4− T1) η = 1 − (Q1/Q2) = 1 − [(T4− T1)/(T3− T2)]
The processes 12 and 34 are adiabatic, with T Vγ−1 = constant. We have V4= V1, V3= V2 Thus
T1V1γ−1= T2V2γ−1.
T3V2γ−1= T4V1γ−1.
Taking the ratio of these equations, we have T2/T1= T3/T4= rγ−1.
Thus
η = 1 − r1−γ
2.10
First note Tb/Ta= Vb/Va= 2.
Work done Heat absorbed a→b Pa(Vb− Va) = PaVa= N kTa CP∆T = CPTa
b→c 0 −CVTa
c→a −R P dV = −NkTaln 2 −NkTaln 2
W (Net work done) = N kTa(1 − ln 2)
Q2 (Heat absorbed) = CPTa=52N kTa η = W Q2 = 2 5(1 − ln 2) = 0.12 In comparison, ηCarnot= 1 − (Tb/Ta) = 0.5. 2.11
First note T2= 4T1. The P, V, T for the points A, B, C, D are as follows:
P V T
A P1 V1= N kT1/P1 T1
B 2P1 2V1 4T1
C 2P1 V1 2T1
9 (a)
Heat supplied along
ACB : CVT1+ CP(2T1) =¡32+ 5¢N kBT1=132N kBT1. ADB : CPT1+ CV(2T1) = ¡5 2+ 3 ¢ N kBT1=112N kBT1. AB : ∆U + ∆W = 3 2N kB(2T1) + 3 2P1V1= 6N kBT1. (b) Heat capacity = ∆Q/∆T = 6N kBT1/3T1= 2N kB. (c)
Work done = P1V1= N kBT1. Heat absorbed = Heat absorbed along ACB
= (13/2)N kBT1.
η = 2 13 2.12
(a)
Since no work is being done, and the temperatures diverge, heat must be transferred from the colder body to the hotter body, with no other effect, and this violates the Clausius statement of the second law.
(b)
The assertion is not true for physical black bodies, because they cannot be point-like but have finite size. Even if the two bodies have identical shapes, their optical images are not reciprocal. That is, the radiation from one body may form an image that is larger than the other body, and thus not completely absorbed by the other body.
Chapter 3
3.1 (a)
For a adiabatic process dS = 0, and the T dS equations give CVdT = −(αT/κT)dV
CPdT = αT V dP
Dividing one by the other, we obtain CP/CV = κT[−V (∂P/∂V )S] = κT/κS
(b)
CVdT + (αT /κT)dV = CPdT − αT V dP T. Put
dT = (∂T /∂P )VdP + (∂T /∂V )PdV .
Equate the coefficients of dP and dV on both sides. One of them gives CP − CV = (αT V /κT)(∂V /∂T )P = α2T V /κT.
(c)
Using U = A + T S, H = G + T S (enthalpy), we have
CV = (∂U/∂T )V = (∂A/∂T )V + S + T (∂S/∂T )V = T (∂S/∂T )V
= −T (∂2A/∂T2)V
CP = (∂H/∂T )P = (∂G/∂T )P+ S + T (∂S/∂T )P = T (∂S/∂T )P
= −T (∂2G/∂T2)P
3.2
The Sacker-Tetrode equation is
S = N kB[(5/2) − ln(nλ3)], where n = N/V , and λ = p 2π~2/mk BT . (a) A = U − T S = (3/2 /)kBT − T S = NkBT ln(nλ3) − NkBT. G = A + P V = N kBT ln(nλ3). (b)
Write ln(nλ3) = ln n + ln λ3. The second term is a function of T only. µ = (∂A/∂N )V,T = kBT ln(nλ3)+N kBT (∂ ln n/∂N )V,T−kBT = kBT ln(nλ3).
µ = (∂G/∂N )P,T = kBT ln(nλ3) + N kBT (∂ ln n/∂N )P,T = kBT ln(nλ3).
3.3
The force on the bead is (P − Pa)A − mg, where
P = pressure in gas, Pa= 1 atm.
The equation of motion for the displacement x is m¨x =(P − Pa)A − mg.
In equilibrium the pressure in the gas is P0= Pa+ (mg/A).
The volume is V0= RT /P0.
Assume adiabatic oscillations: P Vγ= const. This implies dP = −γ(P/V )dV ≈ −γ(P0/V0)Ax.
P = P0+ dP ≈ P0− γ(P0/V0)Ax.
Thus m¨x +¡γA2P2 0/RT
¢ x = 0. The frequency of oscillations is
ω = AP0
p γ/RT
3.4
Let the equilibrium pressure and temperature be P0, T0. Under an
in-finitesimal displacement x, suppose the pressure of compartment 1 changes by dP . Since the process is adiabatic, we have P Vγ = constant, or (dP/P ) + γ(dV /V ) = 0. In terms of the temperature, we have T Vγ−1 = constant, or (dT /T ) + (γ − 1)(dV /V ) = 0.
(a)
For compartment 1, we have to first order dP = −γPL0x dT = −(γ − 1)T0x
L For compartment 2, replace x by −x. (b)
The force acting on the piston is dF = AdP . The equation of motion for x is dF = M ¨x, where M is the mass of the piston. Thus ¨x + (γAP0/M L)x = 0,
and the frequency of small oscillations is ω =pγAP0/M L
(c)
Due to the finite thermal conductivity of the piston, heat flows back and forth between the two compartment, because of the oscillation in the temper-ature difference.Assume that the tempertemper-atures change so slowly that at any moment we regard heat conduction as taking place between two heat reservoirs of fixed temperatures. When an amount of heat dQ flows from 1 to 2, the entropy increase is dS = (dQ/T2) − (dQ/T1). Thus
dS dt = µ 1 T2 − 1 T1 ¶ dQ dt = kB(∆T )2 T1T2 ≈ kB µ ∆T T0 ¶2
13 The temperature difference is
(∆T )2= (T1− T2)2= (2dT )2=4(γ − 1) 2T2 0x2 L2 Hence dS dt = ax 2 where a = 4kB(γ − 1)2/L2. (d)
Energy dissipation, which has so far been ignored, occurs at the rate T0dS/dt =
aT0x2. The time average of this rate is 12aT0x20, where x0 is the amplitude of
oscillation. The energy of oscillation is E = 12M ω2x20. In one period of
oscil-lation, the energy dissipated is ∆E = (2π/ω)1 2aT0x
2
0. This gives a fractional
dissipation per cycle
∆E E = 2πT0 aM ω3 3.5 (a) P = − µ ∂A ∂V ¶ T = a0(v0− v) (b) κT = −v−1(∂v/∂P )T = (a0v)−1 α = v−1(∂v/∂T )P = −v−1(∂P/∂T )V(∂v/∂P )T, by chain rule. α = 1 a0v da0 dT (c) µ = µ ∂A ∂N ¶ V,T = a0(v02− v2) − f 3.6
For this problem it is important to use the entropy expression with arbitrary CV, instead of setting it to (3/2)kB. Write the adiabatic condition as
∆S = ∆S1+ ∆S2= 0, or
(N1+ N2)kBln(Vf/Vi) + (N1CV 1+ N2CV 2) ln(Tf/Ti) = 0.
Thus, Tf/Ti = (Vi/Vf)ς,where ζ = kB(N1+ N2)/(N1CV 1+ N2CV 2).
This means T Vζ = constant. Putting T = P V /N k
BT , where N = N1+N2T, we have P Vξ = constant where ξ = ζ + 1 = N1(CV 1+ kB) + N2(CV 2+ kB) N1CV 1+ N2CV 2 = n1CP 1+ n2CP 2 n1CV 1+ n2CV 2
3.7 (a)
Since the disks are thin, we can assume that their temperatures always remain uniform.
Let the final temperature be T .
The changes in temperatures are respectively ∆T1= T − T1, ∆T2= T − T2.
For simplicity write CP 1= C1, CP 2= C2.
The amounts of heat absorbed are respectively ∆Q1 = C1∆T1, ∆Q2 =
C2∆T2.
Since the system is isolated ∆Q1+ ∆Q2= 0. This gives
T =C1T1+ C2T2 C1+ C2
(b)
Consider the instant when the two temperatures are T0
2, T10, (T20> T10).
When an amount of heat dQ flows from 2 to 1, the entropy increase is dS = (dQ/T0
1) − (dQ/T20).
We can express dQ in terms of the dT ’ through dQ = C1dT10 = −C2dT20.
Thus we can rewrite dS = C1(dT10/T10) + C2(dT20/T20).
∆S = C1 Z T T1 dT0 1 T0 1 + C2 Z T T2 dT0 2 T0 2 = C1ln T T1 + C2ln T T2 3.8
The relations are straightforward mappings from a P V system to a magnetic system.
