Plasma Physics Primer

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 xv and any given instant t. In particular, the number of particles having velocities between

(c1 , c 2 , c 3 ) and ((c1+ dc 1 , c 2 + dc 2 , c 3 + dc 3 ) in a volume dx1, dx 2 , dx 3 centered at xv, at time t is: vv v v dN= f( c , x , t ) dxdc (1) Where v dc= dc1, dc 2 , dc 3 (2) v dx= dx1, dx 2 , dx 3 (3) v [Note: we will often denote a vector b as (b1 , b 2 , b 3 ) .] 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: v v Electron distribution function=fe= f e ( c , x , t ) (4a) v v Ion distribution function=fi= f i ( c , x , t ) (4b) v v Neutral specie distribution function= fo= f o ( c , x , t ) (4c) Or, in general, the distribution function per specie s is: v v fs= f s ( 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. dN Number density (of species) 汉dns f( cv , xv , t ) dc v (5) sdxv s Thus, dns is the number of particles of species s per unit volume having velocities between (c1 , c 2 , c 3 ) and (c1+ dc 1 , c 2 + dc 2 , c 3 + dc 3 ) . The total number density is found from (5) by integration over all possible velocities:

v ns= n s( x , t ) = 蝌 f s dc1 dc 2 dc 3 (6) - v = 蝌 fs dc - v v v Note, physically and mathematically, (6) requires that fs ( c , x , t ) 0 as c  . v v Mass Partial Density of Species s 汉rs r s(x , t ) n s ( x , t ) m s mass Where m is s partical n m Total Mass Density �r= 邋 rs s s (8) s s Temperature of species s is defined based on the kinetic energy of s: (total random kinetic energy)/s 3 Unit volume = n kT (9) 2 s s Where K=Boltzmann’s Constant

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(kT ) = kT . Hence the total kinetic 2s 2 s energy (per unit volume) of specie s @ position xv and time t is:

3 kE n( kT ) = s (10) s2 s volume

We can relate Ts to fs by using definition of kinetic energy: m Kinetic Energy of a single particle= s c2 2 v Since the number of particles (at x and t ) having velocities between (c1 , c 2 , c 3 ) v and (c1+ dc 1 , c 2 + dc 2 , c 3 + dc 3 ) is fs dc , then the kinetic energy of these particles is:

m m (f dcv ) s c2= s c 2 f dc dc dc (12) s2 2 s 1 2 3 Now the total thermal (i.e. translational) kinetic energy of species s at xv and time t (per unit volume) is:

3 ms 2 ns( kT s ) = 蝌 c f s dc1 dc 2 dc 3 (13) 2- 2 Where we have used equation (10) and have integrated equation (12). (Note 2 2 2 2 thatc=( c1 + c 2 + c 3 ).) The average value of any property A (A=(c1 , c 2 , c 3 )) n (@ position xv, time t) per unit volume is defined as:

ns A= 蝌 A( c1 , c 2 , c 3 ) f s dc 1 dc 2 dc 3 (14) - Thus (13) can also be written as: m 3 ns c2 = n( kT ) (13a) s2 s 2 s Fluxes

Let qs be some property of species s [mass, ms ; electric charge,ezs ; momentum,

v m 2 m w; energy, e + s w , where z =free charge (integer), e=charge=1.6(10)-23, s int,s 2 s v w =velocity relative to lab frame, eint,s =internal energy associated with rotation, vibration, electron motion.] The flux of qs For a single particle through an element of area dsv is: v v v qsw o ds = Flux of qs due to motion of single particle through ds The total flux for all particles through dsv is given by:

v v v q v (蝌 (qswf s dw ) ds = total flux of s through ds = ‘flux’ (16) - v v v 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 v v v v of all velocities (where position x in fs ( w , x , t ) is at or near ds ).

