480 CHAPTER 25. CURVILINEAR COORDINATES

The Laplacian of a scalar field is nothing more than the divergence of the gradient. Insymbols, ∆φ ≡ ∇ ·∇φ . From 25.39 and 25.36 it follows

∆φ (x) =1√

g(x)

∂

∂xi

(gik (x)

∂φ (x)

∂xk

√g(x)

). (25.40)

We summarize the conclusions of this section in the following theorem.

Theorem 25.7.1 The following formulas hold for the gradient, divergence and Laplacianin general curvilinear coordinates.

(∇φ (x))r =∂φ (x)

∂xr , (25.41)

(∇φ (x))r = grk (x)∂φ (x)

∂xk , (25.42)

div(F ) =1√

g(x)

∂

∂xi

(F i (x)

√g(x)

), (25.43)

∆φ (x) =1√

g(x)

∂

∂xi

(gik (x)

∂φ (x)

∂xk

√g(x)

). (25.44)

Example 25.7.2 Define curvilinear coordinates as follows

x = r cosθ ,y = r sinθ

Find ∇2 f (r,θ). That is, find the Laplacian in terms of these new variables r,θ .

First find the metric tensor. From the definition, this is

G =

(1 00 r2

),G−1 =

(1 00 r−2

)

The contravariant components of the gradient are(1 00 r−2

)(fr

fθ

)=

(fr

1r2 fθ

)

Then also√

g = r. Therefore, using the formula,

∇2 f (u,v) =

1r

[(r fr)r +

(r

1r2 fθ

)θ

]=

1r(r fr)r +

1r2 fθθ

Notice how easy this is. It is anything but easy if you try to do it by brute force with noneof the machinery developed here.