B.7. GRADIENTS AND DIVERGENCE 351

Adding 2.32 to 2.33 and subtracting 2.34 yields

∂gi j

∂xk +∂gk j

∂xi −∂gik

∂x j = 2{

rik

}gr j.

Now multiplying both sides by g jm and using the fact shown earlier in Theorem B.1.6 thatgr jg jm = δ

mr , it follows

2{

mik

}= g jm

(∂gi j

∂xk +∂gk j

∂xi −∂gik

∂x j

)which proves 2.31. ■

This is a very interesting formula because it shows the Christoffel symbols are com-pletely determined by the metric tensor and its partial derivatives which illustrates the fun-damental nature of the metric tensor. Note that the inner product is determined by thismetric tensor.

B.7 Gradients and DivergenceThe purpose of this section is to express the gradient and the divergence of a vector field ingeneral curvilinear coordinates. As before,

(y1, ...,yn

)will denote the standard coordinates

with respect to the usual basis vectors. Thus

y ≡ ykik, ek (y) = ik = ek (y) .

Let φ : U → R be a differentiable scalar function, sometimes called a “scalar field” inthis subject. Write φ (x) to denote the value of φ at the point whose coordinates are x. Thesame convention is used for a vector field. Thus F (x) is the value of a vector field at thepoint of U determined by the coordinates x. In the standard rectangular coordinates, thegradient is well understood from earlier.

∇φ (y) =∂φ (y)

∂yk ek (y) =∂φ (y)

∂yk ik.

However, the idea is to express the gradient in arbitrary coordinates. Therefore, using thechain rule, if the coordinates of the point of U are given as x,

∇φ (x) = ∇φ (y) =∂φ (x)

∂xr∂xr

∂yk ek (y) =

∂φ (x)

∂xr∂xr

∂yk∂yk

∂xs es (x) =

∂φ (x)

∂xr δrse

s (x) =∂φ (x)

∂xr er (x) .

This shows the covariant components of ∇φ (x) are

(∇φ (x))r =∂φ (x)

∂xr , (2.35)

Formally the same as in rectangular coordinates. To find the contravariant components,“raise the index” in the usual way. Thus

(∇φ (x))r = grk (x)(∇φ (x))k = grk (x)∂φ (x)

∂xk . (2.36)