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 spherically 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 infinitesimal 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 2psin(q), and the incremental width is dq. Also, each part of this ring exerts a force of magnitude 1/r2 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 We've already shown that the integral of this for q = 0 to p vanishes for any x from -1 to +1. (For |x| greater than 1 the integral is simply 4p/x2, confirming the previously stated result for exterior points.) Hence we have If we expand the integrand into powers of x we get 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): As q ranges from 0 to p, 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 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. Symbolically we have 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 This is obviously true for the vectors (1,0) and (0,1), and if we rotate these two vectors through an angle f the transformed coordinates are (cos(f),sin(f)) and (-sin(f),cos(f)), which again leads to the result 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 More generally, in n dimensions, the two vectors (u1,u2,...,un) and (v1,v2,..,vn) are orthogonal if and only if 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 u and v are orthogonal if and only if 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 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 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 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 p, which is simply 2. Hence have the expected result for exterior points 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 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 Notice that the force on an external particle approaches 4p 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 2p in the inward direction, while the force applied by the point of the shell "ahead" of the particle approaches 2p 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 2p 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 4p. As the particle moves further out, the force drops off as 4p(R/L)2. Return to MathPages Main Menu