OK many thanks! I should think that the quartic could be approximated in some way or some other numerical method applied. The classical reduced lagrangian was first given in Note 417(5) Eq. (27), it is

lagrangian = (1/2) m (v squared) / m(r) power 3 / 2 – U

and the reduced hamiltonian is the same as the reduced lagrangian with – U replaced by U. The gravitational potential energy in m space is:

U = – m(r) half mMG / r

Here

v squared = r dot squared + r squared phi dot squared.

417(7): Role of m(r) in the Classical Lagrangian and Hamiltonian

The quartic equation contains the square root in combination with a sum, therefore x^2 also contains square roots. Will look into this further tonight.

What about the Lagrange equations (35)? is the standard v^2 term correct?

Horst

Am 26.10.2018 um 07:41 schrieb Myron Evans:

417(7): Role of m(r) in the Classical Lagrangian and Hamiltonian

Many thanks! I would say that the rest energy in m space is modified by the fact that m(r) is no longer one, so the rest energy is E0 = m(r) half m c squared. The famous Minkowski space rest energy E0 = m c squared is obtained from a work integral : integral F . dr = T2 – T1 where F = dp / dt and p = gamma m v. This procedure gives the Minkowski space relativistic kinetic energy T = (gamma – 1) mc squared = E – m c squared. There is nothing in this method that asserts that E0 must be constant, and as soon as we depart from Minkowski spacetime the rest energy depends on m(r). This is one of the startingly new features of the m theory. At the obsolete Schwarzschild radius the rest energy disappears entirely because m(r) = 0 at r = r0 when m(r) =1 – r0 / r. The m(r) function in the quartic is actually x squared, so your protocol already gives the answer, m(r) = x squared, and there is no need to take square roots. So we can simply plot the four m(r) roots to see how they behave, whether they are real valued or complex valued. If they are complex valued we can use square root (m(r)m(r)*) as is the usual practice in physics.

417(7): Role of m(r) in the Classical Lagrangian and Hamiltonian

This is a remarkable note, giving new experimental methods for determining the function m(r).

Definition (32) for H_0 first seemed a bit arbitrary to me because a non-constant term is subracted from H, but the results are self-consistent.

It should be possible to set up Euler-Lagrange equations for this classical approach of spherical spacetime. The Lagrangian obviously is (35). What has to be inserved for v^2 here? I guess the definition from plane polar coordinates as in (36):v^2 = r dot^2 + r^2 phi dot^2.

The four solutions of the quartic equation (62) are extremely complicated (see attachment) and it is not clear which square roots give real valued results. It might be better to use a numerical solution method like Newton-Raphson and start with m(r)=1.

Horst

Am 25.10.2018 um 10:40 schrieb Myron Evans: