NOTES AND SOLUTIONS TO THERMAL PHYSICS BY CHARLES KITTLE AND HERBERT KROEMER ERNEST YEUNG - LOS ANGELES
A BSTRACT. These are notes and solutions to Kittle and Kroemer’s Thermal Physics. The solutions are (almost) complete: I will continuously add to subsections, before the problems in each chapter, my notes that I write down as I read (and continuously reread). I am attempting a manifold formulation of the equilibrium states in the style of Schutz’s Geometrical Methods of Mathematical Physics and will point out how it applies directly to Thermal Physics. Other useful references along this avenue of investigation is provided at the very bottom in the references. Any and all feedback, including negative feedback, is welcomed and you can reach me by email or my wordpress.com blog. You are free to copy, edit, paste, and add onto the pdf and LaTeX files as you like in the spirit of open-source software. You are responsible adults to use these notes and solutions as governed by the Caltech Honor Code: “No member of the Caltech community shall take unfair advantage of any other member of the Caltech community” and follow the Honor Code in spirit.
S ECOND E DITION. Thermal Physics. Charles Kittel. Herbert Kroemer. W. H. Freeman and Company. New York. QC311.5.K52 1980 536’.7 ISBN 0-7167-1088-9
1. S TATES OF A M ODEL S YSTEM 2. E NTROPY AND T EMPERATURE Thermal Equilibrium. EY : 20150821 Based on considering the physical setup of two systems that can only exchange energy between each other, that are in thermal contact, this is a derivation of temperature. U = U1 + U2 is constant total energy of 2 systems 1, 2 in thermal contact multiplicity g(N, U ) of combined system is X g(N, U ) = g1 (N1 , U1 )g2 (N2 , U − U1 ) U1 ≤U
The “differential” of g(N, U ) is dg =
∂g1 ∂U1
g2 dU + g1 N1
∂g2 ∂U2
dU2 = 0 N2
EY : 20150821 This step can be made mathematically sensible by considering the exterior derivative d of g ∈ C ∞ (Σ), where Σ is the manifold of states of the system, with local coordinates N, U , where U happens to be a global coordinate. Then, ∂ ∂ consider a curve in Σ s.t. it has no component in ∂N , ∂N , and this curve is a “null curve” so that the vector field X ∈ X(Σ) 1 generated by this curve is s.t. dg(X) = 0. With −dU1 = dU2 , 1 ∂g2 ∂ ln g1 ∂ ln g2 1 ∂g1 = =⇒ = g1 ∂U1 N1 g2 ∂U2 N2 ∂U1 N1 ∂U2 N2 Define σ(N, U ) := ln g(N, U ) Then =⇒
∂σ1 ∂U1
=
N1
∂σ2 ∂U2
N2
Date: Fall 2008. I am crowdfunding on Tilt/Open to support basic sciences research: ernestyalumni.tilt.com. If you find this pdf (and LaTeX file) valuable and/or helpful, please consider donating to my crowdfunding campaign. There is also a Paypal button there and it’s easy to donate with PayPal, as I had recently to the Memorial Fund for the creator of Python’s matplotlib. But the most important thing is for anyone, anywhere, and at anytime be able to learn thermodynamics and use it for research and application and so I want to keep this as openly public as possible, in the spirit of open-source software. Tilt/Open is an open-source crowdfunding platform that is unique in that it offers open-source tools for building a crowdfunding campaign. Tilt/Open has been used by Microsoft and Dicks Sporting Goods to crowdfund their respective charity causes. 1
Temperature. T1 = T2 - temperatures of 2 systems in thermal equilibrium are equal. ∂σ [?]. T “must be a function of ∂U N 1 ∂σ =⇒ = kB T ∂U N Experimentally, kB = 1.381 × 10−23 J/K = 1.381 × 10−16 ergs/K. Now 1 ∂σ = τ = kB T τ ∂U N Problems. Solution 1. Entropy and temperature. ∂σ and σ(N, U ) ≡ log g(N, U ). Given g(U ) = CU 3N/2 , (a) Recall that τ1 ≡ ∂U N,V σ(N, U ) = log CU 3N/2 = log C +
3N log U 2
∂σ 3N 1 1 3N = = =⇒ U = τ ∂U 2 U τ 2 (b)
∂2σ ∂U 2
N
<0?
∂2σ ∂U 2
=− N
3N 2
1 U2
<0
Solution 2. Paramagnetism.
U (s) = U1 (s1 ) + U2 (s2 ) = −2mB(s1 + s2 ) = −2mBs or s =
U −2mB
i.e. potential energy U (s) = −2s · mB. For |s| N , then 2
g(N, s) ' g(N, 0) exp −2s /N = g(N, 0) exp σ(N, U ) = ln g(N, U ) = σ0 −
−U 2 2(mB)2 N
U2 1 where σ0 = ln g(N, 0) 2m2 B 2 N
∂σ −U 1 1 = = 2 2 τ ∂U N m B N What is the thermal equilibrium value of this N -spin system of fractional magnetization? If U denotes hU i, thermal average energy, we also get the thermal average spin excess. hU i = h−2mBsi = −2mBhsi =⇒ τ =
m2 B 2 N mBN = −U 2hsi
Solution 3. Quantum harmonic oscillator.
(a) Result from Ch. 1: g(N, n) = Let N − 1 → N =⇒ g(N
(N +n−1)! n!(N −1)! . +n)! + 1, n) = (Nn!N ! .
(N + n)! = ln (N + n)! − ln (n!) − ln (N !) n!N ! ≈ (N + n) ln (N + n) − N − n − n ln n + n − N ln N + N = (N + n) ln (N + n) − n ln n − N ln N σ(N + 1, n) ≡ ln g(N + 1, n) = ln
(b) Let U denote total energy n~ω of oscillators. U U = n~ω or n = ~ω σ(N, U ) = (N + At τ ,
1 τ
=
U U U U ) ln (N + ) − ln − N ln N ω ω ω ω
∂σ ∂U N .
1 1 U 1 U 1 = ln (N + ) − ln = ln τ ω ω ω ω ω 2
Nω Nω + 1 or U = U exp (ω/τ ) − 1
Solution 4. The meaning of “never.”
Suppose 1010 monkeys. (a) Hamlet represents one specific ordering of 1015 with 44 possibilities for each character. The probability of hitting 1 1 100000 = upon Hamlet from a given, random sequence is 44 . Given that log10 44 = 1.64345, then 44100000 101.64345 = 44 or 10−1.64345 = 44−1 so then 100000 1 = 10−164345 44 (b) (age of universe) ·
10 keys second
10 keys second
1 hamlet 5 10 characters
= 1018 s
1019 keys · 1010 monkeys = 1029 keys typed out
= 1019 keys = 1024 possible “Hamlets”
From part (a), the probability that a given, random sequence is Hamlet, 10−164345 (1029 characters)(10−164345 ) = 10−164316 −164345 Note, I think that the probability should be (1029 characters) 1015 Hamlet ) = 10−164321 characters (10 5 Since we are considering the number of “Hamlet”, 10 character sequences. 3. B OLTZMANN D ISTRIBUTION AND H ELMHOLTZ F REE E NERGY cf. Example: Energy and heat capacity of a two state system, pp. 62 of Kittel and Kroemer [1]. Kittel and Kroemer introduces the heat capacity very early, specific to this example. Definition 1. heat capacity CV at constant volume is defined as ∂σ (1) CV := τ ∂τ V Recall the thermodynamic identity (which is introduced many equations later): dU = τ dσ − pdV ∈ Ω1 (Σ) where Σ is a manifold of states of all systems. ∂ i.e. no component Consider local coordinates of Σ, (σ, V ). Consider curve c : R → Σ s.t. c generates a vector field c˙ = σ˙ ∂σ c(τ ) ∈ Σ in the V direction. Notice the prescient choice of parameter τ . Now for internal energy U ∈ C ∞ (Σ), taking the exterior derivative d results in
dU =
∂U ∂U dσ + dV ∂σ ∂V
∂ Then applying dU onto vector field σ˙ ∂σ ,
∂ ∂σ
dU
σ˙
∂σ ∂τ
=
∂U σ˙ = στ ˙ +0 ∂σ
Now,
∂U ∂σ
V
=
V
∂U ∂τ
= V
∂σ ∂τ
τ V
Hence, (2)
CV := τ
∂σ ∂τ
= V
∂U ∂τ
V
EY: 20150825 Why do we need differential geometry? It’s because I always wondered why you could do this: ∂σ ∂U ? ? CV := τ = with τ dσ = dU =⇒ τ ∂σ = ∂U ∂τ V ∂τ V and talk of “differentials.” Definition: Reversible process. EY : 20150824 Mathematically, 1-forms are exact. 3
Pressure. Consider coordinates (σ, V ) ∈ Σ of manifold of thermodynamic states Σ. Imagine a reversible compression of a cube system (so imagine dV < 0; cube’s volume get smaller). σ constant, i.e. dσ = 0 (on this curve in Σ) because as particles in cube gets squeezed, less positions particles could sit in, but they get more kinetic energy, more energetic (more momentum squared). Now U = U (σ,V ) ∈ C ∞ (Σ). ∂U =⇒ dU = ∂U ∂σ V dσ + ∂V σ dV ∂ Again, imagine a curve c : R → Σ, connecting 1 state (σ, V ) ∈ Σ to another state (σ, V + dV ) ∈ Σ s.t. c˙ = V˙ ∂V . ∂U =⇒ dU (c) ˙ = V˙ ∂V σ Introduce 1-form W ∈ Ω1 (Σ) of work done on the cube system from some external agent W = −pdV so W > 0 when dV < 0. Then
W (c) ˙ = −pV˙ = dU (c) ˙ =
=⇒ p = −
(3)
∂U ∂V
∂U ∂V
V˙ σ
σ
Consider another set of coordinates (U, V ) ∈ Σ for manifold Σ. Now entropy σ is a function of U, V , as σ = σ(U, V ) ∈ C ∞ (M ), so that ∂σ ∂σ dσ = ∂U dU + ∂V dV V U ∂ ∂ Consider curve c = (U, V ) ∈ Σ. Then c˙ = U˙ ∂U + V˙ ∂V . For this curve c, σ is constant, meaning dσ(c) ˙ = 0 (it’s a “null curve” of dσ ∂σ ∂σ ˙ U+ V˙ dσ(c) ˙ =0= ∂U V ∂V U Now define Definition 2. 1 := τ
(4)
∂σ ∂U
V
So then we have τ1 U˙ + ∂V U V˙ = 0. For the parameter of curve c, choose the parameter to be V , knowing that σ is constant on this curve, or thermodynamic process. Thus 1 ∂U ∂σ ∂σ −p =− =⇒ =− or τ ∂V σ ∂V U τ ∂V U ∂σ
(5)
p=τ
∂σ ∂V
U
Thermodynamic Let σ = σ(U, V ) ∈ C ∞ (Σ). Then Identity. ∂σ ∂σ dσ = ∂U V dU + ∂V U dV ∈ Ω1 (Σ). ∂σ ∂σ Now recall the quantities we’ve recently used: τ1 := ∂U (this is a definition) and τp = ∂V (it comes from the physics, V U of doing work on the system, by some external agent). Then the thermodynamic identity is obtained: Theorem 1. (6)
τ dσ = dU + pdV
Ideal Gas: A First Look. 4
One atom in a box. one atom of mass M in cubical box of volume V = L3 −~2 2 p = ~i ∇ p2 ψ = ψ 2M ∇ ψ = ψ ny πy nx πx =⇒ ψ(x) = A sin L sin L sin nzLπz π 2 ~2 (n2x + n2y + n2z ) n = 2M L Then the partition function Z1 is X X −n −~2 π 2 2 2 2 Z1 = exp (nx + ny + nz ) = exp τ 2M τ L n (nx ,ny ,nz )
Let α2 =
~2 π 2 ~π or α = 2 2M L τ (2M τ )1/2 V 2/d
Then ∞
Z
∞
Z dnx
Z1 = 0
Z dny
0
In general, Z1 = Now
∞
dnz exp [−α
2
(n2x
+
n2y
+
Z
n2z )]
3
∞
=
dnx exp (−α2 n2x )
0
0
3 1/2 3 1/2 3 1 π π = = α 2 2α
d 1/2
π 2α
Z1 =
π 1/2 V 1/d 2 (2M~π τ )1/2
!d
nQ V = nQ V = d/2 2~π n
=
Mτ
in terms of concentration n = 1/V . M τ d/2 nQ := 2π~ is the quantum concentration. 2 Problems. Solution 1. Free energy of a two state system. (a) Z = 1 + e−/τ F = −τ ln Z = −τ ln (1 + e−/τ ) (b) e−/τ ∂(F/τ ) = ∂τ 1 + e−/τ −/τ e ∂F σ=− = ln (1 + e−/τ ) + τ −/τ ∂τ (1 + e ) U = −τ 2
Solution 2. Magnetic susceptibility
(a) Remember to calculate the multiplicity in the N -spin system (it’s not enough to sum up exp (−s /τ ) factors). Us = −M B = −2smB
M = 2sm
N = N+ + N−
2s = N+ − N− = N+ − (N − N+ ) = 2N+ − N N/2
Z=
X N 2smB = exp N+ τ
s=−N/2
= e−
N mB τ
(1 + e
2mB τ
N/2
X N 2
s=−N/2
)N = 2N coshN
mB τ
N! exp + s ! N2 − s !
2mBs τ
=
N X s=0
N! exp s!(N − s)!
2mB τ
N s− = 2
where it was crucial to use (1 + x)N = neglect dropping the s = 0 term.
