Solution SamSolution to F.F. Chen’s Plasma Physicsple2

7/26/2019 Solution SamSolution to F.F. Chens Plasma Physicsple2

1/31

Solution to

F.F. Chens Plasma Physics

Tao Ye

[email protected]

cby Tao YeAll rights reserved.

Please DO NOT DISTRIBUTE this document.

Instead, link to

yetao1991.wordpress.com

Last Updated: May 2, 2015


2/31

2


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Contents

1 Introduction 5Problem 1-3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Problem 1-8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Problem 1-9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Problem 1-10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Motion of Particle 9Problem 2-2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Problem 2-3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Problem 2-7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Problem 2-10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Problem 2-11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Problem 2-15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Problem 2-16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Problem 2-17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Problem 2-18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Problem 2-19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Problem 2-20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Problem 2-21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Wave in Plasma 17Problem 4-5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Problem 4-9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Problem 4-16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Problem 4-18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Problem 4-23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Problem 4-26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Problem 4-27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Problem 4-37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Problem 4-38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Problem 4-39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Problem 4-40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Problem 4-41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Problem 4-42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3


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4 CONTENTS

Problem 4-43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Problem 4-45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Problem 4-49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Problem 4-50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29


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Chapter 1

Introduction

Problem 1-3

-2

-1

0

1

2

3

4

5

6 8 10 12 14 16 18 20 22 24

log(kT)

log(n)

Problem 1-3

(1)

(2.1)

(2.2)

(3)

(4)

(5)(6)

(7)

1

100

1e4

1e6

1e8

1e-1 1e-2 1e-3 1e-4 1e-5 1e-7

Debye length is constant at the line ofNd is constant at the line of

Figure 1.1: Label (1)(2.1)(2.2)(3)(4)(5)(6)(7) correspond to plasma in differentcondition as the problem describes.

5


6/31

6 CHAPTER 1. INTRODUCTION

According to the definition of the Debye Length

D = ( 0kTene2

)1/2 (1.1)

log(D) = 12

log(0e2

) +1

2log(kTe) 1

2log(n) (1.2)

log(kT) = log(n) + 2 log( D7430

) (kT in eV) (1.3)

Then we can draw the solid straight line in the Figure?? with the Debyelength as parameter ranged from 101 to 107. Points on a certain solid line,named withequi-Debye-length line(analog to equipotential lines in electrostatic), share a same Debye length. Similarily, with the given equation:

ND = 4

3

3Dn (1.4)

= 4

3(7430 k T

n )

3

2 (kT in eV) (1.5)

ND2.7 106 =

(kT)3

2

n1

2

(1.6)

log( ND2.7 106 ) =

3

2log(kT) 1

2log(n) (1.7)

log(kT) = 13

log(n) +2

3log(

ND2.7 106 ) (1.8)

The dot lines in the Fig ?? can be named with equi-ND lines, which meansthey share a same value ofND in a dot lines.

As for the usage of this figure, take the point (4) for example, the point(4)falls in the region enclosed by two solid lines and two dots ones.The two solidlines respectively have the Debye length of 103m and 104m. And the dotslines have the particle numbers ND of 10

4 and 106. That is tantamount to thefact that (4) has Debye length 104m < D < 103m and number of particle104 < ND < 10

6.

Recall for the criteria for plasmas, ND 1 is automatically meet since thesmallest number of particle is larger than 100.

Problem 1-8

The Debye length is

D = 69(T

n)1

2 (T in the unit of K)

= 69 ( 5 107

1033 )

1

2

= 1.54 1011m


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. PROBLEM 1-9 7

Naturally, the number of particles contained in a Debye Sphere is :

ND = 4

3 3D n

= 4

3 (1.54)3

15

Problem 1-9

Since protons and antiprotons have the same inertia, both of them are fixed.Assume that protons and antiprotons follows the Maxwellian distribution.

f(u) = Ae(1

2mu2+q)/kTe ,

whereqequals toe for protons whileqequals to

e for antiprotons. Moreover,

np( ) = nnp( ) = n.

Then we obtain np = nexp(

ekT )

np = nexp(ekT)

The Poissons Equation is

02= 0 2

x2 = e(np np)

Withe/kT 1,

0

2

x2 =

2n

e2

kT

D =

0kT

2ne2 = 0.4879m

Problem 1-10

Regrad it as an isotropic space, which means that has no components of or.

= Aekr

r

2 =

1

r2

r(r2

r) =

2

r

r +

2

r2

r = A kr + 1

r2 ekr

2

r2 =

k2r2 + 2kr+ 2

r3 Aekr


8/31

8 CHAPTER 1. INTRODUCTION

2 = Ak2 ekr

r

2 = ne2

0kBT q0(r a),

whereq0 = 40a0.

ne2

0kBT =k2 k=

ne2

kBT D = 1

k =

0kBT

ne2

Consider the boundary condition :

Aeka

a = 0 A= 0a

eka

So, =

0, r (0, a]0a

e aD

e rD

r , r (r, +) (1.9)


9/31

Chapter 2

Motion of Particle

Problem 2-2

Since A=2, for deuterium ion,

m = 2mp = 3.34 1027kgq = |e| = 1.60 1019Coulomb.

Assume that energy can be entirely converted to kinetic energy, then the mo-mentum can be derived

E= Ek = p2

2m p=

2mEk = 1.46 1020kg m/s

The Larmour radius

rL=mv|q|B =

1.46 10205 1.60 1019 = 0.018m a= 0.6m

So the Larmour radius satisfies the confined-ion condition.

Problem 2-3

To keep a equibrillium in the y direction, the electric forceshould conteract the Lorentz force

Eq= qv B

E= vB = 106V/m

Problem 2-7

Apply the Gauss Law to obtain the magnitude of electric field at r=a

9


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10 CHAPTER 2. MOTION OF PARTICLE

E

l

2a = nel

a2/0

E=enea

2 = 9.04 103V /m

vE=E

B = 4.52 103m/s

(Direction:See in the figure)

Problem 2-10

The mass of a deutron is

md= 1875MeV/c2

The kinetic energy

Ek =1

2mdv

2 v= 1.386 106m/s

v= v cos45= 9.8 105m/s

rLmdv

qB = 0.03m

Problem 2-11

Rm= 4 1

sin2 m = 4 sin m=1

2

m=

6

Since the velocity is isotropic distribution, the direc-tion of velocity should distribute uniformly

d = sin dd

The total solid angle for a sphere is

total= 4

The solid angle for loss cone is

loss= 2

6

0

sin d

20

d= (1

3

2 )4

the fraction of the trapped is32 .


11/31

. PROBLEM 2-15 11

Problem 2-15

Define the displacement in polarization direction isxp.So the work done by the electric field is

W =qE xP

The energy gain rate is

dW

dt =qE

dxpdt

=qE vp.

The change of kinetic energy

dEk

dt =

d

dt(

1

2mv2E) = mvE

dvE

dt

Without loss of generality, let the vEin the same direction of E

vE = EB

dvE

dt = 1

B

dE

dt

dEk

dt = mvE( 1

B

dE

dt) =qE vp

vp = m

qB

1

B

dE

dt

Replace 1

c = m

qB Thus,

vp = 1cB

dE

dt

Problem 2-16

a) The Larmor frequency of electron is

e =eB

m =

1.6 1019 19.109 1031 = 1.76 10

11rad/sec

b) the Larmor frequency of ion is

i =eB

mi=

1.6 1019 11.67 1027 = 9.58 10

7rad/sec

Since 0 = 109rad/sec, e 0 i. The motion of electron is adiabatic,while that of ion not.


12/31


Problem 2-17

Since = mv2

2B is conservative under this condition, it is easy to derive:

mv22B

=mv2

2B

where 12mv2= 1keV, B = 0.1T, B

= 1T, so

1

2mv2 = 10keV.

When collision happens, the direction of motion distorts, so v= v. Then thekinetic energy is

1

2 mv2

= 5keVImplement the adiabatic characteristic of, we know that

mv22B

=mv2

2B 1

2mv2 = 0.5keV

Finally, the energy is

E= E+E =1

2mv2 +

1

2mv2 = 5.5keV

Problem 2-18

a) At a certain moment, we calculate the motion in one periodic circular motionto certify the invariance of. In this period, we assume that the Larmour radiusdoes not change with minor deviation of magnetic field-B. So

s= r2L= m2v2q2B2

d

dt =

dB

dt S= m

2v2q2B2

dB

dt =(induced potential)

So the change of the energy of the particle in one period is

W =q = m2v2

qB2dB

dt

Within this period of gyration, the change of magnetic field is

B =dB

dt=

dB

dt

2m

qB


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14/31


vR+B =2kT

eBr

Area of the top surface of the cyclinder is :

s= [(R+a

2)2 (R a

2)2] = 2Ra

The number of charged particle, which hit the surfaces with in time ofdt is:

N=n v s dt.So the hit rate is

dN

dt =n v s.

SinceR a, the drift velocity with [R a2 , R+ a2 ] region can be considered asuniform. So the accumulation rate is

Racc= n vR+B s e= 4kTnaB

= 20Coulomb/S

Problem 2-20

a)v2+v2is invariant because of the conservation of energy. At z = 0,Bz =B0,

v2 = 23v

2,v2 = 13v

2,

=mv2

2B0=

mv2

2B0.

