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, so drop will evaporate (to gas) spontaneously when R > Rc , ∆G < 0, drop will tend to grow spontaneously (∆G)c =
γ3 16π [ 2 ] 3 nl (∆µ)2
where (∆G)c = ∆G(R = Rc ) assume vapor behaves like ideal gas:
p ∆µ = τ ln peq Eq. (12a), chemical potential of the ideal gas, µ = τ ln nnQ = τ ln ppeq p ≡ vapor pressure in gas phase peq ≡ equilibrium vapor pressure of bulk liquid (R → ∞) erg 2γ −6 use γ = 72 cm cm. 2 , p = 1.1peq , τ = 300 K (H2 O), Rc = n ∆µ = 1 × 10 l Problems. Problem 1. Entropy, energy, and enthalpy of van der Waals gas. (a) Recall the generalized thermodynamic identity; starting with σ = σ(U, V, N ) ∈ C ∞ (Σ) = C ∞ ((U, V, N )), the identity is dU = τ dσ − pdV + µdN . By definition, F = U − τ σ, so that dF = dU − dτ σ − τ dσ = −pdV + µdN − σdτ Then clearly, Now
∂F ∂τ
V,N
= −σ. F (vdW) = −N τ (ln [nQ (V − N b)/N ] + 1) − N 2 a/V
as was postulated about the hardcore repulsion inside a molecule, making V into V − N b. M τ 3/2 Remembering that quantum concentration does depend on τ : nQ ≡ 2π~ 2 Consider ! ! ! 3/2 3/2 ∂ Mτ (V − N b) 3 M (V − N b) 3 ∂τ ln ( ) + 1 = ln τ + ln + 1 − − → 2π~2 N 2 2π~2 N 2τ Thus
nQ (V − N b) 3 nQ (V − N b) 5 ln + 1 + N = N ln + N 2 N 2 V,N F 2 ∂ (b) Recall that F := U − τ σ or U = F + τ σ = F − τ ∂F ∂τ V,N = −τ ∂τ τ V,N σ=−
∂F ∂τ
=N
U = −τ 2
3 N 2a 3N τ N 2a (−N ) + = + =U 2τ V 2 V 43
(c) Give to first order in the van der Waals correction terms a, b. p=
Nτ Nτ Nτ N 2a N 2a Nb N 2a − 2 = − 2 = (1 + )− 2 V − Nb V V (1 − N b/V ) V V V V
where N b V . Thus, H(τ, V ) := U + pV =
5 N 2 bτ 2N 2 a Nτ + − 2 V V
Problem 2. Calculation of dT /dp for water Using the Clausius-Clapeyron equation or vapor pressure equation,
dp L = dτ τ ∆v assume ∆v := vg − vl ≈ vg ≡ or
Vg Ng
=
τ p
Vg Ng
(i.e. the volume of the gas is much bigger than the volume of the liquid), and pVg = Ng τ
(i.e. water vapor acts like an ideal gas)
L L ≈ 2 τ ∆v τ p Looking up fundamental constants and unit conversions manually appears, to me at least, in the 21st century, to be an obsolete exercise (“an exercise in insanity?”). I would like to propose presenting the fundamental constants and unit conversions, sourced from the NIST (National Institute of Standards and Technology) website, and using Python library pandas, as panda dataframe objects, to make the data accessible and computable, and persisting them or databasing the values as panda dataframe objects. Open up thermo.py, which relies on the Physique repository (I’ll put it up on my github, ernestyalumni). N A = F u n d C o n s t [ F u n d C o n s t [ ” Q u a n t i t y ” ] . s t r . c o n t a i n s ( ” Avogadro ” ) ] . l o c [ 4 2 , : ] k BOLTZ = F u n d C o n s t [ F u n d C o n s t [ ” Q u a n t i t y ” ] . s t r . c o n t a i n s ( ” B o l t z m a n n ” ) ] . l o c [ 4 9 , : ] P frmATM = conv [ conv [ ” T o c o n v e r t f r o m ” ] . s t r . c o n t a i n s ( ” atm ” ) ] . l o c [ 1 5 , : ] # 1 atm t o P a s c a l c o n v e r s i o n f o r p r e s s u r e ... ## 2 . C a l c u l a t i o n o f $dT / dp$ f o r w a t e r L = 2260 # J gˆ{−1} b o i l i n g w a t e r t e m p K = KCconv . s u b s ( T C , 1 0 0 ) . r h s # room t e m p e r a t u r e i n K e l v i n t a u b o i l i n g w a t e r = b o i l i n g w a t e r t e m p K ∗ k BOLTZ . V a l u e P = 1 # atm d p d t a u 1 0 0 2 = L ∗ 1 8 . 0 1 5 3 / f l o a t ( N A . V a l u e ) / ( t a u b o i l i n g w a t e r ∗∗2 ) dtaudp 1002 = 1 . / dpdtau 1002 # in J / Pascal dTdp 1002 = d t a u d p 1 0 0 2 / ( k BOLTZ . V a l u e ) # 2 8 . 4 3 4 8 5 3 5 1 1 1 2 6 2 K / atm
dT = 28.43K/atm dp Problem 3. Heat of vaporization of ice Given P (H2 O vapor)(T = −2◦ C) = 3.88 mm Hg P (H2 O vapor)(T = 0◦ C) = 4.58 mm Hg Use the Clausius-Clapeyron equation or vapor pressure equation (for coexistence equilibrium). dp L L = ≈ 2 dτ τ ∆v τ /p dp ∆p p(τ2 ) − p(τ1 ) with ≈ = dτ ∆τ τ2 − τ1 N Now idealizing water vapor as a perfect ideal gas, p = V τ , so p linear in τ , i.e. p = p(τ ). Note that while p is linear in τ , the given temperature values correspond to T , in K, and so one would need this form of the slope equation y = mx + b, where m = N V times a bunch of conversion factors, and one must also include a nonzero b constant, the result of choice of units (celsius versus energy units). dp τ 2 Thus, L = dτ p , with L being the latent heat of vaporization per particle. Note, I treat Avogadro’s number as a units conversion, 1 = NA = 6.02205 × 1023 J/mol, multiply L by NA , LNA . 44
particles mol
. So in the end, for the desired
L1003 = ( 4 . 5 8 − 3 . 8 8 ) / ( 0 − ( −2.) ) ∗ ( ( KCconv . s u b s ( T C , 1 . ) . r h s ) ∗ ∗ 2 ) / ( ( 4 . 5 8 − 3 . 8 8 ) / ( 0 − ( −2)) ∗ ( 1 . ) + 3 . 8 8 ) ∗ k BOLTZ . V a l u e ∗N A . V a l u e # 5 1 7 0 5 . 6 7 5 7 6 4 0 4 8 5
L = 5.17 × 104 J/mol Problem 4. Gas-solid equilibrium
(a) Let’s review the quantum harmonic oscillator (qho). Recall the Hamiltonian in d dimensions 2 b = pb + 1 mω 2 x H b2 2m 2 ∂ where pb = −i ∂x i ei . Suppose for the Hilbert space H, that H = Hx1 ⊗ Hx2 ⊗ Hx3 ,
|ψi ∈ H = Hx1 ⊗ Hx2 ⊗ Hx3 =⇒ |ψi = |ψx1 i ⊗ |ψx2 i ⊗ |ψx3 i ≡ |ψx1 i|ψx2 i|ψx3 i Now b H|ψi =
d X i=1
−1 2m
∂ ∂xi
2
1 + mω 2 (b xi )2 2
! |ψi ≡
−1 2m
∂ ∂xi
2
1 + mω 2 (b xi )2 2
! |ψi
The ladder operators make it much more simpler and straightforward than directly solving the differential equation. Let r r mω i 1 i x bi + pbi ai = (ai + a†i ) x b = 2 mω 2mω r r mω i mω † x bi − pbi pbi = i (ai − ai ) a†i = 2 mω 2 Now let Ni = a†i ai =
