Solution Sample

SOLUTIONS TOF.F. CHENSPLASMAPHYSICS

(VERSION 0.5 )

Tao [email protected]

SJTU

CONTENTS

Preface v

Acknowledgments vii

1 Introduction 1

1.1 1-3 21.2 1-8 31.3 1-9 31.4 1-10 41.5 1-11 4

2 Motion of Particle 5

2.1 2-2 52.2 2-3 62.3 2-7 62.4 2-10 62.5 2-11 62.6 2-15 72.7 2-16 8

iii

iv CONTENTS yetao1991.wordpress.com

2.8 2-17 82.9 2-18 82.10 2-19 92.11 2-20 102.12 2-21 12

3 Wave in Plasma 13

3.1 4-5 133.2 4-9 143.3 4-16 143.4 4-18 153.5 4-23 153.6 4-26 163.7 4-27 163.8 4-37 163.9 4-38 (Need to be confirmed) 183.10 4-39 193.11 4-40 193.12 4-41 193.13 4-42 193.14 4-43 193.15 4-45 193.16 4-49 193.17 4-50 19

PREFACE

Last revised: 11/24/2013 If you have any suggestions or ideas on this book, pleasetell me.My E-mail: [email protected] Personal Homepage: http://yetao1991.wordpress.com

ME

Shanghai,China

November, 2013

v

ACKNOWLEDGMENTS

Thanks All.

T.Y.

vii

CHAPTER 1

INTRODUCTION

1

2 INTRODUCTION yetao1991.wordpress.com

Problem 1-3

-2

-1

0

1

2

3

4

5

6 8 10 12 14 16 18 20 22 24

log(k

T)

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 of Nd is constant at the line of

Figure 1.1 Label (1)(2.1)(2.2)(3)(4)(5)(6)(7) correspond to plasma in different condition asthe problem describes.

According to the definition of the Debye Length

D = (0kTene2

)1=2 (1.1)

) log(D) = 12log(

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 Figure1.1 with the Debye lengthas 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

[email protected] 1-8 3

Debye length. Similarily, with the given equation:

ND =4

33Dn (1.4)

=4

3(7430 kT

n)32 (kT in eV) (1.5)

) ND2:7 106 =

(kT )32

n12

(1.6)

) log( ND2:7 106 ) =

3

2log(kT ) 1

2log(n) (1.7)

) log(kT ) = 13log(n) +

2

3log(

ND2:7 106 ) (1.8)

The dot lines in the Fig 1.1 can be named with equi-ND lines, which means theyshare a same value of ND in a dot lines.

As for the usage of this figure, take the point (4) for example, the point(4) fallsin the region enclosed by two solid lines and two dots ones.The two solid lines re-spectively have the Debye length of 103m and 104m. And the dots lines havethe particle numbers ND of 104 and 106. That is tantamount to the fact that (4) hasDebye length 104m < D < 103m and number of particle 104 < ND < 106.

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

Problem 1-8

The Debye length is

D = 69(T

n)12 (T in the unit of K)

= 69 (5 107

1033)12

= 1:54 1011mNaturally, the number of particles contained in a Debye Sphere is :

ND =4

33D n

=4

3 (1:54)3

15

Problem 1-9

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

f(u) = Ae(12mu

2+q)=kTe ;

4 INTRODUCTION yetao1991.wordpress.com

where q equals to e for protons while q equals to e for antiprotons. Moreover,

np(!1) = n1np(!1) = n1:

Then we obtain (np = n1exp(ekT )np = n1exp( ekT )

The Poissons Equation is

"0r2 = "0 @2

@x2= e(np np)

With e=kT 1,"0@2

@x2=

2n1e2kT

) D =r

"0kT

2n1e2= 0:4879m

Problem 1-10

Problem 1-11

CHAPTER 2

MOTION OF PARTICLE

Problem 2-2

Since A=2, for deuterium ion,

m = 2mp = 3:34 1027kgq = jej = 1:60 1019Coulomb:

Assume that energy can be entirely converted to kinetic energy, then the momentumcan be derived

E = Ek =p2

2m) p =

p2mEk = 1:46 1020kg m=s

The Larmour radius

rL =mv?jqjB =

1:46 10205 1:60 1019 = 0:018m a = 0:6m

So the Larmour radius satisfies the confined-ion condition.

5

6 MOTION OF PARTICLE yetao1991.wordpress.com

Problem 2-3

To keep a equibrillium in the y^ direction, the electric force shouldconteract the Lorentz force

Eq = qvB

) E = vB = 106V=m

Problem 2-7

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

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 ismd = 1875MeV=c

2

The kinetic energy

Ek =1

2mdv

2 ) v = 1:386 106m=s

v? = v cos 45 = 9:8 105m=s

rLmdv?qB

= 0:03m

Problem 2-11

Rm = 4) 1sin2 m

= 4) sin m = 12

) m =

6


Since the velocity is isotropic distribution, the directionof 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

Z 6

0

sin d

Z 20

d' = (1p3

2)4

) the fraction of the trapped isp32 .

