Plasma Physics for Researchers
Plasma Physics for Researchers
                                                                                                                                                          Chapter 1
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                                                                                                                                               Boltzmann equation
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                                                                                                                                                                 1
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                                                                                                                                                          N
                                                                                                                                          F (r, v, t) =         δ(r − ri (t))δ(v − vi (t))             (1.2)
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i=1
Boltzmann equation 3
                                                                                                                     the relation
                                                                                                                                                                            
                                                                                                                                                 3  3            r , v
                                                                                                                                               d rd v =J                         d3 rd3 v                  (1.8)
                                                                                                                                                                    r, v
                                                                                                                     where J is the Jacobian determinant of the transformation written out in
                                                                                                                     full as:
                                                                                                                                                                           
                                                                                                                                          ∂r1 ∂r2 ∂r3                     
                                                                                                                                                                        0 
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                                                                                                                                                             0     0
                                                                                                                                          ∂r1 ∂r1 ∂r1                       
                                                                                                                                                                            
                                                                                                                                                                           
                                                                                                                                          ∂r1 ∂r2                         
                                                                                                                                                     ∂r3
                                                                                                                                                                             
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                                                                                                                                          ∂r2 ∂r2    ∂r 2
                                                                                                                                                             0     0     0   
                                                                                                                                                                            
                                                                                                                                                                           
                                                                                                                                          ∂r1 ∂r2 ∂r3
                                                                                                                                                                           
                                                                                                                                                            0     0     0 
                                                                                                                                          ∂r3 ∂r3 ∂r3
                                                                                                                                                                            
                                                                                                                                     J =                                    
                                                                                                                                                          ∂v1 ∂v2 ∂v3 
                                                                                                                                                                         
                                                                                                                                          0
                                                                                                                                                0     0
                                                                                                                                                           ∂v1 ∂v1 ∂v1 
                                                                                                                                         
                                                                                                                                                                            
                                                                                                                                                                         
                                                                                                                                          0               ∂v  1  ∂v 2  ∂v 3 
                                                                                                                                                0     0                     
                                                                                                                                                          ∂v2 ∂v2 ∂v2 
                                                                                                                                                                            
                                                                                                                                                                         
                                                                                                                                                          ∂v  1  ∂v 2  ∂v 3 
                                                                                                                                          0     0     0                     
                                                                                                                                                           ∂v3 ∂v3 ∂v3
                                                                                                                     This expression for the Jacobian assumes that the after-collision position of
                                                                                                                     a molecule is uncorrelated with the before-collision velocity of the particle,
                                                                                                                     which is the assumption of molecular chaos. Owing to this assumption, the
                                                                                                                     chain of hierarchy of distribution functions can be terminated at the one-
                                                                                                                     particle distribution function, and the Boltzmann equation becomes self-
                                                                                                                     closed with the collision term that is here derived and is expressed by the
                                                                                                                     Boltzmann distribution function itself. Making use of (1.5) and (1.6),
                                                                                                                                                             ∂ai
                                                                                                                                                  J =1+          dt + O(dt)2                               (1.9)
                                                                                                                                                             ∂vi
                                                                                                                     where we note that the determinant J is equal to the product of the two
                                                                                                                     3 by 3 sub-determinants located at the upper-left and lower-right corners.
                                                                                                                     The upper-left determinant is unity. The lower-right determinant is diagonal
                                                                                                                     with elements
                                                                                                                                                          ∂a1
                                                                                                                                                    1+        dt,
                                                                                                                                                          ∂v1
                                                                                                                     and the like. Here the repeated indexes are summed over.
                                                                                                                                                                         
                                                                                                                                                           ∂ai
                                                                                                                                       d3 r  d3 v  = 1 +     dt + O(dt)2 d3 rd3 v                       (1.10)
                                                                                                                                                           ∂vi
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                                                                                                                     which means that the same number of particles are in the new volume
                                                                                                                     element as in the old element except for those gained or lost by interaction
                                                                                                                     among the particles themselves denoted by the last term of (1.11). Expanding
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                                                                                                                     the left hand side (LHS) of (1.11) in a Taylor series about (r, v, t), we have
                                                                                                                                                                                  
                                                                                                                                                            ∂f     ∂f dri   ∂f dvi
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                                                                                                                               LHS(1.11) = f (r, v, t) +        +         +           dt
                                                                                                                                                            ∂t     ∂ri dt   ∂vi dt
                                                                                                                                                                    
