The correct rel. kinetic energy

The correct rel. kinetic energy

Subject: The correct rel. kinetic energy
To: Myron Evans <myronevans123>

Thanks again for this meticulous numerical checking. The lagrangian used in Note 404(4) was the complete lagrangian as defined in Eq. (17) and in Marion and Thornton chapter (14.113) of the third edition. This was one of two methods used to derive the relativistic Leibniz equation (29) and the relativistic angular momentum (31). The two methods gave exactly the same results. The Euler Lagrange variables were r and phi. The calculations are given in Eqs. (21) to (29). The relativistic angular momentum L is given in Eq. (31) and dL / dt gives Eq. (34). Eq. (29) derived in this way as as shown to be the same as Eq. (7), derived from the very fundamental kinematic equation (5). So to obtain the kinematic equation the complete lagrangian (17) must be used with Euler Lagrange variables r and phi. Eq. (7) can be transformed into Eq. (44) using Eqs. (37) and (38). Eq. (44) is the usual form of the relativistic Newton force as given for example in the problem section of Marion and Thornton, third edition, chapter fourteen, problem 38. The new insights given in Note 414(4) are Eqs. (37) and (38). Then Note 414(5) gives a triple cross check using the hamiltonian method. The complete relativistic hamiltonian H must be used, Eq. (1). This is a constant of motion so dH / dt = 0. This gives Eq. (6) of Note 414(5). When this is used in the relativistic force (7), the realtivistic Newton force Eq. (11) is obtained in a third way. so there is a triple cross check. The relativistic force equation has been derived in three ways, each giving the same result:

F = m gamma cubed dv / dt = -mMg / r squared.

As you know, we have used this equation in Cartesian coordinates in several previous UFT papers to give many interesting and original results. Finally the rotating frame method was applied to this equation, giving the relativistic version of paper UFT413. So a whole pile of new physics has emerged.

T = – m c^2 / gamma

or

T = (gamma-1) m c^2 ?

I obtain different results for both. In the second case an additional factor 1/gamma^2 seems to appear in the baseline calculation.

The results in the first case are

and in the second case:

The first case can be simplified with re-inserting gamma.
I will have to check this further.

Horst

Comments are closed.