Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

CHAPTER

1

Chapter 1

1.1

(a) Consider a one-dimensional slab of plasma whose whose bounding surfaces are normal to the x-axis. Suppose that the electrons (whose mass, charge, and number density are me, −e, and ne, respectively) displace a distance δxe parallel to the x-axis, whereas the ions (whose mass, charge, and number density are mi, +Z e, and ni = ne/Z, re- spectively) displace a distance δxi. The resulting charge density that develops on the leading edge of the slab is σ = −e ne δxe + Z e ni δxi = e ne (δxi − δxe).

(1)

An equal and opposite charge density develops on the opposite face of the slab. The (a) directed electric field generated inside the slab is σ e ne E =− =− (δx − δx ).

(2) x i e ǫ0 ǫ0 The equation of motion of an individual electron inside the slab is thus

..

m

e δ xe

= −e Ex =

e2 ne ǫ0

(δxi − δxe).

(3)

Likewise, the equation of motion of an individual ion is

..

Z 2 e2 ni = Z e E = − x (δxi − δxe). i δ xi ǫ0

m

(4)

Let us search for simultaneous solutions of Equations (3) and (4) of the form δxe(t) = δxˆe cos(ω t),

(5)

δxi(t) = δxˆi cos(ω t).

(6)

(ω2 − Π e2) δxˆe + ωp2e δxˆi = 0, Π 2 δxˆe + (ω2 − ω 2 ) δxˆi = 0,

(7) (8)

It follows that

i

pi

where Πe = (e2 ne/ǫ0 me)1/2 and Πi = (Z 2 e2 ni/ǫ0 mi)1/2. The solutions are ω = 0 with δxˆe = δxˆi, and ω2 = Πe 2 + Π 2 with Π 2 δxˆe + Π e2 δxˆi = 0. The former mode i i corresponds to a uniform translation of the slab. The latter mode is a plasma oscillation whose frequency, Π, satisfies Π = Πe2 + Πi 2

1/2

,

(9) 1

2 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises and whose characteristic ratio of ion to electron displacement amplitudes is 2

Πi me δxˆ i δxˆ = − = −Z i . e Π e2 m

(10)

(b) Suppose that the electrons, whose temperature is Te, are distributed according to the Maxwell-Boltzmann law, ne + δne = ne exp(+e δΦ/Te),

(11)

where ne is the equilibrium number density, and δne is the number density perturbation due to the perturbing potential δΦ. Likewise, the ions, whose temperature is Ti, are distributed according to ni + δni = ni exp(−Z e δΦ/Ti).

Thus, in the limit that δΦ is small, we obtain e ne δne = T δΦ, e Z e ni = δΦ. δni — Ti

(12)

(13) (14)

If δΦ is a consequence of a small perturbing charge density, δρext, then the total charge density is !

2 2 e2 ne + Z e ni δρ = δρext + Z e δni − e δne = δρext .

(15) Te Ti

−

Thus, Poisson’s equation,

δρ

2

∇ δΦ = − 0 ,

(16)

ǫ

yields

2

2

– where

λ 2D

δΦ = −

ǫ

0

!2 1 λD

δρ ext

=

1 2

1 +

λD e

!2

(17)

, !2

1 λDi

,

(18)

with λD e = (ǫ0 Te/ne e2)1/2 and λD i = (ǫ0 Ti/ni Z 2 e2)1/2. Comparison of Equation (17) with Eq. (1.14) in the book reveals that λD is the effective Debye length.

1.2 It is reasonable to assume, by symmetry, that the perturbed potential is a function only of the radial spherical coordinate r. In other words, δΦ = δΦ(r). Thus, the governing differential equation becomes !

2 1 d 2 dδΦ r – δΦ = 0 (19) r2 dr dr λ 2D

for r ≠ 0. However, in the limit r → 0 we expect the perturbed potential to approach the Coulomb potential: i.e., q (20) δΦ → 4π ǫ0 r

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

Chapter 1 □ 3 as r → 0. We also expect the potential to be well behaved in the limit r V(r)/r. Equation (19) transforms to give d2V

2

dr

λD2

2 −

→ ∞. Let δΦ =

V = 0.

(21)

The solution that is consistent with the boundary conditions at r = 0 and r = ∞ is V(r) =

q 4π ǫ0 exp —

Thus,

q

δΦ(r) =

(22)

(23)

exp —

4π ǫ0 r

According to Poisson’s equation, the charge density of the shielding cloud is δρ(r) = −ǫ0 ∇2δΦ.