                                                                                                                                                3 3            ∂ai
                                                                                                                                             × d rd vdt 1 +        dt + O(dt)2 .              (1.12)
                                                                                                                                                               ∂vi
                                                                                                                     Thus
                                                                                                                                                                                                    
                                                                                                                                        ∂f   ∂f dri   ∂f dvi    ∂ai                             δf
                                                                                                                                           +        +        +f     dt =                                     .        (1.13)
                                                                                                                                        ∂t   ∂ri dt   ∂vi dt    ∂vi                             δt       c
                                                                                                                                                       Boltzmann equation                                5
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                                                                                                                     Then the number of collisions (per unit time) of the particles in the above
                                                                                                                     flux with the particles of velocity v is (1.15) × f (v, r, t)d3 v:
                                                                                                                                      b db dϕ|v − v1 | f (r, v1 , t)d3 v1 f (v, r, t)d3 v.
                                                                                                                     Then integrating the above equation over all v1 and dividing by d3 v gives
                                                                                                                     the total rate of change of f (v, r, t) due to collisions which scatter v-particles
                                                                                                                     out of the range (v, v + dv):
                                                                                                                                           ∞         2π
                                                                                                                           δf             3
                                                                                                                                   = d v1          db b       dϕ|v − v1 | f (v, r, t)f (r, v1 , t). (1.16)
                                                                                                                           δt out              0          0
                                                                                                                                                                 v + v1 = v + v1 .                                                     (1.20)
                                                                                                                                                               2       2      2     2
                                                                                                                     The collision may be thought of as a linear orthogonal transformation from
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|v − v1 | = |v − v1 |.
d3 v1 d3 v = d3 v1 d3 v.
                                                                                                                     The actual evaluation of this integral for small angle scattering will be
                                                                                                                     performed in §1.7. If the fluid contains several kinds of particles, then each
                                                                                                                     kind of particle has its own distribution function and we readily generalize
                                                                                                                     Boltzmann’s equation to
                                                                                                                                                    ∂fα      ∂fα      ∂fα 
                                                                                                                                                        + vi     + ai     = Cαβ (fα fβ )                                               (1.22)
                                                                                                                                                     ∂t      ∂ri      ∂vi
                                                                                                                                                                                                 β
                                                                                                                     where α and β refer to the particle species, Cαβ (fα fβ ) denotes the collision
                                                                                                                     operator and the summation includes α.
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Boltzmann equation 7
                                                                                                                     Summation invariant. Let Q be any particle property such that the sum
                                                                                                                     of the Q’s of two particles involved in collision remains unchanged in the
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                                                                                                                                                          g = v − v1 ,                                      (1.25)
                                                                                                                         f = f (v, r, t),   f1 = f (v1 , r, t),    f  = f (v , r, t),   f1 = f (v1 , r, t).
                                                                                                                     Hence, ΔQ = 0.
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                                                                                                                     satisfy
                                                                                                                                               
                                                                                                                                         δf
                                                                                                                                               = d3 v1 dσg f  f1 − f f1 = 0              (1.30)
                                                                                                                                         δt c
                                                                                                                     which is satisfied if and only if f  f1 = f f1 . Or
                                                                                                                                       logf (v ) + logf (v1 ) = logf (v) + logf (v1 ).
                                                                                                                     Hence the quantity logf (v) is a summation invariant and it should be
                                                                                                                     expressible as a linear combination of the three summation invariants, 1,
                                                                                                                     mv, and mv 2 /2:
                                                                                                                                                                        m
                                                                                                                                           logf (v) = ma1 + a2 · mv − a3 v 2
                                                                                                                                                                        2
                                                                                                                     where a1 , a2 , a3 are arbitrary constants. By completing the squares, one
                                                                                                                     obtains
                                                                                                                                                                           
                                                                                                                                                          a22     m        a2 2
                                                                                                                                     logf (v) = m a1 +          − a3 v −        .
                                                                                                                                                         2a3      2        a3
                                                                                                                     By a simple transformation, f (v) can be put into the form
                                                                                                                                                   m 32            m(v − V)2
                                                                                                                                          f (v) = N        exp −                         (1.31)
                                                                                                                                                  2πT                  2T
                                                                                                                     where N, V, T are related with a1 , a2 , a3 and defined by
                                                                                                                                                                      
