Solutions to Chen's Plasma Physics

Solutions to Chen’s Plasma Physics
Kalman Knizhnik
1-1. Compute the density (in units of m −3 ) of an ideal gas under the following conditions:
a) At 0 o C and 760 Torr pressure (1 Torr = 1mm Hg). This is called the Loschmidt
number.
b) In a vacuum at 10 −3 Torr at room temperature (20 o C). This number is a useful one
for the experimentalist to know by heart (10 −3 Torr = 1 micron).
a) Avogadro’s number is N A = 6.022 × 10 23 . One mole of gas at STP occupies 22.4 liters. 1 liter is
1 × 10 −3 cubic meters. Thus, the number per cubic meter is N A /n = 6.022 × 10 23 /(22.4 × 10 −3 ) =
2.66 × 10 25 m −3 . Thus, the Loschmidt number is 2.66 × 10 25 ✷
b) Using PV=NkT, we obtain (with R= 1.4 × 10 −23 J K −1 and 1 Torr = 133 Pa):
n = N V = P kT = 10 −3 × 133
1.4 × 10 −23 × (20 + 273) = 3.3 × 1019 m −3 ✷ (1)
1-2. Derive the constant A for a normalized one-dimensional Maxwellian distribution
ˆf(u) = Ae −mu2 /2kT
(2)
such that
This one is straightforward. Just integrate:
1 =
∫ ∞
−∞
∫ ∞
−∞
Ae −mu2 /2kT du = A
ˆf(u)du = 1 (3)
√
2πkT
m
√ m
⇒ A =
2πkT ✷ (4)
1-4. Compute the pressure, in atmospheres and in tons/ft 2 , exerted by a thermonuclear
plasma on its container. Assume kT e = kT i = 20keV , n = 10 21 m −3 and p = nkT , where
T = T i + T e .
This is just unit conversion, albeit with units that nobody really ever remembers. For reference,
1 keV = 1.6 × 10 −19 J, so
p = 10 21 × (20keV + 20keV ) = 4 × 10 22 m −3 keV = 4 × 10 3 m −3 J = 4 × 10 3 N/m 2 (5)
But 1 atm = 10 5 N/m 2 = 1 ton/ft 2 , so (Note: I think there is a mistake in Chen’s solutions here.
If I am mistaken, please let me know).
p = 0.04 atm = 0.04 ton/ft 2 ✷ (6)
1-5. In a strictly steady state situation, both the ions and the electrons will follow the
Boltzmann relation
n j = n 0 e −q jφ/kT j
(7)
For the case of an infinite, transparent grid charged to a potential φ, show that the
shielding distance is given approximately by
λ −2
D
= ne2
ɛ 0
( 1
kT e
+ 1
kT i
) (8)
Show that λ D is determined by the temperature of the colder species.
We’ll use Poisson’s equation
∇ 2 φ = en e − en i
= en 0
(e eφ/kTe − e −eφ/kT i
) ≈ en 0
(1 + eφ − 1 − −eφ ) = e2 n
( φ +
φ ) (9)
ɛ 0 ɛ 0 ɛ 0 kT e kT i ɛ 0 kT e kT i

Now we’ll suppose the φ goes like a decreasing exponential: φ = φ 0 exp(−x/λ D ). Thus, the Laplacian
acting on this is
∇ 2 φ = 1
λ 2 φ = ne2 ( φ +
φ ) ⇒ 1
D
ɛ 0 kT e kT i λ 2 D
= ne2
ɛ 0
( 1
kT e
+ 1
kT i
) ✷ (10)
To show that λ D is determined by the colder species, we suppose first that the electrons are the
colder species: T e ≪ T i . Then,
1
λ 2 D
√
= ne2
kT e ɛ 0
⇒ λ D =
kT e ɛ 0 ne 2 ✷ (11)
Alternatively, if the ions are colder, T i ≪ T e , then a similar analysis yields:
1
λ 2 D
√
= ne2
kT i ɛ 0
⇒ λ D =
kT i ɛ 0 ne 2 ✷ (12)
1-6. An alternative derivation of λ D will give further insight to its meaning. Consider
two infinite, parallel plates at x = ±d, set at potential φ = 0. The space between them
is uniformly filled by a gas of density n of particles of charge q.
a) Using Poisson’s equation, show that the potential distribution between the plates
is
φ = nq
2ɛ 0
(d 2 − x 2 ) (13)
b) Show that for d > λ D , the energy needed to transport a particle from a plate to the
mid plane is greater than the average kinetic energy of the particles.
