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We can use the divergence theorem to transform this into a volume integral:
F
=
Ω
∇·
σ.
Note that the notation
σ
is the accepted short-hand for the vector whose
elements are the divergences of the rows (or columns) of
σ
:
∇·
⎡
⎣
⎤
⎦
∂σ
11
∂x
+
∂σ
12
∂y
+
∂σ
13
∂z
∂σ
21
∂x
+
∂σ
22
∂y
+
∂σ
23
∂z
∇·
σ
=
.
∂σ
31
∂x
+
∂σ
32
∂y
+
∂σ
33
∂z
Ignoring body forces for simplicity, we set this net force equal to the mass
times center-of-mass acceleration:
F
=
M A
=
Ω
ρ
Du
Dt
,
i.e., we have an equality between the two volume integrals:
=
ρ
Du
Dt
Ω
∇·
σ.
Ω
Since this holds for any arbitrary volume, the integrands must be equal:
ρ
Du
Dt
=
∇·
σ.
Adding back in the body force term, we actually have the general momen-
tum equation for a continuum (elastic solids as well as fluids):
Du
Dt
1
ρ
g
+
1
=
ρ
∇·σ.
In the particular case of an inviscid fluid, as we discussed above, the
stress tensor is just the negative of pressure times the identity—see Equa-
tion (8.1). In this case, it's not hard to see that
∇·
σ
simplifies to
−∇
p
,
giving the familiar momentum equation.
For general fluid flow, pressure is still a very important quantity, so we
will explicitly separate it out from the rest of the stress tensor:
σ
=
−
pδ
+
τ,
where
τ
is also a symmetric tensor. We will let the pressure term handle
the incompressibility constraint and model other fluid behavior with
τ
.