When the electron reflects, v = 0, then v2 = v2

= mv2

3B0 Bz = 3

2B0

1 +2z2 z =

2

2

b)

=mv2

2Bz B0(1 +2z2) = 1

2mv2

v2 = v2 v2 =

0B

m 20B

m 2z2

(dz

dt)2 =

0B

m (1

22z2)

dz

dt =

0B0

m

1 22z2

z =

2

2 sin(

20B0

m t+)


15/31

. PROBLEM 2-21 15

And this equation can describe the trajectory of the particle.

c) It is apparent that the gyration frequency is 20B0m

from the the equation

of motion.

d) Claim: =

20B0m t+, z =

2

2 sin

J=

ba

vdz

wherea = 2

2, b=

2

2 . Thus,

a = 2

, b=

2

J= b

a

v

dz = vvdt= 1

m

2B0 b

a

v2

d

where

dt = 1

m

2B0d,

v2 = (dz

dt)2 =

B0m

sin2 .

J = 1

m

2B0

2

2

B0m

sin2 d

=

2B0

2m =constant

Problem 2-21

a) According to the Ampere theorem, the magnetic field can be obtained

2r B = 0I B= 0I2r

In cylinder coordinate:

B = Br

= 0I

2r2

vB =12

vrLBB

,

whererL= mveB , v0 = v.

vB =1

2v0

mv0eB

1

r =

mv200Ie


16/31


Besides,

vR = mv2

e1

RcB

=mv2

e

1

Rc 0I2RC=

2mv20Ie

v= vR+ vB =3mv20

0Ie (z)

(v0 = v0)


17/31


18/31

18 CHAPTER 3. WAVE IN PLASMA

Problem 4-16

If the motion of ion is neglected, the dispersion relation of electron is

c2k2

2 = 1

2p

22 2p2 2h

1) The resonance of X-wave is found by setting k . So the dispersionrelation can be rewrite into

c2k2 =2 2p2 2p2 2h

Differentiate both sides of the equation

2kc2

dk= 2d 2(2p

2h

2 2h 2

pd

So the group velocity is

vg =dw

dk =

kc2

[1 + 2c

2p

(22h)2

]

When k ,which implies that = h.So 1+

2c2

p

(22h)2

, k and 1 + 2

c2

p

(22h)2

has a higher order than k. Thus,

vg = 0 at resonance point.2) The cut-off X-wave is found by setting k=0, then

1 2p2

2

2p

2 2h = R orL

At this point, (1 + 2c

2

p

(22h)2

) is a finite value. As a result

vg = kc2

[1 + 2c

2p

(22h)2

]|k=0 = 0

Q.E.D.

Problem 4-18

For L-wave, c2k2

2 = 1

2p/

2

1 +c/

And

2p =n0e

2

m0, c =

eB

m


19/31

. PROBLEM 4-23 19

The cut-off is found whenk = 0. With the provided condtionf= 2.8GHz,B0=0.3T, the critical density is

n0 = m0

e2 [(2f)2 + 2f

eB

m]

= 3.89 1017m3

Problem 4-23

R-wave

k2 =2

c2

2p

c2(1 c/p)L-wave

k2 = 2

c2

2p

c2(1 +c/p)

Thus

kR=

c

1

2p/

2

1 c

c(1 1

2

2p/2

1 c)

c[1 1

2

2p2

(1 +c

)]

kL=

c

1

2p/

2

1 + c

c(1 1

2

2p/2

1 + c)

c[1 1

2

2p2

(1 c

)]

The difference of the phase is twice of the Faraday rotation angle

= 12

180

(kL kr)

= 180

2

1

c(

1

42)

e2

m0

e

mB(Z)n(z)

= 90

1

c(

1

42)(

0c

)2 e3

m20B(z)n(z)

= 90

e3

42m20c3B(z)n(z)20(degree)

And90

e3

42m20c3 = 1.5 1011degree= 2.62 1013rad

In conclusion

=180

2

L0

(kL kR)dz = 1.5 10110 L0

B(z)n(z)dz

(in the unit of degree)


20/31


Problem 4-26

a)

va = B

0= 2.18 108m/s

b) The Alfven wave represents for phase velocity. And phase velocity did notcarry information. So it does not mean that wave can travel faster than light.

