Proof of the Inverse-Square Law for the Jerk
Updated: Oct 13
Following Newton's method of a polygonal approximation for determining the inverse-square law for the acceleration in configuration space, we may derive the same law for the jerk--only this time in velocity space.
Newton considers a tangent to the curve at a certain point P. At another point Q along the curve, C, there will be a deviation between that point and another point R found along the tangent curve. That distance, RQ, divided by the square of the time interval to get from P to Q will be the acceleration. The radius of curvature of the polar reciprocal, rho*, will be constant, and that condition determines the fact that the acceleration decreases as the inverse of the square of the radius from the source S to the point P where the tangent line meets the curve, C.
Only now RQ will be a velocity, and the distance from Q to P will be the change in velocity, Delta v. The points PRQ will define a triangle. Similarly, we may consider the polar reciprocal of C, C*, where the change in velocity between Q* and P* will be Delta v*. The line P*Q* will be perpendicular to the PS, and, therefore, also to RQ since RQ is parallel to PS. Furthermore, SQ* will be perpendicular to the straight-line extension of PQ, and SP* is perpendicular to RP, which is our original tangent to the curve C.
Hence, the triangle P*Q*S is similar to the triangle RQP so that
RQ/Delta v = Delta v*/SQ*.
Now, Delta v*= rho* Delta phi, where Delta phi is the angle between SQ* and P*S, which is a polygonal approximation to an arc length. So we can write RQ approximately as
RQ ~ Delta v rho* Delta phi/SQ* = Delta v rho* v^2 Delta/ v^2 SQ*,
where we multiplied and divided by v^2. The angular momentum in velocity space is h = v^2 d phi/dt, so that we have
RQ ~ (Delta v/SQ*) X (rho* h Delta t/v^2).
Dividing and multiplying the first factor by Delta t, and writing the acceleration approximately as a~ Delta v/Delta t, we have
RQ ~ (a Delta t/SQ*) X (rho* Delta t/v^2).
In configuration space, SQ*~r*, which is the inverse of the pedal p that intersects the tangent line normally. Instead of the acceleration we would have the velocity v~ Delta s/Delta t so that the product, pv =h, the conserved angular momentum in configuration space.
The corresponding angular momentum in velocity space is h = a r d phi/dt so that SQ*~ 1/r d phi/dt, and inverse (circular) velocity instead of the inverse distance r*. Hence, our expression for RQ is the jerk,
j = RQ/(Delta t)^2 ~ h^2 rho*/v^2,
and rearranging we have
rho* ~ j v^2/h^2.
That the radius of the curvature of the polar duality of C be constant, requires the jerk to decrease as the inverse-square of the velocity.
This is analogous to Newton's inverse-square law for the acceleration in configuration space. Both are based on the conservation of angular momentum in their respective spaces.
As a confirmation of our result, we know that the product of the radii of curvature obey
rho rho* = (a/v d phi/dt)^3 = 1 sin^3 alpha.
Introducing the expression we have found for rho* leaves us with
rho = a/d phi/dt,
where we have used the fact that j= a^2/v, just as the acceleration is always given by a= v^2/r.
Yet we find Newton's condition for the acceleration is given in terms of the radius of curvature,
a sin alpha = v^2/ rho,
where alpha is the angle that the radius r, from the source S makes with the tangent to the point P. The pedal is the the perpendicular distance from the source to the tangent curve at R so the triangle SPR is a right triangle. The expression for the radius of curvature in configuration space is
rho = v/d phi/dt,
and introducing it into Newton's expression results in
sin alpha = v d phi/dt /a.
This is the angle in the product of radii of curvature given above. More generally the sine of Newton's angle, alpha, is
sin alpha = p/r = r d phi/dt/ v = v d phi/dt a,
in reference to configuration, velocity, and acceleration spaces. The latter would be the generalization of the hodograph of the former. The right-triangles are formed by energy the energy conservation
v^2 = (d r/dt)^2 + r^2 (d phi/dt)^2,
and
a^2 = (d v/dt)^2 + v^2 (d phi/dt)^2.