Potential Energy

Yes, I thought about this for some time, and finally decided that the most fundamental equation is Eq. (26) of the note, which uses the Newtonian equivalence principle. Instead of using the reduced mass, the lagrangian can be set up as Eq. (7.1) of Marion and Thornton, “Classical Dynamics”, p. 245, third edition:

lagrangian = (1/2) ( m sup p r bold sib 1 dot r bold sub 1 + m sun c r bold sub 2 dot r bold sub 2) – U(r)


U(r) = – m sub p m sub c / r

r = modulus r bold 1 – r bold 2

However, it is much easier simply to use Eq. (26) of the Note.

To: EMyrone@aol.com
Sent: 14/04/2017 23:02:03 GMT Daylight Time
Subj: PS: Re: 375(5) : Refutation of the Einstein Theory with the Hulse Taylor Binary Pulsar

PS: what happens with the potential energy when the reduced mass is used in the Lagrangian (8)? m_p then does not cancel out in the Lagrange equations.


Am 14.04.2017 um 13:32 schrieb EMyrone:

Using data from the site:


at Stanford it is shown straightforwardly that the Enstein theory is wildly wrong, it gives a precession of 17,891 degrees per earth year, compared with an observed precession of 4.2 degrees per year. The Einstein precession is calculated using the mass of the companion star and the half right latitude of the pulsar’s orbit. This can be calculated from the semi major axis and eccentricity given on the Stanford site. The initial conditions for the solution of the new lagrangian approach (Eq. (29)) are:

r(0) = periastron = 2.6885 ten power eight metres

and the orbital velocity at the periastron:

v(0) = 1.061 ten power five metres per second

calculated from the Stanford data using Eq. (4). There is a huge discrepancy between the Stanford and Cornell data on orbital velocity at the periastron, a factor of three. So I advise using data from the Stanford site. So there can be no confidence whatsoever in claims that the Einstein theory is always precise, and it should be replaced by ECE2 theory.

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