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and use (4.22) to show that

Fgrav = (N 1) Am r (4.26)

rN ;

Note that the perturbation of the centrifugal force does not depend on the number of

spatial dimensions, but that the perturbation of the gravitational force does depend on

N.

Problem g: Using the value of the velocity derived in problem d and expressions (4.25)-

(4.26) show that according to the criterion (4.24) the orbital motion is stable in

less than four spatial dimensions. Show also that the requirement for stability is

independent of the original distance r.

CHAPTER 4. THE DIVERGENCE OF A VECTOR FIELD

40

This is a very interesting result. It implies that orbital motion is unstable in more than

four spatial dimensions. This means that in a world with ve spatial dimensions the solar

system would not be stable. Life seems to be tied to planetary systems with a central star

which supplies the energy to sustain life on the orbiting planet(s). This implies that life

would be impossible in a ve-dimensional world! Note also that the stability requirement

is independent of r, i.e. the stability properties of orbital motion does not depend on the

size of the orbit. This implies that the gravitational eld does not have \stable regions"

and \unstable regions", the stability property depends only on the number of spatial

dimensions.

Chapter 5

The curl of a vector eld

5.1 Introduction of the curl

We will introduce the curl of a vector eld v by its formal de nition in terms of Cartesian

coordinates (x y z) and unit vectors x, y and ^ in the x, y and z-direction respectively:

^^ z

0 1

x y ^ B @y vz @z vy C

^^z ;

curl v = @x @y @z = @ @z vx @xvz A : (5.1)

;

vx vy vz @x vy @y vx

;

It can be seen that the curl is a vector, this is in contrast to the divergence which is a

scalar. The notation with the determinant is of course incorrect because the entries in a

determinant should be numbers rather than vectors such as x or di erentiation operators

^

such as @y = @=@y. However, the notation in terms of a determinant is a simple rule to

remember the de nition of the curl in Cartesian coordinates. We will write the curl of a

vector eld also as: curl v = v:

r

Problem a: Verify that this notation with the curl expressed as the outer product of the

operator and the vector v is consistent with the de nition (5.1).

r

In general the curl is a three-dimensional vector. To see the physical interpretation

of the curl, we will make life easy for ourselves by choosing a Cartesian coordinate

system where the z-axis is aligned with curl v. In that coordinate system the curl

is given by: curl v = (@x vy @y vx )^. Consider a little rectangular surface element

z

;

oriented perpendicular to the z-axis with sides dx and dy respectively, see gure

H

(5.1). We will consider the line integral dxdy v dr along a closed loop de ned by the

sides of this surface element integrating in the counter-clockwise direction. This line

integral can be written as the sum of the integral over the four sides of the surface

element.

Problem b: Show that the line integral is given by Hdxdy v dr =vx(x y)dx+vy (x+

dx y)dy vx (x y + dy)dx vy (x y)dy, and use a rst order Taylor expansion

; ;

I

to write this as

v dr = (@xvy @y vx)dxdy : (5.2)

;

dxdy

41

CHAPTER 5. THE CURL OF A VECTOR FIELD

42

x x+dx

y+dy

dy

y

dx

Figure 5.1: De nition of the geometric variables for the interpretation of the curl.

This expression can be rewritten as:

H v dr

(curl v)z = (@x vy @y vx ) = dxdy

dxdy : (5.3)

;

In this form we can express the meaning of the curl in words:

The curl of v is the closed line integral of v per unit surface area.

Note that this interpretation is similar to the interpretation of the divergence given

in section (4.2). There is, however, one major di erence. The curl is a vector while

the divergence is a scalar. This is re ected in our interpretation of the curl because

a surface has an orientation de ned by its normal vector, hence the curl is a vector

too.

5.2 What is the curl of the vector eld?

In order to discover the meaning of the curl, we will consider again an incompressible

uid and will consider the curl of the velocity vector v, because this will allow us

to discover when the curl is nonzero. It is not only for a didactic purpose that we

consider the curl of uid ow. In uid mechanics this quantity plays such a crucial

role that it is given a special name, the vorticity !:

! r v: (5.4)

To simplify things further we assume that the uid moves in the x y-plane only (i.e.

vz = 0) and that the ow depends only on x and y: v = v(x y).

Problem a: Show that for such a ow

! = r v = (@x vy @y vx )^ :

z (5.5)

;

5.2. WHAT IS THE CURL OF THE VECTOR FIELD? 43

We will rst consider an axi-symmetric ow eld. Such a ow eld has rotation

symmetry around an axis, we will take the z-axis for this. Because of the cylinder

symmetry and the fact that it is assumed that the ow does not depend on z, the

components vr , v' and vz depend neither on the azimuth ' (= arctan y=x) used

p

in the cylinder coordinates nor on z but only on the distance r = x2 + y2 to the

z-axis.

Problem b: Show that it follows from expression (4.15) for the divergence in cylin-

der coordinates that for an axisymmetric ow eld for an incompressible uid

p

(where (r v) =0 everywhere including the z-axis where r = x2 + y2 = 0)

that the radial component of the velocity must vanish: vr = 0.

This result simply re ects that for an incompressible ow with cylinder symmetry

there can be no net ow towards (or away from) the symmetry axis. The only

^

nonzero component of the ow is therefore in the direction of '. This implies that

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