PN
j=1
N j
xj . Note, in changing the sum index, since N is large, we can
mB mB −m ) sinh τ τ τ2 ∂ mB M = −τ 2 ln Z = N m tanh ∂τ τ 2 ∂M Nm mB χ= = sech2 ∂B τ τ
∂τ Z = 2N (N )(coshN −1
5
(b)
For x ≡
M nm
mB N mB F = −τ ln Z = −τ ln (2 cosh ) = −N τ ln (2 cosh ) τ τ 2 −N τ 4 2 √ 2 = tanh mB = ln . . Now 1 − tanh y = sech y. F = −N τ ln 2 τ 2 1−x 1−x2
m2 N τ Solution 3. Free energy of a harmonic oscillator (a) ∞ X −s~ω0 1 = (1 − e−~ω0 /τ )−1 Z= exp = −~ω 0 /τ τ 1 − e s=0 ~ω0 ~ω0 −~ω0 /τ F = −τ ln Z = τ ln (1 − e ) ' τ ln for 1 τ τ (b) ~ω0 ∂F τ ~ω0 −~ω0 /τ −~ω0 /τ σ=− (−e − ln (1 − e− τ ) = −{ln (1 − e−~ω0 /τ ) + ) } = ~ω /τ −~ω /τ 0 0 ∂τ τ 1−e e −1 Solution 4. Energy fluctuations. X e−s β Z= 1 s P X =β s e−s /τ τ ∂β Z = −e−s β U= s = −∂β ln Z ∂ ∂β ∂ 1 ∂ Z s = =− 2 X ∂τ ∂τ ∂β τ ∂β ∂2 Z = 2 e−s β (c) For
mB τ
1, cosh2
mB τ
→ 1. χ =
β
s
s
∂U ∂ ∂β Z = −β 2 U = β 2 ∂β = β2 ∂τ ∂β Z
(∂β2 Z)Z − (∂β Z)2 Z2
! =β
2
∂β2 Z − Z
∂β Z Z
2 ! =
∂U = h2 i − hi2 ∂τ Solution 5. Overhauser effect. System S in energy eigenstate En = n. P (E) = (1)gR (E) dUR dUtot S Note ∆UR = (α − 1). ∆US = . dU d + d = 1 + (α − 1) = α = d in a specific energy eigenstate; gS (n) = 1 While gR (UR ) = multiplicity of reservoir R with UR energy. Now ∂σR 1 = ∂Es τ and gR (UR ) = exp (σR (UR )) R If dU = (α − 1) = ∆U small compared to U R R. d =⇒ τ 2
dUR = UR (n) + (α − 1) d 1 σR (UR ((n + 1))) ' σR (UR (n)) + (α − 1) τ exp (σR (UR (n)) + τ1 (α − 1)) P (ES = (n + 1)) = = exp − (1 − α) P (ES = n) exp (σR (UR (n))) τ Solution 6. Rotation of diatomic molecules. (a) (j) = j(j + 1)0 . g(j) = 2j + 1 Remember that Z is a sum over all states, not over all levels. ∞ ∞ ∞ 2 X X −τ d −(j 2 +j)0 /τ −τ X d −0 /τ j +j −j(j+1)0 /τ (e ) = e Z= (2j + 1)e = dj 0 0 j=0 dj j=0 j=0 UR (ES = (n + 1)) = UR (n) +
(b) For 1
0 τ
τ ZR (τ ) = − 0
Z 0
∞
d −0 /τ x e dx
2
+x
dx = − 6
τ −0 /τ x2 +x τ (e ) − (e−0 /τ )0 = 0 0
(c) For
τ 0
1 ZR (τ ) = 1 + 3e−20 /τ
(d) 0 τ 2 U =τ for 1 U = τ ∂τ ln =τ τ 0 60 e−20 /τ τ 1 20 −20 /τ = for 1, U = τ 2 (3e ) 2 −2 /τ 0 τ 1 + 3e 0 1 + 3e−20 /τ ∂U 0 CV = = 1 when 1 ∂τ V τ −2 /τ ! 20 −20 /τ −20 /τ 0 −20 /τ (1 + 3e ) − (3e ) e 0 e−20 /τ 2 1 2 2 2e τ τ = 120 τ2 (1 + 3e−20 /τ )2 (1 + 3e−20 /τ )2 2∂
CV = 60
For very small
τ 0
ln Z ∂τ
1, CV ≈ 12
e−20 /τ (τ /0 )2
(e) See sketch. Solution 7. Zipper problem.
(a) N links. s = 0 closed, open. Z=
N X
exp (−s/τ ) =
s=0
1 − e−(N +1)/τ 1 − e−/τ
(b) 1 τ . PN Z = s=0 exp (−sB) PN −1 −1 s=0 (−s)e−sβ ∂β ln Z = = hsi = Z 1 − e−/τ −e−(N +1)/τ (−(N + 1))(1 − e−/τ ) − (−e−/τ )(−)(1 − e−(N +1)//τ ) = = 1 − e−(N +1)/τ (1 − e−/τ )2 =
e−/τ (e−N β (N + 1)(1 − e−β ) − 1) e−/τ (e−N β (N + 1)(1 − e−β ) − (1 − e−(N +1)/τ )) ' (1 − e−β ) (1 − e−(N +1)/τ )(1 − e−/τ )
This still does not give the desired approximation. Consider the following: eβ − e−N β 1 − e−(N +1)β = 1 − e−β eβ − 1 β −N β ((e + N βe )(eβ − 1) − (eβ )(eβ − e−N β )) (eβ (eβ + N e−N β − eβ + e−N β ) − eβ − N e−N β ) ∂β Z = = (eβ − 1)2 (eβ − 1)2 Z=
(eβ (N + 1)e−N β − eβ − N e−N β ) (eβ (N + 1)e−N β − e+β − N e−N β ) ∂β Z = ' = β β −N β Z (e − 1)(e − e ) e2β (N e−(N −1)β + e−(N −1)β − eβ − N e−N β ) ((N + 1)e−(N −1)β − (eβ + N e−N β )) = = 2β e2β β βe β N β β β −N e (N +1)e N e +eeN−e (Ne+1)e − eNnβ − eβ − eβ N β β eN β N − eN β = = = = ' e2β e2β e2β eN β eβ
=
'
(−eN β ) = −e−β eN β eβ
=⇒ hsi = e−/τ n πy Solution 8. Quantum concentration. Now Ψ(x, y, z) = A sin nxLπx sin yL sin Ground orbital: nx = ny = nz = 1. 3 π 2 3 π 2 T = hψ0 |ψ0 i = 2m L 2m L 7
nz πz L
. p = 1i ∇,
p2 2m
1 = − 2m ∇2 .
where hψ0 |ψ0 i = 1, ψ normalized. It was normalized in this way: Z 0
∞
Z L 2nx πx n πx nx − sin 1 − cos x L dnx = sin2 dnx = L 2 0
2nx πx L
L 2nx π
2
L = L 2 0
3 L A2 L3 8 2 hψ|ψi = A = = 1 or A2 = 3 2 8 L mτ 3/2 Recall that nQ = 2π~ . 2 Consider the condition that there will be a concentration for which the zero-point quantum kinetic energy is equal to the temperature τ : 3/2 3 2 2/3 2mτ 3 π2 2mτ 2/3 = or n = =⇒ π n = τ or n = 2m L2 2m 3π 2 ~2 3π 2 ~2 3/2 4 4 mτ 3/2 =⇒ n = ( ) = nQ 2 3π 2π~ 3π Solution 9.
Partition function for two systems. X X −E1+2 −E1 −E2 g(E1+2 ) exp Z(1 + 2) = g(E1 + E2 ) exp = exp = τ τ τ E1+2 E1 +E2 =E0 XX −E2 −E1 = g(E1 )g(E2 ) exp exp = Z(1)Z(2) τ τ E1 E2
since systems are independent. Solution 10. Elasticity of polymers. (a) Consider 2s = N+ − N− ; N = N+ + N− , 2s = N+ − (N − N+ ) = 2N+ − N . . For 2s, consider −2s = N+ − (N − N+ ) = 2N+ − N . N+ = −2s+N 2 =⇒ g(N, −s) + g(N, s) =
N 2
N+ =
2s+N 2 .
2N ! + s ! N2 − s !
(b) |s| N l −l 2(N !) = σ(l) = ln g N, + g N, = ln l −l N N 2ρ 2ρ + ! + 2 2ρ 2 2ρ ! L N l N l N l N l N l N + ln + − + + − ln − − − } = ln (2N !) = { 2 2ρ 2 2ρ 2 2ρ 2 2ρ 2 2ρ 2 2ρ where we used ln(x + ∆x) ' ln x + x1 ∆x. l N 1 l N N l N 1 −l N + ln + − + − ln + }= σ(l) = ln (2N !) − { 2 2ρ 2 N/2 2ρ 2 2 2ρ 2 N/2 2ρ 2 N N N N N N l N l 2 l −l l N l2 = ln (2N !) − { ln − + ln − + ln + + + − ln + } 2 2 2 2 2 2 2ρ 2 2ρ N 2ρ 2ρ 2ρ 2 N 2ρ2 ! 2N ! l2 σ(l) = ln − N N N ρ2 2 ! 2 ! (c) ∂σ −2l = ∂l N ρ2
=⇒ f =
2lτ N ρ2
Solution 11. One-dimensional gas.
~2 π 2 2 n = n in one dimension 2m L √ Z ∞ ∞ X −~2 π 2 2 π −α2 n2 Z1 = exp n /τ =⇒ dne = 2m L 2α 0 n=1 8
~2 π 2 2mτ L . where ~π/L α= √ 2mτ 1/2 Recall that σ = ∂F ∂τ , F = U − τ σ, so that α2 =
√
π 2α 2~π/L F = −τ ln Z = −τ N ln = τ N ln √ = τ N ln √ √ = 2α π π 2mτ 1/2 ! √ 1 2π~ 2π~2 = τ N ln = τ N ln √ 2 mL2 τ mLτ 1/2 ! 1 2π~2 τN 1 −2π~2 ∂F = = N ln + 2π~2 ∂τ 2 mL2 τ 2 mL2 τ 2 mL2 τ 1 2π~2 −τ N N 2π~2 = N ln + = ln − 1 2 mL2 τ 2τ 2 mL2 τ 4. T HERMAL R ADIATION AND P LANCK D ISTRIBUTION Problems. Solution 1. Number of thermal photons. We consider a cavity of volume V , and of edge length L (so V L3 ). So then ωn = nπc/L. 1 is the thermal average number of photons in a single mode frequency ω. So then Now exp ~ω ( τ n )−1 X X 1 hsn i = ~ωn exp τ − 1 Consider (nx , ny , nz ) on positive octant, and 2 independent polarization s of em field. Z ∞ Z X (2) ∞ n2 dn 1 1 2 = = π = 4πn dn 8 0 −1 −1 −1 exp ~nπc exp ~nπc exp ~nπc 0 Lτ Lτ Lτ n 3 Z ∞ 2 Lτ x dx V τ 3 =π = N= 2 (2.404) x ~πc e −1 π ~c 0 where I used the substitutions ~πcn Lτ Lτ x =n ~πc Lτ = dn (dx) ~πc x=
Now σ(τ ) = (4π 2 V /45)(τ /~c)3 , so then
σ 1 = N 2.404
4π 4 45
' 3.602
R∞ 2 dx Now how was 0 dx exx −1 evaluated? Solution 2. Surface temperature of a Sun. Given the solar constant of the Earth, the total radiant energy flux density at the Earth from the Sun normal to the incident rays, integrated over all emission wavelengths, solar constant = 0.136 J s−1 cm−2 ,
(49)
(a) 4π(1.49 × 10
11
2
−1
m) · 0.136 J s
cm
−2
102 cm 1m
2
= (4π)(1.49)2 1022 · 0.136 × 104 = 3.8 × 1026 J · s−1
Note that I had used 1.49 × 1011 m as the distance of the Earth from the Sun. (b) Jν = energy flux density or rate of energy emission per unit area. π 2 k4 σB = 60~3Bc2 = 5.670 × 10−8 W m−2 K −4 . Note that W = 1sJ . I will use R◦ = 6.9599 × 1010 cm as the radius of the Sun. 4 × 1026 J · s−1 = J ν = σB T 4 4π(6.9599 × 1010 cm)2 9
2
102 cm 1m
2
4 × 1026 J · s−1 1m 4 4 k = T 10 2 4π(6.9599 × 10 cm) 5.670 × 10−8 J/s T ' 5830 K Solution 3. Average temperature of the interior of the Sun.
(a) R
Z U =−
G
4 3 3 πρr
0
−3GM 2 U= 5R Z 1 RG U =− 2 0
(4πr2 ρdr) 16 = − π 2 Gρ2 r 3
4 3 3 πρr
Z
R
r4 dr =
0
−16 2 2 4 π ρ GR5 ; M = πR3 ρ 15 3
Z −8 2 2 −8 2 2 R 4 r dr = (4πr ρdr) = π Gρ π ρ GR5 = r 3 15 0 2 −8 2 −3 GM 2 M = = π GR5 4 3 = 1.14 × 1041 J 15 10 R πR 3 (b) Using the virial theorem of mechanics, note that 1 kg m2 −11 33 kg · sm2 · kg g 10 2 · 2 × 10 3g −1 3 GM 2 3 6.67 × 10 U= = = 5.72 × 1040 J 2 20 R 20 7 × 1010 cm 1012mcm
2
1 Now hsi = exp (~ω/τ )−1 , is the Planck distribution function, giving the thermal average number of photons in a single mode frequency ω. ~ω thermal average energy hi = hsi~ω = exp (~ω/τ )−1 for τ ~ω, hi ' τ 40 So then 5.72 × 10 J = N hi = N τ .
τ=
5.72 × 1040 J = 4.14 × 106 K 1 × 1057 (1.381 × 10−23 J/K)
Solution 4. Age of the Sun.
