Introduction to Fusion Physics

Introduction to Fusion Physics J. W. Haverkort October 15, 2009 Abstract This is a summary of the first eight chapters from the book Plasma Physics and Fusion Energy by Jeffrey P. Freidberg. Contents 2 The Fusion reaction 2 3 Fusion power generation 3 4 Power Balance in a Fusion Reactor 5 4.1 Steady state.............................. 5 4.1.1 Ideal ignition......................... 6 4.1.2 Ignition............................ 6 4.1.3 Burning plasma........................ 7 4.2 Time dependent power balance................... 8 5 Design of a simple magnetic fusion reactor 9 5.1 Blanket and shield thickness..................... 10 5.2 Plasma Radius and Coil thickness................. 10 5.3 Other quantities........................... 11 6 Overview of Magnetic fusion 12 7 Definition of a fusion plasma 13 7.1 Shielding DC electric fields..................... 13 7.2 Shielding AC electric fields..................... 13 7.3 Low collisionality and collective effects............... 14 7.4 Plasma requirements......................... 15 8 Single particle motion in a plasma 16 1

2 The Fusion reaction The fusion reaction currently under consideration for use in fusion experiments is the D-T reaction: D + T α + n + 17.6MeV (1) with D = 1 D 2 and T = 1 T 2 and α = 2 He 4 deuterium, tritium and Heliumnuclei respectively. Although various alternatives exist, the fusion reaction described by Eq. 1 is the easiest to realize. Deuterium, which is a hydrogen isotope with one additional neutron, is readily available from sea-water which contains one deuterium atom for every 6700 atoms of hydrogen. It can be extracted at very low cost to supply mankind with energy for about two billion years. Because tritium has a half-life of about 12 years there is no natural tritium to be found on earth, so that tritium has to be bred using lithium 3 Li 6 : Li + n(slow) α + T + 4.8 MeV (2) creating some additional energy as a by-product. At present energy consumption rates there is enough lithium for the coming 20,000 years. By that time reactors have to be devised that are able to operate the somewhat more difficult D-D reaction. Adding Eqs. 1 and 2 D + Li 2α + 22.4MeV (3) Note that the neutron produced in the fusion reaction is used in the breeding of tritium. Unavoidable neutron losses call for some form of neutron multiplication in order to keep the reaction going. By conservation of momentum the released energy is spread over the kinetic energy of the produced particles inversely proportional to their masses. The lighter neutrons will therefore carry with 14.1 MeV, most of the 17.6 MeV produced in the D-T reaction of Eq. 1. They will have to be slowed down to cause Lithium to decay through Eq. 2. Of course in a plasma containing D and T also other fusion reactions can occur albeit with a smaller probability. 2

3 Fusion power generation The rate at which fusion reactions take place can be found by imagining a beam of species 1 directed at a sample of species 2. The collision frequency is proportional to the particle densities n 1 and n 2, the relative velocity v and the cross-section σ R 12 = n 1 n 2 σv (4) The cross-section σ, which is usually expressed in barns = 10 28 m 2, approximately the cross-sectional area of a uranium nucleus. It is related to the mean free path λ and the particle density n via λ = 1 (5) nσ Introducing the particle velocities v 1 and v 2, distributed according to the distribution functions f 1 (v 1 ) and f 2 (v 2 ), which are normalized to n 1 and n 2, this can be generalized to a velocity dependent collision rate R 12 = f 1 (v 1 )f 2 (v 2 )σ( v 1 v 2 ) v 1 v 2 dv 1 dv 2 = n 1 n 2 σv (6) For like particles the integral evaluates to R 12 = 1 2 n 1 σv, with the factor 1/2 appearing because the integration over dv 1 dv 2 counts each collision twice. Because the cross-section for Coulomb collisions is much larger than the crosssection for fusion we can assume that the distribution functions f 1 and f 2 will relax to thermodynamic equilibrium on a time-scale much shorter than the nuclear collision time. We can therefore assume a Maxwellian distribution for j = (D or T ) f j (r, vt) = n j ( mj 2πT j ) 3/2 1 2 m j v2 k e B T j (7) Multiplying the reaction rate R 12 of Eq. 6 with the fusion energy E f = 17.6 MeV of Eq. 1, gives the energy production p.u.v. from fusion S f = E f n 1 n 2 σv (8) Because in the fully ionized plasma there are as many electrons as positively charged particles n e = n D + n T. The product n D n T is maximized for a 50% D and 50% T mixture such that S f = 1 4 n2 ee f σv (9) An energy loss is associated with the fact that due to the frequent Coulomb collisions the charged particles change direction and radiate electromagnetic radiation that can escape the plasma. This Bremsstrahlung energy loss per unit time can after some analysis be found to be S B = C B Z eff ne 2 Te with (10) ( ) ( ) 2 e 6 C B (11) 3π 5/2 ɛ 0 c 3 hm 3/2 e 3

