Statistical Mechanics & Thermodynamics 2: Physical Kinetics

Statistical Mechanics & Thermodynamics 2:
Physical Kinetics
Compiled by Inon Sharony
Spring 2006-2008-2009
Class notes from the course by Professor Roman Mints
1.
1Prof. Mints's Homepage
mints@post.tau.ac.il
Oce adress: Shenkar (Physics) Building, Room 414
Oce phone: 9165
1

Contents
1 Brownian Motion 5
1.1 Drunken Walk The Random Walk as a Stochastic Process . . . . . . . . . . . . . . . . . . . . . 5
1.2 Probability Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Probability Distribution for Many Steps (Large N) on a Lattice . . . . . . . . . . . . . . . . . . . 5
1.4 Diusion Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Diusion Equation as a Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Langevin Equation 9
2.1 One-Dimensional Ballistic Motion and Diusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Classical and Quantum Einstein Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2.1 Ideal Classical Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2.2 Degenerate Fermion Gas (DEG) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Poiseuille Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 Diusion in Ambipolar Plasma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.5 Debye-Hückel Screening (1923) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.5.1 The Screening Length I in Dierent Materials . . . . . . . . . . . . . . . . . . . . . . . 17
3 Self Heating Phenomena 19
3.1 Heat Diusion Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 Nonlinear Stationary States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.2.1 Current Carrying Superconducting Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.2.2 Chemical Reactions on the Surface of a Catalyst . . . . . . . . . . . . . . . . . . . . . . . 27
3.2.3 Metal - Dielectric Phase Transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.3 Linear Stability of Nonlinear Stationary States . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.4 Propagation of Single Nonlinear Switching Waves . . . . . . . . . . . . . . . . . . . . . . . . . 31
4 Diusion in Momentum-Space 36
4.1 Electron-Phonon Collisions in Metals at Low Temperatures . . . . . . . . . . . . . . . . . . . . . 36
4.2 Relaxation of Heavy Particles in a Gas of Light Particles . . . . . . . . . . . . . . . . . . . . . . . 37
4.3 Fokker-Planck Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5 Boltzmann Equation 41
5.1 Liouville's theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
5.2 Boltzmann equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.3 collision integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.4 -approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.5 heat conductivity and viscosity of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6 Kinetics of a Degenerate Electron Gas 49
6.1 Electrical and thermal conductivities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
6.2 Wiedemann-Franz law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
6.3 Skin-eect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
6.4 Electrical Conductivity in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
6.5 Onsager relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
6.6 Thermo-Electrical Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
7 Master Equation 66
7.1 Magnetic resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
7.2 Överhauser eect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
2

8 Fluctuation-Dissipation Theorem 71
8.1 Harmonic analyses of Langevin equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.2 velocity-velocity and force-force correlation functions . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.3 power spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.4 classical and quantum limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
8.5 Nyquist formulas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
9 Damping in Collisionless Plasma 76
9.1 Self-consistent eld and collision-less approximations . . . . . . . . . . . . . . . . . . . . . . . . . 76
9.2 Vlasov equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
9.3 plasma waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
9.4 Langmuir plasma frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
9.5 Landau damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3

1 Brownian Motion
1.1 Drunken Walk The Random Walk as a Stochastic Process
A particle moving in one dimension, with probability p to take a step in the right direction during any given
time-step, and q to take a step in the left direction. The step length is a.
m steps are made in the right direction and n steps are made in the left direction. Many steps are taken, so
that m Cn N ) I.
1.2 Probability Distribution
What is the probability that such a course is taken (dened by m and n)?
W @m;nA pmqn N3
n3m3

N
n

pmqn
Nˆ
naH

N
n

pN nqn a @p CqAN a IN a I
Where in the second line we used the expansion for the Newton binomial.
1.3 Probability Distribution for Many Steps (Large N) on a Lattice
The average number of steps right can be designated m pN, and likewise to the left n qN. We can write
m a m Cx
n a n x
Where x is the deviation from the average number of steps. We can also dene the total deviation as `
N @p qACPx.
Assuming x ( n a qN a @I pAN and also x ( m a pN, we will look at
W
H
dqN x| {z }
n
;pN Cx| {z }
m
I
e a qqN xppNCx ¡ N3
@qN xA3@pN CxA3
Using the Stirling formula ln@N3A a N ¡

ln@NA ICO
¢ I
N
£¡
lnW @n;mA a nlnq Cmlnp Cln@N3A ln@n3A ln@m3A
9 nlnq Cmlnp CN lnN N nlnn Cn mlnm Cm
lnn a ln@qN xA
a ln

qN I x
qN
!
a ln@qNACln

I x
qN

a ln@qNA x
qN
I
P

x
qN
P
CO
4
x
qN
Q5
And similarly for the expression lnm. In all we can sum up all the quadratic terms in x, and get the coecients
from normalization
lnW @n;mA 9 xP
PN

I
p
C I
q

a xP
PpqN
W @n;mA G e x2
2pqN
W @n;mA a IpPpqN
e x2
2pqN
5

x hxi a H
xP

xP
a pqN
p
hxPi G
p
N
` h`i a @p qAN
`P

`P
a @p qAP NP RpqN
Note that all the results except ` are symmetric in the switch p 6 q.
If p Ta q, there will be a constant drift of @p qAP NP. For a symmetric random walk (p a q),
`P a Np
`P a
p
N
We dene the total deviation length v ` ¡a , and total diusion time t N ¡, so that
vP a `P ¡aP a N ¡aP a t

¡aP a P¡ aP
P
¡t P¡h ¡t
p
vP G
p
t
Where h a2
P is the dened diusion constant, or diusivity.
1.4 Diusion Equation
The number of particles at point x on a lattice (of step size a) at time t (in discrete time-steps of length ) is
dependent on the amount of particles which occupied the sites to the left and right of x at the previous time-
step, and on the number of particles which may be created or annihilated at x during the current time-step,
which can be described by a source/sink rate function f @x;tA:
n@x;tA a q ¡n@x Ca;t ACp ¡n@x a;t ACf @x;tA¡ (1)
We expand the n terms in series (simultaneously to second order in x and rst order in t)
n@x ¦a;t A 9 n@x;t A¦a ¡ @n@;t A
@
jax C I
PaP ¡ @Pn@;t A
@P jax ¡ @n@x ¦a;sA
@s
jsat
Substituting back in 1
n@x;tA 9 n@x;tAC@q pA¡a ¡ @n@x;tA
@x
C I
PaP ¡ @Pn@x;tA
@xP ¡ @n@x;tA
@t
C ¡f @x;tA
@n
@t
a q p

¡a
| {z }
v
¡@n
@x
C aP
P|{z}
h
¡@Pn
@xP Cf
@n
@t
a v
@n
@x
Ch@Pn
@xP Cf
The resultant diusion equation can be viewed as a continuity equation.
1.5 Diusion Equation as a Continuity Equation
@n
@t
C @j
@x
a H
In order for the diusion equation to be a continuity equation, the current must be dened as
j vn h@n
@x
6

Disregarding the external function f @x;tA for now, we have
@n
@t
a v
@n
@x
Ch@Pn
@xP
Moving to a reference frame which is moving at velocity v, i.e. dening ”n n@x vt;tA
@”n
@t
a h@P”n
@Px
Which is a simple diusion equation.
As an example, we add point Dirichlet initial conditions. The result can serve as a Green's function for
any general initial condition we may wish to apply later.
@n
@t
a h@Pn
@xP C @tA @xA
The solution is gotten by taking the Fourier transform of both sides of the equation
n@x;tA a Ip
Rht
e x2
4ht
In general, for any linear equation we will come across, we will try to solve using a Fourier transform.
diusion with Initial Condition
A gas with uniform density nH occupies a semi-innite space, and is held behind an innite-planar partition at
x a H. At t a H the partition is removed. What is the time-dependent gas density, n@tA?
We will solve the 1-D diusion equation with the boundary condition for an initial point concentration at
x a H. Then we will use this solution to integrate over all contributions from I xH H.
@n
@t
a h@Pn
@xP C @tA @xAnH
Using a Fourier transform
I
I
dte i!t
I
I
dxe ikx @n
@t
a
I
I
dte i!t
I
I
dxe ikx h@Pn
@xP C @tA @xAnH
!
i!n@k;!A a hkPn@k;!ACnH
n@k;!A a nH
hkP Ci!
n@x;tA a I
@PAP
I
I
dtei!t
I
I
dxeikxn@k;!A
a nHp
Rht
e x2=Rht
”n@x;tA a
H
I
n@xH;tAdxH
a
H
I
nHp
Rht
e @x xHA2=RhtdxH

I
xp4ht
nHp
e z2
dz
z x xH
p
Rht
”n@x;tA a nH
P I erf

x
p
Rht
!
The complimentary problem deals with an initial density nH behind some partition, which is lifted at t a H.
What is the density as a function n@tA of time on the originally vacant side of the partition?
7

diusion with Boundary Condition
A particle starting at xH is diusing along the x-axis. A sticky wall is placed at x a H. What is the probability
for the particle to get stuck as a function of t?
The equation to be solved is (diusion equation with point initial condition n@xH;HA a I)
@n
@t
a h@Pn
@xP C @tA @x xHA
with the boundary condition
n@H;tA a H
for all t.
In analogy to electrostatics, the boundary condition can be mathematically represented using the method
of mirror charges, i.e. a mirror particle with density nH which begins diusing at time zero at xH. The
solution is the sum of the real @nHA and mirror @ nHA particles
n@x;tA a nH @x xH;tA nH @x CxH;tA
Where each of the real or mirror particles has a solution of the form of diusion without the boundary
condition
nH @x ¦xH;tA a Ip
Rht
e @x¦x0A2=Rht
The ux at the boundary is
j @H;tA a h@n@x;tA
@x
jxaH
a Ph@nH @x;tA
@x
jxaH
a xH
P
p
h
¡t Q=P ¡e x2
0=Rht
The probability to get stuck is
P @tA a
t
H
dtHj @H;tHA
The complete solution requires numeric integration.
An alternate route is to compute I P @tA a ‚ I
H n@x;tAdx, i.e., via the probability that the particle is not
stuck at time t.
8

