A note on Newton's shell-point equivalency theorem Free

Newton's theorem that an isotropic shell or sphere can be regarded as acting as a point mass located at its center when computing the force it exerts on an external test mass is a cornerstone of his gravitational theory and its application to celestial mechanics. This beautiful result can be regarded as a manifestation of the inverse-square nature of the gravitational force or, equivalently, of its corresponding inverse-first-power potential. Richard Feynman describes this theorem in almost supernatural terms: “When we add the effects all together, it seems a miracle that the net force is exactly the same as we would get if we put all the mass in the middle!”¹ 

Proofs of this theorem and its corollary that a test mass trapped within a uniform spherical shell experiences no net force can be found in Feynman and many mechanics and dynamics texts in common use; these usually proceed by integrating the potential and then extending the argument to a solid sphere.^2–7 An exception is Goldstein's venerable Classical Mechanics, where he probably considered the proof too trivial for this more advanced text.⁸ Griffiths analyzes the equivalent electrostatic case for a uniform shell of electrical charge via both Gauss's law and direct integration of the potential, leaving to an exercise a direct integration of the electric field.⁹ Another approach to the interior of a uniform shell beng force-free is to imagine an off-center test mass equipped with an observer who looks at patches of the shell in opposite directions which subtend the same solid angle. Since the areas of the patches will be proportional to the squares of the observer's distances to them but the forces they exert depend on the inverse squares of these distances, the forces exerted by the two patches are equal and cancel; the argument is then extended to cover the entire sphere. If the force were not inverse-square, this argument would obviously break down and the test mass would experience a net force; this point will be revisited later. This force-free behavior inside the shell is, however, entirely irrelevant to Newton's issue of wanting to prove that the force exerted by the shell on an external test mass acts as if the entire mass of the shell were concentrated at its center; a priori, there is no reason that gravity has to be inverse-square.

More advanced students may be aware that if gravity were instead a linear attractive force, planetary orbits would still be closed ellipses, albeit with the attracting mass being located at the geometric center of the ellipse as opposed to being at one of the foci; this motion also arises with a mass attached to springs that extend off to the x and y directions. While these are obviously quite different scenarios, it is striking that both inverse-square and linear forces give rise to closed elliptical orbits. It can be shown that the gravitational and harmonic potentials are, in fact, the only integer-power ones that can support closed orbits. This behavior is now regarded as a manifestation of Bertrand's theorem, which received its own section in the second edition of Goldstein's book. A paper recently published in this journal offers a more advanced look at the dynamical symmetries underlying this powerful theorem.¹⁰ However, none of these sources point out that the harmonic potential also satisfies shell-point equivalency just like its gravitational cousin, and that it is the only other power-law potential which does so. A proof of this assertion is a straightforward extension of the usual gravitational case and is developed here. This is not a new result, but it seems not to be well known in the physics teaching community. That this is the case first came to my attention when it appeared in a paper on cosmology, although this is likely not the first source to point this out.¹¹ 

The purpose of this paper is to bring this behavior to wider attention. To give this development a motivation for classroom use, pose the following situation to your students: “Imagine that you are Newton, but you can use the full machinery of calculus and potential theory now taught to physics students. You hypothesize that gravity is a central power-law force such that the force exerted by an element of mass dM of a uniform shell on an external test mass m is $F = + Knm ( d M ) r n − 1 r ̂$⁠, where r is the element-to-test mass distance and $r ̂$ is a unit vector from the element to the test mass. The potential energy is then $U ( r ) = − K m ( d M ) r n$⁠, and the potential function is $V ( r ) = U ( r ) / m = − K ( d M ) r n$⁠. Now consider these two questions. First, for what value(s) of the power n does point-shell equivalency hold? Second, of these possible values, which gives a period-distance behavior consistent with Kepler's Third Law, $T 2 ∝ r 3$? For this latter question, assume a circular orbit and invoke a simple centripetal force argument.” Note that if K < 0 and n < 0 (n > 0) the force will be repulsive (attractive) and the reverse for K > 0, but these considerations do not affect the conclusions.

Clearly, no even-j terms will survive because of the term in square brackets.

Shell-point equivalency means that all terms in F must be independent of R. Cases of interest can be analyzed very directly:

• If n = – 1, the only surviving term in the sum will be that for j = 1, giving $F = − ( KMm / z 2 ) z ̂$⁠, the Newtonian result.

• For n = 0, again only j = 1 contributes, but the factor of $( n + 1 − j )$ within the sum vanishes. There is no net force, exactly as one would expect for V = constant.

• For $n = + 1$⁠, there are non-vanishing terms in the sum for j = 1 and j = 3. The first of these is independent of R as in case (a) above, but the second gives a term of the form $R 2 / z$⁠, destroying point-shell equivalency.

• For $n = + 2$⁠, the harmonic potential, the non-vanishing terms in the sum are again j = 1 and j = 3. In this case, however, the contribution from the j = 3 term vanishes because $( n + 1 − j )$ reduces to zero. The j = 1 term yields $F = ( 2 K Mmz ) z ̂$⁠, which possesses point-shell equivalency. If K > 0 as with the Newtonian case, this will be a repulsive force. This might seem strange but can be understood via energy conservation. The potential in this case is $V = − K M z 2$⁠. A repulsive force will endow a test mass with increasing kinetic energy; as z increases, the potential energy must correspondingly become more negative.

• For n > 2, there will be terms arising from $j = 1 , 3 , 5 , …$⁠, but terms from the factor of $R j − 1$ in Eq. (6) will survive, destroying point-shell equivalency.

As for other possibilities, setting n = – 2 in Eq. (3) leads to a logarithmic form for V where terms in R remain; similarly, for $n < − 2$⁠, the terms in the potential cannot be brought to a common denominator without the appearance of offending R-terms. For non-integer values of n, the finite binomial expansions will become infinite binomial series, not all of the terms of which can be forced to vanish through the factors of $( n + 1 − j )$⁠, so point-shell equivalency will be impossible. Only n = – 1 and $n = + 2$ can give equivalency.

Only n = – 1 satisfies the empirical form of Kepler's third law. For the harmonic case where n = 2, the period is independent of the orbital radius, analogous to how the period of a mass-spring system is independent of its initial displacement.

As implied in the Introduction, another point of difference between the harmonic and Newtonian cases is that the net force is not zero within the shell for the harmonic case: $F = − ( 2 K Mmz ) z ̂$⁠. If $K > 0 ( < 0 )$⁠, the force will be attractive toward (repulsive from) the center with the result that the center will be a point of stable (unstable) equilibrium. Worthwhile student exercises might be to check this by direct integration and also to use the solid-angle reasoning to confirm the stability/instability conclusion.

That the harmonic and gravitational potentials both give closed orbits and shell-point equivalency is presumably not a coincidence; I leave it to readers with more advanced command of dynamics to explore what underlying similarities make this so. Sometimes, a fresh look at an old problem reveals new insights.

I am grateful for comments and suggestions from reviewers who considered an earlier version of this paper. In particular, one reviewer encouraged me to consider the development of Eq. (6) more closely.