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 condi-tions:a) At 0oC and 760 Torr pressure (1 Torr = 1mm Hg). This is called the Loschmidtnumber.b) In a vacuum at 10−3 Torr at room temperature (20oC). This number is a useful onefor the experimentalist to know by heart (10−3 Torr = 1 micron).a) Avogadro’s number is NA = 6.022× 1023. One mole of gas at STP occupies 22.4 liters. 1 liter is1× 10−3 cubic meters. Thus, the number per cubic meter is NA/n = 6.022× 1023/(22.4× 10−3) =2.66× 1025 m−3. Thus, the Loschmidt number is 2.66× 1025 2

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× 1019m−3 2 (1)

1-2. Derive the constant A for a normalized one-dimensional Maxwellian distribution

f(u) = Ae−mu2/2kT (2)

such that ∫ ∞−∞

f(u)du = 1 (3)

This one is straightforward. Just integrate:

1 =

∫ ∞−∞

Ae−mu2/2kTdu = A

√2πkT

m⇒ A =

√m

2πkT2 (4)

1-4. Compute the pressure, in atmospheres and in tons/ft2, exerted by a thermonuclearplasma on its container. Assume kTe = kTi = 20keV , n = 1021m−3 and p = nkT , whereT = Ti + Te.This is just unit conversion, albeit with units that nobody really ever remembers. For reference,1 keV = 1.6× 10−19J , so

p = 1021 × (20keV + 20keV ) = 4× 1022m−3 keV = 4× 103m−3 J = 4× 103N/m2 (5)

But 1 atm = 105N/m2 = 1 ton/ft2, 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/ft2 2 (6)

1-5. In a strictly steady state situation, both the ions and the electrons will follow theBoltzmann relation

nj = n0e−qjφ/kTj (7)

For the case of an infinite, transparent grid charged to a potential φ, show that theshielding distance is given approximately by

λ−2D =ne2

ε0(

1

kTe+

1

kTi) (8)

Show that λD is determined by the temperature of the colder species.We’ll use Poisson’s equation

∇2φ =ene − eni

ε0=en0ε0

(eeφ/kTe − e−eφ/kTi) ≈ en0ε0

(1 +eφ

kTe− 1− −eφ

kTi) =

e2n

ε0(φ

kTe+

φ

kTi) (9)

Now we’ll suppose the φ goes like a decreasing exponential: φ = φ0exp(−x/λD). Thus, the Lapla-cian acting on this is

∇2φ =1

λ2Dφ =

ne2

ε0(φ

kTe+

φ

kTi) ⇒ 1

λ2D=ne2

ε0(

1

kTe+

1

kTi) 2 (10)

To show that λD is determined by the colder species, we suppose first that the electrons are thecolder species: Te Ti. Then,

1

λ2D=

ne2

kTeε0⇒ λD =

√kTeε0ne2

2 (11)

Alternatively, if the ions are colder, Ti Te, then a similar analysis yields:

1

λ2D=

ne2

kTiε0⇒ λD =

√kTiε0ne2

2 (12)

1-6. An alternative derivation of λD will give further insight to its meaning. Considertwo infinite, parallel plates at x = ±d, set at potential φ = 0. The space between themis 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 platesis

φ =nq

2ε0(d2 − x2) (13)

b) Show that for d > λD, the energy needed to transport a particle from a plate to themid 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 densityn = 106 m−3 and temperature T = 100oK. What is the Debye length?The Debye length is given by

λD =∑j

ε0kTjnje2j

(14)

Plugging in the numbers:

λD =8.85× 10−12 × 1.4× 10−23 × 100

106 × (1.6× 10−19)2= 0.48 m 2 (15)

As a check, use the SI unit form for the Debye length given in Chen. If T is in Kelvin, and n is incubic meters, then:

λD = 69

√T

nm = 69×

√102

106m = 69× 10−2 m 2 (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 apotential φ0. The electrons remain Maxwellian and move to form a Debye shield, butthe ions are stationary during the time frame of the experiment. Assuming φ0 kTe/e,derive an expression for the potential as a function of r in terms of a, φ0, and λD. (Hint:Assume a solution of the form e−br/r.)Let’s assume a solution of this form: φ = Ae−br/r. Then,

∇2φ =1

r2∂

∂r(r2

∂φ

∂r) = b2φ =

e

ε0(ne − ni) (17)