2 Langevin Equation
We would like to write equations of motion for a particle (of gas for instance, and of radius R) moving in
a viscous environment (another gas, for instance, with viscosity coecient ). Starting from the Newton
equation we write
m•~v a ~p @t;vA ~f @~vAC ~F @tA
~f @~vA a ~v
a TR
Where ~f is a friction force and ~F is a random force, and the friction coecient is related to the viscosity
of the environment according to the Stokes relation (given here for a spherical geometry).
2.1 One-Dimensional Ballistic Motion and Diusion2
In one dimension
mx@tA a •x@tACF @tA
multiplying by x@tA we get
m
P
dP
dtP
¢
xP @tA£
m
d
dt
x@tA
!P
a
P
d
dt
¢
xP @tA£
CF @tAx@tA
Suppose the time between collisions is and we average over times suciently longer than .
hF @tAx@tAi a hF @tAihx@tAi a H¡H
Since the random force is uncorrelated to the particle trajectory, and both have zero mean. Therefore the last
term on the RHS is zero.
From diusion, y @tA I
P

xP @tA
a ht, so
m
P
d2
dt2

xP @tA
a H. Therefore the rst term on the LHS is zero.
The kinetic energy of the particle is
m
P
h¢ d
dtx@tA£Pi
a I
PkBT.
The rst term on the RHS is
P
d
dt

xP @tA
a h.
In all
kBT a h
h a kBT

fkBT
Where f I is named the mobility, rst implied by Galileo, and see also 2.2.
Example:
A particle of radius H:Sm is solvated in water at PHC which has viscosity H:HI with the appropriate
cgs units. The particle time dependent RMS displacement is

rP @tA
a
xP @tA
C
yP @tA
C
zP @tA
a Q¡Pht
p
hrP @tAi a
p
Tht a
r
TkBT

t
p
hrP @RHsAi 9 IHm
Back to the general problem,
y @tA I
P

xP @tA
my @tA m

•xP @tA
| {z }
kBT
a •y @tA
my C•y a kBT
2Appeared in 2006A and 2007B.
9

The exact solution to this equation is (with initial conditions y @HA a •y @HA a H)
y @tA a kBT

h
t m

I e t=m
i
which leads to dene the time scale tH m
. For long times, i.e. t ) tH the exponent falls and
y @tA 9 kBT

¡t a ht
so at long times we recover simple diusion-like behavior. At short times, i.e. t ( tH we expand the exponent
to second order in
t
t0 (all previous orders cancel out) and
y @tA 9 kBT
m
¡tP
which corresponds to ballistic motion.
An assessment for the value of tH for a sphere is
tH a m

a RRQ
TR
G

RP
For the values of the intrinsic properties of water (;) and a particle radius of H:Sm we get tH 9 P¢IH V s a
PHns. This is a very short relaxation time, which means that experiment will show only the diusive behavior.
2.2 Classical and Quantum Einstein Relations
Recall the relations h a kBT
a fkBT.
Particles performing overdamped motion have a very short mean free path, i.e. the time between collisions
is much shorter than any other interesting timescale in the system.
Assume particles distributed with density n@~rA acted upon by a friction force ~f a ~v a I
f~v, but no random
force.
m•~v ( ~f
~v a f~f
The particle current is composed of an external force / potential term, and a concentration gradient term
~j a n~v h~r¡n
a nf~f h~r¡n
a nf~r¡V h~r¡n
where V @~rA is a potential from which the force ~f is derived.
In equilibrium, ~j a H. Also, in equilibrium @~rACV @~rA a const: (here @~rA is the chemical potential).
~r¡V a ~r¡
a @
@n
~r¡n
Where the second line was gotten using the chain rule. Substituting back we have
H a nf@
@n
~r¡n h~r¡n
h a nf@
@n
This is the Einstein relation in its most general form.
Particular examples:
10

2.2.1 Ideal Classical Gas
From combinatorics we have the dependence of @nA. The dependence on temperature is given from the
Sackur-Tetrode formula.
@n;TA a kBT lnn C @TA
@
@n
a kBT
n
h a nfkBT
n
a fkBT (2)
2.2.2 Degenerate Fermion Gas (DEG)
The dependence of the Fermion gas on the density is given, to rst order, by the density dependence of the
Fermi energy (eqn. 6).
9 F @nACO
4
kBT
F
P5
Q D
F @nA G nP=Q
@
@n
a P
Q
F
n
h a nfP
Q
F
n
a P
QfF
Size of a Puddle of Falling Drops
Drops falling under g a W:Vms P from a height h a Im diuse in the horizontal plain as they fall. What is the
RMS radius of the resulting puddle,

rP
? Compare this with the drop size, R. Assume the drops have mass
m a IH IH g and density a H:Wg cm Q, and that the air is at temperature T a QHHK.

rP
a
xP CyP
a P¡Pht a RkBT

t
a TR
m a RRQ
Q
R a

Qm
R
I=Q
Solving the Langevin equation for the vertical motion:
•vz a
m
vz Cg
•vz C
m
vz a g
d
dt

vz @tA¡e

mt¡
a ge

mt
vz @tA¡e

mt vz @HA a
t
H
ge

mtH
dtH
vz @tA¡e

mt a mg

¢
e

mt I£
vz @tA a mg

¢
I e
mt£
The vertical motion is characterized by the time scale
H m

G m
mI=Q a mP=Q ( I
11

Meaning that this time scale is much less than the time scale for free particle motion, characterized by
mv2
P G m.
Therefore the transient eect in the result for vz @tA can be disregarded
vz @tA 9 mg

a fmg
And the time that the drops take to fall is t a h
vz a h
mg.
Putting it all together,

rP
a RkBT
T
¡ h T
mg
a RhkBT
mg
% R¡Im ¡ I:QVH¢IH PQ J K I ¡ QHHK
IH IQ kg ¡ W:Vms P
9 I:TVW¢IH V mp
hrPi % I:Q¢IH R m
R %

Q¡IH IH g
R ¡H:Wg cm Q
I=Q
9 Q¢IH R cm a Q¢IH T m
Since the puddle size,
p
hrPi, is two orders of magnitude greater than the single drop size, R, the eect should
be measureable. The puddle size is also independent on the internal composision of the drops, which would
manifest in the dorp mobility, f, which does not appear in the nal result.
Gas Escaping from a Box to a Vacuum
An atomic gas has density n@tA satisfying n@t a HA nH. The gas occupies a box of volume V and has
temperature T. The gas particles' mass is m, and it escapes through a hole in the box of area A. How much
heat needs to be supplied externally to keep the box at a constant temperature?
The velocity distribution:
f~v a

m
PkBT
Q=P
exp m
PkBT

vP
x CvP
y CvP
z
¡
!
For a spherically symmetric distribution:
fv a RvP

m
PkBT
P=Q
exp mvP
PkBT
!
the average velocity is
v hvi a
I
H
fvvdv a I
P
I
H
fvd

vP¡
a R

m
PkBT
P=Q
PkBT
m
P I
H
te tdt
a P
Q=P
r
PkBT
m
a
r
VkBT
m
We are interested only in the velocities which contribute to a particle leaving the box, i.e. velocities with a
12

positive component along the x-axis (where the x-axis is perpendicular to the hole in the box).
vCx a
I
H
vx
r
m
PkBT
e mv2x
2kBT dvx
a I
P
I
H
r
m
PkBT
e mv2x
2kBT d

vP
x
¡
a I
P
I
H
r
PkBT
m
e tdt
a
r
kBT
Pm
a I
Rv
We next write the continuity equation for the number of particles in the box, N @tA a V ¡ n@tA, where (at low
gas densities) we assume that the current of particles leaving the box through the hole can be approximated as
linearly dependent on the local particle density near the hole:
d
dt
N @tA a V
d
dt
n@tA a A ¡n@tA¡ vCx
a A ¡n@tA¡ I
Rv
The solution is an exponential decrease:
n@tA a nH ¡e Av
4V ¡t a nH ¡e t=
RV
Av
The leaving energy ux
h i a hvCx i
vCx
a PkBT
Each atom has a thermal kinetic energy of
Q
PkBT, which is smaller than the outgoing energy ux per particle.
The box is cooled because the slow particles leave more slowly than the fast particles (the reason for this is the
lack of correlation between dierent particles). To keep the temperature constant, the incoming heat ux needs
to cancel the outgoing energy taken away with the leaving particles.
~P a d
dt
Ek @tA
a I
PkBT ¡ d
dt
N @tA
a I
PkBTV ¡ d
dt
n@tA
a I
PkBTV ¡ Av
RV
¡n@tA
a kBTAvnH
V e Avt
4V
a kBT
P
NH

e t=
The total energy supplied up to time t is
Q@tA a
t
H
dtHP @tHA
a Q
PkBTV ¡ Av
RV
¡
t
H
dtHn@tHA
a Q
PkBTV nH
h
I e AvtH
4V
i
a Q
PNHkBT
h
I e AvtH
4V
i
13

