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# Putting "words" in Newton's "mouth"

Given the nature of Kepler's elliptical orbits of the planets, what was unknown at Kepler's time was the nature of the force producing such orbits. Newton referred to this as the "inverse" problem, namely given the force and the force center, determine the path. Central to the problem is the role of curvature which is caused by the action of a central force on a peripheral body that is orbiting the center of force.

Given Newton's knowledge of curvature, and the assumption of his inverse square law, it has been assumed that Newton was able to give a detailed derivation of the conic section rather than providing a mere outline that could easily be miss represented and miscontrued, And in fact, such is the case

In Harman, Shapiro, eds. "The Investigation of difficult things" there is the article by J Bruce Brackenridge on p. 231 entitled "The critical role of curvature in Newton's developing dynamics" who is paraphrasing Newton's thoughts on what would have been his derivation of the ellipse from the central force of inverse square F=c/r^2. This is the so called solution to the inverse problem.

In the section "A detailed solution of the inverse problem", his expression for his slope z=r'r, where the prime denotes the derivative with respect to the complement \theta of the angle \alpha betten the tangent to the trajectory and the radial force to the center

z=r'/r =\tan\theta=\cot\alpha, (1)

This is the penultimate equation on p. 253 which gives an infinite radius of curvature

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

which is the first equation on p. 252. Instead of going through a complicated derivation to obtain the general equation of a conic

1/r =A+B\cos(\theta+\varepsilon) (3)

which the last equation on p. 255, where A, B, and \varepsilon are arbitary constants where A has been identified as the ratio c/h^2, where c is the numerator of his inverse square law and h is the conserved angular momentum (K in Brachenridge's notation).

It is easilty to integrate Eq. (1) to get

]ln r=-\ln\cos\theta +\ln C, (4)

where C is an arbitrary constant of integration. Since equation (4) describes a striaght line Eq (3) will only be compatible with it provided A\equiv 0, implying that the force vanishes! Obviously, the radius of curvature (2) reflects this since its denominatore,

1+z^2-z' = 1+ \tan^2\theta-\sec^2\theta=0

is a trigonometric identity. A straight line has infinite curvature.

Now there is an ambiguity in the sign of the slope depending on whether it is increasing or decreasing so we could equally as well write instead of (2)

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

if \theta is in the second and fourth quadrants. Now, integration of (4) gives

r= C \cos\theta (5)

whixh is a circle that goes through the orign having radius C/2. The radius of curvature (2) then reduces to

\rho=r\sec^3\theta/2\sec^2\theta=(r/2)\sec\theta, (6)

or \rho=C/2 on the strength of solution (5). This would correspond to uniform circular motion in which the speed is constant, and, also the radius. The expression for the radius of curvature would then reduce to

z' -\sec^2\theta = r^3 F/h^2 . (7)

Since z'=-\sec^2\theta, (7) implies an inverse-fifth force,

F= -2h^2 \sec^5\theta=-2h^2/(C^2 r^5). (8)

This is the central force of a circular orbit passing through the origin, Unlike the inverse-square law and Hooke's laws which are duals to one another, the inverse-fifth law is dual to itself.

The heart of the problem, therefore, lies in the definition of the slope of the curve (1) in relation to the definition of the radius of curvature (2) in polar coordinates. The results expressed thus far pertain exclussively to Euclidean space. In non-Euclidean spaces of constant curvature, where the metric is

ds^2= dr^2 + G^2 d^2\theta, (9)

G would be R \sinh(r/R) in hyperbolic space with R a constant scale factor, R\sin (r/R) in elliptic space that replace G=r in Euclidean space. Distances are measured as tanh(r/R) and \tan(r/R), respectively in boths spaces. The inverse function u is defined as u=\coth r, instead of 1/r as in Euclidean space. Thus, the solution to the Binet equation

u" + u =0, (10)

that would be a straight line

u = C \cos\theta (11)

with C a constant of integration, is now

r=\coth^(-1)(C\cos\theta) (12)

a hyperbola for C>1. So even in the absence of a physical force in Eq. (10), there is curvature to the trajectory in non-Euclidean spaces.

Replacing the definition of slope (1) is now

G^(-1) r' = \tan\theta. (13)

We will explore the consequences of (13) in the next blog.