(a) Consider 4H →42 He. Then 4(1.0078) − 4.0026 = 0.0286 amu. Then 1.6726 × 10−27 kg (3 × 108 m/s)2 = 4.27 × 10−12 J (0.0286 amu) 1.00727647 u Given M = 2 × 103 3 g, 1 1.00727647 u 30 (2 × 10 kg)(0.10) (4.27 × 10−12 J) = 1.28 × 1044 J 4 × (1.0078 amu) 1.6726 × 10−27 kg So 1.28 × 1044 J energy is available. (b) 1.15 × 1045 J 4 × 1026 J · s−1
1 hr 3600 s
1 day 24 hr
1 yr 365 days
= 1.02 × 1010 years
4 Solution 5. Surface temperature of the Earth. JS = σb T is the radiant power per unit area. 2 Total emitted radiation energy of the sun is JS 4πR . 2 4πR JS 2 4πRES
R2
= R2 JS = radiation energy hitting 1 cm2 of Earth’s surface in one second ES Since the Earth is considered a black-body, the rate of absorption must equal the rate of emission: r 2 R 7 × 1010 cm 4 4 4 4 2 σb T = σb Te or Te = T R =⇒ Te = 5800 K = 396.2 K = 123 C 2 RES 1.5 × 1013 cm Solution 6. Pressure of thermal radiation.
(a) sj = number of photons in that mode. Suppose modes of ωj , j = 0, 1, 2, . . . . j = sj ~ωj = total energy in jth mode, sj photons in jth mode. U=
X j
j =
X
sj ~ωj
P =−
j 10
X ∂U ∂ωj =− sj ~ ∂V ∂V j
(b) ωj = jπc/L, V = L3 . So then ωj = jπcV −1/3 . =⇒
dωj −1 −1 ωj = jπcV −4/3 = dV 3 3 V
U (c) p = 3V (d) We want the Kinetic pressure at a concentration of (1 mol/cm3 ). Recalling P = N kVB T , 2 3 6.022 × 1023 10 cm N 1 mol −23 J 1.381 × 10 (2 × 107 K) = 1.663 × 1014 2 P = cm3 1 mol 1m K m
Now for the thermal radiation pressure, p=
U 1 π2 N τ 4 = 4.03 × 1013 2 = 3V 3 15~3 c3 m
where t = 2 × 107 K. For the pressures to be equal, 1 π2 N kB 45(~c)3 N kB 4 4 kB T = T or T 3 = so that T = 3.2 × 107 K 4V 3 3 3 15~ c V π 2 kB Free energy of a photon gas P∞ Q 1 1 Consider Z = s=0 e−s~ω/τ = 1−e−~ω/τ for a single mode. (a) Z = n 1−e−~ω n /τ P nπc −~ωn /τ (b) F = −τ ln Z = τ n ln (1 − e ) ωn = L . Z ∞ Z ∞ X 4πn2 dn (2) ln (1 − e−~nπc/τ L ) = πτ dnn2 ln (1 − e−~nπc/τ L ) = F =τ ln (1 − e−~ωn /τ ) = τ 8 0 0 n Z ∞ Z ∞ 3 −~nπc/τ L n e −~πc n3 ~π 2 c ∞ = πτ (n3 ln (1 − e−~nπc/τ L )) − dn dn ~nπ/τ L = = − −~nπc/τ L τL L 0 1−e e −1 0 0 3 Z ∞ x3 (τ L)3 π 2 τL = − =− ~πc ex − 1 (~2 c3 ) 45 0
Solution 7.
where I used x = ~nπc τL . Solution 8. Heat shields. For Ju = σB (Tu4 − Tl4 ), the thermal flux without the heat shield, in the middle region. Plane m 4 absorbs σB Tu4 + σB Tl4 and emits Jm = σB (Tu4 + Tl4 ) = σTm . 4 4 1/4 Tm = (Tu + Tl ) . The key point is that, by symmetry, plane m emits J2m flux on each side. 4 4 Tu + Tl4 Ju Tu − Tl4 4 Jnet = σB Tu − (σB ) = σB = 2 2 2 Jnet is the same for the other side of the heat shield: Jnet =
−σB Tl4
+ σB
Tu4 + Tl4 2
= σB
Tu4 − Tl4 2
Solution 9. Photon gas in one dimension. E = E0 sin (kx) cos (ωt) is the form of a solution with kL = nπ or k =
since
nπ L ,
nπ ω = v, ωn = v k L 1 = 1 − e−~ω/τ
v 2 Ex x = Et t → v 2 k 2 = ω 2 or Zj =
∞ X
e−s~ω/τ
s=0
where Zj is the partition function for a particular mode frequency ω. P∞ n − s~ω τ d −~ωn −1 −~ω −2 −~ωn −1 −1 s=0 se =Z (1 − exp ) = Z (1 − exp ) exp = hsi = Z d(−~ωn /τ ) τ τ τ exp (−~ωn /τ ) = 1 − exp (−~ωn /τ ) So ~vnπ/L exp ~vnπ −1 exp τ − 1 Lτ n n 2 X X ~vπ n2 exp ~vnπ X ~vπ 2 n2 exp ~vπn ∂U ~vnπ −~vnπ −~vnπ/L Lτ Lτ Lτ Lτ = = = 2 exp 2 ∂τ Lτ Lτ 2 (exp ~nvπ − 1)2 (exp ~vπ exp ~vnπ −1 Lτ Lτ n − 1) n n n Lτ hsi~ωn = hn i =
~ωn
U=
~ωn
11
X
hn i =
X
R∞ P Now n → 0 dn for one-dimensional photon. Let α = ~vπ Lτ . Letting x = αn, Z Z ∞ nvπ 2 Z n2 exp (αn) 1 x2 ex 1 ∂U −x2 xdx dn αn = dx x = { x = + } x−1 ∂τ Lτ (e − 1)2 α (e − 1)2 α (e − 1) e 0 R∞ R ∞ (kx) (kx)3 Coefficient of k term of f (k) is j = 0 exx−1 dx. Now f (k) = 0 sin ex −1 dx, sin (kx) = kx − 3! + . . . , so that Z ∞ x π2 = dx x−1 6 e 0 ∂U Lτ π 2 Lτ π = = = CV ∂τ ~vπ 6 6~v Solution 10. Heat capacity of intergalactic space. Given the density 1 atom m−3 , considering thermal radiation at 2.9 K, then kB T = (1.381 × 10−23 J/K)(2.9 K), ~c = (1.05457 × 10−34 J · s)(3 × 108 m s ). 2 4π 2 3 Recall for radiation, that the energy per unit volume: VU = 15~π3 c3 τ 4 so that ∂U ∂τ = 15~3 c3 τ V . dU 3 3 Assume hydrogen atoms modeled as ideal gas: U = 2 N τ , dτ = 2 N . CVmatter = CVradiation
3 2N 4π 2 15~3 c3
τ 3V
=
45(~c)3 (N/V ) = 2.8 × 10−10 8π 2 (kB T )3
Solution 11. Heat capacity of solids in high temperature limit.
exp
~ωn τ
−1'1+
~ωn + τ
~ωn τ
~ωn τ
=
π~vn Lτ ;
ωn =
πvn L .
For τ ~ωn , 0 ≤ n ≤ nD .
2 3 1 1 ~ωn + + ··· − 1 2 6 τ
By doing long division ~ωn exp
~ωn τ
−1
~ωn
' ~ωn τ
1+
~ωn 2τ
+
)2
(~ωn 6τ 2
=
τ 1+
~ωn 2τ
+
(~ωn )2 6τ 2
=τ +−
~ωn (~ωn )2 + + ... 2 12τ
For np = (6N/π)1/3 Z Z 3π nD ~ωn (~ωn )2 3π nD ~ πvn ~2 π 2 v 2 n2 U= dnn2 (τ − + )= dnn2 (τ − + )= 2 0 2 12τ 2 0 2 L 12τ L2 4/3 ~2 π 2 v 2 1 5 3π 2 ~v 6N ~πv n4D π 6N 3~2 π 3 v 2 (6N/π)5/3 3π 1 3 { nD τ − + n τ − } = + = 2 3 2L 4 12τ L2 5 D 2 π 16L π 120τ L2 So 3π 2 ~v U = 3N τ − 16L Now T = θ, θ = So then
~v kB
6N π
4/3 +
3~2 π 3 V 2 (6N/π)5/3 120τ L2
(6π 2 N )1/3 L
~v(6π 2 N )1/3 3π 2 ~v 61/3 N 4/3 6 15 (6π 2 N )1/3 ~v 15 − = = N k θ B 4/3 L 16L 8 L 8 π U 15 15 6.022 × 1023 particles = N kB = (1.381 × 10−23 J/K) = 15.59 θ 8 8 1 mol which is very close to experimental values. π2 V 4π 2 V 4 3 Solution 12. Heat capacity of photons and phonons. For a photon: U = 15~ ∂τ U = 15~ 3 c3 τ 3 c3 τ . 3π 4 N τ 4 ∂U 12π 4 N τ 3 phonon: U (τ ) = 5(kB θ)3 . ∂τ = 5(kB θ)3 So then 3 12π 4 1022 1 CV = = 2.3 × 1018 5 100 for a phonon. For a photon, 4π 2 (1.381 × 10−23 J/K)3 τ 3 = 220 /K 3 τ 3 15 (1.05457 × 10−34 J · s(3 × 1010 cm/s))3 U = 3N
12
Temperature at which to photon contribution equals to phonon contribution: (220/K 3 )τ 3 = 2.3 × 1018 =⇒ τ = 2.2 × 105 K Solution 13. Energy fluctuations in a solid at low temperatures.
τ2 Recall that
∂U ∂τ
h( − hi)2 i 12π 4 N τ 5 = 2 hi 5(kB θ)3
∂U ∂τ
= h( − hi)2 i V
3π 4 N τ 4 12 π 4 N τ 3 where U = 3 5 (kB θ) 5(kB θ)3 V ! 3 20 1 1 0.068 θ 1 1 3 = (k θ) = B 9π 8 N 2 τ 8 3 π4 N τ3 N T 25(kB θ)6 s 3 0.070 200 = 0.02 F= 1015 10−2
=
Solution 14. Heat capacity of liquid 4 He at low temperatures.
(a) Given v = 2.383 × 104 cm s−1 and accounting for only longitudinal waves (only longitudinal polarization), then the Debye temperature is !!1/3 1/3 1 He ~v (1.05457 × 10−34 J · s)(2.383 × 104 cm/s) 18π 2 0.145 g 1.00727647 u 4.0026u 18π 2 N θ= = = kB V 1.381 × 10−23 J/K cm3 1.67262 × 10−24 g = 28.6 K (b) Recall the derivation for U for phonons in a solid. Account for only longitudinal waves (only longitudinal polarization). Z Z Z ~ πnv π nD 2 ~π 2 v nD n3 π nD n2 ~ωn L = = dn U= dn n dn 2 0 exp (~ωn /τ ) − 1 2 0 exp (x) − 1 2L 0 exp (x) − 1 ~πnv Lτ With ωn = πn L v, x = Lτ or ~πv x = n, then 2 4 Z nD ~ v Lτ x3 U= dx x 2L ~πv e −1 0 18N 1/3 1/3 2/3 1/3 Dv For low temperatures, τ small so take xD = ~πn = ~πv = 18 ~πτ n v to go to ∞. Lτ Lτ π 4 τ π2 V U→ 3 2(~v) 15 2 2 π 3 4 Recall that CV = ∂U ∂τ V . Then CV /V = 15 (~v)3 τ . Recall τB = kB T , and given v = 2.383 × 10 cm/s, then kB (1.381 × 10−23 J/K) = = 5.495 × 106 (1/K · cm) ~v (1.05457266 × 10−34 J · s)(2.383 × 104 cm/s) So if we take CV /V and divide by the given density ρ = 0.145 g/cm3 to get the heat capacity per gram, (and multiply by kB , the Boltzmann constant to get the correct units; Kittel and Kroemer likes using dimensionless formulas ) then 3 2 kB cm3 = 0.0208 × T 3 (CV /V )/ρ = (kB ) π 2 T3 15 ~v 0.145 g Solution 15. Angular distribution of radiant energy flux.