Corollary A similar problem could be where the gas escapes not to a vacuum, but to a surrounding with xed
pressure and temperature, i.e. xed outside particle concentration nout. The solution is gotten by computing
the ux of particles from outside the box into it, and writing the continuity equation in terms of the sum of two
uxes: outgoing and incoming. When the temperature inside and outside the box are equal, the terms A and
v appear equally in both uxes, and the derivative of n@tA is a function of nout n.
d
dt
N @tA a V
d
dt
n@tA a Jout CJin
a A ¡n@tA¡ vCx CA ¡nout ¡ v x
a A ¡‘nout n@tA“¡ I
Rv
Where the last line is valid if
v x a vCx , i.e. when the temperature is equal inside and outside the box
d
dt
n@tA a nout n

Av
RV
Solution to the non-homogenous rst-order ODE
•n@tAC I

n@tA a I

nout
d
dt
h
n@tA¡et=
i
a I

noutet=
n@tA¡et= nH a
t
H
I

noutetH=dtH
n@tA¡et= a nH Cnout
h
et= I
i
n@tA a nHe t= Cnout
h
I e t=
i
Knudsen's Law3
Two containers of volumes VI and VP are connected through a small hole of area A. The containers are held at
a separate temperatures TI and TP, respectively, and we assume that these temperatures remain constant for
the times relevant in the question. What is the ratio of pressures, PI and PP, respectively, which are created in
the containers?
If the hole size scale v
p
A is much larger than the mean free path of the particles, `, then there will
be particle-particle collisions near the hole, and the pressures on both sides of the hole will equilibrate quickly,
such that PI a PP. When we say the hole is small, what is meant is ` ) v. In order to increase the mean
free path, `, we can use a very sparce gas. In this case the pressures equalize much more slowly, and for short
times we can write
PI
PP
a NIRTI
NPRTP
¡ VP
VI
a NITIVP
NPTPVI
a nITI
nPTP
we require that at equilibrium the change in the number of particles in any one of the containers has to be zero.
This change is caused by a current of leaving particles and a current of entering particles, through the hole
H a d
dt
NI a VI
d
dt
nI @tA a JP3I @tA JI3P @tA
JI3P @tA a JP3I @tA
nI @tA¡ vI a nP @tA¡ vP
nI @tA¡
p
TI a nP @tA¡
p
TP
nI
nP
a
r
TP
TI
3Appeared in 2006B.
14

so that in all,
PI
PP
a
r
TI
TP
2.3 Poiseuille Flow4
We are interested in the outgoing ux of a gas escaping from a box to a vacuum through a narrow cylindrical
tube of length v.
We assume that the gas density is low in the sense that the mean free path, `, is much greater than the
cylinder radius, R. This means that gas leaving the box via the tube experiences collisions only with the cylinder
walls
5, and not with other gas particles.
We also assume that the escape process is so slow that the gas density in the box is nearly unchanged, and
remains nH throughout the process. That is, we are only interested in short enough time scales, such that nH
does not change in the time we concern ourselves with.
The particle ux is dened as j a h@n
@x. The diusion coecient is given in terms of the RMS velocity
h a I
P
aP

a I
Pav @TA
vP

vP
a kBT
m
The characteristic length, a, is taken as R.
h a I
PRv
The particle concentration gradient,
@n
@x is estimated as linear in the tube, for lack of better knowledge.
j a h ¡ H nH
v
a I
PRv ¡ nH
v
The particle current through the tube is
J a A ¡j
a RP ¡ I
PRv ¡ nH
v
a
P ¡ nHv
v ¡RQ G RQ
The prefactors are rarely exact, the functional dependence is what's important because it is easily veried
through experiment.
Another approach: Large nH
The pressure in the box is pH a nHkBT, and the pressure outside it is zero (vacuum).
The gas viscosity is
a I
Q`nmv a I
QnmvP
where the last equality is due to ` v. This case was solved by Poisseuille in 1840
J a nPkBTRR
Vv G RR

G RR
`
if we increase the average density of particles in the tube, the mean free path will decrease until it reaches the
order of R and then we will recover J G RQ:
If we were to take ` ) R, the result would be J G RR.
4Appeared in 2007A.
5We ignore ballistic transport, in which a gas particle can travel the entire length of the cylinder without hitting anything.
15

Contribution from Ballistic Transport
Except for the current which comes from diusion, there is a contribution from the current of ballisticly moving
particles (i.e., no collisions with tube or other particles, from box to vacuum).
The particles that can move through the tube without colliding with the tube walls are only those which
have a velocity nearly parallel to the tube itself. This angle is $ R
v small. The RMS speed of these particles is
v, and the ux due to their contribution is j % n ¡ R
v ¡ v. This is the same contribution as the one we got from
diusion, so the dependence on R remains the same, J G RQ (only with a dierent coecient).
2.4 Diusion in Ambipolar Plasma6
We regard a plasma (ionized gas) of electrons and univalent ions (charges e and e, respectively). The Coulomb
interactions are very strong, and unless they are screened, they easily dominate the behavior of the system. The
system therefore very quickly nears a state where it is neutral everywhere, i.e. ne @~rA a ni @~rA n@~rA. Where
by n@~rA we demark the local particle density (of both types of particles).
We interest ourselves in diusion in a system where ~rn Ta H.
~je a he~rne fenee~E
~ji a hi~rni Cfinie~E
The rst term on the RHS of each equation is the diusion term, with the appropriate diusion coecient,
and the second term is the response to the external electric eld, governed by the particle charge (sign) and
the appropriate mobility. In calculating the electric eld, ~E, acting on the ions, we would be wise to use the
Born-Oppenheimer approximation, taking also into account the electronic density and its induced electric
eld.
Requiring that there be no separation of charge we have
~je a ~ji
he~rne fenee~E a hi~rni Cfinie~E
Solving for the local electric force acting on the particles
e~E a he hi
fe fi
¡
~rn
n
Using the classical Einstein relations 2,
he
hi
a fe
fi
which we will use in the calculation of the total particle ux,
~j a~ji C~je a @hi CheA ~rn @fi feA@he hiA
fi Cfe
~rn
a P hihe
hi Che
~rn
The system eective diusion coecient is therefore naturally arrived at.
heff: P hihe
hi Che
note that switching all electrons with ions and vice versa keeps the eective diusion coecient the same.
Since
h $ aP

$ av
where the electron RMS velocity is ve a
q
kBT
me and likewise for the ions.
6Appeared in 2006B, 2007A and 2007B.
16

As for the parameter a, we need to dene the mean free path, `. This is done as a function of the particle
density, n, and the particle-particle interaction cross-section, :
` a I
n
The cross section is the one for Coulomb interaction, and is therefore the same for electrons and ions. Likewise
is the particle density, from the requirement for local charge neutrality. Therefore the mean free paths of the
electrons and ions are equal. In all
hi
he
$ vi
ve
a
r
me
mi
( I
Therefore we have a simpler expression for the eective diusion coecient
heff: % Phi
2.5 Debye-Hückel Screening (1923)7
What would be the eective electrostatic eld at a distance r away from a surplus unit charge added to a neutral
plasma of ions and electrons?
From the Boltzmann equilibrium distribution
ni a n ¡exp

e'
kBT

9 n ¡

I e'
kBT

ne a n ¡exp

e'
kBT

9 n ¡

IC e'
kBT

Where the approximation is valid since only a small charge was added to the neutral system. Small here means
je'j ( kBT.
The Laplace equation:
rP' a R
a e@ni neA
9 PneP'
kBT
So the equation to be solved is
rP' VePn
kBT| {z }
2
' a H
The solution is
'@rA a ¦e
exp@ rA
r
I
r
kBT
VneP
the screened potential is dominated by the Coulomb reciprocal at short range, and falls to zero exponentially
at long range, making it more easily managable, mathematically.
2.5.1 The Screening Length I in Dierent Materials
ePn $ ePl Q a eP
l
¡ I
lP
In the last term, the factor
e2
l plays the role of the energy, and the l P term just serves to give the correct
units overall.
7Appeared in 2006B.
17

For an electron gas, the said energy term is given by the Fermi energy,
l $
r
F
ePn
In metals, the Fermi energy can be approximated (done by Kyoto) using the unit cell size, a $ n I=Q as
F $ ~2
ma2 , and the Coulomb energy as C $ e2
a . Therefore
F
C
$ ~P
maP ¡ a
eP $ I
The screening length in metals is of the order of a unit cell
I $
r
F
ePn
$ a
In semi-conductors
n $ IHIP IHIT cm Q
T a QHHK
I a IH R IH T cm
18

3 Self Heating Phenomena
3.1 Heat Diusion Equation
The continuity equation
@n@~r;tA
@t
C ~r¡~j @~r;tA a H
is a diusion equation when n@~r;tA changes slowly with respect to the average random step.
The ux in the absence of drift (external forces)
~j @~r;tA a h~rn@~r;tA
The heat continuity equation, by analogy, is
@@~r;tA
@t
C ~r¡~q @~r;tA a H
Where ~q @~r;tA is the energy ux.
For small gradients (close to equilibrium), Fourier's heating law states
8
~q a ~rT
writing the energy change in terms of the temperature,
@
@t
a @
@T
@T
@t
C @TA @T
@t
Where C @TA is the heat capacity. It follows that
C @TA @T
@t
a ~r

~rT

a rPT ¡T
More generally, external (time, space and temperature dependent) heat sources can be added in the form
C
@T
@t
a ¡T CQ@~r;t;TA
Adiabatic Heat Propagation
A sound wave is propagating in a media, with frequency ! and velocity vs. The media has heat capacity CV
and heat conductivity . What are the conditions under which the propagation is adiabatic?
In this sense adiabatic means that the heat released by the motion of the wave is redistributed much more
slowly than the motion of the wave. Therefore, the temperature is taken as practically time-constant in relation
to the time dependence of the wave.
C
@T
@t
a ¡T
@T
@t
a
C
¡T
the heat diusion equation gives the velocity of the heat through the media as