(24)

However, according to the governing differential equation, 2

∇ δΦ = for r ≠ 0. Hence, δρ(r) = −

2 λ D2

(25)

δΦ

√

2q 4π r λD

2 exp −

2r

λD

(26)

.

The net shielding charge contained within a sphere of radius r, centered on the origin, is ∫ r ′ ′2 ′ √ 2q ∫ r ′ 2r ′ r exp dr = Q(r) = 4π δρ(r ) r dr .

(27) —

−

0

Thus,

∫ Q(r) = −q

2

λD

√ λD r/ 2 0

λD

0

. √ x D r/ 2

x e−x dx = −q −x e− 0 λ

∫

√

λD r/ 2

+ 0

e− dx , x

(28)

.

√2 r

which reduces to Q(r) = −q 1 − 1 +

λD

exp —

.

(29)

1.3 Consider a one-dimensional slab of plasma whose bounding surfaces are normal to the xaxis. Suppose that the electrons (whose mass, charge, and number density are me, −e, and ne, respectively) displace a distance δxe parallel to the x-axis, whereas the ions remain stationary. The resulting charge density that develops on the leading edge of the slab is σ = −e ne δxe.

(30)

An equal and opposite charge density develops on the opposite face of the slab. The xdirected electric field generated inside the slab is σ e ne E =− = δx .

(31) x e ǫ0 ǫ0

4 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises If E0(t) = Eˆ0 cos(ω t) is the externally generated x-directed electric field then the total electric field inside the slab is e ne E (t) = δx (t) + Eˆ cos(ω t).

(32) 1

ǫ0

e

0

The equation of motion of an individual electron inside the slab is .. m e δx e = −e E1 or

..

δxe + Π δxe = −

(33)

e

Eˆ0 cos(ω t), (34) me where Π = (e2 ne/ǫ0 me)1/2. Let us search for a solution of the form δxe(t) = δxˆe cos(ω t).

We obtain ˆ δxˆe = (e/me) E 0 .

(35) 2 ω −Π2 Thus, writing E1(t) = Eˆ1 cos(ω t), it follows from Equation (32) that ! ! Π2 1 ˆ ˆ E1 = +1E 0 = Eˆ0.

(36) 1 − Π 2/ω2 ω2 − Π 2 2

However, if a dielectric slab is placed in a uniform external field then we expect the internal field to be reduced by a factor ǫ, where ǫ is the relative dielectric constant. Thus, it follows that Π2 ǫ =1− 2.

(37) ω 1.4 Let x measure perpendicular distance between the plates. Suppose that one plate lies at x = d−/ 2 , and the other at x = d/2. Because the spacing between the plates is relatively small, we can assume that the potential, V, is only a function of x. Suppose that the plate at x = −d/2 is held at the potential −V0/2, whereas that at x = d/2 is held at the potential V0/2. According to Eq. (1.14) in the book, the potential between the plates satisfies d2V

2

dx

λD

− 2

V = 0.

(38)

Moreover, V(−d/2) = −V0/2 and V(d/2) = V0/2. It follows that V0 √ V(x) = sinh( 2 x/λD) . √ 2 sinh( 2 d/2 λD ) Hence, the electric field between the plates is E(x) = −

dV dx

V0

(39)

√

cosh( 2 x/λD)

√ . =−√ 2 λD sinh( 2 d/2 λD )

(40)

Now, by Gauss’ law, the charge density on the plate at d = x/2 is σ = −ǫ0 E(d/2) = V0 ǫ0

√

Hence, the charge on the plate is Q = Aσ =

1

√

.

(41)

2 λD tanh(d/ 2 λD)

V0 ǫ 0 A 1 . √ √ 2 λD tanh(d/ 2 λD)

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

(42)

Chapter 1 □ 5 There is an equal and opposite charge on the other plate. Thus, the capacitance is √ Q d/ 2 λD ǫ0 A C= . √ V = d 0 tanh(d/ 2 λD)

(43)

1.5 According to the Maxwell-Boltzmann distribution, taking into account gravitational potential energy, as well as electrostatic potential energy, the electron number density is written e V − me g z ne = n0 exp , (44) T where me is the electron mass, T the plasma temperature, and V(z) the perturbed electrostatic potential. Likewise, the ion number density becomes −e V − mi gz .