1-9. A distant galaxy contains a cloud of protons and antiprotons, each with density
n = 10 6 m −3 and temperature T = 100 o K. What is the Debye length?
The Debye length is given by
λ D = ∑ ɛ 0 kT j
n
j j e 2 (14)
j
Plugging in the numbers:
λ D = 8.85 × 10−12 × 1.4 × 10 −23 × 100
10 6 × (1.6 × 10 −19 ) 2 = 0.48 m ✷ (15)
As a check, use the SI unit form for the Debye length given in Chen. If T is in Kelvin, and n is in
cubic meters, then:
√
√
T
λ D = 69
n m = 69 × 10 2
10 6 m = 69 × 10−2 m ✷ (16)
This is the same order of magnitude so we are ok.
1-10. A spherical conductor of radius a is immersed in a plasma and charged to a
potential φ 0 . The electrons remain Maxwellian and move to form a Debye shield, but
the ions are stationary during the time frame of the experiment. Assuming φ 0 ≪ kT e /e,
derive an expression for the potential as a function of r in terms of a, φ 0 , and λ D . (Hint:
Assume a solution of the form e −br /r.)
Let’s assume a solution of this form: φ = Ae −br /r. Then,
∇ 2 φ = 1 ∂ ∂φ
r 2 (r2
∂r ∂r ) = b2 φ = e (n e − n i ) (17)
ɛ 0

Since the electrons are Maxwellian, they obey n e = n 0 e eφ/kTe ≈ n 0 (1+eφ/kT e ). The ions, however,
are stationary, so n i = n 0 . Thus we have:
b 2 φ = e ɛ 0
(n 0 + n 0
eφ
kT e
− n 0 ) = n 0e 2 φ
ɛ 0 kT e
≡ φ
λ 2 D
⇒ b = 1
λ D
(18)
Thus, so far we have:
φ = A e−r/λ D
r
But we also need to match the boundary condition that φ(a) = φ 0 . So,
(19)
So, finally we have our answer:
φ 0 = A e−a/λ D
a
⇒ A = aφ 0 e a/λ D
(20)
φ(r) = φ 0 e a/λ D
a e−r/λ D
✷ (21)
r
You know what they say: if it satisfies Poisson’s equation and the boundary conditions then it
must be the answer.
2-3. An ion engine (see Fig. 106) has a 1-T magnetic field, and a hydrogen plasma
is to be shot out at an E × B velocity of 1000 km/s. How much internal electric field
must be present in the plasma?
The E × B velocity is given by
v = E × B
B 2 (22)
Plugging in the numbers:
10 6 m/s = |E|
1T
⇒ |E| = 1000 V/m ✷ (23)
2-4. Show that v E is the same for two ions of equal mass and charge but different
energies, by using the following physical picture (see Fig. 2-2). Approximate the right
half of the orbit by a semicircle corresponding to the ion energy after acceleration
by the E field, and the left half by a semicircle corresponding to the energy after
deceleration. You may assume that E is weak, so that the fractional change in v ⊥ is
small.
If the energy of the right part of the orbit is E 1 and the energy of the left part of the orbit is E 2 ,
then we have
E 1 = E 0 + eEr 1 , E 2 = E 0 − eEr 2 (24)
where E 0 is the initial energy and E is the electric field. The velocity is determined by v = √ 2E/m,
so
√
√
2E 0 + 2eEr 1 2E 0 − eEr 2
v 1 =
v 2 =
(25)
m
m
The Larmor radius is determined via r = mv ⊥ /qB, so
r 1,2 = m qB
√
2E 0
m
√
1 ± eEr 1,2
E 0
=
√ √
2mE0
qB (1 ± eEr 1,2 2E 0
) =
2E 0 m
1
ω c
± Er 1,2
2 √ mE 0 ω c
(26)
Thus,
r 1,2 (1 ∓
√ √
√
m E
2E 0 qB ) = 1 2E 0
ω c m (1 ± eE 2E 0
2E 0 ω c m ) (27)

The guiding center moves a distance r 1 − r 2 :
r 1 − r 2 =
√ √
eE 2E 0 1 2E 0
E 0 ω c m ω c m
= 2eE
mω 2 c
(28)
The velocity of the guiding center is
v gc = 2 r 1 − r 2
T
= 2 ω c
2π (r 1 − r 2 ) =
4eE
2πmω c
=
since ω c = eB/m. This is a pretty good approximation.