mω (b pi )2 1 ((b xi )2 + )− 2 2 (mω) 2
where the uncertainty principle commutator relation was used: x bi pbi − pbi x bi = [b xi , pbi ] = i Thus b = (N1 + 1 )ω + (N2 + 1 )ω + (N3 + 1 )ω H 2 2 2 1 1 1 b H|ψi = ((nx + )ω + (ny + )ω + (nz + )ω)|ψi 2 2 2 Consider Zs =
X nx ,ny ,nz
1 1 1 exp −β((nx + )ω + (ny + )ω + (nz + )ω − 0 ) = eβ0 2 2 2
!d X (e−βω )n n
=e
β0
1 1 − e−βω
d
The derivation of p = p(τ ) goes, as above (with Example: Model system for gas-solid equilibrium (pp.285 Ch. 10 “Phase transformations”, Kittel and Kroemer (1980) [1])); recall n p λg = λs , λ g = = nQ τ nQ and so for τ ~ω or 1 βω, 3/2 3/2 M M 1 5/2 −β0 d d 5/2 −β0 −βω d ∼ p= τ e (1 − e ) = τ e β ω 2 2π~ 2π ~3 For d = 3, p=
M 2π
3/2
45
ω 3 −β0 e τ 1/2
(b) From above part (a), dp 1 −1 −1 −1 ) = τ + (−0 dτ p 2 τ2 Now using the approximation that ∆v ≈ vg , and the ideal gas idealization, L=
dp τ 2 τ = 0 − dτ p 2
My explanation: we must excite the atom by its (bounded) ground state energy, 0 . On the other hand, thermal energy of gas, with increasing linear temperature, helps to remove atom from solid to gas. Problem 5. Gas-solid equilibrium (a) Now Z Ns Fs = −τ ln s = −τ (Ns ln Zs − ln Ns !) ' −τ (Ns ln Zs − Ns ln Ns + Ns ) Ns ! where Stirling’s approximation was used and (1/Ns !) factor is for indistinguishability (from quantum mechanics). Recall eβ0 Zs = so ln Zs = β0 − d ln (1 − e−βω ) (1 − e−βω )d from Example: Model system for gas-solid equilibrium (pp.285 Ch. 10 “Phase transformations”, Kittel and Kroemer (1980) [1]). Then Fs = −Ns 0 + Ns τ d ln (1 − e−βω ) + τ Ns ln Ns − Ns τ
(24)
Recall this useful relation (it’s easy to calculate σ from Helmholtz free energy F ): ∂F =σ ∂τ V and so ∂Fs = Ns d ln (1 − e−ω/τ ) + Ns ln Ns − Ns + Ns τ d ∂τ
(25)
= Ns d ln (1 − e−βω ) + Ns ln Ns − Ns − Ns dβω
1 1 − e−βω
(−e
−βω
)(−ω)
−1 τ
=
e−βω =σ 1 − e−βω
Suppose τ ω or 1 βω. Then σ = Ns d ln (1 − e−βω ) + Ns ln Ns − Ns (1 + d). So take the (given) “extreme assumption that the entropy of the solid maybe neglected over the temperature range of interest.” [1] Then comparing σ from Eq. 25 and Fs from Eq. 24, then Fs ' −Ns 0 Ng F = Fs + Fg = −Ns 0 + N τ ln −1 V nQ (b) Now F = −Ns 0 + Ng τ [ln
Ng V nQ
Ng − 1] = −0 (N − Ng ) + Ng τ ln −1 V nQ
Now, taking the partial derivative, ∂F = 0 + τ ∂Ng
Ng 1 ln − 1 + Ng τ V nQ Ng
Taking the minimum free energy with respect to Ng , i.e. set the partial derivative to 0, ∂F Ng −0 = 0 = 0 + τ ln or Ng = nQ V exp ∂Ng V nQ τ (c) Idealize the gas with the ideal gas law. pV = Ng τ =⇒ pg = nQ τ exp
−0 τ
=
46
M 2π~2
3/2 τ
5/2
exp
−0 τ
= pg
