Inverse-Square Forces and Orthogonal Polynomials

A well-known aspect of inverse-square force laws is the fact that
given a spherically symmetrical distribution of charge centered
around the point C, the net force exerted on a test particle located
outside the distribution of charge is equal to the net force that 
would be exerted if all the charge were concentrated at the point C.
On the other hand, the net force exerted by a sherically symmetrical
distribution of matter on a test particle located INSIDE the 
distribution is zero.  Thus the net force exerted by a spherical 
shell of matter on objects in the interior of the shell is zero.

It's easy to see this last fact, simply by noting that the two
regions of a thin spherical shell subtended by a narrow cone 
centered on any point in the interior have linear dimensions
proportional to the distance from the point, and therefore the
areas are proportional to the squares of their respective distances
from the point.  (The regions on the shell also make the same angle
with respect to the ray through the interior point, so the projected
areas are the same.)  Since the shell has the same thickness in both
regions, the quantity of matter in these opposite directions is
proportional to the squares of the distances.  The force is inversely 
proportional to the square of the distance, so the forces exerted by 
these two regions cancel out, and the same is true for the pairs of 
opposing infintessimal regions in every direction, showing that the 
overall net force vanishes.

Another way of formulating this problem is to notice that every point
in the interior of a spherical shell is on a line through the center,
so we can take this line as the x axis of orthogonal coordinates,
and regard the shell as a unit sphere centered at the origin.  We can
now consider the force on a test particle at the point (x,0,0) exerted
by the ring of material at an angle q away from the x axis, as shown
in the figure below.

        

The circumference of the ring of material is 2pi sin(q), and the
incremental width is dq.  Also, each part of this ring exerts a
force of magnitude 1/r^2 on the test particle, where r is the distance
from the ring to the particle.  However, by symmetry, the components
of this force in the y and z directions cancel over the ring, so we
need only consider the component of force in the x direction, which
is the total force times the factor [cos(q)-x]/r.  Consequently the
net differential force dF exerted by the ring of width dq is

                    cos(q) - x                  cos(q) - x
 dF = 2pi sin(q) dq ---------- = 2pi sin(q)---------------------- dq
                        r^3                [1-2xcos(q)+x^2]^(3/2)

We've already shown that the integral of this for q = 0 to pi vanishes
for any x from -1 to +1.  (For |x| greater than 1 the integral is
simply 4pi/x^2, confirming the previously stated result for exterior
points.)  Hence we have

            pi
            /
            |     sin(q) [cos(q) - x]
            |  --------------------------  =  0
            |  [1 - 2xcos(q) + x^2]^(3/2)
            /
           q=0

If we expand the integrand into powers of x we get
         _
        |           /       2     \       3        /       2    \
 sin(q) | cos(q) + ( 3cos(q)   - 1 )x  +  - cos(q)( 5cos(q)  - 3 )x^2
        |_          \             /       2        \            /

         
                 1  /        4            2     \
              +  - ( 35cos(q)  - 30 cos(q)  +  3 ) x^3
                 2  \                           /

                                                                   _
                  5        /        4            2     \            |
               +  - cos(q)( 63cos(q)  - 70 cos(q)  + 15 )x^4  + ... |
                  8        \                           /           _|


Since the integral of this entire expression vanishes for any x 
in the range from -1 to +1, it follows that the integral of the 
coefficient of each individual power of x vanishes.  We also notice
that each coefficient has a factor of sin(q), and if we set is
factor aside we have the sequence of polynomials in s = cos(q):

        1s
        3s^2 - 1
        5s^3 - 3s
       35s^4 - 30s^2 + 3
       63s^5 - 70s^3 + 15s
      231s^6 - 315s^4 + 105s^2 - 5

As q ranges from 0 to pi, the value of s=cos(q) ranges from +1 to -1.
Also, we have ds/dq = -sin(q), so we can replace dq in the integrals
by  ds/sin(q), which cancels the factor of sin(q) in each term of
the expansion listed above.  Thus, the integral of each of these
polynomials from x=-1 to x=+1 vanishes (because the force of gravity
inside a spherical shell is zero).

If we evaluate the above polynomials at s=+1 we get the values 1, 2,
2, 8, 8, 16, and so on.  Therefore, to normalize them, we will divide 
each polynomial by its value with the argument +1.  This gives the
infinite sequence of polynomials

     P1(s)  =   [1s]

     P2(s)  =   [3s^2 - 1]/2

     P3(s)  =   [5s^3 - 3s]/2

     P4(s)  =  [35s^4 - 30s^2 + 3]/8

     P5(s)  =  [63s^5 - 70s^3 + 15s]/8

     P6(s)  = [231s^6 - 315s^4 + 105s^2 - 5]/16

                  etc.

A plot of the first several of these polynomials is shown below.

 

These are known as Legendre polynomials.  Not only is the integral 
of each of these polynomials (from -1 to +1) equal to zero, but the 
integral of the product of any two of these polynomials Pj(s)Pk(s), 
with j not equal to k, is zero.  In other words

             +1
             /
             | Pj(s) Pk(s) ds  =  0
             /
            s=-1

For this reason, these polynomials are said to be orthogonal.  To
understand the reason we apply this geometrical-sounding term to a
pair of polynomials, recall that two vectors u and v with the
components (u1,u2) and (v2,v2) in the two-dimensional plane with
orthogonal axes X1,X2 are perpendicular if and only if their dot
product vanishes.  In other words, the vectors u and v are orthogonal
if and only if

                  u1 v1 + u2 v2  =  0

This is obviously true for the vectors (1,0) and (0,1), and if we
rotate these two vectors through an angle w the transformed coordinates
are (cos(w),sin(w)) and (-sin(w),cos(w)), which again leads to the
result
               -cos(w)sin(w) + sin(w)cos(w)  =  0

The same applies in three dimensions, i.e., two vectors (u1,u2,u3)
and (v1,v2,v3) are perpendicular if and only if their dot product
vanishes, which is to say

               u1 v1  +  u2 v2  +  u3 v3  =  0

More generally, in n dimensions, the two vectors (u1,u2,...,un) and 
(v1,v2,..,vn) are orthogonal if and only if

                   n
                 SUM  uj vj   =   0
                  j=1

Now suppose we regard a continuous function on a fixed interval from,
say, -1 to +1, as defining an infinite-dimensional vector.  Each point 
on the function defines the value of one of its coordinates.  Then we 
can say that two such functions are orthogonal if and only if

                 +1
                 /
                 | u(x) v(x) dx   =   0
                 /
                x=-1

As we've seen, the infinnite set of polynomials Pj(x) is such that 
every pair of distinct polynomials in this set is mutually orthogonal.
Hence, these polynomials collectively constitute a BASIS, just as
two orthogonal vectors in the plane are a basis, and every other
vector in this space can be expressed uniquely as a linear combination
of these basis vectors.

It's worth noting that not only are the polynomials Pj(x) all mutually
perpendicular, but they each have components that "sum to zero", i.e.,
the integral of each individual polynomial from -1 to +1 is zero.
In geometrical terms, this means that all these "vectors" live in
a subspace consisting of the "plane" of points whose coordinates
sum to zero.  In two dimensions this subspace is just the one-
dimensional line x1 + x2 = 0, and in n dimensions it is the one-
dimensional locus with x1+x2+..+xn = 0.  For an infinite-dimensional
space (as with our family of polynomials), the resulting space is
still infinite dimensional, so this isn't really a restriction.