If Kepler's and Newton's Laws Apply to Planar Orbital Motion, then Why is 3 Dimensions Required?

Since Kepler's and Newton's law deal with conic sections, why is it so crucial that their analysis be carried out in 3-dimensions?

Consider the intersection of two different right circular cones that are both tangent to a common sphere as shown in the diagram.

In general, the intersection of two conic surfaces will result in a curve of higher degree. But, because both are tangent to a common sphere, the intersection will be a pair of ellipses. All this evolves in 3 dimensions. More exactly, the intersection of two cones degenerates into a pair of conics iff the conics have a common inscribed sphere.

Then, once reduced to a conic section, why are 3 dimensions necessary for the validity of Kepler's and Newton's laws? Kepler's second law claims that the angular momentum is conserved, and the angular momentum is pointing in the direction normal to the plane. Having taken this into account all else evolves in the plane. And what would happen if we are Poincarites reduced to 2 dimensions, or super-Poincarites living in n-dimensions? (The reference to Poincarites is to the 2-dimensional model of Poincare of the hyperbolic plane.)

Granted that the gravitational acceleration field can be represented as the gradient of the field,

E = (1/2) (grad g + g grad), (1)

where the second term is the transpose that makes the strain tensor, E, symmetric. The stress tensor will be given by

T_(ij) = C_(ijkl) E_(kl) (2)

where C is the elasticity tensor, which we assume is strongly elliptic. We shall restrict our attention to acceleration fields that posses polar symmetric. In spherical coordinates this means that the acceleration vector is given by (g_r,0,0)=(g(r),0,0).

The elastic tensor will then reduces to

C_(rrrr)= 2 mu + lambda, (3)

where lambda is the first Lame parameter, and mu is the shear modulus, which is a measure of rigidity. The stress, then, will be given by

T_(rr) = (lambda+2 mu) div g(r). (4)

We assume that the moduli are constants.

The key to the argument is what we mean by the divergence in (4). It is intended that it be taken in spherical coordinates in which case

div g= (1/r^2) [r^2 g(r)]' = g'-2g/r, (5)

where the comma denotes the differential with respect to r. Since the moduli are constant, the acceleration equation of equilibrium, T_(ij,j)=0 for all i, reduce to

g" + 2(g/r)' =0. (6)

The general solution to (6) is

g =w^2 r - m/r^2, (7)

where we have anticipated the identification of the two arbitrary constants with the square of the angular velocity, w, and the gravitational parameter, m= GM.

Unlike the centrifugal accleration, w^2r, Newton's inverse-square law depends crucially that we are working in 3D. Rather, if we consider cylindrical coordinates (r, phi, z), the divergence is

(1/r)(rg)' = g' + (g/r) = constant, (8)

on account of the equilibrium condition. Now the gravitational acceleration will be given by

g(r) = w^2 r - m/r, (9)

and the constant in (8) will be given by 2w^2.

Whereas the vanishing of the gravitational acceleration, (7), on the curve r=a, gives Kepler's III law,

w^2a^3 =m = constant, (10)

the vanishing of (9) on the same curve will lead to

(wa)^2 = m = constant. (11)

Condition (11) implies constant angular velocity. This is precisely what Newton claimed applied in his proof of the inverse square law. But this is not what he did. His force directed to the focus was

F sin alpha =v^2/p (12)

where the angla alpha is complementary to the angle formed from F and the radius vector to the circle tangent to the motion at any given point, and p is the radius of curvature. Newton didn't claim that v=wr is constant, but, rather, the angular momentum, per unit mass, h=w^2 r sin alpha, was constant. Thus, the central force (12) reduces to

F = const h^2/r^2 (13)

where the constant is

1/p sin^3 alpha. (14)

Luckily for Newton (14) turned out to be a constant for all conics.

In 2D, Kepler's II must be modified to what Newton claimed was uniform circular motion of angular velocity w, and his inverse-square law (2) becomes an inverse law, 1/r.

We can immediately generalize the analysis to n-dimensions. The equilibrium equations are

g" + n (g/r)' =0 (15)

whose general solution is

g(r) = w^2 r - m/r^n. (16)

The vanishing of (16) leads to the generalized Kepler's III:

w^2a^(n+1) = m = constant, (17)

on the curve, r = a.

We can generalize the equilibrium condition, div T =0, by adding a body force, and the second time derivative of g. In the second case,

rho (d^2/dt^2) g = (lambda+2mu) grad(div g), (17)

where we used the fact that

curl curl g = 0 (18)

to replace the laplacian of g by grad div g and rho is the density. Taking the divergence of both sides of (17), gives the d.e.

rho d^2/dt^2 D = (2mu+lambda) D, (19)

where D= div g is the dilatation. This a wave equation for D with phase velocity

v_p= {(2mu+lambda)/rho}^{1/2} (20)

The subscript "p" stands for primus, since these waves were the first to arrive at a seismometer following an earthquake. These are longitudinal waves which are unpolarized!

If we want "s" waves, or secondary waves, we have to take the curl of (17) and we are no longer allowed to replace the laplacian by grad div. In fact (18) is the compatibility condition that there are no rotational waves which would be polarized. This is what general relativity claims, with curl curl g being the linearized Einstein tensor.

Obviously, the whole analysis hinges on adding the "acceleration of the acceleration vector (17) according to Newton's second law. There is no experimental verification that this is justified, and, even it were, it would result in longitudinal waves, like sound waves or ocean waves which are compressional.

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