                                                                                                                                  3          1         3       3     1    m 2 3      m
                                                                                                                        N = f d v, V −            vf d v,        T =        v f d v − V 2.
                                                                                                                                             N                 2     N    2          2
                                                                                                                                                                                         (1.32)
                                                                                                                     Equation (1.31) is a moving Maxwell–Boltzmann distribution function. If the
                                                                                                                     gas has no directional preference (V = 0) and is distributed isotropically,
                                                                                                                                                             m     3
                                                                                                                                                                   2             mv2
                                                                                                                                             f (v) = N                 exp −         .                (1.33)
                                                                                                                                                            2πT                  2T
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Boltzmann equation 9
                                                                                                                                                      F(r) = −e∇φ(r)
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                                                                                                                     This is the kinetic theory version of the entropy, which is in fact the negative
                                                                                                                     entropy, −S(t). We have
                                                                                                                                                 
                                                                                                                                           dH            ∂f
                                                                                                                                               = d3 v        1 + log f (v, t) .
                                                                                                                                            dt           ∂t
                                                                                                                     For a spatially homogeneous gas, using (1.21) in the above equation gives
                                                                                                                                               
                                                                                                                         dH
                                                                                                                             = d v1 d v2 bdb dϕ|v1 − v2 | f2 f1 − f2 f1 (1 + logf1 ).
                                                                                                                                    3    3
                                                                                                                          dt
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                                                                                                                               dt    4
                                                                                                                                                                  
                                                                                                                                     × log (f1 f2 ) − log(fi f2 )
                                                                                                                     where we note that (f2 f1 − f2 f1 ) [log (f1 f2 ) − log(fi f2 )] ≤ 0 always. There-
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Boltzmann equation 11
                                                                                                                     is assumed to be much shorter than the time scale for the moments, i.e.,
                                                                                                                     n(r, t), V(r, t), T (r, t) to change appreciably. Therefore the fluid equations
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                                                                                                                                       ∂n
                                                                                                                                          + ∇ · (nV) = 0          (continuity equation)             (1.41)
                                                                                                                                       ∂t
                                                                                                                                   ∂             ∂
                                                                                                                                      (mnVi ) +     (mnvi vj ) − nFi = 0 (Fi = mai )
                                                                                                                                   ∂t           ∂xj
                                                                                                                                                                  (momentum equation)               (1.42)
                                                                                                                                     ∂ mn 2       ∂               mn 2
                                                                                                                                          v  +                     v vj  − nFi Vi = 0
                                                                                                                                     ∂t 2        ∂xj               2
                                                                                                                                                                  (energy equation)                 (1.43)
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                                                                                                                                                        w = v − V(r, t).
                                                                                                                     Then evidently, w = 0 and vi vj  = Vi Vj + wi wj . Using (1.41) in (1.42)
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                                                                                                                     yields
                                                                                                                                                dVi        ∂Pij
                                                                                                                                             mn       =−         + nFi                        (1.44)
                                                                                                                                                 dt        ∂xj
                                                                                                                     where
                                                                                                                                                      d    ∂         ∂
                                                                                                                                                        =      + Vj
                                                                                                                                                     dt    ∂t       ∂xj
                                                                                                                                                                   
                                                                                                                                              Pij = mnwi wj  = m wi wj f d3 v.                     (1.45)
Boltzmann equation 13
                                                                                                                         So far we have obtained the first three moment equations, (1.41), (1.44),
                                                                                                                     and (1.49). Our purpose is to obtain hydrodynamic equations to govern the
                                                                                                                     hydrodynamic variables n, V, T . The three moment equations contain πij
                                                                                                                     and q, which must be expressed in terms of n, V, T . Since the quantities πij
                                                                                                                     and q involve the integrations of the distribution function f , an approximate
                                                                                                                     solution for f must be found. If f is obtained approximately in terms of
                                                                                                                     (n, V, T ), πij and q can be expressed in terms of (n, V, T ), and a closed
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                                                                                                                     (or perfect).
                                                                                                                         Using (1.57) in (1.56) yields
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                                                                                                                                     dVi   Fi     1    ∂p       2     μ ∂
                                                                                                                                         =    −           − μ∇ Vi −          ∇·V .          (1.58)
                                                                                                                                      dt   m mn ∂xi                    3 ∂xi
                                                                                                                     The energy equation takes the form
                                                                                                                                                                                                