with e = 1.60 10 19 C the elementary charge, ɛ 0 = 8.85 10 12 F/m the electric permittivity of the vacuum, h = 6.63 10 34 Js Planck s constant and m e = 9.11 10 31 kg the electron mass. Z eff = n 1 e j Z jn j denotes the effective charge, in units of e, of the ions in the plasma. Note that the production of Bremsstrahlung radiation is a collisional process, explaining the dependence of Eq. 10 on the square of the particle density. Through σv (see Fig 3.11) the fusion energy production depends very strongly on temperature. This calls for an analysis to determine for which temperature the energy production exceeds the energy losses, as is of course required for a fusion reactor. 4

4 Power Balance in a Fusion Reactor We set-up a 0-D model for the power balance in a fusion device. First for steady state operation and later considering transient effects, assuming 1. A 50%-50% D-T mixture with a negligible concentration of alpha particles, such that 2n D = 2n T = n e n and n α 1. 2. All particles are at equal temperature, i.e. T D = T T = T e T 3. The fuel is assumed in the form of a fully ionized, ideal, gaseous plasma near thermodynamic equilibrium corresponding to Maxwellian distribution functions (Eq. 7). In this case the internal energy density and particle pressures are given by U j = 3 2 n jk B T j p j = n j k B T j (12) such that the total internal energy density and pressure of the fuel are given by U = 3nk B T and p = 2nk B T and U = 3 2 p. The above assumptions are reasonably well satisfied in fusion reactors. 4.1 Steady state The steady state power balance for a fusion plasma can be written with S α + S h = S B + S κ (13) 1. S α : the power supplied to the plasma by the alpha particles, corresponding to the fusion reaction of Eq. 3. The neutrons, due to their charge neutrality, pass through the plasma without losing much energy. Their power, S n, is therefore not included in the plasma power balance. With p = 2nk B T, Eq. 8 for the alpha particles becomes S α = 1 16 E αp 2 σv k 2 B T 2 (14) 2. S h : The external heating of the plasma. This is done primarily by Ohmic heating, i.e. by setting up an electrical current whose dissipation heats up the plasma. Because this current is generated by a transformer process, which is an inherently in-stationary process, this form of heating is not included in the present steady state power balance. 3. S B : Bremsstrahlung, taking power away from the plasma. With p = 2nk B T, Eq. 10 becomes S B = 1 4 C BZ eff p 2 (k B T ) 3/2 (15) 5

4. S κ : Losses due to heat conduction. Losses are often modelled using 1 V q da which, with Fourier s law q = κ T, for a cylindrical plasma of radius a becomes 2κ T r r r=a. The thermal conductivity and temperature gradients are, however, usually difficult to assess. Alternatively, the conduction losses are written in terms of an energy confinement time τ E (p, T ) 1 V The power balance can now be written 4.1.1 Ideal ignition q da = U = 3 p (16) τ E 2 τ E 1 16 E αp 2 σv (k B T ) 2 + S h = 1 4 C p 2 BZ eff (k B T ) + 3 3/2 2 p (17) τ E Ideal ignition is defined as the condition for which alpha power is large enough to overcome Bremsstrahlung losses S α = S B. The pressure dependence cancels out and the condition reduces to σv kb T = 4C B E α 3.8 10 24 m3 /s kev 1/2 k BT 4.4 kev (18) where a high degree of plasma purity is assumed implying that Z eff 1. Note that 1keV 11.6 million Kelvin so that that even for this, very favourable, definition of ignition the plasma is required to be extremely hot. 4.1.2 Ignition Using f α Sα S α+s h the power balance Eq. 13 can be written as S α (1 + 1/f α ) = S B +S κ. The definition of ignition requires the alpha power heating to overcome the Bremsstrahlung and thermal conduction losses without extra heating power, i.e. f = 1: pτ E 3 2 (k BT ) 2 E α 16f α σv C B 4 kb T f α 24k 2 B E α T 2 σv (19) The latter approximation follows when Bremsstrahlung losses are neglected and is known as Lawson s criterion. With pτ E = 2k B T (nτ E ) it can, for f α = 1, be written as nτ E 12 E α k B T σv (20) where nτ E is known as the Lawson parameter. It turns out that the product pτ E has a distinct minimum for T min = 15keV and (pτ E ) min = 8.3atm s (21) To satisfy Eq. 19 in practice one has to make use of this minimum, because for lower temperatures instabilities persist and higher temperatures are too difficult to achieve. Note that the ignition criterion will, in practice, be achieved 6