(a) Recall uω =
ω3
~ π 2 c2
exp ~ω −1 τ is the radiation energy per unit volume per unit frequency range. cuω = energy per unit time, per cross sectional area per unit frequency range. em waves emitted spherically from pt. Q. Suppose em wave comes in at a funny angle other than directly inward. Consider area da that’s from the spherical wave from pt. Q. How much of that goes into solid angle dΩ? =⇒ cuω cos θdA 13
So cuω cos θ is the energy per unit time, per cross-sectional area, per unit frequency range, that enters into some solid angle dΩ. r2 dΩ dΩ = 2 4πr 4π is the fraction of the spectral density that if arrives in solid angle dΩ. dΩ 4π is the spectral density of radiant energy flux that arrives in solid angle dΩ. =⇒ cuω cos θ
(b) Z
Z cos θ sin θdθdϕ = 2π
sin 2θ dθ = −π 2 =⇒
π/2 −1 − 1 cos 2θ = −π =π 2 2 0
cuω 4
Solution 17. Entropy and occupancy. ∞ X
1 1 − e−~ω/τ s=0 ∞ X s~ω ~ω ~ω −~ω/τ −2 −~ω/τ ∂τ Z = e−s~ω/τ = −(1 − e ) (−e ) = Z 2 hsi 2 2 τ τ τ s=0 Z=
e−s~ω/τ =
Then e−~ω/τ 1 hs + 1i = −~ω/τ 1−e 1 − e−~ω/τ ~ω ~ω ∂τ Z = ln hs + 1i + τ 2 hsi = ln hs + 1i + hsi σ = ∂τ (τ ln Z) = ln Z + τ Z τ τ hsi =
Now (nyn´ı)
hs + 1i hsi
hsi ln
= Ze−~ω/τ ln e~ω/τ =
~ω hsi τ
=⇒ σ = hs + 1i ln hs + 1i + hsi ln hsi Solution 18. Isentropic expansion of photon gas. 1/3
(a) τi Vi
1/3
= τf Vf
or Vi Vf
1/3
∆r r
=
r = r(t) = αt so that ∆r = α∆t or
=
τf 2.9 K = 10−3 = τi 3000 K
∆t t
rf − ri ri tf − ti ti =1− = =1− rf rf tf tf Knwowing that (b) Now
ri rf
= 10−3 , then
ti tf
= 10−3 . σ = τ V 1/3 σ 3 =V τ
For constant entropy expansion. 4 U π2 π2 σ π2 σ4 4 = τ U = V = 3 3 3 3 4/3 V 15~ c 15(~c) 15(~c) V 1/3 V ∂U π2 −1 −4/3 4 = σ V ∂V σ 15(~c)3 3 ! ! Z 4/3 Vf π2 π2 σ4 1 1 π 2 Vi τi4 1 τf π 2 Vi τi3 4 −1/3 σ (−V ) = (τi −τf ) W = pdV = − 1/3 = − = 3 3 3 1/3 1/3 1/3 15(~c) 15(~c) 15(~c) 15(~c)3 Vi Vi Vf Vi τi V i 14
Exercise 19. Reflective heat shield and Kirchhoff’s law. For a left plane sheet at τu temperature, right plane sheet at τl temperature
+Ju = σb τu4
−Jl = σb τl4
reflection − rJu = −(1 − a)σb τu4
reflection rJl = (1 − a)σb τl4
absorb (left) aJu = aσb τu4
absorb (left) aJl = aσb τl4
total absorption: a(Ju + Jl ) = aσb (τu4 + τl4 ) total emission: aσb (τu4 + τl4 ). By symmetry, a(Ju2+Jl ) emitted to the left, and to right. e emissivity, where emissivity is defined so radiation flux emitted by the object is e times the flux emitted by a blackbody at the same temperature. By Kirchhoff law, for equilibrium, a = e; object must emit at same rate as it absorbs. −a(Ju + Jl ) −(1 − r)(Ju + Jl ) (1 − r)Ju (1 − r)Jl (Ju − Jl ) − rJu + Ju = + (1 − r)Ju = − = (1 − r) = 2 2 2 2 2 a(Ju + Jl ) (1 − r)(Ju − Jl ) σb (τu4 − τl4 ) = + rJl − Jl = = (1 − r) 2 2 2
Jnet =
5. C HEMICAL P OTENTIAL AND G IBBS D ISTRIBUTION Solution 1. Centrifuge. T = 12 mr˙ 2 + 12 mr2 ω 2 . Consider µext = − 12 M r2 ω 2 (negative so for bigger r away from r = 0 axis, lower chemical potential µ, so to show “centrifugal force” outwards. µtot = τ ln (n/nQ ) − 12 M r2 ω 2 . µtot (r) = µtot (0) for diffusion equilibrium. n n(0) 1 n(r) 1 τ ln − M r2 ω 2 = τ ln =⇒ τ ln = M r2 ω2 nQ 2 nQ n(0) 2 n(r) M r2 ω2 M r2 ω2 = exp or n(r) = n(0) exp n(0) 2τ 2τ Solution 2. Molecules in the Earth’s atmosphere.
M τ 3/2 Recall that for ideal gas, F = −τ [N ln Z1 − ln N !]; Z1 = nQ V = 2π~ V. 2 n ∂F d = −τ [log Z1 − ln N !] = τ ln µ= ∂N τ,V dN nQ 3/2 Mτ where nQ = 2π~ 2 . R Now µ = τ log nnQ µext =
−GMe r
=−
2 gRm r
since g =
GMe Re2 .
µtot = τ ln (n/nQ ) + M gh In equilibrium, this must be independent of r: µtot (r) = µtot (R). M gR2 gR2 M = τ ln (n(R)/nQ ) − τ ln (n(r)/nQ) − R 2 r n(r) R M g R2 n(r) τ ln = Mg − R or exp −R = n(R) r τ r n(R) M g R2 1 =⇒ n(r) = n(R) exp − τ r R so that N = 4πn(R) exp
−M gR τ
Z
∞
R
r2 dr exp
M gR2 τr
Solution 6. Gibbs sum for a two level system. Recall that
Z(µ, τ ) =
∞ X X N =0 s(N )
exp [(N µ − s(N ) )/τ ] =
X ASN 15
exp [(N µ − s(N ) )/τ ];
λ = exp
µ τ
− τ , 1 for N = 0; for N = 0(1)+(1)(λ+λ exp ( − τ )) = λ(1+expZ(−/τ )) hN i = Z exp (−/τ ) hN ()i = 0(1)+(0)(λ)+λ = λ exp Z(−/τ ) Z λ exp (−/τ ) hi = hN ()i = Z
(a) Z = 1 + λ + λ exp (b) (c) (d) (e)
− τ
1, λ + λ exp
. λ for = 0.
Z = 1 + λ + λ exp (−/τ ) + λ2 exp (−/τ ) = (1 + λ)(1 + λ exp (−/τ ) where we considered the possibility the orbital at 0 and at are each occupied by one particle at the same time. So that, for total energy being , exp [(2µ − )/τ ] = (exp µτ )2 e−/τ = λ2 e−/τ . Solution 7. States of positive and negative ionization. −∆
∆
δ
−δ
Z = e 2τ + λe 2τ + λe 2τ + λ2 e 2τ −δ
−∆
∂λ Z = 0 + e∆/2τ + e 2τ + 2λe 2τ λ∂∂ ln Z =
λ(e∆/2τ + e−∆/2τ + 2λe−δ/2τ ) δ
−∆
∆
−δ
e 2τ + λe 2τ + λe 2τ + λ2 e 2τ
=1
λe∆/2τ + λe−∆/2τ + 2λ2 e−δ/2τ = eδ/2τ = λe∆/2τ + λe−∆/2τ + λ2 e−δ/2τ
=⇒
λ2 e−δ/2τ = eδ/2τ λ2 = eδ/τ or 2 ln λ = δ/τ
2τ ln λ = δ Solution 8. Carbon monoxide poisoning.
(a) Z = 1 + λ(O2 )e−A /τ λ(O2 )e−A τ P (O2 ) = = 0.9 or 0.9 = 0.1λ(O2 ) exp 1 + λ(O2 )e−A /τ =⇒ ln A = (kB T ) ln
λ(O2 ) 9
9 λ(O2 )
=
−A ; or τ ln τ
= (8.617 × 10
−5
λ(O2 ) 9
−A τ
= A
eV )(273 + 39) ln k
10−5 9
= −0.3686 eV
(b) Z = 1 + λ(O2 )e−A /τ + λ(CO)e−B /τ
P (O2 ) =
λ(O2 )e−A /τ = 0.1 =⇒ 0.1 + 0.1λ(CO)e−B /τ = 0.9λ(O2 )e−A /τ 1 + λ(O2 )e−A /τ + λ(CO)e−B /τ
ln
9λ(O2 )e−A /τ − 1 λ(CO)
−B = or B = τ ln τ
λ(CO) 9λ(O2 )e−A /τ − 1
= −0.5511 eV
Solution 9. Absorption of O2 in a magnetic field. Recall that 2j + 1 = total number of spin states.
2(1) + 1 = 3. Now U = −m · B. −+µB B τ
A
−(A +µB B)
τ + λ(O2 )e− τ + λ(O2 )e = µB B Z = 1 + λ(O2 )e−A /τ (1 + 2 cosh ) τ λ(O2 )e−A /τ (1 + 2 cosh µBτ B ) µB B −A /τ 0.91 = or 0.91 = 0.09λ(O2 )e (1 + 2 cosh ) τ 1 + λ(O2 )e−A /τ (1 + 2 cosh µBτ B ) 1 91 1 µB B τ 1 91 1 eA /τ − 1 = cosh or B = arccosh eA /τ − 1 2 9 λ(O2 ) τ µB 2 9 λ(O2 )
Z = 1 + λ(O2 )e
The Gibbs sum in the limit of zero magnetic field will differ from that of Problem 8 because there the spin multiplicity of the bound state was neglected. 16
Z = 1 + 3λ(O2 )e−A /τ 3λ(O2 )e−A /τ = 0.9 or 0.9 = 0.3λ(O2 )e−A /τ 1 + 3λ(O2 )e−A /τ 3 λ(O2 ) ln = −A /τ or A = τ ln = −0.6227 λ(O2 ) 3 for T = 300 K, so that τ = 0.049375 eV . P (O2 ) =
τ 0.049375 · 0.59927 0.59927 = = 511.1 T µB 5.7884 × 10−11 × 106 eV /T Solution 10. Concentration fluctuations. P∞ P (a) Recall that Z = N =0 s(N ) exp [(N µ − s(N ) )/τ ]. X N 2 τ 2 ∂2Z ∂2Z = exp [(N µ − s(N ) )/τ ] hN 2 i = 2 ∂µ τ Z ∂µ2 =⇒ B =
ASN
(b) ∂ ∂ hN i = ∂µ ∂µ
τ Z
∂Z ∂µ
!
−
=τ
τ,V
∂Z ∂µ
2 τ,V
Z2
+
∂ hN i = hN 2 i − hN i2 = h(∆N )2 i ∂µ Solution 11. Equivalent definition of chemical potential. Recall that ∂σ ∂σ ∂σ dσ = dU + dV + dN ∂U V,N ∂V U,N ∂N U,V ∂σ µ = −τ (35) ∂N U,V
∂2Z ∂µ2
Z
τ,V
=⇒ τ
(31)
Consider when dσ = dV = 0. =⇒
Note that Using the
dU dN
∂U ∂N σ,V . ∂σ definition, ∂U N,V
∂σ −µ dN + dU = 0 τ ∂U V,N ∂σ µ dU = ∂U V,N dN τ
=
≡ τ1 , so then µ=
∂U ∂N
σ,V
Now F = U − τ σ, by definition. Consider the thermodynamic identity, dU = τ dσ − pdV + µdN . dF = dU − σdτ − τ dσ = τ dσ − pdV + µdN − σdτ − τ dσ =
Likewise, µ =
∂U ∂N σ,V
= −pdV + µdN − σdτ ∂F =⇒ = µ(τ, V, N ) ∂N τ,V ∂F is equivalent to µ(τ, V, N ) = ∂N through the thermodynamic identity as well. τ,V dU = τ dσ − pdV + µdN
σ, V constant. dσ, dV = 0. =⇒
∂U ∂N
=µ σ,V
Solution 12. Ascent of sap in trees. Given: relative humidity r = 0.9. T = 25◦ C 17
n0 = concentration in saturated air that stands immediately above pool of water of water vapor in air. rn0 = actual concentration of water vapor in air at uppermost leaves is rn0 . At pool, µH2 O = µvapor for diffusion equilibrium. Same condition at uppermost leaves, otherwise there’s evaporation: µsap = µvapor (h). µsap = µH2 O (no flow going on in water). Thus, µvapor (h) = µvapor (0) and treat water vapor as an ideal gas. n(0) n(h) + mgh = τ ln τ ln nQ nQ rn0 1 τ 1 τ ln = −mgh =⇒ τ ln = mgh or h = ln n0 r mg r 2 1 10 )(298 K)(kg · m (1.381 × 10−23 K s2 ) ln 9 h= = 147.1 m −27 kg 2) (18 amu) 1.67×10 (9.8 m/s 1 amu Solution 13. Isentropic expansion.
