Plasma Physics Primer Part I
Statistical Behavior of Plasmas
1. Distribution Functions
2. Averages & Fluxes
3. Isotropic Equilibrium Distribution Function
4. Collision Frequency
5. Mean Free Path
6. Boltzmann’s Equation
7. Conservation Laws
1. Distribution Functions: Since there are on the order of 1012 to 1024 gas atoms,
molecules and ions per m3 in a plasma, it is most convenient to look at the
behavior of those particles from a statistical standpoint. The particles in any
given volume in general, have a distribution of velocities that arise due to
numerous collisions between particles. The distribution function f describes
the distribution of particle velocities at any given point x and any given
instant t. In particular, the number of particles having velocities between
(c1, c2 , c3 ) and ( (c1  dc1, c2  dc2 , c3  dc3 ) in a volume dx1, dx2 , dx3 centered at
x , at time t is:
dN  f (c, x, t )dxdc
(1)
Where
(2)
dc  dc1, dc2 , dc3
(3)
dx  dx1, dx2 , dx3
[Note: we will often denote a vector b as (b1, b2 , b3 ) .]
In a plasma, there are several species which interact (collide) with one
another and with other species. These include electrons, ions, and neutral
particles. Each specie will in general have it sown distribution function:
Electron distribution function= fe  fe (c, x, t )
(4a)
Ion distribution function= fi  fi (c, x, t )
(4b)
Neutral specie distribution function= fo  fo (c, x, t )
(4c)
Or, in general, the distribution function per specie s is:
fs  fs (c, x, t )
(4d)
We can either determine the distribution function for all species or we can use
the typically unknown distribution functions to derive useful continuum
conservation laws (conservation of mass, momentum, energy) for each specie
and the overall plasma. fs can only be determined (in closed form) for relatively
simple problems. Typically, we have to use Boltzmann’s equation, which
describes the temporal and spatial evolution of any given fs , to derive continuum
conservation principals.
dNs
Number density (of species)  dns 
(5)
 fs (c, x, t )dc
dx
Thus, dns is the number of particles of species s per unit volume having
velocities between (c1, c2 , c3 ) and (c1  dc1, c2  dc2 , c3  dc3 ) . The total number
density is found from (5) by integration over all possible velocities:
ns  ns ( x, t )  
 

 f dc dc dc
s
1
2
(6)
3


 f dc
s

Note, physically and mathematically, (6) requires that fs (c, x, t )  0 as c   .
Mass Partial Density of Species s  s  s ( x, t )  ns ( x, t )ms
mass
Where ms is
partical
Total Mass Density     s   ns ms
s
(8)
s
Temperature of species s is defined based on the kinetic energy of s:
(total random kinetic energy)/s
3
 ns kTs
Unit volume
2
Where K=Boltzmann’s Constant
(9)
This definition arises from the equipartion of energy theorem which states that
1
each particle degree of freedom has an energy kT associated with it. Thus,
2
since kinetic energy is the energy of translation and since there are 3
translational degrees of freedom (1 for each coordinate direction), then the
1
3
kinetic energy of each particle s is 3( kTs )  kTs . Hence the total kinetic
2
2
energy (per unit volume) of specie s @ position x and time t is:
kEs
3
ns ( kTs ) 
2
volume
We can relate Ts to fs by using definition of kinetic energy:
(10)
ms 2
c
2
Since the number of particles (at x and t ) having velocities between (c1, c2 , c3 )
and (c1  dc1, c2  dc2 , c3  dc3 ) is fs dc , then the kinetic energy of these particles
is:
Kinetic Energy of a single particle=
(fs dc )
ms 2 ms 2
c 
c fs dc1dc2dc3
2
2
(12)
Now the total thermal (i.e. translational) kinetic energy of species s at x and time
t (per unit volume) is:

m
3
ns ( kTs )    s c 2fs dc1dc 2dc3
(13)
2
2

Where we have used equation (10) and have integrated equation (12). (Note
that c 2  (c12  c22  c3 2 ) .) The average value of any property A (A= (c1, c2 , c3 ) ) n
(@ position x , time t) per unit volume is defined as:
ns A  

 A(c , c , c
1
2
3
)fs dc1dc2dc3
(14)