with the help of some external heating, i.e. f α 1. This is due to the fact that for confinement an electrical current is generated in a tokamak whose dissipation adds to the power balance. From Eq. 19 this decreases the minimum pτ E needed for steady state power balance by a factor of f α. 4.1.3 Burning plasma Defining the physics gain factor as Q (P out P in )/P in. With P in = S h V and P out = (S n + S B + S κ )V with V the plasma volume yields using the steady state power balance Eq. 13 Q = S f S h (22) where S f = S n + S α. Therefore Q is also the ratio of the total fusion power and the heating power. Usisng conservation of momentum S n = 4S α, so that S f = 5S α. Neglecting Bremsstrahlung yields with (S κ /S α )pτ E = (pτ E ) I = 24k 2 B T 2 E α σv the value of pτ E required for ignition: Q = Or, with Q/(S + Q) = (pτ E )/(pτ E ) I = f α, 5pτ E (pτ E ) I pτ E (23) f α = Q 5 + Q (24) The burning plasma regime is defined as the balance for which alpha power just equals external power. Alpha power is only one fifth of the total fusion power, but it is the power that remains in the plasma. In this case f α = 1/2 such that Q = 5. External heating decreases f α such that from Eq. 19 it is easier to satisfy the requirements on pτ E. However from Eq. 24 this deteriorates Q. Of course not all of P out can be used, and in order to supply the plasma with P in a larger amount of power is generally needed. A slightly altered definition of Q is based on the total electrical (E) power required, yielding the engineering gain factor Q E (P (E) out P (E) (E) in )/P in. Using typical conversion efficiencies and neglecting Bremsstrahlung, this quantity can be related to Q via In the above analyses the power required to achieve equilibrium has not been taken into consideration. This can, however, be shown to be not large enough to change the power requirements for a large fusion reactor significantly. Q E 2.0 pτ 0.37(pτ) I (pτ) I pτ Q 0.72 (25) 4.0 This shows that for electrical power break even (Q E = 0) a value of pτ E = 0.37(pτ) I is needed corresponding to Q = 2.9. For a commercial fusion power plant an electric power gain factor of Q E = 10 is desirable, such that pτ E = 0.9(pτ) I corresponding to Q 43. The required pτ E is thus not significantly below its ignition value. Finally we note that, from Eq. 25, a physics gain factor of Q = 10 is equivalent to a engineering gain factor of only Q E = 1.8. 7

4.2 Time dependent power balance In our 0-D model, time dependence can be included by using the observation that an imbalance in Eq. 13 leads to a time rate of change of the internal thermal energy U = 3p/2: 3 dp 2 dt = S α + S h S B + S κ (26) or, recalling p = 2nT and assuming constant n = n 0 particle density dt 3n 0 dt = S H(T ) S L (T ) (27) Here the heating and loss terms have been grouped together. Defining T = T 0 as the equilibrium state for which T = 0, the slope dt /dt T0 should be negative for thermal stability. In this case an increase in temperature above T 0 leads to a feedback in which the loss terms S L exceed the heating terms S H such that the temperature is decreased. Evaluating the condition ds L /dt T0 ds H /dt T0 leads, neglecting Bremsstrahlung, to 3n 0 /τ E 1 4 E αn 2 0d σv /dt T0. Or, using the steady state power balance Eq. 13 and the definition of f α, d dt ( T 1/f α ) σv T 0 0 (28) Note upon comparison with Eq. 19 that f α = 1/2 corresponds to the minimum in pτ E and that for a given value of pτ E stability occurs for the high temperature solution of the power balance, i.e. to the right of the minimum in Fig. 4.2. Working at a lower temperature than T min thus requires some form of feedback, called burn control. Including effects of the temperature dependence of σv makes analysing the power balance more difficult, but turns out to substantially improve thermal stability so that the need for burn control is eliminated. 8