(45) ni = n0 exp T Assuming that e V/T , me g z/T , and mi g z/T are all small (which will indeed be the case if the gravitational field is sufficiently weak), the perturbed election and ion number densities become e n0 V m n gz — e 0 , T T m n gz e n0 V — i 0 , δni = − T T

δne =

(46) (47)

respectively. Hence, the perturbed charge density is written δρ = e (δni − δne) ≃ −

2 e2 n 0 V T

−

e mi n 0 g z , T

(48)

where we have made use of the fact that mi ≫ me. Poisson’s equation, d2V

δρ = − , dz2 ǫ0 yields

2 2 d V

λD

dz2

− 2 V = 2 E0 z,

(50)

! ǫ0 T 1/2 e2 n0

(51)

mi .

(52)

where λD = is the Debye length, and

(49)

E = g 0

e By inspection, the solution of Equation (50) that satisfies dV/dz = 0 at z = 0, and is well √ 2z behaved as z → ∞, is λD V(z) = E0 −z − √ exp − .

Dλ 2 The corresponding electric field is E = Ez ez, where dV Ez(z) = −

dz

(53)

√2z

= E0 1 − exp −

. λD

(54)

6 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 1.6 By symmetry, the electric potential takes the form Φ = Φ(x), where Φ(−x) = Φ(x). The ion and electron density distributions are ni = n0 e−e Φ/Ti , ne = n0 e+e Φ/Te .

Now, Φ(x) satisfies Poisson’s equation, ρ e (ni − ne) .

(55) =− ǫ0 ǫ0 Assuming that e Φ/Ti ≪ 1 and e Φ/Te ≪ 1, it is easily demonstrated that, to first order in ! small quantities, 1 1 + ρ ≃ −e2 n0 Φ.

(56) Ti Te d2Φ dx2

=−

Hence, Poisson’s equation becomes

where

d2Φ

1

dx

λD2

2 −

e2 n

1 λ D2

=

Φ = 0, 1

1

0

+

Ti

ǫ0

(57)

!

(58)

.

Te

The most general solution which satisfies the physical boundary condition that Φ → 0 as |x| → ∞ is Φ(x) = V e−|x|/λD .

(59) The boundary condition at x = 0 is " #0+ dΦ = 2 dx 0− which yields

σ dΦ . . =− , dx 0+ ǫ0

2V λD

Thus, Φ(x) =

σ

=

(60)

.

(61)

e−|x|/λD .

(62)

ǫ0 σλ D 2 ǫ0

1.7 Repeating the previous analysis, with Ti = Te = T , we obtain d2Φ e n0 −e Φ/T 2 e n0 Φ/T +e

=− 0 e dx2 ǫ Multiplication by dΦ/dx yields dΦ d2Φ dx dx2

=

−e

=

2 e n0 dΦ ǫ0

dx

ǫ0

sinh(e Φ/T ).

sinh(e Φ/T ).

We can integrate this equation to give !2 dΦ 4n T 0 = [cosh(e Φ/T ) −1] , dx ǫ0

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

(63)

(64)

(65)

Chapter 1 □ 7

FIGURE 1.1 Exercise 1.7.

where use has been made of the physical boundary condition that Φ → 0, dΦ/dx → 0 as |x| → ∞. The boundary condition at x = 0+ is Φ = V, and dΦ σ .

(66) =− dx 2 ǫ0 Thus, the previous two equations can be combined to give ! eV σ2 1+ = .

16 ǫ0 n0 T cosh−1 T

(67)

Now, taking the square-root of (65), we obtain dΦ = dx in the region x > 0. Thus,

or

where λD =

,

— 8 n0 T/ǫ0 sinh(e Φ/2 T )

(68)

, = — 8 n0 T/ǫ0 x,

(69)

∫ Φ

dΦ′

V

sinh(e Φ′/2 T ) ∫ e Φ/2 T

,

e V/2 T

dy = − x , sinh y λ

(70)

D

ǫ0 T/2 e2 n0. Hence, . e Φ/2 T x . ln{tanh(y/2)} e V/2 T = − , λD

(71)

which gives tanh(e Φ/4 T ) = tanh(e V/4 T ) e−|x|/λD.

(72)

8 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Let xˆs = xs/λD and Vˆ = e V/T . It follows that " # tanh(Vˆ/4) xˆs = ln This relationship is plotted in Figure 1.1.

