SOLUTIONS MANUAL for Plasma Physics: An Introduction 2nd Edition by Fitzpatrick Richard

CHAPTER

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, respectively) 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 x-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

me δ xe = −e Ex =

e2 ne (δxi − δxe). ǫ0

(3)

Likewise, the equation of motion of an individual ion is

mi δ xi = Z e Ex = −

Z 2 e2 n i (δxi − δxe). ǫ0

(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 + ω p2 e δxˆi = 0,

(7)

Πi2 δxˆe + (ω2 − ω p2 i) δxˆi = 0,

(8)

It follows that

where Πe = (e2 ne/ǫ0 me)1/2 and Πi = (Z 2 e2 ni/ǫ0 mi)1/2. The solutions are ω = 0 with δxê = δxî, and ω2 = Π 2 + Π 2 with Π 2 δxê + Π 2 δxî = 0. The former mode e e i i corresponds to a uniform translation of the slab. The latter mode is a plasma oscillation whose frequency, Π, satisfies Π = Πe2 + Πi2

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 Π i2 me δxˆi 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).

(12)

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

(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

∇ δΦ = − 0 , ǫ yields

∇ – where 1 λD

λ 2D

1 = 2

(16)

δρext δΦ = − !2 1 λD e

ǫ 0

1 + λD i

(17)

!2 ,

(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 – 2 δΦ = 0 (19) 2 r dr dr λD 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

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

2 − V = 0. dr2 λD2

(21)

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

√2 r q . 4π ǫ0 exp − λD

(22)

δΦ(r) =

√ q 2r . 4π ǫ0 r exp − λD

(23)

Thus,

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) = −

λD2

δΦ

(25) √

2 exp −

4π r λD

2r .

λD

(26)

The net shielding charge contained within a sphere of radius r, centered on the origin, is ∫ √ ′ ∫ r 2q r ′ ′ ′2 ′ ′ 2r r exp − Q(r) = 4π δρ(r ) r dr = − dr . (27) λ D λD 0 0 2

Thus, Q(r) = −q

∫ λD r/√2 0

x e−x dx = .

√ x.λD r/ 2 + −q −x e− 0

√2 r

which reduces to Q(r) = −q 1 − 1 +

λD

∫ λD r/√2 0

e− dx ,

(28)

√ 2r exp −

λD

(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) = Ê 0 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 e 0 ǫ0 The equation of motion of an individual electron inside the slab is

me δ x e = −e E1

(33)

e .. δxe + Π 2 δxe = − me Ê 0 cos(ω t),

(34)

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 δxˆe = (e/me ) Ê 0 . (35) ω2 − Π 2 Thus, writing E 1 (t) = Ê 1 cos(ω t), it follows from Equation (32) that ! ! Π2 1 Ê 1 = Ê 0 . (36) + 1 Ê 0 = ω2 − Π 2 1 − Π 2/ω2 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 dx2

−

2 λD

V = 0.

Moreover, V(−d/2) = −V0/2 and V(d/2) = V0/2. It follows that √ V0 sinh( 2 x/λD) V(x) = . 2 sinh(√2 d/2 λD )

(38)

(39)

Hence, the electric field between the plates is

√ cosh( 2 x/λD) V0 √ . E(x) = − =−√ dx 2 λD sinh( 2 d/2 λD ) dV

(40)

Now, by Gauss’ law, the charge density on the plate at d = x/2 is 1 σ = −ǫ0 E(d/2) = V0 ǫ0 . √ √ 2 λD tanh(d/ 2 λD) Hence, the charge on the plate is V0 ǫ 0 A 1 Q = Aσ = √ . √ 2 λD tanh(d/ 2 λD)

(41)

(42)

Chapter 1 □ 5 There is an equal and opposite charge on the other plate. Thus, the capacitance is √ d/ 2 λD Q 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)

where me is the electron mass, T the plasma temperature, and V(z) the perturbed electrostatic potential. Likewise, the ion number density becomes ni = n0 exp

−e V − mi gz . T

(45)

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 me n 0 g z − , T T e n0 V mi n 0 g z 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

(49)

− 2 V = 2 E0 z,

(50)

! ǫ0 T 1/2 e2 n 0

(51)

where λD = is the Debye length, and

mi g

. (52) e By inspection, the solution of Equation (50) that satisfies dV/dz = 0 at z = 0, and is well √ 2z behaved as z → ∞, is λ E = 0

V(z) = E 0 −z − √ exp − D . λ 2 The corresponding electric field is E = Ez ez, where √ 2z dV E z (z) = − = E 0 1 − exp − D . dz λ

(53)

(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, d2Φ ρ e (ni − ne). =− =− ǫ ǫ dx2 0

(55)

Assuming that e Φ/Ti ≪ 1 and e Φ/Te ≪ 1, it is easily demonstrated that, to first order in ! small quantities, 1 1 2 + ρ ≃ −e n0 Φ. (56) Ti Te Hence, Poisson’s equation becomes

where

d2Φ

λD2

e2 n

1 λD2

2 −

Φ = 0, 1

ǫ0

1 +

(57) ! .

(58)

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 " # σ dΦ 0+ dΦ . =2 . =− , dx 0− dx 0+ ǫ0 which yields

λD Thus, Φ(x) =

(60)

(61)

e−| x|/λD .

(62)

ǫ0

σ λD 2 ǫ0

1.7 Repeating the previous analysis, with Ti = Te = T , we obtain d2 Φ e n0 −e Φ/T 2 e n0 +e Φ/T = sinh(e Φ/T ). =− 0 e −e ǫ0 dx2 ǫ

(63)

Multiplication by dΦ/dx yields dΦ d2Φ dx dx2

2 e n0 dΦ ǫ0

sinh(e Φ/T ).

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

(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 . =− dx 2 ǫ0

(66)

Thus, the previous two equations can be combined to give ! σ2 . 16 ǫ0 n0 T

(67)

, dΦ = − 8 n0 T/ǫ0 sinh(e Φ/2 T ) dx

(68)

, dΦ′ = − 8 n0 T/ǫ0 x, ′ V sinh(e Φ /2 T )

(69)

= cosh−1 1 +

T Now, taking the square-root of (65), we obtain

in the region x > 0. Thus, ∫

∫ e Φ/2 T ,

where λD =

e V/2 T

dy x =− , sinh y λD

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

(70)

(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 x̂ s = x s /λD and V̂ = e V/T . It follows that " # tanh(V̂/4) . x̂s = ln tanh(V̂/4 e)

(73)

This relationship is plotted in Figure 1.1. 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 ∂2Φ 1 ∂ ∂Φ r + 2 = 0, r ∂θ2 r ∂r ∂r

(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 0 =− = cos θ. (76) ∂r r=a− ǫ0 ǫ0 It is fairly clear that Φ(r, θ) = φ(r) cos θ, where ! d dφ r r – φ = 0, dr dr "

and

#r=a+ e n δx 0 dφ = . dr r=a ǫ0

(77)

(78)

−

The solution is

e n0 δx φ=− r 2 ǫ0

for r < a, and

e n0 δx a2

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

σ e n0 δx e =− ex. 2 ǫ0 x 2ǫ0

(79)

(80)

(81)

Thus, the equation of motion of an individual electron is d2δx where

dt2

e =−

Ex = −Π 2 δx,

(82)

Π2 =

e2 n 0

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

(83)

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 me δx = − δx . (84) i e mi 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 (mi/me + 1) e n0 (1 + me/mi) δxi . δxe = − 2 ǫ0 2 ǫ0

(86)

The equation of motion of an electron is d2δxe dt2

=−

e Ex me

=−

e2 n0 (1 + me/mi) δxe.

(87)

e2 n0 (mi/me + 1) =− δxi. 2 ǫ0 mi

(88)

2 ǫ0 me

Likewise, the equation of motion of an ion is d2δxi dt2

e Ex mi

In both cases, the oscillation frequency is Π, where ! e2 n 1 1 0 . Π2 = + 2 ǫ 0 me mi

(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 = 0, (91) sin θ 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. ∂r r=a− ǫ0 ǫ0 It is fairly clear that Φ(r, θ) = φ(r) cos θ, where ! d 2 dφ – 2 φ = 0, r 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=

e n0 δz r − 3 ǫ0

for r < a, and φ=−

e n0 δx a3 3 ǫ0

r2 for r > a. Hence, the electric field inside the sphere is E=

(95)

σ e n0 δz ez = − ez. 3 ǫ0 3ǫ0

(96)

(97)

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

dt2

e =− 2

Ez = −Π 2 δz,

(98)

Π =

e2 n 0

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

(99)

CHAPTER

Chapter 2

2.1

(a) Given that ρ = ρ (− cos γ e1 + sin γ e2) and ρ = u⊥/Ω, it follows that u2 . ⊥ 2 ρρ = ( ) Ω 2 (cos γ) e1e1 − (cos γ sin γ) e1e2 − (cos γ sin γ) e2e1 . 2 +(sin γ) e2e2 .

(1)

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

u⊥2

−(I 2Ω2

bb) ,

(2)

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

∂ ∂xl

Bk)

= e Ω ǫi jk ǫ jαβ (ρα ρl) b β

∂ B k, ∂xl

(3)

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

∂ ∂xl

= e Ω (ρ ρ ) b ∂ B − e Ω (ρ ρ ) b ∂ B k l i k i l k k ∂xl ∂xl = µ (δ – b b ) b ∂ B − µ (δ – b b ) b ∂ B ,

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

k l

⊥

Hence, ∂Bk ∂Bk ∂Bk ∂Bk (u × (ρ · ∇) B)i = µ bi + b i kbl b – b – b i bk bl k ∂xk ∂x ∂x l ∂x i l ∂Bk ∂B = −µ b ∂(B bk) = −µ b = −µ , k k ∂xi ∂xi ∂xi

(5)

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 ρ j

∂ ∂xj

Ei) = e Ω ǫiαβ (ρα ρ j ) bβ

= µ ǫiαβ (δα j – bα b j) b β

∂ ∂x j

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

i jβ

∂t

(7)

∂t

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

∂ ∂xj

Ai) = e Ω ǫiαβ (ρα ρ j ) bβ

= µ ǫiαβ (δα j – bα b j) b β = µǫ

i jβ

∂ ∂x j

∂ Ai ∂x j

b ∂Ai − µǫ b b b ∂Ai iαβ α β j β ∂xj ∂x j

= −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 E1 δ= Ωi B

(9)

parallel to the x-axis, where Ωi = e B/mi. The displacement of the electrons isO (me/mi) smaller, and can, therefore, be neglected. The charge density that develops on the leading edge of the slab is e n e 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 −ǫ = E0 − E1 2 , 0 VA √ where VA = B/ µ0 ne mi. Hence, E0 E = , (12) 1 ǫ where c2 ǫ = 1 + 2. (13) VA 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

Chapter 2 □ 13 x

B B δ r

r′

θ′ A

d/2

FIGURE 2.2 Exercise 2.3

origin. Thus, r is the length AB, and d/2 the length AC. Furthermore, θ′ is the angle BAC, where θ′ = π/2 − θ. Let δ be the angle ABC. According to the cosine rule, r′ 2 = r2 + d2/4 − r d cos θ′. Thus, in the limit d/r ≪ 1, r′ ≃ r 1 − According to the sine rule,

d 2r

(15)

! sin θ .

(16)

sin δ sin θ′ = . d/2 r′

Thus, in the limit d/r ≪ 1, δ≃

(17)

d 2r

cos θ.

(18)

Now, µ0 I sin[(d/2 r) cos θ] µ0 I d cos θ ≃ , 2π r 1 − (d/2 r) sin θ 4π r2 µ0 I cos[(d/2 r) cos θ] B = B cos δ = ≃ µ0 I + µ0 I d sin θ ,

Br = B sin δ = θ

2π r 1 − (d/2 r) sin θ

2π r

4π

(19) (20)

where r, θ, z are cylindrical coordinates. By analogy, the magnetic field generated at point B by the wire at (0, −d/2) is obtained from the previous expressions by making the transformations I → −I and d → −d. Thus, the net magnetic field at Bis µ0 I d cos θ , 2π r2 µ0 I d sin θ Bθ = . 2π r2 Br =

(21) (22)

14 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises ! B = ∇ψ × e = 1 ∂ψ , −∂ψ, 0 . z r ∂θ ∂r

Let us write

(23)

It follows that ψ= 2.4

µ0 I d sin θ . 2π r

(24)

(a) Let ϑ = π/2 − θ. The equation of a field-line is ψ = constant, or µ0 I d cos ϑ 2π

= constant,

(25)

which yields r = req cos ϑ,

(26)

where req ≡ r(ϑ = 0). From Equations (21) and (22), the magnetic field-strength is B(r, ϑ) =

µ0 I d

2π r2

req 2 eq

where

(27)

µ0 I d

Beq ≡ B(ϑ = 0) =

2π r 2eq

(28)

Hence, the variation of the magnetic field-strength along a magnetic field-line is Beq

B(ϑ) =

cos2 ϑ

(29)

Let α be the pitch-angle, and αeq ≡ α(ϑ = 0) the equatorial pitch-angle. From Eq. (2.99) in the book, sin2 α(ϑ) B(ϑ) = , (30) Beq sin2 αeq which yields

sin2 α sin2 αeq

1 =

cos2 ϑ

(31)

At the mirror point α = π/2 and ϑ = ϑm. Hence, cos ϑm = sin αeq.

(32)

Now, sin α = v⊥/v, so v = v sin α = ⊥

v sin αeq cos ϑ

= v cos ϑm , cos ϑ

(33)

and v⊥2

!1/2

vǁ = ±v 1 − 2 = ±v 1 − sin α v cos2 ϑ !1/2 m . = ±v 1 − cos2 ϑ

1/2

= ±v 1 −

sin2 αeq

1/2

cos2 ϑ (34)

Chapter 2 □ 15 However, v = (2 E/m)1/2, so v =

E m

⊥

!1/2

2E vǁ = ± m

cos ϑ m

(35)

cos ϑ !1/2

cos2 ϑm 1− cos2 ϑ

!1/2 .

(36)

(b) An element of field-line length is ds = (dr2 + r2 dϑ2)1/2.

(37)

Given that the equation of a field-line is r = req cos ϑ, it follows that dr = −req sin ϑ dϑ. Hence, ds = req (sin2 ϑ + cos2 ϑ)1/2 dϑ = req dϑ. The bounce period is ∫ ϑm τb = 4 0

(38)

∫ ds m 1/2 ϑm cos ϑ dϑ = 4 r eq 2 E (cos2 ϑ − cos2 ϑ )1/2 . |v | ǁ

Hence,

(39)

√2 π r dζ eq = , (40) τb = 4 req 2)1/2 1/2 (1 − ζ (E/m) 2E 0 ∫1 where ζ = sin ϑ/ sin ϑm. Here, use has been made of the result 0 (1 − ζ2)−1/2 dζ = π/2. (c) From Equations (21), (22), (27), and (28), we have m

1/2

∫ 1

b = sin ϑ er + cos ϑ eθ, and ∇B = −

2B r

(41) (42)

Hence, 2 B cos ϑ e z. r It follows from Eq. (2.110) in the book that the drift velocity can be written b × ∇B =

vd =

2 (v⊥2/2 + vǁ2) cos ϑ Ωeq (B/Beq) r

ez,

(43)

(44)

where Ωeq = e Beq/m. Making use of Equations (26), (29), (35), and (36), we obtain 2E v = (2 cos2 ϑ cos2 ϑ ). (45) m − d e Beq req Now, (vd) =

∫ ϑm

vd ds

dϑ. τb 0 |vǁ| dϑ Hence, it follows from Equations (38), (40), and (45), that ∫ ϑm 4E (2 cos2 ϑ − cos2 ϑm) (vd) = cos ϑ dϑ, e Beq req π 0 (cos2 ϑ − cos2 ϑm )1/2

(46)

(47)

16 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises or

∫ 1

(v ) =

cos θm + 2 sin θm (1 − ζ )

dζ, (48) (1 − ζ2)1/2 where ζ = sin ϑ/ sin ϑ m. However, ∫ 1(1 − ζ2)−1/2 dζ = π/2 and∫ 1(1 − ζ2)1/2 dζ = π/4, 0 0 so we obtain 2E (v ) = . (49) d e Beq req d

e Beq req π

2.5 According to conventional orbit theory, the particle’s gyrocenter moves 2along the magnetic field-line (i.e., in the z-direction) at constant kinetic energy E = (m/2) (v ⊥+ v 2ǁ) and constant magnetic moment µ = (m/2) v 2/B . It follows that ⊥

vǁ(z) = ± (2/m) [E − µ Bz(z)]. Now, it is easily seen that

⊥

sin α =

v2 + v2

⊥ z=0

Hence,

µ B0 E

, vǁ(z) = ± 2 E/m [1 − sin2 α Bz(z)/B0]1/2 , = ± 2 E/m [cos2 α − sin2 α z2/L2]1/2.

(50)

(51)

(52)

It follows that the bounce points (at which vǁ = 0) lie at z = ±z0, where z0/L = cot2 α. The bounce time is thus ∫ z0 ∫ z0 , dz dz 2π L 4L m = τb = 4 = . (53) √ 0 |vǁ(z)| sin α 2 E 0 (z02 − z2)1/2 sin α 2 E/m 2.6 Conservation of the first adiabatic invariant, µ, implies that v2

v 2(x, t) ⊥

B(x, t)

⊥0 ,

(54)

where v⊥ 0 ≡ v⊥(0, t) is a constant of the motion [because B0 ≡ B(0, t) is constant]. Now, vǁ(x, t) = ±(v2 − v⊥ 2)1/2, where v = v(t). Let us define sin α0 =

v⊥ 0

(55) (56)

where α0 = α0(t). It follows that vǁ(x, t) = ±v cos α0 (1 − k2 tan2 α0 x2)1/2.

(57)

The bounce point lies at x = xm(t), where vǁ(±xm, t) = 0. It follows that x (t) = m

1 k tan α0

(58)

and vǁ(x, t) = ±v cos α0 (1 − x2/xm 2)1/2.

(59)

Chapter 2 □ 17 The bounce time is

∫ xm

∫ xm dx 4 dx τb(t) = 4 (1 − x2/x 2 )1/2 |v | = v cos α ǁ 0 0 0 m ∫ 1 dy 2π xm . 4 xm = = v cos α0 0 (1 − y2)1/2 v cos α0

(60)

Thus, making use of Equations (56) and (58), we obtain 2π τ (t) = . b k v⊥ 0

(61)

The second adiabatic invariant takes the form ∫ xm ∫ xm J = 4m |vǁ| dx = 4 m v cos α0 (1 − x2/xm 2)1/2 dx 0

∫ 1

= 4 m v cos α0 xm 0

(1 −y2)1/2 dy = π m v cos α0 xm.

(62)

Again making use of Equations (56) and (58), we obtain π m vǁ 0 J= . k tan2 α0

(63)

Conservation of J implies that " #1/2 tan α0(t) k(0) = . tan α0(0) k(t)

(64)

Thus, it follows from Equation (58) that

" #1/2 k(0) k(0) tan α0(0) xm(t) . = = xm(0) k(t) k(t) tan α0(t)

(65)

Equation (61) yields

τb(t) = k(0) . (66) τb(0) k(t) Finally, it follows from Equations (56) and (64) that " # 1 v⊥2 0 = v 2 1 + v2(t) = ⊥0 tan2 α0(t) sin2 α0(t) # " 1 = v 2 1 + k(t) ⊥0 k(0) tan2 α0(0) " ( )# 1 = v 2 1 + k(t) ⊥0 k(0) sin2 α0 (0) − 1 k(t) . 2 = v2 + v (0) − v 2 . . (67) ⊥0 ⊥0 k(0) Thus, writing E(t) = (1/2) m v2(t), E⊥ 0 = (1/2) m v 2 , and Eǁ 0 = (1/2)m [v2(0) − v2 ], we "

obtain E(t) = E⊥ 0 +

⊥0

k(t) k(0)

⊥0

Eǁ 0.

(68)

18 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 2.7 The equations of perpendicular motion of a particle confined to the current sheet (i.e., with |y| < a) are dvx e B = Ω v y , = vy z 0 y a dt m dvy e y = − v x Bz = −Ω0 v x a , dt m

(69) (70)

where Ω0 = e B0/m. It is convenient to normalize all lengths to a, all times to Ω0−1, and all velocities to a Ω0. The normalized equations of motion become d2 x

dy = y, dt2 dt 2 dy dx = − y. dt2 dt

(71) (72)

These equations can be integrated to give

and

vx = y2/2 − v0,

(73)

v 2 = v 2 − (y2/2 − v0) 2,

(74)

⊥

where v⊥ and v0 are constants. (a) The turning points (at which vy = 0) satisfy y ±2 = 2 (v0 ± v⊥).

