Solutions to Chen’s Plasma Physics Kalman Knizhnik 1-1. 1-1. Compute Compute the densit density y (in units of m−3 ) of an ideal gas under the following conditions: a) At 0o C and 760 Torr pressure pressure (1 Torr = 1mm Hg). This This is called called the Loschmid Loschmidtt number. b) In a vacuum at 10−3 Torr at room temperature (20 (20o C). This number is a useful one for the experimentalist to know by heart (10 (10−3 Torr = 1 micron). a) Avogadro’s number is N is N A = 6. 6 .022 1023 . One mole of gas at STP occupies 22.4 liters. 1 liter is 1 10−3 cubic meters. Thus, the number per cubic meter is N A /n = /n = 6.022 1023 /(22. (22.4 10−3 ) = 2.66 1025 m−3 . Thus, the Loschmidt number is 2. 2.66 1025 b) Using PV=NkT, we obtain (with R= 1. 1 .4 10−23 J K −1 and 1 Torr = 133 Pa):
×
×
×
×
×
n =
N P = = V kT 1.4
×
10−3 10−23
×
× 133 × (20 + 273) = 3.3 × 10
19
×
m−3
(1)
1-2. Derive Derive the constant constant A for a normalized one-dimensional Maxwellian distribution 2
f ( f ˆ(u) = Ae −mu such that
∞
−∞
/2kT
(2)
f ( f ˆ(u)du = 1
(3 )
This one is straightforward. Just integrate: ∞
1=
−∞
2
−mu /2kT
Ae
du = du = A A
2πkT m
⇒
A =
m 2πkT
(4)
1-4. Compute the pressure, pressure, in atmospheres atmospheres and in tons/ft in tons/ft2 , exerted by a thermonuclear plasma plasma on its contain container. er. Assume Assume kT e = kT i = 20 20keV keV ,, n = 1021 m−3 and p = nkT , nkT , where T = T i + T + T e . This is just unit conversion conversion,, albeit with units that nobody really ever ever remembers. remembers. For reference, reference, 19 − 1 keV k eV = 1. 1 .6 10 J , J , so
×
p = p = 1021
22
(20keV + 20keV 20keV ) ) = 4 × 10 × (20keV
m−3 keV = 4
× 10 m− 3
3
J = 4
3
× 10 N/m
2
(5)
But 1 atm 1 atm = = 105 N/m2 = 1 ton/ft2 , so (Note: (Note: I think there is a mistake in Chen’s Chen’s solutions here. here. If I am mistaken, please let me know). p = p = 0.04 04 atm atm = = 0.04 04 ton/ft ton/ft2
(6)
1-5. In a strictly strictly steady state state situation, both the ions and the electrons will will follow the Boltzmann Boltzmann relation relation n j = n0 e−qj φ/kT j (7) For the case of an infinite, transparent grid charged to a potential φ, show that the shielding distance is given approximately by ne2 1 1 ( + ) 0 kT e kT i
2 λ− D =
(8 )
Show that λD is determined by the temperature of the colder species. We’ll use Poisson’s equation
∇ φ = ene − eni = en
0
2
0
0
(eeφ/kT e
− e−eφ/kT ) ≈ en i
0
0
(1 +
eφ kT e
2
φ ) − 1 − −kT eφi ) = e n ( kT φ e + kT i 0
(9 )
Now we’ll suppose the φ goes like a decreasing exponential: φ = φ 0 exp( x/λD ). Thus, the Laplacian acting on this is
−
2
∇ φ = λ1 φ = ne 2
2
(
0
D
φ φ + ) kT e kT i
1 ne2 1 1 = ( + ) 2 0 kT e kT i λD
⇒
(10)
To show that λD is determined by the colder species, we suppose first that the electrons are the colder species: T e T i . Then,
2
1 ne = 2 λD kT e 0
⇒
λD =
kT e 0 ne2
(11)
(12)
Alternatively, if the ions are colder, T i
T e, then a similar analysis yields:
1 ne2 = kT i 0 λ2D
⇒
λD =
kT i 0 ne2
1-6. An alternative derivation of λD will give further insight to its meaning. Consider two infinite, parallel plates at x = d, set at potential φ = 0. The space b etween them is uniformly filled by a gas of density n of particles of charge q . a) Using Poisson’s equation, show that the potential distribution between the plates is nq 2 φ = (d x2 ) (13) 20
±
−
b) Show that for d > λ D , the energy needed to transport a particle from a plate to the mid plane is greater than the average kinetic energy of the particles. 1-9. A distant galaxy contains a cloud of protons and antiprotons, each with density n = 106 m−3 and temperature T = 100o K. What is the Debye length? The Debye length is given by 0 kT j λD = (14) 2 n e j j j