2eE = 2E
πmω c πB ≈ E B ✷ (29)
2-5. Suppose electrons obey the Boltzmann relation of Problem 1-5 in a cylindrically
symmetric plasma column in which n(r) varies with a scale length λ; that is
∂n/∂r = −n/λ.
a) Using E = −∇φ, find the radial electric field for a given λ.
b) For electrons, show that the finite Larmor radius effects are large if v E is as large
as v th . Specifically, show that r L = 2λ if v E = v th .
c) Is (b) also true for ions?
Hint: Do not use Poisson’s equation.
a) We simply solve for φ from the Boltzmann relation for electrons.
Therefore,
n = n 0 e eφ/kTe ⇒ φ = kT e
e ln( n n 0
) (30)
E = −∇φ = − ∂φ
∂r ˆr = −kT e n 0 1 ∂n
e n n 0 ∂r ˆr = kT e ˆr ✷ (31)
eλ
b) We start with the definitions of v E , v th , and r L :
√
v E = E B , v 2kT e
th =
m ,
So, calculating the magnitude of v E :
r L = mv ⊥
eB
v E = kT e
eλB = mv2 th 1
2 eλB = r Lv th
2λ
where in the last step I have assumed that the perpendicular velocity is the thermal velocity. Now,
setting v E = v th , it is easy to see that
r L = 2λ ✷ (34)
c) Sure, why not?
(32)
(33)
2-6. Suppose that a so-called Q-machine has a uniform field of 0.2 T and a cylindrical
plasma with kT e = kT i = 0.2 eV . The density profile is found experimentally to be of
the form
n = n 0 exp[exp(−r 2 /a 2 ) − 1] (35)
Assume the density obeys the electron Boltzmann relation n = n 0 exp(eφ/kT e ).
a) Calculate the maximum v E if a = 1 cm.
b) Compare this with v E due to the earth’s gravitational field.
c) To what value can B be lowered before the ions of potassium (A = 39, Z = 1) have
a Larmor radius equal to a?
We solve for φ:
n 0 exp[exp(−r 2 /a 2 ) − 1] = n 0 exp(eφ/kT e ) ⇒ φ = kT e
e (e−r2 /a 2 − 1) (36)

Thus, the electric field is
and so v E (and it’s maximum) is
E = − ∂φ
∂r ˆr = kT e 2r /a 2ˆr e a 2 e−r2 (37)
v E = E B = 2rkT e
ea 2 B e−r2 /a 2 (38)
∂v E
∂r = 2kT e
ea 2 B e−r2 /a 2 − 4r2 kT e
ea 4 B e−r2 /a 2 = 0 ⇒ r =
So, with a = 1 cm,
√
v E,max = 2kT e a 2 ∣ ∣∣∣a=1
ea 2 B 2 e−1/2 cm,B=0.2 T,kTe=0.2 keV
√
a 2
2
✷ (39)
≈ 8.5 km/sec ✷ (40)
b) If we assume these are potassium ions, we have mg = 39 × 1.6 × 10 −27 × 9.8 = 6.4 × 10 −25 N.
Meanwhile, if we plug in the numbers above into the expression for the electric field (equation 37),
we’ll get that E = 17 V/m. Thus, the force due to the electric field is eE = 1.6 × 10 −19 × 17 =
2.8 × 10 −18 N. Thus the gravitational drift is
F g 6.4 × 10−25
=
F E 2.8 × 10 −18 ≈ 1.5 × 10−7 (41)
times smaller. ✷
c) The Larmor radius is r L = mv th /qB, so, in terms of the constants of v th , we have (setting
r L = a):
√
r L = m √
2kT e
qB m = a ⇒ B = 2mkTe
q 2 a 2 (42)
Plugging in the numbers:
√
2 × 39 × 1.6 × 10
B =
× 0.2 × 1.6 × 10 −19
(17 × 1.6 × 10 −19 ) 2 × (0.1 × 10 −2 ) 2 = 4 × 10 −2 T ✷ (43)
2-8. Suppose the Earth’s magnetic field is 3 × 10 −5 T at the equator and falls off as
1/r 3 , as for a perfect dipole. Let there be an isotropic population of 1−eV protons and
30 − keV electrons, each with density n = 10 7 m −3 , at r = 5 earth radii in the equatorial
plane.
a) Compute the ion and electron ∇B drift velocities.
b) Does an electron drift eastward or westward?
c) How long does it take an electron to encircle the earth?
d) Compute the ring current density in A/m 2 .
