Derivatives of basis vectors and Christoffel symbols

26.18 DERIVATIVES OF BASIS VECTORS AND CHRISTOFFEL SYMBOLS
the outer product of the two tensors, or any contraction of them, is a relative
tensor of weight w1 + w2 . As a special case, we may use ijk and ijk to construct
pseudovectors from antisymmetric tensors and vice versa, in an analogous way
to that discussed in section 26.11.
For example, if the Aij are the contravariant components of an antisymmetric
tensor (w = 0) then
pi = 12 ijk Ajk
are the covariant components of a pseudovector (w = −1), since ijk has weight
w = −1. Similarly, we may show that
Aij = ijk pk .
26.18 Derivatives of basis vectors and Christoffel symbols
In Cartesian coordinates, the basis vectors ei are constant and so their derivatives
with respect to the coordinates vanish. In a general coordinate system, however,
the basis vectors ei and ei are functions of the coordinates. Therefore, in order
that we may differentiate general tensors we must consider the derivatives of the
basis vectors.
First consider the derivative ∂ei /∂u j . Since this is itself a vector, it can be
written as a linear combination of the basis vectors ek , k = 1, 2, 3. If we introduce
the symbol Γkij to denote the coefficients in this combination, we have
∂ei
= Γkij ek .
∂u j
(26.75)
The coefficient Γkij is the kth component of the vector ∂ei /∂u j . Using the reciprocity relation ei · ej = δji , these 27 numbers are given (at each point in space)
by
Γkij = ek ·
∂ei
.
∂u j
(26.76)
Furthermore, by differentiating the reciprocity relation ei · ej = δji with respect
to the coordinates, and using (26.76), it is straightforward to show that the
derivatives of the contravariant basis vectors are given by
∂ei
= −Γikj ek .
∂u j
(26.77)
The symbol Γkij is called a Christoffel symbol (of the second kind), but, despite
appearances to the contrary, these quantities do not form the components of a
third-order tensor. It is clear from (26.76) that in Cartesian coordinates Γkij = 0
for all values of the indices i, j and k.
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TENSORS
Using (26.76), deduce the way in which the quantities Γkij transform under a general
coordinate transformation, and hence show that they do not form the components of a
third-order tensor.
In a new coordinate system
∂ei
,
∂u j
but from (26.69) and (26.67) respectively we have, on reversing primed and unprimed
variables,
Γ
e =
k
∂u k n
e
∂un
k
= e ·
k
ij
and
ei =
∂ul
el .
∂u i
Therefore in the new coordinate system the quantities Γ kij are given by
l ∂u k n ∂
∂u
k
e · j
el
Γ ij =
n
∂u
∂u
∂u i
2 l
∂ul ∂el
∂u k n
∂ u
e ·
e + i j
=
j
i l
n
∂u
∂u ∂u
∂u ∂u
=
∂u k ∂ul ∂um n ∂el
∂u k ∂2 ul
en · el +
e · m
∂un ∂u j ∂u i
∂un ∂u i ∂u j
∂u
∂u k ∂2 ul
∂u k ∂ul ∂um n
+
Γ ,
(26.78)
∂ul ∂u j ∂u i
∂un ∂u i ∂u j lm
n
n
where in the last line we have used (26.76) and the reciprocity relation e · el = δl . From
(26.78), because of the presence of the first term on the right-hand side, we conclude
immediately that the Γkij do not form the components of a third-order tensor. =
In a given coordinate system, in principle we may calculate the Γkij using
(26.76). In practice, however, it is often quicker to use an alternative expression,
which we now derive, for the Christoffel symbol in terms of the metric tensor gij
and its derivatives with respect to the coordinates.
Firstly we note that the Christoffel symbol Γkij is symmetric with respect to
the interchange of its two subscripts i and j. This is easily shown: since
∂ei
∂2 r
∂2 r
∂ej
=
= i j = i,
j
j
i
∂u
∂u ∂u
∂u ∂u
∂u
it follows from (26.75) that Γkij ek = Γkji ek . Taking the scalar product with el and
using the reciprocity relation ek · el = δkl gives immediately that
Γlij = Γlji .
To obtain an expression for Γkij we then use gij = ei · ej and consider the
derivative
∂ei
∂ej
∂gij
= k · ej + ei · k
∂uk
∂u
∂u
= Γl ik el · ej + ei · Γljk el
= Γl ik glj + Γl jk gil ,
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(26.79)
26.18 DERIVATIVES OF BASIS VECTORS AND CHRISTOFFEL SYMBOLS
where we have used the definition (26.75). By cyclically permuting the free indices
i, j, k in (26.79), we obtain two further equivalent relations,
and
∂gjk
= Γl ji glk + Γl ki gjl
∂ui
(26.80)
∂gki
= Γl kj gli + Γl ij gkl .
∂u j
(26.81)
If we now add (26.80) and (26.81) together and subtract (26.79) from the result,
we find
∂gki
∂gij
∂gjk
+
− k = Γl ji glk + Γl ki gjl + Γl kj gli + Γl ij gkl − Γl ik glj − Γl jk gil
∂ui
∂u j
∂u
= 2Γl ij gkl ,
where we have used the symmetry properties of both Γl ij and gij . Contracting
both sides with g mk leads to the required expression for the Christoffel symbol in
terms of the metric tensor and its derivatives, namely
Γmij = 12 g mk
∂gjk
∂gki
∂gij
+
− k
∂ui
∂u j
∂u
.
(26.82)
Calculate the Christoffel symbols Γmij for cylindrical polar coordinates.
We may use either (26.75) or (26.82) to calculate the Γmij for this simple coordinate system.
In cylindrical polar coordinates (u1 , u2 , u3 ) = (ρ, φ, z), the basis vectors ei are given by
(26.59). It is straightforward to show that the only derivatives of these vectors with respect
to the coordinates that are non-zero are
∂eρ
1
= eφ ,
∂φ
ρ
∂eφ
1
= eφ ,
∂ρ
ρ
∂eφ
= −ρeρ .
∂φ
Thus, from (26.75), we have immediately that
Γ212 = Γ221 =
1
ρ
and
Γ122 = −ρ.
(26.83)
Alternatively, using (26.82) and the fact that g11 = 1, g22 = ρ2 , g33 = 1 and the other
components are zero, we see that the only three non-zero Christoffel symbols are indeed
Γ212 = Γ221 and Γ122 . These are given by
Γ212 = Γ221 =
Γ122 = −
1 ∂g22
1 ∂ 2
1
= 2
(ρ ) = ,
2g22 ∂u1
2ρ ∂ρ
ρ
1 ∂g22
1 ∂ 2
=−
(ρ ) = −ρ,
2g11 ∂u1
2 ∂ρ
which agree with the expressions found directly from (26.75) and given in (26.83 ). 967