CV . If vs )
C then the heat
propagation is adiabatic.
8Note the similarity to Ohm's law
~j a ~E
a ~r
~q a ~rT
more on this, and a generalization in 6.5.
19

Melting Ice Sphere
A sphere of ice of initial radius RH is held in an ambient environment, with temperature TI far from the sphere,
and TH is the melting temperature (TI TH). How long will it take the entire sphere to melt?
If the dynamics of heat transfer through the ice is much faster than the heat transfer from the environment,
then the temperature in the ice is always TH.
Since the heat transfer in the environment is very slow, the temperature from the sphere surface to innity
is distributed (radially) according to Gauss's law,
H a @T
@t
G rP @~rA
which under spherical symmetry is
H a rP @rA¡T I
rP
d
dr

krP d
dr

¡T
krP d
dr
T a A
T @rA a B A
kr
where A and B are two integration constants to be determined from the boundary conditions: Since the melting
temperature is known (TH) this must also be the temperature at the edge of the sphere (T @r a RHA TH). Also
we know that T @r a IA TI. In all,
T @rA a TI TI TH
r
RH
The heat current density just inside the sphere, in the normal direction, is
~j a
@T
@r
jR0 ¡ ”r
a
TI TH
RH
”r
but RH is actually time dependent (the sphere shrinks as it melts), so the mass loss rate
dm
dt
a •V a d
dt
R
Q RQ
H @tA
should be proportional to the heat loss rate (with a proportionality constant that is the latent heat, q)
dm
dt
¡q a S”n ¡~j
RRP
H @tA •RH @tA ¡q a RRP
H @tA¡
TI TH
RH @tA
d
dt
R a
q
TI TH
R
d
dt

RP
P

a
q
@TI THA
RP
P a
q
@TI THA¡t CC
with an integration constant C to be determined from initial conditions (R@t a HA a RH). The time it takes the
sphere to melt completely is the time for which R@t a A a H.
H a P¡ @TI THA
q
¡ RP
H
a qRPH
P@TI THA
Where is the time it takes the whole sphere to melt. Note that this time is proportional to the initial sphere
surface area, and not its volume!
20

Melting Ice Sphere Heated Through a Long Cylinder
A lump of ice (of density and latent heat ) at temperature TI is now connected to a long cylinder (of length
L and cross section area A, and has heat conductivity ) which is heated at the far end to a temperate TP.
What is the mass of the lump of ice as a function of time, m@tA?
The temperature prole in the cylinder is calculated by solving the heat equation in one dimention
T @xA a TI C TP TI
L
¡x
the heat current density from the lump to the cylinder is
~q a
@T
@x
jxaH a
L
@TP TIA
the total heat current is ~J a A~q.
The mass loss rate is related to the heat loss rate by the latent heat, :
dm@tA
dt
¡ a A

L
@TP TIA
m@tA a m@HA A@TP TIA
L
¡t
The melting time is (m@t a A a H)
a m@HA¡L
A@TP TIA
Heat Prole in a Coaxial Cylindrical Cavity
A solid cylinder of radius RI is surrounded by a hollow cylinder of radius RP RI, such that their centers
coincide. Given that both cylinders are held at constant and dierent temperatures, i.e. the boundary con-
ditions T @RIA a TI and T @RPA a TP, calculate the temperature prole in the cavity between the cylinders
(T @rA RI r RP).
From symmetry along the cylinders axis, the problem is to solve the 2-D Laplace equation, with two
Dirichlet boundary conditions, in polar coordinates.
H a rPT a I
r
@
@r

r
@
@r
T

r
@
@r
T a A
T a B A
r
T @RIA a TI a B A
RI
T @RPA a TP a B A
RP
RITI a RIB A
RPTP a RPB A
A a RPB RPTP
21

RITI a RIB RPB CRPTP
RITI RPTP a @RI RPAB
B a RITI RPTP
RI RP
A a RP
RITI RPTP
RI RP
RPTP
a RIRPTI RPRPTP RIRPTP CRPRPTP
RI RP
a RIRP
RI RP
@TI TPA
T @rA a RITI RPTP
RI RP
RIRP
RI RP
TI TP
r
Heat Prole in a Half Plane
Given that the temperature outside the half plane (x H) is modulated according to TH @tA a Ta C Tb ™os!t,
nd the temperature prole, T @xA, inside the half plane (for all x ! H).
Eective Heat Conductivity of a Layered (Composite) Material
A composite material is constructed of parallel and innite layers of two component materials, which have heat
conductivities I and P. The fraction of each material in the composite is cI and cP, respectively (cI CcP a I).
What is the eective heat conductivity for heat ow in the direction parallel to the plates and perpendicular to
them?
Perpendicular Eective Heat Conductivity
At steady state the temperature prole is time independent, and therefore so is the heat current:
H a C
@T @x;tA
@t
a c @xA @PT @x;tA
@xP a @~qc
@x
the coecient @xA is either I or P, depending on which layer we looking in. Since the heat current is constant,
the temperature prole should be linear in each of the layers (i), separately. We can dene the temperature
prole in the entire composite as
T @xA a Ti C x xi
xiCI xi
¡@TiCI TiA
a Ti C ¡x
¡`@xA ¡¡T @xA
where Ti is the temperature at the beginning of the i-th layer,¡`@xA is the depth of the layer in which x is
located, ¡x is the depth into that layer where x is, and ¡T @xA is the temperature dierence on that layer.
The heat ux perpendicular to the layers obeys the following equation:
~qc a c
dT
dx
a c ¡ ¡T @xA
¡`@xA
The ux is equal through each of the layers, so
I
¡TI
¡Ì
a ~qc a P
¡TP
¡`P
We can also denote the temperatures at the boundaries of the i'th layer by Ti and TiCI, and the layer length
as ì, and dene the eective perpendicular heat conductivity for the entire composite, c, such that:
I
TP TI
Ì
a P
TQ TP
`P
a c
TH TL
`c
22

Where the total perpendicular length is `c and the total temperature dierence is TH TL. Since the
composite is made of an equal number of layers of type 1 as there are of type 2 (we denote this number by n),
the total perpendicular sample length is given by n times a block of two adjacent layers (one of each type)
`c a n@¡Ì C¡`PA
and the temperature drop over the entire composite is n times the drop after each block of two adjacent layers
(one of each type)
TH TL
n
a ¡TI C¡TP
Thus the eective perpendicular conductivity for the composite is
c a ¡Ì C¡`P
¡TI C¡TP
¡I
¡TI
¡Ì
a
IC ¡`2
¡`1
¡TI C¡TP
¡I¡TI
a
IC ¡`2
¡`1
IC ¡T2
¡T1
¡I
a
IC c2
c1
IC 1
2 ¡ c2
c1
¡I
a PI @cI CcPA
PcI CIcP
c a PI
PcI CIcP
The result is
I
c
a cI
I
C cP
P
where cj @j a I;PA is the fraction of the j material in the composite (cI CcP a I), and we used
c2
c1 a ¡`2
¡`1 and
the fact that the temperate drop on each of the two types of materials is proportional to their fraction in the
composite, and inversely proportional to their conductivity: ¡Tj a cj
j .
Parallel Eective Heat Conductivity
We have already stated in the previous part that at steady state, the heat current is constant along the x axis.
Therefore, the parallel eective conductivity should be equal no matter where it is measured along the x axis in
the sample, and should only change in the parallel direction (i.e. perpendicular to x)
9. The total parallel heat
current, taken over some interval R along the x axis is
~J a RcI ¡I
¡T
`k
CRcP ¡P
¡T
`k
where Rcj is the part of the interval composed of type j layers. The average parallel heat current is then
~j a
~J
R
a cII
¡T
`k
CcIP
¡T
`k
a k ¡ ¡T
`k
and so the eective parallel conductivity is
k a cII CcPP
9This means that the gradient of the heat current (the temperature) is contiuous in the x direction when dealing with heat
current in the parallel direction, and the heat current itself is continuous in the x direction when dealing with heat current in the
perpendicular direction.
23

Heat Transfer in a Kettle
Clean Kettle
Water evaporates from a kettle at a rate a Pg min I. The bottom of the kettle is a copper plate of thickness
` a Qmm and area A a QHHcmP. Copper is known to have heat conductivity Cu a SJ s I cm I K I. The
latent heat of water is a P:PSJ kg I. What is the temperature on either side of the copper plate?
At steady state, the heat current entering the kettle should be equal (due to conservation of energy) to the
heat lost with evaporation: J a a Pg min I ¡PPSHJ g I a RSHHJ min I.
On the other hand, the heat current entering the kettle should obey the heat equation, such that
q G ~rT
J a A
¡T
¡x
¡T a `J
A
¡T a Qmm ¡RSHHJ min I
QHHcmP ¡SJ s I cm I K I
H:Imm I cm
THsmin I
a H:HISK
Kettle with Limescale
The same problem as before, only now a layer of thickness `scale a Imm and heat conductivity scale a
H:HVPSJ s I cm I K I separates the copper plate and the water. What is the temperature of the bottom of
the copper plate and of the water touching the scale?
In the case of the presence of limescale, we would have to use the result for the eective perpendicular heat
conductivity through two layers of two dierent type materials. In that case
¡T a ¡x ¡J
Ac
a `J
A
2
`Cu
`
Cu
C
`scale
`
scale
3
Since `scale $ `Cu but scale ( Cu we could simply disregard the Copper, and calculate for the scale only
¡T 9 `J
Ascale
a Rmm ¡RSHHJ min I
QHHcmP ¡H:HVPSJ s I cm I K I
H:Imm I cm
THsmin I
a I:PIK
The full result, including the eect of the Copper as well as the limescale, would be not much dierent.
3.2 Nonlinear Stationary States
Linear problems can and should be solved by Fourier transforming the equation. We will consider time and
space independent sources only, Q@~r;t;TA a Q@TA, but if these sources have a strongly non-linear dependence
on their argument, T, we will have to resort to other mehtods for a solution. We will further simplify our
problem by discussing rst the I D case a thermally conducting wire (or thin cylinder).
3.2.1 Current Carrying Superconducting Wire
A normal metalic wire with Ohmic resistance is heated when a current with density j is run through it giving
a heat source (heat release) with heat density Q
Q a @TA¡jP
24