3.9 (a)
The desired expression are straightforward mappings of those for a P V sys-tem.
(b)
The first relation is the condition that dA be an exact differential. The second is obtained by using the equation of state M = κH/T .
(c)
The chain rule states (∂T /∂H)S(∂H/∂S)T(∂S/∂T )H= −1.
From (b) we have (∂H/∂S)T = −T2/(κH).
By definition, the heat absorbed at constant H is given by T dS = CHdT .
Thus (∂S/∂T )H= CH/T.
3.10 (a)
The important property to verify is that at constant T the entropy decreases as the magnetic field H increases.
15 Isothermal magnetization: dT = 0.
The heat absorbed is
dQ = CMdT − HdM = −HdM. Therefore ∆Q = − Z H 0 HdM = − κH2 2T0 (c) Adiabatic cooling: dQ = 0. From dQ = CMdT − HdM we obtain dT = (H/CM) dM = ¡
κ/aT2¢M dM . Multiply both sides by T2and
inte-grate: RT1
T0 T
2dT = (κ/a)R0 MM dM.
This gives T13= T03− (κ/2a) M2, or T13= T03−κ
3H2
2aT2 0
This becomes negative when the magnetic field H is sufficiently large. However, the equation becomes invalid long before that happens, for it is based on Curie’s law, which is valid only for weak fields.
Chapter 4
4.1
The system is in contact with a heat reservoir, but initially not in equilibrium with it. Let the stages of the process be labeled A,B,C:. We first calculate the heat absorbed ∆Q, and the entropy change ∆S of the system.
(A) Water cools from 20◦C to 0◦C.
∆Q = CP∆T = −10 × 4180 × 20 J = −8.36 × 105 J. ∆S =R dQ/T = CPRdT /T = CPln(Tf/Ti) = 41800 ln(273/293) = −2.96× 103J/deg. (B) Solidification at 0◦C. ∆Q = −10 × 3.34 × 105J =.−3.34 × 106 J. ∆S = ∆Q/T = −3.34 × 106/273 = −1.22 × 104 J/deg.
(C) Ice cools from 0◦C to -10◦C.
∆Q = C0
P∆T = −10 × 2090 × 10 J = −2.09 × 105 J.
∆S = C0
Pln(Tf/Ti) = 20900 ln(263/273) = −7.80 × 102 J/deg.
Total heat absorbed by system: ∆Qsys= −4.39 × 106J
Total entropy change of system: ∆Ssys= −1.39 × 104J/deg.
The reservoir has a fixed temperature T0= −10◦C..
The total heat absorbed by reservoir equals that rejected by the system: ∆Qres= 4.39 × 106J.
Entropy change of reservoir:
∆Sres= ∆Qres/T0= 4.39 × 106/263 = 1.67 × 104 J/deg.
∆Suniverse= ∆Sres+ ∆Sres= 2.8 × 103 J/deg
4.2
Let P0, T0be the pressure and absolute temperature at the triple point. Let
L be the extensive latent heat (not specific latent heat.) Since the solid-gas 17
transition can be made either via a direct path or a solid-liquid-gas path, we must have
Lsublimation= Lmelt+ Lvap
Vaporization: dP/dT ≈ Lvap/T V = P Lvap/N kBT2.
P = P0exp · Lvap N kBT0 µ 1 −T0 T ¶¸ Melting: dP/dT = Lvap/T ∆V. P = P0+ Lmelt ∆V ln T T0
Sublimation: dP/dT ≈ P (Lvap+ Lmelt)/N kBT2.
P = P0exp · Lvap+ Lmelt N kBT0 µ 1 − TT0 ¶¸ 4.3 dP/dT = /T ∆v = [1.44 J/(18 − 20)cm3]T−1. ∴ dT/dP = −c0T , where c0= 1.39 cm3/J. 4.4 (a)
At a given v > v0, the dashed line lies at a lower free energy than the
solid line. The latter represents a “stretched” that fills the whole volume. The former represent a liquid drop at specific volume v0 that does not fill up the
entire volume. This is therefore the preferred state of the liquid. At v = v0 the
pressure is zero. (b)
Now assume that the liquid coexists with its vapor, treated as an ideal gas. We are in the transition region of a first-order phase transition. At the given temperature, the liquid and gas have fixed densities, which must be consistent with the requirement of equal pressure P and chemical potential µ. Denote quantities for the liquid with subscript 1, and those for the vapor with subscript 2:
P1= a0(v0− v),
µ1= a0(v20− v2) − f,
P2= nkBT,
µ2= kBT ln(nλ3).
where P1, µ1 were obtained in Prob.3.5, and µ2 was given in Prob.3.2, with
λ =p2π~2/mk
BT . Thus, the conditions determining v and n are
a0(v0− v) = nkBT
19 From the first equation, we see that v0− v > 0. It approaches zero as nT → 0.
(c)
Small n corresponds to (v0− v) → 0. The second equation becomes −f ≈
kBT ln(nλ3). Thus nλ3≈ exp(−f/kBT ) 4.5 (a) dP/dT = /[T (v2− v1)] ≈ /T v2= /[T (kBT /P )]. Hence T P dP dT = kBT (b) T (K) (ergs/g) 0.2 8.21×107 0.4 9.37 0.6 10.5 0.8 11.8 1.0 13.1 1.2 14.4 4.6
The accompanying sketch shows G =RV dP . The system skips the closed loop in the graph of G, because it is higher than need be
A A
V
P
P
G
4.7 (a)
A(V, T ) = −RT ln(V − b) − (a/V ) + f(T ) As V → ∞, A(V, T ) → −RT ln V + f(T )
This should approach the ideal gas result (Prob.3.2) RT [ln(nλ3) − 1]. Therefore, up to an additive constant,
f (T ) = −RT µ 1 + 3 2ln T ¶ (b) CV = −T (d2f /dt2) = (3/2)R, which is a constant. 4.8 T dS = CVdT + T (∂P/∂T )V dV = 0. dT /dV = −(T/CV) (∂P/∂T )V = −(RT /CV)(V − b)−1.
Integrating this yields
ln T = −(R/CV) ln(V − b)+ constant.
Thus the adiabatic condition is
T (V − b)R/CV = constant
When a = b = 0, the system reduces to an ideal gas, for which R
CV
= CP− CV
CV = γ − 1
Thus we recover T Vγ−1= constant.
4.9
The second virial coefficient for the van der Waal gas is given by c2 =
b − (a/RT ). A rough fit is
b ≈ 17 cm3/mole
a ≈ 2100R deg cm3/ mole
4.10
Let ∆V = V1− V2 be the difference in volume across the transition line.
Consider variations along the transition line, such as going from a to b, as illustrated in the sketch. The chain rule says
µ ∂∆V ∂T ¶ P µ ∂T ∂P ¶ ∆V µ ∂P ∂∆V ¶ T = −1 This gives µ ∂P ∂T ¶ ∆V = −(∂∆V /∂T )(∂∆V /∂P )P T = α1− α2 κT 1− κT 2
21 If the transition line refers to a second-order phase transition, then across this line ∆V = 0, while the differences in α and κ are nonzero. Thus
dP dT =
∆α ∆κT
Chapter 5
5.1 n = 2.70 × 1019 atoms /cm3 v = 2 × 105 cm/s. N = nv/6 ≈ 1024 s−1 cm−2. 5.2Let tV0be the volume of the room, and V be the volume under consideration
The probability of finding an atom in V is V /V0.
The probability of finding it elsewhere is 1 − (V/V0).
Since there are N independent atoms, the probability of finding none in V is p = µ 1 − VV 0 ¶N = exp µ N ln µ 1 −VV 0 ¶¶
For small V /V0we can use the expansion ln (1 − (V/V0)) ≈ −V /V0. Thus
p ≈ exp(−NV /V0)
Under STP,
N = V0
22.4 liter mole−1 × (6.02 × 10
23 mole−1)
For V0= 27 × 103liter,we have
for V = 1 cm3: p ≈ exp¡−2.7 × 1019¢≈ 10−1019
for V = 1 A3: p ≈ exp¡−2.7 × 10−5¢≈ 1 − 2.7 × 10−5= 0.99997
5.3
Let n = N/V
Probability of finding one atom in dV = ndV. Probability of finding no atom in dV = 1 − ndV. Probability of finding no atom in V = exp(−nV ).
p(r)dr = Prob.(one atom between r, r + dr)×Prob.(no atom in sphere of radius r) p(r) = 4πnr2exp µ −4 3πnr 3 ¶ 5.4
For the beam to remain well-collimated, the atoms should suffer no scattering by the air in the chamber along the flight path of length L. The condition is therefore λ > L, where λ ≈ (nσ)−1 is the mean-free-path, where n is the density
of the air, and σ is the cross section for a collision between atoms in the beam with an air molecule. Thus
n < 1 Lσ
For a rough estimate, take σ ≈ 10−16 cm2. This gives n < 10−15 cm−3. The
estimate can be refined by using a more precise value for σ. 5.5
(a)
The mass density of water is 1 g cm−3. This corresponds to a number density
n = 2 × 1023cm−3. Thus λ = 5 × 1016 cm.