11. B INARY M IXTURES 12. C RYOGENICS 13. S EMICONDUCTOR S TATISTICS 14. K INETIC T HEORY Kinetic Theory of the Ideal Gas Law. Consider molecule strike unit area of wall. Let vz ≡ velocity component normal to plane of wall. Suppose molecules, of mass M , reflected specularly (mirror-like) from wall, ∆pz = −2M |vz | Let a(vz )dvz , number of molecules per unit volume with z-component of velocity between vz and vz + dvz . Z N a(vz )dvz = =n V a(vz )vz dvz number of molecules in (vz , vz + dvz ) velocity range that strike unit area of wall in (per) unit time Z ∞ Z ∞ pressure p = 2M vz a(vz )vz dvz = M vz2 a(vz )dvz = M nhvz2 i −∞
0 1 2 2 M hvz i
=
1 2τ
by equipartition of energy (Ch.3) p = nM hvz2 i = nτ =
Nτ V
Maxwell Distribution of Velocities. cf. Ch.6. distribution function of ideal gas f (n ) = λ exp −τ n Recall, Ch. 6, Sec. “Classical Limit” of Kittel and Kroemer [1], an ideal gas is defined as a system of free noninteracting particles in the classical regime. f () ≡ average occupancy of an orbital at energy ≡ energy of orbital occupied by 1 particle; not energy of system of N particles 1 Fermi-Dirac and Bose-Einstein distribution f () = exp [(−µ)/τ ]±1 In order for f () 1 ∀ orbitals, exp [( − µ)/τ ] 1 ∀ . µ =⇒ f () ' exp [(µ − )/τ ] = λ exp (−/τ ) λ ≡ exp τ f (), average occupancy of orbital of energy , is classical distribution function. 2 1 πn 2 1 −1 2 1 (for, recall 2M ψ = 2M ∂ ψ = Eψ) particle in a box: n = 2M L i∂ number of orbitals in range of quantum number (n, n + dn), probability such orbital is occupied 1 1 ( πn2 dn)f (n ) = πn2 λ exp (−n /τ )dn 2 2 1 1 πn 2 (M L)2 2 M Lv M v2 = or n2 = v or n = 2 2M L π2 π Consider system of N particles in volume V . Let N P (v)dv number of atoms with velocity magnitude in range dv at v 3 1 dn 1 ML −M v 2 N P (v)dv = πn2 λ exp (−n /τ ) dv = πλ v 2 exp dv 2 dv 2 π 2τ 2 3/2 cf. Ch. 6, λ = nnQ = LN3 2π~ Mτ 1 N π 2 L3
2π Mτ
3/2
(26)
=⇒ P (v) = 4π
ML π
M 2πτ
3
3/2
= 4πN
2
v exp
N 2πτ
3/2
−M v 2 2τ
P (v) is Maxwell velocity distribution, P (v)dv is probability particle has speed in dv at v 47
Experimental verification. velocity distribution of atoms which exit from slit of oven. exit beam weighted in favor of atoms of high velocity at expense of those at low velocity. weight factor is velocity component v cos θ normal to plane of hole. π/2 Z 2π 3 sin 2θdθ 2πR3 − cos 2θ = cos θ sin θdθ = ( R ) = 3 2 3 4 0 2πR3 1 + 1 2πR3 1 = = 3 4 3 2 Probability atom leaves hole will have velocity in (v, v + dv) : Pbeam (v)dv Z
1 (cos θ)drrdϕr sin θdθ = ( (2π)R3 ) 3
Z
−M v 2 Pbeam (v) ∝ vPMaxwell ∝ v 3 exp 2τ 2 3/2 M v with, recall PMaxwell = 4π 2πτ v 2 exp −M 2τ Pbeam distribution of transmission through a hole is Maxwell transmission distribution. π/2
Z Z hvout i =
3/2 Z ∞ M −M 2 1 v 3 exp v dv = 2πτ 2 0 2τ 3/2 2 1 1 4τ 1 1 M 21/2 τ 1/2 0− = 4π = 1/2 2 2 −2α α 2πτ 4 M M π
v cos θ sin θdθPMaxwell (v)dv = 4π 0