Plugging in the numbers: λD =
8.85
× 10− × 1.4 × 10− × 100 = 0.48 m 10 × (1.6 × 10− ) 12
23
6
19 2
(15)
As a check, use the SI unit form for the Debye length given in Chen. If T is in Kelvin, and n is in cubic meters, then: T 102 λD = 69 m = 69 m = 69 10−2 m (16) 6 n 10
×
×
This is the same order of magnitude so we are ok. 1-10. A spherical conductor of radius a is immersed in a plasma and charged to a potential φ0 . The electrons remain Maxwellian and move to form a Debye shield, but the ions are stationary during the time frame of the experiment. Assuming φ 0 kT e /e, derive an expression for the potential as a function of r in terms of a, φ 0 , and λ D . (Hint: Assume a solution of the form e−br /r.) Let’s assume a solution of this form: φ = Ae −br /r. Then,
∇ φ = r1 ∂r∂ (r 2
2
2
∂φ e ) = b2 φ = (ne ∂r 0
− ni)
(17)
Since the electrons are Maxwellian, they obey ne = n 0 eeφ/kT e are stationary, so n i = n 0 . Thus we have: b2 φ =
e eφ (n0 + n0 0 kT e
≈ n (1 +eφ/kT e). The ions, however, 0
2
e φ φ − n ) = n kT ≡ ⇒ λ e 0
0
b =
2
0
D
1 λD
(18)
Thus, so far we have: e−r/λD φ = A r But we also need to match the boundary condition that φ(a) = φ 0 . So, φ0 = A
e−a/λD a
(19)
A = aφ 0 ea/λD
⇒
(20)
So, finally we have our answer: e−r/λD r You know what they say: if it satisfies Poisson’s equation must be the answer. φ(r) = φ 0 ea/λD a
and the
(21)
boundary conditions then it
2-3. An ion engine (see Fig. 106) has a 1-T magnetic field, and a hydrogen plasma is to be shot out at an E B velocity of 1000 km/s. How much internal electric field must be present in the plasma? The E B velocity is given by E B v = (22) B2 Plugging in the numbers:
×
×
×
106 m/s =
|E | ⇒ |E | = 1000 V /m 1T
(23)
2-4. Show that vE is the same for two ions of equal mass and charge but different energies, by using the following physical picture (see Fig. 2-2). Approximate the right half of the orbit by a semicircle corresponding to the ion energy after acceleration by the E field, and the left half by a semicircle corresponding to the energy after deceleration. You may assume that E is weak, so that the fractional change in v⊥ is small. If the energy of the right part of the orbit is E 1 and the energy of the left part of the orbit is E 2 , then we have E 1 = E 0 + eEr 1 , E 2 = E 0 eEr 2 (24)
−
where E 0 is the initial energy and E is the electric field. The velocity is determined by v = so 2E 0 + 2eEr 1 2E 0 eEr 2 v1 = v2 = m m
−
2E/m, (25)
The Larmor radius is determined via r = mv⊥ /qB, so m r1,2 = qB
2E 0 m
1
±
eEr 1,2 = E 0
∓
√ 2mE
0
qB
(1
±
eE r1,2 )= 2E 0
2E 0 1 m ωc
, ± 2√ Er mE ωc 12
(26)
0
Thus, r1,2 (1
m E 1 )= 2E 0 qB ωc
2E 0 (1 m
± 2E eE ωc 0
2E 0 ) m
(27)
The guiding center moves a distance r1 r1
−
−r : 2
eE r2 = E 0 ωc
2E 0 1 m ωc
2E 0 2eE = m mωc2
(28)
The velocity of the guiding center is vgc = 2
r1
−r
2
T
=2
ωc (r1 2π
4eE 2eE 2E E = = − r ) = 2πmω ≈ B πmωc πB c
2
(29)
since ω c = eB/m. This is a pretty good approximation. 2-5. Suppose electrons obey the Boltzmann relation of Problem 1-5 in a cylindrically symmetric plasma column in which n(r) varies with a scale length λ; that is ∂n/∂r = n/λ. a) Using E = φ, find the radial electric field for a given λ. b) For electrons, show that the finite Larmor radius effects are large if vE is as large as vth . Specifically, show that rL = 2λ if vE = v th . c) Is (b) also true for ions? Hint: Do not use Poisson’s equation. a) We simply solve for φ from the Boltzmann relation for electrons.