We will assume a relatively simple dependence of the resistance on temperature,
@TA 9
@
SC T Tc
N @TcACT ¡% T Tc
Where SC; N @TcAand % are constants. Tc is the critical temperature of the normal-metal to superconductor
transition, and is well below room temperature. The superconducting wire is cooled by surrounding liquid
Helium.
We are interested in the change of the heat prole thorughout the wire, as a function of time, given an initial
part of the wire being heated (a seed, which may grow or contract). Our problem, therefore, deals with the
time dependence (or dynamics) of the superconductor - normal-metal phase transition. This is one example of
the dynamics of a Landau-Ginzburg phase transition (another is the liquid - gas phase transition in a Van
Der Waals gas).
The heat release function, Q, is a very non-linear function of the temperature, T. Its dependence comes
from the temperature dependence of the resistance, . This means that Q is a function of T , which is itself a
function of the coordinates.
25

The superconducting state satises SC n N, so that an extremely high current density can be reached.
If heating in the superconducting state causes a temperature buildup beyond the critical temperature, Tc, the
sudden switch to normal metal resistance will cause the wire to explode.
We concern ourselves with a 3-D wire, but want to simplify the problem to a 1-D one. This can be done if
the dynamics of the longitudinal heat transfer (along the wire) and the transverse one (in the cross section of
the wire) have widely dierent scales (of energy, time, etc.).
C
@T
@t
a rPT CQ‘T“
a
H
f
f
f
d
@P
@zP C @P
@yP C @P
@xP
| {z }
r2
c
I
g
g
g
e
T CQ‘T“ (3)
Using the guidance of the Fourier heat equation, we assume the heat ux in the surrounding liquid (and near
the surface of the wire) to be linear in the dierence between the temperature at the surface of the wire, Ts,
and the temperature of the environment, TH. This is valid when the two temperatures are not very far apart.
~qliquid
c G Ts TH
We take the liquid heat conductivity as h, and write ~qliquid
c 9 h@Ts THA ¡ ”n, where ”n is the unit vector
directed from the center of the wire outward. From continuity of the heat ux, ~qliquid
c should be equal to both
the heat ux in the liquid and in the wire, close to the surface.
If the temperature within the wire cross section does not change much, we can write the transverse heat ux
in the cross section as
~qwire
c a rcT
%

Tm Ts
b

where Tm is the temperature at the core of the wire (which is higher than Ts), and b is the radius of the wire.
Equating the transverse heat uxes at both sides of the surface of the wire
10

b
@Tm TsA a h@Ts THA
Tm Ts
Ts TH
a hb

If Ts 9 Tm a T we can assume the temperature throughout the corss section does not change much,
in relation to the temperature dierence between the wire and the liquid, T TH. The small parameter
necessary for this approximation to be valid is the Biot number Bi hb
( I. If so, we can approximate
the temperature throughout a cross section at some point along the wire as constant for that cross section, i.e.
T @z;tA I
e
‚‚
dxdyT @x;y;z;tA.
Averaging over each cross section in the heat transfer equation 3 yields
C
@ T @z;tA
@t
a
@P
@zP T @z;tAC
e

dxdyrP
cT @x;y;z;tACQ
¢
T @z;tA£
The averaging of Q‘T“ 9 Q
¢
T
£
is valid for a small Biot number
Q‘T“ 9 Q
¢
T
£
C @Q
@T

T T
¡
C::: a Q
¢
T
£
CO
¢
T T
£
a Q
¢
T
£
CO ‘Bi“
10This is equivalent to expanding Tm as a rst order expansion around Ts:
Tm 9 Ts CrcTjR ¡b C: : :
a Ts C h

@Ts T0A¡b
26

The only thing left is to dene the second term on the RHS as a function of T. This is done using the
divergence theorem:

I
e

dxdyrP
cT @x;y;z;tA
I
e

dxdy~rc ¡
h
~rc ¡T @x;y;z;tA
i
a
I
e
s
d}r”nT @x;y;z;tA
a
I
e ¡g ¡ I
h
j~qj
a Pbj~qj
bPh
a Pj~qj
bh
W
¢
T
£
In all we have
C
@ T @z;tA
@t
a
@P
@zP T @z;tACQ
¢
T @z;tA£
W
¢
T @z;tA£
Where Q describes the heat release in the wire, and W describes the process of cooling to the external environ-
ment. In the case of a wire of nite thickness, the solution has got to be numeric.
11
3.2.2 Chemical Reactions on the Surface of a Catalyst
Other applications of this method are exothermic chemical reactions of oxidation on a metal catalyst, such as
Platinum. Two such examples are:
PCO COP 3Pt
PCOP
RNHQ CQOP 3Pt
PNP CTHPO
CPHR CQOP 3Pt
PCOP CPHPO
Under some temperature, Tc, the catalysis is not eective. Above this temperature, the catalysis is eective,
and the heat released in the reaction itself serves to keep the temperature of the catalyst hight.
The dependence of the heat released in such reactions on the temperature at which the reaction is carried
out is similar to the function Q@TA discussed for heat released in a metalic wire at temperatures below and
above the critical temperature of the normal-metal - superconductor phase transition. At low temperatures, Q
is limited from kinetic considerations, while at high temperatures it is limited from diusive considerations.
12
The catalyst facilitates the burning of exhaust waste after internal combustion. In some experiments, a
focused laser beam is used to spark the burning on the surface of Platinum. The seed (spark) can then
propagate over all the catalyst.
3.2.3 Metal - Dielectric Phase Transition
One more example of such an application is the phase transition from a metal to dielectric in such compounds
as Vanadium Oxide, V nO. The heat released when an electrostatic eld ~E is passed through the compound is
dependent on the compound's specic heat coecient, .
Q@TA a @TA¡

~E

P
The temperature dependence of @TA is similar to the resistance of the normal-metal - superconductor system,
@TA. For Vanadium Oxide, the hight temperature (metallic) conductance is ten orders of magnitude greater
than the low temperature (dielectric) conductance.
11In the fully linear case (i.e., Q is taken as a simple step function), this equation can be solved exactly:
C
@ T @z; tA
@t
a
@2
@z2
T @z; tACQ0 ¡¢
T Tc
¡
g
e ¡

T T0
¡
12See reaction-diusion systems.
27

3.3 Linear Stability of Nonlinear Stationary States13
Returning to the 1-D (longitudnal) heat transfer equation for the superconducting wire, we will look for some
simple solutions. For simplication of notation, we will denote T a T @z;tA.
To begin with, we will concern ourselves with spatially-uniform and time-independent solutions. This means
that the terms
@T
@t a @2T
@z2 a H. A simple approach is to re-write the equation as Q‘T“ a W ‘T“. We plot both Q
and W as functions of the temperature using the same axes. The solutions will simply be the intersects of the
two graphs.
Clearly there are three such intersects:
1. At T Tc i.e. where the metal is a superconductor. We denote this temperature as TSC.
2. At T % Tc i.e. near the superconductor - normal-metal transition. We denote this temperature at Ttrans.
3. At T Tc i.e. where the metal is a normal metal. We denote this temperature as TNM.
13Appeared in 2007B.
28

Furthermore, in the general case for a known material (% is known) diferent current densities, j, may be such
that there are as few as one or as many as three intersects. Note the unique case of two such intersect points.
We now turn to study the stability of these solutions.
14 In general we may check stability under small
perturbations or large perturbations. The latter cannot be done analytically, and computer simulations need to
be performed for specic large perturbations. However, the test for stability under small perturbations is more
crucial, since small perturbations always exists, and a system which is unstable under small perturbations will
not exist in the real world.
Our equation (the heat transfer equation) is a linear equation, and we will assume linear perturbations of
the form
T @z;tA a Ti @z;tAC @z;tA
where Ti is a given equilibrium solution to be checked, and is a small perturbation, i.e.

