Page 320 - Intro to Tensor Calculus

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(i) In the special case φ = m the equation (2.5.86) reduces to the continuity equation for fluids. That is,
equation (2.5.86) becomes
∂
~
(nm)+ ∇· (nmV 1 )= 0 (2.5.96)
∂t
which is the continuity equation
∂ρ
~
+ ∇· (ρV )= 0 (2.5.97)
∂t
~
where ρ is the mass density and V is the mean velocity defined earlier.
~
(ii) In the special case φ = mV 1 is momentum, the equation (2.5.86) reduces to the momentum equation
~ ~
for fluids. To show this, we write equation (2.5.86) in terms of the dyadic V 1 V 1 in the form
∂
~
~ ~
~
nmV 1 + ∇· (nmV 1 V 1 ) − nF =0 (2.5.98)
∂t
or
∂
~
~
~
~
~
~
~
ρ(V r + V ) + ∇· (ρ(V r + V )(V r + V )) − nF =0. (2.5.99)
∂t
~ ~
Let σ = −ρV r V r denote a stress tensor which is due to the random motions of the gas particles and
write equation (2.5.99) in the form
~
∂V ∂ρ
~
~
~
~
~
~
ρ + V + ρV (∇· V )+ V (∇· (ρV )) −∇ · σ − nF =0. (2.5.100)
∂t ∂t
∂ρ
~
~
The term V + ∇· (ρV ) = 0 because of the continuity equation and so equation (2.5.100) reduces
∂t
to the momentum equation
!
~
∂V
~
~
~
ρ + V ∇· V = nF + ∇·σ. (2.5.101)
∂t
~
~
~
~
~
~
~
~
For F = qE + qV × B + mb,where q is charge, E and B are electric and magnetic fields, and b is a
body force per unit mass, together with
3 3
X X
e
e
σ = (−pδ ij + τ ij )b i b j (2.5.102)
i=1 j=1
the equation (2.5.101) becomes the momentum equation
~
DV
~
~
~
~
ρ = ρb −∇p + ∇· τ + nq(E + V × B). (2.5.103)
Dt
~
~
In the special case were E and B vanish, the equation (2.5.103) reduces to the previous momentum
equation (2.5.25) .
2
2
2
(iii) In the special case φ = m ~ ~ m (V +V +V ) is the particle kinetic energy, the equation (2.5.86)
V 1 ·V 1 =
2 2 11 12 13
simplifies to the energy equation of fluid mechanics. To show this we substitute φ into equation (2.5.95)
and simplify. Note that
m h i
2
2
φ = (U r + u) + (V r + v) + (W r + w) 2
2
m h 2 2 2 i (2.5.104)
2
2
2
φ = U + V + W + u + v + w
r
r
r
2