## 13.1 Introduction

The gradient of a scalar and divergence and curl of a vector have a simple form in rectilinear coordinates. This form is maintained if you move the origin or rotate the coordinates, but still use rectilinear coordinates.
Their forms change, however, under if you choose arbitrary coordinates, and they look quite different even in polar coordinates in the plane, and in cylindric and spherical coordinates in three dimensions.
We will derive the form of the divergence and curl in an arbitrary orthogonal coordinate systems by application of Stokes' theorem and the divergence theorem.
The character of a coordinate system that affects the form of these objects is encapsulated in the expression for distance in the coordinate system.
In rectilinear coordinates distance is described by the Pythagorian Theorem:  ds2 = dx2 + dy2 + dz2.
In a general orthogonal coordinate system, with variables w1 , w2 and w3, distance can take on the more general form
ds2 = u12dw12 + u22dw22 + u32dw32, where the u's are each functions of the variables.
Thus,

In polar coordinates with w1 = r and w2 = q, we have

u1 =1 and u2 = r: ds2 = dr2 + r2d2

In spherical coordinates with w1 = , w3 = and w2 =, and r = rsin, we have

u1 =1, u3 = r =sin and u2=: ds2 = d2 + r2d2 + 2d

In cylindric coordinates, we have

ds2 = dr2 + r2d2 + dz2

We now apply the divergence theorem to a little rectangular box of space around the point r oriented with sides parallel to the directions of w1 , w2, and w3, so that the  the coordinates change by dw1, dw2 and dw3 in it. We make these quantities so small that div v is essentially constant inside it, and v is essentially constant on it boundaries.
We get:

(div v(r))V = d(A1v1(r)) + d(A2v2(r)) + d(A3v3(r))

where Ai is the area of the part of the boundary of our box on which wi is constant, and V is the volume of the box.
We can differentiate to get

and  observe that

and V = u1u2u3dw1dw2dw3

Substituting these facts into the last equation  and dividing by V gives us

The key observation here is that both vi and Ai can change across the box, so that one must differentiate the product vi Ai and not just vi .

In two dimensions, there is no w3 and the same equation holds without any last term and with u3 =1.

Thus, in polar coordinates, we get:

In spherical coordinates we find

Exercises