Thus (13) can also be written as:
ms 2
3
(13a)
ns
c  ns ( kTs )
2
2
Fluxes
Let s be some property of species s [mass, ms ; electric charge, ezs ; momentum,
ms 2
w , where zs =free charge (integer), e=charge=1.6(10)-23,
2
w =velocity relative to lab frame,  int,s =internal energy associated with rotation,
msw ; energy,  int,s 
vibration, electron motion.] The flux of s
For a single particle through an element of area ds is:
sw o ds  Flux of s due to motion of single particle through ds
The total flux for all particles through ds is given by:
( 

 ( wf dw )ds  total flux of 
s
s
through ds = ‘flux’
s
(16)

Where we have express fs in terms of w (the relationship between c and w will
be given below). In (16), the term in () is just the sum of all fluxes due to particles
of all velocities (where position x in fs (w , x, t ) is at or near ds ).
From (14), (16) can be expressed as:

Flux= (   (swfs dw )ds  ns sw ds
(17)

We will call ns sw the flux vector, where:
ns sw  

 ( wf dw )
s
s
(18)

Particle flux ( s =1):
  (wf dw ) n
s
s
w l s  Particle flux vector species s
(19)
Mass flux ( s = ms ):
  (m wf dw ) n
s
s
s
wms l s
 ns ms w l s
=Mass flux vector species s
 s w l s
(20)
Since ns ms = s
 ms
Note that total mass flux is given by:
m   ms   s ns w ls
s
(21)
s
We will define average velocity (relative to lab frame) of species s as:
1
vs  ws 
(wfs dw )
ns  
Thus from (21) and (22):
m   ms   s w ls   sv s
s
s
Mass average velocity=
m
v

(22)
(23)
s
m


s
s
s
 v


s
=’bulk velocity’
(24)
s s
s
s
s
Thermal velocities (velocity relative to bulk velocity) we can define a particles
thermal velocity relative to the mass average velocity:
cs  w s  v
(25)
Or relative to the average velocity of species s:
cs  w s  v s
(26)
(Where w s is particles velocity relative to lab frame). The distribution function fs
can be expressed in terms of w, c, or c ; all are equivalent.
Exercise 1:
Show that cs  0 :
cs 

Now
1
ns
1
ns
  c  f (c  , x, t )dc dc  dc 
  (w
s s
s
s
 v s )fs dc (*)
1
2
3
vs  ws 
1
ns
  w fˆ dw
s s
Where fˆs is fs expressed in terms of w (and where cs  w s  v s ).
Now:
dc1  d (w1  v s1 )
Since v s1 is fixed
 dw
1
dc2  dw 2
dc3  dw 3
Thus, (*) can be expressed as:
cs 
1
ns
cs 
  (w
1
ns
s
 v s )fˆs dw
1
  wfˆ dw  n   v fˆ dw
s
s s
s
 w s ls  v s (
1
ns
 
fˆs dw )
 vs  vs
w s ls  v s
cs  0
Since and
 
fˆs dw  ns
Exercise 2: Show that cs is the difference between the species average
velocity ( v s ) and the mass average velocity ( v ).
cs 
1
ns
  cfˆ dc
s

1
ns
  (w  v )fˆ (c  , x, t )dc

1
ns
  (w  v )fˆ (c  , x, t )dw

1
ns
  wfˆ dw  v
s
s
s
cs  v s  v
s
s
s
Note that fˆs (cs , x, t )  fˆs (w  v , x, t )  fˆs (w, x, t )
cs is called the diffusion velocity. This definition recognizes that any difference
between the average velocity of s ( v s ) and the average bulk velocity ( v ) has to
be due to diffusion:
(27)
cs  vs = Diffusion velocity of specie s ( x & t )
Exercise 3: Show that m   sv s  v  s
s
Exercise 4: Show that since m  v  s  v  ,
 v
s s
 0 (refer to exercise 3).
s
Thus the sum of all the specie diffusion fluxes is always zero.
Electrical current= charge flux  s  ezs ; zs = free charge of species s
(example: for Na+: zNa  1, e  1.602(10 19 ) ; for an electron (e-):
ze  1, e  1.602(10 19 ) ; for Ar: zAr  0 ).
Js  nsv s ezs = Charge flux due to motion of species s
(28)
J   Js   nsv s ezs
s
s
= Total charge flux (current) due to motion of all charged species
(29)
Alternative form of (28):
Js  nsvsezs  nsezs (Vs  v )
(28a)
Alternative form of (29):
J   Js   ns ezs (v s )   ns ezsv
s
s
s
J   ns ezs (v s )   ns ezsv
s
s
 j  v e
=current due to diffusion (conduction) + current due to convection
Where:
J   ns ezs (v s ) = conduction current
s
=current due to specie diffusion
e   ns ezs = free charge density
s
Momentum Flux s  msw
(29a)
Exercise 5: Starting with the definition momentum flux of s =
that
  m wwfˆ dw   (vv  v v  vv
s
s
s
s
s
  m wwfˆ dw
s
s
show
)  s (cc )s where the term on the right is a
dyadic.
For simplicity, let’s use shorthand notation to represent the triple integral (   ):

