## Continuing with UFT416

Agreed, your calculations and computations are outstanding, a breakthrough for cosmology. The momentum problem can be solved if the coordinates are (r1, phi), where r1 = r / m(r) power half, then set up the lagrangian with this system of coordinates. It gives the correct answer for momentum, p = partial L / partial r1 dot , where r1 dot = r dot / m(r) power half So all is correct and rigorously self consistent. In the m space, the coordinate r is replaced by r1 in all occurrences, because the metric is spherically symmetric and the spacetime is no longer a flat spacetime. The distance between m and M is r / m(r), and the gravitational potential is – mmG / r1. For small curvatures r1 is about the same as r. The point that you make below very important. The lagrangian used in UFT415 is the only one that conserves angular momentum and the hamiltonian. Now it is possible to proceed in all kinds of ways, limited only by imagination. This method gives L = gamma m r squared phi dot / m(r), and the total energy E = m(r) gamma m c squared. I wIll send over the calculations shortly

Continuing with UFT416

I checked the calculations. For UFT 415, I took the Lagrangian from note 415(8), eq.(6) without the m function. I just saw that you introduced it in eq.(16). I did various tests for paper 415, the only variant that gives conservation of angular momentum and energy is that described in section 415.3 without the m function in the Lagrangian. We will have to check this further.

Horst

Am 05.10.2018 um 07:57 schrieb Myron Evans:

Continuing with UFT416

I need to work some more on the self consistency of the theory and to introduce new methods based on the hamiltonian and geodesic equation. As described on the web the geodesic equation is a way of finding constants of motion. It is also described by Carroll , chapter seven of his online notes. Horst has shown in UFT415 that the lagrangian rigorously conserves total energy and angular momentum, but this lagrangian does not give p self consistently. This is why I weighted it as in Eq. (57) of UFT415. The lagrangian L is defined in hamiltonian / lagrangian dynamics by p = partial L / partial r dot, and so p = gamma m v / m(r) power half. So it must be checked numerically whether the lagrangian of Eq. (57) of UFT415 gives a constant total energy and momentum. The lagrangian method must give p and L self consistently, such that the angular momentum L1 = r x p self consistently, and so that dH / dt = 0, dL1 / dt = 0.