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Solution manual - Chapter 1
Jens Zamanian
March 9, 2014
1.1 Poisson Distribution
a) Divide the time t in small increments dt. For each of these increments the probability of a
collision is dt/⌧ and hence the probability of no collision is 1 dt/⌧ . The number of intervals of
length dt are t/dt. The probability of not having a collision is then the product of the probabilities
for each of the intervals
✓
◆t/dt
2
t
t
dt
Pno coll. (t) = 1
= e dt ln(1 dt/⌧ ) = e dt ( dt/⌧ +O(dt )) = e t/⌧
(1)
⌧
Alternatively, probability of no collision Pno coll. (t) can be calculated as follows. In order for the
particle to not have a collision during the last t + dt seconds it must not have collided in the last
t seconds also it must not collide in the last part dt. (Note, time t is previous time).
✓
◆
dt
Pno coll. (t + dt) = 1
Pno coll. (t)
(2)
⌧
which gives
dPno coll.
Pno coll. (t + dt) Pno coll. (t)
1
(t) =
=
Pno coll. (t).
(3)
dt
dt
⌧
Solution gives Pno coll. (t) = e t/⌧ .
b) The probability that the time between two successive collisions of an electron is in the
interval [t, t + dt] is obtained as follows. Let t = 0 be the time directly after the previous collision.
In order to have a collision in the interval [t, t + dt] it must first not collide in the interval [0, t]
and then collide in the interval [t, t + dt]. The probability for this to happen is now obtained as
the product of the probability of these events
dt
e
⌧
t/⌧
.
(4)
The probability of having a collision in the time interval t to t + dt is usually written P (t)dt where
then
1
P (t) = e t/⌧
(5)
⌧
is the probability density function.
c) We now want to look at the time since the last collision averaged over all electrons. If
we assume that we have N electrons we will on average have N e t/⌧ electrons which did not
collide during the preceding time t. Similarly we have N e (t+dt)/⌧ that did not collide during the
preceding time t + dt. The di↵erence in number comes from the fact that some electrons actually
did collided during the time [ (t + dt), t] (t = 0 is now), the number of such electrons is
✓
◆
e (t+dt)/⌧ e t/⌧
e t/⌧
t/⌧
(t+dt)/⌧
Ne
Ne
=
N
dt = N
dt
dt
⌧
=
N
e
t/⌧
⌧
dt
1
(6)
This is hence the number of electrons that collided in a small interval of time [ (t + dt), t] in
the past. Now we want to find time since last collisions average over all electrons:
Z
1 1
e t/⌧
hti =
tN
dt = ⌧.
(7)
N 0
⌧
d) Subtle di↵erence between this and the last problem! The previous problem was an average
over all electrons. Here we just consider a single electron. For a randomly picked electron it has
a collision in the interval [t, t + dt] with probability (dt/⌧ )e t/⌧ , where t is the time since the
previous collision. The average time between collisions is hence
Z 1
e t/⌧
hti =
t
dt = ⌧.
(8)
⌧
0
e) The time T between the last and next collision average over all electrons is 2⌧ . That the
time is larger than ⌧ can be understood as follows: When we average over the electrons we choose
a specific moment. When we do this there will be more electrons that are in a long time interval
(between collisions) than in a short one.
To derive probability distribution for T we choose a random electron. The fraction of electrons
which will have a collision during the following time intervall [t1 , t1 + dt1 ] is given by e t1 /⌧ dt1 /⌧ .
Out of these a fraction e t2 /⌧ dt2 /⌧ collided in the preceding time intervall [ (t2 + dt2 ), t2 ]. Hence,
the fraction of electrons which collided in [ (t2 + dt2 ), t2 ] and will collide in [t1 , t1 + dt1 ] is hence
e
t1 /⌧
e
t2 /⌧
dt1 dt2
⌧2
.
(9)
Now we are interested in the time T = t1 + t2 between the collisions. Writing the previous result
in terms of this we get
e t1 /⌧ e (T t1 )/⌧ dt1 dT
e T /⌧
=
dt1 dT.
(10)
2
⌧
⌧2
We now want to find the average value for T = t1 + t2 . This is obtained as
Z 1Z 1
dt1 dt2
ht1 + t2 i =
(t1 + t2 )e (t1 +t2 )/⌧
= 2⌧.
(11)
⌧2
0
0
Problem 1.2 - Joule Heating
For thermal averages we have the following properties
⌧
hvi
mv 2
2
=
hvˆ
vi = 0
(12)
=
(T ),
(13)
where v is the velocity directly after a collision. The first condition says that direction is completely
random after a collision. The second says that the speed is dependent of the temperature (at the
point of the collision).
a) The force from the electric field is eE and we hence have ma = eE which gives the
velocity at a time t after a collision
eEt
v = v0
.
(14)
m
After a time t a second collision occurs after which the velocity is v1 . We then get the change in
energy
✓
◆2
mv2
mv12
mv02
m eEt
mv12
E=
=
ev0 · Et +
(15)
2
2
2
2
m
2
2
⌦ ↵ ⌦ ↵
Now we take the thermal average over this. From (13) we get that v12 = v02 and also that the
second term on the right hand side vanishes. Hence we get
h Ei =
(eEt)2
.
