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Problem c: Insert this result in (4.5) and show that the ow eld is given by v(r) =

Ar=r2. Make a sketch of the ow eld.

CHAPTER 4. THE DIVERGENCE OF A VECTOR FIELD

34

The constant A is yet to be determined. Let at the source r = 0 a volume V per unit time

be injected.

Problem d: Show that V = R v dS (where the integration is over an arbitrary surface

around the source at r = 0). By choosing a suitable surface derive that

v(r) = V ^ :

r (4.9)

2r

From this simple example of a single source at r = 0 more complex examples can be

obtained. Suppose we have a source at r+ = (L 0) where a volume V is injected per unit

time and a sink at r; = (;L 0) where a volume is removed per unit time. The total

;V

ow eld can be obtained by superposition of ow elds of the form (4.9) for the source

and the sink.

Problem e: Show that the x- and y-components of the ow eld in this case are given

by:

vx (x y) = 2 (x xL)2L+ y2

V x+L (4.10)

;

(x + L)2 + y2

;

;

V y y

vy (x y) = 2 (x L)2 + y2 (4.11)

(x + L)2 + y2

;

;

and sketch the resulting ow eld. This is most easily accomplished by determining

from the expressions above the ow eld at some selected lines such as the x- and

y-axes.

One may also be interested in computing the streamlines of the ow. These are the lines

along which material particles ow. The streamlines can be found by using the fact that

the time derivative of the position of a material particle is the velocity: dr=dt = v(r).

Inserting expressions (4.10) and (4.11) leads to two coupled di erential equations for x(t)

and y(t) which are di cult to solve. Fortunately, there are more intelligent ways of

retrieving the streamlines. We will return to this issue in section (12.3).

4.3 Sources and sinks

In the example of the uid ow given above the uid ow moves away from the source

and converges on the sink of the uid ow. The terms \source" and \sink" have a clear

physical meaning since they are directly related to the \source" of water as from a tap,

and a \sink" as the sink in a bathtub. The ow lines of the water ow diverge from the

source while they convergence towards the sinks. This explains the term \divergence",

because this quantity simply indicates to what extent ow lines originate (in case of a

source) or end (in case of a sink).

This de nition of sources and sinks is not restricted to uid ow. For example, for the

electric eld the term \ uid ow" should be replaced by the term \ eld lines." Electrical

eld lines originate at positive charges and end at negative charges.

4.4. THE DIVERGENCE IN CYLINDER COORDINATES 35

Problem a: To verify this, show that the divergence of the electrical eld (4.2) for a point

charge in three dimensions vanishes except near the point charge at r = 0. Show also

that the net ux through a small sphere surrounding the charge is positive (negative)

when the charge q is positive (negative).

The result we have just discovered is that the electric charge is the source of the electric

eld. This is re ected in the Maxwell equation for the divergence of the electric eld:

(r E) = (r)="0 : (4.12)

In this expression (r) is the charge density, this is simply the electric charge per unit

volume just as the mass-density denotes the mass per unit volume. In addition, expres-

sion (4.12) contains the permittivity "0 . This term serves as a coupling constant since it

describes how \much" electrical eld is generated by a given electrical charge density. It is

obvious that a constant is needed here because the charge density and the electrical eld

have di erent physical dimensions, hence a proportionality factor must be present. How-

ever, the physical meaning of a coupling constant goes much deeper, because it prescribes

how strong the eld is that is generated by a given source. This constant describes how

strong cause (the source) and e ect (the eld) are coupled.

Problem b: Show that the divergence of the magnetic eld (4.3) for a dipole m at the

origin is zero everywhere, including the location of the dipole.

By analogy with (4.12) one might expect that the divergence of the magnetic eld is

related to a magnetic charge density: (r B) =coupling const: B (r), where B would

be the \density of magnetic charge." However, particles with a magnetic charge (usually

called \magnetic monopoles") have not been found in nature despite extensive searches.

Therefore the Maxwell equation for the divergence of the magnetic eld is:

(r B) = 0 (4.13)

but we should remember that this divergence is zero because of the observational absence

of magnetic monopoles rather than a vanishing coupling constant.

4.4 The divergence in cylinder coordinates

In the previous analysis we have only used the expression of the divergence is Cartesian

v = @xvx + @y vy + @z vz . As you have (hopefully) discovered, the use

coordinates: r

of other coordinate systems such as cylinder coordinates or spherical coordinates can

make life much simpler. Here we derive an expression for the divergence in cylinder

p

coordinates. In this system, the distance r = x2 + y2 of a point to the z-axis, the

azimuth '(= arctan(y=x) ) and z are used as coordinates, see section (3.5). A vector v

can be decomposed in components in this coordinate system:

v =vr^+v''+vz^

r^z (4.14)

where ^, ' and ^ are unit vectors in the direction of increasing values of r, ' and z

r^ z

respectively. As shown in section (4.2) the divergence is the ux per unit volume. Let

CHAPTER 4. THE DIVERGENCE OF A VECTOR FIELD

36

r

}

. dz

vr

dÏ•

{ dr

Figure 4.2: De nition of the geometric variables for the computation of the divergence in

cylinder coordinates.

us consider the in nitesimal volume corresponding to increments dr, d' and dz shown in

gure (4.2). Let us rst consider the ux of v through the surface elements perpendicular

to ^. The size of this surface is rd'dz and (r + dr)d'dz respectively at r and r + dr.

r

The normal components of v through these surfaces are vr (r ' z) and vr (r + dr ' z)

respectively. Hence the total ux through these two surface is given by vr (r + dr ' z)(r +

dr)d'dz vr (r ' z)(r)d'dz.

;

Problem a: Show that to rst order in dr this quantity is equal to @r (rvr ) drd'dz. Hint,

@

use a rst order Taylor expansion for vr (r + dr ' z) in the quantity dr.

Problem b: Show that the ux through the surfaces perpendicular to ' is to rst order

^

@v

in d' given by @' drd'dz.

'

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