Note: The curvature drift is not negligible and will affect the numerical answer, but
neglect it anyway.
a) The grad-B drift is given by
v ∇B = 1 2 v ⊥r L | B × ∇B
B 2 | = 1 2 v ⊥r L | ∇B
B | (44)
We can calculate v ⊥ from the energy, and r L from the magnetic field:
√
√
2E
v ⊥ =
m , r L = mv ⊥
eB
= m 2E
eB m
Thus,
v ∇B = 1 2
(45)
√ √
2E m 2E
m eB m |∇B B | = E
eB |∇B B | (46)

Since B ∼ r −3 , we obtain
So
∇B = ∂B
∂r ˆr = − 3 r 4 = −3B r
v ∇B = 3E
eBr =
⇒
3Er3
eB 0 Rer 3 = 3Er2
eB 0 Re
3
=
| ∇B
B | = 3 r
3E(eV )r2
B 0 R 3 e
where I have used the fact that B = B 0 R 3 e/r 3 . Now we can plug in the numbers:
(47)
(48)
v ∇B,e = 3 × 30 × 103 × (5 × 6 × 10 6 ) 2
3 × 10 −5 × (6 × 10 6 ) 3 = 1.3 × 10 4 m/s ✷ (49)
v ∇B,i = 3 × 1 × (5 × 6 × 106 ) 2
3 × 10 −5 × (6 × 10 6 = 0.42 m/s ✷ (50)
) 3
b) The magnetic field is azimuthal, from north to south, i.e. the −ˆθ direction. The gradient of the
magnetic field is clearly in the radial direction, so we have B × ∇B ∼ ˆr × ˆθ = ˆφ, which is eastward.
This is for the electrons. The ions, which come without the minus sign to cancel the minus sign in
equation 47, will go in - ˆφ, which is westward. ✷
c) Well, it has to travel a distance L = 2π(5R e ), with the velocity v ∇B,e . So,
T =
L 2π × 5 × 6.4 × 106
=
v ∇B,e 1.3 × 10 4 ≈ 4.5 hours ✷ (51)
d) The current density is given by J = nev, so, using the grad-B velocity in this expression, we get
J = nev ∇B,e = 10 7 × 1.6 × 10 −19 × 1.3 × 10 4 = 2 × 10 −8 A/m 2 ✷ (52)
2-10. A 20 − keV deuteron in a large mirror fusion device has a pitch angle θ = 45 o at
the mid plane, where B = 0.7 T . Compute it’s Larmor radius.
The Larmor radius is given by
r L = mv ⊥
(53)
qB
In natural units, a deuteron has m = 2 and q = +1. Furthermore, v ⊥ = vsin(θ). To find v, we
convert the energy to velocity:
√
√
r L = m 2E
eB m sin(θ) = 2 × 1.6 × 10−27 2 × 20 × 10 3 × 1.6 × 10 −19
1.6 × 10 −19 × 0.7 2 × 1.6 × 10 −27 sin(45 o ) = 0.3 m ✷ (54)
2-12. A cosmic ray proton is trapped between two moving magnetic mirrors with
R m = 5 and initially has W = 1 keV and v ⊥ = v || at the mid plane. Each mirror moves
toward the mid plane with a velocity v m = 10 km/sec (Fig. 2-10).
a) Using the loss cone formula and the invariance of µ, find the energy to which the
proton will be accelerated before it escapes.
b) How long will take to reach that energy?
1. Treat the mirrors as flat pistons and show that the velocity gained at each bounce
is 2v m .
2. Compute the number of bounces necessary.
3. Compute the time T it takes to traverse L = 10 10 km that many times. Factor of
two accuracy will suffice.