Problem 2-15

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

W = q ~E ~xPThe energy gain rate is

dW

dt= qE

dxpdt

= qEvp:

The change of kinetic energy

dEkdt

=d

dt(1

2mv2E) = mvE

dvEdt

Without loss of generality, let the vE in the same direction of E

vE = EB

) dvEdt

= 1B

dE

dt

) dEkdt

= mvE( 1B

dE

dt) = qEvp

vp =m

qB

1

B

dE

dt

Replace 1!c = mqB Thus,vp = 1

!cB

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.

Problem 2-17

Since = mv2?

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

mv2?2B

=mv02?2B0

where 12mv2? = 1keV; B = 0:1T;B0 = 1T , so

1

2mv02? = 10keV:

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

1

2mv002? = 5keV

Implement the adiabatic characteristic of , we know that

mv002?2B0

=mv0002?2B0

) 12mv0002? = 0:5keV

Finally, the energy is

E = E? + Ek =1

2mv0002? +

1

2mv002k = 5:5keV

Problem 2-18

a) At a certain moment, we calculate the motion in one periodic circular motion tocertify the invariance of . In this period, we assume that the Larmour radius doesnot change with minor deviation of magnetic field-B. So

s = r2L = m2v2?q2B2


d

dt =

dB

dt S = m

2v2?q2B2

dB

dt= "(induced potential)

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

W = q" = m2v2?qB2

dB

dt

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

B =dB

dt =

dB

dt

2m

qB

So after a period, the energy is

E0k =1

2mv2? + W = (1 +

2m

qB2dB

dt)1

2mv2?

B0 = B +B = (1 +2m

qB2dB

dt)B

0 =E0kB0

=E0B0

=

So is invariant for both ion and electron.b) Assume that 12mv

2? = KT;

12mv

2k = KT at initial moment. Considering

=12mv

2?

B=

12mv

02?

B0;

and B = 13B0, so 12mv

02? = 3KT . Furthermore, we assume that vk remains un-

changed,1

2mv2k = KT

kT? = 3Kev; kTk = 1KeV

Problem 2-19

The magnetic field is B = 0I2r , so the gradient is :

rB = @B@r

=0I

2r2

Apply the equipartition theorem of Maxwellian gas, the total energy is

" =1

2m(p2x + p

2y + p

2z) =

3

2kT

) "? =1

2m(p2x + p

2y) = kT


The perpendicular velocity is :

) v? =r

2kT

m

) rL =mv?eB

) vrB =1

2v?rL

rBB

=kT

eBr;Upwards

* vk =rkT

m) vR =

mv2keBr

=kT

eBr;Upwards

) vR+rB =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 surface s with in time of dt is:

N = n v s dt:

So the hit rate isdN

dt= n v s:

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

Racc = n vR+rB s e = 4kTnaB

= 20Coulomb=S

Problem 2-20

a) v2k + v2? is invariant because of the conservation of energy. At z = 0, Bz = B0,

v2? =23v

2,v2k =13v

2,

=mv2?2B0

=mv2

2B0:

When the electron reflects, vk = 0, then v2? = v2

=mv2

3B0) Bz = 3

2B0

) 1 + 2z2 ) z = p2

2


b)

* = mv2?

2Bz) B0(1 + 2z2) = 1

2mv2?

) v2k = v2 v2? =0B

m 20B

m2z2

) (dzdt

)2 =0B

m(1 22z2)

dz

dt=

r0B0m

p1 22z2

) z =p2

2sin(

r20B0m

t+ )

And this equation can describe the trajectory of the particle.

c) It is apparent that the gyration frequency isq

20B0m from the the equation of

motion.d) Claim: =

q20B0m t+ , z =

p2

2 sin

J =

Z ba

vkdz

where a = p2

2 ; b =p2

2 . Thus,

a = 2; b =

2

J =

Z ba

vkdz =Z

vkvkdt =1

rm

2B0

Z ba

v2kd

where

dt =1

rm

2B0d;

v2k = (dz

dt)2 =

B0m

sin2 :

) J = 1

rm

2B0

Z 2

2

B0m

sin2 d

=

2

rB02m

= constant


Problem 2-21

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

2r B = 0I ) B = 0I2r

In cylinder coordinate:

rB = @B@r

=0I

2r2

) vrB =1

2v?rL

rBB

;

where rL = mv?eB ; v?0 = v?.

