Mathematical Methods for Physics and Engineering : A Comprehensive Guide

VECTOR CALCULUS

∇Φ=

∂Φ

∂ρ

ˆeρ+

1

ρ

∂Φ

∂φ

ˆeφ+

∂Φ

∂z

ˆez

∇·a =

1

ρ

∂

∂ρ

(ρaρ)+

1

ρ

∂aφ

∂φ

+

∂az

∂z

∇×a =

1

ρ

∣

∣∣

∣

∣∣

∣

∣

eˆρ ρeˆφ eˆz

∂

∂ρ

∂

∂φ

∂

∂z

aρ ρaφ az

∣

∣∣

∣

∣∣

∣

∣

∇^2 Φ=

1

ρ

∂

∂ρ

(

ρ

∂Φ

∂ρ

)

+

1

ρ^2

∂^2 Φ

∂φ^2

+

∂^2 Φ

∂z^2

Table 10.2 Vector operators in cylindrical polar coordinates; Φ is a scalar

field andais a vector field.

defined by the vectorsdρˆeρ,ρdφˆeφanddzˆez:

dV=|dρˆeρ·(ρdφeˆφ×dzˆez)|=ρdρdφdz,

which again uses the fact that the basis vectors are orthonormal. For a simple

coordinate system such as cylindrical polars the expressions for (ds)^2 anddVare

obvious from the geometry.

We will now express the vector operators discussed in this chapter in terms

of cylindrical polar coordinates. Let us consider a vector fielda(ρ, φ, z)anda

scalar field Φ(ρ, φ, z), where we use Φ for the scalar field to avoid confusion with

the azimuthal angleφ. We must first write the vector field in terms of the basis

vectors of the cylindrical polar coordinate system, i.e.

a=aρeˆρ+aφˆeφ+azˆez,

whereaρ,aφandazare the components ofain theρ-,φ-andz- directions

respectively. The expressions for grad, div, curl and∇^2 can then be calculated

and are given in table 10.2. Since the derivations of these expressions are rather

complicated we leave them until our discussion of general curvilinear coordinates

in the next section; the reader could well postpone examination of these formal

proofs until some experience of using the expressions has been gained.

Express the vector fielda=yzi−yj+xz^2 kin cylindrical polar coordinates, and hence

calculate its divergence. Show that the sameresult is obtained by evaluating the divergence

in Cartesian coordinates.

The basis vectors of the cylindricalpolar coordinate system are given in (10.49)–(10.51).
Solving these equations simultaneously fori,jandkwe obtain

i=cosφeˆρ−sinφˆeφ

j=sinφˆeρ+cosφˆeφ

k=ˆez.