(b)
The rate of reaction is R = N Iσ, where N is the number of nucleons, I is the neutrino flux, and σ is the reaction cross section. A person of mass 150,kg contains N = 1029 nucleons. Thus R = 5 × 10−10 s−1.
The collision time is τ = R−1= 2 × 109s ≈ 70 yrs.
Thus, one gets hit by a neutrino about once in a lifetime. 5.6
Following the hint, the answer is obtained straightforwardly:
Cn= 2πn/2 Γ¡n 2 + 1 ¢ −→ n→∞ n 2ln π − n 2ln n 2 + n 2 5.7 (a)
From (5.37), Γ(E, V ) = (∂Φ/∂E)∆, where Φ = VN Z p2 1+...+p2n<E dp1· · · dpn with n = 3N . Thus Γ(E, V ) = K0VnΣn ³√ E´= K0VnnCnE(n−1)/2
25 (b) Using S = kBln Γ, we have, up to an additive constant,
S(E) N kB = ln V +ln C3N N + ln E 3/2+ O µ ln N N ¶ = ln³V E3/2´+ O µ ln N N ¶ 5.8 (a)
By the same reasoning as in the last problem, we obtain Γ(E, V ) = K0Σn
³√ E´, where n = 6N . There is no volume dependence in the limit V → ∞, because the particles are confined by the harmonic oscillator potential.
(b) Transcribing the result of the last problem, we have S(E) N kB = ln E3+ O µ ln N N ¶ 5.9
Let the mean-free-path be λ ≈ 10−5 cm. To be away from the origin by a
distance L, a total of (L/λ)2 random steps would have to be taken. Since each step lasts a collision time τ ≈ 10−10s, the total time required is τ (L/λ)2. For L = 1 cm the time is:1 sec. For L = 1 m the time is 104 sec.
5.10
For one coordinate, the probability of return after n collisions is (2πn)−1/2, according to (5.16). For the N -particle state to recur, all 6N coordinates have to return at the same time. When this happens, different particles woud have made different numbers of collisons n. For our order-of-magnitude estimate, we can imagine that all particles have made an average numbers of collisions ¯n, each with probability p = (2π¯n)−1/2 which is a small but finite number. The probability for gas as a whole to return to the initial state is then p6N. That is,
Recurrence time ≈ exp µ
6N ln1 p ¶
in units of the collision time. For N ∼ 1019, this number is of order exp¡1020¢,
which is so large that neither the value of p nor the units used makes any significant difference.
Chapter 6
6.1 Let λ = (2mkBT )−1. h i = R d3p ε f (p) R d3p f (p) = 1 2m R∞ 0 dpp 4exp(−λp2) R∞ 0 dpp2exp(−λp2) =3 2kBT 2®= R d3p ε2f (p) R d3p f (p) = 1 4m2 R∞ 0 dpp 6exp(−λp2) R∞ 0 dpp2exp(−λp2) = 15 4(kBT ) 2 2® − h i2=3 2(kBT ) 2 6.2The energy distribution is defined through P (E)dE = f (p)4πp2dp, where
f (p) is the Maxwell-Boltzmann distribution of momentum. Using E = p2/2m,
we obtain P (E) = c0 √ Ee−E/kBT where c0= nπ−1/2(kBT )−3/2. 6.3
The density is obtained by integrating the distribution function over the momentum. The result is
n(z) = n(0)e−mgz/kBT
6.4
Using the equation of state of the ideal gas, we obtain P(1−γ)/γT = C 0.After
some manipulation this leads to dP P = γ γ − 1 dT T − mg kBT dz = γ γ − 1 dT T 27
Thus T changes with height z according to kB dT dz = − γ − 1 γ mg This can be integrated to yield
kBT (z) = kBT0−γ − 1
γ mgz
For T0= 300 K and γ = 7/5, the temperatures becomes zero at z = 3.17×104m.
(b)
From the above, we find dP
P = − mg kBT
dz
Using the expression for from the last part, we can integrate this to obtain P P0 = µ 1 −γ − 1 γ mgz kBT0 ¶γ/(γ−1) 6.5
There is an effective temperature-dependent potential U (x), given through exp(−U/kBT ) = c0(1 + γx). 6.6 The answer is n1(r) n2(r) = exp£ω2r2(m1− m2)/2kBT ¤ 6.7 (a)
The most probable velocity is that at the maximum of the speed distribution. This will be obtained in (c).
(b)
The pressure is given by
P = Z px>0 d3p2p xvxf (p) = 1 3 Z d3pp p2 p2+ m2f (p)
where we have used vx= px/
p p2+ m2. Write p2 p p2+ m2 = p p2+ m2− m 2 p p2+ m2
29 The second can be neglected in the ultra-relativistic limit p2À m2. Comparing
P with the energy density U/V =R d3ppp2+ m2f (p), we obtain
P V → 1 3U (c)
The velocity distribution f (v) is defined by
f (v)d3v = C exp³−βmv/p1 − v2´d3p Using p = mv/√1 − v2, we obtain f (v) = Cm 3 (1 − v2)5/2 exp µ − m kBT √ 1 − v2 ¶
The non-relativistic limit corresponds to m(1 − v2)−1/2≈ 1
2mv2+ m + O(v2)
Now return to part (a). The most probable velocity v0 corresponds to the
maximum of the speed distribution 4πv2f (v). It is given by the root of the equation (1 − v2)3/2+5 2v 2 (1 − v2)1/2− m 2kBT v2= 0 The non-relativistic and ultra-relativistic limits are (with c restored)
v0 c ≈ r 2kBT mc2 (kBT ¿ mc 2) v0 c ≈ 1 − µ mc2 5kBT ¶2 (kBT À mc2) (d)
Relativistic effects become noticeable when kBT /mc2is appreciable, say, at
10%. For H2this corresponds to kBT = 0.1 × 2 GeV, or T = 2 × 1012K.
6.8 (a)
The distribution is proportional to the velocity distribution exp¡−mv2
x/2kBT¢.
Substitute vx = c (f − f0) /f0 and then normalize the distribution. The result
is P (f ) = µ 2πkBT f0 mc2 ¶−1/2 exp µ − mc 2 2kBT f02(f − f 0)2 ¶ (b) The variance is (f − f0)2= R∞ 0 df (f − f0) 2 exp³−λ (f − f0)2 ´ R∞ 0 df exp ³ −λ (f − f0)2 ´
where λ = mc2/(2k
BT f02). Change the variable of integration to ν =
√
λ(f −f0).
The lower limit of integration becomes −√λf0= −
p mc2/2k
BT . This can be
replaced by −∞ when kBT ¿ mc2, which is true in usual laboratory conditions.
We then obtain (f − f0)2= kBT mc2f 2 0 (c)
The line width is given by the square root of the variance, and thus inversely proportional to√m. The H2 line width is therefore broader than that of O2
by a factorp32/2 = 4. 6.9 (a) f (p)∝ e−βcp, U/N = c¯p = 3kBT, CV = 3N kB. (b) P V = 13U = N kBT. 6.10
Follow the hints and directions given in the problem. 6.11 W = Z vx>v0 d3pvxf (p) = C Z ∞ mv0 dpxvxe−λp 2 x ·Z ∞ −∞ dpyexp ¡ −λp2y ¢¸2 = n r kBT 2πmexp µ − v 2 0 2mkBT ¶ 6.12
(a) The escape velocity is vc=
p
2GM/R ≈ 104m/s. This is to be compared with the most probable speed at STP v0 =
p
2kBT /m ≈ 2.2 × 103 m/s. The
fraction of gas that can escape is f = C n Z ∞ mvc dp4πp2e−p2/2mkBT =√4 π Z ∞ y dxx2e−x2
where C = n(2πmkBT )−3/2 and y = vc/v0. Using the results of Prob.6.10(b),
we obtain f ≈ √2y πe −y2 With y ≈ 4.5, we find f ≈ 5 × 10−8. (b)
31 The time it takes for an atom to go from sea level to the top of the atmosphere through random collisions is
t ≈ L
2
λvc = 3 × 10 11
s ≈ 104 yr
where L = height of atmosphere ≈100 km, λ = mean-free-path ≈3×10−7m.
