Exercises of Plasma Physics MEFT - Master in Engineering Physics Vasco Guerra February 2017 Foreword This collection of exercises is an outcome of my teaching experience as the responsible of the Plasma Physics and Technology course from the Integrated Master in Engineering Physics of Instituto Superior Técnico, Universidade de Lisboa, in the period between 2013 and 2017. This is a one semester introductory course, attended by all students of the Master. As such, it is not intended to be a comprehensive course for students that will follow a plasma physics track. Nevertheless, the course is designed to give a general overview of plasma physics, addressing the basic concepts and the main approaches to study the field, namely single particle motion, fluid descriptions (both two fluid and single fluid) and kinetic theory. Each chapter corresponds to exercises that can be used as a one-week problem sheet. Due to the broad scope of the course, it is impossible to cover in detail all the topics studied within the allocated time. Therefore, various exercises were conceived in order to make students have a glance at some relatively standard phenomena not studied in class, but that can be investigated with the knowledge already acquired, such as the ponderomotive force, the two-stream instability and the derivation of the fluid equations from the moments of Vlasov’s equation. Some of these problems would be rather challenging without guidance, but with the hints included students are expected to succeed in solving them. Various exercises are taken or adapted from the textbook of Francis Chen (F.F. Chen, Introduction to Plasma Physics and Controlled Fusion, Vol. 1, Plenum Press 1984). All these exercises are duly identified along the text. Other exercises were adapted from other books and sources available on the web. The book of Dwight Nicholson (D.R. Nicholson, Introduction to plasma theory, John Wiley & Sons 1983) and the very good online collection by John Howard, from the Australian National University, deserve a special reference. To these problems I have added a significant number of my own. Finally, several exercises used in actual written examinations are included in this collection. They are identified with a (?) mark, which allows assessing the intended level of the course. Lisboa, February 2017 Vasco Guerra 4 Formulae • Constants: Electron mass Electron charge me = 9.1 × 10−31 kg; e = 1.6 × 10−19 C; Vacuum permittivity 0 = 8.854 × 10−12 F/m Boltzmann constant kB = 1.38 × 10−23 J/K Earth’s radius RT ' 6370 km • Conversion factors: 1 u = 1.66 × 10−27 kg; 1 eV = 1.6 × 10−19 J; 1 atm = 760 Torr 5 1 atm = 1.013 × 10 Pa • Mathematical relations: ~a × (~b × ~c) = (~a · ~c)~b − (~a · ~b)~c ~ × (∇ ~ A) ~ =0 ∇ ~ × (∇ ~ × A) ~ = ∇( ~ ∇ ~ · A) ~ − ∇2 A ~; ∇ ~ × (ψ A) ~ = ψ(∇ ~ × A) ~ + (∇ψ) ~ ~; ∇ ×A ˆ ∞ r 2 exp(−Ax )dx = −∞ +∞ ˆ 2 2 π A x exp(−Ax )dx = 0 √ π 4A3/2 1 d2 1 d 2 dφ In spherical coordinates and simmetry, ∇ φ = 2 r = (rφ) r dr dr r dr2 d2 φ 1 dφ In cylindrical coordinates and symmetry, ∇2 φ = 2 + dr r dr 2 6 • Basic relations and fundamental effects Ideal gas law Plasma frequency P V = N kB T q λe = ε0nkeBe2Te q 1 r φ(r) = 4π exp − r λ 0 D q ne e2 ωpe = ε0 me Plasma parameter Λ = ne λ3D Electron Debye length Debye potential (spherical symmtry) Electron cyclotron frequency ωce = Larmor radius eB me v⊥ ωc rL = q vt = kBmT Thermal speed • Drifts ExB drift Grad B drift Curvature drift Fields in vacuum Polarization drift Diamagnetic drift Magnetic moment ~ ×B ~ E 2 B 2 ~ ~ mv⊥ B × ∇B ~vd = 2qB B2 2 ~ mvk ~ur × B ~vd = 2 qB Rc ~ 1 2 1 ~ur × B 2 ~vd = mvk + v⊥ 2 2 qB Rc m d ~ ~vd = E⊥ qB 2 dt ~ ×B ~ 1 ∇P ~vd = − qn B 2 1 mv 2 µ= 2 ⊥ B ~vd = • Waves – Electrostatic electron waves ~ 0 = 0 ou ~k k B ~ 0: ∗ B 2 ω 2 = ωpe + 3k 2 vt2 ; vt2 = kT /m (Langmuir waves) 2 2 ~ 0 : ω 2 = ωpe ∗ ~k ⊥ B + ωce = ωh2 ; (upper hybrid waves) – Ion electrostatic waves ~ 0 = 0 ou ~k k B ~ 0: ∗ B i kB Ti ω 2 = k 2 c2s ; c2s = γe kB Tem+γ (ion acoustic waves) i γe k B T e 1 2 2 γi kB Ti ω =k + mi 1+γe k2 λ2 (ion plasma waves) mi De ~ 0: ∗ ~k ⊥ B ω 2 = k 2 c2s + ωl2 ; 2 ω 2 = k 2 c2s + ωci ωl2 = ωce ωci (lower hybrid oscillations) (ion cyclotronic waves) 7 – Electron electromagnetic waves 2 ~ 0 = 0: ω 2 = ωpe ∗ B + k 2 c2 ~ 0, E ~1 k B ~ 0: ∗ ~k ⊥ B ~ 0, E ~1 ⊥ B ~ 0: ∗ ~k ⊥ B ~ 0: ∗ ~k k B n2 = ; (electromagnetic waves) 2 2 c k ω2 n2 = 1 − n2R,L = 1 − 2 ωpe /ω 2 1∓ωce /ω ; 2 ωpe ω2 ; 2 ω 2 −ωpe 2 ; 2 ω −ωh =1− (ordinary wave) 2 ωpe ω2 (extraordinary wave) (right and left waves) • Transport and MHD ∂ ~ · (ns~vs ) = 0 ns + ∇ ∂t h i ∂~vs ~ v~s = qs ns E ~ + ~vs × B − ∇P ~ s − νs0 ns ms (~vs − ~v0 ) + (~vs · ∇) ns m s ∂t ν = N hσvi ~Γ = nµE ~ − D∇n ~ ρm ∂~v ~ − ∇P ~ = J~ × B ∂t ~Γ = n~v ~ + ~v × B ~ = η J~ E P = Pe + Pi • Maxwell’s equations ~ ·B ~ =0; ∇ ~ ·E ~ = ρ ; ∇ ε0 ~ ~ ×B ~ = µ0 J~ + 1 ∂ E ∇ 2 c ∂t ~ ∂ B ~ ×E ~ =− ∇ ∂t • Kinetic theory ˆ n(~r, t) = f (~r, ~v , t)d3 v ; ~ = h~v i = 1 V n ˆ ~v f (~r, ~v , t)d3 v ∂f ~ r f + q (E ~ + ~v × B) ~ ·∇ ~ v f = 0 (Vlasov eq.) + ~v · ∇ ∂t m 8 CHAPTER 1 Debye shielding and fundamental effects 1. Calculate the number density of an ideal gas at: (a) p = 1 atm and T = 273 K (Loschmidth number) (b) p = 1 Torr and T = 300 K 2. (F. F. Chen) On a log-log plot of n (cm−3 ) vs kB Te (eV) draw lines of constant λD and Λ. On the graph place the following points: (a) Tokamak (Te ' 1 keV, ne ' 1013 cm−3 ) (b) Solar wind near the Earth (Te ' 10 eV, ne ' 10 cm−3 ) (c) Ionosphere, ∼ 300 km above Earth’s surface (Te ' 0.1 eV, ne ' 106 cm−3 ) (d) Laser fusion (Te ' 1 keV, ne ' 1020 cm−3 ) (e) Gaseous electronics (Te ' 1 eV, ne ' 1010 cm−3 ) (f) Interstellar medium (Te ' 1 eV, ne ' 10−1 cm−3 ) (g) Flame (Te ' 0.1 eV, ne ' 108 cm−3 ) 3. Consider Debye’s potential created by a punctual test charge qT that is placed inside an homogeneous plasma. (a) Show that the charge in the shielding cloud exactly cancels qT . Calculate the total charge inside spheres of radii λD /2, λD and 5λD . (b) Determine the electrostatic interaction energy between the test charge and the particles in the plasma and the total mean energy of the plasma particles (assume Te = Ti = T ). 4. Consider a homogeneous plasma with density n0 = 108 cm−3 occupying the region x > 0. Outside the plasma (x < 0) there exists an uniform electric ~ = 100~ux V/cm, which penetrates the plasma. The electron and ion field E temperatures are equal, Te = Ti = 0.1 eV. Show that the plasma shields the 10 Chapter 1. Debye shielding and fundamental effects field and calculate the typical shielding length. Calculate the intensity of the electric field at x = 0.5 cm, assuming that eφ(x)/kB T 1. 5. (F. F. Chen) A spherical conductor of radius R 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 eφ0 kB Te : (a) derive an expression for the potential as a function of r; (b) calculate the charge in the sphere; (c) calculate the sphere capacity for R = 10 cm, Te = 1 keV and n0 = 1014 and 106 cm−3 , and show that for high electron densities the plasma behaves as a dielectric. 6. In the deduction of the electron plasma frequency, suppose the ions are not infinitely massive, but have a mass mi and can move. Modify the discussion to show that the coupled oscillation of the electron and ion “slabs” is made 2 2 + ωpi ). with the total plasma frequency (ωp2 = ωpe 7. (∗) In this problem we want to calculate the plasma oscillation frequency for a spherical plasma, proceeding in a similar way as it was done for the slab configuration in the previous exercise. Consider a spherical plasma of radius R, represented by a uniform positive ion background of density n0 inside the sphere. Assume the ions are infinitely massive. Initially, the electron density has the same volume distribution as that of the ions. The “electron sphere” is then stretched to a radius R+δr and then released. Assume at all instants that the electron density is distributed uniformly on the spherical volume it occupies. 8. An infinite conducting plane is placed inside an homogeneous plasma and charged to a potential φ0 . The electrons move and keep a Boltzmann distribution, with eφ/kB Te 1, while the ions can be considered stationary for the time-scale of the experiment. Consider the xx direction perpendicular to the plane and x = 0 coinciding with the plane. (a) Obtain the potential as a function of x and represent φ(x) (b) Determine the electric field as a function of x and the charge density in the plane. Represent E(x) and compare with the solution in vacuum. (c) Show that the plasma completely shields the charge in the conducting plane. 9. (?) Consider an infinite line, uniformly charged with a linear charge density λ, immersed in a homogeneous plasma. The electrons and ions follow Maxwellian distributions, respectively with temperature Te and Ti . (a) Show that the electrostatic potential can written, in cylindrical co be λ r ordinates, in the form φ(r) = 2πε0 K0 λD , where K0 is the modified Bessel function of second kind of order zero, and determine λD . Derive the expression for the electric field as a function of r. (b) Calculate the total charge around the line, per unit length. 11 The modified Bessel equation of order α is x2 d2 y dy + x − (x2 + α2 )y = 0 . dx2 dx The solution is a linear combination of modified Bessel functions of first and second kind, Iα (x) e Kα (x), which are exponentially growing and decreasing functions, respectively (see figure). ˆ K00 (x) = −K1 (x) +∞ xK0 (x)dx = 1 0 For small x, K0 (x) ' − ln x; K1 (x) ' 1 x 10. (?) Dusty plasmas. In many plasmas there can be found large particles, besides electrons and positive ions, known as dust particles. The aim is to study Debye shielding around a positive test charge qT in dusty plasmas. Assume that electrons and positive ions follow a Boltzmann distribution at temperatures Te e Ti , respectively, while the dust particles are infinitely massive, have a total charge Zd and are uniformly distributed on the volume. (a) Do you expect the charge Zd to be positive, negative, or that it can have any of the signs? Justify. (b) Consider a quasi-neutral plasma where the dust particles are negatively charged. Show that Poisson’s equation can be written in the form ne0 e2 ni0 Te 2 ∇ φ= 1+ φ, 0 kB Te ne0 Ti where ne0 and ni0 are the non-perturbed densities of electrons and ions, i.e., their densities at a large distance from the test charge. (c) Determine the Debye length and tell if it is larger, smaller or equal to the case where there are no dust particles. (d) Obtain the expression for the electrostatic potential φ(r). Historical note: this problem was studied by Lakhsmi, Bharuthram and Shukla in Astrophys. Space Sci. 209 (1993) 213. Dust particles have been observed in asteroid regions, planetary atmospheres (Earth and Titan), comet tails and several laboratory plasmas. 11. (?) Plasma sheaths A plasma is confined between two grounded (φ = 0) parallel plates located at x = 0 and x = l. The ion density is ni (x) = n0 for 0 < x < l. The electron 12 Chapter 1. Debye shielding and fundamental effects density is ne (x) = n0 , 0, if s < x < l − s for 0 < x < s and l − s < x < l . The regions 0 < x < s and l − s < x < l are called the “sheath” regions. (a) Determine the potential φ(x) and the electric field E(x) for 0 < x < l. Find φ0 = φ(x = l/2) and plot the potential and the electric field as a function of x. [Hint: consider the symmetry of the problem and recall that the electric field must be continuous] (b) Does the electric field at the sheaths act to confine the electrons within the bulk plasma or does it tend to destroy the confinement? (c) Chose eφ0 = 5kB Te and find an expression for s. Discuss the plausibility of the value given for φ0 . CHAPTER 2 Single particle motion I 2 1. For particles with the same kinetic energy W = mv⊥ /2, compute the ratio between the Larmor radius of a proton and an electron (mp /me = 1836). 