2m
(16)
b) The average loss per electron and per collision is now (using that the probability of collision
at the interval t to t + dt after the previous collision is dte t/⌧ /⌧ ) is given by
Z 1
e t/⌧ (eEt)2
(eE⌧ )2
dt
=
.
(17)
⌧
2m
m
0
If n is the number of electrons per cubic centimeter we get that the total loss per cubic centimeter
and second is given by
ne2 ⌧ 2 E 2 1
ne2 ⌧ 2
=
E = E2
(18)
m
⌧
m
where 1/⌧ is the number of collisions per second. In a wire of length L and cross section A the
volume is AL and we have the loss per second in the wire
P = AL E 2 =
A2 L( E)2
A2 Lj 2
L
=
= I2
= I 2 R.
A
A
A
(19)
Problem 1.3 - Thomson E↵ect
We now assume that the speed of the electron after a collision vi = |vi | is given by
mvi2
= E(T (ri )),
2
(20)
where T (ri ) is the temperature at point ri where the collision occurs (to simplify the problem we
assume that the speeds are not randomly distributed but only the direction, see not in the end).
We still assume that
hvi i = 0,
(21)
because the direction is random after a collision. Again we have that for a single electron which
collides at one point r0 and then again at r1 at a time t later is
✓
◆2
mv02 (r0 )
m eEt
mv12 (r1 )
E =
ev0 · Et +
2
2
m
2
✓
◆2
m eEt
= E(r0 ) E(r1 ) ev0 · Et +
.
(22)
2
m
Now from regular mechanics we have
r1 = r0 + r = r0 + v 0 t
eEt2
.
2m
(23)
Assuming now that the distance r is small compared to the typical change in thermal energy we
may expand E(r1 ) around r0 to get
E(r1 ) = E(r0 ) + r · rE(r0 ) = E(r0 ) + ( r · rT )
dE
,
dT
(24)
where the last equation is obtained since the thermal energy depends on the position only through
the temperature. Inserting this into Eq. (??) we get
E
=
✓
v0 t
eEt2
2m
◆
dE
· rT
dT
3
m
ev0 · Et +
2
✓
eEt
m
◆2
.
(25)
Now we can take the thermal average and the two terms which are linear in v0 are then vanishes
as all directions are equally probable. The result is then
h Ei
=
dE
(eEt)2
eEt2
· rT
+
.
2m
dT
2m
(26)
Finally we average over time distribution of the collisions to get

Z 1
e t/⌧ eEt2
dE
(eEt)2
eE⌧ 2
dE
(eE⌧ )2
dt
· rT
+
=
· rT
+
⌧
2m
dT
2m
m
dT
m
0
(27)
Finally, the there is one collision per 1/⌧ and the density of electrons is n so the energy loss per
second and volume is
ne⌧ dE
n(eE)2 ⌧
P =
E · rT +
,
(28)
m dT
m
where the first part is due to the thermal gradient.
Note on averages: In the solution above we were a bit sloppy and assumed that the velocity vi
directly after a collision satisfies Eqs. (20) and (21). However, according to Drude’s one really
should assume that not only the direction is random but also energies (and hence alos the |vi |) are
random, i.e.
⌧
mvi2
= E(T (ri ))
(29)
2
hvi i = 0.
(30)
We we do this we run into a bit of problem since we need to take the average to obtain the last
step of Eq. (22). In order to solve the problem now, we have to separate two types of averages,
one over the direction and one over the magnitude of the velocity. If we denote these h·idir. and
h·imag. we then have
⌧
hvi idir.
mvi2
2
mag.
=
0
(31)
=
E(T (ri )).
(32)
ˆ vT where vT is the thermal velocity and v
ˆ is a random
If we average vi over the magnitude we get v
unit vector which vanishes when we take the average over direction. Now, when solving the problem
above we first take the average over magnitude to obtain the last step in Eq. (22). The average
over the direction in order to get from Eq. (25) to (26).
1.4 - Helicon Waves
We have a magnetic field H = Hˆ
z and an electric field E(t) = Re E(!)e i!t perpendicular to
this, i.e. E(!) = (Ex , Ey , 0). We want to consider circularly polarized electric fields so that
E(!) = (Ex , ±iEx , 0).
a) The momentum equation is then given by

dp
p
1
=
e E+
p⇥H .
(33)
dt
⌧
mc
Assuming a function p(t) = Re p(!)e
i!p =
p
⌧
i!t
. Inserting this into the equation we get
eˆ
xEx ⌥ ieˆ
y Ey
4
eH
ˆ
(py x
mc
ˆ)
px y
(34)
If we divide this into components we get the equations
i!px
=
i!py
=
i!pz
=
1
px eEx !c py
⌧
1
py ⌥ ieEx + !c px
⌧
1
pz .