−
−∇
n = n0 eeφ/kT e Therefore, E =
φ =
⇒
kT e n ln( ) e n0
(30)
kT e n 1 ∂n kT e ˆr = − ˆr = ˆr −∇φ = − ∂φ ∂r e n n ∂r eλ 0
(31)
0
b) We start with the definitions of v E , vth , and r L : vE =
E , B
vth =
2kT e mv⊥ , rL = m eB
(32)
So, calculating the magnitude of v E : 2 kT e mvth 1 rL vth vE = = = eλB 2 eλB 2λ
(33)
where in the last step I have assumed that the perpendicular velocity is the thermal velocity. Now, setting v E = v th , it is easy to see that rL = 2λ (34) c) Sure, why not? 2-6. Suppose that a so-called Q-machine has a uniform field of 0.2 T and a cylindrical plasma with kT e = kT i = 0.2 eV . The density profile is found experimentally to be of the form n = n 0 exp[exp( r 2 /a2 ) 1] (35)
−
−
Assume the density obeys the electron Boltzmann relation n = n0 exp(eφ/kT e ). a) Calculate the maximum vE if a = 1 cm. b) Compare this with vE due to the earth’s gravitational field. c) To what value can B be lowered before the ions of potassium (A = 39, Z = 1) have a Larmor radius equal to a? We solve for φ: n0 exp[exp( r 2 /a2 )
−
− 1] = n exp(eφ/kT e) ⇒ 0
φ =
kT e −r (e e
2
/a2
− 1)
(36)
Thus, the electric field is E =
kT e 2r −r /a ˆr = ˆr e − ∂φ ∂r e a 2
2
2
(37)
and so v E (and it’s maximum) is vE = ∂v E 2kT e = 2 e−r ∂r ea B So, with a = 1 cm, 2kT e vE,max = 2 ea B
2
/a2
E 2rkT e −r = e B ea2 B 2
e −r /a e − 4rea kT B 2
4
a2 −1/2 e 2
2
2
/a2
=0
(38)
⇒
a=1 cm,B =0.2 T,kT e =0.2 keV
r =
a2 2
≈ 8.5 km/sec
(39)
(40)
b) If we assume these are potassium ions, we have mg = 39 1.6 10−27 9.8 = 6.4 10−25 N . Meanwhile, if we plug in the numbers above into the expression for the electric field (equation 37), we’ll get that E = 17 V /m. Thus, the force due to the electric field is eE = 1.6 10−19 17 = 2.8 10−18 N . Thus the gravitational drift is
×
×
×
×
×
×
F g 6.4 = F E 2.8
× 10− ≈ 1.5 × 10− × 10−
×
25
7
(41)
18
times smaller. c) The Larmor radius is rL = mvth /qB, so, in terms of the constants of vth , we have (setting rL = a): m 2kT e 2mkT e rL = = a B = (42) qB m q 2 a2
⇒
Plugging in the numbers:
× 2
39 1.6 10−27 0.2 1.6 10−19 =4 (17 1.6 10−19 )2 (0.1 10−2 )2
× × × × × Suppose the Earth’s magnetic field is 3 × 10− B=
× × × ×
× 10−
2
T
(43)
5 2-8. T at the equator and falls off as 3 1/r , as for a perfect dipole. Let there be an isotropic population of 1 eV protons and 30 keV electrons, each with density n = 107 m−3 , at r = 5 earth radii in the equatorial plane. a) Compute the ion and electron B drift velocities. b) Does an electron drift eastward or westward? c) How long does it take an electron to encircle the earth? d) Compute the ring current density in A/m2 . Note: The curvature drift is not negligible and will affect the numerical answer, but neglect it anyway. a) The grad-B drift is given by
