Graebel W.P. Advanced Fluid Mechanics
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344 Appendix
They can be used to find the inverse of A (the inverse is defined such that A
−1
A =)by
A
−1
k
i
=
A
ij
A
jk
−A
k
i
I −
k
i
II
/III
Components of tensors obey all of the familiar associative and distributive laws.
With this somewhat tedious explanation out of the way, we can proceed to the
calculus of tensors. Starting with the vector A = A
i
g
i
, it follows that dA = g
i
A
i
x
j
+
A
i
g
i
x
j
dx
j
.
It must be that the derivatives of the base vectors can be written in terms of the
base vectors, so let
g
i
x
j
=
ij
g
We then have
dA =
g
i
A
i
x
j
+A
i
ij
g
dx
j
=g
i
A
i
x
j
+A
i
j
dx
j
We see that the quantities in parentheses are a valid components of a tensor and therefore
define the covariant derivative of contravariant components by
A
i
j
=
A
i
x
j
+A
i
j
(A.8.18)
By a similar calculation, find that define the covariant derivative of covariant compo-
nents by
A
i
j
=
A
i
x
j
−A
ij
(A.8.19)
Contravariant derivatives are defined by A
i
j
=A
i
k
g
kj
.
The symbols introduced to relate the derivatives of the base vectors to the base
vectors themselves are called Christoffel symbols and are given by
Christoffel symbol of second kind
k
ij
=
k
ji
=
ij
g
k
Christoffel symbol of first kind
ijk
=
jik
=
1
2
g
ik
x
j
+
g
jk
x
i
−
g
ij
x
k
(A.8.20)
Christoffel symbols are not components of any tensor, but when combined with the
partial derivatives as in equations (A.8.18) and (A.8.19), they produce tensor compo-
nents. They are responsible for the additional terms we have seen when working in
cylindrical and spherical coordinates. Since even with symmetry, there are possibly
21 different components to be calculated, when necessary, the wise person resorts to
symbolic processors for their calculation!
With this in hand, representing the Navier-Stokes equations in general coordinates
is elementary: Just replace partial derivatives by covariant or contravariant derivatives.
The continuity equation then becomes
t
+
v
i
i
= 0 (A.8.21)
A.8 Tensors in General Coordinates 345
and the momentum equation is
v
i
t
+v
j
v
i
j
=
i
j
j
+g
i
i
j
=
−p +
v
k
k
i
j
+
v
i
j
+v
j
i
(A.8.22)
It is easy to write equations in general coordinates: Start with the equations in
Cartesian coordinates, replace partial spatial derivatives with covariant derivatives, and
pay attention to being consistent in subscripts and superscripts. And it is relatively easy
to operate on them and manipulate them, but reducing them to any given coordinate
system than the Cartesian one (the Christoffel symbols vanish in Cartesian coordinates)
requires a good deal of patience.
One convenient device for computation of the Christoffel symbols is to start with
equation (A.8.4) and to write it as
T =
ds
dt
2
=g
ij
v
i
v
j
where v
i
=
dx
i
dt
(A.8.23)
Next, compute
f
i
=
1
2
d
dt
T
v
i
−
T
x
i
(A.8.24)
It can be shown that
f
i
=g
ij
d
2
x
j
dt
2
+
jki
v
j
v
k
(A.8.25)
Comparison of the coefficients of v
j
v
k
in equations (A.8.24) and (A.8.25) yield the
Christoffel symbols more conveniently than by computation directly from the definition.
An account of how to determine physical components of tensors can be found in
Truesdell (1953). Briefly, physical base vectors are defined by
gi = g
i
/
g
ii
=g
ij
g
j
/
g
ii
(A.8.26)
where the summation convention is suspended on indices enclosed in parentheses. For
a vector, then, since
V = V
i
g
i
=V
i
g
ii
gi = V
j
g
ij
g
ii
gi (A.8.27)
it follows that the physical components of V are given by
Vi =V
i
g
ii
=V
j
g
ij
g
ii
(A.8.28)
Physical components of higher-order tensors can be computed in a similar manner.
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