Moments of the Boltzmann Equation
As was already discussed, finding solutions to Boltzmann’s equation can be formidably
difficult, even in the simplest of cases. An important manipulation can be achieved by
taking moments (or averages) of pertinent quantities and try to recover fluid-type
conservation equations. On their own right, these fluid equations (like the Navier-Stokes
equations) are also very difficult to solve analytically, unless special cases are treated, but
at least they provide us with a useful physical picture of the system behavior. In what
follows we will ignore the effects of inelastic collisions.
As before, we write Boltzmann’s equation
∂f s Fs
+ w ⋅ ∇f s +
⋅∇ f =∑ ∫
∂t
ms w s r w1
∫ ( f ′f ′ − f f )gσ
s r1
s r1
rs
dΩd 3 w1
Ω
We define a general function φ = φ( x , w,t) , multiply Boltzmann’s equation with it and
then integrate over all velocities w . One by one, the terms of Boltzmann’s equation will
result in
1.
∂f s
⎡∂
∫ ∂t φ d w = ∫ ⎢⎣∂t ( f φ ) − f
3
s
s
∂φ ⎤ 3
⎥d w
∂t ⎦
Recalling the definition of the average of a quantity: φ =
1
n
∫ φfd
3
w , and
exchanging the integral and derivative symbols, we obtain
∂f s
∂
∫ ∂t φ d w = ∂t (n
3
s
2.
φ
s
)−n
s
∂φ
∂t
s
3
3
∫ φw ⋅ ∇f
s d w = ∫ [∇ ⋅ (φwf s ) − φf s∇ ⋅ w − f s w ⋅ ∇φ ]d w
Since w ≠ w( x ) , the second term in the RHS vanishes. Then we have
∫ φw ⋅ ∇f sd 3w = ∇ ⋅ n s φw
s
− n s w ⋅ ∇φ
s
1
Fs
3
3
∇ w ⋅ φFs f s − φf s∇ w ⋅ Fs − f sFs ⋅ ∇ wφ d w
3. ∫ φ ⋅ ∇ w f sd w =
∫
ms
ms
From Liouville’s theorem, F is a conservative and/or magnetic force, therefore
the second term in the RHS vanishes. The first term vanishes as well. To see why,
we use the divergence theorem and write it as a surface integral in velocity space
∫ ∇w ⋅ φ Fs f s d 3w = ∫ φ f sFs ⋅ dSw . The volume integral is evaluated for every
[ (
(
)
)
]
s
possible value in velocity space, therefore the surface boundary is located at
infinity, where there are no particles at all (the distribution function vanishes
there).
1
F
Fs
∫ φ m ⋅ ∇w f sd 3w = −ns ms ⋅ ∇ wφ
s
s
4.
⎛∂ ⎞
∫ φ⎜⎝ ∂ft ⎟⎠
coll
d 3w = ∑ ∫
∫ ∫ φ ( f ′f ′ − f f ) gσ
s r1
s r1
rs
s
dΩd 3 wd 3 w1
r w1 w Ω
We can split this expression into two integrals. In the first one we exchange the
coordinates after the collision with those before it ( w ↔ w ′). And since (as seen
before) the Jacobian of the transformation is unity ( d 3 w1′d 3 w ′ = d 3 w1d 3 w ) and the
magnitude of the relative velocity does not change, then
⎛ ∂f ⎞
∫ φ⎜⎝ ∂t ⎠⎟
coll
d 3w = ∑ ∫
∫ ∫ (φ ′ − φ ) f
f gσ rs dΩd 3 wd 3 w1
s r1
r w1 w Ω
Adding all terms we have the general moment equation
∂
(n φ
∂t s
∂φ
− ns
s)
∂t
Fs
+ ∇ ⋅ n s φw s − ns w ⋅ ∇ φ s − ns
⋅ ∇ wφ
ms
s
=∑ ∫
r
∫
=
s
∫ (φ ′ − φ ) f s f r1 gσ rs dΩd 3wd 3w1
w1 w Ω
Now let us obtain the moments for different values of the function φ = φ( x ,w,t)
a. φ = 1 (Mass)
All the terms where derivatives of φ appear will vanish, leading to
∂n s
+ ∇ ⋅ (n sus ) = 0
∂t
or
∂ρ s
+ ∇ ⋅ ( ρ sus) = 0
∂t
where the expression in the right was obtained after multiplying both sides with
the species mass ms . Finally, adding contributions of all species: ∑ ρ s = ρ (fluid
s
density) and
∑ ρsus = ρu (fluid mass flux), we write the continuity equation
s
∂ρ
+ ∇ ⋅ (ρ u) = 0
∂t
∑ρ u
u=
∑ρ
s s
where
s
s
s
b. φ = msw (Momentum)
Note that ns φ s = ρ sus . Also, since w has no explicit dependence on t or x , then
ns
∂φ
∂t
=0
and
s
2
ns w ⋅ ∇φ s = 0
From the definition of random velocity for species s, w = us + c s , and as c s s = 0
ns φ w s = ρ s ww s = ρ s ( us + c s)( us + c s) = ρ s( usus + c sc s
= ρ susus + Ps′
s)
where Ps′ is the partial pressure of species s across planes moving at us . The
prime is to indicate that the quantity (in this case the pressure) is taken with
respect to the random velocity of species s.
