Fully agreed with this analysis, I will have a look at the velocities. The analysis also holds for the earth sun system, but in that system the reduced mass is essentially the same as the earth’s mass because the mass of the sun (M) is so much bigger than the mass of the earth (m), so

mu = mM / (m + M) is m to an excellent approximation

Rigorously, however, the lagrangian is as you describe here. The earth sun system is no different in principle from the binary pulsar system, in which the masses of the two stars are about the same.

To: EMyrone@aol.com

Sent: 15/04/2017 19:09:30 GMT Daylight Time

Subj: Calculation of double star systems with reduced mass

The introduction of the reduced mass leads to the effect that it does not cancel out in the Euler-Lagrange equations. Insertion of the term for the reduced mass mu leads to an effective potential term

U(r) = -(m_p + m_c) * G/r.

The relativistic Lagrange equations are:

The observations relate to the ellipses of each star partner. Their coordinates are r1 and r2, and not the reduced coordinate r. Therefore the observed values have to be transformed to r according to

bold r = (m_p + m_c) / m_c * bold r_p.

These initial positions can be inserted into the calculation. What happens with the velocities? It seems that because of

bold v = bold r dot

also the velocities have to be transformed.

Horst

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