5 Design of a simple magnetic fusion reactor In the present section we will try to design a nuclear fusion reactor, suitable for commercial energy production. We will leave aside various stability issues and technological challenges and instead focus on the core problem of determining the optimal engineering parameters. A fusion device typically consists of: The plasma - containing the fuel for the fusion reaction of Eq. 1. The first wall - capable of handling an enormous radiation load. The blanket - a material layer interspersed with cooling tubes transporting the absorbed energy away, in order to convert it to electricity. The blanket also has the function of A neutron multiplier: because Eq. 3 is neutron neutral and, as inevitably neutrons escape the reactor, some multiplication is required. A moderator: in which the fast 14.1 MeV neutrons are slowed down to be able to take part in the reaction of Eq. 2 A breeder: where Lithium is converted to Tritium via Eq. 2 A shield - protecting the superconducting magnet which must remain below 100 K, shielded from any form of radiation. Magnets - superconducting coils capable of producing enormous magnetic fields We consider a torus-shaped tokamak with a circular cross-section. In this fusion device we will consider all of the above elements to be present as concentric rings. The design goal is to minimize the costs, while maintaining various constraints, which are taken to be: A power output of P E plant = 1 GW, representative for a large scale power No more than approximately P W = 4 MW/m 2 of neutron power can pass the first wall The largest large-scale magnetic field that can be generated is approximately B max = 13 T The maximum allowable average stress due to magnetic pressure in these superconducting magnets is σ = 300 MPa and From the previous chapter, it is desirable to have a temperature T = 15keV for which σv = 3 10 22 m 3 /s The cross-section for slowing down neutrons is approximately σ sd = 1 barn Once the neutrons are slowed down to an energy of E t = 0.025 ev, tritium has a breeding cross-section of approximately σ br = 950 barns 9

The plasma is contained within a region of radius a, covered by a blanket, first wall and shield of combined thickness b and a magnet of thickness c, see Fig. 5.1. The minor radius of the torus is thus a+b+c. The volume of material is given by V I = (2πR 0 )π((a + b + c) 2 a 2 ) (29) Our goal will be to design a reactor which minimizes the cost per unit energy produced, given the above constraints. The total costs will consist of fixed costs, assumed to be proportional to the electric power output, and nuclear island costs, assumed to be proportional to the volume V I. 5.1 Blanket and shield thickness Neutrons have to be slowed down to take place in the reaction of Eq. 2 which is done mostly in the moderator part of the blanket. It is experimentally observed that at low energies the cross-section of lithium, of which both the moderator part and breeder-part of the blanket are made, is approximately inversely proportional to the neutron velocity, i.e. σ v n E. The neutron energy E decays exponentially E = E n exp x/λ sd over a characteristic slowing down - distance λ sd. Using λ 1/σ (Eq. 5) it follows that the characteristic breeding mean free path is λ = λ br E/Et = λ br En /E t exp x/2λ sd. Assuming the neutron flux Γ n n n v n satisfies a decay-type equation dγ n /dx+γ n /λ we obtain dγ n dx + solving and inverting gives x = 2λ sd ln Et E n exp x/2λ sd λ sd Γ n = 0 (30) [ 1 1 2 ( ) ] En λ br Γn ln (31) E t λ sd Γ n0 Next the moderator and breeder material are assumed to consist of unenriched natural Li (7.5% Li 6 and 92.5% Li 7 ) and we require that Γ n /Γ n0 = 0.99, i.e. 99% of the neutrons have slowed down and underwent a breeding reaction. Due to the double logarithm, any fraction close to one gives approximately the same result. Inserting the numbers gives x 0.88m. The moderator can be considered to have evolved into the breeder when λ sd = λ br E/Et at x 0.79. The breeding thus takes place over a relatively small region. Taking into account some space for neutron multiplication, shielding, cooling tubes, and structural support we take the blanket thickness to be b = 1.2m. 5.2 Plasma Radius and Coil thickness After having obtained the blanket and shield thickness b, we need two more equations in order to find the other two geometrical quantities a and c. The first equation follows from the constraint of the maximum allowable stress in the superconducting magnets. The outward magnetic force is given by the Lorentz force df = BIdl where B varies from B c just inside the coil to 0 on the outside, so we will use the average B c /2. Next we anticipate that the maximum achievable magnetic field strength of B c = B max = 13 T is necessary 10