                                                                                                                                   3 dT                    1                        ∂vi   ∂vj
                                                                                                                                    n   = −λ∇2 T − p∇ · V − πij                         +              (1.59)
                                                                                                                                   2 dt                    2                        ∂xj   ∂xi
                                                                                                                     where λ is the coefficient of thermal conduction: q = −λ∇T .
Boltzmann equation 15
                                                                                                                     or in uniform motion, the three normal stresses have the same value, i.e. the
                                                                                                                     fluid is isotropic:
                                                                                                                     The shear stress πij should be traceless, because, by definition, it makes zero
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                                                                                                                     We also assume that in general flow the shear stress components are propor-
                                                                                                                     tional to the corresponding time rates of angular deformation. Consulting
                                                                                                                     Fig. 1.2, we have the relations
                                                                                                                                                                        
                                                                                                                                                dφ1                  dφ1
                                                                                                                                           dx1       = dV2 ,   π21 ∼
                                                                                                                                                 dt                   dt
                                                                                                                                                                        
                                                                                                                                                dφ2                  dφ2
                                                                                                                                           dx2       = dV1 .   π12 ∼
                                                                                                                                                 dt                   dt
                                                                                                                     So we can write
                                                                                                                                                                                                 
                                                                                                                                    1                          dφ1 dφ2                    dV2 dV1
                                                                                                                             π12   = (π12 + π21 ) = μ              +             =μ           +      .
                                                                                                                                    2                           dt   dt                   dx1   dx2
                                                                                                                     The complete form of πij which is traceless can be easily written down as
                                                                                                                                                                        
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                                                                                                                                                 dVi    dVj    2
                                                                                                                                       πij = μ       +       − ∇ · Vδij .                 (1.66)
                                                                                                                                                 dxj    dxi    3
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Boltzmann equation 17
                                                                                                                     where μ = mmaa+mmb
                                                                                                                                        b
                                                                                                                                          is the reduced mass. Since the kinetic energy is conserved,
                                                                                                                     the above relations give g = g . The relative velocity g is deflected through
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                                                                                                                                                  ∂v                                 
                                                                                                                                                  ax ∂vay ∂vaz ∂vbx ∂vby ∂vbz 
                                                                                                                                                                                            
                                                                                                                                                  ∂vax ∂vax ∂vax ∂vax ∂vax ∂vax 
                                                                                                                                                                                            
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                                                                                                                                                  ∂vax ∂vay
                                                                                                                                                              ∂v      ∂v    ∂v       ∂v   
                                                                                                                                                                az       bx      by     bz  
                                                                                                                                                  ∂v                                        
                                                                                                                                                  ay ∂vay ∂vay ∂vay ∂vay ∂vay 
                                                                                                                                                                                            
                                                                                                                                                                                     
                                                                                                                                                  ∂vax ∂vay  ∂vaz     ∂vbx   ∂vby     ∂vbz
                                                                                                                                                                                            
                                                                                                                                  ∂(va , vb )  ∂vaz ∂vaz ∂vaz ∂vaz ∂vaz ∂vaz 
                                                                                                                             J=                =                                            
                                                                                                                                  ∂(va , vb )  ∂v                                    
                                                                                                                                                  ax    ∂vay ∂v       ∂v     ∂v  by   ∂vbz 
                                                                                                                                                                az       bx
                                                                                                                                                                                             
                                                                                                                                                  ∂vbx ∂vbx ∂vbx ∂vbx ∂vbx ∂vbx 
                                                                                                                                                                                            
                                                                                                                                                                                            
                                                                                                                                                  ∂v                     ∂v  
                                                                                                                                                                                       ∂vbz 
                                                                                                                                                                                          
                                                                                                                                                  ax ∂vay ∂vaz ∂vbx              by
                                                                                                                                                                                            
                                                                                                                                                  ∂vby ∂vby ∂vby ∂vby ∂vby ∂vby 
                                                                                                                                                                                            
                                                                                                                                                                                          
                                                                                                                                                  ∂vax ∂vay ∂vaz ∂v
                                                                                                                                                                           ∂vby ∂vbz  
                                                                                                                                                                         bx                 
                                                                                                                                                  ∂v    ∂vbz ∂vbz ∂vbz ∂vbz ∂vbz 
                                                                                                                                                      bz
Boltzmann equation 19
= (1.75)
                                                                                                                     In the first integral in the above equation, we change the integration variables
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                                                                                                                     where u = m2m    b
                                                                                                                                   a +mb
                                                                                                                                          (g · k̂)k̂ (see (1.73)) and we used d3 va d3 vb = d3 va d3 vb .
                                                                                                                     In the RHS integral, we rename the integration variables by dropping the
                                                                                                                     primes (then g goes to g) to obtain
                                                                                                                                     
