26.20 VECTOR OPERATORS IN TENSOR FORM
SupposeA=[aij],B=[bij]and thatB=A−^1. By considering the determinanta=|A|,
show that
∂a
∂uk=abji∂aij
∂uk.
If we denote the cofactor of the elementaijby ∆ijthen the elements of the inverse matrix
are given by (see chapter 8)
bij=1
a∆ji. (26.93)However, the determinant ofAis given bya=∑
jaij∆ij,in which we havefixediand written the sum overjexplicitly, for clarity. Partially
differentiating both sides with respect toaij, we then obtain
∂a
∂aij=∆ij, (26.94)sinceaijdoes not occur in any of the cofactors ∆ij.
Now, if theaijdepend on the coordinates then so will the determinantaand, by the
chain rule, we have
∂a
∂uk=
∂a
∂aij∂aij
∂uk=∆ij∂aij
∂uk=abji∂aij
∂uk, (26.95)
in which we have used (26.93) and (26.94).
Applying the result (26.95) to the determinantgof the metric tensor, andremembering both thatgikgkj=δijand thatgijis symmetric, we obtain
∂g
∂uk=ggij∂gij
∂uk. (26.96)
Substituting (26.96) into (26.92) we find that the expression for the Christoffelsymbol can be much simplified to give
Γiki=1
2 g∂g
∂uk=1
√
g∂√
g
∂uk.Thus finally we obtain the expression for the divergence of a vector field in ageneral coordinate system as
∇·v=vi;i=1
√
g∂
∂uj(√
gvj). (26.97)LaplacianIf we replacevby∇φin∇·vthen we obtain the Laplacian∇^2 φ. From (26.91),
we have
viei=v=∇φ=∂φ
∂uiei,