Gauss' Theorem

Let's suppose that W is a solid that is contained inside of another surface S, and let's suppose that F(x,y,z) is a vector field such that div( F) = 0 for the solid that is between S and W (i.e, inside of the solid S but outside of the solid W). We often write this solid as S\W, and the boundary of this solid is the union of ¶S and the boundary ¶W with the opposite orientation.

Since div( F) = 0 in S\W, the divergence theorem implies that

ó õ ó õ ¶( S\W)   F·dS =  ó õ ó õ ¶( S\W)   div( F) dV = 0

Since the flux through S\W is given by the flux through S minus the flux through W, we have

ó õ ó õ ¶S  F·dS -  ó õ ó õ ¶W  F·dS = 0    or    ó õ ó õ ¶S  F·dS = ó õ ó õ ¶W  F·dS

That is, if div( F) = 0 between a closed surface W which is inside another closed surface S, then the flux of F through the surfaces is the same.

In particular, let's suppose that F( x,y,z) is an inverse square field, which is of the form

F( x,y,z) =   k á x,y,z ñ ( x2+y2+z2) 3/2

It is called an inverse square field because in spherical coordinates, F is given by

F =   k r2 er

where e_r = á sin( f) cos(q) ,sin( f) sin( q) ,cos( f) ñ is the unit vector in the direction that r changes (i.e., e_r is radial). It is straightforward to show that div( F) = 0 everywhere except at the origin, where F is undefined (see exercises 21-24).       

EXAMPLE 4    What is the flux of an inverse square field through a sphere of radius R centered at the origin?

Solution: Let's use spherical coordinates to define the surface normal and the sphere. To begin with, the sphere is given by r = R, and the unit surface normal is simply n = e_r. A ``rectangle'' on the sphere corresponding to small changes df and dq has ''sides'' of length Rdf and rdq, where r is the radius of the circle at angle f Since r = Rsin( f) is the radius of the circle at angle f, the surface area differential is

dS = Rr dfdq = R2 sin(f) dfdq

so that the surface differential is dS = e_rR²sin( f) dfdq. Thus, the flux is

Flux = ó õ ó õ Sphere  F·dS = ó õ 2p 0  ó õ p 0  k r2 er·er  R2sin( f) dfdq

However, e_r·e_r = 1 since e_r is a unit vector. Moreover, r = R on the sphere, so that

Flux = ó õ 2p 0  ó õ p 0    k R2  R2sin( f) dfdq = 4pk

       

In example 4, we obtain Flux = 4pk, which does not depend on the radius R of the sphere. Thus, the flux of the inverse square field through any sphere centered at the origin is the same, which in turn implies that the flux of the inverse square field through any closed surface containing the origin is the same.

Let's use this idea to obtain an important result about inverse square fields. If we let r = á x,y,z ñ be the vector of coordinate variables, then an inverse square field centered at a point p = ( a,b,c) is of the form

F( r) =   kp || r-p|| 2  up

where k_p is constant and where u_p is the unit vector pointing radially away from p, which is to say that u_p is parallel to r-p. If we define a new coordinate system x^¢ = x-a, y^¢ = y-b, and z^¢ = z-c, then p is the origin of the new x^¢, y^¢, z^¢ coordinate system. Thus, any surface W in x^¢y^¢z^¢ coordinates containing p has a flux of

Flux = 4pkp

and since the surface is independent of the coordinate system used to describe it, the flux is the same in the original coordinates. That is, the flux through any closed surface containing an inverse square field centered at a point p is the same.

Now, let's conclude by extending this to a sum of inverse square fields. Suppose that W is a closed surface with points p₁, p₂,¼,p_n contained inside.

LiveGraphics3d Applet

Gauss Theorem: If F is the sum of inverse square fields centered at p₁,¼,p_n respectively with constants k₁,¼,k_n, respectively, then the total flux of F through a closed surface containing each of these points is 

Flux = n å j = 1  4pkj