                                                                                                                                         Φ(va )fa (va )fb (vb )g b db dϕd3 vb d3 va
                                                                                                                                                      
                                                                                                                                                 =        Φ(va )fa (va )fb (vb )g b db dϕd3 vb d3 va .          (1.76)
                                                                                                                     where
                                                                                                                                                                        
                                                                                                                                                          [[· · ·]] =       (· · ·)b db dϕ.
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                                                                                                                     This is equal to LHS of (1.77) for arbitrary choice of the function Φ, and
                                                                                                                     thus we conclude
                                                                                                                                                                                  
                                                                                                                                               ∂                                3
                                                                                                                             Cab (fa fb ) = −
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Boltzmann equation 21
                                                                                                                     Thus we obtain
                                                                                                                                                  2
                                                                                                                                       2    mb
                                                                                                                        [[ui uj ]] = 4πg
                                                                                                                                         ma + mb
                                                                                                                                                                     
                                                                                                                                             gi gj    4θ   gi gj   2θ 2θ
                                                                                                                                     × b db 2 2 sin + δij − 2 sin cos      .                             (1.83)
                                                                                                                                              g        2    g       2  2
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                                                                                                                                                           2              
                                                                                                                                               2     mb               gi gj         θ
                                                                                                                              [[ui uj ]] = 4πg                 (δij − 2 ) sin2 b db.            (1.84)
                                                                                                                                                   ma + mb             g            2
                                                                                                                     Using the Rutherford scattering formula for Coulomb collisions,
                                                                                                                                                                              
                                                                                                                                    θ     ea eb                        ma mb
                                                                                                                                 tan =          ,    e : charge,  μ =
                                                                                                                                    2    b μ g2                       ma + mb
                                                                                                                     the integral in (1.81) is obtained
                                                                                                                                                 ∞
                                                                                                                                                          θ       ea eb
                                                                                                                                                     sin2 b db =         J                               (1.85)
                                                                                                                                                 0        2       μg2
                                                                                                                                                  ∞                      
                                                                                                                                                        λdλ         b μg2
                                                                                                                                             J=                 λ=           .                           (1.86)
                                                                                                                                                   0   1 + λ2        ea eb
                                                                                                                     Using the above equations in (1.81) and (1.83) gives
                                                                                                                                                                           
                                                                                                                                                     2 2     gi      1    1
                                                                                                                                       [[ui ]] = −4πea eb J            +                                 (1.87)
                                                                                                                                                            ma g 4   mb ma
                                                                                                                                                              4πe2a e2b
                                                                                                                                               [[ui uj ]] =             J (g2 δij − gi gj ).             (1.88)
                                                                                                                                                               g4 m2a
                                                                                                                     Now the collision term (1.79) becomes, upon using the above equations,
                                                                                                                                                                                     
                                                                                                                                          2 2 J      1     1    ∂          gi        3
                                                                                                                        cab (fa fb ) = 4πea eb           +          fa         fb d vb
                                                                                                                                               ma ma mb ∂vai               g3
                                                                                                                                                             2                        
                                                                                                                                            e2 e2     ∂2         g δij − gi gj      3
                                                                                                                                       + 2π a 2b J           fa                f b d  vb . (1.89)
                                                                                                                                             ma ∂vai ∂vaj             g3
                                                                                                                     Use the following identity
                                                                                                                                             2                         2            
                                                                                                                               gi    1 ∂      g δij − gi gj      1 ∂     g δij − gi gj
                                                                                                                                  =                           =−
                                                                                                                               g3    2 ∂vbj        g3            2 ∂vaj       g3
February 19, 2019                                                                                                            14:33   Fundamentals of Theoretical Plasma Physics   9.61in x 6.69in   b3496-ch01   page 22
                                                                                                                     in (1.89) and integrate by parts the result to finally obtain the Landau
                                                                                                                     collision integral
                                                                                                                                                                                      
                                                                                                                                             e2 e2  ∂                 fb ∂fa    fa ∂fb
                                                                                                                           cab (fa fb ) = 2π a b J        d3 vb Uij           −
                                                                                                                                              ma   ∂vai               ma ∂vaj   mb ∂vbj
                                                                                                                                                                                          (1.90)
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                                                                                                                     where
                                                                                                                                                               g2 δij − gi gj
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                                                                                                                                                       Uij =                  .                      (1.91)
                                                                                                                                                                     g3
Boltzmann equation 23
                                                                                                                     Then (1.89) can be put into the Fokker–Planck form of the collision integral:
                                                                                                                                                                              