(73)

. tanh(Vˆ/4 e)

1.8 Let us adopt cylindrical coordinates r, θ, z, where x = r cos θ and y = r sin θ. It is reasonable to suppose that there is no variation of perturbed quantities in the z-direction. (a) Suppose that the electron cylinder is displaced a distance δx in the x-direction. The radial displacement relative to the surface of the ion cylinder at r = a is thus δr = δx cos θ. This displacement gives rise to an effective charge sheet density on the surface of the cylinder of the form σ = −e n0 δr = −e n0 δx cos θ.

Now, the electric potential satisfies Poisson’s equation, !

1 ∂ ∂Φ 1 ∂2Φ r + 2 = 0, r ∂r ∂r r ∂θ2

(74)

(75)

subject to the boundary conditions that Φ be well behaved at r = 0 and r = ∞, and " #r=a+ σ e n δx ∂Φ (76) = − = 0 cos θ. ǫ0 ∂r r=a− ǫ0 It is fairly clear that Φ(r, θ) = φ(r) cos θ, where ! d dφ – φ = 0, r r dr dr and " #r=a+ e n0 δx dφ = . dr r=a ǫ0 — The solution is e n0 δx φ=− r 2 ǫ0 for r < a, and e n δx a2

(77)

(78)

(79)

0

φ=−

2 ǫ0 r for r > a. Hence, the electric field inside the cylinder is σ e n0 δx E= ex. e =− 2 ǫ0 x 2ǫ 0 Thus, the equation of motion of an individual electron is d2δx

e E =

(80)

(81)

2 δx,

(82)

.

2 ǫ0 me It follows that the electrons oscillate at the frequency Π.

(83)

dt2

=−

e

m −Π

where 2

x

Π =

e2 n0

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

Chapter 1 □ 9 (b) Let the displacement of the ion cylinder in the x-direction be δxi, and let that of the electron cylinder be δxe. Assuming that the center of mass of the system remains stationary, we have δx = − δx .

(84) me i

mi e The charge sheet density on the surface of the cylinder is σ = e n0 (δxi − δxe) cos θ = −e n0 (1 + me/mi) δxe cos θ.

(85)

Thus, the x-directed electric field at the surface of the cylinder is Ex =

e n0 (1 + me/mi) e n0 (mi/me + 1) δxe = − δx . i 2 ǫ0 2 ǫ0

(86)

The equation of motion of an electron is d2δxe dt2

=−

e Ex me

=−

e2 n0 (1 + me/mi) 2 ǫ0 me

δxe.

Likewise, the equation of motion of an ion is d2δxi e Ex e2 n0 (mi/me + 1) = =− δxi. mi 2 ǫ0 mi dt2 In both cases, the oscillation frequency is Π, where ! e2 n 1 1 0 . Π2 = + 2 ǫ0 me mi

(87)

(88)

(89)

1.9 Let us adopt spherical coordinates r, θ, ϕ, where z = r cos θ. Suppose that the electron sphere is displaced a distance δz in the z-direction. The radial displacement relative to the surface of the ion sphere at r = a is thus δr = δz cos θ. This displacement gives rise to an effective charge sheet density on the surface of the sphere of the form σ = −e n0 δr = −e n0 δz cos θ.

(90)

Now, assuming that perturbed quantities have no dependence on the azimuthal angle ϕ, the electric potential satisfies Poisson’s equation, ! ! ∂ ∂Φ 1 ∂ 2 ∂Φ 1 sin θ = 0, (91) r + r2 ∂r ∂r r2 sin θ ∂θ ∂θ subject to the boundary conditions that Φ be well behaved at r = 0 and r = ∞, and " #r=a+ σ e n δz ∂Φ = − = 0 cos θ. ǫ0 ∂r r=a− ǫ0 It is fairly clear that Φ(r, θ) = φ(r) cos θ, where ! d 2 dφ r – 2 φ = 0, dr dr and " #r=a+ e n0 δz. dφ = dr r=a−

ǫ0

(92)

(93) (94)

10 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises The solution is φ= for r < a, and

e n0 δz

− 3 ǫ0

r

(95)

e n0 δx a3

φ = − 3ǫ 2 0 r for r > a. Hence, the electric field inside the sphere is E=

σ e n0 δz ez. ez = − 3 ǫ0 3ǫ 0

(96)

(97)

Thus, the equation of motion of an individual electron is d2δz dt2

=−

e

Ez = −Π 2 δz,

(98)

e

m

where Π2 =

e2 n0

.