−
−
∇
1 B B 1 B v∇B = v⊥ rL = v r L ⊥ 2 B2 2 B
| ×∇ |
|∇ |
(44)
We can calculate v⊥ from the energy, and r L from the magnetic field: v⊥ =
2E , m
rL =
mv⊥ m = eB eB
2E m
(45)
Thus, v∇B =
1 2
2E m m eB
2E B E B = m B eB B
| ∇ |
|∇ |
(46)
Since B
∼ r− , we obtain 3
3 B ∇B | = 3 ˆr = − = −3 ∇B = ∂B ⇒ | ∂r r r B r
4
So v∇B =
(47)
3E 3Er 3 3Er 2 3E (eV )r2 = = = eBr eB0 R3e r eB0 R3e B0 R3e
(48)
where I have used the fact that B = B 0 R3e /r3 . Now we can plug in the numbers: 3
3
6 2
× 30 × 10 × (5 × 6 × 10 ) = 1.3 × 10 m/s v∇B,e = (49) 3 × 10− × (6 × 10 ) 3 × 1 × (5 × 6 × 10 ) v∇B,i = = 0.42 m/s (50) 3 × 10− × (6 × 10 ) ˆ b) The magnetic field is azimuthal, from north to south, i.e. the −θ direction. The gradient of the ˆ ˆ magnetic field is clearly in the radial direction, so we have B ×∇B ∼ ˆr × θ = φ, which is eastward. 4
5
6 3
6 2
5
6 3
This is for the electrons. The ions, which come without the minus sign to cancel the minus sign in ˆ which is westward. equation 47, will go in -φ, c) Well, it has to travel a distance L = 2π(5Re ), with the velocity v ∇B,e . So, T =
L v∇B,e
=
2π
6
× 5 × 6.4 × 10 ≈ 4.5 hours 1.3 × 10
4
(51)
d) The current density is given by J = nev, so, using the grad-B velocity in this expression, we get J = nev ∇B,e = 107
× 1.6 × 10− × 1.3 × 10 19
4
=2
× 10−
8
A/m2
(52)
2-10. A 20 keV deuteron in a large mirror fusion device has a pitch angle θ = 45o at the mid plane, where B = 0.7 T . Compute it’s Larmor radius. The Larmor radius is given by mv⊥ rL = (53) qB In natural units, a deuteron has m = 2 and q = +1. Furthermore, v⊥ = vsin(θ). To find v, we convert the energy to velocity:
−
rL =
m eB
10−27
2E 2 1.6 sin(θ) = m 1.6 10−19
× × × × 0.7
× 2
× × 1.6 × 10− × × 10−
20 103 2 1.6
19
27
sin(45o ) = 0.3 m
(54)
2-12. A cosmic ray proton is trapped between two moving magnetic mirrors with Rm = 5 and initially has W = 1 keV and v⊥ = v || at the mid plane. Each mirror moves toward the mid plane with a velocity vm = 10 km/sec (Fig. 2-10). a) Using the loss cone formula and the invariance of µ, find the energy to which the proton will be accelerated before it escapes. b) How long will take to reach that energy? 1. Treat the mirrors as flat pistons and show that the velocity gained at each bounce is 2vm . 2. Compute the number of bounces necessary. 3. Compute the time T it takes to traverse L = 1010 km that many times. Factor of two accuracy will suffice. a) The loss cone formula is sin2 (θm ) = 1/Rm , where θm is the angle of the magnetic mirror and Rm is the mirror ratio Bmax /B0 . We can also write down a formula for sin(θm ) in terms of the parallel and perpendicular velocities: sin(θm ) =