a) The loss cone formula is sin 2 (θ m ) = 1/R m , where θ m is the angle of the magnetic mirror and
R m is the mirror ratio B max /B 0 . We can also write down a formula for sin(θ m ) in terms of the
parallel and perpendicular velocities:
sin(θ m ) = 1 √
Rm
=
v ⊥,f
√
v
2
⊥,f
+ v 2 ||,f
=
1
√
1 + (
v ||,f
v ⊥,f
) 2 (55)

Squaring both sides, and noting that, since µ is invariant, v ⊥,f = v ⊥,i , we have
1
R m
=
1
1 + ( v ||,f
v ⊥,i
) = 1 2 5
⇒ v ||,f = 2v ⊥,i (56)
Thus, we can get the final energy:
W f = 1 2 m(v2 ⊥,f + v 2 ||,f ) = 1 2 m(v2 ⊥,i + 4v 2 ⊥,i) = 5 2 mv2 ⊥,i (57)
But we can’t evaluate this without knowing what the original v ⊥,i is. Fortunately, we know that
initially v ⊥ = v || , so
W i = 1 2 m(v2 ⊥,i + v 2 ||,i ) = 1 2 m(2v2 ⊥,i) = mv 2 ⊥,i = 1 keV (58)
So, finally,
W f = 5 2 W i = 2.5 keV ✷ (59)
b) In the frame of the piston, when the particle bounces off, it’s velocity doesn’t change. In the
piston’s frame, the velocity of the particle as it is coming in is v i − v m , where v m is the velocity of
the piston (it is negative). It’s final velocity is the same but negative. Thus,
v ′ i = v i − v m , v ′ f = v m − v i (60)
where the prime denotes the velocity in the piston’s reference frame. But, in the lab frame,
v f = v f ′ + v m, so we have
v f = 2v m − v i (61)
Thus, the change in velocity on each bounce is 2v m = 20 km/sec. The initial proton velocity is
v i =
√
2W i
m
= √
2 × 1 × 10 3 × 1.6 × 10 −19
1.6 × 10 −27 = 447 km/s (62)
The proton final energy is 2.5 keV . This corresponds to a velocity of
v f =
√
2W f
m
= √
2 × 2.5 × 10 3 × 1.6 × 10 −19
1.6 × 10 −27 = 707 km/s (63)
Thus, the total change in velocity needed is ∆v tot = 707−447 km/s = 260 km/s. This corresponds
to
N bounces =
∆v tot
= 260 = 13 bounces ✷ (64)
∆v 1bounce 20
We can neglect the distance the mirrors move in the time the particle travels the distance, since
v m ≪ v proton . Thus, the time it takes to travel a distance N bounces L is, using ¯v proton = (v f − v i )/2,
T = N bouncesL 2 × 13 × 1010
=
¯v proton 707 − 447
= 10 9 s ≈ 32 years ✷ (65)
2-13. Derive the result of Problem 2-12(b) directly by using the invariance of J.
a) Let ∫ v || ds ≈ v || L and differentiate with respect to time.
b) From this, get an expression for T in terms of dL/dt. Set dL/dt = −2v m to obtain
the answer.
a) The quantity J = ∫ b
a v ||ds is invariant. Thus, is approximate it as v || L, then it’s time derivative
must be 0:
d
dt (v ||L) = L ˙v || + v || ˙L = 0 (66)

) We can solve this expression:
L ˙v || = −v || ˙L ⇒ dv|| L = 2v || v m dt (67)
2-14. In plasma heating by adiabatic compression, the invariance of µ requires that
kT ⊥ increases as B increases. The magnetic field, however, cannot accelerate particles
because the Lorentz force qv × B is always perpendicular to the velocity. How do the
particles gain energy?
Maxwell tells us that an electric field will be induced by a changing magnetic field. The induced
electric field is what accelerates the particles. ✷
4-1. The oscillating density n 1 and potential φ 1 in a “drift wave” are related by
n 1
= eφ 1 ω ∗ + ia
n 0 kT e ω + ia
(68)
where it is only necessary to know that all the other symbols (except i) stand for
positive constants.
a) Find an expression for the phase δ of φ 1 relative to n 1 . (For simplicity, assume that
n 1 is real.)
b) If ω < ω ∗ , does φ 1 lead or lag n 1 ?
a) Solving for φ 1 leads to
φ 1 = ω + ia n 1 kT e
ω ∗ + ia n 0 e = n 1 kT e
n 0 e
(ω + ia)(ω ∗ − ia)
ω∗ 2 + a 2 = n 1 kT e ωω ∗ + a 2 + i(aω ∗ − aω)
n 0 e ω∗ 2 + a 2 (69)
Now, in a drift wave we can suppose that φ 1 ∼ exp(iδ), which in turns tells us that tan(δ) =
Im(φ 1 )/Re(φ 1 ). We have
Thus,
Re(φ 1 ) = n 1 kT e ωω ∗ + a 2
n 0 e ω∗ 2 + a 2 ; Im(φ 1) = n 1 kT e a
n 0 e
ω ∗ − ω
ω∗ 2 + a 2 (70)
δ = tan −1 (a ω ∗ − ω
ωω ∗ + a 2 ) ✷ (71)
b) For ω < ω ∗ , δ > 0. We can set the phase of n 1 to be 0, since all that matters is a phase difference.