(75)

Here, it is assumed that 1 > y+ > y− > 0. Thus, v0 = (y + y )/4, and v⊥ = (y − y )/4. It follows that

−

vx = y2/2 − (y+2 + y−2)/4,

(76)

vy2 = (y+2 − y2) (y2 − − y 2)/4.

(77)

Thus, the mean drift-velocity in the x-direction is 2

(vx) = (y /2) − (y+ + y−)/4, where 2

∫ y+

(y ) =

y−

y2 |vy |

, ∫ y+ dy y−

or 2

(y ) =

1 dy, |vy |

, ∫ y+

∫ y+

y2 dy

2 1/2 [(y2+ − y2) (y2 − y− )]

−

(78)

(79)

dy [(y+2 − y2) (y2 − − y 2)]1/2

(80)

−

A simple change of variables, 2

ζ = y − (y+ + y−)/2 , (y+2 − y−2)/2

(81)

Chapter 2 □ 19 yields

(vx) = (1/2) (y+ + y−) α,

(82)

∫1

where

(1 + κ ζ)1/2 (1 − ζ 2)−1/2 dζ α = ∫ −1 , 1 (1 + κ ζ)−1/2 (1 −ζ 2)−1/2 dζ

(83)

−1

and κ = (y 2 − y 2)/(y 2 + y 2). Thus, +

−

(vx) = −(1/4) (y+ + y−) (1 − α).

(84)

In unnormalized form, this becomes 2

(vx) = −(Ω0/4 a) (y+ + y−) (1 − α).

(85)

Suppose that y± = y0 ± ρ, where ρ/y0 ≪ 1. It follows that y + y κ ≃ 2 ρ/y0 ≪ 1. Thus, where

−

≃ 2 y 2 and 0

(vx) = −(1/2) y0 (1 − α),

(86)

I0 + (1/2) κ I1 − (1/8) κ2 I2 + · · · α≃ , 2 I0 – (1/2) κ I1 + (3/8) κ I2 + · · ·

and

∫ 1 In =

(87)

ζ n (1 −ζ 2)−1/2 dζ.

(88)

−1

However, it is easily demonstrated that I0 = π, I1 = 0, and I2 = π/2. Thus, α≃

1 − κ2/16

≃ 1 − κ /4 ≃ 1 − ρ /y0 , 1 + 3 κ2/16

(89)

which implies that (vx) ≃ −(1/2) ρ2 .

(90)

In unnormalized form, this becomes (vx) ≃ −(Ω0/2 a) ρ2 . In the absence of field-line curvature, and a time-varying electric field, the magnetic drift is entirely made up of grad-B drift. Furthermore, the conventional expression for this drift yields (vx ) =

µ mΩ

e x · e z × ∇Bz = −

µ dBz = − (Ω0 /2 a) ρ2, m Ω dy

(91)

where µ = m v2 /(2 Bz), Ω = e Bz/m, and ρ2 = v 2/Ω 2. ⊥

⊥

(b) The constraint vx(y = 0) = 0 gives v0 = 0. The turning points thus satisfy y = ±y0, where y 2 = 2 v2⊥. Thus, 0

vx = (1/2) y2,

(92)

vy2 = (1/4) (y04 − y4).

(93)

20 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises The mean drift velocity in the x-direction is (vx) = (y 2/2), ∫ y0

where 2

(y ) = ∫ y0

It follows that

−y |vy |

, ∫ y0 dy − y0

−y (y

1 |vy |

, ∫ y0

y2 dy

(y ) =

(94)

− y4)1/2

dy.

(95)

4 4 1/2 −y (y0 − y )

(96)

A simple change of variables gives 2

(y ) = y0 α, ∫ 1

where α=

ζ 2 dζ

(1 − ζ 4)1/2 Thus, in unnormalized units, 0

,∫ 1 0

dζ = 0.4450. (1 − ζ 4)1/2

(vx) = 0.223 (Ω0/a) y02.

(97)

(98)

(99)

CHAPTER

Chapter 3

3.1 We can write ∫

fd v=n

3/2

"∫ ∞

mu2

exp

– 2T "∫ ∞ ! # z2 exp m u duz , – 2 T −∞ 2π T

−∞

where ux = vx − Vx, et cetera. Hence, ∫ f d3v = n ∫ ∞

However,

−∞

Thus, we obtain

We can write ∫

#∫ ∞

1 π1/2

exp −

dux

−∞

whereas

duy (1)

∫ ∞

e−z

−∞

!3 dz .

(2)

ez − dz = π1/2.

(3)

f d3v = n.

(4)

∫

! "∫ ∞ # m u2 3/2 m x 3 vx f d v = n (ux + Vx) exp dux – 2T 2π T −∞ ! # "∫ ∫ ∞ mu2 ∞ m u y2 z exp duz , duy exp − – 2T −∞ 2 T −∞

where ux = vx − Vx, et cetera. Hence, ∫ ∫ ∞ , 1 2T 2 v f d3v = n z + V x e−z dz x 1/2 m π −∞ However,

m uy 2

∫∞ 2 1 e−z dz = 1, 1/2 −∞ π ∫∞ 2 z e−z dz = 0, −∞

1 π1/2

∫ ∞

e−z2 dz

(5)

!2 .

(6)

−∞

(7)

(8) 21

22 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises by symmetry. Hence, we obtain ∫ vx f d3v = n Vx.

(9)

This result generalizes in a straight-forward manner to give ∫ v f d3v = n V.

(10)

We can write ∫

m 3/2 v2 f d3v = n 2π T

∫

m v2

d3v – 2T ! ∫∞ 3/2 m v2 m 2 =n v exp 4π v2 dv, – 2T 2π T 0

which yields

∫ 2

However,

∫ ∞

z 4 e−z dz.

π1/2 m

(12)

0 1/2

3π

z 4 e−z dz =

∫

Hence,

(11)

∫ ∞

8nT

v fdv=

v2 exp

v2 f d3v =

3nT

(13)

(14)

n T.

(15)

! 2π + ms vs2 . 2 Ts s

(16)

m ∫ 1

m v2 f d3v =

2 3.2 We can write

Ts3/2 n

– ln fs = ln

Hence, ∫

3 + ln 2 m T3/2s

ss =

(− ln f s ) f s d v = ln ns ∫ 1 1 + ms v s2 fs d3vs. Ts 2

2π ln

!∫ 3

fs d v s (17)

However, ∫ fs d3vs = ns, ∫

1 2

ms v s f s d v s

Thus,

T3/2s s s = n s ln

3 2

3 + ln 2 m

(18)

n T.

2π s

(19)

3 +

(20)

Chapter 3 □ 23 3.3 We can write

∂ · K12, m1 ∂v1

C12 =

(21)

∫

where K12 = γ12 ∫

The result

w12 · J12 d3v2. ∫

∂ · K12 d3v1 = 0 m1 ∂v1 follows from integration by parts (assuming that f1 → 0 as v1 → ∞, etc.). 3

C12 d v1 =

We can write

∫

(22)

∫ ∂ · K12 d3 v1 = − K12 d3v1 ∂v1 ∫ ∫ = −γ12 w12 · J12 d3v1 d3v2,

(23)

∫

m1 v1 C12 d3v1 =

(24)

where we have integrated by parts. Likewise, ∫ ∫ ∫ m2 v2 C21 d3v2 = −γ21 w21 · J21 d3v1 d3v2.

(25)

However, γ21 = γ12, w21 = w12, and J21 = −J12. Hence, we conclude that ∫ ∫ m1 v1 C12 d3v1 = − m2 v2 C21 d3v2.

(26)

We can write

∫

∫ ∂ · K12 d3v1 = −2 v1 · K12 d3v1 ∂v1 ∫ ∫ = −2 γ12 v1 · w12 · J12 d3v1 d3v2,

∫

m1 v12 C12 d3v1 =

v12

where we have integrated by parts. Likewise, ∫ ∫ ∫ 2 3 m2 v2 C21 d v2 = −2 γ21 v2 · w21 · J21 d3v1 d3v2 ∫ ∫ = 2 γ12 v2 · w12 · J12 d3v1 d3v2. Hence, ∫

m2 v22 C21 d3v2 = −2 γ12

However, Thus, we conclude that

(28)

∫ ∫

∫ m1 v12 C12 d3v1 +

(27)

u12 · w12 · J12 d3v1 d3v2.

u12 · w12 = 0. ∫

(29)

(30)

∫ m1 v 1 C12 d v1 = − 2

m2 v22 C21 d3v2.

(31)

CHAPTER

Chapter 4

∫

4.1 We can write

ms (v − Vs) (v − Vs) fs d3v,

ps =

(1)

which yields ∫ ps =

∫

ms vv fs d3v − ms Vs ∫ + ms VsVs fs d3v.

∫ v fs d3v − ms v fs d3v Vs (2)

But, ∫ Ps =

ms vv fs d3v,

(3)

v fs d3v,

(4)

fs d3v,

(5)

∫ ns Vs = ∫ ns = so we get

ps = Ps − ms ns VsVs,

(6)

Ps = ps + ms ns VsVs.

(7)

or We can write

∫ qs =

2 ∫ 1

ms ws2 (v − Vs) fs d3v ∫ ms ws v fs d v − Vs 2

But,

1 2 ∫

so we get qs =

m w 2 f d 3v = s

1 2

3 2

ms ws fs d v.

(8)

(9)

3 ms ws 2 (v − Vs ) sf d3v − ps V s. 2 2

(10) 25

26 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Now,

∫

1 ms ws2 (v − Vs) fs d3v = 2

However,

∫

1 2 ∫

∫ 1 ∫ m v2 v f d3v − Vs · ms vv fs d3v 2 s s ∫ 1 + m 2 v fs d3v. s Vs 2

(11)

ms v v fs d v = Qs,

(12)

ms vv fs d3v = Ps, ∫ v fs d3v = ns Vs,

(13) (14)

∫

so we get

1 1 ms ws2 (v − Vs ) sf d3v = Qs − Vs · sP + ms ns Vs2 V s. 2 2 It follows from (10) that 1 3 qs = Qs – Vs · Ps + m s ns V s2 Vs – ps Vs. 2 2

(15)

(16)

However, making use of (7), we get qs = Qs – Vs · ps – or

(17)

1 Q =q +p ·V + 3 p V + m n V2V . s s s s s s 2 2 s s s s

(18)

∫ wss′ = ∫ 1

wss′ =

ms n s V s V s –

ps Vs,

4.2 We can write

which gives

1 2

ms (v − Vs) Css′ d v,

(19)

∫ 2

∫

ms v Css′ d v − Vs ·

But,

∫

ms v Css′ d v + Vs

Css′ d v.

(20)

1 m v2 Css′ d 3v, 2

(21)

ms v Css′ d 3v,

(22)

Css′ d 3v.

(23)

wss′ = Wss′ − Vs · Fss′ ,

(24)

Wss′ = wss′ + Vs · Fss′ .

(25)

Wss′ = ∫ Fss′ =

∫ 0= Thus, we obtain or

Chapter 4 □ 27 4.3 The kinetic equation is written ∂ fs + ∇ · (v f s ) + ∇v · (a s f s ) = C s, ∂t where as = (es/ms) (E + v × B). Recall the following definitions: ∫ ns = fs d3v, ∫ ns Vs = v fs d3v, ∫ Ps = ms v v fs d3v, ∫ 1 2 3 Qs = ms v v fs d v, 2

(26)

(27) (28) (29) (30)

∫

and

ms v2 fs d3v = 3 ps + ms ns V s2.

Ps = Furthermore,

(31)

∫ Cs d3v = 0,

(32)

ms v Cs d3v = Fs,

(33)

∫ ∫

1 m v2 Cs d3v = ws + Vs · Fs. 2 s

∫ Let us take (4.35) d3v. We obtain ∫ ! ∫ ∫ ∫ ∂ fs d3v + ∇ · ∇v · (as fs) d3v = v fs d3v + Cs d3v. ∂t

(34)

(35)

Making use of the divergence theorem, and the reasonable assumption that f s → 0 as |v| → ∞ , the third term on the left-hand side is easily seen to be zero. Furthermore, (32) shows that the term on the right-hand side is also zero. The remaining terms give ∂ns

+ (n ∇ · s Vs) = 0, ∂t where use has been made of the definitions (27) and (28). ∫ Let us take ms v (4.35) d3v. We obtain ! ∫ ∫ ∫ ∂ 3 ms v fs d3v + ∇ · ms v ∇v · (as fs) d3v ms v v fs d v + ∂t

(36)

∫

ms v Cs d3v.

The third term on the left-hand side can be integrated by parts to give ∫ ∫ 3 ms v ∇v · (as fs) d v = − ms as fs d3v = −es ns (E + Vs × B).

(37)

(38)

28 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Thus, (37) gives ∂(ms ns Vs) + ∇ · P s – es ns (E + Vs × B) = Fs, ∂t where use has been made of (29) and (33). ∫ Let us take (1/2) ms v2 (4.35) fs d3v. We obtain ∫ ! ∫ 1 ∂ 1 2 3 ms v2 fs d3v + ∇ · ms v v fs d v ∂t 2 2 ∫ ∫ 1 1 2 3 ms v2 Cs d3v. + ms v ∇v · (as fs) d v = 2 2 The third term on the left-hand side can be integrated by parts to give ∫ ∫ 1 2 3 m v ∇v · (as fs) d v = − ms v · as fs d3v = −es ns E · Vs. 2 s

(39)

(40)

(41)

Thus, (40) gives ! ∂ 3 +1m n 2 ps s s V s + ∇ · Qs – es ns E · Vs = ws + Vs · Fs, 2 ∂t 2

(42)

where use has been made of (30), (31), and (34). 4.4 Recall the following definitions: ds ∂ = + V s·, dt ∂t Ps = ps I + πs + ms ns Vs Vs, Qs = q s +

5 1 p s V s + π s · V s + m s ns Vs2 V s. 2 2

(43) (44) (45)

Equation (36) can be written ∂ns + V s · ∇ns + ns ∇ · Vs = 0, ∂t

(46)

ds ns + n s ∇ · Vs = 0 dt

(47)

which reduces to

with the aid of (43). Equation (39) can be written ∂ns ∂Vs V + ∇ps + ∇ · πs + ms (Vs · ∇ns) Vs + ms ∂t s ∂t + ms ns (∇ · Vs) Vs + ms ns (Vs · ∇)Vs − es ns (E + Vs × B) = Fs,

ms n s

(48)

which can be rearranged to give ! dsVs ds ns + ∇p s + ∇ · πs + n s ∇ · V s V s + m s ns dt dt – es ns (E + Vs × B) = Fs.

(49)

Chapter 4 □ 29 It follows from (47) that ms ns

dsVs + ∇p s + ∇ · πs – es ns (E + Vs × B) = Fs. dt

(50)

Equation (42) can be written ∂Vs ∂ns 3 ∂ps 1 + m s V s2 + m s ns V s · + ∇ · qs 2 ∂t 2 ∂t ∂t 5 5 + V · ∇p + p ∇ · V + V · ∇ · π + π : ∇V s s s s s s s s 2 2 1 1 + m V 2 V · ∇n + m n V 2 ∇ · V + m n V · [(V · ∇) V ] s s s s s s s s s s s s 2 2 – es ns E · Vs = ws + Vs · Fs,

(51)

which can be rearranged to give 1 2

2 ms V s

ds ns + n s ∇ · Vs dt

# dsVs + Vs · ms ns + ∇p s + ∇ · πs – es ns (E + Vs × B) − Fs dt 3 ds ps 5 + + p s ∇ · Vs + π s : ∇V s + ∇ · q s = w s . 2 dt 2

(52)

It follows from (47) and (50) that 3 ds ps 5 + p s ∇ · Vs + π s : ∇V s + ∇ · q s = w s . 2 dt 2

(53)

4.5 Recall the definitions ps = ns Ts,

(54) ps 3/2 ,

(55)

, Ts πs : ∇Vs

(56)

ss = ns log

5/2

ss = ss Vs + ws

Θ= s

−

ns qs

−

qs Ts

∇Ts Ts

(57)

The entropy conservation equation ∂ss + ∇ · s s = Θ s, ∂t

(58)

can be written ss ∂ns 3 1 ∂ps −5 ∂ns + ss V · ∇n + 3 1 V · ∇p + s s s s ns 2 Ts ns ∂t 2 Ts ∂t 2 ∂t – 5 V · ∇n + s ∇ · V + ∇ · qs − qs · ∇Ts T 2

30 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises ws πs T: ∇Vs qs ·2∇Ts s = Ts − − . Ts

(59)

Multiplying by Ts and rearranging, we obtain ! ! ss Ts – 5 T ds ns 3 ds ps 5 + ns ∇ · V s + + p s ∇ · Vs ns dt 2 s 2 dt 2 + πs : ∇Vs + ∇ · qs = ws.

(60)

It follows from (47) that 3 ds ps 5 ∇ · V + π : ∇V + ∇ · q = w . + ps s s s s s 2 dt 2 4.6 We can write

! !n ∫ ∞ !n f v d3v = 1 v exp −v2 4π v2 dv n vt π3/2 v t3 0 vt vt2 ∫ ∞ 4 2 = xn+2 e−x dx. 1/2 π 0

(61)

∫

In =

However,

∫∞

2 1 e−x dx = Γ

where Γ(z) is a gamma function. Hence, I = n

! m+1 , 2

! Γ n+3 . 2 π1/2

(62)

(63)

(64)

Now, Γ(1/2) = π1/2, Γ(3/2) = (1/2) π1/2, Γ(5/2) = (3/4) π1/2, and Γ(5/2) = (15/8) π1/2. Hence, I−2 = 2, I0 = 1, I2 = 3/2, and I4 = 15/4. 4.7 We have

∂f

+ v · ∇ f = −ν ( f − f ),

∂t where

(65)

! 2 −|v − V|

. (66) exp vt2 π3/2 vt 3 Let ∇ = L−1∇ˆ , t = (L/vt ) t̂, v = vt v̂, V = vt V̂, ν = vt /l, and ǫ = l/L ≪ 1, where l is the mean-free-path, and L is a typical macroscopic lengthscale. All hatted quantities are designed to be O(1). We can normalize Equation (65) to give ! ∂f ǫ + v̂ ˆ f = f0 – f . (67) ·∇ ∂tˆ f0 =

Let f = f0 + ǫ f1 + O(ǫ2 f0). Assuming that f is time-independent, Equation (67) gives

(68)

f1 = −v̂ · ∇ˆ f0

(69)

f = f0 − ν−1 v · ∇ f0 + O(ǫ2 f0).

(70)

to lowest order in ǫ. Thus,

Chapter 4 □ 31 (a) Suppose that vt and n are spatial constants, but that V = Vy(x) ey. It follows that n } − v 2 + v 2 + [vy − Vy(x)]2 n f0 = 3/2 3 x z . π v exp 2 vt t Hence, if δ f ≡ f − f0 then n } − v 2 + v 2 + [vy − Vy(x)]2 n ν−1 2 vx (vy − Vy) dVy . δ f = − π3/2 v 3 x z dx exp vt2 t 3 vt The viscosity tensor takes the form ∫ π = m (v − V) (v − V) δ f d3v.

(71)

(72)

(73)

It is clear from symmetry that the only non-zero components of this tensor are πxy and πyx. In fact, ∫ 2 2 2 −1 2 m n ν dVy wx + wy + wz 2 2 3 w d w, (74) −x wy exp 2 πxy = πyx = − v t (2π)3/2 v t5 dx where w = v − V. It follows that dVy 2 , (75) πxy = −A m n ν l dx where A = 2 Ax Ay Az, and 1 ∫∞ 2 1 ŵ exp(−ŵ 2 ) dŵ x = , Ax = √ x x 2 π −∞ 1 ∫ ∞ 2 1 2 Ay = √ ŵy exp(−ŵy ) dŵy = , 2 π −∞ ∫ ∞ 1 exp(−ŵz 2 ) dŵz = 1. Az = √ π −∞

(76) (77) (78)

Hence, A = 1/2. (b) Suppose that n is a spatial constant, V = 0, but that T = T (x). It follows that vt = vt(x), and d ln vt/dx = (1/2) d ln T/dx. Hence, ! v2 3 ν−1 dT vx 2 − f. (79) δf = − 2 0 vt T dx ∫

Now,

1 m v2 v δ f d3v. 2 It is clear from symmetry that the only non-zero component of q is qx. In fact, q=

q = A n νl2 − x where

∫ A=

(81)

v2 v x2 v2 3 − 2 v v t4

(80)

f0 n

d3v.