From (14), (16) can be expressed as:

v v v nq wv dsv Flux=(蝌 (qswf s dw ) ds = s s (17) - v We will call nsq s w the flux vector, where: v v v nsq s w= 蝌 ( q s wf s dw ) (18) -

Particle flux (qs =1): v v v 蝌(wfs dw ) = n s w l s = Particle flux vector species s (19) Mass flux (qs =ms ): v v v 蝌(ms wf s dw ) = n s wm s l s v = ns m s w l s =Mass flux vector species s (20) v Since ns m s = rs = rsw l s ˙ = ms Note that total mass flux is given by: m˙ m ˙ n wv l =邋 s = r s s s (21) s s We will define average velocity (relative to lab frame) of species s as: v v1 v v vs= w s = 蝌( wf s dw ) (22) ns Thus from (21) and (22): m˙ m ˙ wv l v v =邋 s =r s s = r s s (23) s s s Mass average velocity= m˙ m˙ s vv = = s r rs s v =’bulk velocity’ (24) rsv s = s rs s Thermal velocities (velocity relative to bulk velocity) we can define a particles thermal velocity relative to the mass average velocity: v v v cs= w s - v (25) Or relative to the average velocity of species s: v v v cs= w s - v s (26) v (Where ws is particles velocity relative to lab frame). The distribution function fs v v can be expressed in terms of w, c , orcv; all are equivalent.

Exercise 1: v Show that cs = 0 : 1 vⅱ v v ⅱv ⅱ cs= 蝌 c s f s( c s , x , t ) dc1 dc 2 dc 3 ns 1 v v =蝌(ws - v s ) f s dc (*) ns Now v v1 vˆ v vs= w s = 蝌 w s f s dw ns ˆ v v v v Where fs is fs expressed in terms of w (and wherecs= w s - v s ). Now:

dc1= d( w 1 - vs 1 )

Since vs1 is fixed = dw1

dc2= dw 2

dc3= dw 3 Thus, (*) can be expressed as:

v 1 v vˆ v cs=蝌( w s - v s ) f s dw ns

v 1vˆ v 1 v ˆ v cs=蝌蝌 wf s dw - 蝌 v s f s dw ns n s v v1 ˆ v =ws l s - v s(蝌 f s dw ) ns v v =vs - v s

v v ws l s= v s and v Since cs = 0 ˆ v 蝌 fs dw= n s v Exercise 2: Show that cs is the difference between the species average v v velocity (vs ) and the mass average velocity (v ). v1 vˆ v cs= 蝌 cf s dc ns 1 v vˆ v v v =蝌(w - v ) fs ( c s , x , t ) dc ns 1 v vˆ v v v =蝌(w - v ) fs ( c s , x , t ) dw ns 1 vˆ v v =蝌wfs dw - v s ns v v v cs= v s - v ˆv v ˆ v v v ˆ v v Note that fs( c s , x , t )= f s ( w - v , x , t ) = f s ( w , x , t )

v cs is called the diffusion velocity. This definition recognizes that any difference v v between the average velocity of s (vs ) and the average bulk velocity (v ) has to be due to diffusion: v v v v cs= v s = Diffusion velocity of specie s ( x& t ) (27) m˙ vv v v Exercise 3: Show that =邋rs s + r s s

v v v ˙ rsv s = 0 Exercise 4: Show that since m= vrs = v r , (refer to exercise 3). s Thus the sum of all the specie diffusion fluxes is always zero.

Electrical current= charge flux� qsez s ; zs = free charge of species s + -19 - (example: for Na : zNa =1, e = 1.602(10 ) ; for an electron (e ): -19 ze = -1, e = 1.602(10 ) ; for Ar: zAr = 0 ). v v Js= n s v s ez s = Charge flux due to motion of species s (28) v v v J=邋 Js = n s v s ez s s s = Total charge flux (current) due to motion of all charged species (29)

Alternative form of (28): vv v v Js= n s v s ez s = n s ez s( V s + v ) (28a) Alternative form of (29): v v v v J=邋 Js = n s ez s( v s ) + n s ez s v s s s v v v J=邋 ns ez s( v s ) + n s ez s v s s v =j + vvr e =current due to diffusion (conduction) + current due to convection (29a)

v v Where: J= ns ez s( v s ) = conduction current s =current due to specie diffusion e n ez r = s s = free charge density s

v Momentum Flux qs= m s w v vˆ v Exercise 5: Starting with the definition momentum flux of s = 蝌ms wwf s dw show

v vˆ v vv v v vv vv that 蝌mrs wwf s dw= r s( vv + v s v + vv s ) = r s ( cc ) s where the term on the right is a dyadic.

For simplicity, let’s use shorthand notation to represent the triple integral ( 蝌 ):

蝌= 蝌 Now recall that cv = wv - v v thus: v vˆ v vv v v ˆ v 蝌mrs wwf s dw= m s( v + c )( v + c ) f s dw Also, ˆv v v ˆ vv v ˆ v v v fs( w- v , x , t ) = f s ( v + c , x , t ) = f s ( c , x , t ) (since v isn’t volume) And v dw= dw1 dw 2 dw 3 = d( v 1 + c 1 ) d ( v 2 + c 2 ) d ( v 3 + c 3 )

= dc1 dc 2 dc 3

v vˆ v vv v v ˆ v 蝌ms wwf s dw= m s( v + c )( v + c ) f s dc vvˆv v v ˆ v v v ˆ v vv ˆ v =蝌ms vvf s dc + m s vcf s dc + 蝌 m s cvf s dc + m s ccf s dc vv vv vˆ v v vv =ms vvn s + m s v c l s n s + m s( cf s dc ) v + m s cc n s

vv vv v v vv rsvv+ r s vv s + r s v s v + r s cc (30*) We can attach physical meaning to each term in the last equation: vv vv rsvv = partial fraction of total mass-average momentum (= rvv ) vv rsvv s = diffusion at moment flux v v rsv s v = moment flux due to diffusion vv rs 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. 骣(c1 c 1 ) ( c 1 c 2 ) ( c 1 c 3 ) vv 琪 rscc=[ P s ] = r s 琪 ( c2 c 1 ) ( c 2 c 2 ) ( c 2 c 3 ) 琪桫(c3 c 1 ) ( c 3 c 2 ) ( c 3 c 3 ) Or in index notation:

Pijs= r s( c i 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 P=( P + P + P ) = P c c s11s 22 s 33 s s i 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= n s kT s (34) Total moment flux is obtained by summing (30*) over all species in plasma: vvvv P T.M.F.= rs= [ s ] (35) s vv rsvv+ [ P] Where [P] = total stress tensor. [J]= Shear stress tensor �[P ] P [ I ] (36) Or

Jij= P ij - Pd ij = r s( c i c j ) (36a) Where [I] is the identity matrix (all off-diagonal terms =0, all on-diagonal terms =1).

1 Energy Flux: q=m w 2 + e s2 s ints

1 2 v v Energy flux = q=(m w + e ) wf dw to obtain (due to motion of s) a s2 sints s physically meaningful expressions replace wv with vv+ cv (and dwv withdcv): 1 =(m ( vv + cv )( v v + c v )( v v + c v ) f dc v +q = e ( v v + c v ) f dc v 蝌2 s s sints s ˆ ˆ v v (fs= f s ( c , x , t )) Energy Flux: 12v 1 2 v v v v v v v vˆ v =ms v vn s + m s v c l s n s + m s v( vg c ) f s dc + m s ( v g c ) cf s dc 2 2 蝌 1v vvˆ v 1 v v v ˆ v v ˆ v v v +ms( cg c ) vf s dc + m s ( c g c ) cf s dc + veint f s dc + e int cfs dc 2蝌 2 蝌 s s 12v 1 v 2 v v vv v vvˆ v =(rsv ) v + v r s v v s + vm s vn s c l s + m s vg ccf s dc 2 2

vms 2 1 v v vˆ v v v v +vms c l s + m s( cg c ) cf s dc + v eint n s + e int cf s dc 2 2 蝌 s s Energy Flux (con’t): 1r 1 =vv(r v2 + r v 2 v v +s c 2 l + e n ) + r v 2 v v + v v [ P ] 2s s s 2 sints s 2 s s 1 +cv( m c2 +e ) fˆ dc v 2 sints s

2 3 Since Ps=[ J s] + P s [ I ] and m c l kT , then R.H.S. becomes: s s2 s 2 e vv v v3kTs P s ints 1 2 v v =rsv( + vg v s + + + ) + r s v v s + v [ J s ] 2 2mr m 2 R.H.S. s s s 1 +cv( m c2 + e ) fˆ dc v 2 sints s v v (Note that vg Ps[ I ] = P s v )

Now, e 3 kTs P sints P s  + + =us + = h s 2 msr s m s r s