                                                                                                                                                  ∂           ∂     1       3
                                                                                                                                cab (fa fb ) = A        −fa             fb d vb
                                                                                                                                                 ∂vai        ∂vai   g
                                                                                                                                                                                       
                                                                                                                                                      ∂2             ∂2              3
                                                                                                                                               +B              fa              g fb d vb
                                                                                                                                                   ∂vai ∂vaj      ∂vai ∂vaj
                                                                                                                     This equation gives the change of the distribution function due to the
                                                                                                                     collisions between the composing particles. The effect of prime and no prime
                                                                                                                     particle collisions are summed over the distribution. The first term is termed
February 19, 2019                                                                                                          14:33   Fundamentals of Theoretical Plasma Physics   9.61in x 6.69in   b3496-ch01   page 24
                                                                                                                     the dynamic friction and the second the diffusion term. This equation,
                                                                                                                     although derived from the Boltzmann collision integral here, is valid for
                                                                                                                     Markovian process of weak scattering and can be derived independently.
                                                                                                                     For this, see References: Krall, N. A. and Trivelpiece, A. W., Principles of
                                                                                                                     Plasma Physics, San Francisco Press, 1986; Sturrock, P. A. Plasma Physics
                                                                                                                     Cambridge University Press (1994). Also the Fokker–Planck equation above
                                                                                                                     can be derived from Lenard–Balescu equation (Chapter 12) by taking an
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                                                                                                                     (1.93) and (1.94) are two coupled equations through the cross-collision terms
                                                                                                                     Cei and Cie . Since the Landau collision terms are appropriate for plasma of
                                                                                                                     frequent Coulomb collisions, Brakinskii’s two-fluid equation is widely used
                                                                                                                     for collisional plasma investigation.
                                                                                                                         In contrast with a simple gas, there are no summation invariants. But
                                                                                                                     we have the following relations:
                                                                                                                                          
                                                                                                                                            Cab d3 v = 0 (a = e, i; b = e, i)                (1.95)
                                                                                                                     which implies that the total number of particles of each species is not changed
                                                                                                                     due to collisions between like particles as well as between unlike particles.
                                                                                                                     We have for like particles
                                                                                                                                                   
                                                                                                                                                     ma vCaa d3 v = 0,                         (1.96)
                                                                                                                                                     
                                                                                                                                                         ma v 2 Caa d3 v = 0.                      (1.97)
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Boltzmann equation 25
                                                                                                                     The above two equations express the conservation of momentum and energy
                                                                                                                     for the total pair of collisions involving unlike particles.
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Re = −Ri . (1.109)
                                                                                                                                            ∞
                                                                                                                                                1
                                                                                                                               ψ(r + a) =          (a · ∇)n ψ(r)
                                                                                                                                            n=0
                                                                                                                                                n!
                                                                                                                                                         ∂        1       ∂ ∂
                                                                                                                                         = ψ(r) + ai        ψ(r) + ai aj         ψ(r) + · · · .
                                                                                                                                                        ∂xi       2      ∂xi ∂xj
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Boltzmann equation 27
                                                                                                                                                           ∂Vαβ     1 ∂ 2 Vαβ  
                                                                                                                                       Uαβ = Vαβ −              vγ +          v v + ···
                                                                                                                                                            ∂vγ      2 ∂vγ ∂vκ γ κ
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                                                                                                                                                                    1 2
                                                                                                                                             where Vαβ =               v δαβ − vα vβ .
                                                                                                                                                                    v3
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                                                                                                                     In the lowest order, the electron and ion distributions are the local
                                                                                                                     Maxwellians
                                                                                                                                                                  3
                                                                                                                                                        ma          2           ma (v − Va (r, t))2
                                                                                                                      fa (v, r, t) = na (r, t)                          exp −                            (a = e, i)
                                                                                                                                                     2πTa (r, t)                     2Ta (r, t)
                                                                                                                                                                                                              (1.113)
                                                                                                                                                  ∂na
                                                                                                                                                      + ∇ · (na Va ) = 0               (1.114)
                                                                                                                                                   ∂t
                                                                                                                                                                              
                                                                                                                                    ∂                                      Va
                                                                                                                           ma n a      + Va · ∇ Va = −∇pa + ea na E +         × B + R(0)
                                                                                                                                                                                     a
                                                                                                                                    ∂t                                      c
                                                                                                                                                                                       (1.115)
                                                                                                                                                        