(a) M τ 3/2 V and (b) Recall that F = −τ [N ln Z1 − ln N !] where Z1 = nQ V = 2π~ 2 ∂F σ = − ∂τ V,W . 3/2 1 ∂F M 3 1/2 = −[N ln Z1 − ln N !] − τ [N τ V]= ∂τ Z1 2π~2 2 1 3 3 = −[N ln Z1 − ln N !] − [N nQ V ] = −[N ln Z1 − ln N !] − N Z1 2 2 3/2 3/2 M With N constant and Z1 = 2π~ τ V , then for an isentropic expansion, τ V 2/3 must remain constant. 2 Solution 14. Multiple binding of O2 . (a) Be wary of the multiplicity, how you count, each of the energy states. 4 2 −2/τ 4 3 −3/τ Z = 1 + 4λe−/τ + λ e + λ e + λ4 e−4/τ = (1 + λe−/τ )4 2 3 P () =
4λe−/τ (1 + λe−/τ )4
(b) P (4) =
λ4 e−/τ e−/τ = 4 −/τ 4 (1 + λe ) 1 λ+e−/τ
6. I DEAL G AS Reversible Isothermal Expansion. Q = 0, insulated gas, no heat flow to or from the gas (“adiabatic”) σ constant in system isolated from reservoir, if expansion reverisble (slowly) What is the pressure after expansion? Remember Cp = CV So let
5 2N 3 2N
=
5 =γ 3
3 1 = γ−1 2 1
(7)
σ(τ, V ) = N (ln τ γ−1 + ln V + constant )
(8)
ln τ γ−1 V = constant or τ γ−1 V constant
(9)
=⇒ τ1γ−1 V1 = τ2γ−1 V2
1
1
1
1
18
(63)
(61) (62)
for ideal monoatomic gas. Use pV = N τ γ
γ
τ γ−1 τ γ−1 =⇒ 1 = 2 p1 p2
(10) 1
γ
1
(64)
γ
or p1γ−1 V1γ−1 = p2γ−1 V2γ−1 or p1 V1γ = p2 V2γ I will recap Problem 10 and my solution. Isentropic relations of ideal gas. C
(a) γ = CVp . For isentropic process, pV γ = pi Viγ . Then, essentially and equivalently, take the exterior derivative: V γ dp + γpV γ−1 dV = 0 =⇒
dp γdV + =0 p V
For τ V γ−1 = τi Viγ−1 then taking d: dτ V γ−1 + τ (γ − 1)V γ−2 dV = 0 =⇒ For
γ−1 dτ + dV = 0 τ V
γ
γ
τ 1−γ p = τi1−γ pi then taking d: γ
dpτ 1−γ +
2γ−1 γ dp γ dτ τ 1−γ dτ p = 0 =⇒ + =0 1−γ p 1−γ τ
(b) isentropic bulk moduli Bσ = −V since p =
∂p ∂V
=γ σ
pi Viγ = γp Vγ
pi Viγ Vγ
Bσ = γp EY : 20150606 I think when one consider small, linear longitudinal perturbations of the gas system, with pressure being the external restoring force, then sound waves propagate (correct me if I’m wrong) and this is the way to derive Bσ . ∂p isothermal bulk moduli Bτ = −V ∂V = nτ V =p τ 1/2 1/2 = γp velocity of sound in gas is c = Bρσ ρ for ideal gas of molecules of mass M , pV = N τ Manifold interpretation; I’m using Chapter 5 Applications in Physics, Section A Thermodynamics of [2]. Consider a 2dimensional manifold of equilibrium states M, s.t. dimM = 2. Local coordinates are (U, V ) with U being a global coordinate. Now U = CV τ so use τ as a global coordinate, i.e. the local coordinates of M can be (τ, V ). ∂ ∂ Consider a curve in M parametrized by λ: (τ (λ), V (λ)). The corresponding tangent vector X = τ˙ ∂τ + V˙ ∂V ∈ X(M) has components in coordinate vector basis τ˙ = τ˙ (λ) V˙ = V˙ (λ) Recall dτ γ−1 + dV = 0 τ V Consider applying this 1-form onto X: γ−1 τ˙ γ−1 ˙ d dτ − dV (X) = + V = 0 =⇒ (τ V γ−1 ) = 0 τ V τ V dλ Then the equalities of the endpoints of this curve (τ (λ), V (λ)) are the equalities above. The interpretation is that the isentropic process draws out a curve in M and can be written as a curve or as a tangent vector field (specifically a section of T M). 19
τ1 V1γ−1 = τ2 V2γ−1
(11)
γ 1−γ
τ1
(12)
γ 1−γ
P1 = τ2
P1 V1γ
(13) T1 = 300 K ,
V1 /V2 =
=
(66)
P2
P2 V2γ
(67) (68)
1 2,
2/3 1 T2 = (300 K) = 189 K 2
(14)
(69)
The gas is cooled in expansion process by T1 − T2 = 300 K − 189 K = 111 K
(15)
(70)
Expansion at constant entropy is important e.g. methods of refridgeration. What is the change in energy in the expansion? U2 − U1 = CV (τ2 − τ1 ) =
1 N (τ2 − τ1 ) γ−1
Problems. Solution 1. Derivative of Fermi-Dirac function. Recall f =
1 exp [(−µ)/τ ]+1
∂ f = −1(e(−µ)/τ + 1)−2 (e(−µ)/τ ) ∂ f ( = µ) =
= (e(−µ)/τ + 1)−1
1 τ
−1 4τ
Solution 2. Symmetry of filled and vacant orbitals. = µ + δ
1 e−δ/τ −1 = = 1 + −δ/τ = 1 − f (µ − δ) +1 1 + e−δ/τ e +1 Solution 3. Distribution function for double occupancy statistics. (a) η = 1 + λe−/τ + λ2 e−2/τ where λ = eµ/τ . f () = f (µ + δ) =
eδ/τ
hN i = λ∂λ ln ζ =
λ −/τ λe−/τ + 2λ2 e−2/τ (e + 2λe−2/τ ) = ζ ζ
(b) ζ = 1 + 2λe−/τ + λ2 e−2/τ 2λe−/τ + 2λ2 e−2/τ 2λ(e−/τ )(1 + λe−/τ ) = ζ ζ P −p/tau Solution 4. Energy of gas of extreme relativistic particles. For p ' p, s e = Z. With the factor 2 for the 2 possible polarizations, Z Z ∞ ∞ Z ∞ 4π ∞ 2 p dpe−p/τ = π{ p2 e−p/τ (−τ ) − Z = (2) 2pe−p/τ (−τ )} = 2τ π pe−p/τ dp = 8 0 0 0 0 ∞ Z ∞ −p/τ −p/τ 3 == 2τ π{ pe (−τ ) − e (−τ )dp} = 2τ π hN i = λ∂λ ln λ =
0
0
∂ ln Z = τ 2 ∂τ {ln (2πτ 3 )} = τ 2 ∂τ (3 ln τ ) = U = 3τ ∂τ Solution 5. Integration of the thermodynamic identity for an ideal gas. For constant N , recall dU pdV 1 ∂U 1 ∂U pdV dσ = + = dτ + dV + τ τ τ ∂τ V τ ∂V τ τ ∂U CV = ∂τ V , and for an ideal gas pV = N τ . Z Z 1 ∂U dσ = σ = CV ln τ + N ln V + dV + σ1 τ ∂τ τ ∂U Now U = 23 τ for an ideal gas, so ∂V = 0. τ U = τ2
=⇒ σ = CV ln τ + N ln V + σ1 σ1 independent constant of τ and V . Solution 6. Entropy of mixing. 20
Solution 7. Relation of pressure and energy density. ∂U ∂V N ,
(a) Recall that U = U (σ, V, N ). p = − U=
−s /τ s s e Z
and
∂s ∂V
P
X ∂U = ∂V s
∂s ∂V
e−s /τ + s
N
−1 τ
∂U ∂V N .
e−s /τ
/Z −
N
X ∂s 1 X − e−s /τ s2 e−s /τ /Z 2 ∂V τ s s 2
Now p = − So if the system is in state s; then ps = −
∂s ∂V N .
P =⇒ hpi = p = (b) Now s =
2 2
π n ~ L2 2M
=
s
−
∂s ∂V N
e−s /τ
Z
2 2
π n ~ V 2/3 2M
∂s ∂V
N
2 πn2 ~2 2 = − V −5/3 = − s 3 2M 3V
(c) −2s −s /τ s 3V e
P p= Time for a large fluctuation. nQ 5 (a) Recall σ = N [ln n + 2 ]. nQ = n 2V + 52 ] σf = N [ln QA N n V + 52 ] σi = N [ln QA N Now
Z
=
2 3
1 V
U
Solution 8.
M τ 3/2 . 2π~2
N 5N NQ nQ N 5N g ∼ eσ ∼ exp ln + =e 2 n 2 n 2 2 −4 1c (938 M eV /c )((0.8617 × 10 eV )/K)(300 K) 1 M eV = 7.88 × 1030 /m3 nQ = 4 × 2π(6.582122 × 10−22 M eV · s)2 106 eV 3 × 108 m/s Now P V = N τ . Then 2 5 N/m2 (1 atm)( 1.013×10 ) kg·m/s 1 atm 1N N P = = = 2.445 × 1025 /m3 τ V 1.381 × 10−23 J/K(300 K) Now 1 L = 10−3 m3 and so for 0.1 L, 1 )(10−4 m3 ) = 2.445 × 1021 m3 7.88 × 1030 /m3 = 3.22 × 105 2.445 × 1025 /m3
(2.445 × 1025 With
5
21
=⇒ g ∼ e 2 (2.445×10
)
21
(3.22 × 105 )2.445×10
(b) (c) Solution 9. Gas of atoms with internal degree of freedom. For an ideal monatomic gas, assume noninteracting.
(a) λint = λ. λ = exp (µ/τ ) ideal gas. µ = τ ln (n/nQ );
nQ =
M τ 3/2 . 2π~2
λext = exp (−∆/τ ) or 1 M τ 3/2 Z1 is the usual canonical partition function, Z1 = nQ V where nQ = 2π~ 2 X −s −∆ Z1 = exp )Z1 = λ(1 + e−∆/τ )Z1 (λ exp (−s /τ )(1) + λ exp τ τ s (Z1 )N = (λ(1 + e−∆/τ )Z1 )N 21
(b) (c) Solution 10. Isentropic relations of ideal gas.
(a) Isentropic process, so pV γ = pi Viγ . V γ dp + γpV γ−1 dV = 0 =⇒
dp γ + dV = 0 p V
Dealing with an ideal gas, pV = N τ still applies. τ V γ−1 = τi Viγ−1 dτ V γ−1 + τ (γ − 1)V γ−2 dV = 0 =⇒
dτ (γ − 1) + dV = 0 τ V
Using pV = N τ again, note that p1−γ τ γ = constant . (1 − γ)p−γ dpτ γ + p1−γ γτ γ−1 dτ = 0 =⇒ (b) Using p =
dp γ dτ + =0 p 1−γ τ
pi Viγ Vγ ,
∂p −γpi Viγ = ∂V V γ+1 γpi Viγ ∂p = = γp =⇒ −V ∂V Vγ So that Bσ = −V (∂p/∂V )σ = γp, the isentropic bulk moduli. ∂P nτ Bτ = −V = =p ∂V τ V since pV = nτ nτ p= V ∂p −nτ = ∂V V2 7. F ERMI AND B OSE G ASES Problems. Problem 1. Density of orbitals in one and two dimensions. (a) Show that the density of orbitals of a free electron in one dimension is D1 () = (L/π)(2m/~2 )1/2 ,
(16)
where L is the length of the line. (b) Show that in two dimensions, for a square of area A, D2 () = Am/π~2
(17) independent of . p2 2m 2 2 2 ~2 (nx +ny )π 2m L2
Solution 1. Recall, for the free electron: H = 2
n2 π 2
=
−~2 2 2m ∇
~ x =⇒ s = 2m for 2-dim. L2 for 1 dim., s = If F = Fermi energy, energy of the highest filled orbital, πn2 1-dim: N = 2nF . 2-dim.: N = (2) 14 (πn2F ) = 2F 2 factor for 2 possible spin states.
1-dim: F = 2-dim: F =
~2 2m ~2 2m
1-dim.: D() =
N 2 π2 1 ~πN 2 1 1 ~m 2 2 n N= 2 L2 = 2m 2 V 2 = 2m 2 π 2 mV F 2N ~2 N π 2 ~2 N π ~2 π = m π V = m V = m n N = π~2 π L dN d
=
q
2m
2V 2 1 −1/2 ~π 2 (F )
=
N 2
=
L π
2m 2 1/2 ~
2-dim.: N = Am/π~2 22
q
2mF
2V 2 ~π
Energy of relativistic Fermi gas. For electrons with an energy mc2 , where m is the rest mass of the electron, the energy is given by ' pc, where p is the momentum. For electrons in a cube of volume V = L3 the momentum is of the form (π~/L), multiplied by (n2x + n2y + n2z )1/2 , exactly as for the nonrelativistic limit. Problem 2.
(a) Show that in this extreme relativistic limit the Fermi energy of a gas of N electrons is given by F = ~πc(3n/π)1/3 ,
(18)
where n = N/V . (b) Show that the total energy of the ground state of the gas is 3 N F . 4 The general problem is treated by F. J¨uttner, Zeitschrift f¨ur Physik 47, 542 (1928). U0 =
(19)
Solution 2.
(a) ' pc =
~nπ L c,
Recall, for 3-dim.: N F =
π~ L
~nF π L c. = (2) 81
F =
3N 1/3 π
= ~πc
4 3 3 πnF
=
π 3 3 nF .
nF =
3N 1/3 . π
3n 1/3 π
(b) Z nF Z ~π 2 c 4 ~π 2 c nF 1 2 ~nπc dnn dnn3 = = n = n = 2 ∗ ∗ 4π U0 = 2 8 L L 4L F 0 0 n≤nF 4 ~π 2 c F L ~π 2 c 3N F L 3 = = = N F 4L ~πc 4L π ~πc 4 X
Problem 3. Pressure and entropy of degenerate Fermi gas.
(a) Show that a Fermi electron gas in the ground state exerts a pressure 5/3 (3π 2 )2/3 ~2 N (20) p= 5 m V In a uniform decrease of the volume of a cube every orbital has its energy raised: The energy of an orbital is proportional to 1/L2 or to 1/V 2/3 . (b) Find an expression for the entropy of a Fermi electron gas in the region τ F . Notice that σ → 0 as τ → 0. Solution 3. 2
~ (a) Recall U0 = 53 N 2m (3π 2 N )2/3 V −2/3
3 ~2 ∂U = N (3π 2 N )2/3 ∂V 5 2m
−2 3
V
−5/3
−1 ~2 = (3π 2 )2/3 5 m
N V
5/3
So then 5/3 −∂U0 1 ~2 N 2 2/3 p= = (3π ) ∂V 5m V 2 2/3 ~2 3π N and that the heat capacity of an electron gas is Cel = 12 π 2 N ττF = (b) Recall that F ≡ τF = 2m V helps directly with finding the entropy. Z τ Z τ 1 11 2 τ π2 N σ(τ ) − σ(τ0 ) = dU = π N dτ = τ τF 2τF τ0 τ τ0 τ 2
∂U ∂τ ,
which
Let σ(τ0 = 0) = 0, σ(τ ) =
π2 N τ 2τF
Problem 4. Chemical potential versus temperature. Explain graphically why the initial curvature of µ versus τ is upward
for a fermion gas in one dimension and downward in three dimensions (Figure 7.7). Hint: The D1 () and D3 () curves are different, where D1 is given in Problem 1. It will be found useful to set up the integral for N , the number of particles, and to consider from the graphs Rthe behavior of the integrand between zero temperature and a finite temperature. F Solution 4. Recall, N = 0 dD(). 23
1-dim: 1/2 Z L 2m L (2m)1/2 F d dD1 () = N= d−1/2 = = π ~2 π ~ 0 0 0 F L (2m)1/2 1/2 L (2m)1/2 1/2 = (2 ) = 2F π ~ π ~ Z
F
Z
F
0
3-dim: F
Z N= 0
V d 2 2π
2m ~2
3/2
1/2
V = 2π 2
2m ~2
3/2
2 3/2 3
3/2 F V 2m 3/2 F = 3π 2 ~2 0
Note the difference in the concavity of the N () curves. Solution 5.