2

Since the electrons are Maxwellian, they obey ne = n0eeφ/kTe ≈ n0(1+eφ/kTe). The ions, however,

are stationary, so ni = n0. Thus we have:

b2φ =e

ε0(n0 + n0

eφ

kTe− n0) =

n0e2φ

ε0kTe≡ φ

λ2D⇒ b =

1

λD(18)

Thus, so far we have:

φ = Ae−r/λD

r(19)

But we also need to match the boundary condition that φ(a) = φ0. So,

φ0 = Ae−a/λD

a⇒ A = aφ0e

a/λD (20)

So, finally we have our answer:

φ(r) = φ0ea/λDa

e−r/λD

r2 (21)

You know what they say: if it satisfies Poisson’s equation and the boundary conditions then itmust be the answer.

2-3. An ion engine (see Fig. 106) has a 1-T magnetic field, and a hydrogen plasmais to be shot out at an E × B velocity of 1000 km/s. How much internal electric fieldmust be present in the plasma?The E×B velocity is given by

v =E×B

B2(22)

Plugging in the numbers:

106 m/s =|E|1T

⇒ |E| = 1000 V/m 2 (23)

2-4. Show that vE is the same for two ions of equal mass and charge but differentenergies, by using the following physical picture (see Fig. 2-2). Approximate the righthalf of the orbit by a semicircle corresponding to the ion energy after accelerationby the E field, and the left half by a semicircle corresponding to the energy afterdeceleration. You may assume that E is weak, so that the fractional change in v⊥ issmall.If the energy of the right part of the orbit is E1 and the energy of the left part of the orbit is E2,then we have

E1 = E0 + eEr1, E2 = E0 − eEr2 (24)

where E0 is the initial energy and E is the electric field. The velocity is determined by v =√

2E/m,so

v1 =

√2E0 + 2eEr1

mv2 =

√2E0 − eEr2

m(25)

The Larmor radius is determined via r = mv⊥/qB, so

r1,2 =m

qB

√2E0

m

√1± eEr1,2

E0=

√2mE0

qB(1± eEr1,2

2E0) =

√2E0

m

1

ωc± Er1,2

2√mE0ωc

(26)

Thus,

r1,2(1∓√

m

2E0

E

qB) =

1

ωc

√2E0

m(1± eE

2E0ωc

√2E0

m) (27)

3

The guiding center moves a distance r1 − r2:

r1 − r2 =eE

E0ωc

√2E0

m

1

ωc

√2E0

m=

2eE

mω2c

(28)

The velocity of the guiding center is

vgc = 2r1 − r2T

= 2ωc2π

(r1 − r2) =4eE

2πmωc=

2eE

πmωc=

2E

πB≈ E

B2 (29)

since ωc = eB/m. This is a pretty good approximation.

2-5. Suppose electrons obey the Boltzmann relation of Problem 1-5 in a cylindri-cally 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 vE is as largeas vth. Specifically, show that rL = 2λ if vE = vth.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.

n = n0eeφ/kTe ⇒ φ =

kTeeln(

n

n0) (30)

Therefore,

E = −∇φ = −∂φ∂r

r = −kTee

n0n

1

n0

∂n

∂rr =

kTeeλ

r 2 (31)

b) We start with the definitions of vE , vth, and rL:

vE =E

B, vth =

√2kTem

, rL =mv⊥eB

(32)

So, calculating the magnitude of vE :

vE =kTeeλB

=mv2th

2

1

eλB=rLvth

2λ(33)

where in the last step I have assumed that the perpendicular velocity is the thermal velocity. Now,setting vE = vth, it is easy to see that

rL = 2λ 2 (34)

c) Sure, why not?