3 ǫ0 me It follows that the electrons oscillate at the frequency Π.

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.

(99)

CHAPTER

2

Chapter 2

2.1

(a) Given that ρ = ρ (− cos γ e1 + sin γ e2) and ρ = u⊥/Ω, it follows that u2 . ⊥ ρρ = (cos2 γ) e1e1 − (cos γ sin γ) e1e2 − (cos γ sin γ) e2e1 ( ) Ω2 . +(sin2 γ) e2e2 .

(1)

However, (cos2 γ) = (sin2 γ) = 1/2, and (cos γ sin γ) = 0. Hence,

(ρρ) =

u ⊥2 2Ω

(I − 2

(2)

bb) ,

because I = e1e1 + e2e2 + bb. (b) We can write e (u × (ρ · ∇) B)i = e ǫi jk (u j ρl

∂xl

Bk) ∂

(ρα ρl) b

= e Ω ǫi jk ǫjαβ

because u = Ω ρ × b. Thus,

∂

β ∂x l

Bk,

(3)

∂

Bk ∂xl = e Ω (ρ ρ ) b ∂ B − e Ω (ρ ρ ) b ∂ B k l i i l k ∂xl k ∂xl k = µ (δ – b b ) b ∂ B − µ(δ – b b ) b ∂ B , (4) k kl k l il i l i ∂xl k ∂xl k where use has been made of Equation (2), as well as the fact that µ = e u 2/(2 Ω).

(u × (ρ · ∇) B)i = e Ω (δkα δiβ − δkβ δiα) (ρα ρl) bβ

⊥

Hence,

(u × (ρ · ∇) B) = ∂Bµk b

∂xk – bi bk bl ∂x l ∂Bk = −µ b ∂(B = −µ b

i

∂Bk

∂Bk

i

– bk

bk) k

∂xi

k

∂xi

+b b b ikl

∂x i

=

∂B

−µ

,

∂Bk !

∂xl (5)

∂xi

because ∂Bk/∂xk = ∇ · B = 0, and bk ∂bk/∂xi = (1/2) ∂(bk bk)/∂xi = 0 (because bk bk = 1). Thus, (u × (ρ · ∇) B) = −µ ∇B. (6) 11

12 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises (c) We can write ∂

e (u · (ρ · ∇) E) = e (ui ρj

Ei) = e Ω ǫiαβ (ρα ρj ) bβ

∂

∂

∂x j

∂xj = µǫ

iαβ

(δ

αj

– b b )b α

j

Ei

Ei

β

∂xj ∂E i =µǫ b − µǫ b b b ∂Ei i jβ β iαβ α β j ∂xj ∂xj = −µ (∇ × E) = µ b · ∂B = µ ∂B. β

b

β

∂t

(7)

∂t

(d) We can write e (u · (ρ · ∇) A) = e (ui ρj = µǫ

iαβ

∂

Ai) = e Ω ǫiαβ (ρα ρj ) bβ

∂xj

(δ – b b ) b αj

α

j

∂ β

∂

Ai ∂x j

Ai

∂xj = µ ǫ b ∂Ai − µǫ b b b ∂Ai i jβ β iαβ α β j ∂xj ∂xj = −µ bβ (∇ × A)β = −µ b · B = −µ B.

(8)

2.2 Suppose that E1 is the electric field inside the slab. The resulting displacement of the ions is E δ= 1 Ωi B

(9)

parallel to the x-axis, where Ωi = e B/mi. The displacement of the electrons i Os (me/mi) smaller, and can, therefore, be neglected. The charge density that develops on the leading edge of the slab is e ne E1 .

(10) σ = e ne δ Ωi B = An equal and opposite charge density develops on the opposite face of the slab. Thus, we can write c2 σ (11) E1 = E0 −ǫ0 = E0 − E1 V 2 , A

√

where VA = B/ µ0 ne mi. Hence, E =

E0

1

where ǫ =1+

,

(12)

c2 .

VA2

(13)

ǫ

2.3 Consider Figure 2.2. The magnetic field B generated at a general point B by the wire at C = (0, d/2) is of magnitude µ0 I B= , (14) 2π r′ and is directed at right-angles to CB, as shown. Here, r′ is the length CB. Let A lie at the

Solutions Manual for Plasma Physics an Introduction 2nd Edition by Fitzpatrick Richard.