3/2 M 2πτ for (doing the integration by hand)
= 4π
2
(e−αv )0 = −2αve−αv
2
2
2
(v 2 e−αv )0 = −2αv 3 e−αv + 2ve−αv
2
2
2
(v 2 e−αv +
2 e−αv 0 ) = −2αv 3 e−αv α
Particle Diffusion. pp. 399 of Kittel and Kroemer [1] Consider a system. One end in diffusive contact with reservoir at chemical potential µ1 Other end in diffusive contact with reservoir at chemical potential µ2 Constant temperature τ . If µ1 > µ2 , particle flow through system from reservoir 1 to reservoir 2; 1 → 2. ni ≡ particle concentration in i Take jn = −Dgradn
(27)
which is Fick’s law, and where D ≡ particle diffusion constant or diffusivity. Mean free path l. Particles freely travel over l. Assume in a collision at z, particles come into local equilibrium at local chemical potential µ(z), local concentration n(z). 1 At z, particle flux density in positive z direction 2 n(z − lz )cz particle flux density in negative z direction − 12 n(z + lz )cz Note n(z − lz ) is particle concentration at z − lz Jnz =
1 dn [n(z − lz ) − n(z + lz )] cz = − cz lz 2 dz
where cz = c cos θ lz = l cos θ R hcz lz i = cl
hemisphere
R π/2
cos2 θdS
R hemisphere
dS
= cl
0
R 2π dθ 0 dϕ cos2 θ sin θdθ 1 = cl R π/2 R 2π 3 dθ sin θ 0 dϕ 0
Comparing with Fick’s law, Jnz =
−1 dn 1 cl or JN = − cl∇n 3 dz 3
For diffusivity D is then D = 31 cl. 48
Now recall that l, the mean free path, was derived from kinetic theory: 1 l= nπd2 where d is the diameter of the particle. The mean thermal velocity was derived from the Maxwell distribution: 1/2 8τ c= Mπ Problems. Problem 1. Mean speeds in a Maxwellian distribution. v2 M 3/2 2 v exp −M (a) P (v) = 4π 2πτ 2τ Now 0 Z ∞ Z ∞ Z Z ∞ 2 exp (−αv 2 ) 3 2 3 4 2 2 exp (−αv ) v dvv exp (−αv ) = = 0 − 3v = v exp (−αv 2 ) = −2α −2α 2α 0 0 0 !0 " √ Z Z −αv2 # 2 π 3 e−αv 3 e 3 √ = v = 0− = 2α −2α 2α −2α 4α2 2 α So 2
∞
Z
2
hv i =
v P (v) = 4π 0
M 2πτ
√ π 1 3 1 3τ = M = q M 2 2 2 M M 4 2τ 2τ
3/2
3
2τ
so vrms =
3τ M
1/2
Now hv 2 i = 3hvx2 i =
τ 1/2 3τ so hvx2 i1/2 = M M
(b) ∂P (v) 2P (v) = + P (v) ∂v v
−M τ
r 2 M 2τ v = P (v) − v = 0 if vmp = 0 or vmp = v τ M
where mp stands for most probable. r vmp =
2τ < M
r
3τ = vrms M
(c) !0 3/2 3/2 Z ∞ M v2 −M v 2 M M e− 2τ 3 2 c= dvvP (v) =: hvi = v exp = 4π = dv4π dvv −M 2πτ 2τ 2πτ 0 0 0 τ # ! ∞ # 3/2 " 3/2 " 3/2 2 Z ∞ M v2 M v2 M e− 2τ M 2τ e− 2τ M 2τ = 4π 0− dv2v −M = 4π = 4π = M 2πτ 2πτ M 2πτ M2 −τ 0 τ 0 r r 1 τ 8τ = = 8π 3/2 M πM (2π) Z
∞
Z
∞
Note vrms = c
3τ 1/2 M 8τ 1/2 πM
=
3π 8
1/2
(d) |vz | = |v cos θ| = v| cos θ|. For for a fixed θ, v cos θ is 1 point on a sphere, whereas v| cos θ| could be 2 pts on a sphere. Now Z π Z π/2 Z π Z π/2 Z π sin 2θdθ − sin 2θdθ | cos θ| sin θdθ = cos θ sin θdθ + − cos θ sin θdθ = + = 2 2 0 0 π/2 0 π/2 π/2 π cos 2θdθ cos 2θ 1 1+1 = + = + =1 −4 4 2 4 0 π/2 49