√ R1
m
=
v⊥,f
2 2 v⊥,f + v||,f
=
1
v
1 + ( v⊥||,f )2 ,f
(55)
Squaring both sides, and noting that, since µ is invariant, v ⊥,f = v ⊥,i , we have 1 1 1 = v||,f 2 = Rm 5 1 + ( v⊥,i )
⇒
v||,f = 2v⊥,i
(56)
Thus, we can get the final energy: 1 1 5 2 2 2 2 2 W f = m(v⊥,f + v||,f ) = m(v⊥,i + 4v⊥,i ) = mv⊥,i 2 2 2
(57)
But we can’t evaluate this without knowing what the original v⊥,i is. Fortunately, we know that initially v ⊥ = v || , so 1 1 2 2 2 2 m(v⊥,i + v||,i ) = m(2v⊥,i ) = mv ⊥,i = 1 keV 2 2
W i =
(58)
So, finally,
5 W f = W i = 2.5 keV (59) 2 b) In the frame of the piston, when the particle bounces off, it’s velocity doesn’t change. In the piston’s frame, the velocity of the particle as it is coming in is v i vm , where v m is the velocity of the piston (it is negative). It’s final velocity is the same but negative. Thus,
−
vi = v i
vf = v m
− vm,
− vi
(60)
where the prime denotes the velocity in the piston’s reference frame. But, in the lab frame, vf = v f + vm , so we have vf = 2vm vi (61)
−
Thus, the change in velocity on each bounce is 2vm = 20 km/sec. The initial proton velocity is vi =
2W i = m
× 2
1
× 10 × 1.6 × 10− 1.6 × 10− 3
27
19
= 447 km/s
(62)
The proton final energy is 2.5 keV . This corresponds to a velocity of
×
× 10 × 1.6 × 10− = 707 km/s (63) 1.6 × 10− Thus, the total change in velocity needed is ∆vtot = 707 − 447 km/s = 260 km/s. This corresponds vf =
2W f = m
to
N bounces =
2
2.5
3
19
27
∆vtot 260 = = 13 bounces ∆v1bounce 20
(64)
We can neglect the distance the mirrors move in the time the particle travels the distance, since vm v proton . Thus, the time it takes to travel a distance N bounces L is, using v¯ proton = (vf vi )/2,
−
T =
N bounces L 2 13 1010 = = 109 s v¯ proton 707 447
× × −
≈ 32 years
(65)
2-13. Derive the result of Problem 2-12(b) directly by using the invariance of J . a) Let v|| ds v|| L and differentiate with respect to time. b) From this, get an expression for T in terms of dL/dt. Set dL/dt = 2vm to obtain the answer. a) The quantity J = ab v || ds is invariant. Thus, is approximate it as v || L, then it’s time derivative must be 0: d (v L) = Lv˙ || + v|| ˙L = 0 (66) dt ||
≈
−
b) We can solve this expression: Lv˙ || =
−v|| ˙L ⇒
dv|| L = 2v|| vm dt
(67)
2-14. In plasma heating by adiabatic compression, the invariance of µ requires that kT ⊥ increases as B increases. The magnetic field, however, cannot accelerate particles because the Lorentz force q v B is always perpendicular to the velocity. How do the particles gain energy? Maxwell tells us that an electric field will be induced by a changing magnetic field. The induced electric field is what accelerates the particles.