                                                                                                                                       3      ∂
                                                                                                                                         na      + Va · ∇ Ta + na Ta ∇ · Va = 0.       (1.116)
                                                                                                                                       2      ∂t
                                                                                                                                             me n e        (0)
                                                                                                                                      R(0)
                                                                                                                                       e =−         u = −Ri : f rictional f orce,                             (1.117)
                                                                                                                                              τe
                                                                                                                                                        √      3/2
                                                                                                                                                       3 me Te
                                                                                                                               u = Ve − Vi , τe = √                   : collision time,
                                                                                                                                                     4 2π ni J Z 2 e4
                                                                                                                     where Z is ionic charge number and J is given by (1.86). The energy equation
                                                                                                                     (1.116) is the adiabaticity: the entropy change, dsa /dt = 0.
                                                                                                                         As the higher order solutions, the momentum and energy equations take
                                                                                                                     the forms of (1.101) and (1.102), respectively. But the frictional force is
                                                                                                                     reduced by a factor of 0.51 and, in addition, a thermal force RT , which
February 19, 2019                                                                                                          14:33   Fundamentals of Theoretical Plasma Physics   9.61in x 6.69in   b3496-ch01   page 28
                                                                                                                                                        Re = Ru + RT                              (1.118)
                                                                                                                                                    me n e
                                                                                                                                                Ru = −     (0.51u + u⊥ )                         (1.119)
                                                                                                                                                     τe
                                                                                                                                                              3 ne B0
                                                                                                                                       RT = −0.71ne ∇ Te −               × ∇Te .                 (1.120)
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                                                                                                                                                              2 ωce τe B0
                                                                                                                     The symbols and ⊥ are referred to the direction of the ambient magnetic
                                                                                                                     field (B0 ). If the plasma is unmagnetized, the ⊥-components are discarded.
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                                                                                                                     The thermal forces are given rise to, due to the temperature gradients. The
                                                                                                                     heat Qa is a plasma-proper term which represents the heat generated by
                                                                                                                     collisions between electrons and ions:
                                                                                                                                                   3me ne
                                                                                                                                             Qi =          (Te − Ti ) ≡ QΔ .             (1.121)
                                                                                                                                                    mi τ e
                                                                                                                     The heat acquired by the electrons is obtained by using (1.121) in (1.112).
                                                                                                                     Another point worthy of attention in Braginskii’s solution is that a heat
                                                                                                                     flux (qeu ) is given rise to by the relative motion between electrons and ions,
                                                                                                                     in addition to the usual heat flux(qeT ) generated by temperature gradients:
                                                                                                                     qe = qeu + qeT .
                                                                                                                                                                   3 ne Te B0
                                                                                                                                          qeu = 0.71ne Te u +                 × u.               (1.122)
                                                                                                                                                                   2 ωce τe B0
                                                                                                                     The ion heat flux qi takes the familiar form ∝ ∇Ti . Viscosity πij , heat flux q,
                                                                                                                     heat Q, and frictional and thermal force R all contribute to the production
                                                                                                                     of entropy in the plasma. Define the entropy per electron as
                                                                                                                                                      3
                                                                                                                                               se =     lnTe − ln ne + const.                     (1.123)
                                                                                                                                                      2
                                                                                                                     The electron entropy equation is written in the form
                                                                                                                                                                   
                                                                                                                                     ∂                           qe     QΔ
                                                                                                                                       (ne se ) + ∇ · se ne Ve +      +    = ϑe                   (1.124)
                                                                                                                                    ∂t                           Te     Te
                                                                                                                     where ϑe is the entropy production per unit volume:
                                                                                                                                                                                    
                                                                                                                                                       1 e ∂Vei ∂Vej      2
                                                                                                                         Te ϑe = −qe · ∇lnTe − R · u − πij         +     − δij ∇ · Ve .
                                                                                                                                                       2      ∂xj    ∂xi  3
                                                                                                                                                                                      (1.125)
Boltzmann equation 29
                                                                                                                     that the electron and ion distributions become the same after infinite time.
                                                                                                                     Derive the momentum equation for the fluid velocity Ve (r, t). Use (1.93)
                                                                                                                     for the left hand side. This example demonstrates the necessary algebra.
                                                                                                                     Multiply (1.93) by v and integrate over d3 v.
                                                                                                                                                       
