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Date: 9-12-2020
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Divergence
Let us start with a vector field A. Consider A . dS over some closed surface S, where dS denotes an outward pointing surface element. This surface integral is usually called H the flux of A out of S. If A is the velocity of some fluid, then A . dS is the rate of flow of material out of S. If A is constant in space then it is easily demonstrated that the net flux out of S is zero:
(1.1)
since the vector area S of a closed surface is zero. Suppose, now, that A is not uniform in space. Consider a very small rectangular volume over which A hardly varies. The contribution to A . dS from the two faces normal to the x-axis is
(1.2)
where dV = dx dy dz is the volume element. There are analogous contributions
from the sides normal to the y and z-axes, so the total of all the contributions is
(1.3)
The divergence of a vector field is defined
(1.4)
Divergence is a good scalar (i.e., it is coordinate independent), since it is the dot product of the vector operator ∇ with A. The formal definition of divA is
(1.5)
This definition is independent of the shape of the infinitesimal volume element. One of the most important results in vector field theory is the so-called divergence theorem or Gauss' theorem. This states that for any volume V surrounded by a closed surface S,
(1.6)
where dS is an outward pointing volume element. The proof is very straightforward.
We divide up the volume into lots of very small cubes and sum ∫A . dS over all of the surfaces. The contributions from the interior surfaces cancel out, leaving just the contribution from the outer surface. We can use Eq. (1.3) for each cube individually. This tells us that the summation is equivalent to ∫div A dV over the whole volume. Thus, the integral of A . dS over the outer surface is equal to the integral of divA over the whole volume, which proves the divergence theorem. Now, for a vector field with div A = 0,
(1.7)
for any closed surface S. So, for two surfaces on the same rim,
(1.8)
Thus, if div A = 0 then the surface integral depends on the rim but not the nature of the surface which spans it. On the other hand, if div A ≠ 0 then the integral depends on both the rim and the surface.
Consider an incompressible fluid whose velocity field is v. It is clear that v . dS = 0 for any closed surface, since what flows into the surface must flow out again. Thus, according to the divergence theorem, ∫div v dV = 0 for any volume. The only way in which this is possible is if div v is everywhere zero. Thus, the velocity components of an incompressible fluid satisfy the following differential relation:
(1.9)
Consider, now, a compressible fluid of density ρ and velocity v. The surface integral is the net rate of mass flow out of the closed surface S. This must be equal to the rate of decrease of mass inside the volume V enclosed by S, which is written -( ∂/∂t)(∫V ρ dV ). Thus,
(1.10)
for any volume. It follows from the divergence theorem that
(1.11)
This is called the equation of continuity of the fluid, since it ensures that fluid is neither created nor destroyed as it flows from place to place. If ρ is constant then the equation of continuity reduces to the previous incompressible result div v = 0. It is sometimes helpful to represent a vector field A by ''lines of force" or ''field lines". The direction of a line of force at any point is the same as the direction of A. The density of lines (i.e., the number of lines crossing a unit surface perpendicular to A) is equal to |A|. In the diagram, |A| is larger at point
1 than at point 2. The number of lines crossing a surface element dS is A . dS. So, the net number of lines leaving a closed surface is
(1.12)
If divA = 0 then there is no net flux of lines out of any surface, which mean that the lines of force must form closed loops. Such a field is called a solenoidal vector field.
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