Ti ( I. The
linearity of both the equation and the proposed solution gives a new heat transfer equation for :
C
@
@t
a
@P
@zP C

@Q‘T“
@T
@W ‘T“
@T
'
¡
since
Q‘T“ W ‘T“ 9 Q‘Ti“ W ‘Ti“C

@Q‘T“
@T
@W ‘T“
@T
'
¡ CO
4

T
P5
and the non--dependent terms cancel with terms from the rest of the equation (assuming that Ti is indeed
an equilibrium solution of the heat transfer equation).
The heat transfer equation for is a linear equation, and we guess a Fourier type solution @z;tA
H exp@ikz C
tA (i.e., Fourier transform in space and a Laplace transform in time of both sides of the
equation). Clearly, the condition for stability is
H, where the equality is meta-stable.
14In all problems of a current carrying wire, we can tell experimentally if a state is unstable if the I V curve has a negative
slope, i.e. negative resistance. The negative resistivity can be measured by adding a normal resistor in parallel such that the total
measured resistance is posititve, bu the normal resistor's resistance can be changed in order to measure the region of voltages for
which the wire resistance in itself is negative.
29

The transformed equation for is
C
a kP C

@Q‘T“
@T
@W ‘T“
@T
'
¡

a kP
C
C I
C
¡

@Q‘T“
@T
@W ‘T“
@T
'
H
The rst term on the RHS of the last line contributes to stability. This term stems from heat dissipation in
the wire, away from the heated region, this heat counterow increases as k increases. To better understand the
second term, we will treat the exterme case of k a H, where the stability condition is simplied to
@Q‘T“
@T
@W ‘T“
@T
H
@Q‘T“
@T
@W ‘T“
@T
Turning, again, to the graphic solution, we compare the slopes of Q and W at the three possible intersection
points. The conclusion is that the solutions at TSC and at TNM are stable, whereas the transition solution at
Ttrans: is unstable. The TSC solution corresponds to the whole wire being in the superconducting state with a
low enough temperature throughout. The TNM solution corresponds to the whole wire being in a normal-metal
state. Can we derive solutions where the wire is partly at TSC and partly at TNM? How will these solutions
change in time?
Universality with the time-independent Schrödinger Equation
We had
C
a kP C

@Q‘T“
@T
jTi @W ‘T“
@T
jTi
'
¡
let's rename ©@x;tA @z;tA
C
© a
dP
dxP©C

@
@T
fQ WgjTi

¡©
E© a ~P
Pm
dP
dxP©CV ©
The mapping to the Schrödinger equation is
C
U 3 E
U 3 ~P
Pm
@
@T
fW QgjTaTi U 3 V
The requirement of a solution to be stable is
H, which in the case of the time-independent Schrödinger
equation translates to it having a non-negative ground state energy, so that all the energy spectrum is non-
negative E ! H (C is always positive). Let's take a closer look at the limiting case of E a H.
For
a H we had
H a THH
i @zACQ‘Ti @zA“ W ‘Ti @zA“
taking the dirivative by z
H a
dP
dzPTH
i @zAC @
@Ti
fQ‘Ti @zA“ W ‘Ti @zA“g¡TH
i @zA
which is an example of the time-independent Schrödinger equation with E a H, where © a THi. Let's examine
the solutions to this equation, and see whether they indeed are the ground state solutions, or if there could be
solutions with E H, which entail system instability.
15
15The switching wave having an energy E a H can be explained by saying that the switching wave front is leading a macroscopic
(say half) part of the sample. This is not a localized phenomenon, and therefore we expect it to have a low energy.
30

The spatial derivative fo the switching wave soutions (which look like step functions) look like a delta
function.
16 This function has no nodes and therefore we conclude that it is indeed a ground state solution to
the relevant Schrödinger equation. That is, since we got this solution for E a H, all the solutions obey E ! H,
and this solution is stable.
For a domain type solution (i.e. the region between two opposing fronts) the derivative looks like
This function has a node at z a H, so clearly it cannot be a ground state solution. If this solution was gotten
for E a H, than clearly there exists some other solution with E H, making the domain solution unstable.
3.4 Propagation of Single Nonlinear Switching Waves
If we were to analytically solve the linear problem of the step function resistance (Q@TA a QH ¡ ¢@Tc TA),
we would solve for T Tc and for T Tc separately, and then match the two solutions at the point with
T @z;tA a Tc. The location of this seam between the two regions (superconducting, and normal-metal) is time
16The amplitude of the derivative function may seem large, but remember that all solutions can be scaled as we wish it is only
the functional form we are after.
31

dependent, therefore we have a moving front in the wire. This is a switching wave which travels through the
wire at some velocity, v. We now try to nd wave solutions of the form T @z;tA a T @z vtA T @A to the
non-linear heat transfer equation
Cv
dT
d
a
dPT
dP CQ‘T“ W ‘T“
We rewrite this equation as
THH a CvTH fQ Wg
This equation has similar form to the well-known Newtonian equation for a driven and damped particle, or,
the high friction limit of the Langevin equation
17
my a
•y CF
The mapping is
T U3 y
U3 t
U3 m
Cv U3
fQ Wg U3 F
Note that the friction coecient,
a Cv, can have either positive or negative sign, depending on the direction
of motion, v.
A dierent mapping can nd universality between our equation and the Schrödinger equation.3.3
To gain physical intution as to the solution of this problem we concentrate on the last mapping.
The potential energy surface on which the particle is moving is the integral over the force
rU @yA a F @yA
U @yfA U @HA a
yf
H
F @yAdy U3
Tf
H
fQ‘T“ W ‘T“gdT
The pseudo-force, F a fQ Wg looks like
17Note that the friction coecient
here is not the same as in the time dependence power e
t we dealt with in the previous
section.
32

(minus) The integral over which is the pseudo-potential energy surface, U ‘T“18
18Note that shading the areas in the graph for f which contribute to the integral or splitting the integration interval into two
‘TSC; Tc“ and ‘Tc; TNM“, we have Maxwell's Equal Area Rule as it appears in the phase transition analysis of a Van Der Waals
gas, where it is called a Maxwell construction.
33

A direct solution of the heat transfer equation can be helped along using integration of the entire equation:
H a
I
I
dzTH @zAf THH @zACCvTH @zA ‘Q@TA W @tA“g
a @TH @zAAP jI
I CCv
I
I
dz @TH @zAAP
I
I
dzTH @zA‘Q@TA W @tA“
a Cv
I
I
dz @TH @zAAP
I
I
dz
dT @zA
dz
‘Q@TA W @tA“
v a
‚ TNM
TSC dT ‘Q@TA W @tA“
C
‚ I
I dz @TH @zAAP
Alternatively, we can observe that a freely moving particle on such a potential energy surface should obey
34

conservation of energy, including the dissipated energy via the pseudo-friction term CvTH
U ‘TNM“ U ‘TSC“ a
TNM
TSC
CvTHdT
U ‘TNM“ a Cv
I
I
THdT
dz
dz
I
C
U ‘TNM“‚ I
I ‘TH“P dz
a v
Where we took U ‘TSC“ H as a reference energy. The result is the velocity of a wave discribing the temperature
prole in the wire. The prole has tails at the two stable temperatures, TSC and TNM, and the disturbance
itself is a switching wave from one of the temperatures to the other.
The sign of v denes the direction in which the switching wave propagates. It is also possible to get a
standing wave with v a H. This case maps to the case of
a H, for which the energy conservation criterion gives
another interesting insight: The transition time from one turning point (on the potential surface) to the other
(i.e. TSC and TNM) grows to innity as the energy of the particle travelling in the potential well between these
points nears the well depth. This is because the particle feels an increasingly anharmonic well as it investigates
the walls of the well away from its minimum. At rst these anharmonic eects only cause small deviations from
harmonic oscillations of the particle in the well, but as the particle energy grows and it investigates increasingly
higher energies of the well, the approach towards each of the turning points becomes assymptotic.
19 A particle
starting at TSC and rolling down the well towards TNM will never reach it, due to pseudo-energy considerations.
That is, unless
H in which case the pseudo-friction force can actually push the particle up-hill, even past
TNM. All that's needed is the correct wave velocity, and a heating wave can actually switch the temperature
at a point on the wire form one stable value to the other.
A combination of two such wavefronts leads to a temperature prole of a single domain of temperature TSC
(TNM) implanted in a wire at temperature TNM (TSC, respectively). When the initial heating of the wire creates
such a prole, we call it a seed. Depending on the velocities of the two wavefronts (call them vL and vR)
the dierent temperature seed, or domain, may shrink or expand. If vL a vR we have a dierent temperature
domain of xed size moving through the sample. This wave is called a soliton.
19 This is a requirement on the pseudo-potential, needed so that the pseudo-force will be continuous even at the transition points.
35

4 Diusion in Momentum-Space
4.1 Electron-Phonon Collisions in Metals at Low Temperatures20
In a harmonic metalic solid with no defects, the mechanism for electron scattering is via interactions with the
phonons. The electrons in the metal have characteristic energy of the Fermi energy, F , and Fermi momentum,
pF . Phonons in the solid have frequency !, and momentum ~q a ~~k, where ~k is the wavevector.
The phonon spectrum is linear at small wavevector and energy: limj~kj3H !~k a vs

~k

, where the proportion-
ality factor vs is the speed of sound in the solid. The low-frequency density of states is D @!A G !P, the Debye
model assumes this can be extended until some cuto frequency, !D, which is dened from the normalization
condition on the sum of modes. Experimentally, this checks out to be fair.
There is an interaction between the electrons and phonons, but the phonon energy is much smaller than
the electron energy
~!D
F ( I, where !D is the maximal phonon frequency in the Debye model of a solid.
Therefore the momentum transfer in an elecron-phonon interaction is much smaller than the fermi momentum:
j¡~pj a j~pf ~pij a j~qj ( pF . The small momentum transfer means that we can look at a single collision as
a small change in the direction of the electron momentum. For a suciently small change, we can dene the
small angle change as
j'Hj 9 j~qj
j~pj j~pf ~pij
j~pij
Because the phonons are Bosons, the characteristic phonon average energy is proportional to the temper-
ature, ~! $ kBT. For temperatures much lower than the Debye temperature, i.e. kBT ( ~!D kB¢D, only
the lower frequency phonons are populated. This is the source for validity of the use of the linear dispersion
releation !~k a vs

~k

.
vs j~qj a ~! $ kBT
j~qj $ kBT
vs
The electron Fermi energy is of the order F $ ~2
Pma2 , and the Fermi momentum is of the order pF $ ~
a $ qD
(where a is the distance between nearest neighbors, and qD is the Debye phonon momentum, which corresponds
to the Debye phonon frequency !D). Note that the Fermi momentum, which is the characteristic momentum
of the interacting electrons, is temperature independent.
The temperature dependence of the small parameter 'H a q
p:
'H @TA a q @TA
p@TA $ q @TA
pF
$ kBT
vspF
$ kBT
~!D
a T
¢D
( I
Therefor our constraint on the temperature (that it be much smaller than the Debye temperature) is consistent
with the requirement that the single collision direction change be small.
We conclude that electron scattering by phonons consists of small collisions (where the momentum transfer
is small with relation to the initial and nal electron momenta). This can be treated as a diusion process in
momentum space (where each collision is a small step with step size of the momentum transfer).

