Begin2.DVI

Figure 8-5. Tube of field lines.

The sides of the tube are composed of field lines, and at any point on a field

line the direction of the tangents to the field lines are in the same direction as the

vector field F
at that point. The unit normal vector at a point on one of these field

lines is perpendicular to the unit tangent vector and therefore perpendicular to the

vector F
so that the dot product F
·eˆn= 0 must be zero everywhere on the sides of

the tube. The sides of the tube consist of field lines, and therefore there is no flux

of the vector field across the sides of the tube and all the flux enters, through S 1 ,

and leaves through S 2 .In particular, if F
is solenoidal and div F
= 0,then

∫∫

S 1

F
·dS
+

∫∫

Sides

F
·dS
+

∫∫

S 2

F
·dS
= 0 or

∫∫

S 1

F
·dS
=−

∫∫

S 2

F
·dS

If the vector field is a velocity field then one can say that the flux or flow into S 1

must equal the flow leaving S 2.

Physically, the divergence assigns a number to each point of space where the

vector field exists. The number assigned by the divergence is a scalar and represents

the rate per unit volume at which the field issues (or enters) from (or toward) a point.

In terms of figure 8-5, if more flux lines enter S 1 than leaves S 2 ,the divergence is

negative and a sink is said to exist. If more flux lines leave S 2 than enter S 1 ,a source

is said to exist.

Example 8-4. Consider the vector field

V
 =√ kx

x^2 +y^2

ˆe 1 +√ ky

x^2 +y^2

ˆe 2 (x, y )= (0,0), k a constant

A sketch of this vector field is illustrated in figure 8-6.