                                                                                                                                       3  ∂fe        ∂                  ∂
                                                                                                                                me d vv        = me         vfe d3 v = (me ne Ve (r, t))
                                                                                                                                           ∂t        ∂t                ∂t
                                                                                                                                                                               
                                                                                                                                    3     ∂fe             3        ∂fe       ∂
                                                                                                                            me d vv v ·        = me d vvi vj           = me        d3 v vi vj fe
                                                                                                                                           ∂r                     ∂xj       ∂xj
                                                                                                                                                          ∂
                                                                                                                                                  = me       (ne vi vj )                           (1.127)
                                                                                                                                                         ∂xj
                                                                                                                     where v = vi (in the following ‘i’ is the free index indicating the vector
                                                                                                                     component). We define the random velocity
                                                                                                                                                     wi = vi − Vi (r, t).                            (1.128)
                                                                                                                     Then we have
                                                                                                                                ∂                                    ∂
                                                                                                                            =      (me ne Vi Vj + me ne wi wj ) =     (me ne Vi Vj + Pij )
                                                                                                                               ∂xj                                  ∂xj
                                                                                                                                                                 
                                                                                                                                              3           1           ∂fe 3
                                                                                                                                     −e d vv E + v × B ·                 d v
                                                                                                                                                          c           ∂v
                                                                                                                                                                          
                                                                                                                                                                 1             ∂fe
                                                                                                                                        = −e d3 vvi Ej + ejlm vl Bm                .
                                                                                                                                                                 c             ∂vj
                                                                                                                     Integrating by parts and using ∂vi /∂vj = δij give
                                                                                                                                                                          
                                                                                                                                     3            Bm
                                                                                                                              = e d v Ej δij +       ejlm (vi δlj + vl δij )        (ejlm δlj = 0)
                                                                                                                                                   c
                                                                                                                                                                       
                                                                                                                                                            1
                                                                                                                                              = ene E + Ve × B .                                     (1.129)
                                                                                                                                                            c
February 19, 2019                                                                                                          14:33      Fundamentals of Theoretical Plasma Physics          9.61in x 6.69in   b3496-ch01   page 30
−νme ne (Ve − Vi ).
                                                                                                                              ∂t               ∂xj                                c
                                                                                                                                   = −νme ne (Ve − Vi ).
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                                                                                                                     Using the continuity equation in the above yields [Krall and Trivelpiece 1986
                                                                                                                     p. 89]
                                                                                                                                                                                    
                                                                                                                                        ∂                                      Ve
                                                                                                                               me n e      + Ve · ∇ Ve + ∇ · Pij + ene E +         ×B
                                                                                                                                        ∂t                                      c
                                                                                                                                     = −νme ne (Ve − Vi ).                                                  (1.130)
                                                                                                                                                 ∂Pij       ∂Pji
                                                                                                                     Note that ∇ · Pij =         ∂xj    =   ∂xj .
                                                                                                                     Example. Derive the energy equation by taking the moments of (1.93) and
                                                                                                                     (1,94) with 12 mα vα2 . The first two terms give
                                                                                                                                                                                               
                                                                                                                                       ∂        1                        ∂        1
                                                                                                                                                  mα nα v 2 α       +                        2
                                                                                                                                                                                    mα nα vj v α .
                                                                                                                                       ∂t       2                       ∂xj       2
                                                                                                                     Collecting the above results gives for the energy equation in the form
                                                                                                                                                                   
                                                                                                                          ∂ 1           2        ∂    1            2
                                                                                                                                mα nα v α +           mα nα vj v α − eα nα E · Vα = Qα .
                                                                                                                          ∂t 2                  ∂xj 2
                                                                                                                                                                                          (1.131)
Boltzmann equation 31
                                                                                                                     References
                                                                                                                     Braginskii, S. I. Transport process in a plasma, in Reviews of Plasma Physics Vol. 1:
                                                                                                                          Consultant’s Bureau, New York, 1965.
                                                                                                                     Braginskii, S. I. Transport phenomena in a completely ionized Two-temperature plasma,
                                                                                                                          Soviet Phys. JETP, 6 (1958) p. 358.
                                                                                                                     Bhatnagar, P. L., Gross, E. P. and Krook, M. Phys. Rev. 94 (1954) p. 511
                                                                                                                     Chapman, S. and Cowling, T. G. The Mathematical Theory of Nonuniform Gases,
                                                                                                                          Cambridge University Press, 1952.
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