⌧
(35)
(36)
(37)
The last equation gives pz = 0. The two other we write in matrix form
✓
◆✓
◆
✓
◆
i! + ⌧1
!c
px
1
=
eE
x
py
±i
!c
i! + ⌧1
(38)
Using the inverse of a two-by-two matrix (switch places on diagonal, switch sign on the other two
and divide by the determinant) we get
✓
◆
✓
◆✓
◆
eEx
px
i! + ⌧1
!c
1
=
(39)
py
±i
!c
i! + ⌧1
( i! + 1/⌧ )2 + !c2
From this we get
px
=
py
=
✓
◆
eEx
1
i! + ⌥ i!c
( i! + 1/⌧ )2 + !c2
⌧
✓
✓
◆◆
eEx
1
!c ± i
i! +
= ±ipx .
( i! + 1/⌧ )2 + !c2
⌧
Now we have the relation
(40)
(41)
nep
m
j=
(42)
so we get
jx
=
=
=
=
1
ne2 Ex
⌧ ⌥ i!c
=
2
m
i! + ⌧1 + !c2
⇥1
⇤
i(! ± !c )
ne2 Ex
⌧
⇥1
⇤⇥
m
i(! !c ) ⌧1 i(! +
⌧
ne2 ⌧ Ex
1
ne2 Ex
m
m
i! +
1
0
1
⇥1
1
⌧
⌧
!c )
⇤
i! ⌥ i!c
⇤⇥
i! + i!c ⌧1 i!
i!c
⇤
i(! ⌥ !c )⌧
i(! ⌥ !c )⌧
Ex .
(43)
Note that this is a correction to the conductivity tensor 0 which is now given by the term above,
i.e.
0
=
(44)
1 i(! ⌥ !c )⌧
b) Looking for a solution of the form E = E0 (1, ±i, 0)eikx
the book
!2
r2 E = 2 ✏(!)E,
c
we get that k and ! must satisfy the equation
k 2 c2 = ! 2 ✏(!).
i!t
we insert this into Eq. (1.34) in
(45)
(46)
Furthermore we have (from the book) that
✏(!) = 1 +
5
4⇡i
.
!
(47)
Inserting our conductivity tensor we have
✏(!) = 1 +
4⇡i
! 1
0
=1+
i(! ⌥ !c )⌧
ne2 ⌧
=1
i(! ⌥ !c )⌧ )
4⇡i
! m(1
!p2
1
,
! ! ⌥ !c + i/⌧
(48)
where !p2 = 4⇡ne2 /m is the plasma frequency (SI-units !p2 = ne2 /(✏0 m)).
c) Now we wish to solve the equation to see what the solutions are. Choosing Ey = +iEx (the
upper sign) we have the equation
k 2 c2 = ! 2
!p2
!
!c + i/⌧
!
(49)
We now need some relevant values. One condition that is wanted is that eH/m = !c ⌧
1. This
implies that the magnetic field should be strong and that the collision time is not too short. In
the book they have ⌧ ⇠ 10 14 s which is much too short since it would require a field of a few
100 T.
However, by lowering the temperature to a really low value the collision time can be increased
significantly. In potassium (K) at 4.2 K the collision time turns out to be around 10 10 s (we have
used n ⇠ 1022 /cm3 and that the resistivity is ⇢ = 1/ = 2.0 ⇥ 10 11 ⌦m). If we further assume
that we have a magnetic field of around 10 T we get !c ⌧ ⇠ 100 so that our condition is fulfilled.
In CGS-units we use (SI-units within parenthesis)
n
c
=
e
=
m
H
1022 /cm3
=
(1028 / m3 )
10
4.80 ⇥ 10
9.11 ⇥ 10
=
5
=
10 gauss
10
28
(50)
(3.00 ⇥ 108 m/s)
3.00 ⇥ 10 cm/s
(9.11 ⇥ 10
g
19
(1.60 ⇥ 10
esu
34
(51)
C)
(52)
kg)
(53)
(10T)
(54)
For these typical values we have
!p
=
!c
=
⌧
=
5.6 ⇥ 1015 /s ⇡ 1016 /s
(55)
1.8 ⇥ 1012 /s ⇡ 1012 /s
10
10
(56)
s
(57)
From Eq. (46) we note that a real k (imaginary k implies damping) is obtained if ✏(!) > 0. First
we consider ! > !p . Using that !p
!c and that !⌧ > !p ⌧
1 we have
✏(!) = 1
!p2
! !
1
!c + i/⌧
!p2
! !⌧
=1
⌧
⇡1
!c ⌧ + i
!p2
!2
(58)
and we see that this is indeed larger than unity. For ! < !c we have
!p2
! !
✏(!) = 1
1
!c + i/⌧
d For ! ⌧ !c we have
✏(!) ⇡
which yields
!p2
k c =!
!c
2 2
or
For our values together with a wavelength of
⇡1+
!p2 1
> 1.
! !c !
!p2
,
!!c
! = !c
(60)
✓
k 2 c2
!p2
◆
.
3
(61)
= 1 cm (k = 2⇡/ ) this gives
! ⇠ 0.1/s.
For 10 kgauss = 0.1T we get ! ⇠ 10
(59)
/s. Really slow oscillations!
6
(62)