(a) For 3 He, given I = 1/2, density of liquid 0.081 g cm−3 , we want to find vF , F , τF . 2 ~ F = (3π 2 n)2/3 = 2m 3 2/3 (6.582 × 10−22 M eV · s)2 1 kg 1u 1 He 2 0.081 g 3π = = 2 cm3 103 g 1.67 × 10−27 kg 3u 1c 2 · 3 · 938 M eV /c2 3×1010 cm/s = 4.24 × 10−10 M eV = 4.24 × 10−4 eV Now suppose we have a nonrelativistic gas. Then 12 mvF2 or vF2 = cm vF = 1.675 × 104 sec TF =
2F m
.
4.24 × 10−4 eV = 4.92 K 0.8619 × 10−4 eV /K
(b) π 2 3N π2 τ π2 D(F )τ = N τ= = 1.003kB T N 3 3 2τF 2 τF Solution 6. Mass-radius relationship for white dwarfs. (a) Z Z G 4 πr3 ρ 2 M 4 1 M −3GM 2 r dr = 4π 4 3 G π R5 4 3 = U = ρ(r)φ(r)r2 4πdr = −4π ρ 3 r 3 5 5R 3 πR 3 πR (b) 2 ~ F = (3π 2 n)2/3 2m is the Fermi energy. With V = 34 πR3 , 2/3 1 h2 3π 2 N (3π 2 )2/3 ~2 N 5/3 (3π 2 )2/3 ~2 N 5/3 2 Ttot = N mv = N F = N = = = 2 2m V 2 m V 2/3 2 m 4 π 2/3 R2 3 9π 2/3 2 5/3 5/3 ~ N 2 (M/MH ) 4 = ' ~ 2 mR2 mR2 Cel =
since N =
M MH
since MH m.
(c) ~2 M 5/3 5/3 mMH R2
=
GM 2 R
(6.582 × 10−22 M eV · s)(1.05457 × 10−34 J · s) 3 g 5/3 0.511 M eV /c2 ((1.67 × 10−27 kg) 10 1 kg )
=⇒ M 1/3 R '
~2 /G 5/3
mMH
103 g 1 kg
2
/(6.67 × 10−11
m3 /s2 ) kg
Dealing with units and dimensions, J = kg ·
m2 ; s2
N · m2 m3 /s2 m3 /s2 = kg · = kg 2 kg 2 kg 24
(3 × 108 m/s)2 1 c2
(6.582 × 10−22 M eV · s)(1.05457 × 10−34 kg · (0.511 M eV /c2 )(1.67
×
10−24
g)5/3 (6.67
×
m2 s )
102 cm 1m
3 2 10−11 mkg/s
2
103 g 1 kg
2
102 cm 3 ) 1m
×
(3 × 1010 cm/s)2 1 c2
1020 g 1/3 cm
(d) ρ=
M 4 3 3 πR
M
= 4 3π
1020 g 1/3 cm M 1/3
3 =
(2 × 1033 g)2 3 × 1066 g 2 M2 = = 4 60 3 π1060 g cm3 g cm3 3 π10 g cm
4 60 3 π10
106
g cm3
(e) M 1/3 R '
~2 /G
1017 g 1/3 cm
5/3 mMH
=⇒ R =
1017 g 1/3 cm = 7.937 km (2 × 1033 g)1/3
Solution 7. Photon condensation. Ne = 2.404V τ 3 /π 2 ~3 c3 . The condition is that N = Ne :
2.404V τ 3 π 2 ~3 c3 2 3 3 1/3 π ~ c N τ= 2.404 V
N = Ne =
With a concentration of 1020 cm−3 , T = 1.7 × 106 K (the critical temperature in K below which Ne < N . Solution 8. Energy, heat capacity, and entropy of degenerate boson gas. Recall that the distribution function for bosons is 1 f (, τ ) = exp [( − µ)/τ ] − 1 Consider N noninteracting bosons of spin zero. 1 = 0 for ground state. Thus, f (0, τ ) = exp −µ . Recall that ( τ )−1 3/2 Z ∞ Z ∞ 2M 1 V N (τ ) = 1/2 −1 dD()f (, τ ) = d 2 2 4π ~ λ exp 0 0 τ −1 Recall, U =0
1 exp (−µ/τ ) − 1
∞
Z
Z
∞
D()d = e(−µ)/τ − 1 e(−µ)/τ 0 3/2 Z ∞ 2M 3/2 V = 2 2 (−µ)/τ 4π ~ e −1 0
+
0
V − 1 4π 2
2M ~2
3/2
τ < τ , so λ ≈ 1, for N0 to be sufficiently large. =⇒ U = Using
V 4π 2
2M ~2
3/2 Z
∞
0
3/2 −1
e/τ
x = /τ dx = d/τ
It’s true that ∂τ (e/τ ) = e/τ
V =⇒ U = 4π 2 −
2M ~2
3/2
τ 5/2
Z 0
∞
x3/2 dx = B0 τ 5/2 C0 ex − 1
τ2
3/2 Z ∞ 3/2 V 2M x dx 5 3/2 τ CV = 2 4π 2 ~2 ex − 1 0 Now
1 τ
=
∂σ ∂U V
. V B0 25 τ 3/2 C0 dτ 5 = V B0 C0 τ 2 τ 5 3/2 dU =V B0 τ C0 dτ 5 2 −−−−−−−2−−−−−−→= V B0 C0 τ 3/2 2 3 0 Z
σ(U ) − σ(U0 ) =
1 dU = τ
Z
25
Z
τ 1/2 dτ =
1/2 d =
5 5 U 3/5 5 = (V B0 C0 )2/5 U 3/5 = σ(U ) = V B0 C0 τ 3/2 = V B0 C0 3/5 3 3 (V B0 C0 ) 3 3/2 Z ∞ 3/2 !2/5 x dx 2M 5 3/5 V U = x−1 4π ~2 e 3 0 3/5 = τ 3/2 . where we had used U = V B0 τ 5/2 C0 or V BU0 C0 Solution 9. Boson gas in one dimension. In one-dim., s = =⇒ n =
~2 π 2 n 2 ~2 π 2 n 2 2mV 2 = or s = n2 2m L2 2m V 2 ~2 π 2 2mV 2 ~2 π 2
1/2
1/2 =⇒
dn = d
2mV 2 ~2 π 2
1/2
1 21/2
Note the difference with 3-dim.: For spinless bosons, 1/2 4πn2 dn π 2mL2 2mL2 1 = d = s 8 2 ~2 π 2 ~2 π 2 21/2 3/2 π 2mL2 = 1/2 d = D()d 4 ~2 π 2
D(n)dn =
Z
∞
N (τ ) =
∞
Z
dD()f (, τ ) = 0
d 0
=
1 2
2
2mV ~2 π 2
1/2 Z
∞
0
2mV 2 ~2 π 2
1/2
1 1 exp −1= −1 1/2 τ 2 λ d −1 (λ exp − 1) τ 1/2
For → 0, N (τ ) → ∞ which is not characteristic of a Boson. Solution 10. Relativistic white dwarf stars. ' pc λ = 2π~/p. Virial theorem: 2hT i = −hU i 2
3 N F 4
=
3 4/3 N ~πc 2
2hT i = khU i
1/3 3 1 3GM 2 3GM 2 =− − = π L 5R 5R
F = ~πc
3n π
1/3
Approximating the sphere as a box, 4 L3 = πR3 3 1/3 4 L= π 1/3 R 3
N
4/3
=
GM 2 5R
2L ~π 2/3 c31/3
=⇒ N =
GM 2 5R
3 4π
1/3 L=R
2L ~π 2/3 c31/3
3/4
=
2GM 2 5
41/3 ~32/3 π 1/3 c
3/4
Now M = N mH , where mH is the mass of hydrogen, so 1= =⇒ N =
5~(32/3 )π 1/3 c 41/3 (2Gm2H )
3/2 =
2Gm2H 5
41/3 2/3 ~3 π 1/3 c
3/4
N 1/2
(1.05457 × 10−34 J · s)(3 × 108 m/s) m3 −27 kg)2 (6.67 × 10−11 kg·s 2 )(1.67 × 10 26
!3/2 3/2 5(32/3 )π 1/3 · = 2.2 × 1058 41/3 2
8. H EAT AND W ORK Energy and Entropy transfer: Definition of Heat and Work. Consider Σ, a manifold consisting of points representing thermodynamic states of a single system. For instance, for global coordinates (U, V ), (U, V ) ∈ Σ. Consider W, Q ∈ Ω1 (Σ), 1-forms on Σ. Now define Q as Q ≡ τ dσ ∞ with σ = σ(U, V ) ∈ C (Σ) τ = τ (U, V ) ∈ C ∞ (Σ) Recall energy conservation in this form:
dU = W + Q Consider pure heat and, so, no work. Now Q = τ dσ ∂ (U is suited Heat Engines: Conversion of Heat into Work. Consider curve c : R → Σ s.t. c generates vector field c˙ = ∂σ c(t) ∈ Σ for this). ∂ Act on this vector field ∂σ ∈ X(Σ) with Q, i.e. ∂ Q =τ ∂σ This is what’s meant when it’s said “reversible heat transfer accompanying 1 unit of entropy is given by temperature τ ” [1]. Consider Figure 8.1 on page 229 of Kittel and Kroemer [1]. Roughly it looks like this:
τ = τh
dσh = Qh /τh
Qh
τ
W
τ = τl
dσl = Ql /τl
Ql
But what’s really going on? Consider Qh = dU , the initial heat input at high temperature (I’ll show that later) τh . ∂ Consider a curve c0 ∈ Σ s.t. c˙0 = σ˙ 0 ∂σ . Then ∂U Qh (c˙0 ) = dU (c˙0 ) = = τ σ˙ 0 ≡ τh σ˙ 0 ∂σ V We can integrate the 1-form dU ∈ Ω1 (Σ) for 2 reasons: mathematically, it is an exact form. Physically, we are considering a reversible process, passing through thermodynamic states of the system, starting with the system being in energy U0 and ending up with energy U1 , and all the energy states (∈ R) in between. Z 1 =⇒ U1 − U0 = τh dσ 0
If this is conducted all at temperature τh during the whole process, then U1 − U0 = τh that dσ is an exact form and can be integrated over that curve c0 .
R1 0
dσ = τh (σ1 − σ0 ). It’s in this case
Legendre transforms revisited. Let’s recall 2 of our favorite thermodynamic potentials, U , and Helmholtz free energy F . They are related by Legendre transformations that transform 1 coordinate into its conjugate coordinate, somewhat like how the Legendre transform transforms that Lagrangian in canonical coordinates into a Hamiltonian written with the conjugate momentum. However, I do want to point out that, for Lagrangians and Hamiltonians, the Legendre transformation is a fiber derivative between tangent bundle to the cotangent bundle on the manifold. In our current case, we want a mundane Legendre transformation between convex function to another convex function, a coordinate transformation by a C ∞ function, not a morphism between vector spaces. Recall F . It’s defined as such: F ≡ U − τ σ, so dF = dU − τ dσ − σdτ = −σdτ − pdV 27
Consider curve c : R → Σ c(t) ∈ Σ Consider 2 curves that generate vector fields: c˙ = τ˙
∂ ∂ or c˙ = V˙ ∂τ ∂V
Now, in general, mathematically, dF = +
∂F ∂τ
dτ + V
∂F ∂V
dV τ
Thus,
∂F = −σ ∂τ V ∂F = −p ∂V τ
Maxwell relations are easily derived: ∂2F ∂2F = so ∂V ∂τ ∂τ ∂V
∂σ ∂V
= τ
∂p ∂τ
V
So-called natural coordinates for F are τ, V . So Σ 3 (τ, V ) (i.e. after a Legendre transformation, the coordinates become (τ, V ) for each thermodynamic state. Recall U as a thermodynamic potential. Using energy conservation and how Q is defined, dU = Q + W = τ dσ + pdV Natural coordinates are σ, V for U . So Σ 3 (σ, V ). heat engine ideal heat engine: (σ1 , V0 )
1 Qh = τh dσh
Ql = τl dσl
Qh = τh dσh
W
0
(σ0 , V0 )
2
W + Ql = Qh or W = Qh − Ql τh − τl W = Qh − Ql ≤ Qh = ηC Qh τh Carnot efficiency ηC ≡
τh −τl τh
Ql = τl dσl (σ2 , V1 )
σl = σh so
Qh Ql = τh τl
is the ratio of the work generated to the heat added, in the reversible process.