'P @tA
a h' ¡t
h' a

'PH

where

'P @tA
is the mean square of the direction change of a scattering electron after time t. Also,

'PH

is the
mean square of the direction change in a single collision, and is the characteristic time between collisions.
The macroscopic eect of the diusion process will become appearent (and contribute to such macroscopic
quantities as resistance and heat conductivity) only when
p
h'P @tAi $ Iradian $ IVH. We dene the time
this takes as
t'

'P @t'A
h'
I
h'
a
h'PHi a

kBT
~!D
P a

~!D
kBT
P
¡ )
20Appeared in 2006B.
36

The temperature dependence of @TA is inversly proportional to the phonon density, which (since they are
Bosons) is proportional to TQ (a sphere in momentum space, with radius kBT)
G I
nph:
G I
TQ
In all
t' a

~!D
kBT
P
¡ G

I
T
P
¡ I
TQ a I
TS
This result is attributed to Bloch.
The Drude result for conductivity is Drude a ne2
m , where n is the carrier density, e is their charge, m is
their mass, and is the time between collisions. In our case, we will put a t', so a ne2
m t' G T S.
4.2 Relaxation of Heavy Particles in a Gas of Light Particles
A small number of heavy gas particles with mass M (we denote their momenta as ~p) is surrounded in a large
number of light gas particle of mass m (we denote their momenta as ~q).
A collison (in one frame of reference) is depicted as follows: A heavy gas particle begins with momentum ~pi
and ends up with momentum ~pf. The participating light gas particle gains momentum ~q a ~pi ~pf.
Conservation of energy and momenta:
pPi
PM
a pPf
PM
C qP
Pm
~pi a ~pf C~q
The small parameter requirement for the validity of the treatment as a difusive process in momentum space:
j~qj
j~pij a Pm
MC T m
9 Pm
M
( I
4.3 Fokker-Planck Equation21
Let's write an equation for the change in the fraction of particles with momentum ~p at time t, f @~p;tA. This
change is due to particles with momentum ~p C ~q losing momentum ~q (gain of particles with ~p), and particles
with momentum ~p losing momentum ~q (loss of particles with ~p). The rates of these processes are written as
W @~p C~q;~qA and W @~p;~qA, respectively. Assuming rst order (linear) dynamics, i.e.
@f
@t G f, we get:
@f @~p;tA
@t
a

W @~p C~q;~qAf @~p C~q;tAd~q

W @~p;~qAf @~p;tAd~q
We will assume, as previously, j~pj ) j~qj, and expand the integrands in series:
W @~p C~q;~qAf @~p C~q;tA 9 W @~p;~qAf @~p;tAC
Qˆ
aI
q
@
@p
‘W @~p;~qAf @~p;tA“
C
Qˆ
;

‘W @~p;~qAf @~p;tA“
From now on we will employ the Einstein summation convetion: ab a €Q
aI ab.
Pluggin back into the kinetics equation we get
@f @~p;tA
@t
a @
@p

Af @~p;tAC @
@p

f @~p;tA“
'
A

qW @~p;~qAd~q
B

W @~p;~qAd~q
21Appeared in 2006A, 2006B and 2007B.
37

Where we identied the constants A and B

as something akin to the rst and second moments of the
rate W @~p;~qA, with respect to the second argument, ~q, only.
We can also dene the respective current in momentum space,
j Af @~p;tAC @
@p

f @~p;tA“
Such that
@f
@t
a @j
@p
A @f
@t
C @j
@p
a H
is the momentum space distribution continuity equation.
Equilibrium boundary condition:
At equilibrium we expect stationarity,
@f
@t a @
@p j a H, and we denote the equilibrium momentum distribution
f a fH.
H a j a

A C @B

In a system with no external electromagnetic elds, we have the equilibrium momenta distribution: The Boltz-
mann distribution, fH G exp

p2
PMkBT

@fH
@p

MkBT
fH a H
This gives us a relationship between A and B

which holds for all fH @~pA.
A a B

Putting this back into the kinetics equation we have the Fokker-Plank equation, to which we added an
external force term
22
@f
@t
a @
@p

B

!'
@f
@~p
~F
Scattering of light particles o heavy particles which is isotropic and independet of the heavy
particle's momentum
W @~p;~qA a W @j~qjA a W @qA
Since
B

B
B I
T

qPW @qAd~q
because we have from isotropy that
qP a qxqx Cqyqy Cqzqz a Q¡qxqx
22 @f
@t a @f
@~p
@~p
@t a @f
@~p
•~p a @f
@~p
~F
38

Substituting into the Fokker-Plank equation,
@f
@t
a B
@
@~p

~p
MkBT
f @~p;tAC @f @~p;tA
@~p
'
@f
@~p
~F
Assuming a simple form of W @qA G ¢@q qHA we can write an identity which connects the time between
collisions, , and the scattering rate, W:
I

W @qAd~q
I a ¡ R
Q qQ
H
we already saw that f a q2
0
IH , which means B can be thought of as some scattering length (a one dimentional
cross section).FIND REFERENCE FOR THIS IN LANGEVIN PART!!
Mobility
The mobility is dened as h~vi a B ¡ ~F, where the average velocity is dened (as are all the dynamic variables)
by the (momentum) distribution function h~vi
‚
f @~pA¡~v ¡d~p.
The deviation (response) of the distribution function from the equilibrium distribution (fH) due to a small
external force (perturbation) should be small, and linear in the disruption (perturbation).
f a fH C
( fH
G

~F

plugging this into the Fokker-Plank equation, neglecting terms which are second order (in

~F

A and seeing
that the equilibrium terms cancel out,
B
@
@~p

~p
MkBT
C @
@~p
'
a @fH
@~p
~F
The solution is
a ~p ¡ ~F
B
fH
39

5 Boltzmann Equation
5.1 Liouville's theorem
We discuss a system with N particles in phase space using TN coordinates (positions qi and momenta pi
for i a ‘I;QN“ particles and spatial components). A point in phase space completely describes the entire
conguration of the system. The system is dened using a Hamiltonian, r@qi;piA, and the equations of
motion are the Hamilton equations:
•qi a @r
@pi
•pi a @r
@qi
We would like to write an expression for the probability for the system to be in the innitisimal neighborhood
of a certain point in our TN dimentional phase space at a certain point in time, f @qi;pi;tA. This equation is a
continuity equation
df
dt
Cr@f~vA a H
where ~v is the generalized phase space velocity, and f~v is the phase space current. More explicitly,
H a @f
@t
C
QNˆ
iaI
@
@qi
@f •qiAC @
@pi
@f •piA
!
From the Hamilton equations we have
H a @
@qi
@r
@pi
@
@pi
@r
@qi
a @ •qi
@qi
C @ •pi
@pi
which we can arbitrarily choose to add to the phase space equation of motion
H a @f
@t
C
QNˆ
iaI
@
@qi
@f •qiAC @
@pi
@f •piACf

@ •qi
@qi
C @ •pi
@pi
!
to get
H a @f
@t
C
QNˆ
iaI
@f
@qi
•qi C @f
@pi
•pi
!
what we actually got on the right hand side (RHS) of the equation is the total derivative,
df
dt. This means
that if we follow an innitisimal volume along its trajectory in phase space, we will see no change in the phase
space distribution f inside the volume. This is equivalent to comparing the phase space distribution to the
density of an incompressible uid in hydrodynamics.
The spatial density of particles is simply the spatial part of the phase space distribution
n@~r;tA a

d~pf @~r;~p;tA
the energy current
~q @~r;tA a

d~p~v@~r;~p;tAf @~r;~p;tA
a

d~p
~p
m
@~r;~p;tAf @~r;~p;tA
the electric current (particle charge times particle current)
~j @~r;tA a e

d~p~vf @~r;~p;tA
a e
m

d~p~pf @~r;~p;tA
In our discussion so far, we did not take into account any possible collisions. We will do so now.
41

5.2 Boltzmann equation
Collisions are non-local hops of the phase space distribution in time. For example, consider a binary collision
between a particle with initial momentum ~p and another particle with initial momentum ~pI where their mo-
menta after the collision are changed to ~pH and ~pHI, respectively. The rate of such a process may be written as
W @~p;~pIY~pH;~pHIA.
Since this process should be time-reversible, we expect it to have the symmetry that if t is changed to t and
all momenta are given the opposite sign (e.g. ~p becomes ~p), nothing should be changed. Another symmetry we
expect this process to obey is invariance under spatial inversion: If all positions are inversed, and all momenta
are inversed, nothing is changed (~r becomes ~r, and ~p becomes ~p).
Time reversal symmetry means W of one process should be symmetric to its reverse process with ~pi 3 ~pi
(i.e., taking time and all momenta with opposite signs). Applying time-reversibility to W @~p;~pIY~pH;~pHIA gives
W @~p;~pIY~pH;~pH
IA a W @ ~pH; ~pH
IY ~p; ~pIA
Symmetry under spatial inversion means that W should be the same if all ~pi are exchanged with ~pi. Applying
spatial inversion symmetry to the result (from local isotropy) gives
W @~p;~pIY~pH;~pH
IA a W @~pH;~pH
IY~p;~pIA
The binary collision process gives the following phase space distribution time dependence: The density at a
certain point in phase space is increased due to collisions which send particles to this point in phase space, and
decreased due to collisions which remove particles from this point.
df @~pA
dt
a