(82)

32 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises It follows that A = I 6 I02 + 6 I4 I2 I0 + 2 I23 − where

3 I4 I0 2 − 32I 20I , 2

1 ∫∞ exp(−x2) dx = 1, I0 = √ π −∞ 1∫ ∞ 1 2 x2 exp(−x ) dx = , I2 = √ 2 π − 3 ∞ 1∫ 4 = , 2 ∞ x exp I4 = √ (−x ) dx 4 π − 1∫ 15 ∞ x6 exp 2 I6 = √ (−x ) dx = 8 . ∞ π

(83)

(84) (85) (86) (87)

−

Thus, A = 5/2.

∞

(c) Suppose that V = 0, and n = n(x) and T = T (x), but that p = n T is constant. It follows that vt = vt(x), and d ln vt/dx = (1/2) d ln T/dx, and d ln n/dx = −d ln T/dx. Hence, ! ν−1 dT v2 5 vx 2 − f. (88) δf = − 2 0 vt T dx ∫

Now,

1 m v2 v δ f d3v. 2

(89)

It is clear from symmetry that the only non-zero component of q is qx. In fact, q = A n νl2 − x where

∫ A=

(90)

v2 v x2 v2 5 − 2 v v t4

It follows that

dT !

d3v.

(91)

A = I I 2 + 6 I I I + 2 I 3 −5 I I 2 − 5 I 2 I , 4 0 6 0 4 2 0 2 2 0 2

(92)

where I2 =

I0 = 1, I = 4

I = 6

, 2 15

(93) .

(94)

Thus, A = 5/4. 4.8 We have

e E · ∇v fe = fe − f0, m e νe

where f0 =

2! exp −v . vt2e π3/2 vt3e

(95)

(96)

Chapter 4 □ 33 Let It follows that

fe = f0 + f1.

(97)

e E · ∇v( f0 + f1) = f1. m e νe

(98)

Assuming that ∇v ∼ vt−e 1, we find that f1

∼

E E + Ec

(99)

where Ec = me νe vt e/e. Thus, if E ≪ Ec then f1 ≪ f0, and (98) reduces to e E · ∇v f0 = f1, m e νe which implies that

f = f + e

Now,

Hence,

v 0

2v ∇v f0 = − 2 f0. vt e

∫ j = −e

so ji = ∫

(101)

(102)

f = − 2e E·v f . 1 0 me νe vt2e

Now,

However,

E·∇ f .

me νe

(100)

(103)

∫ v fe d3v = −e

2 e2 n e m e νe vi v j

∫ Ej f0

v f1 d3v,

vi v j f0 d3v. 2 vt e n

d 3v =

vt2e ne

1 2

δ .

(104) (105)

(106)

Hence, we get j = σ E, where σ=

(107)

e2 n e me νe

(108)

CHAPTER

Chapter 5

5.1 For fields varying as exp[ i (k · r − ω t)] the equations ∇ · E = ρ/ǫ0 and ∇ · B = 0 yield ρ ik ·E = , (1) ǫ0 i k · B = 0,

(2)

respectively. However, the equation of charge conservation, ∂ρ

j = 0,

(3)

−ω ρ + k · j = 0.

(4)

∂t yields

∇·

Hence, (1) and (2) give ω c2

k · E = −µ0 k · j,

(5)

k · B = 0.

(6)

However, these two expressions are also obtained from the dot products of Eqs. (5.4) and (5.5) in the book with k. 5.2 The linearized ion and electron equations of motion are −i ω mi Vi = +e (E + Vi × B0),

(7)

−i ω me Ve = −e (E + Ve × B0),

(8)

where B0 = (0, 0, B0). The current is written j ≡ n e (Vi − Ve) = σ · E.

(9)

Finally, the dielectric permittivity is defined K≡I+

iσ ǫ0 ω

(10)

Consider an electrostatic wave: E = (0, 0, Ez), Vi = (0, 0, Viz), Ve = (0, 0, Vez), j = (0, 0, jz). The equations of motion yield −i ω mi Vi z = e Ez,

(11) 35

36 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises −i ω me Ve z = −e Ez. It follows that

eE jz = n e

−i ω mi

eE +

! z

−i ω me

iǫ =

(12)

Π 2 + Π 2 Ez, i

(13)

giving σ = jz = i ǫ0 Π 2 + Π 2 . zz i e Ez ω Hence, P ≡ Kzz = 1 −

Π2

(14)

Π2 − 2e . ω

i ω2

(15)

Consider a right-hand circularly polarized electromagnetic wave: E = (Ex, i Ex, 0), Vi = (Vi x, i Vi x, 0), Ve = (Ve x, i Ve x, 0), j = ( jx, i jx, 0). The equations of motion yield −i ω mi Vi x = e Ex + e B0 i Vi x,

(16)

−i ω me Ve x = −e Ex − e B0 i Ve x,

(17)

−i ω mi (1 + Ωi/ω) Vi x = e Ex,

(18)

−i ω me (1 + Ωe/ω) Ve x = −e Ex.

(19)

It follows that

Π2

Πe

jx = i ǫ0

i +

ω+Ω

ω+Ω e

Ex,

(20)

and R ≡ Kxx = 1 −

Πi2 ω (ω + Ωi)

–

Π e2 ω (ω + Ω e)

(21)

Consider a left-hand circularly polarized electromagnetic wave: E = (Ex, −i Ex, 0), Vi = (Vi x, −i Vi x, 0), Ve = (Ve x, −i Ve x, 0), j = ( jx, −i jx, 0). The equations of motion yield −i ω mi Vi x = e Ex − e B0 i Vi x,

(22)

−i ω me Ve x = −e Ex + e B0 i Ve x,

(23)

−i ω mi (1 − Ωi/ω) Vi x = e Ex,

(24)

−i ω me (1 − Ωe/ω) Ve x = −e Ex.

(25)

It follows that

Πi 2 jx = i ǫ0

and

Πe

+ Ex, ω−Ω i ω−Ω e Πi2 Π2 e L ≡ Kxx = 1 − – . ω (ω − Ωi) ω (ω − Ωe )

(26) (27)

Chapter 5 □ 37 5.3 Now S = D=

R+L

, 2 R−L

(28) ,

(29)

2 so R = S + D,

(30)

L = S − D.

(31)

R L = (S + D) (S − D) = S 2 − D 2.

(32)

. S − n2 cos2 θ, −i D, n2 cos θ sin θ . . . . = 0. . i D, S − n2, 0 2 . n2 cos θ sin θ, . 0, P − n2 sin θ

(33)

It follows that 5.4 The dispersion relation is

Multiplying out the determinant, we get 0 = (S − n2 cos2 θ) (S − n2) (P − n2 sin2 θ) − D2 (P − n2 sin2 θ) – (n 2 cos θ sin θ)2 (S − n2 ),

(34)

or 0 = [S 2 − D 2 − n2 S (1 + cos2 θ) + n4 cos2 θ] (P − n2 sin2 θ) – n 4 (S − n2 ) cos2 θ sin2 θ, or

. . 0 = (S 2 − D 2) P − n2 S P (1 + cos2 θ) + (S 2 −D 2) sin2 θ + n4 P cos2 θ + S sin2 θ .

Making use of the identity, S 2 − D 2 = R L, this reduces to . . 0 = R L P − n2 S P (1 + cos2 θ) + R L sin2 θ + n4 (P cos2 θ + S sin2 θ). 5.5

(35)

(36)

(37)

Let

We have

A = S sin2 θ + P cos2 θ,

(38)

B = R L sin2 θ + P S (1 + cos2 θ),

(39)

C = P R L.

(40)

F 2 = B 2 − 4 A C,

or F 2 = R 2 L 2 sin4 θ + P 2 S 2 (1 + cos2 θ)2 + 2 P R L S sin2 θ (1 + cos2 θ)

(41)

38 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 2 – 4 P R L S sin 2 θ − 4 P2 (S2 − D2 ) cos θ,

(42)

or F 2 = R 2 L 2 sin4 θ − 2 R L P S sin4 θ + P 2 S 2 sin4 θ + 4 P 2 D 2 cos2 θ. This reduces to

F 2 = (R L − P S )2 sin4 θ + 4 P 2 D 2 cos2 θ.

(43) (44)

5.6 The dispersion relation (5.44) in the book can be written n2 (S n2 − R L) s2 + P n2 (n2 − S ) c2 = P (S n2 − R L),

(45)

where s ≡ sin θ and c ≡ cos θ. Let t ≡ tan θ. It follows that c2 = 1/(1+t 2) and s2 = t 2/(1+t 2). Hence, we obtain n2 (S n2 − R L) t 2 + P n2 (n2 − S ) = P (S n2 − R L) (1 + t 2),

(46)

which can be rearranged to give (n2 − P) (S n2 − R L) t 2 = −P (n4 − 2 S n2 + R L).

(47)

However, S = (R + L)/2, so the previous expression yields P (n2

R) (n2 L) – − t =− 2 . (n − P) (S n2 − R L)

(48)

Π2 Π2 R = 1 − e (1 + ω/Ωe) −1 − i (1 + ω/Ωi) −1 ω Ωe ω Ωi

(49)

5.7 In the limit ω → 0,

reduces to

Πe R≃1−

But,

Ω

Π2 1

Πe

+ e

Ω

Π e2

iω

Πi 2

Ωe Hence, as ω → 0, R → 1+

+ . Ω 2 Ωe 2

Π2 i

(50)

= 0.

(51)

Π2 . Ω

(52)

Ωi Π2 + Ω

2 i

Because R → L as Ωe → −Ωe and Ωi → Ωi, it follows from the previous analysis that as 2 ω → 0, Π2 eΠ i L → 1+ + . (53) Ω Ω e2

2 i

Because S = (R + L)/2 and D → (R − L)/2, it follows from the previous analysis that as ω → 0, Π e 2 Πi 2 S →1+

+ , Ω2 e Ω2 i

(54)

Chapter 5 □ 39 D → 0.

(55)

Finally, it follows from the definition Π2 − i ω2 ω2

(56)

Π2 Π2 P→− e − i . ω2 ω2

(57)

P=1− that as ω → 0,

Π e2

5.8 It follows from the analysis in Exercise 5.1 that ω Vi − i Ωi Vi × b = i (e/mi) E,

(58)

ω Ve − i Ωe Ve × b = −i (e/me) E.

(59)

ω Vi x − i Ωi Vi y = i (e/mi) Ex,

(60)

ω Vi y + i Ωi Vi x = i (e/mi) Ey.

(61)

Hence,

Dividing the previous two equations, we obtain (i Vi x/Vi y) + (Ωi/ω) , (i Ex/Ey) = 1 + (Ω i/ω) (i V i x /Vi y )

(62)

which can be rearranged to give (i Ex/Ey) − (Ωi/ω) . 1 − (Ωi/ω) (i Ex/Ey)

(i Vi x /Vi y ) =

(63)

Finally, because Vi → Ve as mi → me, Ωi → Ωe, and E → −E, it follows from the previous analysis that (i Ex/Ey) − (Ωe/ω) (i Ve x /Ve y ) = . (64) 1 − (Ωe/ω) (i Ex/Ey) For a RH/LH circularly polarized wave (i Ex/Ey) =±1. Circular motion in the electron/ion cyclotron directions corresponds to counter-clockwise/clockwise rotation, respectively, in the x-y plane (with the z-axis coming out of the paper). Thus, circular motion in the electron/ion cyclotron directions corresponds to (i Vx/Vy) = ±1. However, it follows from the previous expressions that (i Vx/Vy) = ±1 whenever (i Ex/Ey) = ±1. Hence, a RH/LH circularly polarized wave is associated with circular motion in the electron /ion cyclotron directions, respectively. 5.9 The linearized species s equation of motion is −i ω ms Vs + νs ms Vs = es (E + Vs × B0), where B0 = (0, 0, B0). The current is written X es Vs = σ · E. j≡n

(65)

(66)

Finally, the dielectric permittivity is defined K

≡

I+

iσ

. ǫ0 ω

(67)

40 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises (a) Consider an electrostatic wave: E = (0, 0, Ez), Vs = (0, 0, Vsz), j = (0, 0, jz). The equation of motion yields −i ω ms Vs z + νs ms Vs z = es Ez, It follows that

n es2 Ez

jz =

= i ǫ0

X Π 2 Ez s

(−i ω + νs) ms

ω + i νs

(68)

where Π 2 = n e 2/ǫ0 ms, giving s

σzz =

jz Ez

X = i ǫ0 s

Hence, P ≡ Kzz = 1 −

X s

Π s2 ω + i νs

Π s2 ω (ω + i νs)

(69)

(70)

Consider a right-hand circularly polarized electromagnetic wave: E = (Ex, i Ex, 0), Vs = (Vs x, i Vs x, 0), j = ( jx, i jx, 0). The equation of motion yields −i ω ms Vs x + νs ms Vs x = es Ex + es B0 i Vs x,

(71)

−i (ω + i νs + Ωs) ms Vs x = es Ex,

(72)

or where Ωs = es B0/ms. It follows that X jx = i ǫ 0

ω + i νs + Ω s

and R ≡ Kxx = 1 −

Π s2 Ex

(73)

Π s2 ω (ω + i νs + Ωs)

(74)

Consider a left-hand circularly polarized electromagnetic wave: E = (Ex, −i Ex, 0), Vs = (Vs x, −i Vs x, 0), j = ( jx, −i jx, 0). An analogous calculation to that given previously yields X Π s2 . (75) L = 1− s ω (ω + i νs − Ωs) It is clear, given the definitions of Πs and Ωs, that the previous expressions for P, R, and L can be obtained from the standard expressions by making the substitution ms → ms (1 + i νs/ω). (b) The dispersion relation of a transverse EM wave is n2 = Hence, k2 =

k2 c2 = P. ω2

X 1 − 2

(76)

ω2

Π2

ω (ω + i νs)

(77)

Chapter 5 □ 41 In the limit ω ≫ νs, Πs, this gives X Π2 ω k≃

Thus,

≃

2 (ω2 + ν 2) s

Πs

−

Πe

ω ≃

2 ω2

(78)

2 ω (ω2 + ν 2)

1 −

Πs

− s

2 ω (ω + i νs)

ω kr ≃

1−

X Π 2 (ω − i νs)

1− c

2 ω2

(79)

because Πe ≫ Πi. Also, ki ≃

1X c

1 X Π 2s νs

Π 2 νs

s ≃ 2 (ω2 + ν 2s)

ω2

(80)

X ω2 =

(81)

Π s2 1 + i νs/ω

(82)

Assuming that ω ∼ Πe ≫ νs, the previous dispersion relation yields ! X Π 2 νs X i νs 2 2 s 2 . ω ≃ ≃ Π −i 1 + Πs e ω ω s s Thus, ω = Πe 1 − i

X Π 2 νs s

2 s 2Πe ω

(83)

(84)

It is clear that, to lowest order, ω = Πe. Hence, the previous expression gives ω ≃ Πe − i Γ, where Γ=

X Π 2 νs s

2Π2 .

(85)

(86)

Thus, the temporal variation of the wave is exp(−i ω t) = exp(−i Πe t) exp(−Γ t).

(87)

Clearly, collisions cause the oscillation to damp away. 5.10 The linearized perturbed equations are ∂n1 + n 0 ∇ · V1 = 0, ∂t ∂V1 = e n E – ∇p , me n 0 1 − 0 1 ∂t p1 n0−Γ − Γ p0 0n −Γ−1 n1 = 0,

(88) (89) (90)

42 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises ǫ0 ∇ · E1 = −e n1.

(91)

Let all perturbed quantities vary as exp[i (k · r − ω t)]. It follows that −i ω n1 + i n0 k · V1 = 0,

(92)

−i me n0 ω V1 = −e n0 E1 − i p1 k,

(93)

p1 n0 − Γ p0 n1 = 0,

(94)

i ǫ0 k · E1 = −e n1.

(95)

Hence, k · V1 = (ω/n0) n1, p1 = Γ (p0/n0) n1, k · E1 = i (e/ǫ0) n1. Thus, −i me n0 ω k · V1 = −e n0 k · E1 − i p12k ,

(96)

ω2 n1 = Π 2 n1 + k 2 c 2 n1,

(97)

yields

where Πe = (n0 e2/ǫ0 me)1/2 is the plasma frequency, and cs = (Γ p0/me n0)1/2 the electron sound speed. The dispersion relation is thus ω2 = Π 2 + k2 c 2. e

The phase-velocity is

v =

(98)

(99)

. 1/2 k 1 − Πe 2/ω2 whereas the group-velocity takes the form p

vg =

dω

= c s = cs 1 − Πe2/ω2 dk vp

1/2

(100)

Obviously, the wave only propagates when ω > Πe. Furthermore, in the cold plasma limit, cs → 0, the wave does not propagate at any frequency.

CHAPTER

Chapter 6

6.1 Maxwell’s equations in an electrically neutral medium take the form ρ ∇·E = , ǫ0 ∇ · B = 0, ∂B ∇×E = − , ∂t ∂E . ∇ × B = µ 0 j + ǫ 0 µ0 ∂t If P = ǫ0 (n2 − 1) E, and ρ = −∇ · P, ∂P j= , ∂t

(1) (2) (3) (4)

(5)

(6) (7)

then Equations (1)–(4) become ∇ · (n2 E) = 0,

(8)

∇ · B = 0,

(9)

∂B ∇×E = − , ∂t

(10)

2 ∂E ∇ × B = ǫ0 µ0 n ∂t .

(11)

Let ∂/∂t → −i ω. It follows that ∇ · (n2 E) = 0,

(12)

∇ · c B = 0,

(13)

∇ × E = i k0 c B,

(14)

∇ × cB = −i k0 n2 E,

(15)

where k0 = ω/c. Here, we have used ǫ0 µ0 = 1/c2. Note that Equations (12) and (13) are redundant, because they are implicit in Equations (14) and (15). 43

44 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 6.2 Starting from Equations (14) and (15), if E = Ey(z) e−i ω t ey,

(16)

B = Bx(z) e−i ω t ez,

(17)

then we obtain dEy

= i k c B x, − 0 dz d (c Bx) = − i k 0 n2 Ey. dz

(18) (19)

The previous two equations can be combined to give d2Ey

2 2 Ey = 0.

(20)

+ k0 n dz2 6.3 Starting from Equations (14) and (15), if E = Ey(z) e i (kx xω t) ey, B = Bx(z) e

i (kx x−ω t)

(21)

ex + Bz(z) e

i (kx x−ω t)

ez,

(22)

then we obtain dEy dz

ik − 0 c B x,

(23)

kx Ey = k0 c Bz,

(24)

d (c Bx) 2 − i kx c Bz = −i k0 n Ey. dz The previous three equations can be combined to give

(25)

d2Ey

2 2 Ey = 0, + k 0 q dz2 d (c Bx) + i k0 q2 Ey = 0, dz

(26) (27)

where q2 = n2 − S 2,

(28)

and S = kx/k0. Equations (26) and (27) have the same mathematical form as Eqs. (6.3) and (6.4) in the book, with n2 replaced by q2. Hence, reusing the analysis in the book, the WKB solutions becomes [see Eqs. (6.17) and (6.18) in the book] ! ∫z Ey(z) ≃ q−1/2 exp ±i k0 q dz′ , (29) 0

c Bx(z) ≃ ∓q

1/2

exp ±i k0

∫z

! q dz′

(30)

Furthermore, the criterion for the validity of these solutions is [see Eq. (6.16) in the book]. !2 . . 1 . 3 1 dq 1 d 2q . (31) k02 . 4 q2 dz – 2 ≪ 1. 2 q3 dz .

Chapter 6 □ 45 6.4 Starting from Equations (14) and (15), if E = Ex(z) e i (kx x−ω t) ex + Ez(z) e i (kx x−ω t) ez, B = By(z) e

i (kx x−ω t)

e y,

(32) (33)

then we obtain dEx dz

− i k x E z = i k 0 c B y,

(34)

d (c By) = i k 0 n 2 E x, dz

(35)

kx c By = −k0 n2 Ez.

(36)

The previous three equations can be combined to give dEx dz 2

d (c By) where

dz2

1 −

− i k0

dn2 d (c By)

n2 dz

q2 c By = 0,

(37)

2 2 c By = 0,

(38)

+ k0 q

q2 = n2 − S 2,

(39)

and S = kx/k0. Let us search for a solution to Equation (38) of the form

It follows that

c By(z) = e i φ(z).

(40)

!2 dφ d 2φ i dn2 dφ = k 02 q2 + i 2 − . dz dz n2 dz dz

(41)

We expect the final two terms on the right-hand side of the previous equation to be small. Hence, to first order, we can write !2 dφ ≃ k20 q2 , (42) dz which yields dφ = k q, dz ± 0 d2φ dq = ±k . 0 dz2 dz

(43) (44)

To second order, Equation (38) yields !# 1/2 " dφ dq 2 q dn . − = ± k02 q2 ± i k0 dz dz n dz

(45)

Making use of the binomial theorem, we obtain dφ dz

≃ ±k0 q +

i dq 2 q dz

−

i dn n dz

(46)

46 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises which can be integrated to give φ(z) ≃ ±k0

∫z

q dz′ + i ln q 1/2 − i ln n.