2. Consider a particle of charge q > 0 and mass m, initially at rest at (x, y, z) = ~ = B0 ~uz and E ~ = E0 ~uy , (0, 0, 0), in the presence of a static magnetic field B with E0 , B0 > 0. (a) Sketch the orbit of the particle. (b) Derive an exact expression for the orbit of the particle. (c) Show that the orbit can be separated into an oscillatory term and a constant drift term. After averaging in time over the oscillatory motion, is there any net acceleration? If not, how are the forces balanced? (d) In a neutral plasma, with positive and negative particles and ions of different masses, would there be any net current? (e) Suppose the electric field were replaced by a gravitational force in the yy direction, would there be a net current? 3. (F. F. Chen) An electron beam with density ne = 1014 m−3 and radius R = 1 ~ = B0 ~uz , where B0 = 2 cm crosses a region with a uniform magnetic field B T and the zz axis is aligned with the direction of propagation of the beam. ~ ×B ~ drift at r = R (note Determine the direction and magnitude of the E ~ that E is the electrostatic field created by the charge of the beam). 4. (F. F. Chen) Suppose electrons obey the Boltzmann relation in a cylindrical symmetric plasma column, ne (r) = n0 exp(eφ/kTe ). The electron density varies with a scale length λ, i.e., ∂ne /∂r ' −ne /λ. ~ = −∇φ, ~ find the radial electric field for given λ. (a) Using E ~ ×B ~ (b) For electrons, show that rL = 2λ when the E p drift velocity, vE , is equal to the thermal speed, defined here as vt = 2kTe /m (this means 14 Chapter 2. Single particle motion I ~ ×B ~ drift that the finite Larmor radius effects are important if the E velocity is of the order of the thermal speed).1 5. (F. F. Chen) Suppose the earth’s magnetic field is 3 × 10−5 T at the equator and falls off as 1/r3 as in 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 equator plane. ~ drift velocities. (a) Compute the ion and electron ∇B (b) Does an electron drift eastward or westward? (c) How long does an electron take to encircle the earth? (d) Compute the current ring density in A/m2 . Note: the curvature drift is non-neglible... but neglect it anyway. 6. (?) Hall thrusters are widely used in space propulsion. A stationary plasma thruster (SPT) is schematically represented in the figure.2 The thruster has a cylindrical shape, with an open chamber defined by inner (Ri ) and outer (Re ) radii and height L, where the anode is placed. In the chamber there is a magnetic field, pointing from Ri to Re . An axial electric field points outwards from the anode. The thruster experiences a propelling force if it is able to eject positive ions along the direction of the electric field. Xenon is injected into the chamber, and electrons coming from the cathode ionize the Xe atoms, creating new electron/ion pairs. Our purpose is to study the motion of electrons and ions created by ionisation of Xe in the thruster. Consider Re = 5 cm, Ri = 3 cm and L = 30 cm. The fields in the chamber are approximately E = 5kV /m and B = 5 mT, and the mass of Xe is 131.3 u (1u = 1.66 × 10−24 g). (a) Assuming that the electrons are created with a speed perpendicular to ~ ~v⊥ , describe qualitatively their motion and draw schematically their B, ~ trajectory (neglect any possible curvature, ∇B and centrifugal force drifts). (b) A simple image of the thrust operation can be obtained by calculating the electron and ion Larmor radii, rLe and rLi . Assuming the velocity of ions and electrons to be, respectively, v⊥,i = 100 m/s (ions are formed from ionisation of the injected neutral Xe atoms) and v⊥,e = 1 × 104 m/s (electrons coming from the cathode are accelerated from the E 1 Notice the definition of the thermal speed from Chen’s book, a factor of defined in this text. 2 M. Keidar and I. I. Beilis, IEEE Trans. Plasma Sci. 34 (2006) 804 √ 2 larger than 15 field), calculate rLe and rLi and compare it with the relevant thruster dimensions. Does the thruster experience a propelling force in these conditions? (c) Solve the equations of motion for the electrons created in an ionising collision, assuming they are created with zero speed. Consider a plane geometry (i.e., solve the equations in cartesian axes), with the xx-axis ~ Determine the amplitude of oscillation in the xx direction. along E. (d) In the conditions of c., what would be the amplitude of oscillation of the positive ions in the xx direction, if they were created withe zero speed? Comment the results of c. and d.. Note: a tutorial on the physics and modelling of Hall thrusters is presented by J. P. Boeuf, J. Appl. Phys. 121 (2017) 011101. 7. (?) Consider a particle of charge q, initially at rest and placed at xi = qE0 ~ − mω 2 , that moves under the effect of a high-frequency electric field, E = E0 cos(ωt)~ux . (a) Solve the equation of motion and describe the trajectory. (b) Assume now that the amplitude of the electric field slowly varies in space, E(x, t) = E0 (x) cos(ωt), where E0 is a growing function of x. i. Describe qualitatively the trajectory. [Suggestion: think on the values of the field E0 on the turning points of the trajectory] ii. To solve for the trajectory, decompose the motion on a slowly varying component, x0 , denoted as oscillation center, and a highfrequency component x1 , where x1 is measured from the oscillation centre, x = x0 + x1 (note that this is analogous to the decomposition of motion in the guiding centre motion and an oscillating component, used to describe motion on a magnetic field). Expand E0 (x) around x0 and: A. define condition of validity of the expansion and show that the equation of motion can be written in the form dE0 (x0 ) m(ẍ0 + ẍ1 ) = q E0 (x0 ) + x1 cos(ωt) ; dx B. start from the equation of motion to justify that, approximately, ẍ1 = qE0 cos(ωt) m and solve for x1 (t); C. start from the equation of motion to show that the oscillation centre feels a force Fp = − q2 d 2 E . 4mω 2 dx 0 [Suggestion: start by taking the temporal average of the equation of motion on the sort time 2π/ω and note that hxi = x0 ] 16 Chapter 2. Single particle motion I Note: the ponderomotive force is very important in several applications and basic research phenomena, such as the study of the interaction of intense lasers with dense plasmas and plasma acceleration. CHAPTER 3 Single particle motion II 1. (F. F. Chen, Fermi acceleration of cosmic rays). A cosmic ray proton is trapped between two moving magnetic mirrors with mirror ratio Rm = 5. Initially its energy is W = 1 keV and v⊥ = vk at the mid-plane. Each mirror moves toward the mid-plane with a velocity vm = 10 km/s and the initial distance between the mirrors is L = 1010 km. (a) Using the invariance of µ, find the energy to which the proton is accelerated before it escapes. (b) How long does it take to reach that energy? [Suggestions: i) suppose that the B field is approximately uniform in the space between the mirrors and changes abruptly near the mirrors, i.e., treat each mirror as a flat piston and show that the velocity gained at each bounce is 2vm ; ii) compute the number of bounces necessary; iii) assume that the distance between the mirrors does not change appreciably during the acceleration process.] 2. (F. F. Chen) A plasma with an isotropic distribution of speeds is placed inside a magnetic mirror with mirror ratio Rm = 4. There are no collisions, so that the particles in the loss cone escape, while the others remain trapped. Calculate the fraction of particles that remains trapped. 3. (F. F. Chen) The magnetic field along the axis of a magnetic mirror is B( z) = B0 (1 + α2 z 2 ), where α is a constant. Suppose that at z = 0 an electron has 2 . velocity v 2 = 3vk2 = 32 v⊥ (a) Describe qualitatively the electron motion. (b) Determine the values of z where the electron is reflected. (c) Write the equation of motion of the guiding center for the direction ~ and show that there is a sinusoidal oscillation. Calcule the parallel to B frequency of the motion as a function of v. 18 Chapter 3. Single particle motion II ~ = −Ėt~uy , 4. Consider a particle moving in a time-dependent electric field E ~ where Ė is a constant, and a uniform magnetic field B = B0 ~uz . ~ ×B ~ drift. (a) Calculate the E (b) Relate the resulting accelerated drift with a force and verify that the drift due to that force is the polarization drift. 5. (F. F. Chen) A plasma is created in a toroidal chamber with average radius R = 10 cm and square cross section of size a = 1 cm. The magnetic fiel is generated by an electrical current I along the symmetry axis. The plasma is Maxwellian with temperature kT = 100 eV and density n0 = 1019 m−3 . There is no applied electric field. ~ field, for both positive (a) Sketch the typical drift orbits in the non-uniform B ions and electrons with vk = 0. (b) Calculate the rate of charge accumulation (Coulomb per second) due to the curvature and gradient drifts on the upper part of the chamber. The magnetic field in the center of the chamber is 1 T and you can use the approximation R a if necessary. ~ = E0 exp(iωt) ~ux , 6. (?) Consider an electron moving in an oscillating electric field, E ~ perpendicular to a constant and uniform magnetic field, B = B ~uz . (a) Calculate the drifts existing on the particle motion and describe qualitatively the motion. (b) Try now to confirm the results you have already obtained, starting directly from the equations of motion. In particular, show that you can indeed recover the results from a) for low frequencies of the field, i.e., ω ωce [Suggestion: i) search for solutions of the form ~v = ~vk +~vL +~vD exp(iωt), where ~vL is the velocity of the cyclotron motion and ~vD is constant and ~ ii) verify you can obtain an equation for ~vD in the perpendicular to B; ~ 0 − ev~D × B; ~ iii) make the cross product with B ~ form iωm~vD = −eE ~ and eliminate ~vD × B]. CHAPTER 4 Fluid drifts 1. (F. F. Chen 3.6) An isothermal plasma is confined between the planes x = ~ = B0 ~uz . The density distribution is n(x) = ±a in a magnetic field B 2 2 n0 1 − x /a . (a) Derive an expression for the electron diamagnetic drift velocity, as a function of x. (b) Draw a diagram showing the density profile and the direction of the ~ points out electron diamagnetic drift on both sides of the midplane, if B of the paper. (c) Evaluate vD at x = a/2, if B = 0.2 T, kTe = 2 eV and a = 4 cm. ~ field 2. (F. F. Chen 3.7) A cylindrically symmetric plasma in a uniform B column 2 has n(r) = n0 exp − rr2 0 e ni = ne = n0 exp eϕ kTe . ~ ×B ~ (~vE ) and electron diamagnetic drifts (~vDe ) ãare (a) Show that the E equal in magnitude and have opposite directions. (b) Show that the plasma rotates as a rigid body. (c) In the reference frame that rotates with velocity ~vE there are drift waves that propagate with speed vϕ = 0.5vDe . What is vϕ in the laboratory frame? Represent on a r − θ diagram the directions and relative magnitudes of ~vE , ~vDe and ~vϕ in the lab frame. (d) Obtain the diamagnetic current, J~D , as a function of r. (e) Calculate JD for B = 0.4 T, n0 = 1016 m−3 , kTe = kTi = 0.25 eV and r = r0 = 1 cm. 3. A cylindrical plasma column of an isothermal plasma of radius R = 8 mm and equal ion and electron temperatures, kB T = 5 eV, is immersed on on a magnetic field B = 0, 6 T, aligned with the cylinder axis (coincident with the zz axis). The density has a profile n(r) = n0 J0 2, 4 Rr , where J0 is 20 Chapter 4. Fluid drifts 12 −3 the Bessel function of first kind of order zero and n0 = 10 0 cm . Assume eϕ you can consider ni = ne = n = n0 exp kT . Note: J0 (x) = −J1 (x), J0 (1, 2) ' 0, 67; J1 (1, 2) ' 0, 49. (a) Obtain the expressions for the ion and electron diamagnetic drift as a function of r. Justify qualitatively the direction of the drifts. (b) Calculate the diamagnetic current density at r = R/2 (value and direction). CHAPTER 5 Waves in non-magnetized plasmas 1. (F. F. Chen 4.6) Compute the effect of collisional damping on the propagation of Langmuir waves, by adding a term −mnν~v to the electron equation of motion and rederiving the dispersion relation for Te = 0 (plasma oscillations). Show that the wave is damped in time. 2. (F. F. Chen 4.13) An 8 mm microwave interferometer is used on an infinite plane-parallel plasma slab 8 cm thick. (a) What is the plasma density if a phase shift of 1/10 fringe is observed? Assume a uniform density and note that one fringe corresponds to a 360o phase shift. (b) Show that if the phase shift is small, then it is proportional to the density. 