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and an antisymmetrical part
(
)
2
A
≡
A
ij
= ∂
U
i
∂
x
j
− ∂
U
j
∂
x
i
The terms on the left-hand side of equation [2.29] can
easily be decomposed into
⎡
⎣
⎤
∂
a
ij
∂
U
k
∂
a
ij
∂
DS
ij
Dt
+
x
k
+
a
ik
a
kj
=
+
A
ik
A
kj
+
S
ik
S
kj
+
⎢
⎥
t
⎦
[2.34]
⎡
⎣
⎤
DA
ij
Dt
+
A
ik
S
kj
+
S
ik
A
kj
⎢
⎥
⎦
The first term in square brackets on the right of the above
equality is symmetrical, and the second term below it is
antisymmetrical. However, the Hessian
x
j
is
symmetrical. By identifying the symmetrical and
antisymmetrical parts of the transport equation [2.30] after
decomposition, we successively come to
2
P
−∂
∂
x
i
∂
2
S
ij
DS
ij
Dt
2
P
∂
1
ρ
∂
[2.35]
+
A
ik
A
kj
+
S
ik
S
kj
=−
x
j
+ ν
∂
x
i
∂
∂
x
k
∂
x
k
and
2
A
ij
DA
ij
Dt
∂
[2.36]
+
A
ik
S
kj
+
S
ik
A
kj
= ν
∂
x
k
∂
x
k
We can easily show that the antisymmetrical tensor
A
ij
is
linked to the vorticity component
k
by
Ω
= Ω
+ ω
k
k
1
2
ε
ijk
Ω
k
[2.37]
A
ij
=−
The antisymmetrical tensor
ε
ijk
is defined such that:
-
0
if two of the indices are arbitrarily identical
ε
=
ijk
(
0
);
ε
112
= ε
122
= ε
313
=
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