P (v)dv is probability that particle has speed or velocity magnitude in (v, v + dv). R π/2 It counts the event of v cos θ and v cos (π − θ), so it counts v| cos θ| twice. P (v)dv = 0 dθ2P (v| cos θ|) 1/2 Z πZ ∞ P (v)dv 2τ v =⇒ cz = h|vz |i = h|v cos θ|i = hv| cos θ|i = | cos θ|dθ = 2 πM 0 0 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
(28)
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 jk ∗ dxk = gkl j l k dxi1 ∧ · · · ∧ dxin = ij voln+1 (n + 1)! i1 ...in
ρ ∗ dt = ρ
so thus ∗J = ρvoln + ij voln+1 Hence d∗J =
(29) Special case:
√ ∂ g ∂t
∂ √ 1 ∂ √ 1 (ρ g) √ voln+1 + ( gjk ) √ voln+1 = d(ρvoln ) + dij voln+1 ∂t g ∂xk g
=0 ∂ ∂jk ∂ρ n+1 √ vol + ln g jk voln+1 + voln+1 = 0 d∗J = k ∂t ∂x ∂xk ∂ρ ∂ ∂jk √ =⇒ + ln g jk + =0 ∂t ∂xk ∂xk
√ ∂jk Special case: if g constant, ∂ρ ∂t + ∂xk = 0 Let j ≡ −Ddρ (j is a closed form on M ) where dρ =
For flat metric,
√
∂ρ i ∂xi dx
i = 1 . . . n, D constant
d ∗ J = dρvoln + −Dd ∗ dρ = 0 g constant, ∂ρ ∂2ρ =D ∂t (∂xk )2
For relativistic case, 50
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 −ρ 0i1 ...in dxi1 ∧ · · · ∧ dxin + ji i dxµ1 ∧ · · · ∧ dxµn (n + 1)! (n + 1)! µ1 ...µn √ √ ∂(ρ g) 1 ∂(ji g) 1 d∗J =− √ voln+1 + √ 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 b ∂τ + ∂(ju )k = 0 C ∂t ∂xk
(30)
∂2τ ∂τ = Dτ ∂t (∂xk )2
(31)
(5)
b Dτ ≡ K/C
(6)
Propagation of Sound Waves in Gases. pressure associated with sound wave δp = δp0 exp [i(kx − ωt)]
(32)
(27)
Suppose ideal gas: pV = N τ or p = ρτ /M
(33)
(28)
Consider “solid ball” or “billiard ball” particle (extended particle, not pt. particle, but no internal structure) ρ=
NM V
Force on particle Z Z Z dP d M M ∂u = ρvoln 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 F =
(34)
dU + pdV = τ dσ
define fractional deviations s, θ
(35)
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)] 51
(5)
(37)
kτ kτ ωu − s− θ=0 M M ωs − ku = 0 (40)
(38)
bV θ − ps = 0 or C bV θ − ns = 0 τC
(36)
(39)
(41)
kτ n ωu = 1+ s bV M C kτ n k ω= 1+ b M CV ω So ω=
(39)
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) =
(40)
θ(x, t) =
(41)
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
(42)
1 θ(x, t) = 2π
(43)
Z
(10)
∞
dk exp [ikx − Dk 2 t]
(60)
−∞
and so, doing the Gaussian integral, 1 θ(x, t) = √ exp 4πDt
(44)
−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) 52
R EFERENCES [1] [2] [3] [4] [5]
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. Ed Groth. Physics 301, Thermal Physics, Fall 1999, Fall 2002, Fall 2003, Fall 2004, Fall 2005, http://grothserver.princeton.edu/ ˜groth/phys301f99/
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. E-mail address:
[email protected] URL: http://ernestyalumni.wordpress.com
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