Carnot cycle. 3
W23
(σH , τl )
2
W34
−W12 = Qh 4
W41
1
W23
(σH , τh )
W34 (σL , τl )
−W12 = Qh (σL , τh )
W41
H The total work is as such: dU = 0 for 2 reasons: mathematically, the integration of an exact 1-form around a closed curve is 0, and physically, we return the system back to its original state, as this is a reversible process. I I I I I dU = 0 = τ dσ − pdV =⇒ − W = τ dσ = [τh (σH − σL ) + 0 + τl (σL − σH ) + 0] = (τh − τl )(σH − σL ) 28
Example: Carnot cycle for an ideal gas. W23
3
(σH , τl , V3 )
2
W34 = −Ql
−W12 = Qh
(σL , τl , V4 )
1
W41
(σH , τh , V2 )
W34 = −Ql
−W12 = Qh 4
W23
(σL , τh , V1 )
W41
with isothermal expansion Qh = −W12 =
2
Z
pdV = N τh ln
1
adiabatic expansion W23 = −
Z
V2 V1
3
τl V3γ−1
dU = U (τh ) − U (τl ) = CV (τh − τl )
=
τh V2γ−1
2
isothermal compression − Ql =
W34
Z
4
−pdV = N τl ln
= 3
V3 V4
EY : 20150911 I don’t have a good reason why CV which is defined for constant V , that CV ≡ isentropic (i.e. adiabatic) expansion from 2 → 3. The total work done is V2 W = N (τh − τl ) ln V1 Energy Conversion and the Second Law of Thermodynamics. W1 = η1 Qh 11 Qh
τh τl
1 γ−1
V4 τh 1 = ( ) γ−1 V1 τl
adiabatic compression W41 = CV (τh − τl )
Ql1 = (1 − η1 )Qh
V3 or = V2
∂U ∂τ V
, can be used in the
Wout = η2 Qh − η1 Qh 12 Qh
Qh
01
Ql2
02
Q(in) = (η2 − η1 )Qh Ql2 = (1 − η2 )Qh
Ql2 waste heat W1
11
Wout
12
Qh
01
02
Q(in) Ql2
So with Q(in) heat in, Wout net work can be done. But that’s a decrease in overall entropy. This violates the law of increasing entropy. 29
Define H = U + pV . H ∈ C ∞ (Σ), where Σ is the manifold of equilibrium (and non-equilibrium) states of the system. Path Dependence of Heat and Work. Mathematically, Q and W are not necessarily exact 1-forms. So they are path-dependent. EY : 20150911 That Q, W are not necessarily exact 1-forms would imply that Σ has some nontrivial, interesting topological features. Heat and Work at Constant Temperature or Constant Pressure. isothermal work.
dU = W + Q = W + τ dσ F = U − τσ
dF = dU − τ dσ − σdτ = W − σdτ If dτ = 0, on an isothermal curve, dF = W , W becomes an exact 1-form, with potential function F , the Helmholtz free energy. isobaric heat and work. e.g. boiling of liquid. When liquid boils under atmospheric pressure, vapor pressure displacing atmospheric odes work against atmospheric pressure. isobaric process. Consider this change of volume: dx = dV A . Now peq = vapor pressure. F = peq A = patm A (force equilibrium) (σ1 , V1 ) W = −patm Adx = −patm dV (σ0 , V0 ) W = −patm dV ≡ −pdV = −d(pV ) is part of total work done on system. If −d(pV ) > 0, work provided by environment and is “free”. If −d(pV ) < 0, work delivered to environment and not extractable from system for other purposes. W + d(pV ) = dU − Q + d(pV ) = dH − Q Recall that for enthalpy H = U + pV , dH = dU + V dp + pdV = dU − W + V dp = τ dσ + V dp σ, p are natural coordinates of H. dH − Q = W + d(pV ) An isobaric curve s.t. dp = 0, dH = Q + W + d(pV ) so Q + W is an exact 1-form of H − pV =⇒ d(H − pV ) = W + Q. 2 classes of constant pressure processes: (a) W + d(pV ) = 0 dH = Q e.g. liquid evaporation from open vessel, because no effective work is done. heat of evaporation is enthalpy difference between vapor phase and liquid phase (b) constant temperature and constant pressure. G = F + pV = U − τ σ + pV dG = dF + V dp + pdV = dU − τ dσ − σdτ + V dp + pdV = V dp − σdτ dG = W − σdτ + d(pV ) = W + d(pV ) − σdτ with natural variables are p, τ at constant temperature, W + d(pV ) is exact 1-form, dG 30
F IGURE 1. Problem 8.1(c) Problems. Solution 1. Heat pump. (a) For a heat pump, input: σh = Qτhh output: σl = Qτll Reversible condition: σh = σl = Qτhh = Qτll so that Qh = ττhl Ql . l Qh − Ql = Qh − ττhl Qh = τhτ−τ Qh net heat inputted to pump heat. h Thus, τh − τl W = ηc = Qh τh If heat pump is not reversible, σh > σl , so that
Qh τh
>
Ql τl
or
τl τh Qh
> Ql ,
τl τh Qh
Qh − Qh − Ql W = < = ηc, ideal Qh Qh Qh (b) Qh = electricity consumed by reversible heat pump. hh Carnot engine: W = (τhh − τl )(σhh − σl ), with σhh = Qτhh , and σl = Qτll Condition that electricity consumed by reversible heat pump: Qhh Ql W = (τhh − τl ) − = Qh τhh τl Note we let σl = Qτll since both heat pump andCarnot engine are reversible. Qhh Qh Qh Qhh 1 Qhh τhh (τh + τhh − τl ) 1 =⇒ − = =⇒ = Qh + =⇒ = τhh τh τhh − τl τhh τhh − τl τh Qh τh (τhh − τl ) For Thh = 600 K, Th = 300 K, Tl = 270 K, 600(300 + 600 − 270) Qhh = = 3.82 Qh 300(600 − 270) (c) See Figure (1). Solution 2. Absorption refrigerator.
(a) See Figure (2). (b) Given τhh > τh , by energy conservation: Qhh + Ql − Qh = 0 reversible refrigerator: σhh + σl − σh = 0, Ql Qh Qhh + − =0 =⇒ τhh τl τh Qhh Ql Qhh + Ql 1 1 1 1 + = or Qhh − = Ql − τhh τl τh τhh τh τh τl Ql Ql 1 1 1 1 τh − τhh τl − τh τhh − τh τl = − / − = / = = Qhh τhh τh τh τl τh τhh τh τl τh − τl τhh Qhh Note that Ql − Qh = Ql − (Qhh + Ql ) = −Qhh ; we’ve removed Qhh heat from refrigerator’s inside. 31
F IGURE 2. Problem 8.2(a) Photon Carnot engine. Recall, photons are relativistic: = pc. Recall p =
~ i ∇.
=⇒ s = pc = ~ks c = ~c. Recalling that there are 2 polarization states for a photon in 3-dim. space, Z Z ∞ nπ nπ π 2 ~c ∞ 3 −π~c 1 − L ~c/τ 2 ~c e = n dnexp n = U = (2) (4π) n dn 8 L L Lτ 0 0 2 ∞ Z ∞ Lτ −π~c Lτ π ~c −π~c 2 3 n − 3n exp n = { n exp dn} = L Lτ −π~c 0 Lτ −π~c 0 2 Z ∞ π ~c −Kτ −π~c n dn} = = {(−1) 3 n2 exp L π~c Lτ 0 2 ∞ Z ∞ −π~c π ~c −Lτ −π~c Lτ Lτ − 2nexp = {(−1) 3{ n2 exp n n dn = L π~c Lτ −π~c 0 Lτ −π~c 0 2 2 Z ∞ π ~c −Lτ Lτ −π~c 2 = (−1) 3(2) n dn = nexp L π~c Lτ −π~c 0 2 3 2 4 ∞ Z ∞ π ~c −Lτ π ~c −Lτ −π~c −π~c = (−1)3 3(2)(1) n dn = (−1)3 3(2)1 exp n = exp L π~c Lτ L π~c Lτ 0 0 3 L V =6 τ4 = 6 2 3 τ4 = U π 2 ~c (π ~c) ∂σ = τ1 , and using this is usually the most direct way to obtain entropy. To get the entropy, recall, ∂U V
Solution 3. ns π L
Z
dU = τ
=⇒ dσ =
Z
6V (π 2 ~2 c)3
4τ 3
dτ 6V 1 8V τ 3 = 2 2 3 τ 3 =⇒ σ(τ ) = 2 2 3 τ (π ~ c) 3 (π ~ c)
Consider Isothermal expansion: Helmholtz free energy F is needed. F = U − τσ = Then p=−
∂F ∂V
= τ, N
W12 = p(V2 − V1 ) =
6V 8V τ 3 2V τ 4 4 τ − τ = − (π 2 ~2 c)3 (π 2 ~2 c)3 (π 2 ~2 c)3
2τ 4 (π 2 ~2 c)3
2τh4 (V2 2 (π ~2 c)3
Isentropic expansion: =⇒ V2 τh3 = V3 τl3 or V3 = V2 So for this isentropic process,
V2 τh3
∆Q12
− V1 )
τh τl
8τh3 (V2 − V1 ) (π 2 ~2 c)3 τ 8τ 4 = σ = 2 2h 3 (V2 − V1 ) ∆ (π ~ c)
σ12 =
3
.
3
=Vτ , U=
6V (π 2 ~2 c)3
V2 τh3 V
4/3 = 32
6(V2 τh3 )4/3 −1/3 V (π 2 ~2 c)3
2(V2 τh3 )4/3 −4/3 V (π 2 ~2 c)3 V3 Z Z 2(V2 τh3 )4/3 −4/3 −6(V2 τh3 )4/3 2(V2 τh3 )4/3 −1/3 = pdV = = V dV = (−3V ) (π 2 ~2 c)3 (π 2 ~2 c)3 (π 2 ~2 c)3 V2 τl 6V2 τ 4 = 2 2h 3 1− (π ~ c) τh p=
W23
Isothermal compression: W34 = σ34 =
8τl3 (π 2 ~c)3 (V4
− V3 ) =
−∂U 6(V2 τh3 )4/3 =− ∂V (π 2 ~2 c)3
2τl4 (π 2 ~2 c)3 (V4
8τh3 (π 2 ~c)3 (V1
− V3 ) =
∆W =
6(V4 τl3 )4/3 = (π 2 ~2 c)3
6V2 τ 4 2τh4 (V2 −V1 )+ 2 2 h 3 2 2 3 (π ~ c) (π ~ c)
−1 −4/3 V 3
2τh3 τl (π 2 ~2 c)3 (V1
=
1 1/3
−
V3
1 1/3
! =
V2
− V2 ).
− V2 ).
Isentropic compressiong: V4 τl3 = V1 τh3 or V4 = V1 W41
τh τl
3
1 1/3
V4
. 1
−
1/3
V1
! =
−6V1 τh4 (π 2 ~2 c)3
1−
τl τh
τl −6V1 τ 4 τl τl 2τ 3 τl 8τ 4 (V2 − V1 ) 1− + 2 h2 3 (V1 −V2 )+ 2 2 h3 1 − = h 2 2 3 τh (π ~ c) (π ~ c) τh (π ~ c) τh
1−
Qh =
∆W 8τh4 (V2 − V1 ) τl =⇒ =1− (π 2 ~c)3 Qh τh
Solution 4. Heat engine-refrigerator cascade. Consider the heat engine as a Carnot cycle.
W + Wr = (τh − τl )σh where Wr = work consumed by refrigerator. Qh Ql = = σl τh τl This must be true for any heat engine undergoing Carnot cycle; furthermore, we can say it’s the most efficient heat engine possible. reversible refrigerator: QL + Wr = QH , (by E-consv.) σL = σH = QτLL = QτHH , (by reversible condition) Note, Ql is energy transfer from heat engine to τl reservoir. QL is energy transfer from τl reservoir to refrigerator. QL ≥ Ql , otherwise, no cooling, no thermal energy extracted from τl resevoir to lower its temperature. QL = Ql at equilibrium; no further cooling, τr reached. Note that τl is given as the environmental temperature. Assume refrigerator throws out QH heat into the environment. → τH = τl . Since QL heat inputed into refrigerator from a τl reservoir now lowered to τr , τl → τr . τl τl Wr = QL − QL = − 1 QL τr τr σh =
since for a reversible refrigerator, σL = σH = W =⇒ = Qh
τr 1− τh
−
QL τL
=
QH τH .
τl QL τr τl τr τl −1 = 1− − −1 = 1− τr Qh τh τr τh τh
Combinations of reversible systems = reversible system. Tl = 20◦ C Solution 5. Thermal pollution. Given . Consider a Carnot cycle. Th = 500◦ C Ql τh 500 W = (τh − τl )σl = (τh − τl ) = − 1 Ql = − 1 1500 M W = 36000M W τl τl 20 If improvements in hot-steam technology would permit raising Th by 100◦ C, 600 W = − 1 1500 M W = (29)(1500 M W ) = 43500 M W 20 There was a 17.2 % increase in output. Solution 6. Room air conditioner. 33
(a) Ql τh W = (τh − τl ) = − 1 Ql τl τl τh dQl τh P P = −1 = − 1 A(τh − τl ) =⇒ τl = (τh − τl )(τh − τl ) = τh2 − 2τh τl + τl2 τl dt τl A =⇒ τl2 − 2τh τl − P − τl = τh + 2A
P τl + τh2 = 0 A
r (τh +
P 2 ) − τh2 2A
(b) For Tl = 17◦ C = 290 K, Th = 310 K, A=
(2 kW )(290 K) 580 × 103 W W P τl = = = 1450 (τh − τl )2 (310 − 290)2 400 K K
Solution 7. Light bulb in a refrigerator
Carnot refrigerator draws 100 W. For any Carnot cycle, Ql W = (τh − τl ) = τl
τh τl − 1 Ql = 1 − Qh τl τh
Carnot refrigerator expels Qh thermal energy to hot τh environment and inputs Ql thermal energy from τl reservoir. Ql + W = Qh Work W must be drawn by Carnot refrigerator to do work. Suppose Carnot cycle part of the refrigerator must input in heat from light bulb to cool down its inside, i.e. consider Carnot refrigerator in equilibrium with light bulb, now inputting in heat from light bulb Qext , and drawing in work to expend out Qh thermal energy into the environment. =⇒ Qext = Ql ˙ = Q˙ l in this case, so W
τh τh − 1 Ql − Ql = 0 or − 2 Ql = 0 τl τl =⇒ τl =
300 K τh = = 150 K 2 2
Solution 8. Geothermal energy.