6.13 (a)
The number of atoms with momentum magnitude between p and p + dp is V 4πp2f (p)dp, where V is the volume, and f (p) is the Maxwell-Boltzmann
distribution. Thus ∆N = 4πV Z ∞ p0 dpp2f (p) ∆E = 4πV Z ∞ p0 dpp2 p 2 2mf (p)
Using the results of Prob 6.10(b), and .N = nV , and E = 32N kBT , we obtain
the fractional changes
∆N N = 2πy 1/2e−y ∆E E = 4π 3 y 3/2e−y for y = 0/kBT À 1.. (b)
From kBT = 23E/N , we obtain kB∆T = 23[(∆E/N ) − E(∆N/N2)], hence
∆T T = ∆E E − ∆N N ≈ 2πy 1/2 µ 2 3y − 1 ¶ e−y
Taking the logarithm of the equation for ∆N/N,we have ln(N/∆N ) = y − ln(2πy1/2), which gives
y ≈ ln(N/∆N) + ln³2π ln(N/∆N )1/2´= ln(πN/∆N ) + ln ln(N/∆N ) We then find, to leading order,
∆T T ≈ 4 3 ∆N N r ln∆N N
6.14 (a)
Let the axis along the needle be labeled 1, and a perpendicular axis 2. The moments of inertia about these axes are I1, I2, with I2 À I1. By the
equipartition of energy we have J2 1 2I1 = J 2 2 2I2 = kBT 2
where Jiis the components of angular momentum along the axis i. Thus
J2 J1 = r I2 I1 À 1
That is, the angular momentum is nearly parallel to the axis of the needle. (b)
The equipartition of energy states 1 2CV 2=1 2kBT This givespV2= 6.5 µV . 6.15 (a)
Take 1 mole of N2. The mass is 28 g.
For v = 7 km/s, the kinetic energy is K.E. = 12M v2= 0.5(0.028)(7000)2= 686 kJ.
When this energy is converted into heat, the temperature rise is of the order of
∆T =K.E./kB= 5500 K.
Thus, the astronauts would be fried. (b)
A constant deceleration a is equivalent to the application of a potential mxa, where x is the distance, and m is the mass of an air molecule. The Boltzmann factor gives a relative density distribution
∆n n = exp µ −mxa kBT ¶
which equals the fractional change in pressure ∆P/P at constant temperature. The difference in pressure between the points x = 0 and x = L is therefore ∆P = P0[(1 − exp(−mxL/kBT )].
(c)
Let the total mass of air be M = N a, where N is the total number of air molecules. The force is
F = A∆P =AP0maL kBT
= P0V N kBT
33 The stopping time is t = v/a, which corresponds to a distance 12at2 = v2/2a.
The work done is therefore
W = F v 2 2a = 1 2M v 2
Thus the translational kinetic energy is completely converted to mechanical work, and ∆T = 0..
(d) The translational velocity of the air must be, at all times, much smaller than sound velocity, relative to the walls of the container.
Chapter 7
7.1 Particle flux : IN=Rvx>0d3pvxf (p) Energy flux: IE= R vx>0d 3pv xf (p)mv2/2Average energy of an escaped particle IE IN =m 2 R∞ 0 dvv 5exp¡ −mv2/2kBT ¢ R∞ 0 dvv3exp (−mv2/2kBT ) = 2kBT
Thus, the escaped atoms come to thermal equilibrium at a temperature T1, with 3
2kBT1 = 2kBT . Hence T˙ 1 = 4
3T . This assumes that the total amount of gas
escaped is so small that the temperature of the original system is unchanged. 7.2
Let n1, n2 denote the densities of U-238 and U-235 respectively. Flux
= npkBT /2πm.
After one stage of effusion, n1 n2 = µ n1 n2 ¶ 0 r m2 m1 After k stages, µ n1 n2 ¶ k = µ n1 n2 ¶ 0 µ m2 m1 ¶k/2
Find k such that n1= n2.
k = 2 ln (n1/n2)0 ln (m1/m2)
=2 ln (99.27/0.75) ln (238/235) = 775 7.3
In an adiabatic process P Vγ = constant. Using the equation of state, we find that P (N kBT /P )γ = constant, or P1−γTγ = constant. Differentiating this
relation with respect to P we obtain µ ∂T ∂P ¶ S = γ − 1 γ T P 35
The particle density is n = P/kBT . Thus µ ∂n ∂P ¶ S = 1 kBT γ Hence κS = 1 nkBT γ c = r kBT γ m 7.4
Test the condition K/c ¿ CP. Using the data given, we find K/c = 1.44 ×
10−5, CP = 0.24T, in the mixed unit given. Thus the condition is well-fulfilled,
and shows that sound propagated adiabatically. 7.5
From (7.19) ∂2ρ/∂t2 + ρ∇ · ∂u/∂t = 0. Instead of the Euler equation
ρ∂u/∂t = −∇P , use the Navier-Stokes equation.(7.48). Then in first-order approximation (7.21) is replaced by .∂2ρ ∂t2 − ∇ 2P +4ν 3∇ 2 (∇ · u) =0
where ν is the viscosity. Use the continuity equation ρ∇ · u = − ∂ρ/∂t, and convert ∇2P to ∇2ρ as in (7.22). The result is
.∂2ρ ∂t2 − 1 c2∇ 2 ρ −4ν3ρ∇2∂ρ∂t=0
For a sinusoidal wave ρ = ρ0+ ρ1exp(ikx −iωt), the last term is i(4νkω/3ρ0)ρ1. Thus the damping coefficient is 4νkω/3ρ0.
7.6
The one-dimensional diffusion equation has solution
n(x, t) = √N 4πDtexp µ − x 2 4πDt ¶
The gas is characterized by the diffusion constant D. Suppose the detector has spatial resolution ∆x. We want to find the time t at which n(L, t)∆x = 1. That leads to the implicit equation
t = L2 4πD
1
37 In the first approximation, we put t = L2/4πD on the right side. This gives
t = L2 4πD
1 ln (N ∆x/L) The logarithm is not very sensitive to ∆x.
7.7
The insulating power η is the inverse of the coefficient of thermal conduc-tivity. Thus η ∝ σ√m, where σ is the collision cross section, and m the mass of the molecule of the gas. Assuming that the molecular diameter increases like m1/3 we have
η∝ m5/6
To double η, we need to increase m by a factor 26/5 = 2.3. To double the
insulating power of air, we would need a gas of molecular weight 69. 7.8
This is a hypothetical exercise, since we are ignoring an important heat source, the radiation from the sun. (See Prob.17.3). The total rate of heat generated is 43πR3W , and this must equal the rate of heat radiated 4πR2σT4
i,
where T1 is the surface temperature, and σ is Stefan’s constant. This give the
surface temperature T1= µ RW 3σ ¶1/4
In the interior, the rate of heat generation per unit volume is ρW , and this equals ∇ · q, where q is the heat flux vector. Using q = −κ∇T , we have the equation for the temperature distribution ∇2T = −ρW/κ. Assuming that T is spherically symmetric, and using spherical coordinates, we obtain
d drr 2dT (r) dr = − ρW κ r 2
Integration of this equation, observing that T cannot be singular at r = 0, gives T (r) = T0−
ρW 6 r
2
where T0 is the temperature at r = 0. The surface temperature is
T1= T0− ρW 6 R 2 Thus T0= 1 6ρW R 2+ µ RW 3σ ¶1/4
7.9 (a)
The heat absorbed by per unit volume is dQ = −∇ · qdt, which defines the heat flux vector q. Putting dQ = T ds, we have
∂s ∂t +
1
T∇ · q = 0 (b)
Consider the heat flux due to heat conduction q = −κ∇T. Write (∇·q)/T = ∇ · (q/T ) − q · ∇(1/T ). The last equation can be rewritten
∂s ∂t = −∇ · q T + κ µ ∇T T ¶2
The second term, which is always positive, is the rate of irreversible entropy production.
7.10
Suppose the thickness of the ice sheet is x. Consider a unit square of the ice sheet. The mass of the sheet increases at the rate ρdx/dt, and generates heat at the rate ρdx/dt. This must equal the heat flux, which we can represent as
qx=
κ∆T x for a small thickness x. Thus
dx dt =
κ∆T ρx
Chapter 8
8.1 (a)
The number of ways to choose the n atoms to remove from N sites is N !/ [n! (N − n)!]
(b)
The number of ways to choose the n interstitials out of M is M !/ [n! (M − n)!]
(c)
The total energy is E = n∆.The phase space volume is Γ(n) = N !M !
n! (N − n)!n! (M − n)! Using the Stirling approximation, we obtain the entropy S kB = ln Γ(n) = n lnN n − (N − n) ln ³ 1 −Nn´+ n lnM n − (M − n) ln ³ 1 −Mn´ The temperature is defined through
1 kBT = 1 kB ∂S ∂E = 1 ∆ ∂ ln Γ(n) ∂n This gives ∆ kBT = ∂ ∂nln Γ(n) = ln µ N n − 1 ¶ + ln µ M n − 1 ¶ (d)
The previous equation can be rewritten as n2 (N − n) (M − n) = exp µ − ∆ kBT ¶ 39
The low- and high-temperature limits are n ≈ √N M exp (−∆/2kBT ) (kBT ¿ ∆) 1 n ≈ 1 N + 1 M (kBT À ∆) (e) n N ≈ exp (−∆/2kBT ) For T = 300 K: n/N ≈ e−20= 2 × 10−9. For T = 1000 K: n/N ≈ e−6= 2.5 × 10−3. 8.2 (a)
Since each link can be pointed left or right independently, the number of ways to choose N+ links to point right is Γ = N !/[N+!(N − N+)!]. We must
have N− = N − N+. The entropy is S = kBln Γ, which leads to
S
N kB = −r ln r − (1 − r) ln(1 − r)
where r = N+/N is the fraction of right-pointing links.
(b)
The internal energy is independent of N+, and we can set it to zero. Thus
the free energy is A = −T S, where T is just a constant scale factor. (c)
The tension τ can be obtained from dU = 0 = T dS + τ dL, where L is the length of the chain:
L = a(N+− N−) = a (2N+− N) = aN(2r − 1) We obtain τ kBT = − 1 2aN kB ∂S ∂r = 1 2aln 1 r(1 − r)
where kBT is just a scale factor. The tension is never zero. It is minimum when
r = 1/2, and goes to infinity when the chain is fully stretched to the right (r = 1) or to the left (r = 0).
In this model, “temperature” is not a relevant concept, since energy is irrel-evant. The factor T in T dS is an arbitrary scale factor.
If we give each left-pointing link an energy , then the total energy would be E = N−= N (1−r). The temperature would be given by T−1= − ln [r(1 − r]],
apart from a scale factor 8.3
(a)
Assume that a link can be up or down independently. The partition function is the product of the partition functions of the individual links. The possible
41 energies are 0 and mga. Thus QN = [1 + exp (−βmga)]N. We have ignored the
fact that the energy of the nth link depends on its height, and therefore on the states of the preceding links. We have also ignored is the restriction that the links cannot go above the ceiling.
(b)
U = −∂ ln Q∂βN = N mga exp (βmga) + 1
The length of the chain is L = (N − N0)a, where N0= U/(mga) is the number of up links. Thus
L = N a 1 + exp (−βmga) (c)
Since U = mga[N − (L/a)], the force constant is mg. 8.4
(a)
The possible states are labeled by the number of open links n = 0, 1, 2, · · · , N. The energy with n open links is En = n∆. The partition function is
QN= N X n=0 e−βn∆=1 − e−β( ¯ N +1)∆ 1 − e−β∆ (b)
The average number of open links is ¯
n = −∆1 ∂ ln Q∂β N = e−β∆ 1 − e−β∆ −
(N + 1) e−β( ¯N +1)∆
1 − e−β( ¯N +1)∆
The second term is negligible for large N. At low temperatures β∆ À 1 we have
¯
n ≈ e−β∆ 8.5
(a)
There are 6 sites in each hexagon, but each site is shared by 3 hexagons. Thus we can assign 2 sites to a hexagon. On the other hand, each hexagon is associated with one interstitial site. Thus, in an infinite lattice, there are half as many interstitial sites as lattice sites.
(b)
The entropy is given by S kB
= ln Γ(E) = ln Γvacancy+ ln Γinterstitial
Γvacancy = N ! M !(N − M)! Γinterstitial = (N/2)! M !(N/2 − M)!
The energy is E = M ∆, and the volume fixed, and proportional to N . Thus, this gives S(E, V ).Using the Stirling approximation, we obtain
S kB = {N ln N − M ln M − (N − M) ln(N − M)} + {N → N/2} = 2M ln√N 2M − (N − M) ln µ 1 −M N ¶ − µ N 2 − M ¶ ln µ 1 − 2M N ¶ (c)
From T−1 = ∂S//∂E we obtain the relation
µ E N ∆ ¶2 = 1 2 µ 1 − E N ∆ ¶ µ 1 − 2E N ∆ ¶ e−∆/kBT
This can be easily solved, but we only give the high- and low-temperature limits: E N ∆ ≈ ½ 21/2exp(−∆/2k BT ) (kBT À ∆) 1/3 (kBT ¿ ∆)
The above is equal to the average interstitial fraction M/N at a given temper-ature.
8.6 (a)
The partition function for N non-interacting particles is QN = QN1, where
Q1 is that for a single particle:
Q1= 3 X n=1 exp (−β n) = 2e−β(bx 2 −cx/2)+ e−β(bx2 +cx)
The free energy per particle is a(x, T ) = −kBT ln Q1.
(b)
We find the equilibrium value of ¯x by minimizing a(x, T ) with respect to x, or maximizing Q1.Assume that ¯x is small, and expand the exponential to order
x2. The condition Q01= 0 gives two roots:
¯ x =
½
0
(4kBT /c)£1 − (4bkBT /c2)¤
Since ¯x cannot be negative, the nontrivial root is acceptable only when T < Tc,
where
kBTc =
c2
4b Examining the sign of Q00
1 shows that when T < Tc the nontrivial roots
corre-sponds to a maximum, while ¯x = 0 corresponds to a minimum. For T > Tc, the
only solution is ¯x = 0, which corresponds to a maximum. Thus there is a phase transition at T = Tc.
43 8.7
For a classical relativistic gas,
QN(V, T ) = Z d3Np d3Nq N !h3N exp " −β N X i=1 p (cpi)2+ (mc2)2 # = V N N ! 1 h3NI N(β) where I(β) = Z d3p exph−βp(cp)2+ (mc2)2i
Using the Sterling approximation to write N ! ≈ NN, we obtain
AN(V, T ) = −NkBT · ln µ V N h3 ¶ + 1 + ln I(β) ¸
In the nonrelativistic limit kBT ¿ mc2 we have
I(β) ≈ Z d3p exp · −β µ mc2+ p 2 2m ¶¸ = e−mc2/kBT(2πmk BT )3/2 AN(V, T ) ≈ N · mc2+ kBT ln µ λ3N V ¶ − kBT ¸ µ ≈ mc2+ kBT ln ¡ nλ3¢
In ultra-relativistic situations kBT À mc2 we can neglect the rest energy, and
take I(β) ≈ Z d3p e−βcp= 4π Z ∞ 0 dp p2e−βcp= 8π µ kBT c ¶3 AN(V, T ) ≈ NkBT " ln à π2N V µ ~c kBT ¶3! − 1 # µ ≈ kBT ln ¡ nL3¢ where L = π2/3 ~c kBT 8.8
The partition function is QN= ξN, where ξ is the partition function for one
particle: ξ =X k e−β~2k2/2m= V (2π)3 Z d3ke−β~2k2/2m= V λ3 where λ =p2π~2/mk
BT . The free energy is A = −kBT ln QN= −NkBT ln
¡ V /λ3¢, The equation of state is P = −∂A/∂V = NkBT /V.
8.9 (a)
The partition is QN = ξN, where
ξ = 1 τ Z ∞ −∞ dpe−βp2/2m Z ∞ −∞ dqe−βmω2q2/2= 2πkBT τ ω (b)
The free energy is
A = −kBT ln QN = −NkBT ln µ 2πkBT τ ω ¶ Thus S = −∂A ∂T = N kB · 1 + ln µ 2πkBT τ ω ¶¸ U = A + T S = N kBT CV = ∂U ∂T + N kB 8.10 (a) QN =¡eβµ0B+ e−βµ0B¢ N (b) hMi = −β1∂B∂ ln QN = µ0N eβµ0B− e−βµ0B eβµ0B+ e−βµ0B (c) M2®− hMi2= 1 β2 ∂2 ∂B2ln QN= 4µ2 0N (eβµ0B+ e−βµ0B)2
Chapter 9
9.1 (a) Q(z, T ) = N0 X N =0 µ N0 N ¶ zNe−βN = N0 X N =0 µ N0 N ¶¡ ze−β ¢N=¡1 + ze−β ¢N0 (b) hNi N0 = z N0 ∂ ∂zln Q = 1 z−1eβ + 1 (c) U = − ∂ ∂β ln Q = N0 z−1eβ + 1 C = ∂U ∂T = N0kB(β )2eβ z (z−1eβ + 1)2 9.2 (a)The grand partition function for the O2 lattice gas is
Q1(z, T ) = N X N1=0 µ N N1 ¶¡ z1e−β 1¢ N1 =¡1 + z1e−β1¢ N
The fraction of occupied sites is hN1i N0 = z1 N0 ∂ ∂z1ln Q1 = 1 z−11 eβ 1+ 1 45
Setting the above to f = 0.9, with z1= 10−5 and T = 310 K, we find,
1= kBT ln
z1(1 − f)
f ≈ −0.37 eV (b)
The grand partition function is now given by
Q(z, T ) = N X N1=0 NX−N1 N2=0 µ N N1 ¶µ N − N1 N2 ¶¡ z1e−β1¢ N1¡ z2e−β2¢ N2 = N X N1=0 µ N N1 ¶¡ z1e−β 1¢ N1¡ 1 + z2e−β 2¢ N−N1 = ¡1 + z1e−β 1+ z2e−β 2¢ N
The fraction of sites occupied by O2is
hN1i N = z1 N ∂ ∂z1ln Q = z1e−β 1 ! + z1e−β 1+ z1e−β 2
Set this to 0.1 and solve for 2. With 1 from (a), we obtain 2= −0.55 eV 9.3 (a) E(M ) = − M Γ(M ) = µ N M ¶ (b)
The grand partition function of the adsorbed gas is
Q(z, T ) = N X M =0 µ N M ¶¡ zeβ ¢M =¡1 + zeβ ¢N
where z = eβµ.The average fraction of occupied sites can be obtained either by
maximizing the summand using the Stirling approximation: ln ·µ N M ¶¡ zeβ ¢M ¸ ≈ M ln¡zeβ ¢+ N ln N − M ln M − (N − M) ln (N − M) or by calculating the grand canonical average:
¯ M N = z N ∂ ∂zln Q = 1 z−1e−β + 1
47 (c)
The chemical potential for an ideal gas is given in Prob.3.2: µ = kBT ln
¡ nλ3¢, where λ = p2π~2β/m, and n = βP . In equilibrium, the chemical potntail of
the adsorbed gas must equal that of the surrounding gas. Thus ¯ M N = λ3βP e−β + λ3βP (d) M2− ¯M2= z ∂ ∂zz ∂ ∂zln Q = ze−β (e−β + z)2 9.4 (a)
The equation of state is µ V −13 ¶ µ P + 3 V2 ¶ = 8 3T
Differentiating both sides with respect to P at constant T , we find
−∂V∂P = V − 1 3 P +V32 −V63 ¡ V −13 ¢ = ¡ V −13 ¢2 8 3T − 6 V3 ¡ V −13 ¢2
Near the critical point we put V = 1 and let T → 1+. Thus
κT = − 1 V ∂V ∂P ≈ 1 6 (T − 1)
(b) The fractional density fluctuation near the critical point diverges: n2− ¯n2 ¯ n2 = T VκT ≈ 1 6 (T − 1) 9.5
The condition for equilibrium is ln¡n+L3
¢ +ln¡n−L3¢= 0, or n +n−= L−6. Given n − n+= n0, we find-n+ n0 = 1 2 "s 4 (n0L3)2 + 1 − 1 # n− n0 = 1 2 "s 4 (n0L3)2 + 1 + 1 #
9.6 (a)
Let Ni be the number of molecules of type Xi present. The reaction
con-sumes νimolecules of type Xi. Thus the change in Niis proportional to νi, with
the same proportionality constant for all i. Hence δN = δNi/νi is independent
of i. (b)
Minimizing the free energy, we have 0 = δA =X i ∂A ∂Ni δNi= X i µiνiδN = 0
Since δN is arbitrary, we havePiµiνi= 0.
9.7 (a)
In a fixed volume, the densities obey the relations δn1
2 = δn2= − δn3
2 Hence A = n1− 2n2 and B = n1+ n3remain constant.
(b)
The chemical potential for a classical ideal gas is, according to Prob.3.2, µ = kBT ln¡nλ3¢, where λ =
p
2π~2/mk
BT . The condition for chemical
equi-librium is 2 ln¡n1λ31 ¢ + ln¡n2λ32 ¢ −2 ln¡n3λ33 ¢
= 0 where the λiare independent
of thus densities. Thus
n2 1n2 n2 3 = K0 where K0= (4/9)3 ¡ mkBT /π~2 ¢3/2
. Two other conditions are n1+ n3 = n0 n1 = 2n2 These imply n31= 2K0(n0− n1)2 High-temperature limit K0→ ∞ : n1≈ n0 µ 1 − r n 0 2K0 ¶ Low-temperature limit K0→ 0 n1≈ (2K0n0)1/3 9.8
49 The density depends on the power series
y =
∞
X
=1
b z
We seek an expansion of the equation state in the form P nkBT = P∞ =1b z P∞ =1 b z = 1 + a2y2+ a3y3+ · · · To the lowest two orders, we have
P∞ =1b z P∞ =1 b z = 1 + a2 Ã∞ X =1 b z !2 + a3 Ã∞ X =1 b z !3 + · · ·
We expand both sides to order z3, obtaining
z + b2z2+ b3z3+ · · · = ¡z + 2b2z2+ 3b3z3+ · · · ¢ h 1 + a2 ¡ z + 2b2z2+ · · · ¢2 + a3(z + · · · )3+ · · · i = z + (2b2+ a2) z2+ (4b2a2+ a3+ 3b3) z3+ · · ·
Equating the coefficients of z2 and z3 on both sides, we obtain a2 = −b2
Chapter 10
10.1
The energy residing in a mode of frequency ω of the transmission line is E = ~ω eβ~ω− 1 = ~ω β~ω +12(β~ω) 2 + · · · ≈ kBT µ 1 − 2k~ω BT ¶
The second term above gives the first quantum correction. As a estimate use the fundamental mode ω = πc/L, where c is the velocity of light, and L the length of the transmission line. The Nyquist theorem becomes
V2= 4k BT R∆ν µ 1 − 2Lkπ~c BT ¶
For L = 1 mm, the correction amounts to approximately 1% at T = 300 K. 10.2
The accompanying sketch illustrates the construction that would lead to a fractal of dimension 2.
(a) Start with a straight line of unit length.
(b) Halve the step size, and double the path length by taking more steps. The way to do this is not unique. Pick one of the ways.
(c) In the next iteration, each previous segment is independently replaced by a path of twice the length with half the step size.The path length L depends on the step size τ according to L∝ τ1−D, with D = 2.
10.3
Ignoring the possibility that two suspended particles collide with each an-other, we can regard the suspension as an ideal gas in equilibrium with. the medium, which acts as a heat reservoir. Therefore its partial pressure obeys the ideal gas law.
10.4
It is straightforward to show that n(x, t) = √ 1
4πDtexp ¡
−x2/4Dt¢
satisfies the diffusion equation. To show the initial condition, note n(x, t) −→
t→0
½
0 (x 6= 0) ∞ (x − 0) and, for all t 6= 0,
Z ∞ −∞ dxn(x, t) = 1 Therefore n(x, t) −→ t→0δ(x) 10.5 (a)
For the Brownian particles: D = 4 × 10−9cm2/s.
For O2: D ≈ 0.1cm2/s
Thus an O2 molecule will travel
q
1
4 × 1010ρ ≈ 10 cm.
(b)
From Einstein’s relation η = D/kBT.
F = u/η = kBT u/D ≈ 10−5dyne.
10.6
Perrin obtained A0= 7.05 × 1023, which would have led to
kB = 8.32 × 10 6 7.05 × 1023 = 1.18 × 10−16 cgs (Modern value:1.381 × 10−16) e = 2.9 × 10 14 7.05 × 1023 = 4.14 × 10−10 cgs (Modern value: 4.803 × 10−10) 10.7 (a)
Substitute j = −D∇n, into the continuity equation ∇· j + ∂n/∂t = 0 to obtain
−D∇2n +∂n ∂t = 0
53 (b)
With a drift current produced by a uniform constant external force Fext,
The total particle current is
j= −D∇n +n ηFext
where η is the mobility. Thus the diffusion equation generalized to −D∇2n +1
ηFext· ∇n + ∂n
∂t = 0 (c)
The absorption,contributes a term −V (r)n to the rate of change of the par-ticle density. From this point of view, the Schr¨odinger equation describes a diffusion in imaginary time, with absorption, of the wave function ψ. What makes quantum mechanics distinctive is that ψ is a complex probability ampli-tude, and not a probability.
Chapter 11
11.1
If the showers are distributed at random, the probability that one occurred on Tuesday would be 1/7, and the probability that it did not occur would be 6/7. The probability that none of the 12 showers occur on Tuesday would be (6/7)12= 0.157. Better bring the umbrella.
11.2
If parking tickets were issued at random, the probability of getting 12 tickets on two days of the week would be (2/7)12= 3 × 10−7. This is so small that we
must reject the assumption that tickets were given out at random, and advise the student to use a parking lot on those days. Of course, this assumes that the police maintains the same tactic.
11.3
What determines whether the man goes north or south is the correlation between northbound and southbound trains, as illustrated in the sketch. If he enters the station during the interval x, he goes north. Otherwise he goes south. Since he went north 70% of the time, we conclude x = 0.7.
11.4
Generate a sequence of random number between 0 an 1. Divide the interval (0,1) into say 10 equal bins, and keep a running score of the number of random numbers in each bin, as they are being generated. At the end of the run, plot
a histogram of the numbers in each bin. If the sequence is truly random, the histogram should fit a Poisson distribution..
11.5
The current-voltage characteristic of the device is shown in the accompa-nying sketch.
Let the probability of finding the voltage to have a value between V and V + dV be P (V )dV.
Let the probability of finding the current to have a value between I and I + dI be Q(I)dI.
The current is never negative. So Q(I) = 0 for I < 0. For I > 0, we have Q(I) = P (V )dV dI = P (V ) V0 I + I0 (I > 0) where V = V0ln µ 1 + I I0 ¶
In the range V ≤ 0, we must have Z 0
−∞
dIQ(I) = Z 0
−∞dV P (V ) ≡ α
Thus Q(I) should contain a term αδ(I). The complete result is
Q(I) = ½ P (V ) V0 I+I0 + αδ(I) (I ≥ 0) 0 (I < 0) . 11.6
Let the probability density for y be Q(y). We have Z y −∞ dy0Q(y0) = Z x −∞ dx0P (x0)
where x = √y/b. The integrands on both sides are zero for negative arguments. Thus Z y 0 dy0Q(y0) = Z x 0 dx0x 0 a exp µ −x 02 2a ¶ = exp³− y 2ab ´ − 1 Differentiating this gives
57 Q(y) = 1 2abexp ³ − y 2ab ´ (y ≥ 0) 11.7
Chapter 12
12.1 (a) S(ω) = Z ∞ −∞ dτ eiωτG(τ ) = ν Z ∞ −∞ dτ eiωτ Z ∞ −∞ dtf (t)f (t + τ ) + 2πδ(ω)I2 = ν¯¯fω2¯¯ + 2πδ(ω)I2where I = νR−∞∞ dtf (t)I = νR−∞∞ dtf (t) by Campbell’s theorem (11.29).. (b)
fω=R0∞dteiωτ−λt= (iω − λ)−1.
I = νR0∞dte−λt= ν/λ.
S(ω) = ν ω2+ λ2 +
2πν2
λ2 δ(ω)
The white-noise component is ν/λ2, the first term in the limit λ À ω. 12.2
We can use the result of Prob.12.1(a), with fω=R−∞∞ dteiωtϕ(t) −→ ω→0 R∞ −∞dtϕ(t) ≡ q. I = νR−∞∞ dtϕ(t) = νq. S(ω) −→ω →0νq + 2πν 2q2δ(ω) 12.3 (a)
I(t) and I(t + τ ) are the same if there are an even number of sign changes during τ and equal but opposite if there are an odd number of sign changes. Thus
hI(t)I(t + τ )i = a2(Peven− Podd)
(b)
The probability that there are k crossing in the time interval τ > 0 is given by the Poisson distribution
P (k; ν) = (ντ )
k
k! e
−ντ
The probability there are an even and odd number of crossings are given re-spectively by Peven = e−ντ Ã 1 +(ντ ) 2 2! + (ντ )4 4! + · · · ! Podd = e−ντ Ã ντ +(ντ ) 3 3! + (ντ )5 5! + · · · ! Thus for τ > 0 hI(t)I(t + τ )i = a2e−ντ Ã 1 − ντ +(ντ ) 2 2! − (ντ )3 3! + · · · ! = a2e−2ντ
If τ < 0, then hI(t)I(t + τ)i = hI(t − |τ|)I(t)i = hI(t)I(t + |τ|)i by invariance in time translation. Thus the general answer is obtained by replacing τ by |τ|.
(c) S(ω) = Z ∞ −∞dτ hI(t)I(t + τ)i = a 2Z ∞ 0 dτ eiωτ−ντ+ a2 Z 0 −∞ dτ eiωτ +ντ = 2a2Re Z ∞ 0 dτ eiωτ−ντ = 2νa 2 ω2+ ν2 12.4 From (12.2) W3(3, 1, 2) = Z dx4W4(3, 1, 4, 2) From (12.24) and (12.29) W3(3, 1, 2) = W2(3, 1)P (3, 1|2) = W2(3, 1)P (1|2) W4(3, 1, 4, 2) = W3(3, 1, 4)P (3, 1, 4|2) = W3(3, 1, 4)P (4|2) = W2(3, 1)P (1|4)P (4|2)
Substituting these into the last equaton we obtain W2(3, 1)P (1|2) = Z dx4W2(3, 1)P (1|4)P (4|2) P (1|2) = Z dx4P (1|4)P (4|2)
61 12.5
The problem one faces in a computer simulation of a phase transition is “critical slowing down”. It is easy to obtain a rough value for Tc, but very
difficult to attain precision. This is because on a finite lattice the transition will not be sharp, but increasing the lattice size also increases the time to reach thermal equilibrium. It becomes increasingly difficult for large blocks of spins to flip, since all spins have to flip at the same time, by chance. Indeed, this is why the largest block, namely the whole lattice, does not flip at all, leading to spontaneous magnetication. To speed up the simulation, one has to improve the algorithm by making trial flips of not just single spins, but blocks of spins of random sizes.
12.6 (a)
Since the diffusion equation is invariant under translations in space and time, the solution (10.28) can be generalized to a starting position x0 and starting
time t0by by replacing x, t by x − x0, t − t0, respectively.
(b)
Denote the transition probability from step i to step j by P (i|j) = p 1 4πD(ti− tj) exp à −(xi− xj) 2 4D(ti− tj) !
To begin, show that for n = 2, P (2|0) =
Z
dx1P (2|1)P (1|0)
The right side is X = 1 4πDp(t2− t1) (t1− t0) Z ∞ −∞ dx1exp à −4D1 à (x2− x1)2 t2− t1 − (x1− x0)2 t1− t0 !!
The exponent can be wrtiten as − 1 4D µ x2 2 t2− t1 + x 2 0 t1− t0 ¶ − A 4D µ x21+2B A x1 ¶ where A = 1 t2− t1 + 1 t1− t0 B = x2 t2− t1 + x0 t1− t0
Performing the integral, we obtain X = p 4πD/A 4πDp(t2− t1) (t1− t0) exp à −4D1 à (x2− x1)2 t2− t1 − (x1− x0)2 t1− t0 − B2 A !! = p 1 4πD(t2− t0) exp à −(x2− x0) 2 4D(t2− t0) !
Next show that the result for n − 1 implies that for n, where n > 2. The integrals one has do is similar to the one above. This will complete the proof by induction.
12.7
Chapter 13
Solutions are given in the text.
Chapter 14
14.l
The relativistic energy is E =p(pc)2+ (mc2)2. In the ultra-relativistic
do-main we can neglect the mass term and thus E = pc. The deBroglie wavelength is h/p = hc/E. The thermal wavelength is therefore proportional to hc/kBT .
14.2 14.3 U = 3 2P V ≈ 3 2N kBT ³ 1 ± 2−5/2nλ3´ CV = ∂U ∂T ≈ 3 2N kB ³ 1 ∓ 2−7/2nλ3´ The upper sign is for fermions, lower sign is for bosons. 14.4 (a) N =X λ nλ= z X λ exp (−β λ) = zQ 65
(b)
The internal energy per particle is defined by U N = P λ λexp (−β λ) P λexp (−β λ) Thus, −∂ ln Q ∂β = 1 Q X λ λexp (−β λ) = U (c) Q = X αβγ
exp¡−β¡ transα + rotβ + vibγ ¢¢= QtransQrotQvib
U
N = − ∂
∂β(ln Qtrans+ ln Qrot+ ln Qvib) cV =
1 N
∂U
∂T = ctrans+ crot+ cvib
14.5 Qtrans = V (2π)−34π Z ∞ 0 dkk2exp¡−β~2k2/2m¢= V /λ3 U N = −∂ ln λ −3/2/∂β = 3 2kBT ctrans kB = 3 2 . 14.6 (a) ln Qrot ≈ ln ¡ 1 + exp¡−β~2/I¢¢≈ exp¡−β~2/I¢ U N = − ∂ ln Qrot ∂β = ~2 I exp µ −β~ 2 I ¶ crot kB ≈ 3 µ β~2 I ¶2 exp µ −β~ 2 I ¶ (b) Qrot ≈ Z ∞ 0 d 2 exp¡−β~2 2/2I¢ ∝ .β−1 U/N ≈ kBT
67
crot
kB ≈ 1
(c)
The internal energy rises exponentially from T = 0 to approach a linear behavior. The qualitative behaviors are as shown in the accompanying sketch.
U kT kT C rot rot h /I2 k 14.7 (a) Qvib = ∞ X n=0 exp (−β~ω(n + 1/2)) = eλ/2¡eλ− 1¢−1 U N = − ∂ ln Qvib ∂β = ~ω 2 eλ+ 1 eλ− 1 (λ=β~ω) cvib kB = e−β~ω µ β~ω 1 − e−β~ω ¶2 kT Cvib hω k (b) ¿ n +1 2 À = −∂λ∂ ln Qvib = 1 2 eλ+ 1 eλ− 1 *µ n +1 2 ¶2+ − ¿ n +1 2 À2 = ∂ 2 ∂λ2ln Qvib= eλ (eλ− 1)−2 where λ=β~ω.
14.8
kBTvib ≈ ~ω
kBTrot ≈ ~ 2
I For H2, Tvib= 6100 K, Trot= 85.4 K.
From T = 0, the specific heat rises to 3kB/2 before it reaches Trot, then
increases by kB around T = Trot, and increases by kB again around T = Tvib.
14.9 n = γn+ bσn γn = ~ω µ n +1 2 ¶ σn = ~ω µ n +1 2 ¶2 h i = P n(γn+ bσn) e−β(γn+bσn) P ne−β(γn+bσn)
Expanding this to first order in b, we have h i
~ω ≈ ¯ ~ω + bν
2− bβ~ω³ν3− ν2ν¯´
where ν = ¯n + 1/2, and a bar denotes average with respect to the unperturbed system with b = 0.
−∂ ∂λν
2= ν3− ν2¯ν
From Prob.14.7(b) we have
ν2= e
2λ+ 6eλ+ 1
4 (eλ− 1)2
from which we obtain
ν3− ν2¯ν = − ∂ ∂λν 2=e λ¡3eλ+ 4¢ 2 (eλ− 1)3 where λ=β~ω. Thus h i − ¯ b~ω ≈ e2λ+ 6eλ+ 1 4 (eλ− 1)2 − λeλ¡3eλ+ 4¢ 2 (eλ− 1)3
Chapter 15
15.1
The cross section for a partially polarized beam is σpol= |α|2σ1+ |β|2σ2= Tr (ρσ)
where |α|2+ |β|2= 1. With respect to the present basis (i.e., the spin states 1 and 2) we have ρ = µ |α|2 0 0 |β|2 ¶ σ = µ σ1 0 0 σ2 ¶
The matrix trace is indepedent of the basis. 15.2 (a) Qclassical= Z dpdq τ e −βH = 1 τ Z ∞ −∞ dp exp¡−βp2/2m¢ Z ∞ −∞ dq exp¡−βmω2q2/2¢= 2π βτ ω (b) Qquantum= ∞ X n=0 e−βEn= ∞ X n=0 exp · −β~ω µ n +1 2 ¶¸ = e−β~ω/2 1 − e−β~ω (c) Qquantum−→ β→0 1 β~ω = 2π βhω Comparison with Qclassical gives
τ = h which is Planck’s constant.
15.3
Postulate the form
e−β(K+V )= e−βKe−βVe−12β 2
X
to second order in β. Expand both sides to second order: e−β(K+V )≈ 1 − β (K + V ) +1 2β 2 (K + V )2 e−βKe−βVe−β2X ≈ · 1 − βK +12β2K2 ¸ · 1 − βV +12β2V2 ¸ · 1 − 12β2X ¸ ≈ 1 − β (K + V ) +12β2¡K2+ V2+ 2KV − X¢ ≈ 1 − β (K + V ) +12β2h(K + V )2+ KV − V K − Xi Therefore X = KV − V K = [K, V ] 15.4 (a) QN =¡eβµ0B+ e−βµ0B¢ N (b) hMi = −β1∂B∂ ln QN = µ0N eβµ0B− e−βµ0B eβµ0B+ e−βµ0B (c) M2®− hMi2= 1 β2 ∂2 ∂B2ln QN= 4µ2 0N (eβµ0B+ e−βµ0B)2 15.5 (a) Nb = N z−1e−β + 1 Nf = X k i z−1eβk+ 1 where k = ~2k2/2m. (b) The condition is Nb+ Nf = N , or 1 z−1e−β + 1 + 1 nλ3f3/2(z) = 1
71 where λ =p2π~2/mkT .
(c)
For small z, the condition becomes
zeβ + z/(nλ3) = 1 Thus
z = nλ3¡1 − nλ3eβ ¢ This is valid for nλ3¿ 1.
(d)
For high temperatures β → 0. Thus nλ3¿ 1, and z ¿ 1. From Nf = (V /λ3)f3/2(z), or nfλ3= f3/2(z), we obtain nfλ3 ≈ z = nλ3¡1 − nλ3eβ ¢ nf n ≈ 1 − nλ 3 eβ
For low temperatures we expect most particles to be in one of the bound states, and thus nf/n → 0.For β → ∞, the condition for z becomes
zf3/2(z) = nλ3e−β
This means that z is small, so the condition reduces to z2= nλ3e−β .Thus nf n ≈ 1 √ nλ3e −β /2 15.6 hnki = 1 Q X {n1,n2,··· } nkexp [−β ( 1n1+ 2n2+ · · · ) + βµ] = − 1 β ∂ ∂ kln Q
Differentiating both sides with respect to p with p 6= k, we obtain
−β1∂∂ phn ki = 1 Q X {n1,n2,··· } nknpexp [−β (1n1+ 2n2+ · · · ) + βµ] −1 β µ ∂ ∂ p 1 Q ¶ X {n1,n2,··· } nkexp [−β ( 1n1+ 2n2+ · · · ) + βµ] = hnpnki − hnpi hnki We know that hnki = ¡
z−1eβ k± 1¢−1 does not depend on
p. Thus the above
is zero, or
15.7 hσ2i − hσi2=X k∈G X p∈G [hnknpi − hnki hnpi] = X k∈G h hn2ki − hnki2 i
The last relation follows from the fact that terms with k 6= p do not contribute, as shown in the last problem. By (15.35), hn2
ki − hnki2 = hnki ∓ hnki2. This
Chapter 16
16.1
The fraction of electrons that can excited is of the order of kBT / F. Hence
the effective density is nkBT / F, where n is the electron density. The
mean-free-path iswhere σ is the collision cross section.
λ ≈ F nσkBT
16.2
n = 4.35 × 1027cm−3 F = 24.6 MeV
Av. energy per nucleon = 3
5 F = 14.8 MeV
16.3 (a)
The Fermi wave number kF is given through (2s + 1)V (4π/3)k3F = N .
Thus, kF = [3n/4π(2s + 1)]1/3. pF = ~kF F = q p2 Fc2+ m2c4 (b) U = 2V Z p hp (cp)2+ (mc2)2− mc2i P = 2 Z p pxvx= 2 3 Z p(p · v) whereRp=R|p|<p Fd 3p/h3. (c) 73
For n1/3<< mc/~, particles near the Fermi surface are non-relativistic: p (cp)2+ (mc2)2≈ mc2+ p2 2m We have v = p/m. Hence U ≈ V Z p p2 2m P V ≈ 23V Z p p2 m = 2 3U
For n1/3>> mc/~, particles near Fermi surface are ultra-relativistic:
p (cp)2+ (mc2)2≈ cp We have v =cp/p. Hence U ≈ 2V Z p cp P V ≈ 2 3V Z p cp = 1 3U (d) F = 6 × 10−5 eV. 16.4 (a)
Let p± be the Fermi momenta of the spin-up and spin-down gases. The energy of an atom of up(down) spin is
(H) = p 2 ± 2m∓ µH Thus N±= V h3 4π 3 p 3 ± = 4πV 3h3 (2m) 3/2 [ (H) ± µH]3/2 (b)
For complete polarization, we have N−= 0, hence (H) = µH, and N+=
4πV
3h3 (4mµH) 3/2
The total density is now n = N+/V . The minimum field is
Hmin= µ 3π2 4 ¶2/3 ~2n2/3 µm
75 16.5
(a)
Consider a shell of thickness dr in the gas. Let the pressure differential be dP . The inward force acting on a patch of the shell of area dA is −dAdP . In hydrostatic equilibrium this must equal the gravitational attraction due to the mass at the center. Thus
−P dA = γMρ(r)r−2dAdr dP dr = − γM ρ(r) r2 (b) P = 2 5n F ∝ n 5/3∝ ρ5/3 Thus dρ ρ1/3 = −K dr r2 Assuming ρ(∞) = 0, we have ρ(r) = C0 r3/2 16.6 (a) Nb = N z−1e−β + 1 Nf = X k i z−1eβ k+ 1 where k= ~2k2/2m. (b) The condition is Nb+ Nf = N , or 1 z−1e−β + 1+ 1 nλ3f3/2(z) = 1 where λ =p2π~2/mk BT . (c)
For small z, the condition becomes
zeβ + z/(nλ3) = 1 Thus
z = nλ3¡1 − nλ3eβ ¢ This is valid for nλ3¿ 1.