Hence, c By(z) = e

i φ(z)

= n q −1/2 exp

±i k0

∫z

(47) !

q dz′

(48)

Substituting the previous expression into Equation (37), and only keeping first-order terms, we get ! ∫z 1 1/2 − ′ Ex(z) = ±n q exp ±i k0 q dz . (49) Substituting (48) into (38), we find that the former is a good solution of the latter provided !2 . that ! 1 1 dq 1 d2q 1 dn 2 . . 1 1 d 2n 3 . (50) + − 2 n dz . ≪ 1. k0 2 . 4 q2 dz – . 2 q3 dz2 q2 n dz2 6.5 (a) According to Eq. (6.85) in the book, the group-velocity of an electromagnetic wave, of angular frequency ω, propagating vertically through a vertically stratified, unmagnetized, collisionless plasma of plasma frequency Πe(z), where z represents altitude above ground level, is " #1/2 Π 2(z) uz(ω, z) = c 1 −e 2 . (51) ω The signal is reflected at height z0, where Πe(z0) = ω. Hence, the net travel time of a pulse that is launched from height z = 0, reflected at height z0, and returns to height z = 0, is ∫ z0(ω) ∫ dz 2 z0(ω) ω = τ(ω) = 2 dz. (52) 2 − Π 2(z)]1/2 [ω u (ω, z) c z 0 0 e Thus, the equivalent height is ∫ z0(ω) c ω h(ω) = τ(ω) = dz. (53) 2 0 [ω2 − Πe 2(z)]1/2 (b) Let ω2 = v and Π e2(z) = u(z). Equation (53) becomes ∫ z0(v 1/2) dz v −1/2 h(v 1/2) = , 0 [v − u(z)] 1/2

(54)

where u(z0) = v, and u(z) < v for 0 < z < z0. Let us multiply both sides of the previous equation by (w − v) −1/2/π and integrate from v = 0 to w. We obtain ∫ w ∫ w ∫ z (v 1/2 ) dz 0 1 1 dv. v −1/2 (w −v)−1/ h(v 1/2 ) dv = 2 π 0 1/2 1/2 π 0 0 (w − v) (v − u) (55) As illustrated in Figure 6.3, provided that z0(v 1/2) is a monotonically increasing function of v 1/2, we can swap the order of integration in the double integral on the righthand side of the previous expression to give # ∫ ∫ 1/2 "∫ w dv 1 w v −1/2 (w − v)−1/2 h(v 1/2) dv=1 z0 (w ) dz. 1/2 (v − u)1/2 π 0 π 0 u(z) (w − v)

Chapter 6 □ 47 (56) Let v = w cos2 θ + u sin2 θ. It follows that ∫ w dv

∫ π/2 =

2 dθ = π.

(57)

v −1/2 (w−v)−1/2 h(v 1/2) dv.

(58)

u(z) (w − v)1/2 (v − u)1/2

Hence, we obtain z0(w 1/2) = 1 π

∫ w 0

Making the substitutions v = w sin2 α and w 1/2 = ω, we get ∫ 2 π/2 z0(ω) = h(ω sin α) dα. π 0

(59)

However, by definition, ω = Πe when z = z0. Hence, the previous equation can also be written ∫ 2 π/2 z(Πe) = h(Πe sin α) dα. (60) π 0 (c)

If ω Π0

h(ω) = h0 + δ then Π z = h0 + 2 δ e π Π0 However,

∫ π/2 sinp α dα = 2p−1 0

!p (61)

!p ∫ π/2 sinp α dα.

(62)

Γ(1/2 + p/2) Γ(1/2 + p/2) . Γ(1 + p)

(63)

Hence, we obtain " Πe(z) =

π Γ(1 + p) Γ(1/2 + p/2) Γ(1/2 + p/2)

#1/p

!1/p Π 0 z − h0 . 2 δ

(64)

Of course, this expression only holds for z ≥ h0 (it cannot hold for z < h0 because it would give a complex plasma frequency). For z < z0, we have Πe = 0. 6.6

(a) According to Eq. (6.127) in the book, the horizontal range of the pulse is ∫ z0(S ) dz , , x0 = 2 S 0 n2(z) − S 2

(65)

where S = sin θ, and n(z0) = |S |. If n2 = 1 − α (z − h0) for z ≥ h0, and n2 = 1 for z < h0, ∫ h ∫ h +C 2 /α then 0 0 dz dz + , (66) , x0 = 2 S 2 − α (z − h ) 0 C C h0 0 where C = cos θ. It follows that " # ∫ 1 dζ x0 = 2 S h 0 + C , (67) C α 0 (1 − ζ)1/2

z→

48 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises

z = z0(v1/2)

v = u(z) v→

FIGURE 6.3 Exercise 6.5

which reduces to x0 = 2 h0 tan θ +

sin 2θ.

(68)

Now, dx0 dθ

2 h0

cos 2θ α " # α h0 2 cos4 θ −cos2 θ + = 2 α cos2 θ 4 = (cos2 θ − cos2 θ ) (cos2 θ − cos2 θ ), − + α cos2 θ cos2 θ 4

(69)

√ 1∓ 1− 4αh 0 . (70) cos θ± = 4 Thus, it is clear that if α h0 < 1/4 then x0(θ) has a maximum at θ = θ−, and a minimum at θ = θ+, where 0 < θ− < θ+ < π/2. It follows that if a root of Equation (68) lies in the range θ− < θ < θ+ then there are two other values of θ in this range that also satisfy the equation. On the other hand, if α h0 > 1/4 then x0(θ) is a monotonically increasing function of θ in the range 0 < θ < π/2, and Equation (68) only possesses a single root. The critical case, α h0 = 1/4, cor√responds to θ− = θ+ = π/3. Moreover, when θ = θ− then Equation (68) yields x0 = 6 3 h0. where

(b) According to Eq. (6.124) in the book, the travel time, t0, of an obliquely propagating radio pulse is related to its horizontal range, x0, as x0 . (71) c t0 = sin θ Thus, it follows from Equation (68) that 2 h0 4 ct = + cos θ. 0 cos θ α

(72)

Chapter 6 □ 49 Hence, d (c t0) 2 h0 4 =− 2 + , d cos θ cos θ α d2 (c t0) 4 h0 = . d cos θ 2 cos3 θ

(73) (74)

It follows that c t0 attains a minimum value when θ = θ0, where cos θ0 =

! h0 α 1/2 . 2

(75)

However, this is clearly only possible provided α h0 < 2. According to Equation (68), the corresponding horizontal range is !1/2 2 x0(θ0) = 4 h0 . (76) –1 α h0

CHAPTER

Chapter 7

7.1 The plasma wave dispersion relation, (7.23) in the book, can be written ! 2 = 0, 1 + I − e i π ∂F0 ǫ0 me k 2 ∂u u=ω/k ∫∞ ∂F 0/∂u e2 P du, ǫ0 m e k −∞ ω − k u ! n m u2 e F0(u) = exp – . 2 Te (2π Te/me)1/2

where

and

Let

1 ω−ku

It follows that

≃

1 ω

ku ω

(2)

(3)

! k2 u2 k3u3 + 3 + ··· . ω2 ω

! 1 −2 n u k u k2 u2 k3u3 π1/2 (2 T /m )3/2 ω 1+ ω + + + · · · ω2 ω3 0 e e e −∞ 2 ! me u × exp – 2 Te du.

e2 I= ǫm k

(1)

(4)

∫ ∞

(5)

By symmetry, this reduces to ! 2 k2 T e 2 Πe 2 I=− J2 + J 4+ · · · , me ω 2 π1/2 ω 2 where Jm ≡

(6)

∫∞ ζ m exp(−ζ 2) dζ = Γ[(m + 1)/2],

−∞

(7)

and Πe = (n e2/ǫ0 me)1/2. Thus, J2 = Γ(3/2) = π1/2/2, and J4 = Γ(5/2) = (3/2) J2, giving Π2 Te Π 2 e I = − e − 3 k2 . (8) ω2 me ω 4 Thus, the dispersion relation (1) becomes Π2 T Π2 e2 e 2 e e iπ 1− −3k − 2 4 ω me ω ǫ0 me k 2

! ∂F0 ∂u

= 0.

(9)

u=ω/k

52 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 7.2 The previous dispersion relation can be written 1−

Πe 2

– 3 (k λD) 2

Πe 4

π 1/2 e2 n m1/2 ω e ǫ0 Te3/2 k3

−

m ω2

= 0.

(10)

2 Te k2

Let x = (ω/Πe)2 and y = k λD, where Πe = (n e2/ǫ0 me)1/2 and λD = (Te/me Π 2e)1/2. It follows that 1 3 y2 1 − − 2 + i ǫ(x, y) = 0, (11) x x where ! π 1/2 x1/2 x ǫ(x, y) = (12) exp − 2 . 2y 2 y3 Writing x = 1 + δx, where |δx| ≪ 1, we have δx 2 − (1 + 3 y2) δx + 3 y2 − i ǫ1 = O(δx 3), where ǫ1 = ǫ(1, y) = Hence,

3 y2)2 + 4 i ǫ ]1/2 2 − 1 ≃ 3 y – i ǫ1. 2 7.3 The dispersion relation (7.23) in the book is written ∫ ∞ e2 n u m exp( m u2/2 T ) e du. e −e 1− 1/2 ǫ0 me k (2π Te/me) ω−ku −∞ Te δx =

(1 + 3 y2)

! 1 exp − 2 . 2y y3

π 1/2 1

–

[(1

(13)

(14)

(15)

(16)

Let t = (me/2 Te)1/2 u. It follows that 1+

1 2 (k λD ) π1/2

∫ ∞

t e−t

dt,

(17)

−∞ t − ζ

√ where ζ = (ω/k) (me/2 Te)1/2 = (ω/Πe)/(k λD)2/ 2, where Πe = (n e2/ǫ me)1/2. Hence, we obtain Z′(ζ) 1− = 0. (18) 2 (k λD )2 Assuming that |ζ| ≫ 1, and that Im(ζ) ≪ 1/|ζ|, the appropriate asymptotic expansion of Z(ζ) ! is 2 1 1 +O 1 . Z(ζ) = i π1/2 e−ζ (19) – − 3 5 ζ 2ζ ζ It follows that ! 1 3 2 Z′(ζ) = −2 i π1/2 ζ e−ζ + − +O . (20) 1 ζ2 Thus, the dispersion relation (18) yields −2 i π

1/2

ζe

2 −ζ

+ 2 ζ

2ζ4

ζ6

! 3 1 + 4 + O 6 = 2 (k λ D )2 ζ 2ζ

(21)

It can be seen that the ordering |ζ| ≫ 1 corresponds to the ordering k λD ≪ 1. Writing ζ=√

1 2 (k λD)

(1 + δ),

(22)

Chapter 7 □ 53 where |δ| ≪ 1, (21) gives −i π 2

1/2

" # 1 exp − 1 + (1 + δ)−2 + 3 (k λD)2 (1 + δ)−4 = 1, (k λD)3 2 (k λD)2

which can be solved to give δ≃

3 2

i π 1/2 (k λD)2 − 2 2

1 (k λD)3

" exp −

2 (k λD)2

which implies that # " √ 2 (k λD) ζ ≃ 1 + 3 (k λD)2 − i π 1/2 1 exp − 1 . 2 2 2 (k λD)3 2 (k λD)2 7.4 The dispersion relation (7.49) in the book takes the form ∫ ∞ ∫ ∞ e2 ∂F /∂u e2 ∂F 0 i /∂u du = 0, oe du + 1+ ǫ0 me k −∞ ω − k u ǫ0 mi k −∞ ω − k u ! where m u2 n s F0 s = exp – . 1/2 2 Ts (2π Ts/ms) Let Πs = (n e2/ǫ0 ms)1/2, λD s = (Ts/ms Π s2)1/2, and ∫ ∞ e2 ∂F /∂u os du. Is = ǫ0 ms k −∞ ω − k u It follows that

!3/2 ∫ ∞

(24)

(25)

(26)

(27)

(28)

2! ms u exp – du. 2 Ts −∞ ω − k u

2Π ms Is = − s 1/2 π k 2 Ts

(23)

(29)

Let t = (ms/2 Ts)1/2 u. We obtain Is = −

2 Π 2 m !1/2 ∫ ∞ s

π1/2 k

2 Ts

t e−t2

−∞ ω − k (2 Ts/ms)

or 1 1 Is = 2 2 (k λD s) π1/2

∫ ∞

−t2

2 t e dt, −∞ t − ζs

dt,

(30)

1/2 t

(31)

where ζs = (ms/2 Ts)1/2 (ω/k). Hence, Z′(ζs) Is = − . 2 (k λD s)2

(32)

The dispersion relation (26) has the form 1 + Ie + Ii = 0, which yields 1−

Z′(ζe) 2 (k λD e

)2 i

−

Z′(ζi) 2 (k λD )2

(33)

= 0.

(34)

54 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Using the expansions

√ 2 1 Z′(ζ ) ≃ −i 2 π ζ e−ζi + , ζ i

√ Z′(ζe) ≃ −i 2 π ζe

(35)

− 2,

(36)

Te 2 1 −i 2π 1/2 ζi e −i + 2 = 2 ( k λD e)2. ζ Ti ζi

(37)

e−ζe

we obtain 1/2

−i 2π ζe e −e − 2 + 2

Now, ζe = (me/mi)1/2 (Ti/Te)1/2 ζi. Let y = (Ti/Te)1/2 ζi. It follows that ! ! !3/2 me 1/2 T e T 1 1/2 e + exp − y y −i 2π − 2 + 2 = 2 (k λD e )2, mi Ti Ti y

(38)

assuming that ζe ≪ ζi. Assuming that me/mi ≪ 1, and writing y = yr + i yi, where yr and yi are real, and |yi | ≪ yr , we find that, to lowest order, 1 2 −2 + 2 = 2 (k λD e) , (39) yr which implies that 1 yr = √ . (40) 2.1/2 . 2 1 + (k λD e) To next order, we find that !3/2 ! !1/2 Te yi 1 (π/8)1/2 Te me − . + exp (41) = −. .3/2 m 2 Ti 1 + (k λD e)2 . . Ti i yr 1 + (k λD e)2 Now, ω 2 Ti = k mi

!1/2 ζi =

2 Te mi

!1/2 y.

(42)

! ωr Te 1/2 1 ≃ . .1/2 , k mi 1 + (k λD e) 2

(43)

Hence, writing ω = ωr + i γ, we have

and

(π/8)1/2

≃ −.

ωr

1 + (k λD e

. 3/2

me mi

!1/2

+ Te Ti

!3/2

exp − . 2 Ti 1 + (k λD e) 2

! . .

(44)

7.5 We have k = (k⊥, 0, kǁ),

(45)

v = (v⊥ cos θ, v⊥ sin θ, vǁ),

(46)

B = ω −1(k × B) = ω −1 −kǁ Ey, kǁ Ex − k⊥ Ez, k⊥ Ey

and

∫ t

e − 1s f

= ms ω

Jn Jm e i [(n Ωs+kǁ vǁ−ω) (t −t)+(m−n) θ]dt′, ′

(C cos χ + S sin χ + T ) −∞

n,m

(47)

(48)

Chapter 7 □ 55 where

! " ∂ f0 s ∂ f0 s # C = (ω − k v ) +k v E + k v ∂ f0 s − k v ∂ f0 s E , z ǁ ǁ ǁ ⊥ x ⊥ ǁ ⊥ ⊥ ∂v ∂v ∂v ∂v ⊥ ǁ " ǁ # ∂ f⊥0 s S = (ω − k v ) + k v ∂ f0 s E , ǁ ǁ

T =ω

∂ f0 s ∂vǁ

ǁ ⊥

∂v⊥

∂vǁ

(49) (50) (51)

Here,

χ = −Ωs (t′ − t) + θ,

(52)

and the arguments of the Bessel functions are as = k⊥ v⊥/Ωs. The dispersion relation is ∫ i X e s (K · E − E)x = v⊥ cos θ f1 s d3v, (53) ω ǫ0 s ∫ i X e s (K · E − E)y = v⊥ sin θ f1 s d3v, (54) ω ǫ0 s ∫ i X e s vǁ f1 s d3v, (55) (K · E − E)z = ω ǫ0 s ∫∞

∫

where

dv=

I v⊥ dv⊥

∫ ∞ dθ

Now,

I I

dθ 2π

dvǁ.

(56)

−∞

e i (m−n) θ = δ(m −n),

cos χ e i (m−n) θ =

1 2

(57)

δ(m −n + 1) e−i Ωs (t −t) ′

1 ′ δ(m − n − 1) e i Ωs (t −t), 2 I dθ i ′ i (m−n) θ sin χ e = − δ(m − n + 1) e−i Ωs (t −t) 2π 2 i ′ + δ(m − n − 1) e i Ωs (t −t), 2 I dθ 1 1 cos θ e i (m−n) θ = δ(m − n + 1) + δ(m − n − 1), 2π 2 2 I dθ i i i (m−n) θ sin θ e = − δ(m − n + 1) + δ(m − n − 1), 2π 2 2 I dθ 1 ′ cos θ cos χ e i (m−n) θ = δ(m − n + 2) e−i Ωs (t −t) 2π 4 1 . . + δ(m n) e −i Ωs (t′−t) + e i Ωs (t′−t) − 4 1 ′ + δ(m − n − 2) e i Ωs (t −t), 4 +

(58)

(59) (60) (61)

(62)

56 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises I dθ 1 ′ sin θ sin χ e i (m−n) θ = − δ(m − n + 2) e−i Ωs (t −t) 2π 4 1 . . + δ(m n) e −i Ωs (t′−t) + e i Ωs (t′−t) − 4 1 ′ – δ(m − n − 2) e i Ωs (t −t), 4 I dθ i ′ cos θ sin χ e i (m−n) θ = − δ(m − n + 2) e−i Ωs (t −t) 2π 4 i . . −i Ωs (t′−t) i Ωs (t′−t) – δ(m − n) e −e 4 i ′ + δ(m − n − 2) e i Ωs (t −t), 4 I

i ′ sin θ cos χ e i (m−n) θ = − δ(m − n + 2) e−i Ωs (t −t) 2π 4 i . . ′ −i Ω (t′−t) – e i Ωs (t −t) + δ(m − n) e s 4 i ′ + δ(m − n − 2) e i Ωs (t −t). 4

(63)

(64)

dθ

(65)

So, performing the θ and t′ integrals yields ∫ X e2 X s 3 H x, d v (K · E − E)x = − ω2 ǫ m s

(66)

∫ X e2 X s Hy, d3v (K · E − E)y = − ω2 ǫ m s

X e2 s (K · E − E)z = − ω2 ǫ0 ms

∫

X d3v

where

(67)

Hz,

(68)

" ! # Jn Jn+2 v⊥ C Jn Jn−2 1 1 2 Hx = + Jn + + 4 S (n −1) S (n −1) S (n + 1) S (n + 1) " ! # Jn Jn+2 1 1 i v⊥ S Jn Jn−2 2 − − – + Jn S (n −1) S (n + 1) S (n + 1) 4 S (n − 1) # " v⊥ T Jn Jn−1 Jn Jn+1 , + + 2 S (n) S (n) Hy = −

i v⊥ C

Jn Jn−2 2

− Jn

Jn Jn+2 − S (n + 1) S (n − 1) S (n + 1)! # 1 1 Jn Jn+2 + + S (n −1) S (n + 1) S (n + 1) #

S (n − 1) " J J v⊥ S 2 n n−2 − Jn – 4 S (n −1) " i v⊥ T Jn Jn−1 Jn Jn+1 , − – S (n) S (n) 2 4

−

(69)

(70)

Chapter 7 □ 57 " # vǁ C Jn Jn−1 Jn Jn+1 Hz = + 2 S (n −1) S (n + 1) " # Jn Jn+1 i vǁ S Jn Jn−1 − – 2 S (n − 1) S (n + 1) J2 + vǁ T n , S (n) and S (m) =

(71)

1 . m Ω s + kǁ v ǁ − ω

(72)

This can be rewritten X (K · E − E)x = −

e s2

∫

, n Ωs + k ǁ vǁ − ω ∫ X X Fy e2 s 3 d v , (K · E − E)y = − ω 2 ǫ 0 ms s n n Ω s + kǁ v ǁ − ω ∫ X X F e s2 z 3 , (K · E − E)z = − d v 2ǫ m ω 0 s n Ω + k ǁ vǁ − ω s s n s

d3v

ω 2 ǫ0 m s

(73)

(74) (75)

where F = x

v⊥ C 4

2 + J2 + Jn+1 Jn−1 n+1 Jn− + J n+1 n− 1 1

2 i v⊥ S 2 J – J J + J n+1 n− − 4 n−1 – Jn−1 Jn+1 n+1 1 v⊥ T + (J J + J J ) , n n+1 n n− 2 1

(76)

i v⊥ C 2 + J2 Fy = − J n+1 Jn− – Jn+1 – Jn−1 Jn+1 n− 4 1 1

v⊥ S − 4 Jn+1 Jn− – Jn+1 – Jn−1 + Jn−1 Jn+1 1 i v⊥ T − 2 (Jn Jn− – Jn Jn+1 ) , 1 F = z

(77)

vǁ C

(J J + J J ) n+1 n n−1 n 2 i vǁ S − 2 (J n+1 Jn – Jn−1 Jn) + vǁ T Jn2.

