Inverse-Square Forces and Orthogonal Polynomials
A well-known aspect of Newtonian gravity (and more generally of any
inverse-square force) is the fact that given a spherically symmetrical
distribution of matter centered around the point C, the net force
exerted on a test particle located outside the distribution of matter
is equal to the net force that would be exerted if all the matter
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 (i.e., closer to the center
than any of the gravitating matter) 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 this
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
429s^6 - 693s^4 + 315s^2 - 35
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 infinite 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.
Incidentally, if we define the functions
F0(q) = sin(q) [1]
F1(q) = sin(q) [cos(q)]
F2(q) = sin(q) [3cos(q)^2 - 1]
F3(q) = sin(q) [5cos(q)^3 - 3cos(q)]
F4(q) = sin(q) [35cos(q)^4 - 30cos(q)^2 + 3]
F5(q) = sin(q) [63cos(q)^5 - 70cos(q)^3 + 15cos(q)]
etc.
then the integral for the inward force on a point inside a spherical
shell of matter of radius R (with mass equal to the surface area) at
a distance L from the center is
pi
/ 3 4
f(x) = 2pi | -F1(q) x^2 - F2(q) x^3 - - F3(q) x^4 - - F4(q) x^5 - ... dq
/ 2 2
q=0
where x = L/R < 1. We've seen that the integral of each term vanishes,
so the net force is zero at any internal point. On the other hand,
for external points, i.e., with x = L/R > 1, the integrand of the
force integral can be expanded in powers of X = R/L < 1, and we have
pi
2piR^2 / 3 4
f(X) = ------ | F0(q) + 2 F1(q) X + - F2(q) X^2 + - F4(q) X^3 + ... dq
L^2 / 2 2
q=0
This shows how closely related are the facts that the internal force
vanishes and the external force is inversely proportional to the square
of the distance from the center. We know that the integrals of F1,
F2, ... etc., all vanish, so the only contributing term is the integral
of F0(q) = sin(q) for q from 0 to pi, which is simply 2. Hence have
the expected result for exterior points
4 pi R^2
f(X) = --------
L^2
where the numerator is just the surface area of the sphere, which (by
our choice of units) equals to total mass of the shell.
The magnitudes of the coefficients in the expansions of the integrands
are
interior: 1/1 2/1 3/2 4/2 5/8 6/8 7/16 8/16 9/128 ...
exterior: 0/1 1/1 2/2 3/2 4/8 5/8 6/16 7/16 8/128 ...
The remaining case to be considered is when R/L = 1, which means the
point is precisely on the shell, neither outside nor inside. In this
case the integral is
pi
/ sin(q) (1-cos(q)) / |pi \
f(1) = 2pi | ------------------- dq = 2pi ( sin(q/2)| ) = 2pi
/ [2(1-cos(q)]^(3/2) \ |q=0 /
q=0
Notice that the force on an external particle approaches 4pi as R/L
approaches 1 from the outside, so it is exactly half of this limit
on the shell itself, and then drops to zero in the interior. This
implies that as a particle approaches the shell from the inside, the
force applied by the entire shell minus the point of contact approaches
2pi in the inward direction, while the force applied by the point of
the shell "ahead" of the particle approaches 2pi in the outward
direction. As long as the particle is in the interior, these two
cancel out. When the particle is AT the boundary, the force applied
by that single point is null, and the net force is just 2pi inward.
As soon as the particle exits the surface of the shell, the force
toward that point of contact becomes inward, adding to the total,
which becomes 4pi. As the particle moves further out, the force
drops off as 4pi(R/L)^2.
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