Now, taking the electromagnetic force Fs = qs E + w × B , we have
(
F
⋅ ∇ wφ
ns
ms
)
= n sqs E + w × B ⋅ I
(
)
= n sqs E + us × B
(
s
s
)
For the collision part, we write φ ′ − φ = ms( w′ − w) . In terms of the center of mass
and relative velocity, g = w1 − w
w=G−
then
mr g
mr + ms
and
w′ = G −
mr g′
mr + ms
msmr ms (w ′ − w ) =
( g − g ′) = μsr ( g − g′)
mr + ms
and the collision integral can be written as
M rs = μrs ∫
∫
w1 w
g′
χ
f s f r1 gd 3 wd 3 w1 ∫ σ rs ( g − g ′)dΩ
Ω
g − g′
g
g − g′
ϕ
⎛ g ⎞
⎜
(g − gcos χ )⎜ ⎟⎟
⎝ g⎠
We observe from the diagram above, that the component of the g − g′ vector
perpendicular to g will cancel out after integrating over all angles ϕ . What
survives is just the component parallel to g . Given this, and the fact that the
magnitude of the relative velocity is invariant, we write
M rs = μrs ∫
∫
w1 w
f s f r1 gd 3 wd 3 w1 ∫ σ rs g(1− cos χ )dΩ
Ω
3
Separating the angular contribution we obtain the momentum transfer cross
section
(1)
Qrs (g) = ∫ σ rs ( χ ,g)(1− cos χ )dΩ
Ω
The collision integral becomes
M rs = μrs ∫
∫
1
f s f r1 ggQ(rs) (g) d 3 wd 3 w1
w1 w
which is the rate of momentum transfer from species s to species r per unit
volume due to collisions. Adding up all terms,
∂
(ρsus ) + ∇ ⋅ ( ρsusus ) + ∇ ⋅ Ps′− n sqs E + us × B = ∑ M rs
∂t
r
(
)
Since we averaged absolute momentum, this is the Eulerian form, with
∇ ⋅ (momentum flux).
c. φ = ms( w − us ) (Momentum with deviations from the species s mean velocity)
Instead of repeating the process all over again, we recognize that such function
can be separated in two additive terms, the first one φ1 = msw will lead to the
same momentum equation found in (b) while the second φ 2 = msus is proportional
to the result in (a) since us is already an averaged quantity. Using the result in (a)
for φ 2 = msus we obtain
⎡ ∂n
⎤
msus⎢ s + ∇ ⋅ (n sus)⎥
⎣ ∂t
⎦
then we subtract this from the result in (b)
∂
⎡∂n
⎤
(ρsus ) + ∇ ⋅ ( ρsusus ) + ∇ ⋅ Ps′− n sqs E + us × B − msus⎢ s + ∇ ⋅ (nsus )⎥ = ∑ M rs
⎣ ∂t
⎦ r
∂t
(
)
For the second term in this expression, using tensorial notation
∂u
∂
∂
∇ ⋅ (ρ s usu s ) =
ρ s uiu j ) = ρ su i j + u j
( ρsui ) = ρs (us ⋅ ∇)us + us ∇ ⋅ (ρsu s )
(
∂x i
∂x i
∂xi
After regrouping we obtain
⎡∂ ⎤
ρ s⎢ us + ( us ⋅ ∇ ) us⎥ + ∇ ⋅ Ps′− n sqs E + us × B = ∑ M rs
⎣ ∂t
⎦
r
(
4
)
or
Dsu s
ρs
+ ∇ ⋅ Ps′ − nsqs E +us × B = ∑ M rs
Dt
r
(
)
This is the Lagrangian form of the momentum
equation for species s. Note
that
since we averaged deviations from us , we get substantial derivatives at us .