Given ∆Qh = −M CdTh , Tl lower reservoir temperature stays constant. τh decreasing, dτh < 0. ∆Qh τl dτh ∆W = (τh − τl ) = 1− (−M C) τh τh kB MC MC τi τ =⇒ W = − (τh − τl ln τh )|τfi = − τl ln − (τi − τf ) kB kB τf For M = 1017 g, C = 1 J/g · K, Tl = 20◦ C = 293 K, Ti = 600◦ C = 873 K, Tf = 110◦ C = 383 K W = 2.486 × 1019 J 3600 sec
Note that 1014 kW h = 1017 Js · h = 3.6 × 1020 J. 1h Solution 9. Cooling of nonmetallic solid to T = 0. Recall that C = aT 3 = since τl decreasing. For the refrigerator: Ql + W = Qh . dW = (τh − τl )
Ql =− τl
∂U ∂T V
. Then dQl = aTl3 dTl . Now dτl < 0
τh 1 −a τh −a − 1 (aτl3 dτl ) = − 1 τl3 dτl = 4 (τh τl2 − τl3 )dτl 4 4 τl kB kB τl kB 0 −a 1 1 aTh3 W = 4 τh τl3 − τl4 = =W kB 3 4 12kB τh 34
9. G IBBS F REE E NERGY AND C HEMICAL R EACTIONS Solution 1. Thermal expansion near absolute zero
(a)
∂G ∂G = −σ =V ∂τ N, p ∂p τ ∂σ ∂V 2 =− =⇒ 2 ∂ G ∂V ∂τ p ∂p τ ∂ G ∂σ = =− ∂τ ∂p p ∂τ p ∂p∂τ τ ∂p τ ∂G ∂G =µ =V ∂N p ∂p τ ∂µ ∂V 2 =⇒ = 2 ∂V ∂ G ∂N p ∂p N ∂ G ∂µ = = ∂N ∂p ∂N ∂p∂N N ∂p N p p 2 2 ∂µ ∂ G ∂σ ∂ G ∂µ ∂σ = =− =⇒ =− ∂τ ∂N N ∂τ N ∂N ∂τ τ ∂N τ ∂τ N ∂N τ −1 ∂σ = 0 as τ → 0 since σ → constant as τ → 0 by third law of thermodynamics. (b) α = V1 ∂V ∂τ p = V ∂p τ
Solution 2. Thermal ionization of hydrogen.
(a) Given e + H + H, note that e + H + − H = 0. Recall Y ν [e][H + ] = K(τ ) = nQjj exp [−νj Fj (int)/τ ] [H] j M τ 3/2 where nQ = 2π~ V. 2 For dissocation of H into e− + H + choose zero of internal energy of each composite particle (here H) to concide with energy of dissociated particles (here H + , e− ) at rest; place energy of ground state of composite particle H at −I, I is energy required in reaction to dissociate composite particle into its constituents and is taken to be positive, i.e. the ionization energy. K(τ ) = (ne− )1 exp [−Fint (e− )/τ ] · (nH + )1 exp [−Fint (H + )/τ ](nH )−1 exp (−(−1)(Fint (H)/τ )) Note that nH + ' nH . Let ne− = nQ . Importantly, note Fint (e− ) + Fint (H + ) − Fint (H) = I Fint (H) is at a lower free energy than e− and H + . =⇒ K(τ ) = nQ e−I/τ =⇒
[e][H + ] = nQ e−I/τ [H]
(b) By charge conservation, [e] = [H + ], so that 1/2
[e] = [H]1/2 nQ exp (−I/2τ ) Given [H] ' 1023 cm−3 , me = 0.511 M eV /c2 , T = 5000 K, I = 13.6 eV ionization energy, [e] = (1023 cm−3 )1/2 (2.92 × 1010 1/cm3/2 )1/2 exp (−13.6 eV /2kB 5000 K) = 1.3 × 1015 cm−3 Note that H(exc) and H are just two different states of atomic hydrogen. Their concentrations must therefore be proportional to the probability of occurrence of these states, and the ratio of probabilities is the ratio of the respective Boltzmann [H(exc)] p(H(exc)) = [H] p(H) If H(exc) is the internal energy of the first excited state and H is the internal energy of the ground state of atomic hydrogen, we are given that H(exc) − H = 34 I. We also need to take into account the fact that the first excited electronic state of hydrogen is 4-fold degenerate i.e. one 2s-orbital and three 2p-orbitals.1 Therefore, [H(exc)] p(H(exc)) 4e−H(exc) /τ = = = 4e−3I/4τ [H] p(H) e−H /τ [H(exc)] = 4[H]e−3I/4τ = 2.092 × 1013 cm−3 1(from solutions to Homework 8, Ph12c, Caltech, June 6, 2008, by Prabha Mandayam, Heywood Tam) 35
[e] = 62 [H(exc)] Solution 4. Biopolymer growth.
Recall that G(N, p, τ ) P = N µ(p, τ ), since G was chosen to be an extensive quantity (it scales with size). For more than one chemical species G = j Nj µj . dF = 0 for equilibrium, for constant P, τ . µj = chemical potential of species j, µj = (∂G/∂Nj )τ,p . P Given i νj Aj , e.g. H2 + Cl2 = 2HCl, P ˆ where dNj = νj dN ˆ , dG = 0 → P νj µj = 0. dG = ( j νj µj )dN j Recall the mass action law derivation: assume constituents act as ideal gases; µj = τ (ln nj − ln cj ), nj concentration of species j; cj ≡ nQj Zj (int). X
νj ln nj =
j
X
νj ln cj =⇒
j
X
ν
ln nj j =
j
ν
ln cj j = ln
j
ν
Y
X
nj j = K(τ )
Y
ν
nj j = ln K(τ )
j
mass action law
j
(a) By mass action law,
[ monomer][N mer] [(N +1)mer]
=
[1][N ] [N +1]
= KN .
[1]3 [1]2 [1][2] = = K1 K2 [2] [3] [3]
[1][1] = K1 [2]
=⇒ [N + 1] = [1]N +1 /K1 K2 K3 . . . KN
j j+1 Y [1]j+2 [1]j+1 [1][j + 1] Y = Kl Kj+1 = = Kl [j + 1] [j + 2] [j + 2] l=1
(b) Recall that K(τ ) = KN
Q
l=1
ν
j
nQjj exp [−νj Fj (int)/τ ]
nQ (N )nQ (1) −FN F1 −(FN + F1 − FN +1 ) FN +1 nQ (N )nQ (1) = exp − exp exp = nQ (N + 1) τ τ τ nQ (N + 1) τ
3/2 Nτ where nQ (N ) = M and MN is the mass of N mer molecules, FN is the free energy of one N mer molecule. 2π~2 (c) Assume N 1 so nQ (N ) ' nQ (N + 1). Assume [1] = 1020 cm−3 . Assume ∆F = FN +1 − FN − F1 = 0, meaning zero free energy change in the basic reaction step. We’re given the molecular weight of the monomer to be 200. M1 τ 3/2 We want [N[N+1] . ] at room temperature. Now KN ' nQ (1) = 2π~2 [1][N ] = nQ (1) = [N + 1] Note that
M1 τ 2π~2
3/2 or
[1] [N + 1] = = [N ] nQ (1)
2π(6.582×10−22 M eV ·s)2 200(938M eV /c2 )(0.8617×10−4 eV /K)(298 K)
=⇒
3×1010 cm/s 1c
2 3/2
2π~2 M1 τ
3/2
= 0.3627 × 10−27 cm.
[N + 1] = 3.627 × 10−8 [N ]
(d) We want the condition 1<
[N + 1] [1] = exp [N ] nQ (1) =⇒ ∆F < τ ln
−∆F τ
or ln
nQ (1) −∆F < [1] τ
[1] = −0.44 eV nQ (1) 36
1020 cm−3
10. P HASE T RANSFORMATIONS 11. B INARY M IXTURES 12. C RYOGENICS 13. S EMICONDUCTOR S TATISTICS 14. KINETIC T HEORY 15. P ROPAGATION Heat Conduction Equation. nonrelativistic case: Let manifold N = R × M , with dimM = n. Let J ∈ X(N ) = X(R × M ) be a vector field in N . ∞ Let ρ ∈ C (N ) be a smooth function on N . ρ = ρ(t, x) locally Let J ∈ Ω1 (N ) be a 1-form on N that is isomorphic to J (Tangent-Cotangent isomorphism theorem), i.e.
J = J[ Ji = gij Jj with gij being the metric on N (not just M !) Note that as N = R × M , g0j = δ0j The local form of J is the following: J=ρ
∂ ∂ + ji i ∂t ∂x
i = 1...n
So J = Ji dxi = gij Jj dxi = ρdt + jk dxk
k = 1...n
Thus J = ρdt + jk dxk
(21)
k = 1...n
Now do the Hodge star operator, resulting in a n-form ∗J ∈ Ωn (N ) and so ∗J = ρ ∗ dt + jk ∗ dxk = ρvoln + ij voln+1 as
√
g 0 dxi ∧ · · · ∧ dxin i1 . . . in ∈ {1 . . . n} n! i1 ...in√ 1 g k dxi1 ∧ · · · ∧ dxin = ij voln+1 jk ∗ dxk = gkl j l (n + 1)! i1 ...in
ρ ∗ dt = ρ
so thus ∗J = ρvoln + ij voln+1 Hence d∗J =
(22) Special case:
√ ∂ g ∂t
∂ √ 1 ∂ √ 1 (ρ g) √ voln+1 + ( gjk ) √ voln+1 = d(ρvoln ) + dij voln+1 k ∂t g ∂x g
=0 ∂ ∂jk √ ln g jk voln+1 + voln+1 = 0 k ∂x ∂xk ∂ ∂jk ∂ρ √ + ln =0 g jk + =⇒ ∂t ∂xk ∂xk
∂ρ n+1 d∗J = vol + ∂t
√ ∂jk Special case: if g constant, ∂ρ ∂t + ∂xk = 0 Let j ≡ −Ddρ (j is a closed form on M ) where dρ =
∂ρ i ∂xi dx
i = 1 . . . n, D constant
d ∗ J = dρvoln + −Dd ∗ dρ = 0 37
For flat metric,
√
g constant, ∂ρ ∂2ρ =D ∂t (∂xk )2
For relativistic case, Consider manifold M , with dimension dimM = n + 1 J=ρ
∂ ∂ + ji i ∂t ∂x
Let J = J[ . Jµ = gµν Jν For special case of flat Minkowski space, J = −ρdt + ji dxi ∗J = −ρ ∗ dt + ji ∗ dxi √ √ g g 0 i1 in −ρ dx ∧ · · · ∧ dx + ji i dxµ1 ∧ · · · ∧ dxµn (n + 1)! i1 ...in (n + 1)! µ1 ...µn √ √ ∂(ρ g) 1 ∂(ji g) 1 n+1 d∗J =− + √ vol √ voln+1 = 0 ∂t g ∂xi g √ √ ∂(ρ g) ∂(ji g) =0 =⇒ − + ∂t ∂xi Fick law (14.19) for particle flux density, j = −Dn dn where Dn n
particle diffusivity constant particle concentration
J = ndt + j b thermal conductivity; homogeneous medium C heat capacity per unit volume. ju = −Kdτ b dt + ju J = Cτ (23)
(24)
b ∂τ + ∂(ju )k = 0 C ∂t ∂xk ∂τ ∂2τ = Dτ ∂t (∂xk )2
(5)
b Dτ ≡ K/C
(6)
Propagation of Sound Waves in Gases. pressure associated with sound wave (25)
δp = δp0 exp [i(kx − ωt)]
(27)
Suppose ideal gas: (26)
pV = N τ or p = ρτ /M
(28)
Consider “solid ball” or “billiard ball” particle (extended particle, not pt. particle, but no internal structure) NM ρ= V Force on particle Z Z Z dP d M M ∂u n F = = ρvol u = L ∂ +u u = + [u, u] ∂t dt dt V V ∂t Suppose [u, u] = 0 (certainly for flat spaces; what about for curved spaces? [u, u] 6= 0? Possibly? I don’t know. EY: 20150317 (27)
dU + pdV = τ dσ
define fractional deviations s, θ (28)
ρ = ρ0 (1 + s) τ = τ0 (1 + θ)
where ρ0 , τ0 are density and temperature in absence of sound wave. assume u, s, θ have form of traveling exp [i(kx − ωt)] 38
(5)
(30)
kτ kτ ωu − s− θ=0 M M ωs − ku = 0 (40)
(31)
bV θ − ps = 0 or C bV θ − ns = 0 τC
(29)
(39)
(41)
kτ n ωu = 1+ s bV M C kτ n k ω= 1+ b M CV ω So ω=
(32)
γτ 1/2 M
k
(42)
bV + n bp C C = bV bV C C ∂ω γτ 1/2 vs = = ∂k M γ=
Problems. Problem 1. Fourier analysis of pulse t=0
θ(x, 0) = δ(x) =
(33)
θ(x, t) =
(34)
1 2π
Z
1 2π
Z
∞
dk exp (ikx)
(58)
dk exp [i(kx − ωt)]
(59)
−∞
∞
−∞
Given a dispersion relation at this form: Dk 2 = iω
(35)
1 θ(x, t) = 2π
(36)
Z
(10)
∞
dk exp [ikx − Dk 2 t]
(60)
−∞
and so, doing the Gaussian integral, 1 θ(x, t) = √ exp 4πDt
(37)
−x2 4Dt
(14)
Problem Diffusion in two and three dimensions.
(a) θ2 r 2 θ2 ∂θ2 =− + ∂t t 4 Dt2 2 ∂ θ2 1 θ2 1 θ2 1 θ2 =− − + (x2 + y 2 ) (∂xi )2 2 Dt 2 Dt 4 D2 t2 =⇒
∂θ2 ∂ 2 θ2 =D ∂t (∂xi )2
(b) (c) 39
R EFERENCES [1] [2] [3] [4]
Charles Kittel, Herbert Kroemer, Thermal Physics, W. H. Freeman; Second Edition edition, 1980. ISBN-13: 978-0716710882 Bernard F. Schutz, Geometrical Methods of Mathematical Physics, Cambridge University Press, 1980. ISBN-13: 978-0521298872 Paul Bamberg and Shlomo Sternberg, A Course in Mathematics for students of physics: 2, Cambridge University Press, 1990. T. Frankel, The Geometry of Physics, Cambridge University Press, Second Edition, 2004.
There is a Third Edition of T. Frankel’s The Geometry of Physics [4], but I don’t have the funds to purchase the book (about $ 71 US dollars, with sales tax). It would be nice to have the hardcopy text to see new updates and to use for research, as the second edition allowed me to formulate fluid mechanics and elasticity in a covariant manner. Please help me out and donate at ernestyalumni.tilt.com or at subscription based Patreon, patreon.com/ernestyalumni. E-mail address:
[email protected] URL: http://ernestyalumni.wordpress.com
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