Very interesting as usual. In UFT415 the Lagrangian is defined in Eq. (57) to give the relativistic momentum p as in Eq. (55). So the lagrangian is defined to give the correct relativistic linear momentum. The angular momentum is then L = r x p. The position vector r is calculated from first principles in Eqs. (1) to (11). The kinematics are very fundamental and in UFT414 the total energy and angular momentum were conserved. Numerical experimentation for Section 3 would be most interesting and very important. For a stationary metric m(r) must be time independent as in Carroll’s lecture notes. If so, the conservation of angular momentum means that d (gamma m r squared phi dot / (m(r)) = 0 , i.e. d (gamma m r squared phi dot) = 0. This is Eq. (67). It follows that conservation of angular momentum constrains the choice of m(r). It is already known that m(r) = 1 conserves angular momentum. Some more information is available from dH / dt = 0, where H = gamma m c squared – mMG / r. H is the hamiltonian and is of course conserved. Using a time dependent m(r) means a non stationary metric, and it may be that a non stationary metric is needed for conservation of H and L in m space. That would be a highly interesting result. Since no one has done this type of work before, the numerical work is very important. I also recommend having a look at Carroll’s lecture notes to find out exactly what he means by m(r). Finally, Steve Crothers’ metric is more general than that of m theory. Another clue is that if m = 1 – 2MG / r c squared, the obsolete Schwarzschild metric is obtained and presumably that conserves angular momentum and total energy.
: Fwd: Double Cross Check of the m Theory
To: Myron Evans <myronevans123>
Meanwhile I did some tests. With the m(r)-dependent gamma and the Lagrangian (in my earlier email a minus sign was missing)
L = – m*c^2/gamma – m*M*G/r
(without explicit m(r)) the angular momentum
L1 = gamma / m(r) * m r^2 phi dot
is conserverd, but oly if the time dependence of m(r) is respected. I guess this is similar to the potential which, considered as a field in space, is time-independent. However, the Lagrange mechanism computes time curves (trajectories) in space, therefore the coordinates in the equations of motion relates to a mass point in motion. This then holds for all occurences of r in the equations. The equations do not become significantly more complicated. There is a summand
in the equation for d^2r/dt^2 which requires that m(r) has to compensate c^2 in the transition to the non-relativistic case. This is fulfilled for m(r)=1, but may require that m(r) does not drop too steeply for r–>0.
I just saw that you sent over UFT 415. I can add these discussions to section 3 but it may mean that the cross checks are not complete if we agree that the time dependence of m(r) has to be added.
Am 28.09.2018 um 12:10 schrieb Horst Eckardt:
Obviously you arrived at a different expressions for gamma and the Lagrangian as before. Do I see it right that now is
L = sqrt(m(r))*m*c^2/gamma – m*M*G/r,
With these definitions, the Lagrange formalism now gives the constant of motion
where the factor 1/m(r)^1/2 is contained the first time.
With the definitions I used before, none of angular momentum or total energy was conserved. I will check this with the new definitions.
Am 26.09.2018 um 15:21 schrieb Myron Evans:
Double Cross Check of the m Theory
This note shows that with the lagrangian (25) the kinematic and lagrangian methods give exactly the same result, Eq. (29), so the orbit is found by integrating Eqs. (30) and (31). This is a precise double cross check. From the kinematic method the conserved angular momentum is Eq. (32). Eq. (31) shows that dL / dt = 0 as required. Also, this note gives a detailed explanation of the meaning of Eq. (8), the vector Euler Lagrange equation with Lagrange variable r bold. In quantum field theory four vectors are used as Lagrange variables. There is freedom of choice of the Lagrangian, the kinematic results are obtained by choosing the lagrangian (25). So everything is precisely self consistent. The kinematic method is very fundamental, more so than the lagrangian method, which in turn is more powerful than the Newtonian method. Horst’s powerful integrator routine can now be used with Eqs. (30) and (31). So now I will proceed to writing up Sections 1 and 2 of UFT415.