3. (Exam 2015/2016) The international space station (ISS) orbits approximately 400 kms above the surface of the Earth. The average profile of the electron density in the ionosphere is shown in the figure. If the astronauts in the ISS want to communicate with the Earth, in which range of frequencies shall they tune their radios? Note: Anyone can communicate by radio with the ISS astronauts. The details and the frequencies actually used can be found in NASA’s webpage (http://spaceflight.nasa.gov/station/reference/radio/) 22 Chapter 5. Waves in non-magnetized plasmas 4. (F. F. Chen 4.10) Hannes Alfvén (Nobel Prize in Physics in 1970) has suggested that perhaps the primordial universe was symmetrical between matter and antimatter. Suppose that the universe was at one time a uniform mixture of protons, antiprotons, electrons and positrons, each species having a density n0 . (a) Obtain the dispersion relation for high-frequency electromagnetic waves in this plasma, neglecting collisions, annihilations and thermal effects. (b) Obtain the dispersion relation for ion waves. Use Poisson’s equation, neglect Ti (but not Te ) and assume that all leptons follow the Boltzmann relation. 5. (Exam 2013/2014) We want to study electrostatic longitudinal waves in a non-magnetized plasma. Consider a unidimensional problem and that you can neglect the thermal motion of the positive ions, but not of the electrons. (a) Show that the dispersion relation can be written in the form 2 2 ω 2 = ωpi + ωpe ω2 ω 2 − γe vt2 k 2 (b) Obtain and discuss the limiting case Te → 0 (c) Obtain and discuss the limiting case M → ∞ (d) Obtain and discuss the limiting case m → 0 (e) Obtain and discuss the limiting case m → 0 e λD k 1 6. (Exam 2013/2014, two-stream instability ) Consider the one-dimensional propagation of waves in a cold (Te = Ti = 0) non magnetized plasma, where the ions are initially stationary (i.e., the zeroth order term of the ionic velocity is zero), but the electrons travel at speed v0 (i.e., the zeroth order term of the electronic speed is v0 ). Neglect the effect of the collisions. 23 (a) Use the two-fluid equations and the Poisson’s equation to show that the dielectric constant of the plasma can be written in the form (k, ω) = 1 − 2 2 ωpi ωpe − . ω2 (ω − kv0 )2 (b) Verify that the dispersion relation is a polynomial function of fourth order (so that for each real value of k there are four solutions for ω). ω2 ω2 pe Sketch approximately the function f (ω) = ωpi2 + (ω−kv 2 for a fixed k 0) and mark on the graph where the four roots are ω [you do not need to give the exact values, we are only interested in understanding the form of the function]. (c) In some situations the dispersion relation has only two real roots, which happens for small enough kv0 (convince yourself this is the case, by looking at the graph you have just drawn). In that case, one of the imaginary roots corresponds to an unstable wave, growing exponentially in time. Show that, if kv0 = ωpe ω, the instability growth rate is 1/3 √ me 1 ωpe s−1 [Suggestion: start by expanding the given by 23 21/3 mi last term of (k, ω) to the first order in ω/ωpe ]. (d) From the general relations in a) e b), derive the dispersion relation in the limit mi → ∞. Then obtain the limit v0 → 0 (while keeping mi → ∞). Comment the results. 7. (Exam 2015/2016) Consider a plasma formed by electrons and two species of positive ions, a light species (a) and a heavy species (b). We want to study the propagation of low-frequency electrostatic waves in this plasma. As the plasma is quasi-neutral, the non-perturbed electron and ion densities verify the relation ne0 = na0 + nb0 . (a) Justify why you can use the plasma approximation, neglect the electron inertia and consider isothermal electrons. (b) Write the relevant fluid equations and linearize them, keeping only the terms up to first order. (c) Show that the first order perturbations of the electron and ion-a densities are related by r n , na1 = ω2 ma Ta e1 k2 kB Te − γa Te where r = na0 /ne0 . (d) What is the relation between the first order perturbation on the electron and ion-b densities? (e) Show that the dispersion relation can be written as 1= ω2 k2 r + ma Ta kB Te − γa Te a (1 − r) m mb ω2 k2 ma kB Te a Tb − γb m mb Te (f) Verify that on the limit Ta = Tb = 0 we have ion acoustic waves, corresponding to an ion of effective mass M given by 1 r 1−r = + . M ma mb 24 Chapter 5. Waves in non-magnetized plasmas If both ion species have similar densities, which one determines the plasma behaviour? Comment the result. [Historical note: These waves were experimentally observed by Nakamura and Saitou, Plasma Phys. Control. Fusion (2003) 45 759; the case Ta , Tb 6= 0 gives two solutions, a fast acoustic wave and a slow acoustic wave and it is much more complex to analyse.] 8. (Exam 2016/2017) Consider an electromagnetic wave in a cold, non-magnetised plasma. Neglect the ion motion but consider the effect of collisions between electrons and neutrals, assuming a constant collision frequency ν. (a) Write the relevant system of equations to study this system. (b) Linearize the system of equations in the usual form and keep only the first order terms. (c) Show that the dispersion relation can be written in the form 2 ωpe c2 k 2 = 1 − . ω2 ω(ω + iν) (d) Assuming ν/ω 1, show that the skin depth (attenuation distance) is given by !1/2 2 ωpe 2c ω 2 1− 2 . 2 ν ωpe ω [Suggestion: use the dispersion relation and take a real frequency ω and an imaginary wavenumber k = iα + β.] 9. (Exam 2017/2018) In this problem we study classical and quantum ion acoustic waves. (a) Consider first classical ion acoustic waves, i.e., those studied in class. i. The dispersion relation for ion acoustic waves was derived under the plasma approximation. Explain what does this approximation mean in the study of linear waves in plasmas and why the dispersion relations obtained in this way should be valid for small values of k. ii. Show that the dispersion relation of ion plasma waves does reduce to that of ion acoustic waves in the limit k → 0. Explain which of the two terms on the r.h.s. of the dispersion relation must dominate for the expression to be valid. (b) We turn now our attention to the quantum case. Quantum hydrodynamic models (QHM) generalize the fluid equations for plasmas with the inclusion of a quantum correction term to the force balance equation. It is possible to show that, in the limit me /mi → 0 and for TF i < Ti < Te TF e , where TF s denotes the Fermi temperature of species s and the remaining quantities have their usual meaning, this equation leads to the following relationship between the potential φ and the electron density: 2 2 n2 eφ 1 1 cs ∂ √ = − + e2 − √ H 2 ne , (5.1) 2kB TF e 2 2n0 2 ne ωpi ∂x2 25 where the quantum sound speed is given by cs = 2kB TF e mi 1/2 and H= ~ωpe 2kB TF e is a dimensionless parameter characterizing the importance of quantum diffraction effects (it is the ratio between the plasmon energy and the Fermi energy). The relevant fluid equations are the ion continuity and momentum conservation equations, Poisson’s equation and equation (5.1). Neglect the ion thermal agitation. i. Write the four fluid equations in terms of the four unknowns ni , vi , ne and φ. ii. Show that the linearizarion of equation (5.1) leads to " 2 # 1 2 cs k eφ1 ne1 = 1+ H , 2kB TF e 4 ωpi n0 where n0 is the unperturbed electron (and ion) density. iii. Use the ion equations to derive ni1 k 2 eφ1 = 2 . n0 ω mi iv. Obtain the dispersion relation for the quantum ion-acoustic waves, 2 c2 k 2 2 ωpi 1 + H4 ωs 2 pi k2 . ω 2 = ω2 pi H 2 c2s k2 2 1 + 4 ω2 c2 + k s pi v. Discuss the limits of small and large wave numbers in when H → 0. [Historical note: the dynamics of a quantum electron gas is an important issue for a variety of physical systems, such as ordinary metals, semiconductors and astrophysical systems under extreme conditions (e.g., white dwarfs). This problem was studied by F. Haas et al., Phys. Plasmas 10 (2003) 3858.] 26 Chapter 5. Waves in non-magnetized plasmas CHAPTER 6 Waves in magnetized plasmas 1. (F. F. Chen 4.7) For the upper hybrid oscillations, show that the elliptical orbits are always elongated in the direction of ~k (hint: derive an expression for vx /vy ). 2. (F. F. Chen 4.22) Faraday rotation of an 8-mm wavelength microwave beam in a uniform plasma in a 0.1 T magnetic field is measured. The plane of polarization is found to be rotated 90o after traversing 1 m of plasma. What is the density? 3. (F. F. Chen 4.21) Show that in a positronium plasma, i.e., a neutral plasma of electrons and positron, there is no Faraday rotation [suggestion: write the system of linearized equation in matrix form, Ax = 0, and ask Mathematica .. for help to calculate det(A) ^]. 4. (F. F. Chen 4.25) A microwave interferometer employing the ordinary wave cannot be used above the cutoff density. To measure higher densities, one can use the extraordinary wave. (a) Write an expression for the cutoff density for the X wave. (b) On a vφ2 /c2 vs. ω diagram, show the branch of the X-wave dispersion relation on which such interferometer would work. 5. (Exam 2014/2015) We want to study electrostatic waves in an electronegative plasma, formed by electrons, positive ions and negative ions. The plasma is quasi-neutral, i.e., the non perturbed densities are n+0 = n0 , ne0 = (1 − ε)n0 e n−0 = εn0 , respectively for the positive ions electrons and negative ions. Assume that initially there is no electric field and all the fluid velocities are zero. The plasma is immersed on a constant and uniform magnetic field, ~ = B0 ~uz . Consider only longitudinal perpendicular waves, with the wave B ~ 1 , aligned along the xx axis. electric field, E (a) Write the fluid equations relevant to study this problem. 28 Chapter 6. Waves in magnetized plasmas (b) Linearize the equations for the electrons, keeping only the first order terms. Show that ωce vex , vey = i ω where (obviously) the speeds vex and vey are the components of the first order correction to the electron velocity. (c) Still using only the electron equations from c), show that ne1 = −i k(1 − ε)n0 eE1 , me Ω2e 2 where Ω2e = ω 2 − ωce − γe k 2 c2se and c2se = kB Te me . (d) Defining Ω+ e Ω− similarly toá Ωe , what are the expressions for n+1 e n−1 ? (e) Show that, in the plasma approximation, the dispersion relation can be written in the form ε me 2 2 m− me m− 2 2 Ω Ω + (1 − ε)Ω2+ Ω2− + Ω− Ωe = 0 . m+ + e m+ m2+ (f) Obtain the dispersion relation in the limit ε = 0 and me m+ . Comment the result. Historical note: these waves were predicted theoretically by N. D’Angelo, IEEE Trans. Plasma Sci. 20 (1992) 658 and detected experimentally by T. An, R. L. Merlino e N. D’angelo, Phys. Fluids 5 (1993) 1917. 6. (Exam 2015/2016) Consider an electromagnetic mode with E1 ⊥ B0 and k k B0 . (a) Write the system of linearised (vectorial) equations that leads to the dispersion relation for these waves, neglecting ion motion (mi → ∞), electron thermal motion (Te → 0) and collisions. (b) It can be shown that the system you just wrote leads to Ex (ω 2 − c2 k 2 − α) + Ey iαωce /ω 2 2 2 Ey (ω − c k − α) − Ex iαωce /ω where α= =0 =0 ωp2 . 2 ω2 1 − ωce Continue from here to obtain the dispersion relation for this wave (in the form given in the formulae for the exam). (c) Show (briefly) that the modes are right and left hand circularly polarized, and identify which is which. (d) Define and obtain the cutoff frequencies. Comment the results. 7. (Exam 2017/2018) We want to study with generality wave propagation in ~ 0 = B0 ~uz ). cold, collisionless, homogenous, infinite, magnetized plasmas (B Assume the ions to be stationary. 