  
Now recall that c  w  v thus:
 mswwfˆsdw   ms (v  c )(v  c )fˆsdw
Also,
And
fˆs (w  v , x, t )  fˆs (v  c, x, t )  fˆs (c, x, t ) (since v isn’t volume)
dw  dw1dw 2dw 3  d (v1  c1 )d (v 2  c2 )d (v 3  c3 )
 dc1dc2dc3
 m wwfˆ dw   m (v  c )(v  c )fˆ dc
  m vvfˆ dc   m vcfˆ dc   m cvfˆ dc   m ccfˆ dc
 m vvn  m v c l n  m (  cfˆ dc )v  m cc n
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
vs
svv  svvs  svsv  s cc
(30*)
We can attach physical meaning to each term in the last equation:
svv = partial fraction of total mass-average momentum (= vv )
svvs = diffusion at moment flux
sv sv = moment flux due to diffusion
s cc =see below
Note: word definitions of 2nd and 3rd terms are somewhat arbitrary. These will
drop out when we sum all specie momentum fluxes to obtain overall moment
flux.
The last term is the momentum flux component due to the random motion of
species s; it is thus interpreted as the partial stress tensor of species s.
 (c1c1 ) (c1c2 ) (c1c3 ) 
s cc  [Ps ]  s  (c2c1 ) (c2c2 ) (c2c3 ) 
 (c c ) (c c ) (c c ) 
3 2
3 3 
 3 1
Or in index notation:
Pijs  s (ci c j )
We will identify the trace of [Ps ] with the partial mechanical pressure Ps (this will
be shown to be true when we derive conservation of momentum equation).
Thus:
1
1 3
Ps  (P11s  P22s  P33s )  Ps  ci c j
(33)
3
3 i 1
s
If we assume that species s behaves as an ideal gas (i.e. no forces between
individual particles negligible overall particle volume) then:
Ps  ns kTs
(34)
Total moment flux is obtained by summing (30*) over all species in plasma:
T.M.F.= svv   Ps 
(35)
Where P  = total stress tensor.
svv  P 
s
J  = Shear stress tensor
 [P ]  P [I ]
Or
Jij  Pij  P ij  s (ci c j )
(36)
(36a)
Where [I] is the identity matrix (all off-diagonal terms =0, all on-diagonal terms
=1).
1
msw 2   ints
2
1
Energy flux = s   ( msw 2   ints )wfs dw to obtain (due to motion of s) a
2
physically meaningful expressions replace w with v  c (and dw with dc ):
1
  ( ms (v  c )(v  c )(v  c )fs dc  s    ints (v  c )fs dc
2
(fˆ  fˆ (c, x, t ))
Energy Flux: s 
s
s
1
1
msv 2vns  msv 2 c l s ns  msv  (v c )fs dc  ms  (v c )cfˆs dc
2
2
1
1
 ms  (c c )vfˆs dc  ms  (c c )cfˆs dc  v   ints fˆs dc    ints cfs dc
2
2
Energy Flux: 
1
1
 ( sv 2 )v  v sv 2v s  vmsvns c l s  msv  ccfˆs dc
2
2
m
1
v s s c 2 l s  ms  (c c )cfˆs dc  v  ints ns    ints cfs dc
2
2
Energy Flux (con’t):

1
1
 v ( sv 2  sv 2v s  s c 2 l s   ints ns )  sv 2v s  v [P ]
2
2
2
1
  c ( ms c 2  ints )fˆs dc
2
3
Since Ps  Js   Ps [I ] and ms c 2 l s  kTs , then R.H.S. becomes:
2
 sv (
R.H.S.
 ints
v2
3 kTs Ps
1
 v vs 


)  sv 2v s  v [Js ]
2
2 ms s
ms
2
1
  c ( ms c 2   ints )fˆs dc
2
(Note that v Ps [I ]  Psv )
Now,
 ints
P
3 kTs Ps


 us  s  hs
2 ms s
ms
s