W @~pH;~pH
IY~p;~pIAf @~pHAf @~pH
IAd~pId~pHd~pH
I

W @~p;~pIY~pH;~pH
IAf @~pAf @~pIAd~pId~pHd~pH
I
a

W @~pH;~pH
IY~p;~pIA‘f @~pHAf @~pH
IA f @~pAf @~pIA“d~pId~pHd~pH
I
St:ffg
The more explicit equation (known as the Boltzmann equation) is written as (in terms of a single particle's
coordinates)
@f
@t
C @f
@~r
•~r C @f
@~p
•~p a @f
@t
C @f
@~r
~v C @f
@~p
~F a St:ffg
5.3 collision integral
The RHS of the Boltzmann equation is called the collision integral, or Stosszahl, and is physically problematic
since it suggests non-local eects have non-zero importance (i.e. the integration is done over all phase space,
while we know that this cannot be right we could put this information in the process rate, W). In general,
the collision integral may include many body collisions or interactions.
At equilibrium, we expect
df
dt a H. From energy conservation for elastic collisions,
@pHAP
Pm
C @pHIAP
PmI
a pP
Pm
C pPI
PmI
so we see that the equilibrium condition is met since the phase distribution at equilibrium fH @~pA G exp

p2
PmkBT

is such that fH @~pHAfH @~pHIA fH @~pAfH @~pIA a H.
Away from equilibrium, we have no idea how to solve the Boltzmann equation.
5.4 -approximation
A simple model we can use instead of solving the collision integral is to assume a simple relaxation time, , to
the equilibrium phase distribution
df
dt
a @f
@t
C @f
@~r
~v C @f
@~p
~F a f fH

42

5.5 heat conductivity and viscosity of gases
Assuming a constant temperature gradient on the system of interest, we will consider the system at steady state
(
@f
@t a H), and disregard any other external forces ( ~F a H). From the original, -approximated Boltzmann
equation, we are left with
@f
@~r
~v a f fH

we will, again, consider the linear response regime (f a fH C , ( fH). Substituting this into the previous
equation,
@fH
@~r
~v C @
@~r
~v a
we use the chain rule to write
@fH
@~r
a @fH
@T
@T
@~r
a @fH
@T
¡ ~rT
we substitute back and get
@fH
@T
¡ ~rT ¡~v C @
@T
¡ ~rT ¡~v a
however G ~rT and we will disregard the second term on the LHS since it is quadratic in the perturbation,
O
h
@rTAPi
.
In order to discuss pure conduction, we need to disregard convection (radiation does not appear in this
problem since we do not have a radiation eld). To remove convection we need all the particle currents in the
system to be zero, H a j a ‚
d~p~vf. However, if we simply take j a rT a H, we will also have rT a H and
then ~q a H which is a trivial and uninteresting answer, which we know is not the general result we are looking
for. We can get a clue by looking at an ideal gas and noticing that the particle currents are zero if the pressure
is spatially uniform. In the case of an ideal gas,
P a nkBT
H a ~rP a ~rn ¡kBT CnkB ¡ ~rT
H a
~rn
n
C
~rT
T
~rn G ~rT
Therefore, we will expand our discussion from the system of equations
~q a ~rT
~j a ~rT
to the system
~q a ~rT C

~rn
~j a ~rT C~rn
The phase space distribution function with respect to a system governed by the two thermodynamic variables,
T and n leads to the following equation (via the chain rule)
df
d~r
a df
dT
~rT C df
dn
~rn
we will regard the phase distribution function in the linear response regime (f a fH C ), where the
equilibrium phase distribution function is given by the Boltzmann distriubution for an ideal gas,
fH a n
@PmkBTAQ=P ¡exp

pP
PmkBT

Further, for a system under no external forces, and in steady state
df
d~r
~v a f fH

dfH
dT
~rT C dfH
dn
~rn

¡~v a
43

from the denition of fH we also have
dfH
dn
a fH
n
dfH
dT
a fH
T

pP
PmkBT
Q
P

For an ideal gas we have (from the equation of state)
~rn
n
a
~rT
T
let's try to guess a simple generalization of this for some non-ideal gas
~rn
n
a
¡
~rT
T
The relevant Boltzmann equation is now
dfH
d~r
¡~v a fH
T
¡ pP
PmkBT
Q
P
!
¡

~rT

¡~v a
(4)
we will denote a Q
P C
23.
Particle Current
The particles phase space current density is
~j a

~vfd~p (5)
where it is to be expected that at equilibrium the current is zero
~jH a

~vfHd~p a H
therefore we only get a contribution from the term of the distribution function. Instead of plugging into
5, we use 4 to write
~j a

~v
fH
T
pP
PmkBT

!
~v~rT

d~p
or, in component notation (and using the Einstein summing convention for the index k)
ji a

vi
fH
T
pP
PmkBT

!
vk
@T
@xk

d~p
The integral is anti-symmetric with respect to the indeces i and k, i.e. we only have a contribution from the
three cases of i a k (a x;y;z). Therefore,
ji a
@T
@xk
ik
in the isotropic case, the non-directional term, ;24 is taken where vPi a I
QvP (from isotropy) and is therefore
independent of i:
a I
Q
I
H
vP ¡ I
T
n
@PmkBTAQ=P exp

pP
PmkBT

¡ pP
PmkBT

!
¡RpPdp
a R
Qp
I
T
n
@PmkBTAQ=P
I
H
vP ¡exp

pP
PmkBT

¡ pP
PmkBT

!
¡pPdp
a R
Qp
I
T
n
@PmkBTAQ=P
I
H
pP
mP ¡pP ¡ pP
PmkBT

!
¡exp

pP
PmkBT

dp
23c.f. Heat capacity ratio (or adiabatic index).
24c.f. Thermal expansion coecient
44

we perfom the change of variables uP p2
PmkBT
a V
Qp
nkB
m
I
H
uR ¢
uP
£
e u2
du
a V
Qp
nkB
m
V
bbb`
bbbX
I
H
uTe u2
du
| {z }
I@u6A

I
H
uRe u2
du
| {z }
I@u4A
W
bbba
bbbY
The integrals are known (or can be worked out) and equal
I

uR¡
a Qp
V
I

uT¡
a ISp
IT
Therefore
a nkB
m
S
P
!
a nkB
m
‘I
“
From which we see that for
a I we get zero particle current, for all temperature proles.
Energy Current
~q a

~vfd~p
Where we will assume a monoatomic, free gas, a p2
Pm. Since the energy current is zero at equilibrium, we only
get a contribution from (and not from fH)
~q a

~v d~p
a

~v

fH
T
pP
PmkBT

!
~v~rT
'
d~p
a
T

~v
pP
Pm
fH
pP
PmkBT

!
~v~rT

d~p
where in the second line, we have again used 4. Writing the result in component form
qi a
T

vi
pP
Pm
fH
pP
PmkBT

!
vk
@T
@xk

d~p
a
@T
@xk
ik
again, from the anti-symmetry of the integrand with respect to the indeces i and k. The non-directional term
K in the isotropic case is
a P
Qp
n
mT @PmkBTAQ=P
I
H
vPpR pP
PmkBT

!
exp

pP
PmkBT

d~p
a V
Qp
nkPBT
m
I
H
uT ¢
uP
£
e u2
du
with the same change of varialbes as for the particle current. We not also need the result of another integral,
I

uV¡
a IS¡U
QP
p
45

and we use the previous result (for the cancellation of particle current),
a I to get
a S
P
nkPBT
m
So we see that for
a I, a H and Ta H, i.e. there is heat ow but no net particle ow.
The mean time between collisions, , can be dened from the thermal average velocity, hvi, and the mean
free path, `:
9 `
hvi
whereas the mean free path is related to the particle density, n, and the scattering cross section,
` 9 I
n
so that
9 I
n hvi
this means that
a S
P
kPBT
m hvi G
p
T
the heat conductivity is proportional to the square root of the temperature, and is independent of the particle
density.
It should be emphasized that the entire derivation is only valid where the temperature gradient is small on
the length scale of the mean free path
¡T
L
( ¡T
`
where L is the length scale over which the gradient is computed.
Qualitative Explanation of the Dynamics
The particle ux through a point, xH, in the positive x direction is
qCx @xHA a n
T hviC ‘T @xH À“
where, in the isotropic case, the chance that a particle moving through xH is moving in the positive x direction
is
I
T. Further, we allowed the heat capacity to have a temperature dependence, and its eect is taken over a
particle comming into xH from ` away, and travelling with thermal speed hvi.
q x @xHA a n
T hviC ‘T @xH CÀ“
The net energy current in the x direction is
qx @xHA a nhvi
T fC ‘T @xH À“ C ‘T @xH CÀ“g
Expanding the temperature prole in Taylor series,
T @xH ¦À a T @xHA¦`
@T
@x
jx0 CO

@PT
@xP
!
The rst term in the expansion is cancelled in the net current
qx @xHA a nhvi
T

P¡ dC
dT
`
dT
dx
'
a nhvi`
Q
dC
dT| {z }

dT
dx
9 hvi
Q
dC
dT
G
p
T
and we have the same dependence of on temperature and independence of particle density that we got before.
46

Statistical Mechanics & Thermodynamics 2: Physical Kinetics

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