×
4-1. The oscillating density n1 and potential φ1 in a “drift wave” are related by n1 eφ1 ω∗ + ia = n0 kT e ω + ia
(68)
where it is only necessary to know that all the other symbols (except i) stand for positive constants. a) Find an expression for the phase δ of φ1 relative to n1 . (For simplicity, assume that n1 is real.) b) If ω < ω ∗ , does φ1 lead or lag n1 ? a) Solving for φ 1 leads to φ1 =
ω + ia n1 kT e n1 kT e (ω + ia)(ω∗ ia) n1 kT e ωω∗ + a2 + i(aω∗ = = ω∗ + ia n0 e n0 e ω∗2 + a2 n0 e ω∗2 + a2
−
Now, in a drift wave we can suppose that φ1 Im(φ1 )/Re(φ1 ). We have Re(φ1 ) =
− aω)
(69)
∼ exp(iδ ), which in turns tells us that tan(δ ) =
n1 kT e ωω∗ + a2 n1 kT e a ω∗ ω ; Im(φ1 ) = 2 2 n0 e ω∗ + a n0 e ω∗2 + a2
−
Thus, δ = tan −1 (a
ω∗ ω ) ωω∗ + a2
−
(70)
(71)
b) For ω < ω∗ , δ > 0. We can set the phase of n1 to be 0, since all that matters is a phase difference. Thus, φ 1 lags n1 . 4.2 Calculate the plasma frequency with the ion motions included, thus justifying our assumption that the ions are fixed. (Hint: include the term n1i in Poisson’s equation and use the ion equations of motion and continuity. We will use Gauss’s Law, Fourier transforming the field and charge perturbations into plane waves of the form x = x0 + x1 , where x is any quantity, vector or scalar. The subscript 0 indicates the equilibrium value, and the subscript 1 indicates the perturbation. We only keep terms of to first order in small quantities. ρ e(ni ne ) E = ik E1 = (72) 0 0
∇·
⇒
−
·
Similarly, the equation of motion for the electrons, me
dve = dt
mi
dvi = eE dt
ions,
−eE ⇒ ⇒
iωme ve = eE1
iωmi vi =
−eE
1
(73)
(74)
and continuity equation for electrons, ∂n e + ∂t
∇ · (neve) = 0 ⇒ −ωne + ne k · ve = 0 1
(75)
0
and ions
∂n i + (ni vi ) = 0 ωn i1 + ni0 k vi = 0 (76) ∂t We now have 9 equations and 9 unknowns. I will skip the boring algebra. Solving for ω yields
∇·
⇒ −
ω 2 = ω p2 + Ω p2
·
(77)
2
ni e where Ω p = m is the ion plasma frequency. Clearly, omitting the ion plasma frequency from i the calculation is justified since m i me . 0
4.4 By writing the linearized Poisson’s equation used in the derivation of simple plasma oscillations in the form (E) = 0 derive an expression for the dielectric constant applicable to high-frequency longitudinal motions. Fourier transform Poisson’s equation:
∇ ·
·
ρ e = n1 0 0
iωn1
− n ik · v = 0 ⇒
ik E =
(78)
We also know that ∂n 1 + n0 ∂t
∇ · v = 0 ⇒ 1
and we know Newton’s Law: m
∂ v = ∂t
0
−eE ⇒
v = i
n1 = n0
e E mω
k v ω
·
(79)
(80)
Thus, equation (78) gives us k E =
·
−
ie e2 n0 n0 k v = k E 0 ω 0 ω 2 m
·
·
We can do a trick here, and pull everything over to the left side. Writing ω p2
2
(81)
≡ e n /m
k E(1
·
−
ω p2 ) =0 ω2
⇒ ∇ · {(1 −
and thus we obtain = 1
−
ω p2 ω2
0
0
ω p2 )E = 0 ω2
(82)
}
(83)
4.6 a) Compute the effect of collisional damping on the propagation of Langmuir waves (plasma oscillations), by adding a term mnν v to the electron equation of motion and rederiving the dispersion relation for T e = 0. b) Write an explicit expression for Im(ω) and show that its sign indicates that the wave is damped in time.
−
a) The cold electron equations of motion are mne and
∂n e + ∂t
∂ v = ∂t
−eneE − mneν v ⇒
∇ · (nev) = 0 ⇒
iωne
iωmv =
−eE − mν v
− inek · v = 0 ⇒
k v = ω
·
(84)
(85)
We also have Gauss’s law:
∇ · E = − nee ⇒ 0
k E = i
·
ne e m = i ω p2 0 e
(86)
We will dot equation (84) with k: k E = k v
·
·
(iωm
− mν )
e
(87)
Plugging in equations (86) and (87), we obtain m 2 (iω 2 m mων ) i ω p = e e
−
⇒
ω p2 = ω 2 + ων
(88)
So we see that if we include collisions, the oscillation frequency is different from the plasma frequency. b) Lets let ω = ω R + iωI . Then expression (88) becomes 2 ω p2 = ω R
2
− ωI + 2iωRωI + iωI ν + ωRν
This means that 2ωR ωI + ωI ν = 0
⇒
ωI =
− ν 2
(89)
(90)
Now we suppose a plane wave solution for the field quantities, i.e.
∝ e−iωt
E
(91)
we obtain
∝ e−iω
E
R t+ωI t
⇒
∝ e−iω
E
Thus, the wave is exponentially attenuated in time.
Rt
e−
νt 2
(92)
