86
6 Tensor Analysis
d
V
ds
=
V
i ;k
dx
k
ds
e i =
V
i ;k t
k
e i , t
k
≡
dx
k
ds
.
(6.17)
Here t
k are the components of the tangent vector to the curve. The object in the last
parentheses is then the component array of the curve derivative of the vector.
The last expression for the curve derivative may also be written in an informative
canonical form. First note that the components t
k may be expressed in terms of the
abstract vector
t as we see from the relations
t = t
n
e n , so ˜
dx
k
t
= t
n ˜
dx
k
( e n ) = t
n
δ
k
n = t
k
.
(6.18)
We substitute this for t
k into (6.17) and find
d
V
ds
= V
i ;k
e i ˜
dx
k
t
= V
i ;k
e i ⊗ ˜
dx
k
−,
t
.
(6.19)
This displays the curve derivative in the direction
t in terms of the basis vectors and
forms.
Having obtained the covariant derivative of a vector we see it is fairly obvious
how to infer the necessary definition for the covariant derivative of a general tensor;
the logic is much the same as used for (6.10) in Sect. 6.1. We consider the special
case of a (2,0) tensor which is the direct product of two vectors,
T =
V ⊗
W = (V
i
e i ) ⊗ (W
n
e n ) = V
i W
n
( e i ⊗ ⊗
e n ).
(6.20)
We then impose the product or Leibniz rule for derivatives and after some algebra
find a relation analogous to (6.16),
∇T = ∇
V ⊗
W +
V ⊗ ∇
W = (V
i ;k W
n
+ V
i W
n ;k )
e i ⊗ ⊗
e n ⊗ ˜
dx
k
=
T
in ;k
e i ⊗ ⊗
e n ⊗ ˜
dx
k
.
(6.21)
From this it is clear that the covariant derivative of any tensor must be given in terms
of its components and the basis by
∇T =
T
i...
j...;k
( e i ⊗ . . .)( ˜
dx
j
⊗ . . . ˜
dx
k
).
(6.22)
That is, given the component array for the covariant derivative discussed in Sect. 6.1
the covariant derivative of the abstract tensor obeys the same sort of equation as
(4.61).
The special case of the covariant derivative of the metric tensor is worth
mentioning due to its importance. We have
∇g = g i j;k
˜
dx
i
⊗ ˜
dx
j
⊗ ˜
dx
k
.
(6.23)
6 Tensor Analysis
d
V
ds
=
V
i ;k
dx
k
ds
e i =
V
i ;k t
k
e i , t
k
≡
dx
k
ds
.
(6.17)
Here t
k are the components of the tangent vector to the curve. The object in the last
parentheses is then the component array of the curve derivative of the vector.
The last expression for the curve derivative may also be written in an informative
canonical form. First note that the components t
k may be expressed in terms of the
abstract vector
t as we see from the relations
t = t
n
e n , so ˜
dx
k
t
= t
n ˜
dx
k
( e n ) = t
n
δ
k
n = t
k
.
(6.18)
We substitute this for t
k into (6.17) and find
d
V
ds
= V
i ;k
e i ˜
dx
k
t
= V
i ;k
e i ⊗ ˜
dx
k
−,
t
.
(6.19)
This displays the curve derivative in the direction
t in terms of the basis vectors and
forms.
Having obtained the covariant derivative of a vector we see it is fairly obvious
how to infer the necessary definition for the covariant derivative of a general tensor;
the logic is much the same as used for (6.10) in Sect. 6.1. We consider the special
case of a (2,0) tensor which is the direct product of two vectors,
T =
V ⊗
W = (V
i
e i ) ⊗ (W
n
e n ) = V
i W
n
( e i ⊗ ⊗
e n ).
(6.20)
We then impose the product or Leibniz rule for derivatives and after some algebra
find a relation analogous to (6.16),
∇T = ∇
V ⊗
W +
V ⊗ ∇
W = (V
i ;k W
n
+ V
i W
n ;k )
e i ⊗ ⊗
e n ⊗ ˜
dx
k
=
T
in ;k
e i ⊗ ⊗
e n ⊗ ˜
dx
k
.
(6.21)
From this it is clear that the covariant derivative of any tensor must be given in terms
of its components and the basis by
∇T =
T
i...
j...;k
( e i ⊗ . . .)( ˜
dx
j
⊗ . . . ˜
dx
k
).
(6.22)
That is, given the component array for the covariant derivative discussed in Sect. 6.1
the covariant derivative of the abstract tensor obeys the same sort of equation as
(4.61).
The special case of the covariant derivative of the metric tensor is worth
mentioning due to its importance. We have
∇g = g i j;k
˜
dx
i
⊗ ˜
dx
j
⊗ ˜
dx
k
.
(6.23)