(78)

In other words, i v⊥ S 2 v⊥ C v⊥ T 2 (Jn+1 − n− (Jn+1 + J )2 − )+ J n (J n+1 + J n− ), 4 2 1 1 J n− 4 1 i v⊥ C 2 v⊥ S i v⊥ T F = (J − J2 ) + (J – J )2 + J (J – J ), y n+1 n n+1 n+1 n− n−1 n− 4 4 2 Fx =

(79) (80)

58 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Fz =

vǁ C i vǁ S Jn (Jn+1 + J n− ) − Jn (J n+1 – J n− ) + vǁ T Jn2. 2 2 1 1

It follows that

X Ki j = δi j + s

e s2 ω 2 ǫ 0 ms

S ij

, ω − kǁ vǁ − n Ωs

(81)

(82)

where v⊥ (J n+1 + Jn−1 )2 U, 4v ⊥ S = −i (J 2 − J 2 ) U, 12 n+1 n−1 v⊥ 4 v⊥ S = (J J (J + J )2 X + + J ) Y, 13 n n+1 n+1 n− n−1 4 2 1 v⊥ 2 S =i (J − J 2 ) U, 21 n−1 4 n+1 v⊥ S 22 = (J n+1 − Jn−1)2 U, 4v v⊥ ⊥ S =i (J 2 − J 2 ) X + i J (J – J ) Y, 23 n n+1 n−1 n− 4 n+1 2 S 11 =

(83) (84) (85) (86) (87) (88)

S 31 =

vǁ

J (J + J ) U, n+1 n−1 n 2 vǁ S 32 = −i Jn (Jn+1 – Jn−1 ) U, 2 vǁ J (J + J ) X + vǁ J 2 Y, S 33 = n+1 n−1 n n 2

(89) (90) (91)

and ∂ f0 s

U = (ω − k v ) ǁ ǁ

X = k⊥ v ǁ Y=ω

∂ f0 s ∂v⊥

∂ f0 s

+k v

ǁ ⊥

∂v⊥ − k⊥ v⊥

∂ f0 s

(92)

∂vǁ ,

(93)

∂vǁ

(94)

∂vǁ Now, 1

(J n+1 + J

Jn, 2 as 1 (J ′ 2 +1 – Jn−1) = −Jn , n−1) =

(95) (96)

so 2

S 11 = v⊥ (n Jn/as) U,

(97)

S 12 = i v⊥ (n Jn /a s ) Jn′ U,

(98)

S 13 = v⊥ (n Jn/as) Jn W,

(99)

S 21 = −i v⊥ (n Jn /a s ) Jn′ U,

(100)

Chapter 7 □ 59

where

S 22 = v⊥ Jn′ 2 U,

(101)

S 23 = −i v⊥ Jn Jn′ W,

(102)

S 31 = vǁ (n Jn/as) Jn U,

(103)

S 32 = i vǁ Jn′ Jn U,

(104)

S 33 = vǁ J n2 W,

(105)

X+Y =

n Ωs vǁ ∂ f0 s v⊥

+ (ω − n Ω ) s

∂v⊥

∂ f0 s

7.6 The Maxwellian velocity distribution is −ms (v⊥ +2 vǁ ) 2 ns f0 s = (2π T /m )3/2 exp . 2T s

(106)

∂vǁ

(107)

Hence, 2

−ms (v⊥ + vǁ ) ns ω v ⊥ U=− exp , 3/2 2Ts (2π T s /m s ) (T s/m s) 2

−ms (v⊥ + vǁ ) ns ω3/2vǁ 2T V = −(2π T /m ) (T /m ) exp . s

Consider K11. We have

X n e2

(108)

(109)

X∫ ∞

2π v dv ⊥ ⊥ K11 = 1 − 2ǫ m 3/2 (T /m ) ω (2π T /m ) 0 s 0 s s s s n ∫ ∞s n2 Ωs2 (ω/k2⊥) Jn2 −ms (v⊥2 + vǁ 2 ) dv exp . ǁ – ω − kǁ vǁ n Ωs 2 Ts −∞ s s

(110)

Let !1/2 2 Ts vs = , ms v⊥ x= , vs as k⊥ vs h= = , x Ωs T k 2s h2 ⊥ , = λs = ms Ω s2 2 ω − n Ωs ξ = , n k v t=

vǁ

(112) (113) (114) (115)

ǁ s

vs Πs =

(111)

ns e 2s ǫ0 m s

(116) !1/2 .

(117)

60 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises It follows that K

=1+

2 X∫ ∞ XΠ s s

x dx J

∫∞

However,

∫ (h x) e−x

dx x J 2(h x) e−x = n

∫ ∞

e−λs I (λ ), n

(120)

(119)

−t 2

X Π 2 e−λs X n2 In Z ω k ǁ vs

(118)

e dt. −∞ t − ξ n

Z(ξn) = π−1/2

K11 = 1 +

4 n2

. 1/2 t − ξ h2 k v π n ǁ s −∞

and

Hence,

e−t 2

∞

λs

(121)

Similarly, K12 reduces to ∫ ∞ dt e−t2 4 i n X Π 2 X∫ ∞ 2 2 x ′(h x) e− s x dx J (h x) J . 12K = n n 1/2 ω 0 t − ξn h kǁ vs −∞ π s n

(122)

However, differentiation of Equation (119) with respect to h yields ∫ ∞ . . 2 dx x2 J (h x) J′(h x) e−x = h e−λs I′ (λ ) − I (λ ) n n n s . n s 4 0

(123)

Hence, K12 =

X Π 2 e−λs X s i n (I′ − In) Z. n ω k ǁ vs n s

(124)

For K13 we get ∫ ∞ X Π 2 X∫ ∞ dt 2 t e−t2 2 n 2 s x dx J (h x) J (h x) e . ′ −x n 13K = n 1/2 ω π t − ξn h kǁ vs 0 −∞ s n However, ′

Z (ξn) = −π so we get K13 =

−1/2

(125)

∫ ∞ 2 t e −t 2 dt,

(126)

X Π 2 e−λs X (−) n In Z s /2 . 1 ω k v (2 λ ) ǁ s s s n

(127)

−∞

t − ξn

It is clear that K21 = −K12. For K22 we get

X Π 2 X∫ ∞ . 2 . s K =1+ x3 dx J′ (h x) 2 e−x 22 n ω n 0 s

However, ∫ ∞

. 2 1 x dx . J′ (h x) 2 e−x = 3

e−

" 2 n In (λ s)

∫ ∞

e−t2

4 . 1/2 t − ξ k v n ǁ s −∞ π

# + 2 λs In(λs) − 2 λs I′ (λs) ,

λs 0

λs

(128)

(129)

Chapter 7 □ 61 ! X Π 2 e− X n 2 I s n ′ K22 = 1 + ω λs λ + 2 λs In − 2 λ s In Z. k v

ǁ s

(130)

For K23 we get K23 =

X Π 2 X∫ ∞ s

∫ x2 dx Jn (h x) Jn′(h x) e−x

Hence, K23 =

X Π 2 e−λs X i λ 1/2 s s

ω k ǁ vs

2 1/2

dt 2 t e−t 2 2 (−i) . 1/2 t − ξn kǁ vs −∞ π

(131)

(I′ − In) Z′.

(132)

∞

It is clear that K31 = K13 and K32 = −K23. Finally, K33 gives ∫ 2 X Π 2 X∫ ∞ dt 2 t 2 e−t 2 ∞ s 2 K =1+ x dx J (h x) J ′(h x) e −x , 33 n n 1/2 π ω 0 t − ξ k v −∞ ǁ s n s n or K33 = 1 +

X Π 2 e−λs X s (−) In Z′ ξn. ω k ǁ vs n s

(134)

7.7 In the limit vs → 0, we have λs ∝ v 2s and ξn ∝ vs. Thus, X Π 2 Ti j s Ki j ≃ δi j + , ω k ǁ vs s where

(133)

(135)

! ! 1 1 1 1 kǁ vs , ≃ −1 + + =− 11 2 ω −Ωs ω + Ωs 2 ξ1 ξ−1 ! ! kǁ vs 1 −1 , T ≃ − i 1 − 1 = −i 12 2 ξ1 ξ− 2 ω − Ωs ω + Ωs T

(136) (137)

T13 ≃ 0, ! ! 1 1 1 kǁ vs , T ≃ i 1 − − =i 21 2 ω −Ωs ω + Ωs 2 ξ1 ξ−1 ! ! 1 1 1 1 1 k v ǁ s , T ≃− + + =− 22 2 ξ1 ξ− ω + Ω s 2 ω − Ωs

(138) (139) (140)

T23 ≃ 0,

(141)

T31 ≃ 0,

(142)

T32 ≃ 0,

(143)

T33 ≃ −

1 ξ0

kǁ vs . − ω

(144)

Hence, K11 = K22 = 1 − 1 2

X Π2 s

s ω2

ω ω − Ωs

ω ω + Ωs

! ,

(145)

62 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises i X Π2

K12 = −K21 = −

2 X Π2 s , K33 = 1 − ω2

s ω2

ω ω − Ωs

−

ω + Ωs

! ,

(146) (147)

and K13 = K31 = K23 = K32 = 0. 7.8 In the limit k⊥ → 0, we have λs → 0. Thus, Ki j ≃ δi j +

X Π s2 Ti j

(148)

. Z(ξ1) + Z(ξ−1) ,

(149)

ω k ǁ vs

where T11 ≃

2 i. . T12 ≃ Z(ξ1) − Z(ξ−1) , 2 T13 ≃ 0, . i. T21 ≃ − Z(ξ1) − Z(ξ−1) , 2 1. . T22 ≃ Z(ξ1) + Z(ξ−1) , 2 T23 ≃ 0,

(150) (151) (152) (153) (154)

T31 ≃ 0,

(155)

T32 ≃ 0,

(156)

T33 ≃ −Z′(ξ0) ξ0.

(157)

Now, c2 c 2 k2 ki k j − 2 δi j + Ki j. 2 ω ω

(158)

" ! !# k 2 ǁc2 1 X Π s2 ω − Ω ω + Ω s s , =1− + Z +Z ω2 2 s ω kǁ v s kǁ vs kǁ v s

(159)

Mi j = Thus, k2ǁ c2 M11 = K11 − 2 ω

and M12 = K12

" ! !# X Π2 ω + Ωs i ω − Ω – s s , Z = Z 2 s ω kǁ v s kǁ vs kǁ v s

(160)

and M13 = K13 = 0,

(161)

Chapter 7 □ 63 and

M21 = K21 = −K12 = −M21,

(162)

k2 c2 k2 c2 ǁ ǁ M22 = K22 − = K11 − ω2 = M11, ω2

(163)

M23 = K23 = 0,

(164)

M31 = K31 = 0,

(165)

M32 = K32 = 0,

(166)

and

and and and and

M33 = K33 = 1 −

X s

Πs 2

(kǁ vs)2

Z′

! .

(167)

kǁ v s

7.9 In the limit kǁ → 0, we have ξn → ∞. Thus, Ki j ≃ δi j +

X Π 2 e−λs X s Ti j, ω k v ǁ s s n

(168)

where n2 In(λs) kǁ vs T11 ≃ − ξ λ = − s (ω − n Ω ), n s s λ [In′ (λ s ) − In (λ s )] [In′ (λ s ) − In (λ s )] kǁ v s T 12 ≃ −i = −i , ξn (ω − n Ωs) n2 In(λs)

T13 ≃ 0,

(170) (171)

T21 ≃ −T12, T22 ≃ −

(169)

(172)

In (λ s )/λs + 2 λ s In (λ s ) − 2 λs In′ (λ s )

ξn [n2 In(λs) + 2 λ 2 In(λs) − 2 λ 2 I′ (λs)] kǁ vs s s n , =− λs (ω − n Ωs)

(173)

T23 ≃ 0,

(174)

T31 ≃ 0,

(175)

T32 ≃ 0, ≃−

T 33

(176) In(λs) ξn

In(λs) kǁ vs

=− . (ω − n Ωs)

(177)

Now, Mi j = Ki j, except for M22 = K22 − k2 c2/ω2 and M33 = K33 − k2 c2/ω2. Thus, we ⊥

obtain M11 = 1 −

X Π 2 e−λs X n2 In(λs) s

s , ω λs n=−∞,∞ ω − n Ωs

⊥

(178)

64 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises M12 = −M21 = −i

X s

. . 2 Π s −λs X n I′n(λs) − In(λs) e , ω ω − n Ωs n=−∞,

(179)

∞

k⊥2 c2 ω2 X Π 2 e−λs X n2 In(λs) + 2 λ 2 In(λs) − 2 λ 2 I′(λs) s s n s , − ω λ s ω − n Ωs s n=−∞,

M22 = 1 −

∞ 2 XΠ s M33 = 1 − − e− λ s 2 ω ω s

k 2⊥c2

n=−∞,

(180)

In(λs)

, ω − n Ωs

(181)

∞

and M13 = M31 = M23 = M32 = 0. 7.10 Equation (7.118) in the book is written X e2 ∫ t ∫ ∂f ∂f ! X k v ! k v ! 0s 0s ⊥ ⊥ s ⊥ ⊥ 2 k cos χ + k Jm k =i Jn ⊥ ǁ ǫ m Ω Ωs s 0 s ∂v −∞ ǁ ∂v ⊥ s n,m=−∞,∞ . . .> ′ 3 ′ exp i (n Ωs + kǁ vǁ − ω) (t − t) + (m − n) θ d v dt , (182) where χ = −Ωs (t′ − t) + θ, and ∫∞ I ∫ ∞ ∫ 3 dθ dvǁ. dv= v⊥ dv⊥ 0

Now,

−∞

(183)

dθ i (m n) θ e − = δ(m − n), 2π I dθ 1 ′ cos χ e i (m−n) θ = δ(m −n + 1) e−i Ωs (t −t) 2 2π 1 ′ + δ(m − n − 1) e i Ωs (t −t). 2

(184)

(185)

Hence, Equation (182) yields X e2 ∫ t ∫ ( ∂f "1 ′ 0s s δ(m −n + 1) e−i Ωs (t −t) k⊥ ǫ 2 0 ms −∞ ∂v ⊥ s ) # 1 ∂ f0 s i Ωs (t′−t) + δ(m − n − 1) e + kǁ ∂v δ(m − n) 2 ǁ ! ! X k⊥ v ⊥ J m k ⊥ v⊥ Jn Ω s Ωs n,m=−∞,∞ . 3 ′ ′ . exp i (n Ωs + kǁ vǁ − ω) (t – t) d v dt ,

k2 = i

which can be integrated to give " # X e 2 ∫ X ( ∂ f0 s Jn J n+1 Jn J n−1 s k k2 = + ⊥ ∂v⊥ 2 S (n − 1) 2 S (n + 1) s ǫ 0 ms n=−∞, ∞ ∂f ) 2 J 0s n d3v, +kǁ ∂vǁ S (n)

(186)

(187)

Chapter 7 □ 65 where

S (n) = n Ωs + kǁ vǁ − ω,

(188)

and the argument of the Bessel functions is as = k⊥ v⊥/Ωs. Thus, we obtain " ∂f J X e2 ∫ X 1 s k2 = k⊥ 0 s n (J n+1 + J n−1) ǫ0 m s n Ω s + k v − ω ∂v⊥ 2 ǁ ǁ s n=−∞,∞ ∂ f0 s 2 # 3 +k J d v. ǁ ∂vǁ n However,

1 2

n (Jn+1 + Jn−1) =

Thus, we obtain X ∫ J 2(k v /Ω ) X e2 s s n ⊥ ⊥ s 1+ 2 k ǫ0 ms n= −∞, ω − kǁ v ǁ − n ∞ Ωs 7.11 Given that

= 0s

(189)

Jn.

nΩ ∂f s

(190)

v⊥ ∂v⊥

∂f ! +k

ms (v⊥2 + v 2)ǁ

exp

−

(2π T /m )3/2 s s

∂vǁ

d3v = 0.

(191)

(192)

2 Ts

Equation (191) yields " # X ∫∞ X e2 n dv 2π v J 2(a ) m sv 2⊥ ⊥ ⊥ n s s s exp 0 =1− −2 T k 2 ǫ0 ms n=−∞,∞ 0 (2π Ts/ms)3/2 (Ts/ms) s ∫ ∞ s dv (n Ω + k v ) m v2 ǁ s ǁ ǁ s ǁ exp − −∞ ω − kǁ vǁ – n Ωs 2 Ts

(193)

Let vs = (2 Ts/ms)1/2, x = v⊥/vs, h = as/x = k⊥ vs/Ωs, λs = h 2/2 = k 2 v 2/(2 Ω 2), ξn = ⊥

(ω − n Ωs)/(kǁ vs), t = vǁ/vs, and Πs = (ns e s2/ǫ0 ms)∫1/2. It follows that ! dt e−t n Ω X 4Π2 X ∞ x s 2 2 dx x J n(h x) e− +t . s 0= 1/2 t − ξ kǁ vs k 2 v s2 π n −∞ s n=−∞,∞ ∫∞

However,

dx x J 2(h x) e−x = n

∫ ∞ −1/2

Z′(ξn) = −π −1/2

X Π2

k v s

−∞, 2 2s n=∞

(194)

(195)

e−t2

−∞ t − ξ n

∫ ∞

dt,

(196)

2 t e−t dt. −∞ t − ξn

Thus, we obtain 0=1+

e−λs I (λ ),

and Z(ξn) = π

" e−λs In(λs)

2 n Ωs kǁ v s

(197)

# Z(ξ ) − Z′(ξ ) . n

(198)

66 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises However,

Z′(ξn) = −2 [1 + ξn Z(ξn)].

It follows that

X 2Π2

s e−λs In(λs) k 2 v s2 n=−∞,∞

0=1+ s

Finally, because

(199)

. . 1 + ξ0 Z(ξn) .

(200)

e−λs In(λs) = 1,

(201)

n=−∞,∞

and k 2 = k 2 + k 2, we find that ⊥

⊥

7.12 We have

2 Π s2 1 + ξ e0−λs (k 2 + k 2) v 2

0=1+

X 2Π2 s s

In (λ s ) Z(ξn ) .

(202)

n=−∞,∞

1 − ξ 0 e −λ s

(k⊥ v )2 s

In(λs)

= 0.

(203)

ξn

n=−∞, ∞

However, because

e−λs In(λs) = 1,

(204)

! ω = 0, e−λs In(λs) 1 − (k⊥ vs)2 n=−∞, ω − n Ωs

(205)

n=−∞,∞

this expression can be written 1+

X 2Π2 s s

or 1−

X 2Π2

2Π

(k⊥ v )2 s

e−λs In(λs)

n= −∞, ∞

e−λs In(λs)

n Ωs ω − n Ωs

= 0,

(206)

n Ωs (ω − n Ωs + n Ωs) = 0,

(207)

! X 2 Π s2 e−λs In(λs) n Ωs 1 + n Ωs = 0. ω (k⊥ vs)2 n=−∞, ω − n Ωs

(208)

ω (k⊥ vs)2 n=−∞,

ω − n Ωs

∞

1−

1− or

∞

X s

∞

However, because I−n(λs) = In(λs), it follows that X n In(λs) = 0.

(209)

n=−∞,∞

Thus, we obtain 1−

e−λs In(λs) n Ω s = 0, ω (k⊥ vs)2 n=−∞, ω − n Ωs s

2Π2

(210)

∞

X Πs 2 e−λs X n2 In(λs) 1− = 0. ω λs n=−∞,∞ ω − n Ωs s

(211)

Chapter 7 □ 67 7.13 We have

u g(u2) 2 n e ve , F 0′ (u) = − π [v 2 + (u − V) 2] 2 [v 2 + (u + V) 2] 2

(212)

g(u2) = u4 + 2 (ve2 + V 2) u2 + ve4 − 2 ve 2 V 2 − 3 V 4.