d. φ = ms ( w − u ) (Momentum with deviations from the fluid mean velocity)
Start by noting that ns φ s = ρ s (us − u ) = ρ sVs , where Vs is the diffusion velocity
of species s. Then, one by one, the terms of the moment equation will be
∂
(n φ
∂t s
and
∂φ
ns
∂t
s
) = ∂∂t ( ρ V )
s s
∂u
= −ρ s
(remember that w does not depend explicitly on t).
∂t
s
Also
∇ ⋅ns φ w s = ∇ ⋅ ρ s ( w − u )w
(
s
)
Defining the fluid random velocity c = w − u , and noting that c
the last term can be written as
∇ ⋅ns φ w s = ∇ ⋅ ρ s c (c + u )
(
s
)
= ∇ ⋅ (ρ s c c s + ρ s c u
s
= u s − u = Vs ,
)s = ∇ ⋅ Ps + ρsVs u
(
)
after defining the pressure tensor Ps = ρ s c c s , which can be written in terms of
the partial pressure Ps′. To see this, we use the definition
of the random velocity
of species s, c s = w − u s , so that c − cs = u s − u = Vs and c = cs + Vs , therefore
Ps = ρ s cs + Vs c s + Vs
(
)(
)
s
= ρ s c sc s s + ρ s VsVs
s
and we get Ps = Ps′ + ρ s VsVs s . In many cases the mean velocity of individual
species is not that different from the mean fluid velocity, so that the diffusion
contribution to the pressure tensor is usually small.
Now, for the remaining parts of the moment equation
ns w ⋅∇φ
s
= −ρ s (u + c ) ⋅ ∇ u
s
= −ρ s u ⋅ ∇u − ρ sVs ⋅ ∇u
and the force term is similar to what we obtained in (b)
5
ns
Fs
⋅∇ wφ
ms
= ns Fs ⋅ I
s
s
= n sqs E + us × B
(
)
Using the definition of the diffusion velocity, we rewrite the term as
Fs
ns
⋅∇ wφ
ms
= ns qs E + u + Vs × B = nsq s E + u × B + Vs × B = ns qs [ E′ +Vs × B ]
( (
s
) )
[(
)
]
Where E′ is the electric
field as seen in the frame of reference of the fluid moving
at a mean velocity u . The collision integral in this case is the same as the one
found in (b), therefore the complete moment equation is
⎡ ∂u ⎤
∂
ρ sVs + ∇ ⋅ Ps + ρ sVsu + ρ s ⎢ + u ⋅∇ u + Vs ⋅∇ u⎥ − nsqs [E ′ + Vs × B ] = ∑ M rs
⎣ ∂t
⎦
∂t
r
(
(
)
)
Now we add for all species, such that the fluid density, charge and pressure are
∑ρ
s
s
=ρ
∑n q
s s
∑ Ps = P
= ρ ch
s
s
The summation
of
collision terms cancels out given de symmetry of momentum
transfer M rs = − Msr , while the one over the diffusion velocities is zero by
definition
∑∑ M
s
r
rs
=0
∑ ρ V = ∑ ρ (u
s s
s
s
s
− u ) = ρ u − ρu = 0
s
We also define the diffusion current density as j D = ∑n sqsVs . The total current
s
density
be this plus the contribution of charges moving with the fluid
would
j = j D + ρ ch u . The momentum moment equation for the fluid is finally
⎡ ∂u ⎤
ρ ⎢ + u ⋅ ∇ u⎥ + ∇ ⋅ P = ρ ch E ′ + j D × B
⎣ ∂t
⎦
or in the absence of net charge, ρ ch = ∑ ns qs = 0
s
⎡ ∂u ⎤
ρ ⎢ + u ⋅ ∇ u⎥ + ∇ ⋅ P = j × B