) vrB =1

2v?0

mv?0eB

1

r=

mv2?00Ie

Besides,

vR =mv2ke

1

RcB

=mv2ke

1

Rc 0I2RC=

2mv2k0Ie

) ~v = ~vR + ~vrB =3mv2k00Ie

(z^)

(vk0 = v?0)

CHAPTER 3

WAVE IN PLASMA

Problem 4-5

According to the dispersion relation

!2 = !2p +3KTem

k2

Now we calculate the unknown value in the equation

!p 2pn = 5:62 109rad=sec

! = 2f = 6:908 109rad=sec3kTem

=3 100 1:6E19

9:109 1031 = 5:27 1013

) k2 =!2 !2p

3kTem

= 3:05 105 ) k = 552:377

) = 2k

= 1:14 102m

13

14 WAVE IN PLASMA yetao1991.wordpress.com

Problem 4-9

The critical density is

n0 =m"0!

2

e2= 1:12 1015m3

Problem 4-16

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

c2k2

!2= 1 !

2p

!2!2 !2p!2 !2h

1) The resonance of X-wave is found by setting k ! 1. So the dispersion relationcan be rewrite into

c2k2 = !2 !2p!2 !2p!2 !2h

Differentiate both sides of the equation

2kc2dk = 2!d! 2!(!2p !2h

!2 !2h!2pd!

So the group velocity is

vg =dw

dk=

kc2

![1 +!2c!

2p

(!2!2h)2]

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

!2c!2p

(!2!2h)2! 1; k ! 1 and 1 + !

2c!

2p

(!2!2h)2has 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 !2p

!2!2 !2p!2 !2h

) ! = !R or !L

At this point, !(1 +!2c!

2p

(!2!2h)2) is a finite value. As a result

vg =kc2

![1 +!2c!

2p

(!2!2h)2]jk=0 = 0

) Q:E:D:


Problem 4-18

For L-wave,c2k2

!2= 1 !

2p=!

2

1 + !c=!

And

!2p =n0e

2

m"0; !c =

eB

m

The cut-off is found when k = 0. With the provided condtion f = 2:8GHz;B0 =0:3T , the critical density is

n0 =m"0e2

[(2f)2 + 2feB

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

s1 !

2p=!

2

1 !c! !

c(1 1

2

!2p=!2

1 !c!) !

c[1 1

2

!2p!2

(1 +!c!)]

kL =!

c

s1 !

2p=!

2

1 + !c! !

c(1 1

2

!2p=!2

1 + !c!) !

c[1 1

2

!2p!2

(1 !c!)]

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

=1

2

180

(kL kr)

=180

2

1

c(

1

42)e2

m"0

e

mB(Z)n(z)

=90

1

c(

1

42)(0c)2

e3

m2"0B(z)n(z)

=90

e3

42m2"0c3B(z)n(z)20(degree)


And90

e3

42m2"0c3= 1:5 1011degree = 2:62 1013rad

In conclusion

=180

2

Z L0

(kL kR)dz = 1:5 10110Z L0

B(z)n(z)dz

(in the unit of degree)

Problem 4-26

a)

va =Bp0

= 2:18 108m=s

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

Problem 4-27

= n0M = 1:67 1019

vA =Bp0

= 2:18 104m=s

Problem 4-37

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

m@ve@t

= eE mve

Linearize the equation

im!ve = eE mve ) j1 = n0eve = n0e2E1

im(! + i)

The equation of the transverse wave is

(!2 c2k2)E1 = i!j1="Insert j1 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 !

When ! or ! 1,then

(1 + i

!)1 1 i

!

So the dispersion relation turns to

c2k2 = !2 !2p + i!2p!

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 = Im(!)i

So the damping rate:

= Im(!) = !2p

2j!j2

c) With previous conclusion of a), we get

k2 =!2 !2p

c2+ i

!2p

c2!

Let k = e+ di, then k2 = e2 d2 + 2edi, we get8


Problem 4-38 (Need to be confirmed)

The loss part of energy will heat up the electron to oscillation. So we firstly need toderive the vibration motion due to the microwave. The wave-length of the microwaveis = 0:3m So the frequency is

! =2c

= 6:28 109rad=sec

The collision frequency is

= nn = 102=sec

So the equation of motion of electron is

m@ve@t

= eE mve

Linearize the equation to be

i!mve = 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(kr!t) = E0ei(Re(k)r!t)eIm(k)r

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

A =E2outE2ini

= (eIm(k)R1

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

And R1 R2 = 100km = 105m

!2p =nee

2

"0m= 3:17 1014

Im(k) =

2c

!2p!2

1q1 !2p!2

= 1:34 1012

) A = e2:69107 = 0:999999732So the loss fraction is

1A 0


Problem 4-39

Unchecked.

Problem 4-40

Problem 4-41

Problem 4-42

Problem 4-43

Problem 4-45

Problem 4-49

Problem 4-50

Unchecked.

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