(213)

where Note that F 0′ (0) = 0. Thus, there is an extremum of the function F0 (u) at u = 0. The roots of g(u2) = 0 lie at

u 2 = −(ve 2 + V 2) ± 2 V (V 2 + ve 2)1/2.

(214)

In order for one of these roots to be positive, we require that (v 2 + V 2)2 < 4 V 2 (V 2 + v 2),

(215)

v 4 − 2 V 2 v 2 − 3 V 4 < 0,

(216)

(v 2 − 3 V 2) (v 2 + V 2) < 0,

(217)

√

which implies that ve <

√

3 V.

(218)

Thus, for ve > 3 V there is a single extremum of the function F0(u) at u = 0. Because F0(0) > 0 and F0(u) √→ 0 as |u| → ∞, we deduce that this extremum is a maximum. On the other hand, for ve < 3 V, there are two additional extrema on either side of the extremum at u = 0. These additional extrema satisfy g(u2) = 0. Given that F0(u) ≥ 0 and F0(u) → 0 as |u| → ∞, we deduce that the additional extrema are maxima, and that the extremum at u = 0 is now a minimum. " # 1 ve 1 + F0(u) = n e , 2π ve2 + (u − V) ve2 + (u + V) 2

7.14 We have

and

F (0) = n 0

Hence, F0(u) − F0(0) = ne

∫ ∞ I=

(220)

2π ve2 + V 2

ve u2 (3 V 2 − ve 2 − u2) , π [v 2 + (u − V) 2] [v 2 + (u + V) 2] (v 2 + V 2) e

and

(219)

(221)

F (u) − F (0) 0

−∞

u2 ∫

1 = ne πe v 2 + V 2 e

du (3 V 2 − v 2 − u2) du

∞ −∞ [v

2 + (u − V) 2e] [v 2 + (u + V) 2]. e

(222)

Let x = u/ve and v = V/ve. It follows that I= where

∫ ∞

ne J , − 2 π ve 1 + v2

3 v2) dx − J= . 2 2 − [1 + (x − v) ] [1 + (x + v) ] ∞

(223)

(x2 + 1

(224)

68 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Now,

∫ ∞(

J= − ∞

) 1 1 4 v2 dx, + − 2 1 + (x − v)2 2 1 + (x − v)2 [1 + (x − v)2] [1 + (x + v2)]

which yields

J = π − 4 v2 K, ∫

where

∞

K= ∫

Now, K=

− ∞

∞

(225)

(226)

dx [1 + (x − v)2] [1 + (x + v)2]

(227)

−∞ (x − v − i) (x + v − i) (x − v + i) (x + v + i)

(228)

We can convert K into a complex integral that runs along the real axis, and is completed in the upper half-plane. The residues from the roots at z = v + i and z = −v + i are 1

R = 1

8 v i (v + i)

and R = 2

(229)

(230)

1 8 v i (v − i)

respectively. Hence, "

# 1 1 π 1 . K = 2π i + = 8 v i (v + i) 8 v i (v − i) 2 1 + v2 ! 1 − v2

It follows that

and

(231)

J = π − 4 v2 K = π 1 + v2 ,

(232)

J ne (v2 − 1) = n (V 2 − ve2) . I = − ne e = π ve 2 1 + v2 ve2 (v2 + 1)2 (V 2 + ve2) 2

(233)

7.15 The appropriate dispersion relation follows from Eq. (8.125) in the book: ∫ ∞ F0 2 ∫ ∞ e2 ∂F 0/∂u du, k2 = du = e ǫ0 me −∞ u − ω/k ǫ0 me −∞ (u − ω/k)2 where we have integrated by parts. Given that " # 1 ve 1 F0(u) = n e + , 2π ve2 + (u − V) ve2 + (u + V) 2

(234)

(235)

we obtain k2 = Π 2 e

ve 2π

"∫ ∞

2 2 + (u − V) 2] −∞ (u − ω/k) [v e

∫

∞

−∞ (u − ω/k)

du 2 [v 2 + (u + V) 2] e

(236)

Chapter 7 □ 69 where Πe = (ne e2/ǫ0 me)1/2. The previous expression can also be written "∫ ∞ ve du k2 = Π 2 e 2π −∞ (u − ω/k)2 (u − V − i ve) (u − V + i ve) # ∫ ∞ du . (237) + 2 − (u − ω/k) (u + V − i ev ) (u + V + iev ) ∞ Thus, given that ω/k lies in the upper half of the complex plane, the first integral in the square brackets has a single pole in the lower half-plane at u = V − i ve whereas the second integral has a single pole at u =–V − i ve. We can evaluate these integrals by closing them in the lower half-plane and making use of the residue theorem. The residual at the former pole is 1 R = , (238) 1 (V − i ve − ω/k)2 (−2 i ve) whereas the residual at the latter pole is 1 R = . (239) 1 (−V − i ve − ω/k)2 (−2 i ve) Hence, the first integral evaluates to 1 I = −2π i R = π , (240) 1

ve (V − i ve − ω/k)2

and the second integral evaluates to 1 I = −2π i R = π . 2 2 ve (V + i ve + ω/k)2 Thus, Equation (237) yields ve k2 = Π 2 (I + I ), 1 2 e 2π # or " 1 1 2 2 = Πe , + (k V − ζ) (k V + ζ) 2

(241)

(242)

(243)

where ζ = ω + i k ve. The previous expression can be rearranged to give ζ 4 − [Π 2 + 2 (k V)2] ζ 2 + (k V)2 [(k V)2 − Π 2] = 0. e

It follows that

(244)

. .1/2 2 ζ 2 = Π e2 + 2 (k V)2 ± Πe Π e2 + 8 (k V)2 .

(245)

In the limit that Πe ≫ k V, this reduces to

. . 2 ζ 2 ≃ Π e2 + 2 (k V)2 ± Π e2 + 4 (k V)2 .

Hence,

" ζ ≃ ±Πe 1 +

3 (k V)2 2 Πe2

(246)

# ,

(247)

or ζ ≃ ±i k V. The most unstable mode corresponds to ζ ≃ i k V, which implies that

(248)

ω = i k (V − ve),

(249)

γ ≡ −i ω ≃ k (V − ve).

(250)

70 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 7.16 Equation (8.136) in the book can be written ∫ ve ∞ due k2 = Π 2 e 2 2 + (u − U) 2 π −∞ (ue − ω/k) [v e e ∫∞ dui + Π 2 vi . i π −∞ (ui − ω/k)2 (v 2 + u 2) i

However, we saw in the previous exercise that ∫ ∞ du

2 2 −∞ u − ω/k)2 [v e + (u − V) ]

(251)

ve (V − i ve – ω/k)

(252)

Hence, we deduce that 1=

Π e2

(ω − k Ue + i k ve)2

7.17 We have X Π2

Π i2

(253)

(ω + i k vi)2 2 !

∫∞

dv⊥ 2π J 2(as) exp – v⊥ n 2 3/2 π v s ⊥ vs ǁ 0 v s2⊥ ∫ ∞ s dv [n Ω + (T /T ) k v ] v 2ǁ , ǁ s s⊥ sǁ ǁ ǁ exp − ω − kǁ vǁ – n Ωs −∞ v2

ǫ =1−

s k2

(254)

sǁ

1/2

where Πs = (ns e s/ǫ0 ms) , as = k⊥ v⊥/Ωs, vs ⊥ = (2 Ts ⊥/ms)1/2, and vs ǁ = (2 Ts ǁ/ms)1/2. Let x = v⊥/vs ⊥ and t = vǁ/vs ǁ. It follows that X 4Π2 ∫ ∞ 2 s dx x Jn2(h x) e−x ǫ =1+ 2 (k v ǁ) s 0 s ! ∫ ∞ 2 dt e−t n Ωs Ts ǁ + t , (255) 1/2 (t − ξ ) k v n −∞ π ǁ s ǁ Ts ⊥ where h = k⊥ vs ⊥/Ωs and ξn = (ω − n Ωs)/(kǁ vs ǁ). However, ∫ ∞ 1 2 dx x J 2(h x) e−x = e−λs I (λ ), n s n 2 0

(256)

where λs = h2/2 = k 2 v 2 /2 Ω 2. Furthermore, ⊥ s⊥

∫ ∞

e−t dt, −∞ t − ∫ ζ ∞ 2t 1/2 ′ − Z (ζ) = π e−ζ 2 dt. −∞ t − ζ Z(ζ) = π−1/2

It follows that X

(257) (258)

! ! k2 v2 k2 v2 ⊥ s⊥ ⊥ s ⊥ In exp − 2 (k vs ǁ)2 n= −∞, 2 Ω s2 −2 Ωs Π2

ǫ =1+ s

2 n Ωs Ts ǁ Z k ǁ vs ǁ T s ⊥

∞

! !# ω − n Ωs – Z′ ω − n Ωs . k ǁ vs ǁ k ǁ vs ǁ

(259)

Chapter 7 □ 71 Writing Z(ζ) = Zr(ζ) + i Zi(ζ),

(260)

ǫ = ǫ r + i ǫ i,

(261)

and where Zr(ζ) and ǫr are the principal parts of Z(ζ) and ǫ, respectively, we obtain ! ! X Π2 X k2 v2 k2 v2 s ⊥ s⊥ ⊥ s⊥ I ǫr = 1 + exp − n − (k vs ǁ)2 n= −∞, 2 Ω s2 2 Ω s2 s "

∞

! !# 2 n Ωs Ts ǁ ′ ω − n Ω ω − n Ω s s –Z , Zr r kǁ v s ǁ Ts ⊥ k ǁ vs ǁ k ǁ vs ǁ

and X

ǫi =

(k vs ǁ)2 n= "

respectively.

Π2

k2 v2

exp

s −∞, ∞

⊥ s⊥ 2 −2 Ωs

k2 v2

(262)

⊥ s⊥ 2 −2 Ωs

! !# 2 n Ωs Ts ǁ ω − n Ω s – Z′ ω − n Ωs , Zi i kǁ v s ǁ Ts ⊥ kǁ v s ǁ kǁ v s ǁ

(263)

CHAPTER

Chapter 8

8.1 The modified MHD equations are ∂ρ + V · ∇ρ + ρ ∇ · V = 0, ∂t ∂V ρ + ρ (V · ∇)V = −∇p + µ −1(∇ × B) × B + µ ∇2V, 0 ∂t ∂B = ∇ × (V × B) + (µ σ) −1 ∇2B, 0 ∂t ! ! ∂ p +V ·∇ = 0. ρΓ ∂t

(1) (2) (3)

(4)

Suppose that ρ = ρ0 + ρ1, p = p0 + p1, V = V1, B = B0 +B1, where all equilibrium quantities are constants. Assuming that all perturbed quantities vary as exp[ i (k · r − ω t)], we obtain −ω ρ1 + ρ0 k · V1 = 0, −ω ρ0 V1 + p1 k − µ0

−1

(k × B1) × B0 − i µ k2 V1 = 0,

ω B1 + k × (V1 × B0) + i (µ0 σ) −1 k2 B1 = 0, ! ρ1 p1 = 0. −ω p – Γ 0 ρ0

(5) (6) (7) (8)

Now, (5) gives ρ1 = (ρ0/ω) k· V1. So, (8) yields p1 = (Γ p0/ρ0) ρ1 = (Γ p0/ω) k ·V1. But, (7) gives (B0/ω) k × (V1 × b) B1 = − , (9) 1 + i k 2/(µ0 σω) whereas (6) yields −ω2 [1 + i µ k2 /(ω ρ0)] V1 + (Γ p0/ρ0) (k · V1) k = (B0 µ0/ρ0) ω (k × B1) × b.

(10)

The last two equations can be combined to give α β ω2 V1 − β VS2 (k · V1) k = V A2 (k × [k × (V1 × b)]) × b,

(11)

where V 2 = Γ p0/ρ0, V 2 = µ0 B 2/ρ0, α = 1 + i µ k 2/(ω ρ0), and β = 1 + i k 2/(µ0 σω). The S

above equation can be rearranged to give 2 [α β ω2 − (k · b)2 VA 2] V1 = (k · V1) [(β VS 2 + VA 2) k − (k · b) V A b]

74 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises – (b · V1) (k · b) VA2 k.

(12)

This expression is equivalent to the standard expression, Eq. (8.37) in the book, provided that VS2 is multiplied by a factor β, and ω2 by a factor α β. Now, for parallel propagation, k = b, and the above expression becomes 2 (α β ω2 − k 2 VA 2) V1 = (k · V1) (β VS 2 − V A ) k.

A shear-Alfvén wave is characterized by k · V1 = 0, so we obtain √ αβω k= . VA

(13)

(14)

In the limit that µ is small and σ is large, the above dispersion relation gives k = ω/VA to lowest order. Hence, µω α 1/2 ≃ 1 + i , (15) 2 ρ0 VA2 and

β 1/2 ≃ 1 + i

ω . 2 µ0 σ V A2

Thus, to first order, the dispersion relation yields ω2

ω k=

2 VA3 µ0 σ

ρ0

(16) ! .

(17)

8.2 According to Eqs. (8.41), (8.43), and (8.44) in the book, V+ and V− are the larger and smaller roots, respectively, of V 4 − V 2 (V 2 + V 2) + V 2 V 2 cos2 θ = 0. A

(18)

It follows that V 2 + V 2 = V 2 + V 2, −

V 2 V 2 = V 2 V 2 cos2 θ. +

−

(19)

(20)

Now, 2 (V+2 − VS 2 cos2 θ) (V−2 − V cos2 θ) = +V 2 − V 2 −SV 2 cos2 θ +(V 2 +−V 2) +SV 4 cos4 θ. (21) S

Making use of Equations (19) and (20), we obtain 2 2 (V+2 − VS 2 cos2 θ) (V−2 − VS 2 cos2 θ) = V cos2 θ −SV 2 cos2 θ A(V 2 +SV 2) +SV 4 cos4 θ, (22) A V S

which reduces to 2 4 (V+2 − VS 2 cos2 θ) (V−2 − V cos2 θ) = −V cos2 θ sin2 θ. S S

(23)

Hence, (assuming that θ ≠ 0 and θ ≠ π/2), we deduce that (V 2 − V 2 cos2 θ) (V 2 − V 2 cos2 θ) < 0. +

−

(24)

Because, by definition, 2 V+2 − VS 2 cos2 θ > V−2 − V cos2 θ, S

(25)

Chapter 8 □ 75 it follows that V+2 − VS 2 cos2 θ > 0, V 2 − V 2 cos2 θ < 0, –

(26) (27)

which implies that V+ > VS cos θ,

(28)

V− < VS cos θ.

(29)

8.3 Equation (8.65) in the book takes the form 1 du

u − 2

u dr

! =

−

G M⊙

(30)

mp r

However, r =

G M⊙ m p

(31)

4T 2T u2 = , c mp Hence, (30) becomes

(32)

4T rc 1 du 2 − = , 1 u − uc mp r r u dr

(33)

1 du2 u2− u 2 4 T 1 −rc , = c 2 2 u dr mp r r

(34)

1 du2 2 u2 dr

du2 dr

u2 − uc = 2

2 u c2 r

rc 1− , r

! 4 u c2 r u2 1− 2c = 1− c . u r r

(35)

(36)

The previous equation yields ∫ u2

! ! ∫ r u2 1 rc − 2 dr, 1− 2c du2 = 4 u c2 r r u

(37)

u2 − u 2 ln u2 = 4 u 2 ln r + rc + A, c c r where A is a constant. Thus, ! !2 !2 r rc u u = 4 ln + 4 + C, − ln rc r uc uc

(38)

(39)

where A = uc2 (C − ln uc 2 − 4 ln rc).

(40)

76 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Let r = 1 + x, rc !2 u = α + β x + γ x2. uc

(41) (42)

It follows that rc r ln

≃1−x+x ,

(43)

x2 , 2

(44)

≃ x−

rc ! !2 1 β2 u β γ – 2 ln ≃ ln α + x + 2 x . uc α α 2 α

(45)

Hence, (39) becomes α − ln α + β

− α

x+ γ

γ 1 β2 ! + x2 = 4 + C + 2 x2, − α 2 α2

(46)

which yields C = α − ln α − 4, β β − = 0, α γ 1 β2 γ− + = 2. α 2 α2

(47) (48) (49)

It follows that either β = 0, in which case γ = 2 α/(α − 1), or α = 1, in which case β = ±2. Hence, we either have 2 u 2x2 u2 = u 2 1 + 2 c 2 + O(x3 ) , (50) 0 u −u 0

where u 0 ≠ 1 is arbitrary, or

. . u2 = uc 2 1 ± 2 x + O(x2) .

(51)

It is clear from (36) that du2/dr can only change sign at r = rc (i.e., at x = 0). Thus, for the solution (50), if u02 > u c2 then du2/dr is negative for r just less than rc (i.e., x < 0), and positive for r just greater than rc (i.e., x > 0). It follows that du2/dr is negative for all r < rc, and positive for all r > rc. Hence, u2 attains its minimum value at r = rc. Because this value is greater than u 2, it follows that u2 > u 2 for all r. This is a Class 4 solution. c

Similarly, for the solution (50), if u 2 < u 2 then u2 < u 2 for all r. This is a Class 1 solution. 0

For the solution (51), with the plus sign, du /dr > 0 for r just less than rc (i.e., x < 0) and for r just greater than rc (i.e., x > 0). It follows that du2/dr is positive for all r. Thus, u2 is a monotonically increasing function of r with u2 ≶ u 2c as r ≶ rc. This is a Class 2 solution. Similarly, for the solution (51), with the minus sign, du2/dr is negative for all r. Thus, u2 is a monotonically decreasing function of r with u2 ≷ u 2c as r ≶ rc. This is a Class 3 solution.

Chapter 8 □ 77 8.4 Assuming a common exp(γ t) time dependence, Eqs. (8.107)–(8.110) in the book yield Φ1 = L I + M J,

(52)

Φ2 = M I + L′ J,

(53)

Ω Φ2 − R I, 2π γ Φ2 = −R′ J, γ Φ1 =

(54) (55)

respectively. Hence, we obtain γL− M

Ω 2π

Ω

′ + R I + γ M −L

! J = 0,

(56)

. γ M I + γ L ′ + R ′ J = 0.

(57)

2π

Eliminating I and J between the previous two equations gives ! Ω . γ2 L L′ − M 2 + γ L R′ + L′ R + −M + R R′ = 0. 2π Solving the quadratic equation, we get , −(L R′ + L′ R) ± (L R′ + L′ R) 2 + 4 R′ (L L′ − M 2) (M Ω/2π − R) . γ= 2 (L L′ − M 2)

(58)

(59)

8.5 We have Hence, and

B = ∇ψ × ez + Bz ez.

(60)

µ0 j = ∇ × B = ∇Bz × ez − ∇2ψ ez,

(61)

µ0∇ × j = −∇(∇2ψ) × ez − ∇2 Bz ez.

(62)

Given that ∂/∂z = 0, we can write ∇×E =

∂Ez ∂y

ex −

∂Ez ∂x

ey +

or ∇ × E = ∇Ez × ez +

∂E y ∂x

∂E y ∂x −

−

∂Ex ∂y

∂Ex

∂y

ez,

(63)

! ez.

(64)

Thus, the poloidal component of the Faraday-Maxwell equation, ∂B ∇×E = − , ∂t yields

(65)

∇Ez × ez = −∇ψ × ez, or E = z

∂ψ . − ∂t

(66) (67)

78 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises The z-component of the MHD Ohm’s law,

yields or

E + V × B = η j,

(68)

Ez + (∇φ × ez) · (∇ψ × ez) = η jz,

(69)

∂ψ η 2 + ∇φ × ∇ψ · ez = − ∇ ψ, ∂t µ0 ! ! η ∂2ψ ∂2ψ ∂ψ ∂ψ ∂φ ∂ψ ∂φ + + − = . ∂t ∂x ∂y ∂y ∂x µ0 ∂x2 ∂y2 –

(70) (71)

Multiplying both sides of the previous equation by ψ and integrating over the whole x-y plane, we obtain ! ∫ ∫ ∫ ∂ψ ∂φ ∂ψ ∂φ 1d 2 ψ dx dy + ψ dx dy 2 dt ∂x ∂y − ∂y ∂x ! 2 ∫ ∫ η ∂2ψ ∂ ψ = ψ + dx dy. (72) µ0 ∂x2 ∂y2 The second term on the left-hand side of the previous equation can be integrated by parts to give !# ! ∫ ∫ " ∂ ∂φ ∂ ∂ψ −φ ψ dx dy = 0. (73) ∂y ∂x + φ ∂x ψ ∂y Here, we have neglected any surface terms on the assumption that ψ and φ are both zero at large x and y. The term on the right-hand side can be integrated by parts to give !2 !2 ∫∫ ∫ ∫ ∂ψ η ∂ψ η 2 dx dy. (74) |∇ψ| dx dy = − − – µ0 µ0 ∂x ∂y Hence, Equation (72) yields ∫ ∫ ∫ d ψ 2 dx dy = 2 η |∇ψ|2 dx dy. − dt µ0

(75)

By inspection, the solution of this equation is such that |∇ψ| → 0 for all x and y, as t → ∞. Hence, Bp ≡ ∇ψ × ez → 0 as t → ∞. In other words, a two-dimensional poloidal magnetic field cannot be maintained against ohmic decay by dynamo action. The curl of the MHD Ohm’s law, combined with the Faraday-Maxwell equation, yields –

∂B ∂t

+ ∇ × (V × B) = η ∇ × j.

