22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2: The Moment Equations
Boltzmann-Maxwell Equations
1. Recall that the general coupled Boltzmann-Maxwell equations can be written as
a.
∂f j
∂t
+ v ⋅ ∇f j +
∇×E = −
mj
(E + v × B ) ⋅ ∇v f j
=
∑ C jk
k
+ sj
∂B
∂t
∇ × B = μ0 J +
∇ ⋅E =
qj
1 ∂E
c 2 ∂t
σ
∈0
∇ ⋅B = 0
b. The particle and electromagnetic equations are coupled by
σ=
∑ q j nj = ∑ q j ∫
j
j
J=
∑ q j n ju j = ∑ q j ∫ vf j dv
j
j
f j dv
c. The collision operators Cjk arise from elastic collisions and satisfy a
corresponding set of conservation relations.
2. Derivation of fluid equations – take moments as follows:
a. mass
⎡ df j
C jk
∫ ⎢ dt − ∑
k
⎣
⎤
− s j ⎥ dv = 0
⎦
b. momentum
⎡ df j
C jk
∫ mj v ⎢ dt − ∑
k
⎣
⎤
− s j ⎥ dv = 0
⎦
c. energy
m jv 2 ⎡ df j
⎤
C jk − s j ⎥ dv = 0
∫ 2 ⎢ dt − ∑
k
⎣
⎦
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 1 of 10
3. Introduce macroscopic quantities nj (r, t), uj (r, t), pj (r, t).
4. The moment equations become a set of coupled, time dependent PDE’s relating the
various macroscopic quantities.
5. There initial Boltzmann equation is a single scalar equation in fj: fj = fj (r, v, t). There
are seven independent variables.
6. The resulting moment equations contain six fluid variables: nj, uj, Tj, pj, all functions
of (r, t). There are four independent variables.
7. The fluid equations are far simpler to solve.
8. At a basic mathematics level the moment method appears to be ill conceived. You
cannot solve a partial differential equation by integrating over several independent
variables and then solving a reduced equation.
9. Example consider ψ = ψ ( r , θ , t ) satisfying
a.
∂ψ 1 ∂
∂ψ 1 ∂
∂ψ
=0
=
rf ( r , θ )
+ 2
g ( r,θ )
∂t
r ∂r
∂r r ∂θ
∂θ
b. Integrated over θ assuming periodicity: define G = (1 2π ) ∫ Gdθ
∂ψ
∂
1 ∂
ψ =
r f
∂r
∂t
r ∂r
c. This equation is a correct expression, but not a useful one since two different
averages appear: ψ , f ∂ ψ ∂r . Integrating over θ leads to a single, simpler
reduced equation, but with two unknowns.
10. This is a general property of moment equations. When we take moments of the
Boltzmann equation, we will obtain a set of correct relations, but there will be more
unknowns than equations.
11. How do we resolve this problem? We will close the set of equations by solving the
Boltzmann equation and then evaluating some of the higher order, additional
unknowns.
12. This would seem to make the entire procedure circular. If we are going to solve the
Boltzmann equation anyway why bother with the moment equations?
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 2 of 10
13. There is method to this madness.
a. First, even if we know the solution to the Boltzmann equation the moments
represent more useful information in that they describe the measurable
physical quantities in the system.
b. Second, and equally important, we are not just going to “simply solve” the
Boltzmann equation for the extra unknowns. The full equation is enormously
complicated to solve.
c. Instead, we shall solve the Boltzmann equation by means of various
expansions ( e.g. me mi , rL a, ω Ω, etc ) .
d. Each order in the expansion is exponentially more painful to calculate than
the previous order.
e. By having a carefully defined set of moment equations, we can determine
beforehand exactly how many terms are needed in the expansions. In
addition we can rewrite the moment equations in such a way as to further
minimize the number of terms required.
f.
These two reasons (physical variables, minimum algebra) are strong
motivation for using the moment procedure.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 3 of 10
Moment Equations
1. Procedure to be followed:
a. Calculate exact moments: mass, momentum, energy.
b. Introduce random velocity v=uj (r, t) + w.
c. Define physical variables nj, uj, Tj, pj…
d. Do some algebraic bookkeeping.
e. Arrive at a set of moment equations (with more unknowns than equations).
2. Now focus on the momentum equation from which we derive the MHD equilibrium
equation. Show that this equation is valid for both MHD stability analysis and
transport phenomena. Motivation: almost no knowledge of higher order unknowns is
required.
3. Spend half a semester investigating applications of the MHD equilibrium equation.
4. Derive in detail the collision operators Cjk.
5. Spend half a semester investigating applications of the transport equations in
cylindrical geometry.
