Newton, in the early 1670's, derived an expression for the radius of curvature, rho. Yet, he did more than that for, in fact, he derived two radii of curvature, on for his original curve, C, and the second for its polar reciprocal. The polar reciprocal, C*, is a unit circle centered at the source, and hence, has constant curvature, rho*. In doing so, he was able to derive the inverse-square law.
Several facts concerning polar reciprocals are: 1. If the original curve, C, is a conic section, so, too, will be its polar reciprocal, C*; 2. by duality the dual of the polar reciprocal, (C*)*=C, is the original curve; and 3. If r is the radial distance to a point on a curve, measured from the source, the pedal p will be the the distance of the source to the point on the tangent line through that point that is perpendicular to it, then r* and p* will be the analogous quantities with respect to the polar reciprocal where r -> r*= 1/p and p-> p*= 1/r, by the property of duality.
We are now going to show that the polar reciprocal C* is the curve C in velocity space. We start with the definition of the pedal:
1/p^2 = u"^2 + u^2, (1)
where u=1/r, and the primes denote differentiation with respect to the angle, theta. Realizing that (1) can be written as
1/p^2 = v^2/h^2, (2)
where h=pv = r^2 d theta/dt, the angular momentum (per unit mass). Two is none other than the expression for the conservation of energy,
h^2/p^2 = v^2 = 2 m/r + 2E (3)
in the case of the Kepler problem, where m is the gravitational parameter, GM, and E is the total energy. Starring (3), we have
h^2/p*2 = 2m/r* + 2E (4)
and applying duality gives the reciprocal polar as
h^2 r^2= 2 m p +2E (5)
which is a circle.
It is evident that the pedal is related to rotational motion. If we transfer to velocity space, it is seen that
r/p = v/r d theta/dt (6)
on the strength of the conservation of angular momentum. So in velocity space r -> v and the pedal p 0 -> r d theta/dt, the angular velocity. Analogous to (1), the pedal is defined as
1/(r d theta/dt)^2 = w"^2 + w^2 (7)
Extending (6) we have
v/r d theta/dt = a/v d theta/dt (8)
where a is the acceleration. (8) now implies the conservation of angular momentum (per unit mass) in velocity space
h = v^2 d theta/dt = a r d theta/ dt (9)
so that a = v^2/r, generally. Thus, (7) follows from
h^2/(r d theta/dt)^2 = a^2 = (d v/dt)^2 + v^2 (d theta/dt)^2 (10)
by dividing through by h^2.
The radius of curvature of the polar reciprocal is
rho* = w" + w, (11)
and since the curve is a circle it must be constant. Multiplying both sides by w' and integrating leads to
rho*/v + C = (1/2)(w'^2 + w^2), (12)
where C is an arbitrary constant of integration. Introducing (12) into (7) results in
1/(r d theta/dt)^2 = 2 rho*/v + 2 C. (13)
Parenthetically, (11) appears in Newton's expression for the radius of curvature of C, as
rho = (1+r'^2/r^2)^(3/2)/ rho*.
Extracting r d theta/dt from the numerator allows it to be written as
rho rho* = (v/r d theta/dt)^3.
By (6) and (8), the product of the radii can be expressed as
rho rho* = (r/p)^3 = (v/r d theta/dt)^3 = (a/v d theta/dt)^3 = (j/a d theta/dt)^3,
where we added the jerk, since j = a^2/v, a third-order time derivative which is linear in the velocity. Even if we introduce the acceleration a=v^2/r, the jerk is still odd in the velocity, j=v^3/r, as it must be.
Returning to (13), in order to make contact with the polar reciprocal (5) we have to interpret (13) in configuaration space, where h is conserved and rho* = m/h^2. Thus, (13) becomes
h r^2 = 2m p + 2E, (14)
where E=C h^3.
Equation (14) is to within a factor h, on the left-hand side the polar reciprocal (5). We may attribute this to duality which, not only, messes up dimensions allowing for the introduction of constant factors, by more important, either (5) or (14) says that the polar reciprocal curve requires
v ~ m/ r^2 (15)
v^2 ~ m/r (16)
as the original curve would imply. Since v is not time-reversal invariant (15) breaks the time reversal symmetry implied by Kepler's third law (16). Whereas Kepler's third law states that the square of the period is proportional to the cube of the radius, its polar reciprocal, (15), would have the period itself proportional to the cube of the radius.