(76)

Assuming that ∇ψ = 0,

Hence,

V × B = (∇φ × ez) × Bz ez = −Bz ∇φ.

(77)

∇ × (V × B) = −∇Bz × ∇φ

(78)

Chapter 8 □ 79 Thus, the z-component of Equation (76) yields ∂Bz

η 2 − ∇Bz × ∇φ · ez = − ∇ Bz, 0µ ∂t ! ! 2 ∂B ∂B ∂φ ∂B ∂φ ∂2 B η z z ∂ Bz z z . − = + + ∂t ∂x ∂y ∂y ∂x µ0 ∂x2 ∂y2 –

(79) (80)

By analogy with Equations (71) and (75), we deduce that ∫ 2 2 η∫ ∫ d B dx dy = − |∇B |2 dx dy. z z µ0 dt

(81)

By inspection, the solution of this equation is such that |∇Bz | → 0 for all x and y, as t → ∞. Hence, Bt ≡ Bz ez → 0 as t→ ∞. In other words, a two-dimensional axial magnetic field cannot be maintained against ohmic decay by dynamo action. 8.6 In cylindrical geometry, ∂Br Aθ Bθ ∂Br Aθ ∂Br + + Az – , ∂z ∂r r ∂θ r ∂B A B θ θ r + . [(A · ∇)B] = A ∂Bθ + Aθ ∂Bθ + A [(A · ∇)B]r = Ar θ

∂r

r ∂θ

(

Hence, if

∂z

(82) (83)

for r ≤ a , for r > a

(84)

B(r, θ, z, t) = B(r) exp[ i (m θ − k z) + γ t],

(85)

(0, r Ω, U) 0 where Ω and U are constants, and V=

then (temporarily allowing Ω to vary with r) [(B · ∇)V − (V · ∇)B]r = −i (m Ω − k U) Br, dΩ [(B · ∇)V − (V · ∇)B]θ = r Br − i (m Ω − k U) Bθ. dr

(86) (87)

In cylindrical geometry, (∇2A) r = ∇2 A r –

2 ∂Aθ

−

, r2 ∂θ r2 (∇2A) = ∇2 A + 2 ∂Ar − Aθ , where 2

∇ f=

∂2 f

1 ∂f

∂r2

r ∂r

Hence, 2

d2 B r

(∇ B)r = dr2 2

1 dBr

r2 ∂θ +

1 ∂2 f

∂2 f

r2 ∂θ2 + ∂z2 .

(m2 + k2 r2 + 1) Br

−

(89)

− r r2 dr 1 dBθ (m2 + k2 r2 + 1) Bθ

d Bθ 2 (∇ B)θ = dr2 + r

(88)

(90) i 2 m Bθ

− +

(91)

r2 i 2 m Br r2

(92)

80 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises It follows that the r-component of ∂B η = (B · ∇) V − (V · ∇) B + ∇2 B ∂t µ0

(93)

is γ Br = −i (m" Ω − k U) Br

# d Br 1 dBr (m + k r + 1) Br i 2 m Bθ , + − − dr 2 r dr r2 r2 2

η µ0

2 2

whereas the θ-component is dΩ B – i (m Ω − k U) Bθ γ Bθ = r dr " r # 2 2 2 2 η d Bθ 1 dBθ (m + k r + 1) Bθ i 2 m Br . + + − + µ0 dr 2 r dr r2 r2

(94)

(95)

8.7 We have γ Br = −i (m Ω − k U) Br " # η d2Br 1 dBr (m2 + k2r 2 + 1) Br i 2 m Bθ , + + − − r dr r2 r2 µ0 dr 2

(96)

and γ Bθ = −i (m" Ω − k U) Bθ

# 2 2 2 2 η d Bθ 1 dBθ (m + k r + 1) Bθ i 2 m Br , + + − + µ0 dr 2 r dr r2 r2

(97)

as well as B± = Br ± i Bθ, r y= , a µ0 a 2 , τR = η

(98) (99) (100)

q2 = k2 a2 + γ τR + i (m Ω − k U) τR, 2

(101)

s = k a + γ τR.

(102)

It follows that 2

q Br = 2

q Bθ =

d2 B r dy2 d2Bθ dy2

+ +

1 dBr

(m2 + 1) Br

− y y2 dy 2 1 dBθ (m + 1) Bθ y

−

i 2 m Bθ −

y2 i 2 m Br

(103)

(104)

(105)

Forming (103) +i (104), we obtain 2

q B+ =

d2 B + dy2

1 dB+ y

(m2 + 1) B+ −

2 m B+ −

Chapter 8 □ 81 Forming (103) −i (104), we obtain d2B−

q B− = dy2 + y

dy The previous two equations can be written ± 2

d2B dy

(m2 + 1) B−

1 dB−

+y 2

−

2 m B− +

. dB± . 2 + q2 y2 B± = 0. (m ± 1) − dy

(106)

(107)

Of course, when y > 1 then Ω = U = 0. Therefore, q → s, and the previous equation becomes . . d2B dB± 2 2 y±2 +y (m ± 1) + s y2 B± = 0. (108) − 2 dy dy 8.8 We have B±(y) = C±

Im±1(q y) Im±1(q)

(109)

B (y) = C

Km±1(s y)

(110)

for y < 1, and ±

Furthermore,

" dB± #y=1+ dy

It follows that where

Km±1(s)

= ±i Ω τR

y=1−

B+ + B− 2

Ω τR −G± C± = ±i (C+ + C−), 2 I ′ (q) K ′ (s) G = q m±1 − s m±1 . ±

Im±1(q)

(111)

(112) (113)

Km±1(s)

Hence, Ω τR G− (C+ + C−), −G+ G− C+ = i 2 Ω τR G + (C + + C −). −G+ G− C− = 2 − i

(114) (115)

Adding the previous two equations, we obtain G+ G− = 8.9 We have

i Ω τR (G+ − G− ). 2

" !# 2 1 π −z 1 + 4 m − 1+ O e Km(z) = , z5/2 2 z1/2 3/2 8 z " !# 1 1 1 4 m2 − 1+ +z Im(z) = √ e O z5/2 . − 2π 8 z3/2 z1/2

(116)

(117) (118)

82 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises Hence,

! !−1 ! 4 m2 1 1 2 z dKm − 1 + 4m − 1 +O 1 , = −z − 1 + 3 Km dz 2 8z 8z z2 ! 2 ! ! −1 4m 1 1 1 , z dIm 4 m2 − 1 − + O 1 − 3 1 − =z− Im dz 2 8z 8z z2

which reduces to

! ! 4 m2 1 1 − +O 1 , = −z − 1 + Km dz 2 4z z2 ! 2 ! 1 4m 1 z dIm − +O 1 . 1− =z− Im dz 2 4z z2 z dKm

Now, G± =

dIm±1

Im±1

(120)

(121) (122)

# z

dKm±1

Km±1

– z=q

(119)

(123)

z=s

which reduces to

! 4 (m ± 1)2 − 1 + 1 1 + 4 (m ± 1)2 − 1 1 G ±= q + s − 1− ! 4s 2 2 ! 4q +O

1 q2

1 s2

(124)

! ! 2 1 3 1 1 1 m . G =q+s+ ±m+ + O( + + ± 2 8 q s 2 2 q s

(125)

G+ G− ≃ (q + s) 2,

(126)

To lowest order, and

G + − G− ≃ 2 m

1 +

2 m (q + s) =

(127)

Hence, the dispersion relation G G = +

−

Ω τ (G − G )

−

(128)

reduces to (q + s) q s ≃ i m Ω τR.

(129)

x = k2 a2 + γ τR.

(130)

q2 = x + i µ,

(131)

s2 = x.

(132)

8.10 Let It follows that

Thus, the dispersion relation

(q + s) q s ≃ i m Ω τR

(133)

Chapter 8 □ 83 becomes

. . (x + i µ)1/2 + x1/2 (x + i µ)1/2 x1/2 = i m Ω τR.

However, if µ is small then (x + i µ)

1/2

≃x

1/2

! iµ . 1+ x

(134)

(135)

Hence, the dispersion relation becomes ! 2x

3/2

! iµ 1+ ≃ i m Ω τR, 2x

iµ 1+ 4x

or x3/2

(136)

! i m Ω τR 3iµ 1+ ≃ . 4x 2

(137)

Let x = x 0 + x 1, where x1/x0 ∼ µ/x0. It follows that x3/2 0

3 x1 1+ 2 x0

(138)

! 3iµ = i m Ω τR, 1+ 4 x0

(139)

Hence,

3 x1 2 x0

x3/2 = i m Ω τR, 0 ! 3iµ = 1, 1+ 4 x0

(140) (141)

which implies that m Ω τ !2/3 x0

= e i π/3

x =

iµ . − 2

x ≃ x0 + x1

= e i π/3

γ τR ≃ e i π/3

(142) (143)

Thus,

giving

m Ω τR 2

!2/3 –

iµ 2

! iµ m Ω τR 2/3 – k2 a2 − . 2 2

(144)

(145)

8.11 The MHD equations take the form dρ

V = 0,

(146)

+ ∇p − j × B = 0,

(147)

E + V × B = 0,

(148)

dt dV ρ

+ρ

∇·

84 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises ! d p dt ρ Γ

= 0,

(149)

Maxwell’s equations are written ∇ · B = 0,

(150)

∂B ∇×E = − , ∂t ∇ × B = µ0 j,

(151) (152)

where we have neglected the displacement current. Equation (150) is identical to Eq. (8.169) in the book. Taking the curl of Equation (148), and making use of Equation (151), we obtain ∂B ∂t

− ∇ × (V × B) = 0,

(153)

which is identical to Eq. (8.170) in the book. Equation (146) can be written

However, so we obtain

∂ρ + V · ∇ρ + ρ ∇ · V = 0. ∂t

(154)

∇ · (ρ V) = V · ∇ρ + ρ ∇ · V,

(155)

∂ρ

+ (ρ V) = 0, ∇· ∂t which is identical to Eq. (8.171) in the book. Equation (8.172) in the book can be written ∂Ti j ∂ (ρ V ) + = 0, i ∂t ∂x j where

! Bi B j Bk Bk Ti j = ρ Vi V j + p + δi j – . 2 µ0 µ0

(156)

(157)

(158)

It follows that ∂Vi ∂ρ ∂Vi ∂ρ ∂p +ρ V + ρ V ∂Vj Vi + ρ + Vi V j + j i ∂t ∂t ∂x j ∂x j ∂x j ∂xi B j ∂B j ∂Bi B j Bi ∂B j + − − =0 µ0 ∂xi ∂x j µ0 µ0 ∂x j

(159)

However, Equation (150) is equivalent to ∂B j

= 0.

(160)

∂x j Moreover, Equation (146) can be written ∂ρ ∂ρ + ρ ∂Vj = 0, +Vj ∂xj ∂t ∂x j

(161)

Chapter 8 □ 85 which implies that

∂Vj ∂ρ ∂ρ = 0. + ρ V Vi + Vi V j i ∂x j ∂t ∂x j

(162)

Equations (159), (160), and (162) can be combined to give ∂Vi ∂p B j ∂B j ∂Bi B j ∂Vi ++ + − = 0, ρ + ρV j ∂t ∂x j ∂xi µ0 ∂xi ∂x j µ0

(163)

which is equivalent to ρ

∂V ∂t

However,

d dt

and

1 2 ∇(B /2) − (B · ∇) B = 0. µ0 µ0 1

+ ρ (V · ∇) V + ∇p +

≡

∂

+ V · ∇,

∂t

(165)

(164)

∇(B /2) − (B · ∇) B = 0 B × (∇ × B) = −j × B µ0 µ0 µ where use has been made of ∇ · B = 0. Hence, Equation (164) yields dV ρ

+ ∇p − j × B = 0,

(166)

(167)

which is identical to Equation (147). Hence, we deduce that Eq. (8.172) in the book is equivalent to the MHD equation of motion, (147). Equation (8.173) in the book can be written ∂U

∂t where U=

1 2

and u= i

= 0,

(168)

∂xi

ρV V + j

ρV V V + j

∂ui

Bj Bj + , Γ − 1 2 µ0 p

pV +

Γ−1

B j B j Vi µ0

−

(169) B j V j Bi

(170)

µ0

Hence, 1 ∂ρ V V + ρ V ∂Vj 1 ∂p B j ∂B j 1 ∂ρ V V V + + + j j i j j j 2 ∂t ∂t µ0 ∂t 2 ∂xi Γ − 1 ∂t + ρ V V ∂Vj + 1 ρ V V ∂Vi + Γ V ∂p + Γ p ∂Vi j j i j i ∂xi ∂xi 2 Γ − 1 ∂xi Γ − 1 ∂xi 2 Vi B j ∂B j B j B j ∂Vi V j Bi ∂B j Bi B j ∂Vj B j V j ∂Bi + + − − − = 0. µ0 ∂xi µ0 ∂xi µ0 ∂xi µ0 ∂xi µ0 ∂xi Making use of Equations (160) and (161), the previous equation reduces to ∂Vj ρ Vj

∂t

∂p

Γ − 1 ∂t

B j ∂B j µ0 ∂t

(171)

86 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises + ρ V V ∂Vj + Γ V ∂p + Γ p ∂Vi i j ∂xi Γ − 1 i ∂xi Γ − 1 ∂xi +

2 Vi B j ∂B j B j B j ∂Vi V j Bi ∂B j Bi B j ∂Vj + − − = 0. µ0 ∂xi µ0 ∂xi µ0 ∂xi µ0 ∂xi

(172)

Equations (150) and (153) yield ∂B + (∇ · V) B − (B · ∇) V + (V · ∇) B = 0, ∂t

(173)

∂B j ∂Vi ∂B j ∂Vj = 0, +Bj + Vi − Bi ∂t ∂xi ∂xi ∂xi

(174)

which implies that B

∂B j

∂Vj ∂B j + B B ∂Vi – B B +V B = 0, j j i j i j j ∂xi ∂xi ∂xi ∂t

(175)

Equations (172) and (175) can be combined to give ∂Vj

∂Vj + ρ Vi V j ∂xi ∂t Γ − 1 ∂t V j Bi ∂B j ∂Vi Vi B j ∂B j Γ Γ p + − = 0. + V ∂p + i ∂x µ ∂x µ ∂x i 0 i 0 i Γ − 1 ∂xi Γ − 1

ρ Vj

∂p

(176)

Equation (163) yields ∂Vj ρ Vj

∂t

∂p

∂Vj + ρ Vi V j

∂xi

+ Vi

∂xi

−

V j Bi ∂Bj µ0

∂xi

Vi B j ∂B j µ0

∂xi

= 0.

(177)

The previous two equations can be combined to give 1

∂p

Γ − 1 ∂t

1 Γ−1

∂p ∂xi

∂Vi p = 0. Γ − 1 ∂xi Γ

The previous equation can be combined with Equation (161) to give ! ! ∂p ∂p ∂ρ ∂ρ = 0, ρ + Vi −Γ p + Vi ∂t ∂x i ∂t ∂x i or

dp ρ

−Γ p

(178)

(179)

dρ

= 0, (180) dt dt which is equivalent to Equation (149). Hence, we deduce that Eq. (8.173) in the book is equivalent to the MHD equation of state, (149). 8.12 For a parallel MHD shock, Vy = By = 0. Hence, the MHD Rankine-Hugoniot relations, (8.180)–(8.185) in the book, reduce to [Bx]21 = 0,

(181)

[ρ Vx]21 = 0, [ρ V + p]2 = 0,

(182) (183)

Chapter 8 □ 87 "

ρV3 + x

#2 Γ p Vx = 0. Γ−1 1

(184)

In region 1, Bx = B1, ρ = ρ1, p = p1, and Vx = V1. Likewise, in region 2, Bx = B2, ρ = ρ2, p = p2, and Vx = V2. Hence, we obtain B1 = B2,

(185)

ρ1 V1 = ρ2 V2, ρ1 V + p1 = ρ2 V 2 + p2, 1 2 1 Γ 1 Γ 3 ρ V + p V = ρ V3 + p V. 1 1 2 2 1 1 2 2 2 2 Γ−1 Γ−1

(186) (187)

Equation (185) yields

= 1.

(188)

(189)

B1 If we define r= then Equation (186) gives

ρ2 (190)

ρ1

= r −1.

(191)

V1 It we define R=

p2 (192)

then Equation (187) yields ρ1 V 2 + p1 = r −1 ρ1 V 2 + R p1,

(193)

Γ M 2 + 1 = r −1 Γ M 2 + R,

(194)

where M1 = V1/VS 1, and VS 1 = (Γ p1/ρ1)1/2. The previous equation can be rearranged to give R = 1 + Γ M 12 (1 − r −1). (195) Equation (188) yields

−2

R r −1

, M + = r Γ −1 2 1 Γ−1 2 which can be rearranged to give ! Γ−1 R=r+ M12 (r − r −1). 2 M1 +

(196)

(197)

Finally, eliminating R between Equations (195) and (197), we obtain r=

(Γ + 1) M 12 2 + (Γ − 1) M12

(198)

88 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises 8.13 The downstream Mach number is ρ !1/2 2

M2 = V2

= r −1 V1

Γ p2

r ρ1

!1/2

Γ R p2

(r R)1/2

(199)

It follows from Eq. (8.191) in the book that r R = r + Γ M 12 (r − 1).

(200)

Making use of Eq. (8.190) in the book, we get (Γ − 1) + 2 Γ (M 2 − 1)

1/2

r R = M12

2 + (Γ − 1) M12

(201)

Equations (199) and (201) can be combined to give 2 + (Γ − 1) M 2

1/2

M2 =

2 Γ M12 − (Γ − 1)

(202)

The condition for M2 < 1 is 2 + (Γ − 1) M 2 < 2 Γ M 2 − (Γ − 1),

(203)

M12 > 1.

(204)

which reduces to Thus, if M1 > 1 then M2 < 1. 8.14 For a perpendicular MHD shock, Vy = Bx = 0. Hence, the MHD Rankine-Hugoniot relations, (8.180)–(8.185) in the book, reduce to [Vx By]21 = 0,

(205)

[ρ Vx]21 = 0, By2 2 = 0,

(206)

ρ Vx + p + 1

2µ 2 2 By Vx

ρ Vx +

(207)

0 1

Γ−1

p Vx +

µ0

= 0.

(208)

In region 1, By = B1, ρ = ρ1, p = p1, and Vx = V1. Likewise, in region 2, By = B2, ρ = ρ2, p = p2, and Vx = V2. Hence, we obtain V1 B1 = V2 B2,

(209)

ρ1 V1 = ρ2 V2,

(210)

B12

ρ1 V 2 + p1 + 1

1 2

ρ1 V13 +

Γ−1

p1 V 1 +

2 µ0 B 2 V1 1

µ0

= ρ2 V 2 + p 2 +

B22

, 2 µ0 1 B 2 V2 Γ = p 2 V2 + 2 . ρ2 V23 + Γ−1 2 µ0 2

(211) (212)

Chapter 8 □ 89 If we define

ρ2

(213)

ρ1 then Equation (210) gives

= r −1,

(214)

V1 whereas Equation (209) yields

B1 If we define

= r.

(215)

V2 p2

(216)

p1 then Equation (211) yields Γ M 2 + 1 + β −1 = r −1 Γ M 2 + R + r2 β −1, 1

(217)

where M1 = V1/VS 1, VS 1 = (Γ p1/ρ1)1/2, and β1 = 2 µ0 p1/B 21. The previous equation can be rearranged to give R = 1 + Γ M 2 (1 − r −1) + β −1 (1 − r2). (218) 1

Equation (212) yields 1 Γ 1 Γ Γ M2 + + 2 β −1 = Γ M 2 r −2 + r −1 R + 2 β −1 r, 1 1 1 1 2 2 Γ−1 Γ−1

(219)

which can be rearranged to give r −1 R = 1 + 2

! 1 2 Γ − 1 2 − M (Γ − 1) (1 − r ) + β −1 (1 − r). Γ 2 1 1

(220)

Eliminating R between Equations (218) and (220) yields 2

Γ M1 2

! Γ − 1 M (Γ − 1) (r + 1) − β −1 r2, Γ 2 1 1

− β −1 r (r + 1) = r + 1

which can be rearranged to give . . 2 2 (2 − Γ) r 2 + Γ 2 (1 + β1) + (Γ − 1) β1 M 1 r − Γ (Γ + 1) β1 M1 = 0.