Problem 4-27

= n0M= 1.67 1019

vA= B

0= 2.18 104m/s

Problem 4-37

a)Consider the elastical collision from ion, the equation of eletrons motion

mvet

= eE mve

Linearize the equation

imve = eE mve j1 = n0eve= n0e2E1

im(+ i)

The equation of the transverse wave is

(2 c2k2)E1 = ij1/

Insertj1 into the wave equation, we get

(2 c2k2)E1= + i

2pE1

2 c2k2 = 2p

+ i

c2k2

2 = 1

2p

(+ i)

Q.E.D.

b) Apply the previous result

c2k2 =2 2p

1 + i


21/31

. PROBLEM 4-38 21

When or 1,then

(1 + i )1 1 i

So the dispersion relation turns to

c2k2 =2 2p+ i2p

Assume that = a+bi, then

2p+c2k2 = (a2 b2) + 2abi + i

2p

a+bi

= (a2 b2) + 2pvb

a2 +b2+ i(2ab+

2pa

a2 +b2)

The image part should be zero, that is

ib=2pv

2(a2 +b2)i= I m()i

So the damping rate:

= Im() = 2p

2||2

c) With previous conclusion of a), we get

k2 =2 2p

c2 + i

2p

c2

Letk = e+di, then k2 =e2 d2 + 2edi, we get e2 d2 =

22pc2

2ed = 2p

c2

By solving the simultaneous equations, it is easy (Uh huh!)to obtain that

= 1

Im(k)=

1

d= (

2c

2

2p)(1

2p

2)1/2

Problem 4-38

The loss part of energy will heat up the electron to oscillation. So we firstlyneed to derive the vibration motion due to the microwave. The wave-length ofthe microwave is = 0.3mSo the frequency is

=2c

= 6.28 109rad/sec


22/31


The collision frequency is

= nn = 102

/sec

So the equation of motion of electron is

mvet

= eE mve

Linearize the equation to be

imve = eE mve ve = eEim(+ i)

Then the dispersion raltion is

k2 =2 2p

c2 +

2p

2ci

k= Re(k) +Im(k)i

And microwave in plasma is

E= E0ei(krt) =E0ei(Re(k)rt)eIm(k)r

The term eIm(k)r means the wave decay when penetrate the ionosphere. Theratio of outgoing energy v.s. incident energy is

A=E2outE2ini

= (eIm(k)R1

eIm(k)R2)2 =e2Im(k)(R1R2)

AndR1

R2 = 100km = 105m

2p =nee

2

0m = 3.17 1014

Im(k) =

2c

2p2

11 2p2

= 1.34 1012

A = e2.69107

= 0.999999732

So the loss fraction is1 A 0

Problem 4-39

a) When k , the resonance happens, which indicates:

22(1 2p

2) 2csin2 c

2csin

4 + 44(1 2p

2)2 cos2

1/2= 0


23/31


24/31


25/31

. PROBLEM 4-42 25

Problem 4-42

Boltzman relation is

ne1 = n0e1kTe

The plasma approximation

ZnA1+nH1 = ne1

The equation of continuity

inA1 = nAikvA1

inH1 = nHikvH1

And the equation of motion

MA(i)vA1 = MA vA1t

=Z eE1 = Z e(ik1)

MH(i)vH1 = MHb vH1t

=eE1 = e(ik1)Then we get

2

k2 =

kTen0

(Z2nA

MA+

nHMH

)

The phase velocity is

v=

k =

kTen0

(Z2nA

MA+

nHMH

)

Problem 4-43

The Poisson Equation

0 E1 = n1+e Zn1eAnd the continuity equation

n1+ = k

n0v1+

n1 = k

n0v1

Since kT = 0, B0 = 0, there is no collision and magnetic term in the equationof motion.

M+v+1

t = eE1 v+1 = eE1

iM+

Mv1

t = ZeE1 v1 = ZeE1

iM


26/31


Then we get

i0kE1 =

k

n0e2E1

iM+ Zk

n0

Ze2E1

iMThat is

=

n0+e2

M+0+

Z2n0e2

M0

Problem 4-45

Assume that the ionosphere is extremly cold, we can presume that

kTi = 0

So the sonic ion wave velocity is

vs =

kTeM

And the Alfvenic velocity is

vA= B0Mn0

And it indicates that super sonic wave is not super-Alfvenic. That is ,

v >

kTeM

Te< Mv2

k = 1.2 107K.

This is the upper limit of temperature.

v l

r >0

when < L r


31/31

. PROBLEM 4-50 31

c)

p = (1 ra )2c,

2 = 2l(r= 0) = m

M 42c =

4m

5M2c ,

L = c

2 [1 +

1 + 16(1 r

a)2],

2l = m

M

16(1 ra )21 + 16(1 ra)2

2c .

Similarly, when > l, r >0 ; when < L, r

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