for good confinement. Otherwise our optimization would yield a trivial result for B c 0. The current flowing in the superconductor follows from Ampere s law, yielding B c = I/2πµ 0 R 0. The opposing inward tensile force is constrained by material properties, yielding a first relation between a, b and c. A second relation is obtained from minimizing the reactor volume V I (Eq. 29) per unit of electric power produced. Solving these two equations for the two unknowns a and c yields a = 2 m and c = 0.79 m. Using the constraint on the maximum wall loading, the major radius can be found to be R 0 = 5 m. These numbers constitute, with a volume of V P 400 m 3 and V P + V I 1570 m 3, quite a large reactor in comparison with a fission reactor. This large size is partly due to the fact that the magnets have to be large in order to withstand the high stresses due to the high magnetic fields, but mostly due to the fact that all the produced heat has to be absorbed in the blanket. Cooling tubes cannot be inserted into the reactor as in the case of a nuclear fission reactor. A fusion reactor must have a large volume, since heat can be removed only through the relatively small bounding surface area. This also implies a relatively low plasma power density. From the obtained plasma volume and output electricity, the power density S f = (P α + P n )V P 4.9MW/m 3, which is some twenty times lower than that of a fission reactor. This might be favourable in terms of safety but leads to an inherently larger, and therefore costlier, device. This downside has to be balanced by the advantages of a lower level of radioactivity, ease of waste disposal and reduced risk of proliferation. 5.3 Other quantities The ideal plasma temperature has already been calculated to be T = 15 kev. The number density n follows from the plasma power density S f through Eq. 9 and the pressure results from the equation of state yielding p = 7.2 atm and n = 1.5 10 20 m 3. Note the very low number density of only some 10 20 particles per cubic meter. A hydrogen gas at room temperature has, with 2 g/mol, some 500 mol m 3 or n 3 10 26 m 3. The amount of plasma in our reactor is thus a mere 10 20 V P (2g/mol)/N A 0.1g. From the previously derived criterion pτ E = 8.3atm s we obtain an energy confinement time of τ E = 1.2 s. Lastly, we calculate an important plasma physics quantity, the plasma β defined as the ratio between the pressure and the magnetic pressure: p β = B0 2/2µ 0 (32) We choose to use the magnetic field B 0 at the centre of the plasma at R = R 0. As Ampere s law dictates that the field falls off with 1/R, the magnetic field at R = R 0 is (R 0 a b)/r 0 = 0.36 times the maximum magnetic field of 13 T. This yields a significantly lower value of B 0 = 4.7 T, and a value of β = 0.082, approximately the maximum value ever reliably obtained in a tokamak. 11

6 Overview of Magnetic fusion One can look at plasma physics from several alternative viewpoints 1. The motion of single charged particles in external electric and magnetic fields (Ch. 8 and 9) 2. Two-fluid models in which the macroscopic fields are treated self-consistently (Ch. 10) 3. Magnetohydrodynamics (MHD) which is a self-consistent single-fluid model (Ch. 11 and 12) 4. A kinetic description, using a Boltzmann-type of equation (not considered in this book) Topics of further interest include several possible fusion devices (Ch. 13), plasma (heat) transport (Ch. 14), heating and current drive (Ch. 15), and the future of fusion research (Ch. 16). 12