2-6. Suppose that a so-called Q-machine has a uniform field of 0.2 T and a cylindricalplasma with kTe = kTi = 0.2 eV . The density profile is found experimentally to be ofthe form

n = n0exp[exp(−r2/a2)− 1] (35)

Assume the density obeys the electron Boltzmann relation n = n0exp(eφ/kTe).a) Calculate the maximum vE if a = 1 cm.b) Compare this with vE due to the earth’s gravitational field.c) To what value can B be lowered before the ions of potassium (A = 39, Z = 1) havea Larmor radius equal to a?We solve for φ:

n0exp[exp(−r2/a2)− 1] = n0exp(eφ/kTe) ⇒ φ =kTee

(e−r2/a2 − 1) (36)

4

Thus, the electric field is

E = −∂φ∂r

r =kTee

2r

a2e−r

2/a2 r (37)

and so vE (and it’s maximum) is

vE =E

B=

2rkTeea2B

e−r2/a2 (38)

∂vE∂r

=2kTeea2B

e−r2/a2 − 4r2kTe

ea4Be−r

2/a2 = 0 ⇒ r =

√a2

22 (39)

So, with a = 1 cm,

vE,max =2kTeea2B

√a2

2e−1/2

∣∣∣∣a=1 cm,B=0.2 T,kTe=0.2 keV

≈ 8.5 km/sec 2 (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

FgFE

=6.4× 10−25

2.8× 10−18≈ 1.5× 10−7 (41)

times smaller. 2

c) The Larmor radius is rL = mvth/qB, so, in terms of the constants of vth, we have (settingrL = a):

rL =m

qB

√2kTem

= a ⇒ B =

√2mkTeq2a2

(42)

Plugging in the numbers:

B =

√2× 39× 1.6× 10−27 × 0.2× 1.6× 10−19

(17× 1.6× 10−19)2 × (0.1× 10−2)2= 4× 10−2 T 2 (43)

2-8. Suppose the Earth’s magnetic field is 3 × 10−5 T at the equator and falls off as1/r3, as for a perfect dipole. Let there be an isotropic population of 1−eV protons and30− keV electrons, each with density n = 107 m−3, at r = 5 earth radii in the equatorialplane.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/m2.Note: The curvature drift is not negligible and will affect the numerical answer, butneglect it anyway.a) The grad-B drift is given by

v∇B =1

2v⊥rL|

B×∇B

B2| = 1

2v⊥rL|

∇BB| (44)

We can calculate v⊥ from the energy, and rL from the magnetic field:

v⊥ =

√2E

m, rL =

mv⊥eB

=m

eB

√2E

m(45)

Thus,

v∇B =1

2

√2E

m

m

eB

√2E

m|∇BB| = E

eB|∇BB| (46)

5

Since B ∼ r−3, we obtain

∇B =∂B

∂rr = − 3

r4= −3

B

r⇒ |∇B

B| = 3

r(47)

So

v∇B =3E

eBr=

3Er3

eB0R3er

=3Er2

eB0R3e

=3E(eV )r2

B0R3e

(48)

where I have used the fact that B = B0R3e/r

3. Now we can plug in the numbers:

v∇B,e =3× 30× 103 × (5× 6× 106)2

3× 10−5 × (6× 106)3= 1.3× 104 m/s 2 (49)

v∇B,i =3× 1× (5× 6× 106)2

3× 10−5 × (6× 106)3= 0.42 m/s 2 (50)

b) The magnetic field is azimuthal, from north to south, i.e. the −θ direction. The gradient of themagnetic field is clearly in the radial direction, so we have B×∇B ∼ r× θ = φ, which is eastward.This is for the electrons. The ions, which come without the minus sign to cancel the minus sign inequation 47, will go in -φ, which is westward. 2

c) Well, it has to travel a distance L = 2π(5Re), with the velocity v∇B,e. So,

T =L

v∇B,e=

2π × 5× 6.4× 106

1.3× 104≈ 4.5 hours 2 (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 = 107 × 1.6× 10−19 × 1.3× 104 = 2× 10−8 A/m2 2 (52)

2-10. A 20− keV deuteron in a large mirror fusion device has a pitch angle θ = 45o atthe mid plane, where B = 0.7 T . Compute it’s Larmor radius.The Larmor radius is given by

rL =mv⊥qB

(53)

In natural units, a deuteron has m = 2 and q = +1. Furthermore, v⊥ = vsin(θ). To find v, weconvert the energy to velocity:

rL =m

eB

√2E

msin(θ) =

2× 1.6× 10−27

1.6× 10−19 × 0.7

√2× 20× 103 × 1.6× 10−19

2× 1.6× 10−27sin(45o) = 0.3 m 2 (54)