(221)

(222)

8.15 In Eq. (8.204) in the book, if we exchange the identities of regions 1 and 2, so that M1 → M2 and β1 → β2, then r → r −1. Thus, we obtain . . 2 (2 − Γ) r −2 + Γ 2 (1 + β2) + (Γ − 1) β2 M2 2 r −1 − Γ (Γ + 1) β2 M2 2 = 0, (223) which can be rearranged to give . . 2 2 2 F(r) = 2 (2 − Γ) + Γ 2 (1 + β2) + (Γ − 1) β2 M 2 r − Γ (Γ + 1) β2 M2 r = 0.

(224)

If r1 and r2 are the two roots of the previous quadratic relation then it is clear that 2 (2 − Γ) r r =− < 0. (225) + M2 1 2 Γ (Γ 1) β2 2 Thus, only one of these roots is positive. We require the positive root to be such that r > 1.

90 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises otherwise the second law of thermodynamics is violated. However, F(0) > 0. Thus, we need F(1) > 0, or . . 2 (2 − Γ) + Γ 2 (1 + β2) + (Γ − 1) β2 M2 2 – Γ (Γ + 1) β2 M22 > 0, (226) which can be rearranged to give

M2 < 1 +

. (227) Γ β2 However, M2 = V2/VS 2 and 2 V 2 /(Γ β2) = V 2 , so the previous equation is equivalent to 2

V2 < V+ 2,

(228)

, where V+ 2 =

VS22 + V A22 .

8.16 Our basic equations are Vx 1 By 1 − Vy 1 Bx 1 = 0,

(229)

Vx 2 By 2 − Vy 2 Bx 2 = 0,

(230)

B x 1 = B x 2,

(231)

ρ1 Vx 1 = ρ2 Vx 2,

(232)

ρ1 V x21 + p 1 + ρ1 Vx 1 Vy 1 − 1

ρ1

= ρ2 V x22 + p 2 +

2 µ0 Bx 1 By 1

= ρ2 Vx 2 Vy 2 −

µ0

Vx 1 + Vy 1

By21

Vx 1 +

p1 V x 1 =

Γ −1 2 It follows from Equation (231) that

1 2

Bx 2

B y22

, 2 µ0 Bx 2 By 2

µ0

(234) Γ

Vx 2 + Vy 2

ρ2

(233)

Vx 2 +

= 1,

Γ−1

p1 Vx 2.

(235) (236)

Bx 2 and from Equation (232) that

Vx 2

ρ1

ρ2

Vx 1

(237)

(238)

Equations (229) and (230) yield Vy 1 = and V

= y1

Hence,

Vx 2 By 2

Vx 1 By 1 Bx 1 = r −1

Vx 1 By 2

Vy 2 Vy 1

= r −1

(239)

Bx 1

Bx 2 By 2

(240)

By 1

Equation (234) gives 2

By 1 V x 1 −

B2 x1 µ0 ρ 1

−1

= By 2 r

Vx 1 −

B 2x 1 µ0 ρ1

(241)

Chapter 8 □ 91 √ Let v1 = Vx 1, Bx 1 = cos θ1 B1, and VA 1 = B1/ µ0 ρ1. It follows that = r v 22 − cos2 θ2 V 2 B By 2 v1 − r cos θ1 VA 12 . y1

Vy 2

v 2 − cos2 θ V 2

(242)

Hence, Equation (240) yields

Vy 1

v − r cos2 θ1 1 2

VA21

(243)

Now, By 1 = sin θ1 B1, so Equation (233) gives 2 y 2!2 B 21 sin2 θ , ρ v 2 + p + B1 sin2 θ = r −1ρ v 2 + p B 1 1 1 1 1 2 1 1 By 1 2 µ0 2 µ0

p2 p1

=1+ V

(1 − r −1 ) v 2 +

=1+

1−

VS21 r

Γ v1 (r − 1)

2 y1

S 12 2

1 −B 2 y 2 V 2 B

which yields

A 1 sin

(244)

(245) θ1 ,

r VA 1 sin θ1 [(r + 1) v1 − 2 r VA 1 cos θ1] 2

2 (v12 − r VA21 cos2 θ 1) 2

(246)

Finally, Equation (235) gives v2 +

2 VS21 cos 2θ1 = r

Γ−1

−2

cos2 θ +

V y22

2 sin y1

θ1 v1 + r

2 2 −1 p2 2 VS 1 cos θ1

Γ−1

(247)

which reduces to 0 = (v 2 − r cos2 θ1 V 2 )2 [(Γ + 1) − (Γ − 1) r] v 2 − 2 r V 2 1

A1 2

1 2

(248)

S1 2

– r sin θ1 v1 VA 1 [Γ + (2 − Γ) r] v1 − [(Γ + 1) − (Γ − 1) r] rcos θ1 VA 1 .

CHAPTER

Chapter 9

9.1 Equation (9.9) in the book yields ω ≡ ∇ × V = ∇ × (∇φ × ez) = ∇ × ∇ × (φ ez) = −U ez,

(1)

U = ∇2φ.

(2)

where

Here, use has been made of ∂/∂z = 0. Equations (9.8) and (9.10) in the book give µ0 j ≡ ∇ × B = ∇ × (∇ψ × ez) = ∇ × ∇ × (ψ ez) = −J ez,

(3)

J = ∇2ψ.

(4)

where Equation (9.4) in the book can be written

! ∂V 1 ρ + ∇V2 + ω × V + ∇p = j × B. ∂t 2

Taking the curl of this equation, we obtain " # ∂ω ρ + (V · ∇) ω − (ω · ∇) V = (B · ∇) j − (j · ∇) B, ∂t

(5)

(6)

where we have made use of ∇ · ω = ∇ · V = ∇ · B = ∇ · j = 0. Taking the z-component of the previous equation, we get ! ∂ωz ρ + V · ∇ω z = B · ∇ j z , (7) ∂t because Vz = Bz = 0. But, V · ∇ωz = ∇φ × ez · ∇ωz = [ωz, φ] = [φ, U]. Likewise, B · ∇ jz = ∇ψ × ez · ∇ jz = [ jz, ψ] = −[J, ψ]/µ0. Hence, the previous equation yields ∂U ρ = ρ [φ, U] + µ−1 [J, ψ]. (8) 0 ∂t Equations (9.7) and (9.10) in the book yield

∇ × E = −∇ × (ψ ez), or

∇ × (E + ψ ez) = 0,

(9) (10) 93

94 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises which implies that

E = −ψ ez − ∇Φ.

(11)

The z-component of the previous equation yields

Ez = −ψ,

(12)

E + (∇φ × ez) × (∇ψ × ez) = η j,

(13)

E + [φ, ψ] ez = η j.

(14)

because ∂/∂z = 0. Equation (9.5) in the book gives

or Taking the z-component of the previous equation, we get ∂ψ

= [φ, ψ] +

∂t

(15)

µ0

We can obtain a J0 term by assuming that ∇Φ has a uniform z-component that drives the equilibrium current. 9.2

(a) We have

( F(x̂) =

F′(0) xˆ F′(0) sgn(xˆ)

|xˆ| < 1 , |xˆ| ≥ 1

(16)

as well as

d2ψ ˆ 2 ψ F′′ ψ = 0, (17) −F − k dx̂ 2 subject to the constraints ψ( –xˆ) = ψ(xˆ), and ψ(xˆ) → 0 as |xˆ| → ∞ . Given that ψ(xˆ) is even in xˆ, we need only solve (17) in the region xˆ ≥ 0. Now, F′′ = 0 everywhere, except at xˆ = 1. Hence, Equation (17) reduces to d2ψ

ˆ 2 ψ = 0. − k 2

(18)

ψ(xˆ) = C exp[−kˆ (xˆ − 1)]

(20)

dx̂ The most general solution of the previous equation that satisfies the boundary condition at infinity is sinh(kˆ xˆ) cosh(kˆ xˆ) (19) ψ(xˆ) = A +B sinh(k̂) cosh(k̂) in the region 0 ≤ xˆ < 1, and

in the region xˆ > 1. The first matching condition at xˆ = 1 is the continuity of ψ (because the normal component of a magnetic field cannot be discontinuous across an interface). This condition yields A + B = C. (21) The second matching condition is obtained by integrating Equation (17) across xˆ = 1: "

# ′ x̂=1+ dψ xˆ=1+ = [F ] x̂=1− ψ(1). dx̂ x̂=1 F(1) −

(22)

Chapter 9 □ 95 However,

+ [F ′ ] x̂=1 x̂=1−

so we obtain

F(1) #xˆ=1

(23)

+ dψ = −ψ(1), dx̂ x̂=1 −

which yields

= −1,

−kˆ C − kˆ A tanh(k̂) +

(24) #

= −C.

tanh(k̂)

(25)

Combining this expression with Equation (21), we get B 1 − k̂ − k̂ tanh(k̂) = . A k̂ + k̂/ tanh(k̂) − 1

(26)

Now, because ψ(xˆ) is an even function of xˆ, " #xˆ=0+ . ∆′ = 1 dψ = 2 dψ . ψ dx̂ x̂=0 ψ(x̂) dx̂ .

(27)

x̂=0

−

Thus, it follows from Equation (19) that ∆′ = Hence, we get ∆ = (b) In the small-kˆ limit,

2kˆ

′

2 kˆ

tanh(k̂) A

(28)

# k̂ + k̂ tanh(k̂) − 1

tanh(k̂) 1 − k̂/ tanh(k̂) −

k̂ ˆ 3 ˆ ˆ k ˆ5 tanh(k) = k − + O(k ). 3

(29)

(30)

Hence. ∆′ =

2 kˆ

kˆ + kˆ 2 − 1 2

kˆ (1 − kˆ /3) 1 − (1 − k̂

or ∆′ =

1−

kˆ

1 /3)−

2 1 − kˆ + O(k̂), = − k̂ k̂ 1 + k̂/3

! 2 8 k̂ + O(k̂) = − + O(k̂). 3 kˆ 3 4

(31)

(32)

In the large-kˆ limit, tanh(k̂) ≃ 1 − 2 exp(−2 k̂). Hence, ∆′ = or

2 k̂

k̂ + k̂ [1 − 2 exp(−2 k̂)] − 1

, 1 − 2 exp(−2 k̂) 1 − k̂ [1 − exp(−2 k̂)]−1 − k̂

. . ′ ˆ ˆ 2 k̂ − 1 − 2 k̂ exp(−2 k̂) , ∆ = 2 k 1 + 2 exp(−2 k) 1 − 2 k̂ − 2 k̂ exp(−2 k̂)

(33) (34)

(35)

96 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises or

. ∆′ = − ˆ . ˆ 1 − [2 k̂/(2 k̂ − 1)] exp(−2 k̂) , 2 k 1 + 2 exp(−2 k) 1 + [2 k̂/(2 k̂ − 1)] exp(−2 k̂)

(36)

. . ∆′ = −2 k̂ 1 + 2 exp(−2 k̂) − [4 k̂/(2 k̂ − 1)] exp(−2 k̂) " # 2 = −2 kˆ 1 – (37) exp( −2 k̂) , ˆ 2k−1 " or !# 1 1 ′ ˆ ∆ = −2 k + 2 1 + (38) + O ˆ 2 exp(−2 k̂). k 2 kˆ (c) Now, expression (29) for ∆′ can only change sign if the numerator or the denominator of the term in square brackets passes through zero. However, the denominator is only zero when kˆ = 0. Hence, the expression can only change sign at finite kˆ if the numerator passes through zero: i.e., at k = kc, where k̂c [1 + tanh(k̂c )] = 1.

(39)

It is easily demonstrated graphically that the previous transcendental equation only has a single root at positive k̂. Given the ∆′ > 0 for small k̂, and ∆′ < 0 for large k̂, it follows that ∆′ > 0 for k̂ < k̂c . The previous equation can be solved iteratively to give k̂c = 0.639. 9.3

(a) Recalling the analysis of Exercise 9.1, we now have ! ∂V 1 ρ + ∇V2 + ω × V + ∇p = j × B + µ ∇2V. ∂t 2 Taking the curl of this equation, we get " # ∂ω ρ + (V · ∇) ω − (ω · ∇) V = (B · ∇) j + (j · ∇) B + µ ∇2ω. ∂t Taking the z-component of the previous equation, we get ! ∂ωz ρ + V · ∇ωz = B · ∇ j z + µ ∇2ω z. ∂t

(40)

(41)

(42)

But, ωz = −U, V · ∇ωz = [φ, U], and B · ∇ jz = −[J, ψ]/µ0. Thus, we obtain ∂U = ρ [φ, U] + µ−01 [J, ψ] + µ ∇2U. (43) ∂t (b) The only term that differs between our equations and the reduced-MHD equations is the term involving µ in the vorticity equation. This yields an additional term in the linearized equations of the form !2 d2 2 − k µ ∇2U 1 = µ ∇2 ∇2φ 1 = µ φ1. (44) dx2 ρ

Thus, we obtain γ ψ1 = i k B0 F φ1 +

! 2 − ψ1, k 2

µ0 dx

(45)

! ! !2 d2F/dx2 d2 d2 d2 2 1 2 2 − γ ρ 2 − k φ1 = i µ 0 k B0 F 2 – k − − k ψ1 + µ φ1. (46) dx dx2 F dx

Chapter 9 □ 97 (c) Let x = a x̂, k = k̂/a, γ = γ̂/τH , ψ1 = −a B0 ψ̂, φ1 = i (γ a/k) φ̂, S = τR /τH , and P = τR/τM. Here, τH = k−1/(B2/µ ρ)1/2, τR = µ0 a2/η, and τM = ρ a2/µ. Direct 0 0 substitution into the previous two equations yields ! d2 2 – k̂ S γ̂ ψ̂ − F φ̂ = ψ̂, (47) dx̂2 ! ! !2 2 ′′ 2 F d d2 2 d 2 2 2 −1 P ˆ γ̂ k̂ ψ̂ + γ̂ S – k̂ φ = −F – k̂ − φˆ. (48) dx̂2 − F dx̂2 dx̂2 (d) Within the resistive layer, we can assume that d/dx̂ ≫ k̂. Moreover, we can write F(x̂) = x̂. The previous two equations yield 2

d ψ̂ ˆ ˆ S γ̂ (ψ − x̂ φ) ≃ , dx̂ 2 2 2 d ψ̂ d 4 φ̂ 2 d φ̂ γ̂ ≃ −x̂ 2 + γ̂ S−1P 4 . 2 dx̂ dx̂ dx̂

(49) (50)

(e) Fourier transformation is equivalent to the substitutions d dx̂

→ iS

xˆ →

1/3

(51)

. S 1/3 dp

(52)

Hence, Equations (49) and (50) yield Q ψ̂

Q dφ̂ i = p2 ψ̂, – S 1/3 dp − d 2 1/3 ˆ −Q 2 p2 φˆ = p i S ψ + Q P p 4 φˆ, dp

(53) (54)

where Q = γ̂ S 1/3 . It follows from Equation (53) that i S 1/3 ψˆ =

Q dφˆ . − Q + p2 dp

(55)

Substitution into Equation (54) yields ! d p2 dφˆ – (Q p2 + P p4) φˆ = 0. dp Q + p2 dp

(56)

(f) Suppose that we neglect the final term on the left-hand side of Equation (56) to give ! d p2 dφˆ – (Q p2 + P p4) φˆ ≃ 0, (57) dp Q + p 2 dp which can be solved to give φˆ = B –

Q p

! + p + C + O(p2).

(58)

The characteristic variation lengthscale of this solution is p ∼ Q 1/2. When p ∼ Q 1/2

98 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises the first term on the left-hand side of (56) is O(Q −1), the second term is O(Q 2), and the third term is O(P Q 2). Thus, the neglect of the latter two terms is valid provided Q ≪ 1, P −1/3.

(59)

When p ≫ Q 1/2, Equation (54) reduces to d 2 φ̂

4 ˆ + P p ) φ = 0. − (Q p 2

dt2 Suppose that we neglect the third term on the left-hand side. We obtain d 2 φ̂ dt2

− Q p2 φˆ = 0.

(60)

(61)

√ The solution to this equation that is well behaved as → p ∞is U(0, 2 Q 1/4 p), where U(a, x) is a parabolic cylinder function. The small argument asymptotic expansion of this function yields " # Γ(3/4) 1/4 2 1 2 Q p + (p ) φ̂(p) = A – . (62) O Γ(1/4) Matching to the solution (58) reveals that " # Γ(3/4) Q 5/4 φˆ = A 2 + 1 + O(p) . Γ(1/4) p

(63)

when p ≪ Q 1/2. Finally, comparison with Eqs. (9.47) and (9.49) in the book yields ∆ = 2π

Γ(3/4)

S 1/3 Q 5/4.

(64)

Γ(1/4) The characteristic lengthscale of the solution (62) is p ∼ Q −1/4. Thus, the neglect of the final term on the left-hand side of Equation (60) is valid provided that Q ≫ P Q −1/2, or Q ≫ P 2/3. (65) Combining the inequalities (59) and (65), the criterion for the validity of Equation (64) is 1 ≫ Q ≫ P 2/3. (66) Suppose that we neglect the second term on the left-hand side of Equation (60). We obtain d 2 φ̂ (67) − P p4 φˆ = 0. dp2 The solution to this equation that is well behaved as p → ∞ is φˆ = p1/2 K1/6 P 1/2 p3/3 ,

(68)

where Kν(x) is a modified Bessel function. The small argument asymptotic expansion of this function yields " # 2/3 Γ(5/6) 1/6 2 1 6 P p + (p ) φ̂(p) = A – . (69) O Γ(1/6)

Chapter 9 □ 99 Matching to the solution (58) reveals that " # Γ(5/6) Q P 1/6 φˆ = A 6 2/3 + 1 + O(p) . p Γ(1/6)

(70)

when t ≪ Q 1/2. Finally, comparison with Eqs. (9.47) and (9.48) in the book yields ∆ = 6 2/3 π

Γ(5/6)

S 1/3 Q P 1/6.

(71)

Γ(1/6) The characteristic lengthscale of the solution (69) is p ∼ P −1/6. Thus, the neglect of the second term on the left-hand side of Equation (60) is valid provided that Q ≪ P P −1/3, or Q ≪ P 2/3. (72) Combining the inequalities (59) and (72), the criterion for the validity of Equation (71) is Q ≪ P −1/3, P 2/3. (73) 9.4 As before, the inflow velocity is v0 ∼

(74)

η B∗ E z ∼ µ0 δ .

(75)

B∗

The z-component of Ohm’s law yields ∗

Continuity of plasma flow inside the sheet gives L v0 ∼ δ∗ v∗ ,

(76)

assuming incompressible flow. In the presence of strong plasma viscosity, the magnetic pressure gradient within the reconnecting layer is is balanced against the viscous force (µ ∇2V), instead of inertia, so µ v∗ B ∗2 (77) ∼ 2 . L µ0 δ∗ Finally,

dΨ dt

∼ v 0 B ∗.

(78)

Let B∗ = B0 B̂∗ , Ψ = a B0 Ψ̂ , t = τR tˆ, δ∗ = a δ̂∗ , v0 = (a/τR ) v̂0 , v∗ = (a/τR ) v̂∗ , where τR is specified in Eq. (9.25) in the book. Equations (74) and (75) yield vˆ0 ∼ Equation (77) gives B2 ∼ ∗

1 δˆ ∗

π k̂ v̂∗ P δˆ 2 S 2

(79)

(80)

∗

where S = τR/τH, kˆ = k a, L = π/k, and τH is specified in Eq. (9.24) in the book. The continuity equation (76) implies that δˆ ∼ π vˆ0. ∗ kˆ vˆ∗

(81)

100 □ Plasma Physics: An Introduction (Second Edition): Solutions to Exercises The previous three equations can be combined to give B̂∗ S

vˆ∗ ∼

, kˆ P1/2

vˆ0 ∼

B̂∗ S 1/2 πP

δ̂∗ ∼

π P1/2

(82) !1/2 ,

(83)

(84)

!1/2

B̂∗ S

Let us make the estimate B̂∗ = 2 Ψ̂ 1/2 . Equation (78) gives

dΨ dtˆ

∼ 2 v̂0 Ψ̂

1/2

(85) .

(86)

Hence, Equation (83) gives dΨ̂

23/2 S 1/2 ˆ 3/4 = √ Ψ . dtˆ π P1/4

(87)