7 Definition of a fusion plasma A plasma is qualitatively defined as an ionized gas whose behaviour is dominated by collective effects and which possesses a very high electrical conductivity. The high conductivity is ensured by a very low collisionality of the charged particles, allowing a plasma with about 10 8 times fewer electrons than e.g. copper to have a conductivity approximately 40 times higher. In this chapter more quantitative criteria for defining a plasma are derived. The first two of these criteria relate to the conductivity, which has to be large enough to effectively shield the plasma from DC electric fields and AC electric fields. A third criterion is a condition on the collisionality and a fourth says something about the size of a plasma 7.1 Shielding DC electric fields Any deviation from a non-zero electric DC field in a good conductor would imply an electric force on the free charges, displacing them so that they will cancel the electric field. In a thin layer at the surface of the conductor, however, the electric field will be non-zero due to the fact that thermal fluctuations make the charge spread out over a finite region. The steady state force balance yields an equality between the pressure force p = (nk B T ) and the electric force ±ene with E φ. Assuming T e = T i = cst, this yields, in one dimension, n e = n 0 exp eφ/k B T e and n i = n 0 exp eφ/k B T e. Using Poisson s equation d 2 φ/dx 2 = (e/ɛ 0 )(n e n i ) one obtains d 2 φ/dx 2 = (en 0 /ɛ 0 )(exp eφ/k B T e exp eφ/k B T e ). Assuming eφ k B T e,i the right hand side can be linearised to obtain the solution d 2 φ dx 2 = e2 n 0 ɛ φ(x) e x/λ D with 1 λ 2 D ( 1 T e 1 T i ) φ (33) = 1 λ 2 + 1 De λ 2 = e2 n 0 + e2 n 0 (34) Di ɛ 0 k B T e ɛ 0 k B T i The characteristic size of the layer over which the electric field decays is called the Debye length λ D. In order to shield DC fields effectively this size must me much smaller than the characteristic size of the plasma. If this is the case the plasma will be very nearly neutral in the bulk, a condition sometimes called quasi-neutrality. 7.2 Shielding AC electric fields next we investigate the effect of a sinusoidally varying external electric field on a plasma. As the electrons are much lighter than the ions, and can therefore respond much faster to a time-varying electric field than the ions, we will assume the ion density is fixed at n i = n 0. The one-dimensional continuity, momentum, and Poisson equations are written v e t + v v e e x = n e t + n ev e x = 0 (35) e φ m e x T e n e m e n e x (36) 13

2 φ x 2 = e ɛ (n e n 0 ) (37) Linearising in the small quantities n 1 = n e n 0, φ and v e assuming harmonic time dependences e iωt yielding an equation for φ n 1 = i n 0 dv e (38) ω dx v e = i T ( e d eφ n ) 1 (39) ωm e dx T e n 0 d 2 φ dx 2 = e n 1 (40) ɛ 0 d 2 φ dx 2 1 ω2 /ω pe φ = 0 with ω pe = n 0e 2 = 2 λ De (41) m e ɛ 0 v T e λ 2 De with ω pe the plasma frequency and v T e = 2k B T the thermal velocity. Note that this equation is the same as Eq. 33, only with a modified effective Debye length ˆλ 2 = λ 2 De /(1 ω2 /ω 2 pe) showing that AC fields are shielded less effectively than DC fields (ω = 0). For ω > ω pe the plasma can no longer be shielded from an AC field, the inertia of the electrons is too large for a timely response. 7.3 Low collisionality and collective effects In order for a smooth macroscopic electric field to be a valid description, the plasma particle density has to be sufficiently low so that two charges rarely get close enough to each other for the two-particle Coulomb force to dominate the long-range electric field generated by the smoothed out distribution of the entire charge population. This requires that the number of particles Λ D = 4π 3 n eλ 3 De = 4π 3 ɛ 3/2 0 Te 3/2 e 3 1 (42) n 1/2 e within a Debye sphere, a sphere of radius λ De, is much larger than one. This can be seen in a number of ways: For a continuum description the plasma must be sub-dividable into a large number of fluid elements, each containing many particles. In order to have a good spatial resolution, the size of a fluid elements should be much smaller than λ De, from which it follows that Λ D 1 The Coulomb interaction is insignificant when on average the inter-particle distance b is so large that the Coulomb potential energy e 2 /4πɛ 0 b is negligible compared to the particle kinetic energy O(k b T e ) so that b n 1/3 e b Coulomb e 2 /4πɛ 0 k b T e or rearranging Λ D O(1) The mean free path λ ee = 1/n e σ ee (Eq. 5) and collision frequency ν ee v e /λ ee corresponding to e-e collisions, have a characteristic cross-section σ ee πb 2 Coulomb = e4 /4πɛ 2 0m 2 ev 4 such that ν ee ω pe /Λ D and λ ee = 14