2-12. A cosmic ray proton is trapped between two moving magnetic mirrors withRm = 5 and initially has W = 1 keV and v⊥ = v|| at the mid plane. Each mirror movestoward the mid plane with a velocity vm = 10 km/sec (Fig. 2-10).a) Using the loss cone formula and the invariance of µ, find the energy to which theproton 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 bounceis 2vm.2. Compute the number of bounces necessary.3. Compute the time T it takes to traverse L = 1010 km that many times. Factor oftwo accuracy will suffice.

a) The loss cone formula is sin2(θm) = 1/Rm, where θm is the angle of the magnetic mirror andRm is the mirror ratio Bmax/B0. We can also write down a formula for sin(θm) in terms of theparallel and perpendicular velocities:

sin(θm) =1√Rm

=v⊥,f√

v2⊥,f + v2||,f

=1√

1 + (v||,fv⊥,f

)2(55)

6

Squaring both sides, and noting that, since µ is invariant, v⊥,f = v⊥,i, we have

1

Rm=

1

1 + (v||,fv⊥,i

)2=

1

5⇒ v||,f = 2v⊥,i (56)

Thus, we can get the final energy:

Wf =1

2m(v2⊥,f + v2||,f ) =

1

2m(v2⊥,i + 4v2⊥,i) =

5

2mv2⊥,i (57)

But we can’t evaluate this without knowing what the original v⊥,i is. Fortunately, we know thatinitially v⊥ = v||, so

Wi =1

2m(v2⊥,i + v2||,i) =

1

2m(2v2⊥,i) = mv2⊥,i = 1 keV (58)

So, finally,

Wf =5

2Wi = 2.5 keV 2 (59)

b) In the frame of the piston, when the particle bounces off, it’s velocity doesn’t change. In thepiston’s frame, the velocity of the particle as it is coming in is vi − vm, where vm is the velocity ofthe piston (it is negative). It’s final velocity is the same but negative. Thus,

v′i = vi − vm, v′f = vm − vi (60)

where the prime denotes the velocity in the piston’s reference frame. But, in the lab frame,vf = v′f + vm, so we have

vf = 2vm − vi (61)

Thus, the change in velocity on each bounce is 2vm = 20 km/sec. The initial proton velocity is

vi =

√2Wi

m=

√2× 1× 103 × 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

vf =

√2Wf

m=

√2× 2.5× 103 × 1.6× 10−19

1.6× 10−27= 707 km/s (63)

Thus, the total change in velocity needed is ∆vtot = 707−447 km/s = 260 km/s. This correspondsto

Nbounces =∆vtot

∆v1bounce=

260

20= 13 bounces 2 (64)

We can neglect the distance the mirrors move in the time the particle travels the distance, sincevm vproton. Thus, the time it takes to travel a distance NbouncesL is, using vproton = (vf − vi)/2,

T =NbouncesL

vproton=

2× 13× 1010

707− 447= 109 s ≈ 32 years 2 (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 = −2vm to obtainthe answer.a) The quantity J =

∫ ba v||ds is invariant. Thus, is approximate it as v||L, then it’s time derivative

must be 0:d

dt(v||L) = Lv|| + v||L = 0 (66)

7

b) We can solve this expression:

Lv|| = −v||L ⇒ dv||L = 2v||vmdt (67)

2-14. In plasma heating by adiabatic compression, the invariance of µ requires thatkT⊥ increases as B increases. The magnetic field, however, cannot accelerate particlesbecause the Lorentz force qv×B is always perpendicular to the velocity. How do theparticles gain energy?Maxwell tells us that an electric field will be induced by a changing magnetic field. The inducedelectric field is what accelerates the particles. 2

4-1. The oscillating density n1 and potential φ1 in a “drift wave” are related by

n1n0

=eφ1kTe

ω∗ + ia

ω + ia(68)

where it is only necessary to know that all the other symbols (except i) stand forpositive constants.a) Find an expression for the phase δ of φ1 relative to n1. (For simplicity, assume thatn1 is real.)b) If ω < ω∗, does φ1 lead or lag n1?a) Solving for φ1 leads to

φ1 =ω + ia

ω∗ + ia

n1n0

kTee

=n1n0

kTee

(ω + ia)(ω∗ − ia)

ω2∗ + a2

=n1n0

kTee

ωω∗ + a2 + i(aω∗ − aω)

ω2∗ + a2

(69)

Now, in a drift wave we can suppose that φ1 ∼ exp(iδ), which in turns tells us that tan(δ) =Im(φ1)/Re(φ1). We have

