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Is Newton's Derivation of the Conic From His Inverse Square Law Correct?

It appears that Newton did not do the calculation, but from his earlier work on curvature, was capable of doing it. For a comprehensive discussion, I refer the reader to J Bruce Brackenridge in The Investigation of difficut things, edited by P M Harman and A E Shapiro.


In the analysis, Newton supposedly derived the conic on the assumption of a central force law given by his inverse-square law. In the early 1670's Newton derived a series of formulas for the radius of curvature. In Cartesian coordinates (x,y) he found the radius of curvature given by

\rho= (1+y^2)^(3/2)/y", (1)

where the prime denotes differentiation with respect to x. In polar coordinates, he found

\rho= r)1+z^2)^(3/2)/[(1+z^2)-z'], (2)

where now the prime denotes differentiation with respect to the angle \theta, and Newton's slope is given by z=r'/r.

Newton considered an instantaneous osculating circle to the curve of the trajectory. The radius of curvature \rho is the radius of the instantaneous osculating circle. At the point P where the tangent to the curve touches the rim of the circle it makes an angle \alpha with respect to the line along which the center of force acts. The source S is at one of the foci. Rather, the force to the origin F_0 of the instantaneous center of the osculating center makes an angle \theta, called the complementary angle to \alpha, with respect to the force directed to the center of force. The force F_0 is given by the centrifugal force v^2/\rho, where \rho is the radius of curvature. Newton assumes that the "motion along an incremental arce of the general curve at the point P with uniform circular motion along an incremental arce of curvature at the same point". The forces are consequently related by

F_0=F_S \cos\theta (= F_S \sin\alpha)=v^2/\rho.. (3)


The angle \theta can be expressed in terms of the slope z, viz.,

z=r'/r=\tan\theta, (4)

and thus

(1+z^2)=\sec^2\theta. (5)


It is here that we first take acception to the derivation. Without further ado, we can integrate (4) to obtain,

r=1/costheta=sec\theta (6)

apart from an arbitrary constant of integration. This is a staight line, and the "proof" falls dead. To see this z'=sec^2\theta, and the radius of curvature (2) is infinite supporting the fact we are dealing with a straight line.

It is therefore quite incredible that by assuming the conservation of angular momentum h=vr cos\theta, he would have ever come out with the expression

F_S= (h^2/r^3)[(1+z^2)-z'], (7)

for it vanishes identically. The proof also employs the fact that since F_Sr^2=const., \rho\cos^3\theta=const., and vice-versa. This would identify the inverse-square law as the central force, would it not be for the fact that it vanishes by way of (7).


Moreover, (2) is derived from (1) by expressing the radius of curvature as

\rho = (x'^2+y'^2)^(3/2) /|x'y"-y'x"|, (8)

where the prime now stands for differentiation with respect to time. Expression (8) is

\rho = v^3/|v X a| (9)

where a is the acceleration vector. So \rho contains both the velocity and its rate of change, thereby accounting for the curvature. However, the derivation of (1) rests on the assumption of a flat metric

ds=\sqrt{dx^2+dy^2}=dx\sqrt{1+y'^2} (10)

where the prime now stands for differentiation with respect to x. Now using the fact that

y'= \tan\theta

and taking the derivative with respect to x

y"= \sec^2\theta d\theta/\dx

and solving for d\theta/dx gives the radius of curvature,

\rho= ds/d\theta= ds/dx dx/d\theta,

as (1).


My concern is that you are beginning with a flat metric (10) and looking for the reason why there should be a radius of curvature. The philosophy of general relativity that curvature lies in the non constancy of the metric coefficients. So why would some one start with a flat metric and come out with a finite radius of curvature? The fact that Newton's slope (4) gives a straight line testifies that there is a basic inconsistency in the derivation. One should expect that curvature results from a metric given by

ds^2 =E(x,y)dx^2+G(x,y)dy^2,

where the metric coefficients are space-dependent. This would lead to Newton's slope be relpaced by

z= f(r) r',

instead of (4) where f(r) is some power of r.

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