3λ De Λ D. For long range effects to dominate, λ ee should be long compared to the Debye length and the collision frequency has to be much smaller than the plasma frequency, resulting again in the condition Λ O(1) 7.4 Plasma requirements We have found that a plasma, in order to effectively shield AC and DC electric fields should have a Debye length much smaller than the system size and a plasma frequency much larger than any other occuring frequency in the system. In order for collective effects to dominate, there should be many particles within one Debye sphere. Finally the Larmor radius or gyroradius (to be discussed in the next chapter) should be much smaller than the system size, in order to obtain good plasma confinement. When discussing plasmas, these criteria will tacitly always assumed to be valid. Note that in these criteria always characteristic frequencies and lengths of the plasma are compared with macroscopic frequencies and lengths. The same criteria also hold with respect to mean free paths and collision frequencies. For fusion plasma parameters, however, the resulting conditions are much less strict. Namely, due to the low collisionality in a fusion device the mean free path is much larger than the system size and the thermal transit frequency is much higher than the collision frequency. 15

8 Single particle motion in a plasma Consider a single charged particle in homogeneous background electric and magnetic fields E and B. The equations of motion m d2 r = q(e + v B) (43) dt2 show a distinct difference between motion parallel dr/t = qe/m and perpendicular to the magnetic field dr/t = q(e + v B)/m. We will in what follows always refer to these as simply the parallel and perpendicular directions. A parallel electric field E will causing accelerating motion a = qe /m parallel to the magnetic field lines. Perpendicular to the magnetic field, any initial velocity v perpendicular to the field lines will be bent into a circular motion by the magnetic force. The characteristic frequency and radius of this rapid circular motion is found from either solving the perpendicular component of the above equations of motion or from the centrifugal-magnetic force balance mv 2 /r L = qv B. This yields the Larmor- or gyro-frequency and radius qb ω c m and r L = mv qb = 2mkB T (44) qb where in the last step the perpendicular velocity is assumed to be thermal in nature so that 1 2 mv2 = kt. From this we see that the electron gyro-frequency is a factor m i /m e faster than that of the ions and the ion gyroradius is a factor mi /m e larger than the electron gyroradius. Now we split the charged particle motion in this rapid circular motion and a much slower average guiding centre motion, which in turn varies much faster in time than Coulomb collisions and fusion events, i.e. ω c ω g ν Coul ν fus (45) Apart from a parallel electric field, which accelerates charged particles, other effects can change the location of the guiding centre. A perpendicular force F leaves the perpendicular equations of motion for the variable v = v F B/qB 2 unchanged, leading to a gyrocentre drift velocity V F = 1 q F B B 2 (46) Physically this drift is due to the fact that in the part of the gyrocyclus of decreased velocity the gyroradius decreases (in order to generate sufficient centifugal force to balance the magnetic force) and the radius is increased in the portion of the cycle of increased velocity, resulting in a sidewards motion relative to the applied force. Note that when the force is in the same direction for electrons and ions, their drift velocities will be in opposite direction causing a net current to flow. We consider the following drifst of the guiding centre. With F = q the drift V E = E B B 2 the E cross B drift. is independent of charge and is called A rotating charged particle constitutes a magnetic dipole with its magnetic moment µ = qv 2πr L πrl 2 = 1 2 qv r L = mv 2 /2B parallel to the magnetic field so that in an inhomogeneous magnetic field with B/ B r L a 16

force µ B is exerted on the dipole. This results in a drift velocity V B = v2 B B 2qB B. Note that V 2 B /v T r L /a 1. When the charged particles move along curved magnetic field lines with radius of curvature R c, a centrifugal force F = mv 2 R c/rc 2 is exerted resulting in a drift force V κ = mv2 q R c B R 2 c B2 For a vacuum field B B = BR c B/R 2 c, and the B and curvature drifts can be combined to V B,κ = m q R c B R 2 cb 2 (v + 1 2 v2 ) (47) 17