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3.5. CYLINDER COORDINATES 29

Problem c: Derive these properties directly using geometric arguments.

CHAPTER 3. SPHERICAL AND CYLINDRICAL COORDINATES

30

Chapter 4

The divergence of a vector eld

The physical meaning of the divergence cannot be understood without understanding what

the ux of a vector eld is, and what the sources and sinks of a vector eld are.

4.1 The ux of a vector eld

To x our mind, let us consider a vector eld v(r) that represents the ow of a uid that

has a constant density. We de ne a surface S in this uid. Of course the surface has an

orientation in space, and the unit vector perpendicular to S is denoted by n. In nitesimal

^

elements of this surface are denoted with dS ndS. Now suppose we are interested in

^

the volume of uid that ows per unit time through the surface S, this quantity is called

. When we want to know the ow through the surface, we only need to consider the

component of v perpendicular to the surface, the ow along the surface is not relevant.

Problem a: Show that the component of the ow across the surface is given by (v n)^ ^n

and that the ow along the surface is given by v;(v n)^ . If you nd this problem

^n

di cult you may want to look ahead in section (10.1).

Using this result the volume of the ow through the surface per unit time is given by:

ZZ ZZ

(v n)dS =

^ v dS

= (4.1)

this expression de nes the ux of the vector eld v through the surface S. The de nition

of a ux is not restricted to the ow of uids, a ux can be computed for any vector eld.

However, the analogy of uid ow often is very useful to understand the meaning of the

ux and divergence.

Problem b: The electric eld generated by a point charge q in the origin is given by

r

E(r) = 4 q^r2 (4.2)

"0

in this expression ^ is the unit vector in the radial direction and "0 is the permittivity.

r

Compute the ux of the electric eld through a spherical surface with radius R with

the point charge in its center. Show explicitly that this ux is independent of the

radius R and nd its relation to the charge q and the permittivity "0 . Choose the

coordinate system you use for the integration carefully.

31

CHAPTER 4. THE DIVERGENCE OF A VECTOR FIELD

32

Problem c: To rst order the magnetic eld of the Earth is a dipole eld. (This is

the eld generated by a magnetic north pole and magnetic south pole very close

together.) The dipole vector m points from the south pole of the dipole to the north

pole and its size is given by the strength of the dipole. The magnetic eld B(r) is

given by (ref. 31], p. 182):

B(r) = 3^(^ m) m :

rr (4.3)

;

r3

Compute the ux of the magnetic eld through the surface of the Earth, take a

sphere with radius R for this. Hint, when you select a coordinate system, think

not only about the geometry of the coordinate system (i.e. Cartesian or spherical

coordinates), but also choose the direction of the axes of your coordinate system

with care.

4.2 Introduction of the divergence

In order to introduce the divergence, consider an in nitesimal rectangular volume with

sides dx, dy and dz, see g (4.1) for the de nition of the geometric variables. The

{

vx

{

dz

dy

{

dx

Figure 4.1: De nition of the geometric variables in the calculation of the ux of a vector

eld through an in nitesimal rectangular volume.

outward ux through the right surface perpendicular through the x-axis is given by

vx (x + dx y z)dydz, because vx(x + dx y z) is the component of the ow perpendicular

to that surface and dydz is the area of the surface. By the same token, the ux through

the left surface perpendicular through the x-axis is given by x (x y z)dydz, the sign

;v ;

is due to the fact the component of v in the direction outward of the cube is given by x .

;v

(Alternatively one can say that for this surface the unit vector perpendicular to the sur-

face and pointing outwards is given by n = x.) This means that the total outward ux

^ ;^

through the two surfaces is given by vx (x + dx y z)dydz vx (x y z)dydz = @vx dxdydz.

@x

;

The same reasoning applies to the surfaces perpendicular to the y- and z-axes. This means

4.2. INTRODUCTION OF THE DIVERGENCE 33

that the total outward ux through the sides of the cubes is:

d = @vx + @vy + @vz dV = (r v) dV (4.4)

@x @y @z

where dV is the volume dxdydz of the cube and (r v) is the divergence of the vector

eld v.

The above de nition does not really tell us yet what the divergence really is. Dividing

(4.4) by dV one obtains (r v) = d =dV . This allows us to state in words what the

divergence is:

The divergence of a vector eld is the outward ux of the vector eld per unit

volume.

To x our mind again let us consider a physical example where in two dimensions uid

is pumped into this two dimensional space at location r = 0. For simplicity we assume

that the uid is incompressible, that means that the mass-density is constant. We do not

know yet what the resulting ow eld is, but we know two things. Away from the source

at r = 0 there are no sources or sinks of uid ow. This means that the ux of the ow

through any closed surface S must be zero. (\What goes in must come out.") This means

that the divergence of the ow is zero, except possibly near the source at r = 0:

(r v) = 0 r =0:

for (4.5)

6

In addition we know that due to the symmetry of the problem the ow is directed in the

radial direction and depends on the radius r only:

v(r) = f(r)r: (4.6)

Problem a: Show this.

This is enough information to determine the ow eld. Of course, it is a problem that we

cannot immediately insert (4.6) in (4.5) because we have not yet derived an expression for

the divergence in cylinder coordinates. However, there is another way to determine the

ow from the expression above.

p

Problem b: Using that r = x2 + y2 show that

@r = x (4.7)

@x r

and derive the corresponding equation for y. Using expressions (4.6), (4.7) and the

chain rule for di erentiation show that

df

v =2f(r) + r dr (cilinder coordinates): (4.8)

r

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