6. Consider next the conservation of mass equation: we derive this carefully, carrying
out all the steps.
a.
⎡ ∂f j
∫ dv ⎢⎢ ∂t
+ v ⋅ ∇f j +
⎣
b. Define n j =
nju j =
∂f j
∫ f j dv
∫ vf j dv
qj
mj
(E + v × B ) ⋅ ∇v f j − ∑ C jk
k
particle number density
particle flux
∂n j
∂
dv f j =
∫
∂t
∂t
c.
∫ dv
d.
∫ dv v.∇f j = ∫ dv ⎡⎣∇ ⋅ ( vf j ) − f j ∇ ⋅ v ⎤⎦
∂t
=
⎤
− sj ⎥ = 0
⎥⎦
=0
(
= ∇ ⋅ ∫ dv v f j = ∇ ⋅ n j u j
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
)
Lecture 2
Page 4 of 10
qj
⎡
∂f j
∂f
∂f ⎤
E ⋅ ∫ dv ⎢e x
+ ey
+ ez
⎥=0
mj
∂vy
∂v z ⎥⎦
⎢⎣ ∂v x
e.
∫ dv mj E ⋅ ∇V f j
f.
∫ dv m j v × B ⋅ ∇v f j
=
qj
qj
=
qj
mj
⎡
∂f j
∫ dv ⎢(vy Bz − v z By ) ∂v x
⎣
⎤
+ ⋅ ⋅ ⋅⎥ = 0
⎦
g. − ∫ dv ∑ C jk = −∑ ∫ dvC jk = 0 (conservation of particles)
k
k
h. − ∫ dvs j ≡ −Snj
(source of density)
7. Combine terms: note one equation, four unknowns nj, uj
∂n j
∂t
(
)
+ ∇ ⋅ n j u j = Snj
8. Similar procedure for the momentum equation yields
∂
n j m j u j + ∇ ⋅ n j m j vv
∂t
(
9. a. Here
)
l
∑k
(
) − q j n j (E + u j × B ) = ∑ kl ∫ mj vC jk dv + Spj
denotes k ≠ j (due to conservation of momentum in like particle
collisions)
b. S pj ≡
∫ dvm j vs j
= 0 (source of momentum, zero for practical applications)
c. Q = ∫ Qf j dv n j
10. Similar procedure for energy equation yields
2
mjv
∂ 1
1
l
mj n j v 2 + ∇ ⋅ mj n j v 2v − q j n j u j ⋅ E = ∑ k ∫
C jk dv + SEj
∂t 2
2
2
(
)
11. a. Here SEj ≡ ∫ dv m j v 2 2 s j
(sources of energy, say due to rf).
b. Also, k=j vanishes from collision term because of conservation of energy in like
particle collisions
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 5 of 10
Plasma Bookkeeping
1. The moment equations can be written in more physical terms by introducing the
random velocity and defining various physical quantities in addition to nj and uj.
2. The random velocity w: this is a change of independent variables from v to w defined
by
3. By definition dv=dw and w = 0
4. Then
vv = u j u j + u j w + wu j + ww = ww + u j u j
=0
=0
5. Define
where
pj =
1
n j m j ω2
3
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 6 of 10
HJG
I
1
Π j = n j m j ww − ω2 Ι
3
6. Similarly
3p j
v 2 = u2j + 2w ⋅ u j + ω2 = u2j +
nj mj
=0
v 2v =
(u
2
j
)
(
)
+ 2w ⋅ u j + ω2 u j + u2j + 2w ⋅ u j + ω2 w
=0
=
u2juj
+
3p j u j
nj mj
+2
=0
HJG
u j ⋅ Pj
nj mj
+
2h j
mj n j
where
hj ≡
1
n j m j ω2w
2
is the heat flux, the flux of heat due to random motion.
7. Now define
∫ mj (u j + w )C jk dv = ∫ m j wC jk dw ≡ R jk
=0
`
∫
mj
2
(
)
u2j + 2u j ⋅ w + ω2 C jk dv = u j ⋅ R jk +
∫
m j ω2
2
C jk dw
=0
≡ u j ⋅ R jk + Q jk
where Rjk is the average momentum transferred due to unlike collisions and Qjk is the
heat generated due to unlike collisions.