29 (a) Let us start by rewritting Maxwell’s equations in an appropriate form. i. Linearize Faraday’s and Ampère’s laws and show that plane waves can be described by the general equation 2 ~k(~k · E ~ 1 ) − k2 E ~1 + ω · E ~1 = 0 , 2 c (6.1) where the dielectric tensor is related to the conductivity tensor σ by ic2 µ0 =I+ σ, ω σxx σxy σxz Ex ~ 1 = σyx σyy σyz Ey I is the identity matrix and J~1 = σ · E σzx σzy σzz Ez ii. Equation (6.1) can be written using the dispersion matrix D = kxx kxy kxz 2 {~k~k − k 2 I + ωc2 }, where ~k~k is the tensor kyx kyy kyz , in the kzx kzy kzz form ~1 = 0 . D·E Assuming D is known, how would you derive the dispersion relation? (b) To use the procedure delineated above to study waves in plasmas it is necessary to calculate the conductivity tensor. i. Linearize the force equation for the electrons to show that e iωEx + ωce Ey 2 m ω 2 − ωce e iωEy − ωce Ex vy = − 2 m ω 2 − ωce e iEz vz = − m ω vx = − ii. Show that the conductivity tensor is σ⊥ σ× −σ× σ⊥ 0 0 given by 0 0 , σk where σk = ne2 i me ω ; σ⊥ = ne2 iω 2 me ω 2 − ωce ; σ× = ne2 ωce . 2 me ω 2 − ωce (c) The above formalism provides the dispersion relation for any cold, magnetized plasma (!!!) (inclusion of ion motion is straightforward). It is ~ the dispersion matrix takes the form possible to show that when ~k ⊥ B S −iD 0 2 2 , 0 D = iD − kωc2 + S k2 c2 0 0 − ω2 + P where S = S(ω, ωpe , ωce ), D = D(ω, ωpe , ωce ) and P = 1 − 2 ωpe ω2 . 30 Chapter 6. Waves in magnetized plasmas i. Show that the system comprises two independent dispersion relations. Obtain one of them explicitly and the other one in terms of S and D. ii. Which wave corresponds to the latter dispersion relation? CHAPTER 7 Diffusion and transport in weakly ionized plasmas 1. The cross section for electron-neutral momentum transfer in Ar can be approximated by the relation σ(u) = αu[eV ], with α = 1.37×10−20 m2 /eV (see figure, σ[10−16 cm2 ] vs. u [eV]). Calculate the mean collision frequency for momentum transfer in an argon plasma at p = 5 Torr and Tg = 20 o C, characterized by an electron temperature kTe = 1 eV, i.e, assuming a Maxwellian distribution of velocities (pay attention to the definition and normalization of the distribution!), 3/2 mv 2 m 2 f (v) = 4πv exp − . 2πkTe 2kTe Compare with the value you would obtain if the cross section were constant, with the value corresponding to the mean energy. 4.50E+01 4.00E+01 3.50E+01 3.00E+01 2.50E+01 2.00E+01 1.50E+01 1.00E+01 5.00E+00 0.00E+00 0 5 10 Real 15 20 25 30 35 Linear Approxima8on 2. (F. F. Chen, 5.1) The electron-neutral collision cross section for 2 eV electrons in He is about 6πa20 , where a0 = 0.53 × 10−8 cm is the radius of the first 32 Chapter 7. Diffusion and transport in weakly ionized plasmas Bohr orbit of the hydrogen atom. A positive column with no magnetic field has p = 1 Torr of He (at room temperature), and kTe = 2 eV. (a) Compute the electron diffusion coefficient in m2 /s, assuming that hσvi is equal to the product σv for 2 eV electrons. (b) If the current density along the column is 2 kA/m2 and the plasma density is 1016 m−3 , what is the electric field along the column? 3. (Exam 2013/2014) Suppose that the electron distribution function in a homogeneous plasma can be approximated by a superposition of two Maxwellians at different temperature, i.e., f (~v ) = α1 f1 (~v ) + α2 f2 (~v ), where fj (~v ) = n m 2πkB Tj 3/2 mv 2 exp − , 2kB Tj with j = 1, 2, α1 + α2 = 1 e v = |~v |. (a) Verify that the distribution function is correctly normalized, i.e., n. ´´´ f (~v )d3 v = (b) Show that the average value q of the absolute value of the velocity of each 8k T B j of the maxwellians is hvj i = πm and calculate the average value of the absolute value of the velocity of the distribution. (c) The cross section for electron-neutral momentum transfer can be approximated by σm (u) = βm u, where βm is constant and u is the electron energy. Show that the mean collision frequency for momentum transfer associated to each Maxwellian is νm = 2N βm kB Tj hvj i and calculate the average value of the momentum collision frequency of the distribution. (d) The ionization cross section of the same gas can be approximated by σi (u) = 0, if u < ui , and σi (u) = βi , if u ≥ ui , where ui is the ionization threshold. Show that the ionization frequency associated with ui ui each Maxwellian is νi = N βi hvj i kTj + 1 exp − kTj and calculate the mean ionization frequency of the distribution. (e) Calculate the values in items c. and d. for kB T1 = 1 eV, kB T2 = 16 eV, α1 = 0.99, α2 = 0.01, βm = 10−20 m2 /eV, βi = 10−20 m2 , ui = 15 eV and N = 1023 m−3 . Comment the results. (f) (to answer in the las problem sheet) Suppose that an electronic wave is excited on a plasma with an initial distribution with a shape similar to the one used in this problem. Is there a region of wavelengths where, at least in principle, these waves are unstable. If yes, can you define an interval of phase speeds where to search for these waves? Useful integrals: ˆ ∞ √ 2 2 x exp −Ax dx = 0 ˆ ∞ 0 x 2 π 4A3/2 2 exp (−Ax) dx = A3 ˆ ∞ 3 2 x exp −Ax dx = 1 2A2 2 (Ax2 3 2 i + 1) exp(−Axi ) x exp −Ax dx = xi 2A2 0 ˆ ∞ 33 4. (Lieberman and Lichtenberg 5.2) A steady-state argon plasma is created at high pressure between two parallel plates located at x = ±L/2 by illuminating the region between the plates with ultraviolet radiation (which ionizes the neutrals). The radiation creates a uniform number of electron-ion pairs per unit volume and pe unit time, G0 (m−3 s−1 ), everywhere within the plates. The electrons and ions are lost by ambipolar diffusion to the walls. (a) Show that in the limit Ti Te the ambipolar diffusion coefficient is given by Da ' µi (kTe /e). (b) Assuming Da uniform in space and constant in time, obtain the stationary plasma profile, n(x), and the value of the density at the center, n0 , assuming you can impose the boundary condition n(x) ' 0 at the walls. 5. (Exam 2015/2016) Consider an axisymmetric cylindrical weakly-ionized plasma ~ = E~ur , B ~ = B~uz and ∇P ~ i,e = ∂Pi,e /∂r~ur . Neglect the convective with E term and consider the stationary case. Assume neutrality, the same temperature for electrons and ions and that you are on a reference frame where the average velocity of the neutrals is zero. (a) Write the expressions for the r and θ components of the two fluid force equations. (b) Solve the previous equations for vr and vθ and verify that: i. for the r component, ver = −µer E − Der 1 ∂n n ∂r where µer = µe 1+ , 2 ωce νe2 Der = De 1+ 2 ωce νe2 and µe = e me νe , De = kB Te ; me νe ii. for the θ component veθ = vE + vD 1+ νe2 2 ωce where vE = − E B , vD = − kB Te 1 ∂n . eB n ∂r (c) Find the expression of E that ensures ambipolarity along the radial direction. (d) Obtain the expression of Der for very intense B-fields (ωce νe ) and verify which is the length scale of the associated “random walk” motion. Comment the result. Note: it is interesting to compare the results of this exercise with exercise 4 from the problem sheet 8. 34 Chapter 7. Diffusion and transport in weakly ionized plasmas 6. (Exam 2016/2017) A steady-state nitrogen plasma is created between two parallel plates at high pressure by an external electric field. The plasma contains electrons and two main types of positive ions, N2+ and N4+ . The electrons and ions are lost by ambipolar diffusion to the walls. (a) Assuming a strongly collisional regime (high pressure), a constant electronneutral collision frequency and an isothermal plasma (γ = 1), obtain the expressions for the mobility and diffusion coefficients of species s, as well as for the ratio Ds /µs . Justify all the approximations you make. (b) Write the quasi-neutrality and congruency hypotheses in this case. (c) Show that the ambipolar electric field is ~ ~ ~ ~ = D1 ∇n1 + D2 ∇n2 − De ∇ne , E µ1 n1 + µ2 n2 + µe ne where the indexes 1 and 2 represent each the two positive ions and e the electrons. (d) Further assuming the proportionality hypothesis, ~ 1 ~ 2 ~ e ∇n ∇n ∇n ' ' , n1 n2 ne ~ s for all species, where the ambipolar diffusion show that ~Γs = −Das ∇n coefficients for the positive ions (s = 1, 2) are Das = Ds − µs D1 n1 + D2 n2 − De ne . µ1 n1 + µ2 n2 + µe ne (e) Show that, in the limit Te Ti (i = 1, 2), Dsi ' Di TTei . (f) Within the conditions of the problem, justify that we must have ne Dse = n1 Ds1 + n2 Ds2 . [Historical note: the expressions for the ambipolar diffusion coefficients for a plasma comprised of electrons and several positive ions are given, e.g., in V. Guerra, P. A. Sá and J. Loureiro, Eur. Phys. J. Appl. Phys. 28 (2004) 125] 7. (Exam 2017/2018) Consider a weakly ionized plasma diffusing in the ambipolar regime. Assuming Da uniform in space and constant in time, obtain the radial profile of the fundamental diffusion modes for a cylindrical plasma column of radius R and show that the characteristic decay time for this mode is R2 τ0 ' 2.405 2D . a 8. (Exam 2017/2018) Consider a stationary weakly ionized plasma in a strongly ~ = B~uz . collisional regime, in the presence of an external magnetic field, B Assume you can neglect thermal effects. (a) Show that the equation for the perpendicular velocity of the electrons is given by µe 1 ~ ~ve⊥ = − ~v , 2 E⊥ + ωce ν2 E 1 + ν2 1 + ω2e e where ~vE mobility. ce ~ ×B ~ drift velocity and µe = is the E e m e νe is the electron 35 ~ (b) Show that the electron conductivity tensor, σ e , defined by J~e = σ e · E, can be written in cartesian coordinates in the form σ⊥ −σH 0 σ⊥ 0 , σ e = σH 0 0 σk and give the expressions for the longitudinal conductivity σk , perpendic2 ular conductivity σ⊥ and Hall conductivity σH in terms of σ0 = mnee νe , ωce and νe . [Suggestion: start by defining J~e⊥ and J~ek in terms of ~v⊥ and ~vk ] (c) Discuss the limiting cases ωce = 0 and νe = 0. 36 Chapter 7. Diffusion and transport in weakly ionized plasmas CHAPTER 8 Diffusion and transport in fully ionized plasmas 1. (F. F. Chen 5.9) Suppose the plasma in a fusion reactor is in the shape of a cylinder 1.2 m in diameter and 100 m long. The 5 T magnetic field is uniform, except for short mirror regions at the ends, which we may neglect. Other parameters are kTi = 20 keV, kTe = 10 keV and n(r = 0) = 1021 m−3 . The density profile is found experimentally to be approximately as sketched in the figure. (a) Assuming classical diffusion, calculate D⊥ at r = 0.5 m (b) Calculate dN/dt, the total number of electron-ion pairs leaving the central region radially per second. (c) Estimate the confinement time, τ by τ ' −N/(dN/dt). 2. (F. F. Chen 5.11) A cylindrical plasma column has a density distribution n = n0 1 − r2 /a2 , where a = 10 cm and n0 = 1019 m−3 . If kTe = 100 eV, kTi = 0 and the axial magnetic field is B0 = 1 T, what is the ratio between the Bohm and the classical diffusion coefficients perpendicular to B0 ? 3. (F. F. Chen 5.18) If a cylindrical plasma column diffuses at the Bohm rate, calculate the steady-state radial density profile, n(r), ignoring the fact that it 38 Chapter 8. Diffusion and transport in fully ionized plasmas may unstable. Assume the density is zero at r = ∞ and has the value n0 at r = r0 . ~ = 4. (F. F. Chen 5.15) Consider an axisymmetric cylindrical plasma with E ~ = B~uz and ∇P ~ i = ∇P ~ e = ∂P/∂r~ur . Neglect the convective term Ee ~ur , B and consider the stationary case. (a) Write the two-fluid equations. (b) From the θ components of these equations, show that vir = ver . (c) From the r components, show that vsθ = vE + vDs (s = i, e). (d) Find an expression for vir and show it does not depend on Er . 5. (Exam 2014/2015) (a) Use the MHD equations to derive the expression ρm ∂~v ~ × B) ~ + σ0 (~v × B) ~ ×B ~ − ∇P ~ , = σ0 ( E ∂t where σ0 is the plasma conductivity (σ0 = 1/η). ~ in (b) Solve the equation for the velocity components perpendicular to B ~ the case E = 0 and P = const., to show that the characteristic time for diffusion across the magnetic field is τ= ρm , σ0 B 2 i.e., ~v⊥ (t) = ~v⊥ (0) exp (−t/τ ). 6. (Exam 2015/2016) Consider a fully ionised plasma where the density varies ~ = B0 (x)~uz . slowly along ~ux and where the magnetic field is given by B (a) Use the MHD equations to show that, in stationary regime, ∂P ∂x = Jy B0 . (b) The MHD equations provide a macroscopic image of the plasma. Explain the physical meaning of the expression obtained. (c) On a more microscopic image, since the positive ions are typically heavier and colder than the electrons, the electric current density calculated in a) is carried essentially by the electrons. Calculate the electron velocity associated with that current. Comment the result. 