Re(φ1) =n1n0

kTee

ωω∗ + a2

ω2∗ + a2

; Im(φ1) =n1n0

kTea

e

ω∗ − ωω2∗ + a2

(70)

Thus,

δ = tan−1(aω∗ − ωωω∗ + a2

) 2 (71)

b) For ω < ω∗, δ > 0. We can set the phase of n1 to be 0, since all that matters is a phase difference.Thus, φ1 lags n1. 2

4.2 Calculate the plasma frequency with the ion motions included, thus justifying ourassumption that the ions are fixed. (Hint: include the term n1i in Poisson’s equationand use the ion equations of motion and continuity.We will use Gauss’s Law, Fourier transforming the field and charge perturbations into plane wavesof the form x = x0 + x1, where x is any quantity, vector or scalar. The subscript 0 indicates theequilibrium value, and the subscript 1 indicates the perturbation. We only keep terms of to firstorder in small quantities.

∇ ·E =ρ

ε0⇒ ik ·E1 =

e(ni − ne)ε0

(72)

Similarly, the equation of motion for the electrons,

medvedt

= −eE ⇒ iωmeve = eE1 (73)

ions,

midvidt

= eE ⇒ iωmivi = −eE1 (74)

8

and continuity equation for electrons,

∂ne∂t

+∇ · (neve) = 0 ⇒ −ωne1 + ne0k · ve = 0 (75)

and ions∂ni∂t

+∇ · (nivi) = 0 ⇒ −ωni1 + ni0k · vi = 0 (76)

We now have 9 equations and 9 unknowns. I will skip the boring algebra. Solving for ω yields

ω2 = ω2p + Ω2

p 2 (77)

where Ωp =√

nie2

miε0is the ion plasma frequency. Clearly, omitting the ion plasma frequency from

the calculation is justified since mi me.

4.4 By writing the linearized Poisson’s equation used in the derivation of simple plasmaoscillations in the form ∇ · (εE) = 0 derive an expression for the dielectric constant εapplicable to high-frequency longitudinal motions.Fourier transform Poisson’s equation:

ik ·E =ρ

ε0=

e

ε0n1 (78)

We also know that

∂n1∂t

+ n0∇ · v1 = 0 ⇒ iωn1 − n0ik · v = 0 ⇒ n1 = n0k · vω

(79)

and we know Newton’s Law:

m∂v

∂t= −eE ⇒ v = i

e

mωE (80)

Thus, equation (78) gives us

k ·E = − ie

ε0ωn0k · v =

e2n0ε0ω2m

k ·E (81)

We can do a trick here, and pull everything over to the left side. Writing ω2p ≡ e2n0/mε0

k ·E(1−ω2p

ω2) = 0 ⇒ ∇ · (1−

ω2p

ω2)E = 0 (82)

and thus we obtain

ε = 1−ω2p

ω22 (83)

4.6a) 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 andrederiving the dispersion relation for Te = 0.b) Write an explicit expression for Im(ω) and show that its sign indicates that thewave is damped in time.

a) The cold electron equations of motion are

mne∂v

∂t= −eneE−mneνv ⇒ iωmv = −eE−mνv (84)

and∂ne∂t

+∇ · (nev) = 0 ⇒ iωne − inek · v = 0 ⇒ k · v = ω (85)

9

We also have Gauss’s law:

∇ ·E = −neeε0

⇒ k ·E = inee

ε0= i

m

eω2p (86)

We will dot equation (84) with k:

k ·E = k · v(iωm−mν)

e(87)

Plugging in equations (86) and (87), we obtain

im

eω2p =

(iω2m−mων)

e⇒ ω2

p = ω2 + ων 2 (88)

So we see that if we include collisions, the oscillation frequency is different from the plasma fre-quency.

b) Lets let ω = ωR + iωI . Then expression (88) becomes

ω2p = ω2

R − ω2I + 2iωRωI + iωIν + ωRν (89)

This means that2ωRωI + ωIν = 0 ⇒ ωI = −ν

22 (90)

Now we suppose a plane wave solution for the field quantities, i.e.

E ∝ e−iωt (91)

we obtainE ∝ e−iωRt+ωI t ⇒ E ∝ e−iωRte−

νt2 2 (92)

Thus, the wave is exponentially attenuated in time.

10