8. As they now stand, the moment equations can be written as
∂n j
∂t
(
)
+ ∇ ⋅ n ju j = snj
I
∂
n j m ju j + ∇ ⋅ n j m ju ju j + ∇ ⋅ P j − q j n j E + u j × B =
∂t
(
)
(
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
)
(
) ∑ kl R jk
Lecture 2
Page 7 of 10
HJG
1 ∂
3 ∂p j
3
⎡1
⎤
n j m j u2j +
+ ∇ ⋅ ⎢ n j m j u2j u j + p ju j + u j ⋅ Pj + h j ⎥ − q j n j u j ⋅ E =
2 ∂t
2 ∂t
2
2
⎣
⎦
(
)
∑ k (u j ⋅ R jk + Q jk ) + sEj
l
Algebraic simplification
1. Define the convective derivative, moving with the fluid
dQ j
dt
=
∂Q j
+ u j ⋅ ∇Q j
∂t
2. Define the temperature
Tj = pj nj
3. The mass equation is OK as is
4. Momentum equation simplifications:
∂
n j m ju j + ∇ ⋅ n j m ju ju j
∂t
(
)
= mj nj
= mj n j
∂u j
∂t
du j
dt
(
+ m ju j
∂n j
∂t
)
(
)
+ m j u j ∇ ⋅ n j u j + m j n j u j ⋅ ∇u j
+ m ju j Snj
5. The momentum equation becomes
mj nj
du j
dt
I
+ ∇ ⋅ P j − q j nj E + u j × B =
(
) ∑ kl R jk − mju j Snj
6. Energy equation simplifications
a.
2
n j m j u2j
m j n j ∂ 2 m j u2j ∂n j m j u2j
mj n j
∂ n j mj u j
uj +
+∇⋅
uj =
+
∇ ⋅ nju j +
u j ⋅ ∇u2j
∂t
2
2
2 ∂t
2
∂t
2
2
=
b.
m j n j d 2 m j u2j
uj +
Snj
2 dt
2
∂n j
⎞
⎞ 3 ⎛ ∂T j
3 ⎛ ∂p j
+ ∇ ⋅ p j u j ⎟⎟ = ⎜⎜ n j
+ Tj
+ T j ∇ ⋅ n j u j + n j u j ⋅ ∇T j ⎟⎟
⎜⎜
2 ⎝ ∂t
∂t
∂t
⎠
⎠ 2⎝
=
⎞
3 ⎛ dT j
+ T j Snj ⎟⎟
⎜⎜ n j
dt
2⎝
⎠
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 8 of 10
c. Energy equation becomes
I
mj n j d 2 3
dT j
u j + nj
+ ∇ ⋅ u j ⋅ P j + h j − q j n j uj ⋅ E =
dt
2 dt
2
(
∑ k (u j ⋅ R jk
l
)
⎛ mj 2 3 ⎞
+ Q jk + SEj − ⎜⎜
u j + T j ⎟⎟ Snj
2 ⎠
⎝ 2
)
d. Note that u j ⋅ (momentum equation) is equal to
I
mj nj d 2
l
u j + u j ⋅ ∇ ⋅ P j − q j n ju j ⋅ E = ∑ k u j ⋅ R jk − m j u2j Snj
2 dt
e. Now subtract (d) from (e)
I
I
dT j
3
nj
+ ∇ ⋅ u j ⋅ P j + hj − u j ⋅ ∇ ⋅ P j =
dt
2
(
f.
)
∑
l
k
Q jk + SEj
⎛ m j u2j 3 ⎞
+⎜
− T j ⎟ Snj
⎜ 2
2 ⎟
⎝
⎠
Note the following identity (recalling that by definition Pij = Pji)
I
I
∂
∂
∇ ⋅ u⋅P −u⋅∇ ⋅P =
ui Pij − ui
Pij
∂x j
∂x j
(
)
(
= Pij
)
∂
ui
∂x j
I
= P : ∇u
g. The energy equation becomes
dT j I
3
nj
+ P j : ∇u j + ∇ ⋅ h j =
dt
2
∑
l
k
Q jk + SEj
⎛ m j u2j 3 ⎞
+⎜
− T j ⎟ Snj
⎜ 2
2 ⎟
⎝
⎠
7. Summary of fluid moments
dn j
dt
+ n j ∇ ⋅ u j = Snj
mj nj
du j
dt
I
+ ∇ ⋅ P j − q j nj E + u j × B =
(
dT j I
3
nj
+ P j : ∇u j + ∇ ⋅ h j =
dt
2
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
) ∑ lk R jk − mju j Snj
∑
l
k
⎛ m j u2j 3 ⎞
Q jk + SEj + ⎜
− T j ⎟ Snj
⎜ 2
2 ⎟
⎝
⎠
Lecture 2
Page 9 of 10
I
8. Assuming the collisional terms are known, the fluid unknowns are n j , u j , P j , T j ,
and hj. (17 unknowns, 5 equations)
9. Even with a scalar pressure, there are still 9 unknowns.
10. The moment equations above are exact, if not particularly useful. They do, however,
accurately describe both MHD and transport phenomena.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 2
Page 10 of 10