7. (Exam 2016/2017) As seen in class, the generalised Ohm’s law can take the form ~ ~ + ~v × B ~ = η J~ + 1 J~ × B ~ + 1 ∇P ~ e + me ∂ J . E 2 en ne en ∂t During a substorm in the nightside magnetotail (disturbance in the magnetosphere) the following values have been measured: E ' 0.1 mV/m; v ' 100 km/s; B ' 1 nT; J ' 1 nA/m2 ; n ' 1 cm−3 ; Pe ' 0.1 nPa. In these circumstances, the characteristic length scale is L ' 104 km, the characteristic time scale is τ ' 10 s and the effective resistivity is less than 1 mS−1 . Compare the magnitudes of the various terms in Ohm’s law in this case. Comment the results. 39 8. (Exam 2016/2017) MHD equations can be used to investigate the origin of the magnetic fields in stars, planets and the universe. (a) Consider Ohm’s law in its common form (i.e., where the r.h.s contains only the resistivity term) and assume that η is spatially uniform. Further assume the displacement current can be neglected in Ampère’s law. Derive the following closed equation for the magnetic field, ~ ∂B ~ × (~v × B) ~ + χ∇2 B ~ , =∇ ∂t where χ = η/µ0 is the magnetic diffusivity. (b) The previous equation does not explain the origin of magnetic fields in ~ the a medium initially non-magnetised [as the equation is linear in B, ~ = 0) = 0 implies B(t ~ > 0) = 0]. initial condition B(t Repeat the previous question keeping as well the electron pressure gradient term in Ohm’s law, to show that the equation for the temporal evolution of the magnetic field has now a source term creating a magnetic field in the direction perpendicular to the gradients of density and electron temperature. [Historical note: this source term is known as the Biermann battery and ~ b = kB Te ∇n. ~ can be conveniently expressed in terms of the field E The en Biermann battery provides the first seeds of the magnetic field, which are ~ ~ then efficiently amplified by the dynamo associated with the ∇×(~ v × B) term. The historical reference is L. Biermann, Z. Naturforsch. 5a (1950) 65.] (c) Knowing that in Earth’s core χ ' 2 m2 s−1 and that the Earth’s core radius is R ' 3.5×106 m, make a rough order of magnitude estimation of the decay time of Earth’s magnetic field due only to magnetic diffusion. 9. (Exam 2017/2018) In the deduction of the MHD equations, it is argued the mi me n ∂ J~ electron inertia term can often be neglected in comparison e ∂t n ~ Establish and discuss the conditions of with the Hall term (mi − me )J~ × B. validity of this approximation. ∂ 10. (Exam 2017/2018) Consider a cylindrically symmetric plasma column ( ∂z = ∂ ∂ 0; ∂θ = 0) of radius R, in equilibrium ( ∂t = 0), confined by a magnetic field. (a) Show that the radial component of the MHD force equation can be written as ∂P (r) = Jθ (r)Bz (r) − Jz (r)Bθ (r) . ∂r (b) [2.5 val] Use Ampere’s law neglecting the displacement current to eliminate J and show that the former equation can take the form ∂ 1 2 1 2 1 Bθ2 (r) P (r) + Bz (r) + Bθ (r) = − , ∂r 2µ0 2µ0 µ0 r which allows defining a magnetic pressure ~ force. J~ × B 1 2 2µ0 B associated with the 40 Chapter 8. Diffusion and transport in fully ionized plasmas (c) Imagine a situation with Bθ = 0 and where the plasma density decreases radially. Interpret and discuss the meaning of the previous equation in this case. [Suggestion: start by drawing a typical profile of n(r), then draw Bz (r) and mark all forces acting on the plasma.] CHAPTER 9 Kinetic theory I 1. Derive the continuity equation from the Vlasov’s equation (integrate in d3 v). 2. Derive the force equation from Vlasov’s equation (multiply by ~v and integrate in d3 v). The most laborious term is the one involving the gradient in configuration space, which makes appear the average value of the tensor ~v~v . Calculate the explicitly this term when: (a) the thermal agitation is negligible; (b) the average velocity is zero (i.e., the fluid is at rest and there is only thermal agitation); (c) in the general case where the velocoty can be decomposed as ~v = ~u + w, ~ where ~u is the average velocity of the fluid and w ~ corresponds to the thermal agitation. 3. (Exam 2015/2016) Consider a stationary plasma, without magnetic field, un~ = −∇φ. ~ der the effect of an electrostatic field E We want to obtain the electrostatic potential due to a test charge placed in the plasma. We shall look for a stationary solution of Vlasov’s equation on the separable form fs (~r, ~v , t) ≡ fs (v, ~r) = f0s (v)ψs (~r), where f0 is the Maxwellian distribution, f0 (~r, ~v , t) ≡ f0 (v) = n0 m 2πkB Te 3/2 mv 2 exp − 2kB Te , where v = |~v | and s = e, i. (a) Show that the distribution f0 is properly normalised, i.e., n0 . (b) Use Vlasov’s equation to show that çã ~ s (~r) ~ r) qs ∇φ(~ ∇ψ =− ψs (~r) kB Ts ´ f0 (v)d3 v = 42 Chapter 9. Kinetic theory I (c) Solve the previous equaiton and show that qs φ(~r) fs (v, ~r) = f0 (v) exp − , kB Ts where n0 in the Maxwellian distribution is the plasma density faraway from the test charge, i.e., in a region where φ(~r) = 0. (d) Write Poisson’s equation using the distributions fs and show that eφ eφ en0 2 exp − exp − =0. ∇ φ− 0 kB Te kB Ti Comment the result. 4. (Exam 2016/2017) The kinetic study of the behaviour of electrons in a plasma can be made using a general kinetic equation of the form ∂f ~ r f + q (E ~ + ~v × B) ~ ·∇ ~ v f = ∂f + ~v · ∇ , ∂t m ∂t c where the r.h.s. ´ term represents the influence of collisions. Assume the normalisation f (~r, ~v , t)d3 v = ne (~r, t). One of the simpler expressions for the collision term is given by the relaxation time approximation, also known as the Krook model, which takes collisions into account using ∂f = −νc (f − f0 ) , ∂t c where νc is a constant collision frequency and f0 (~r, ~v ) is the equilibrium distribution of the electrons. (a) Show that for a homogeneous plasma in the absence of external fields the difference between f and f0 decays exponentially with time. (b) Consider now electrons in an unmagnetized, homogeneous, time-independent plasma in a weak constant electric field, E~1 . Linearise the distribution function, f (~r, ~v , t) ≡ f (~v ) = f0 (~v ) + f1 (~v ) , where f0 is the (uniform and stationary) unperturbed distribution, assumed to be a Maxwellian, and f1 is a first order perturbation. i. Show that e2 J~ = − νc m ˆ ~ ·∇ ~ v f0 ~v d3 v . E ii. Show that the electrical conductivity is given by σc = ne e2 . mνc [Note: This is one of many examples of deriving familiar macroscopic results from underlying kinetic equations.] CHAPTER 10 Kinetic theory II 1. (F. F. Chen 7.2) An electron plasma wave with 1 cm wavelength is excited in a 10 eV plasma with n = 1015 cm−3 . The excitation is then removed and the wave Landau damps away. How long does it take for the amplitude to fall by a factor of e? 2. (Exam 2014/2015) Consider a one dimensional electron distribution function of the form 2 nb np u (u − ub )2 , f0 (u) = 1/2 exp − 2 + 1/2 exp − vt vb2 π vt π vb resulting from the injection of an electron beam of average speed ub and q kB T density nb on a Maxwellian plasma of density np , where vt = m is the thermal electron speed. Suppose as well that vt ∼ vb and ub vt , nb np , and neglect ion motion. ´ +∞ (a) Calculate −∞ f0 (u)du to verify that the distribution function is correctly normalised. (b) Sketch f0 (u) and show where do you expect that unstable waves may exist. (c) In which interval of phase speeds would you search for unstable waves? [Suggestion: determine where the two components of f0 give the same contribution] (d) Determine the frequency, wave number and growth rate for the fastest growing mode. 3. (F. F. Chen 7.3) An infinite, uniform plasma with fixed ions has an electron distribution function composed of (1) a Maxwellian distribution of “plasma electrons” with density np and temperature Tp at rest in the laboratory frame, and (2) a Maxwellian distribution of “beam electrons” with density nb and temperature Tb centered at ~v = V ~ux . If nb np , plasma oscillations in 44 Chapter 10. Kinetic theory II the x-direction are Landau damped. If nb is large, there will be a two-stream instability. The critical density for the onset at the instability can be estimated by setting the slope of the total distribution function to zero, as follows: (a) write expressions for fp (v) and fb (v), using the abbreviations v = vx , 2kB Tp B Tb and b2 = 2km ; a2 = m (b) assuming that the value of the phase velocity vϕ will be the value of v at which fb (v) has the largest positive slope, find vϕ and fb0 (vϕ ); (c) find fp0 (vϕ ) and set fp0 (vϕ ) + fb0 (vϕ ) = 0; (d) para V b show thatthe beam critical density is given approximately √ T V nb V2 b by np = 2e Tp a exp − a2 . 4. (Exam 2015/2016, Gardner’s theorem) We want to study the propagation of Langmuir waves starting from Vlasov’s equation. As it has been shown in class, if we assume immobile positive ions (mi → ∞), the dispersion relation can be written in the form 2 ˆ +∞ ωpe 1 ∂g du = 0 (k, ω) = 1 − 2 k ∂u u − ω/k −∞ where g(u) is the unidimensional distribution function ˆ +∞ ˆ +∞ 1 g(u) = f0 (u, vy , vz )dvy dvz . n0 −∞ −∞ (a) Justify that if g is Maxwellian and the wave phase speed is much larger than the electron thermal speed we can, on a first approximation accounting only for the contribution of the electrons of the body of the distribution, neglect the pole on the integral. Obtain the dispersion relation in this case ´ +∞ [Suggestion: recall that for a Maxwellian and u vϕ , −∞ g(u)/(u − vϕ )2 du ' 1/vϕ2 + 3vt2 /vϕ4 ] (b) In fact ω can be complex. There are unstable modes if the imaginary part of the frequency is positive. We want to show Gardner’s theorem, establishing that a single-humped velocity distribution is always stable. The proof can be made by contradiction. Consider ω = ωr + iγ in the expression from a), where ωr and γ are the real and imaginary parts of the frequency, respectively. Assuming γ > 0 the integral in the dispersion relation can be made along the real axis u, since the pole is above that axis. g(u)" u" v0" i. Show that the dispersion relation can be written in the form ∂g ωr 2 ˆ +∞ ωpe ∂u u − k r (k, w) = 1 − 2 2 2 = 0 k −∞ u − ωkr + γk 45 2 ωpe γ i (k, ω) = − 2 k k ˆ +∞ u− −∞ ∂g ∂u ωr 2 k + γ 2 k =0 ii. Show that 2 ωpe 1+ 2 k ˆ +∞ −∞ u ∂g ∂u (v0 − u) 2 2 − ωkr + γk =0, where v0 is the value of u corresponding to the hump in the distribution function (see figure). [Suggestion: consider the linear combination r − i (kv0 − ωr )/γ] iii. Show that the expression from the previous question can never be satisfied and conclude about the stability of single-humped distributions. 5. (Exam 2016/2017) Consider longitudinal oscillations of electrons in the absence of a magnetic field. Collisions with neutrals are taken into account by the Krook model (relaxation time approximation), so that electrons are described by the kinetic equation ∂f ~ rf − e E ~ ·∇ ~ v f = −νc (f − f0 ) , + ~v · ∇ ∂t m where νc is a constant collision frequency and f0 (v) an unperturbed velocity distribution corresponding to n0 particles per unit volume. The dynamics of the ion motion are neglected, the ions act merely as a uniform background of positive charge. (a) Linearize the equation in the usual way and show that f1 = 1 eE1 ∂f0 , i(kv − ω − iνc ) m ∂v where v ≡ vx . (b) Show that the dispersion relation can be written in the form ˆ +∞ k2 dg 1 − dv = 0 , 2 ωpe dv v − ω/k − iνc /k −∞ ´ where g(vx ) = n10 dvy dvz f0 (vx , vy , vz ). (c) Determine the damping rate of the wave for small collision frequencies. Comment the results, referring the conditions where Landau damping can be observed, if any. [Suggestion: Recall that without collisions you have the same dispersion relation as in 5b, with νc = 0 and where ω is complex, ω = ωr + iωi ; in that case the result is ωi ' 3 πωpe dg 2k2 du (u = ωr k ).] 6. (Exam 2017/2018) The purpose of this problem is to make a kinetic study of ~ 0 = 0). A complete delinear, transverse waves in non-magnetized plasmas (B scription of the plasma is given by the Vlasov equation and the Maxwell equations. The linearization is made about equilibrium zero order, space and timeindependent, isotropic distribution functions, fs (~r, ~v , t) = f0s (v)+f1s (~r, ~v , t), where s ∈ {e, i}. Assume the unperturbed plasma is quasi-neutral. 