Surface Area, The Divergence Theorem

When Divergence = 0

Remember that the divergence of a field f at a point p measures the tendency of the underlying substance, or energy, to expand at the point p. If f is the velocity of a fluid, is the fluid expanding (positive divergence), just passing through (0 divergence), or compressing (negative divergence)? Note that f could cause water to move slower, but over a larger area, and still give a divergence of 0. Let water rush into the neck of a funnel at high speed, and move out the wide mouth at a slower speed. A depiction of f might present long arrows clustered together, pushing the water quickly into the neck of the funnel, and short arrows spread apart around the mouth of the funnel. It is possible for f to have divergence 0 throughout. At any point p, water comes in quickly from behind, and leaves at a slower speed and in several directions. Water does not compress or expand at the point p; it is merely passing through. In this sense, referring to divergence as "spreading out" is somewhat misleading. The water does indeed spread out in space, but it does not actually expand. It moves slower as it spreads out, to compensate. The same amount of material is involved from start to finish.

Here is another 0 divergence example. Consider the electric field, emanating from an electron at the origin. This is an inverse square field. This can be written as some constant (which we will ignore) times x y or z over r³, where r is the radial distance sqrt(x²+y²+z²). The partial with respect to x is 1/r³ - 3x²/r⁵. Do the same for y and z and add them up, giving 3/r³ - 3r²/r⁵, or 0. The electric field gets weaker as you move away from its source, but it gets weaker in proportion to the surface area of the expanding shell. This is the inverse square law, and it implies a divergence of 0 everywhere, except at the origin, where 1/r is not defined. Electric charge bursts into existence at the origin, and charge is conserved everywhere else.

An inverse square law may be necessary for the consistency of our universe. Any other relationship would imply a nonzero divergence, as though electric or gravitational energy could be created or destroyed. That's not suppose to happen. In Flatland, gravity is proportional to 1/r, and in 4 dimensional space, gravity is proportional to 1/r³. That's what we think would happen, if those universes were well behaved.

Fluid flow equations are simpler when the fluid is, well, a fluid, rather than a gas. Water is virtually incompressible. Assuming a constant temperature and salinity, the density of water at the bottom of the ocean is only a tweak higher than the density of water at the top. When water moves about, the divergence can be set to 0, thus simplifying the equations. In contrast, air is easily compressed. Squeeze a column of air inside a piston, and the divergence is everywhere negative, as molecules crowd together.

The Divergence Theorem

Now for the divergence theorem, which is also called Gauss' theorem, in honor of its discoverer, Gauss. (biography) The theorem is somewhat intuitive, after you've had a chance to digest it. Consider a closed region in space. The net amount of water flowing out of that region is equal to the total divergence, or expansion, of water within that region. In some places, water might be expanding. In some places it might be contracting. If we add all these up, using a triple integral, the result is the total amount of water that is pushed out through the surface, which is the surface normal integral. The triple integral of ∇.f equals the surface normal integral of f. That's it in a nutshell.

Recall the proof of Green's theorem. We broke the region up into small squares, and/or partial squares, and proved the theorem for each square, whence the theorem was valid for the entire region. This proof works the same way. If the divergence theorem is true for two adjacent cubes, sharing a common wall, then it holds for the larger, rectangular box. The two triple integrals (from the two cubes) combine to form the triple integral throughout the entire rectangular box, and the two surface normal integrals sum to the surface normal integral on the outside of the box, since the surface normal integrals cancel on the common wall. Water flowing out of the left cube, through the common wall, has to flow into the right cube; a positive and a negative, and they cancel each other out. Therefore, we only need prove the theorem for arbitrarily small, manageable regions in space.

The simplest shape is a box, as described above. Let's look at f₃, the third component of f, in the z direction. This runs parallel to the walls of the box, hence the walls contribute nothing to the surface normal integral. The flow into the box, from below, is the double integral of f₃ on the floor. The flow out of the box is the double integral of f₃ on the ceiling. The difference is the surface normal integral, as far as f₃ is concerned. At the same time, restrict the triple integral of ∇.f to f₃, whence the integrand becomes the partial of f₃ with respect to z. Integrate this as z runs from floor to ceiling. Apply the fundamental theorem of calculus, and the result is f₃(z₁) - f₃(z₀). In other words, the difference of f₃ at the ceiling and at the floor. Integrate this over the area of the box to complete the triple integral. This is precisely the same formula as the surface normal integral.

Do the same thing for left and right, and front and back, and add up the three equations. The result is the triple integral of the divergence, and the surface normal integral over the entire box. These are equal per component, and they are equal in aggregate. The divergence theorem holds for every rectangular box of any size.

Next consider a partial box, whose ceiling curves downward. This tiny cell lies at the boundary of the region of interest in 3 space. For convenience, let the floor lie at z = 0, and let z = h(x,y) determine the ceiling. The triple integral has not changed. It still becomes the difference of f₃, at the ceiling and at the floor, integrated over the area of the cell. The flow into the cell from below is still the double integral of f₃ on the floor. The flow out of the cell, through the ceiling, is equal to the double integral of f₃ on the ceiling, thanks to the umbrella theorem. Nothing has changed, and the divergence theorem holds.

If the cell has a curved front wall, rather than a curved ceiling, integrate with respect to y, rather than z. The result is the same. Similarly, cells with curved back walls, or side walls, behave as expected. Almost any shape can be partitioned into cells and/or partial cells of this type, and that completes the proof.

Divergence and Green

Green's theorem is implied by the divergence theorem. Draw a curve in the plane, and pull it upward to form the walls of a cylinder of height 1. Let the component functions of our vector field be q(x,y), -p(x,y), and 0. (There is no force or motion in the z direction.) The triple integral of ∇.f becomes the double integral of q∂x-p∂y. At the same time, the surface normal integral is a line normal integral times 1 (for the height). The line normal integrand is the velocity vector, rotated 90 degrees clockwise (thus normal to the curve), and dotted with f. Instead, let's leave the velocity vector alone, rotate f 90 degrees counterclockwise, and dot the two vectors. This gives a traditional line integral. When f is rotated by 90 degrees, its components are p and q. Therefore the line integral of p,q around c equals the double integral of q∂x-p∂y throughout the interior, and that is Green's theorem.