Thus, φ 1 lags n 1 . ✷
4.2 Calculate the plasma frequency with the ion motions included, thus justifying our
assumption that the ions are fixed. (Hint: include the term n 1i in Poisson’s equation
and use the ion equations of motion and continuity.
We will use Gauss’s Law, Fourier transforming the field and charge perturbations into plane waves
of the form x = x 0 + x 1 , where x is any quantity, vector or scalar. The subscript 0 indicates the
equilibrium value, and the subscript 1 indicates the perturbation. We only keep terms of to first
order in small quantities.
∇ · E = ρ ɛ 0
⇒ ik · E 1 = e(n i − n e )
ɛ 0
(72)
Similarly, the equation of motion for the electrons,
ions,
m e
dv e
dt = −eE ⇒ iωm ev e = eE 1 (73)
m i
dv i
dt = eE ⇒ iωm iv i = −eE 1 (74)

and continuity equation for electrons,
∂n e
∂t + ∇ · (n ev e ) = 0 ⇒ −ωn e1 + n e0 k · v e = 0 (75)
and ions
∂n i
∂t + ∇ · (n iv i ) = 0 ⇒ −ωn i1 + n i0 k · v i = 0 (76)
We now have 9 equations and 9 unknowns. I will skip the boring algebra. Solving for ω yields
√
ni e 2
ω 2 = ω 2 p + Ω 2 p ✷ (77)
where Ω p =
m i ɛ 0
is the ion plasma frequency. Clearly, omitting the ion plasma frequency from
the calculation is justified since m i ≫ m e .
4.4 By writing the linearized Poisson’s equation used in the derivation of simple plasma
oscillations in the form ∇ · (ɛE) = 0 derive an expression for the dielectric constant ɛ
applicable to high-frequency longitudinal motions.
Fourier transform Poisson’s equation:
We also know that
and we know Newton’s Law:
Thus, equation (78) gives us
ik · E = ρ ɛ 0
= e ɛ 0
n 1 (78)
∂n 1
∂t + n k · v
0∇ · v 1 = 0 ⇒ iωn 1 − n 0 ik · v = 0 ⇒ n 1 = n 0
ω
(79)
m ∂v
∂t = −eE ⇒ v = i e
mω E (80)
k · E = − ie
ɛ 0 ω n 0k · v = e2 n 0
ɛ 0 ω 2 m k · E (81)
We can do a trick here, and pull everything over to the left side. Writing ω 2 p ≡ e 2 n 0 /mɛ 0
and thus we obtain
k · E(1 − ω2 p
ω 2 ) = 0 ⇒ ∇ · {(1 − ω2 p
)E} = 0 (82)
ω2 ɛ = 1 − ω2 p
ω 2 ✷ (83)
4.6
a) Compute the effect of collisional damping on the propagation of Langmuir waves
(plasma oscillations), by adding a term −mnνv to the electron equation of motion and
rederiving the dispersion relation for T e = 0.
b) Write an explicit expression for Im(ω) and show that its sign indicates that the
wave is damped in time.
a) The cold electron equations of motion are
and
mn e
∂v
∂t = −en eE − mn e νv ⇒ iωmv = −eE − mνv (84)
∂n e
∂t + ∇ · (n ev) = 0 ⇒ iωn e − in e k · v = 0 ⇒ k · v = ω (85)

We also have Gauss’s law:
∇ · E = − n ee
ɛ 0
⇒ k · E = i n ee
ɛ 0
= i m e ω2 p (86)
We will dot equation (84) with k:
(iωm − mν)
k · E = k · v
e
(87)
Plugging in equations (86) and (87), we obtain
i m e ω2 p = (iω2 m − mων)
e
⇒ ω 2 p = ω 2 + ων ✷ (88)
So we see that if we include collisions, the oscillation frequency is different from the plasma frequency.
b) Lets let ω = ω R + iω I . Then expression (88) becomes
ω 2 p = ω 2 R − ω 2 I + 2iω R ω I + iω I ν + ω R ν (89)
This means that
2ω R ω I + ω I ν = 0 ⇒ ω I = − ν 2 ✷ (90)
Now we suppose a plane wave solution for the field quantities, i.e.
we obtain
E ∝ e −iω Rt+ω I t
Thus, the wave is exponentially attenuated in time.
E ∝ e −iωt (91)
⇒
E ∝ e −iω Rt e − νt
2 ✷ (92)