46 Chapter 10. Kinetic theory II ~ r, t) (a) Write the expressions that allow the calculation of ρ(~r, t) and J(~ from the distribution functions and show they are first order quantities. (b) Looking for plane waves, show that the first order distributions are given by qs /ms ~ ~ f1s = E1 · ∇v f0s (v) . i(ω − ~k · ~v ) ~ 0 = 0 and B~0 = 0, but that in principle E ~1 = 6 0 and Note: recall that E ~ B1 6= 0. (c) Use Maxwell’s equations to show that for transverse waves 2 ~1 . ~ 1 = iωµ0 J~1 + ω E k2 E c2 (d) Show that ~ 1 (ω 2 − k 2 c2 ) = −i ω E 0 X e2 ˆ 1 ~ ·∇ ~ v f0s (v)~v d3 v . E ~k · ~v ) 1 m s i(ω − s ~ 1 = E1 ~uy and ~k = k~ux and neglect the ion motion (mi → ∞). (e) Assume E Simplify the expression above to derive the dispersion relation ˆ +∞ g(vx ) 2 2 2 2 ω − k c = ω ωpe dvx , ω − kvx −∞ where 1 g(vx ) = n0 ˆ +∞ ˆ +∞ f0e (v)dvy dvz . −∞ −∞ [Suggestion: integrate by parts in dvy ] (f) For non-relativistic plasmas the phase velocity of the wave is always much larger than the thermal speed. Therefore, approximate ω − kvx ' ω and obtain the dispersion relation of these waves. (g) Would you expect these waves to be significantly damped? SOLUTIONS TO CHAPTER 1 Debye shielding and fundamental effects 1. From P V = N kB T and n = N/V , n[m3 ] = 1.013 × 105 p[Torr] 1 p[Torr] ' 9.66 × 1024 . −23 1.38 × 10 760 T [K] T [K] Substituting, (a) n ' 2.69 × 1025 m−3 = 2.69 × 1019 cm−3 (Loschmidth number); (b) n ' 3.22 × 1022 m−3 = 3.22 × 1016 cm−3 . 2. Using the definitions of λD and Λ, λD [m] '7.44 kB T [eV] n0 [cm−3 ] 1/2 Λ '4.12 × 108 n0 [cm−3 ] kB T [eV] n0 [cm−3 ] 3/2 . The different values obtained are summarized in the table below and represented in figure 1.1 a) b) c) d) e) f) g) λD (m) 7.4 × 10−5 7.4 2.4 × 10−3 2.4 × 10−8 7.4 × 10−5 24 2.4 × 10−4 Λ 4.1 × 106 4.1 × 109 1.3 × 104 1.3 × 103 4.1 × 103 1.3 × 109 1.3 × 103 48 Solutions to chapter 1. Debye shielding and fundamental effects ne [cm-3] 10 10 10 20 Laser fusion 16 12 10 Tokamak Gaseous electronics Flame 8 Ionosphere 10 4 Solar wind 10 Interstellar medium 0 10-1 100 101 kTe [eV] 102 103 104 Figure 1.1: Problem 2: (—) constant Λ; (– –) constant λD 3. (a) The Debye potential created by a punctual test charge qT inside an homogeneous plasma is given by qT 1 r φ(r) = exp − . 4π0 r λD Therefore, from Poisson’s equation, ∇2 φ = − ρ0 and using spherical coordinates, ∇2 φ(r) = 1 d2 r dr 2 (rφ), 1 d2 qT r exp − r dr2 4π0 λD 1 = 2 φ(r) . λD ∇2 φ = Hence, for all points except the origin, r 1 qT ρ(r) = − 2 exp − , λD 4π0 r λD which corresponds to the charge density of the shielding cloud. Note: it is of course possible to obtain the result following the standard 49 derivation of the Debye length, ρ = e(ni − ne ) eφ eφ = n0 e exp − − exp + kB Ti kB Ti eφ eφ ' n0 e 1 − −1− kB Ti kB Te n0 e2 1 1 1 = + ≡ 2 φ0 , kB Ti Te λD where the usual assumptions of i) potential energy much smaller than kinetic energy; ii) electrons and ions in equilibrium with the electrostatic field (“adiabatic response” of both electrons and ions); and iii) quasineutrality [ni (r → ∞) = ne (r → ∞)] have been considered. Note as well that the test charge can be considered to get the total charge density with the additional term qT δ(~0). The charge in the shielding cloud inside a sphere of radius R centred in qT is (to get the total charge it is necessary to add qT ) ˚ ˆ Q(r ≤ R) = R r2 ρ(r) dr ρ(r) dV = 4π r≤R =− qT λ2D ˆ 0 0 R r r exp − dr . λD Integrating by parts, with u = r, du = dr, v = exp − λrD and dv = − λ1D exp − λrD , ( R ˆ R ) r r qT r exp − − exp − dr Q(r ≤ R) = λD λD 0 λD 0 R R R =qT exp − + exp − −1 λD λD λD Substituting values, Q(r ≤ λD /2) = − 0.09 qT Q(r ≤ λD ) = − 0.26 qT Q(r ≤ 5λD ) = − 0.96 qT lim Q(r < R) = − qT R→∞ This example shows that the Debye length is a characteristic length and not the distance for a perfect shielding. Alternative solution:1 1 Thanks due to my former student to Ricardo Barrué 50 Solutions to chapter 1. Debye shielding and fundamental effects The electrostatic field can be readily obtained from the potential, 1 dφ qT r 1 ~ + E(r) = − ~ur = exp − dr 4π0 λD r2 rλD Using Gauss’ law for the charge enclosed by a spherical surface of radius r = R, ‹ R R qT 2 ~ exp − 1+ , E · ~n dS = E(R) · 4πR = 0 λD λD Q(r ≤ R) + qT ≡ , 0 where Q(r ≤ R) is defined as before. Hence, R R R + exp − −1 . Q(r ≤ R) = qT exp − λD λD λD (b) The electrostatic energy of interaction in a system of point charges is 1 X qi qj W = , 4π0 i<j rij where rij is the distance between charges i and j. We may then consider that the interaction energy between charge i and all the other charges is X qj 1 1 Wi = qi , 2 4π0 rij j6=i so that the total interaction energy is W = of only two point charges). P i Wi (think on a system Note: in this view the interaction energy of a system of two equal point charges is distributed equally among both charges [see, e.g., B. Jayaram and A. Das, J. Mol. Struct. 543 P (2001) 123]. Different definitions can be used, such as W = 12 i Wi , and the interaction energy of charge i with all others is twice that defined above; this difference in the definitions makes a difference of a factor of two in the resolution of the exercise and is completely irrelevant for the point being made. With the definition above, the interaction energy between the test charge qT and the remaining charges is ˚ 1 1 ρ(r) Wq = qT dV 2 4π0 r ˆ +∞ 1 1 qT r 1 2 = qT 4πr − 2 exp − dr 8π0 λ 4πr λ r D 0 D =− qT2 . 8π0 λD The mean kinetic energy of the test charge is 23 kT , so that its total energy is 3 qT2 Etot = kT − . 2 8π0 λD 51 Considering a typical case with qT = +e and Λ ∼ n0 λ3D 1 (many particles inside a Debye sphere), it is immediate to conclude that the interaction potential energy of the particle is much smaller than its kinetic energy, as it should be from the very definition of a plasma: n0 λ3D 1 kT 0 λD 1 n0 n0 e2 e2 0 λD qT2 3 kT . 2 8π0 λD kT 4. The configuration of the problem is depicted in figure 1.2 plasma E x=0 Figure 1.2: Problem 4 Inside the plasma (x > 0), Poisson’s equation ∇2 φ = − ρ0 reads − d2 φ(x) e [ni (x) − ne (x)] . = 2 dx 0 Since Te = Ti = T , the ion and electron densities can be written as eφ(x) eφ(x) ni (x) = n0 exp − ' n0 1 − kB T kB T eφ(x) eφ(x) ne (x) = n0 exp + ' n0 1 + , kB T kB T 52 Solutions to chapter 1. Debye shielding and fundamental effects and 1 d2 φ(x) = 2 φ(x) , 2 dx λD 1/2 0 kB T . with the Debye length defined as λD = 2n 2 e 0 The solution is x φ(x) = φ0 exp − λD x + φ1 exp + λD , where φ0 and φ1 are constants. The condition lim φ(x) = 0 x→+∞ implies φ1 = 0. φ0 can be determined from the electric field outside the plasma, assuming continuity at x = 0, φ0 x dφ = exp − , E(x ≥ 0) = − dx λD λD i.e., E(x = 0) = λφD0 ≡ E0 . Moreover, the expression above shows that the plasma shields the external electric field, within a characteristic distance λD . Substituting values, λD ' 1.66 × 10−2 cm and 0.5 E(x = 0.5 cm) ' 100 exp − ' 9 × 10−12 V/cm . 0.0166 We verify that for x = 0.5 cm∼ 30λD the electric field already decays about 13 orders of magnitude. 5. (a) In the conditions of the problem, ni = n0 and ne = n0 exp kBeφTe , so that eφ ρ = e(ni − ne ) = n0 1 − exp kB Te eφ n0 e2 ' n0 e 1 − 1 − =− φ. kB Te kB Te Poisson’s equation takes then the form ∇2 φ = − with λD = 0 kB Te no e 2 1/2 1 ρ = 2 φ, 0 λD . Writing the Laplacian in spherical coordinates and assuming spherical symmetry, φ(~r) = φ(r), 1 d2 1 (rφ) = 2 φ . r dr2 λD Further defining ϕ = rφ, d2 1 ϕ= 2 ϕ, dr2 λD 53 whose solution is r ϕ(r) = C1 exp − λD r + C2 exp + λD , where C1 and C2 are constants. The two boundary conditions are lim ϕ(r) = 0 ⇒ C2 = 0 , r→+∞ φ(r = R) = ϕ(R) = φ0 ⇒ C1 = Rφ0 exp R R λD . Finally, the electrostatic potential is given by r−R R . φ(r) = φ0 exp − r λD Notice that when λD R we have exp λRD ' 1 and the usual expression for the Debye potential created by a point charge is recovered, whereas for R < r λD the potential is approximately the same as created by a point charge in vacuum (as it should be!). (b) The charge in the sphere, Q, can be calculated from the condition of total shielding by the plasma (see problem 3), ˚ Q=− ρ(r) dV . Alternatively, as the charge in a conductor is distributed on its surface, we may use the discontinuity condition for the electric field, E(r = R+ ) − E(r = R− ) = σ , 0 where σ is the charge density on the surface of the conducting sphere (σ = Q/4πR2 ) and E(r=R− ) is the electric field inside the conductor and hence it is zero. ~ = −∇φ ~ = − dφ ~ur , In the plasma (r > R), E dr R d R r E(r) = − φ0 exp exp − dr r λD λD R 1 1 r−R = φ0 + exp − . r r λD λD Substituting, 1 1 + R λD 1 1 σ = 0 φ0 + R λD R Q = 4π0 φ0 R 1 + . λD E(r = R+ ) = φ0 54 Solutions to chapter 1. Debye shielding and fundamental effects (c) The capacity is given by C= Q R = 4π0 R 1 + . φ0 λD Since the capacity of a conducting sphere immersed in a dielectric of relative permittivity r is C = 4π0 r R, the plasma behaves as a dielectric of relative permittivity R . r = 1 + λD In the limit λD R, when the characteristic length for the shielding is much larger than the dimension of the sphere (i.e., poor shielding for distances comparable to the radius of the conductor), we obtain the result for a conducting sphere in vacuum. In the opposite limit, λD R, corresponding to a very efficient shielding of the sphere by the plasma, r ' R/λD . For Te = 1 keV the Debye length is λD ' 2.35 m and 2.35 × 10−5 m, respectively for n0 = 106 cm−3 and n0 = 101 4 cm−3 . The Debye length is to be compared with R = 10 cm, showing the two values of the density correspond to the two limiting cases discussed above. For completeness, the capacity of the sphere is C0 ' 1.1 × 10−11 F in vacuum, C ' 1.6 × 10−11 F∼ C0 for n0 = 106 cm−3 , and C ' 4.7 × 10−8 F C0 for n0 = 1014 cm−3 . 6. Consider two slabs of length L and cross sectional area ∆S, one “holding” singly charged positive ions with mass mi and number density n0 and the other “holding” electrons of mass me and the same number density n0 (to ensure quasi-neutrality). The slabs can move along the x-axis, perpendicular to their cross section. The deviation of the ion and electron slabs in relation to their equilibrium position is defined by the displacements xi and xe , respectively (see figure 1.6). The region of width |xi − xe | where the net charge density is different from zero has a total charge ∆Q = n0 e(xe −xi )∆S. The configuration corresponds to a parallel plate capacitor, for which the electric field is uniform and along the x-axis, ~ = ∆Q/∆S ~ux = n0 e (xe − xi )~ux . E 0 0 The total charge in the electron slab is Qe = −(n0 e∆SL), so that the force on it is 2 ~ = − (n0 e) (xe − xi )∆SL~ux . F~e = Qe E 0 Similarly, the force on the ion slab is 2 ~ = + (n0 e) (xe − xi )∆SL~ux . F~i = Qi E 0 The equations of motion for each slab are (note that the total mass of the electron slab is Me = me n0 ∆SL; similarly for the ion slab) F~e = Me~ae = me n0 ∆SLẍe ~ux , 55 Figure 1.3: Problem 6 F~i = Mi~ai = mi n0 ∆SLẍi ~ux . Equating both sets of equations, ẍe = − n0 e2 2 (xe − xi ) = −ωpe (xe − xi ) , me ε0 n0 e2 2 (xe − xi ) = +ωpi (xe − xi ) , mi ε0 2 1/2 n0 e and ωpi = m . i ε0 ẍi = + with ωpe = n0 e 2 me ε 0 1/2 Finally, defining χ = xe − xi , we have, successively, χ̈ = ẍe − ẍi 2 2 χ̈ = −ωpe (xe − xi ) − ωpi (xe − xi ) 2 2 χ̈ = −(ωpe + ωpi )χ χ̈ = −ωp2 χ 2 2 with ωp2 = ωpe + ωpi . 7. The total number of electrons in the plasma, N , is N = 43 πR3 n0 . Therefore, the electron density when the plasmas occupies a sphere or radius R + δr is ne (δr ) = N R = n0 . (R + δr )3 + δr )3 4 3 π(R 56 Solutions to chapter 1. Debye shielding and fundamental effects The electron field inside the plasma can be obtained from Gauss’s law, applied to a spherical surface of radius r < (R + δr ), ‹ ~ · ~n dS = Qint , E 0 where 4 Qint =e(n0 − ne ) πr3 3 4 3 R3 =en0 1 − πr . (R + δr )3 3 Hence, 4 3 en0 1 4πr E(r) = 1− πr , 3 0 (1 + δr /R) 3 1 en0 1 r 1− . E(r) = 0 3 (1 + δr /R)3 2 Expanding in Taylor series f (x) = 1 (1+x)3 for small x, f (x) ' 1 − 3x, δr en0 r en0 1 r 1−1+3 δr . = E(r) ' 0 3 R 0 R The equation of motion for an electron placed at the surface of the plasma sphere reads d2 me 2 (R + δr ) = −eE(r=R+δr ) , dr leading to e2 n0 d2 δ r R + δr me 2 = − δr dr 0 R 2 e n0 '− δr 0 Finally, d2 δr 2 = −ωpe δr , dr2 q e 2 n0 where ωpe = me 0 is the usual plasma frequency. The latter equation corresponds to an harmonic oscillator of angular frequency ωpe , proving the result. 8. (a) The electron and ion densities are given by eφ eφ ' n0 1 + ne =n0 exp kB Te kB Te ne =n0 . 57 For all points except the plane x = 0, ∇2 φ = − ρ , 0 2 2 where ∇2 φ ≡ ddxφ2 and ρ = e(ni −ne ) ' keBnT0e φ . Substituting, Poisson’s equation reduces to 1 d2 φ = 2 φ, dx2 λD q with the Debye length given by λD = 0ek2Bn0Te . The general solution is (see also problem 4) x x φ(x) = C1 exp − + C2 exp + , λD λD where C1 and C2 are constants. The regions x > 0 and x < 0 have to be solved separately. For x > 0, the first boundary condition is lim φ(x) = 0 , x→+∞ implying C2 = 0. The additional condition lim φ(x) = φ0 x→0+ further determines C1 = φ0 . Hence, for x > 0 we have x φ(x) = φ0 exp − . λD Due to the symmetry of the problem, φ(−x) = φ(x), as represented in figure 1.4. (b) The electrostatic field is readily obtained as ~ = − dφ ~ux = 1 φ0 exp − x ~ux , E dx λD λD for x > 0. In the limit x → 0+ we should recover the field created by an infinite plane charged with surface density σ in vacuum, lim E(x) = x→0+ σ . 20 This condition establishes the relation between φ0 and σ, φ0 = Accordingly, for x > 0 φ0 x E(x) = exp − . λD λD σλD 20 . For x < 0 we have E(−|x|) = −E(|x|), as represented in figure 1.5. The discontinuity of the electric field at x = 0 is evidently σ/0 , as required by Maxwell’s equations. 58 Solutions to chapter 1. Debye shielding and fundamental effects ϕ/ϕ0 1 0,5 -5 -2,5 0 2,5 5 x/λD Figure 1.4: Problem 8 - electrostatic potential. E σ/2ε0 -5 -2,5 0 2,5 5 x/λD -σ/2ε0 Figure 1.5: Problem 8 - electrostatic field. While the electric field created by an infinite conducting plane in vacuum is constant and does not decay away from the plane, the presence of the plasma shields the charge in the plane and the electric field (as well as the potential),which decays with a characteristic decay length given by the Debye length. Both the electrostatic field and potential are only significantly different from zero in the vicinity of the conducting plane. 59 (c) From Poisson’s equation (see the beginning of the exercise), 1 ρ(x) φ(x) = − λ2D 0 and so, for x > 0, ρ(x) = − φ0 x σ x exp − = − exp − . λ2D λD 2λD λD The total charge in the plasma, facing a section of area A of the plane and in the positive-x region, is given by ˚ ˆ +∞ Q+ = ρ(x) dV = A ρ(x) dx 0 ˆ +∞ x σ exp − dx =−A 2λD λD 0 x=+∞ σ x σ =−A −λD exp − = −A . 2λD λD x=0 2 Q+ σ =− . A 2 Similarly, the charge in the plasma in the negative-x region is Therefore, Q+ + Q− = −σ . A 9. (a) Proceeding in a similar fashion as in the previous exercise, ρ(r) 0 ρ(r) =e[ni (r) − ne (r)] eφ(r) ni (r) =n0 exp − ' n0 1 − kB Ti eφ(r) ne (r) =n0 exp + ' n0 1 + kB Te no e2 1 1 ρ(r) = − + φ(r) . kB Ti Te ∇2 φ(r) = − eφ(r) kB Ti eφ(r) kB Te Defining 1 n0 e2 n0 e2 1 1 = 2+ 2 = + , 2 λD λe λi 0 kB Ti 0 kB Ti Poisson’s equation takes the usual form ∇2 φ(r) = 1 φ(r) . λ2D In cylindrical coordinates and with cylindrical symmetry, ∇2 φ = d2 φ 1 dφ + . dr2 r dr Q− A = − σ2 . 60 Solutions to chapter 1. Debye shielding and fundamental effects Multiplying by r2 , r2 dφ d2 φ r2 + r − φ=0. dr2 dr λ2D dφ dξ dξ and noting that dφ dr = dξ dr and dr = d dφ 1 dφ 1 (ξλD )2 + (ξλD ) −ξ 2 φ = 0 , | {z } dr dξ λD | {z } dξ λD | {z } | {z } r r2 Further defining ξ = r λD , 1 λD , dφ dr dφ dr ξ 2 λD dφ d2 φ 1 +ξ − ξ2φ = 0 , 2 dξ λD dξ dφ d2 φ − ξ2φ = 0 , ξ2 2 + ξ dξ dξ which is the modified Bessel equation of order zero. The solution is φ(ξ) =AI0 (ξ) + BK0 (ξ) , r r φ(r) =AI0 + BK0 , λD λD where A and B are constants. As I0 grows exponentially, the boundary condition φ(r → +∞) → 0 implies A = 0. To find B, we impose lim E(r) = Evac. (r) , r→0 where λ 2π0 r is the electrostatic field created by an infinite wire with linear charge density λ in vacuum. The electric field is dφ d r dK0 (ξ) dξ E(r) = − = −B K0 = −B dr dr λD dξ dr 1 1 r = − B[−K1 (ξ)] = BK1 . λD λD λD Since for small ξ, K1 (ξ) ' 1ξ , i.e., K1 λrD ' λrD , for small r Evac. (r) = E(r) ' B . r λ The boundary condition as r → 0 leads to B = 2π , so that 0 r λ r λ K1 ; φ(r) = K0 . E(r) = 2π0 λD λD 2π0 λD 61 (b) From ∇2 φ = − ρ0 = 1 φ, λ2D λ r ρ(r) = − K0 . 2πλ2D λD The total charge in the plasma, around a length l of the wire, is ˚ ˆ +∞ rρ(r) dr ρ(r) dV = 2πl 0 ˆ +∞ λ r =2πl r− K0 dr 2πλ2D λD 0 ˆ +∞ λ λD ξ K0 (ξ) λD dξ =−l 2 |{z} | {z } λD 0 Q= ˆ = − lλ |0 r dr +∞ ξK0 (ξ) dξ = −lλ . {z } =1 As Q = −λ , l the plasma perfectly shields the charge in the wire. Since the Bessel functions K0 (ξ) and Ki (ξ) decay with a characteristic decay length of the order of 1 (see the figures in the formulation of the problem), the potential and the electric field are shielded efficiently for distances r λD , as expected. The shielding is complete only when r → ∞, but is already nearly perfect for distances of a few Debye lengths. Like in the previous exercise, it is possible to verify the importance and efficiency of the Debye shielding of the plasma and the meaning of quasineutrality in the very definition of plasma. As an example, when the problem is solved in vacuum the potential decreases very slowly and it is not even possible to impose the condition φ(r → ∞) = 0, contrary to what happens in the present case. 10. (a) As the electrons are lighter and more mobile than the ions, they tend to arrive first at the dust particles, which get negatively charged. The mechanism is somewhat analogous to the negative charging of the walls of a reactor, at the origin of the phenomenon of ambipolar diffusion (cf. chapter 7). Typically, Zd ∼ 100. (b) The electron, ion and dust particle densities are given, respectively, by eφ eφ ' ne0 1 + kB Te kB Te eφ eφ ni =ni0 exp − ' ni0 1 − kB Ti kB Ti ne =ne0 exp nd =nd0 . 62 Solutions to chapter 1. Debye shielding and fundamental effects In turn, Poisson’s equation is ρ e = − (ni − ne − Zd nd ) 0 0 e eφ eφ '− ni0 1 − − ne0 1 + − Zd nd0 0 kB Te kB Te e eφ eφ = ni0 + ne0 + ne0 + Zd nd0 − ni0 . 0 kB Ti kB Te ∇2 φ = − In this case, the quasi-neutrality condition is written as ne0 + Zd nd0 = ni0 . Substituting in the previous expression, eφ eφ ne0 e2 ni0 Te e ni0 + ne0 = 1+ φ. ∇2 φ = 0 kB Ti kB Te 0 kB Te ne0 Ti (c) The usual expression ∇2 φ = is obtained defining λD = 0 kB Te ne0 e2 1 φ, λ2D 1/2 where λD has dimensions of length, 1 1+ ni0 Te ne0 Ti 1/2 = λDe 1 1+ ni0 Te ne0 Ti 1/2 , where λDe is the electron Debye length in the absence of dust particles. Without dust particles, ni0 /ne0 = 1; on the other hand, in the presence of dust particles, ni0 /ne0 > 1, leading to a smaller Debye length. (d) The electrostatic potential can be obtained as in the standard derivation of the Debye length. It follows problem 5, with the exception of the second boundary condition, which should be replaced by lim φ(r) = r→0 1 qT , 4π0 r corresponding to recovering the potential created by a point charge in vacuum when we are very close to the charge (no shielding). With the qT notation of problem 5, this means setting C1 = 4π and, accordingly, 0 φ(r) = qT r exp − , 4π0 r λD the usual expression of the Debye potential, but with λD as defined above. SOLUTIONS TO CHAPTER 2 Single particle motion I 1. From rLs = v⊥s /ωcs and ωcs = eB/ms , where s = {e, i} denotes electrons and protons, respectively, rLi v⊥i mi = . rLe v⊥e me Furthermore, the condition on the kinetic energy corresponds to √ me v⊥e . Hence, r √ rLi mi = ' 1837 ' 43 . rLe me √ mi v⊥i = 2. (a) The particle is initially accelerated by the electric field on the positive y-direction. Its orbit is turned by the magnetic field, so that the resulting ~ ×B ~ drift in the motion is a combination of cyclotron motion with an E ~ ~ positive x-direction. The E × B drift is the outcome of the increase of the Larmor radius with v⊥ (the velocity in the plane perpendicular ~ in this case v⊥ is the velocity in the x − y plane), which takes to B; place while vy > 0. From energy conservation it is clear that when the particle returns to y=0 it has zero velocity. Moreover, since ~v (t=0)=0, vk = vz is always zero, as there are no forces in the z-direction. The orbit is schematically sketched in figure 2.1. ~ = B0 ~uz and E ~ = E0 ~uy , the compo(b) In cartesian coordinates, with B ~ + ~v × B) ~ are nents of the Lorentz force F~ = q(E max = qvy B0 may = −qvx B0 + qE0 maz = 0 . The last equation implies vz = cte. Using the initial condition vz (t=0)=0, it comes vz (t) = 0. 64 Solutions to chapter 2. Single particle motion I y Larger rL vy=0 Smaller rL x t=0; vy=0 y vd x Figure 2.1: problem 2 The two first equations can be written as q v̇x = m vy B 0 q v̇y = − m vx B0 + qE0 . (2.1) Differentiating the first equation and replacing v̇y , v̈x = −ωc2 vx + where ωc = Defining qB m , q 2 E 0 B0 , m which is the equation of a driven harmonic oscillator. ζ = ωc2 vx − q 2 E 0 B0 , m so that ζ̈ = ωc2 v̈x , and replacing in the equation of the driven harmonic oscillator, ωc−2 ζ̈ = − ζ | {z } v̈x ζ̈ = −ωc2 ζ . (2.2) 65 This is the equation of a simple harmonic oscillator, with solution ζ(t) = ζ0 cos(ωc t + ϕ) , where the amplitude ζ0 and the initial phase ϕ have to be determined from the initial conditions. From (2.2) the expression for vx (t) is vx (t) = E0 1 ζ0 cos(ωc t + ϕ) + , ωc2 B0 while from (2.1) vy (t) = m 1 v̇x (t) = − 2 ζ0 sin(ωc t + ϕ) . qB0 ωc Determination of ζ0 and ϕ: vy (t = 0) = 0 ⇒ ϕ = 0 ζ0 E0 vx (t = 0) = 0 ⇒ 2 + =0 ωc B0 q 2 ζ0 = − E0 B0 . m The particle velocity is thus given by E0 [1 − cos(ωc t)] B0 E0 sin(ωc t) . vy (t) = B0 vx (t) = Finally, the orbit is obtained as ˆ t x(t) = vx (τ ) dτ 0 τ =t E0 1 = τ− sin(ωc τ ) B0 ωc τ =0 E0 1 = t− sin(ωc t) ; B0 ωc ˆ t y(t) = vy (τ ) dτ 0 E0 τ =t [cos(ωc τ )]τ =0 B 0 ωc E0 = [1 − cos(ωc t)] , B0 ωc =− where the initial conditions x(t = 0) = y(t = 0) = 0 were used. (c) From the previous item, ~v = vx ~ux + vy ~uy E0 E0 E0 =− cos(ωc t) ~ux + sin(ωc t) ~uy + ~ux B0 B0 B0 ≡ ~vc + ~vd , 66 Solutions to chapter 2. Single particle motion I where ~vc = − E0 E0 cos(ωc t) ~ux + sin(ωc t) ~uy B0 B0 corresponds to an oscillatory motion (cyclotron motion) and ~vd = E0 ~ux B0 ~ ×B ~ drift). to a constant drift (E Taking the time average over several gyroperiods, h~v i = ~vd = cte., since the time average of the sinusoidal functions is zero (hcos(· · · )i = hsin(· · · )i = 0). Therefore, h~ai = 0 and there is no average acceleration. ~ cancels the electric force q E. ~ The average force q~vd × B (d) The drift velocity ~vd is independent of the charge and the mass. Hence, all the particles drift with the same velocity and there is no net current in a neutral plasma. The particle’s trajectory for q < 0 is depicted in figure 2.2 y x vd Figure 2.2: trajectory for q < 0 (problem 2). (e) In this case the drift velocity could be obtained simply by replacing E = F/q in the previous result for ~vd , and would be ~vd = F mg ~ux = ~ux , qB qB which does depend on the charge. Accordingly, the drift velocity would be different for particles of different mass and would have opposite directions for particles of different charge, resulting in a net electrical current. 