PS: Re: The correct rel. kinetic energy

OK, many thanks, the complete lagrangian must be used so that is why the constant of motion in this second case does not agree with the very fundamental kinematic method. Both the kinematic and lagrangian methods with complete lagrangian give

L = gamma m r squared phi dot

for the relativistic angular momentum. Then dL / dt = 0 with this relativistic angular momentum gives the same result as obtained form the fundamental kinematics in any coordinate system.

The correct rel. kinetic energy

PS: here is the protocol of both versions of kinetic energy. Observe the constant of motion in the second case, it is gamma^3*m*r^2*omega, for which reasons ever.


Am 10.09.2018 um 17:51 schrieb Horst Eckardt:

Did we use the relativistic kinetic energy

T = – m c^2 / gamma


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.



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