⎣ ∂t
⎦
6
1
e. φ = msw 2 (Kinetic energy)
2
As before, start by noting that
moment equation
∂
(n φ
∂t s
∂ ⎡ ρs
s
) = ∂t ⎢⎣ 2
∂φ
= 0 and ∇φ = 0 . For the first term of the
∂t
⎤ ∂ ⎡ρ ⎤
⎤ ∂ ⎡ρ
2
2
w 2 s⎥ = ⎢ s (c s + us) ⋅ (c s + us ) s⎥ = ⎢ s c s s + us ⎥
⎦
⎦ ∂t ⎣ 2
⎦ ∂t ⎣ 2
(
3
1
2
ms c s = kTs′ we obtain
s
2
2
and from the definition of temperature
∂
(n φ
∂t s
∂ ⎡ ρs
s
)
) = ∂t ⎢⎣ 2 u
2
s
⎤
3
+ nsk Ts′⎥
⎦
2
Recall that the prime denotes that the quantity (in this case the temperature) is
taken with respect to the random velocity of species s. For the next term we have
1 2
ρ
2
2
∇ ⋅ ns φw s = ∇ ⋅ ρ s w w = ∇ ⋅ s (c s + us + 2c s ⋅ us)(c s + us) s
2
2
s
⎡ρs 2
ρ 2 ρ 2
⎤
= ∇ ⋅ ⎢ c s c s s + s c s s us + s us us + ρ s c s(c s ⋅ us) s⎥
⎣2
⎦
2
2
Defining the heat flux (also with respect to the random velocity of species s)
ρ 2
qs′ = s c s c s
2
s
and noting that (using index notation)
ρ s c s (c s ⋅ u s ) s = ρ s c i c j u j
s
= ρ s c ic j
s
u j = Pij′ u j = Ps′us
Therefore we have
⎡
3
⎤
ρ
∇ ⋅ n s φw s = ∇ ⋅ ⎢qs′ + n s us kTs′ + s us2 us + Ps′us⎥
⎣
⎦
2
2
For the force term of the moment equation, we have
7
Fs
⋅ ∇ wφ
ns
ms
= n s Fs ⋅ w
s
= nsqs E + w × B ⋅ w
(
)
s
= nsqs E ⋅ w = nsqsE ⋅ us = E ⋅ js
s
s
Where j s is the mean current carried by species s. Finally, for the collision term
we observe that (keep in mind that the magnitude of the relative velocity vector
does not change)
2
2⎤
⎡
1 ⎢⎛⎜ m r ⎞ ⎥
1
mr ⎞⎟ ⎛⎜ 2
2
g′⎟ − G −
φ ′ − φ = ms[ w′ − w ] = ms⎢⎜ G −
g⎟
2
2 ⎣⎝
mr + ms ⎠ ⎜⎝
mr + ms ⎟⎠ ⎥⎦
msmr =
G ⋅ (g − g′) = μsr G ⋅ ( g − g′)
mr + ms
We write the collision integral as
E rs = μrs ∫
∫f
f gd 3 wd 3 w1 ∫ σ rsG ⋅ (g − g ′)dΩ
s r1
Ω
w1 w
Following the same reasoning as in case (b), we obtain for the momentum transfer
cross section
Qrs (g) =
(1)
∫ σ ( χ,g)(1− cos χ )dΩ
rs
Ω
So that the collision integral reduces to
E rs = μrs ∫
∫f
f gG ⋅ gQrs(1) ( g)d 3 wd 3 w1
s r1
w1 w
Putting all the terms together we find the Eulerian form of the kinetic energy
moment equation
⎤
⎡
3
∂ ⎡ ρs 2 3
ρs 2 ⎤ us + n s kTs′⎥ + ∇ ⋅ ⎢q′s + n s us kTs′ + us us + Ps′us ⎥ − E ⋅ j s = ∑ E rs
⎦
⎣
⎦
2
∂t ⎢⎣ 2
2
2
r
Rearranging to put it in a more interesting way
⎡ ⎛1
∂ ⎡ ⎛1
3 ⎞⎤
3 ⎞⎤
⎡ ⎤ 2
2
⎢n s ⎜ ms us + kT ′⎟ ⎥ + ∇ ⋅ ⎢n s us ⎜ ms us + kTs′⎟⎥ + ∇ ⋅ ⎢⎣q′s + Ps′us ⎥⎦ = E ⋅ j s + ∑ E rs
⎝2
∂t ⎣ ⎝ 2
2 ⎠s⎦
2 ⎠⎦
⎣
r
In this way, the two terms in the RHS can be considered as “inputs” to the energy
equation described in the LHS.
8
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http://ocw.mit.edu
16.55 Ionized Gases
Fall 2014
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