3. The electrostatic field created by the electron beam is readily obtained from Gauss’ law. Considering a cylinder of radius r > R and length l, as shown in 67 figure 2.3, ‹ ~ · ~n dS = Qint E 0 ene πR2 l E(r)2πrl = − 0 en R2 e ~ ~ur E(r) = − 2r0 ~ = R) = − ene R ~ur E(r 20 Bz uθ R v l Figure 2.3: problem 3. ~ ×B ~ drift velocity is The E ~vd = ~ ×B ~ E E(R)B0 ene R = ~uθ = ~uθ . B2 B02 20 B0 Substituting values, vd ' 4.5 × 103 m/s. ne e 4. (a) Since ne (r) = n0 exp kBeφTe and ∂n ∂r ' − λ , ∂ne n0 e ∂φ eφ = exp ∂r kB Te ∂r kTe ne n0 eφ '− = − exp ; λ λ kTe e ∂φ 1 =− kB Te ∂r λ ∂φ kB Te =− . ∂r eλ Hence, ~ = −∇φ ~ = − ∂φ ~ur = kB Te ~ur . E ∂r eλ 68 Solutions to chapter 2. Single particle motion I (b) Assuming the electrons have a Maxwellian velocity distribution, their average kinetic energy is K= 1 3 mhv 2 i = kB Te , 2 2 where each component of the velocity contributes with 2 (equipartition theorem). Since v⊥ = vx2 + vy2 , 2 hv⊥ i= 1 2 kB Te to K 2kB Te ≡ vt2 . m Hence, the average Larmor radius is hrL i = mhv⊥ i mvt = , eB eB or, equivalently, vt = eBhrL i . m On the other hand, E kB Te = B eλB 2kB Te m m = = vt2 m 2eλB 2eλB vt hrL ieB m = hrL i . = vt m } 2eλB 2λ | {z vE = vt The last expression can be re-written as vE hrL i = 2λ . vt When vE = vt the Larmor radius is hrL i = 2λ. Notice that when the magnetic field is strong enough to impose vE vt , the Larmor radius is much smaller than the characteristic distance for the variation of the electron density ne , rL λ. In this case, within a gyroperiod the electron sees a nearly homogeneous plasma and one need not bother with the fast Larmor motion, which can be averaged out to leave the slower guiding center drift. This procedure corresponds to the study of phenomena that occur on time scales τ = 1/ω such ~ ×B ~ that ω ωce (τ τce ). The former picture is modified as the E drift increases, since the Larmor radius becomes comparable with the typical length λ and the perturbation approach may become questionable (in particular the very use of the orbit theory in constant electric and magentic fields, as the sources of the fields are changing significantly within a Larmor orbit). 5. (a) At the equatorial plane the magnetic field points along the z-direction (see figure 2.4) and, as it falls as 1/r3 , 3 RT ~ B = B(r)~uz = B0 ~uz , r 69 with B0 = 3 × 10−5 T and where RT ' 6370 km is the radius of the Earth. uz B Earth N B uθ Figure 2.4: problem 5. Since the gradient drift is given by ~vd = 2 ~ ~ 1 mv⊥ B × ∇B , 2 q 2B B it is enough to calculate the gradient in the direction perpendicular to the magnetic field, ~ ⊥= (∇B) B0 3 ∂B ~ur = −3 4 RT3 ~ur = − B(r)~ur . ∂r r r 2 ~ × ∇B/B ~ Accordingly, |B | = 3/r and vd = 2 1 mv⊥ 3 . |q| 2B r 2 For an isotropic velocity distribution, hv 2 i = 32 hv⊥ i (see exercise 4) and 3 2 so the kinetic energy is K = 4 mhv⊥ i or, alternatively, r v⊥ = 4K . 3m Substituting values, and noting that K/e is the kinetic energy in eV and B(r = 5RT ) = B0 /25, vd = K 2 K 50 = , e B(r = 5RT )5RT e B0 RT 1 it comes, vde ' 7.85 × 103 m/s for the electrons and vdi = ved 30×10 3 ' 2.61 × 10−1 m/s for the positive ions. ~ points in the direction −~ur , so that B ~ × ∇B ~ points (b) The gradient ∇B in the direction −~uθ (see figure 2.4). Taking into account the charge of the particles in the expression for ~vd , the electrons drift eastwards (i.e., along ~uθ ) while the positive ions drift westwards (along −~uθ ). 70 Solutions to chapter 2. Single particle motion I (c) At the distance r = 5RT an electron has to cover a distance 2π(5RT ). T The time it takes to orbit the Earth is therefore T = 10πR ' 7 h. vd (d) J = ne(vdi − vde ) ' nevde = 1.25 × 10−8 A/m2 . 6. (a) If there would be no electric field, the electrons would simply describe cyclotron motion (uniform circular motion around the magnetic field ~ ×B ~ drift is lines). Since there is an electric field, to this motion a E added. The resulting motion is depicted in figure 2.5. ExB drift L Cyclotron motion B Re E E B Ri 2Ri Re E Re B Cyclotron motion ExB drift ExB drift Figure 2.5: simplified electron trajectory in the thruster (problem 6). (b) The electron and ion Larmor radii are given by, respectively, v⊥e me ' 1.14 × 10−5 m eB v⊥i mi = ' 2.72 × 10−2 m eB RLe = RLi Since rLe L, the electrons are magnetized (i.e.,trapped in the magnetic field lines). For the positive ions, on the contrary, 2rLi . L, so that a part of the positive ions can espace the confinement, by moving to regions where the magnetic field vanishes (see figure 2.6). These ions are ejected and communicate an acceleration to the thruster. In this rough estimation the thruster operation does not seem very efficient, since 2rLi /L ' 0.2. Notice, however, that the ion acceleration by the electric field was not taken into account. Moreover, there is another design for the thrusters, the thruster with anode layer (TAL), with a short acceleration channel, L ∼ 3 cm, for which the current estimation gives 2rLi ' L. 71 L B≃0 B≠0 E 2rL Figure 2.6: ciclotron motion in the thruster (problem 6). (c) This exercise is very similar to problem 2, so that just an outline is given ~ and defining the y−axis along B ~ (cf. here. Taking the x−axis along E figure 2.6), the Lorentz force equation reads evz B v̇x = − eE mv̇x = −eE + evz B m + m mv̇y = 0 v̇y = 0 −→ . mv̇z = −evx B v̈z = − eB v̇ m x The equation for vy implies vy (t) = cte. Since vy (0) = 0, vy (t) = 0. E 2 , the vz − ωc2 B Substituting v̇x in the last equation and defining ζ = ωce 2 simple harmonic oscillator equation ζ̈ = −ωce ζ is recovered. Therefore, ζ(t) = ζ0 cos(ωce t + ϕ), where ζ0 and ϕ are constants to be determined from the initial conditions. Moving back to the original variables and imposing vx (0) = 0 and vz (0) = 0, the solution for the velocities is E sin(ωce t) B E vz (t) = − [cos(ωce t) − 1] . B vx (t) = The trajectory is obtained by integration the velocities on time, E 1 cos(ωce t) B ωce E E 1 sin(ωce t) + t , z(t) = − B ωce B x(t) = − E eE where we identify a cyclotron motion with radius rLe = Bω = m eB 2 ce and a constant velocity drift along zz. Substituting values, rLe = 1.14× 10−3 m L = 30 cm. E 2 iE (d) Similarly, rLi = Bω =m eB 2 = 2.7 × 10 m L. ci On the one hand, the results confirm that the electrons are magnetized, confined the to chamber, because rLe L. On the other hand, since rLi L, the positive ions can be accelerated and leave the chamber, producing a thrust and pushing system, as the qualitative analysis of 6b somehow anticipated. However, the effect is significantly larger than 72 Solutions to chapter 2. Single particle motion I ~ estimated in that simple analysis, due to the ion acceleration on the E field. ~ 7. (a) From the force equation, F~ = q E, mẍ = qE0 cos(ωt) ÿ = 0 z̈ = 0 ; qE0 dvx dt = m cos(ωt) vy (t) = vy (t = 0) = 0 vz (t) = vz (t = 0) = 0 . Hence ˆ t qE0 qE0 cos(ωτ ) dτ = sin(ωt) , m mω 0 ˆ t qE0 qE0 −qE0 qE0 x(t) = xi + − cos(ωt) + , sin(ωτ ) dτ = 2 2 2 mω mω mω mω 0 | {z } vx (t) = xi qE0 x(t) = − cos(ωt) . mω 2 The charge as an harmonic oscillatory motion around x = 0, with fre0 quency ω and amplitude |q|E mω 2 , as shown in figure 2.7. x qE0/mω2 t -qE0/mω2 Figure 2.7: oscillatory trajectory in problem 7a. (b) ~ is stronger in the positive direction of the x-axis. i. The electric field E This means the “restoring force” is stronger on the upper part of the trajectory than on the lower part and, accordingly, the center of mass drifts to the regions of smaller electric field. The trajectory is schematically represented in figure 2.8. 73 x t Figure 2.8: qualitative sketch of the trajectory in problem 7b. A. Following the suggestions, we have, successively, x = x0 + x1 dE0 E0 (x) ' E0 (x0 ) + x1 (x0 ) dx dE0 (x0 ) cos(ωt) . E(x, t) ' E0 (x0 ) + x1 dx The expansion is valid if x1 dE0 (x0 ) |E0 (x0 ) dx ; dE0 E0 (x0 ) (x0 ) , dx x1 which means that the amplitude of the high-frequency oscillatory component is much smaller than the characteristic distance of the variation of the electric field. The result follows directly from the force equation, mẍ = qE, which reads dE0 m(ẍ0 + ẍ1 ) = q E0 (x0 ) + x1 (x0 ) cos(ωt) . dx B. Since x1 is a quickly oscillating function, ẍ1 ẍ0 and |E0 (x0 )| 0 x1 dE dx (x0 ) , the equation of motion in the previous item reduces to mẍ1 ' qE0 cos(ωt) . This equation was solved in 7a, with solution x1 (t) = − qE0 cos(ωt) . mω 2 74 Solutions to chapter 2. Single particle motion I C. Taking the temporal average of the equation of motion†on the short time interval 2π/ω (which is analogous to taking the tem~ ×B ~ poral average along a gyroperiod in the calculation of the E ~ drifts), or the ∇B dE cos(ωt) . mhẍ0 i + mhẍ1 i = qhE0 cos(ωt)i + q x1 dx Furthermore, from the previous step, mhẍ1 i = qhE0 cos(ωt)i, so these two terms cancel in the previous equation, which reduces to dE dE qE0 2 cos (ωt) mẍ0 ' q hx1 cos(ωt)i = q − dx dx mω 2 q2 dE dE0 q2 2 = − E E0 hcos (ωt)i ; = − 0 2 2 | {z } mω dx 2mω dx 1/2 2 ẍ0 = − q dE0 q2 d E0 =− 2 2 E02 , 2 2 2m ω dx 4m ω dx d 0 where the equality dx (E02 ) = 2E0 dE dx was used. Finally, the charge feels a ponderomotive force q2 d E02 , 2 4mω dx which acts in the direction of decreasing E-field, as described qualitatively in the beginning of the exercise. Fp = mẍ0 = − SOLUTIONS TO CHAPTER 3 Single particle motion II 1. (a) Let the indexes 0 and R denote the mid-plane of the mirrors and the reflection points, respectively. Since µ= 1 2 2 mv⊥ B , the conservation of µ reads 1 2 2 mv⊥0 B0 = 1 2 2 mv⊥R BR , where the magnetic field takes the minimum value at the center, B0 = bmin . The conservation of energy implies 1 1 1 2 2 2 mv⊥0 + mvk0 = mv⊥R , m m m where the condition of zero parallel velocity at the reflection point was used (vkR = 0). The limit when the proton escapes corresponds to BR = Bmax . (b) How long does it take to reach that energy? Suggestions: i) suppose that the B field is approximately uniform in the space between the mirrors and changes abruptly near the mirrors, i.e., treat each mirror as a flat piston and show that the velocity gained at each bounce is 2vm ; ii) compute the number of bounces necessary; iii) assume that the distance between the mirrors does not change appreciably during the acceleration process. 2. (F. F. Chen) A plasma with an isotropic distribution of speeds is placed inside a magnetic mirror with mirror ratio Rm = 4. There are no collisions, so that the particles in the loss cone escape, while the others remain trapped. Calculate the fraction of particles that remains trapped. 76 Solutions to chapter 3. Single particle motion II 3. (F. F. Chen) The magnetic field along the axis of a magnetic mirror is B( z) = B0 (1 + α2 z 2 ), where α is a constant. Suppose that at z = 0 an electron has 2 . velocity v 2 = 3vk2 = 32 v⊥ (a) Describe qualitatively the electron motion. (b) Determine the values of z where the electron is reflected. (c) Write the equation of motion of the guiding center for the direction ~ and show that there is a sinusoidal oscillation. Calcule the parallel to B frequency of the motion as a function of v. ~ = −Ėt~uy , 4. Consider a particle moving in a time-dependent electric field E ~ where Ė is a constant, and a uniform magnetic field B = B0 ~uz . ~ ×B ~ drift. (a) Calculate the E (b) Relate the resulting accelerated drift with a force and verify that the drift due to that force is the polarization drift. 5. (F. F. Chen) A plasma is created in a toroidal chamber with average radius R = 10 cm and square cross section of size a = 1 cm. The magnetic fiel is generated by an electrical current I along the symmetry axis. The plasma is Maxwellian with temperature kT = 100 eV and density n0 = 1019 m−3 . There is no applied electric field. ~ field, for both positive (a) Sketch the typical drift orbits in the non-uniform B ions and electrons with vk = 0. (b) Calculate the rate of charge accumulation (Coulomb per second) due to the curvature and gradient drifts on the upper part of the chamber. The magnetic field in the center of the chamber is 1 T and you can use the approximation R a if necessary. ~ = E0 exp(iωt) ~ux , 6. (?) Consider an electron moving in an oscillating electric field, E ~ = B ~uz . perpendicular to a constant and uniform magnetic field, B (a) Calculate the drifts existing on the particle motion and describe qualitatively the motion. (b) Try now to confirm the results you have already obtained, starting directly from the equations of motion. In particular, show that you can indeed recover the results from a) for low frequencies of the field, i.e., ω ωce [Suggestion: i) search for solutions of the form ~v = ~vk + ~vL + ~vD exp(iωt), where ~vL is the velocity of the cyclotron motion ~ ii) verify you can obtain an and ~vD is constant and perpendicular to B; ~ 0 − ev~D × B; ~ iii) make the equation for ~vD in the form iωm~vD = −eE ~ and eliminate ~vD × B]. ~ cross product with B