## Precession from the Minkowski Type Orbital Equation

Agreed.

To: EMyrone@aol.com

Sent: 05/05/2017 18:07:25 GMT Daylight Time

Subj: Re: Precession from the Minkowski Type Orbital EquationGood point that the precession depends on the position of the observer. The direction of the orbiting mass has to be taken into account. As far as I know, the precession for Mercury is positive. For the S2 star we would have to know if it orbits in positive or negative angular direction, seen from us.

Horst

Am 05.05.2017 um 11:41 schrieb EMyrone:

Many thanks. A positive precession means an anticlockwise rotation of the perihelion of the ellipse, a negative precession means a clockwise rotation of the perihelion of the ellipse. The sense of rotation depends where the observer is situated. To an observer above the orbital plane the sense of rotation is opposite to an observer below the plane. If one imagines a point going to the right looking at the computer screen, the same point goes to the left if one imagines looking at it from behind the computer screen. The important thing is the modulus of the precession, its absolute value. The factor gamma fourth is always less than one. It would be interesting to find the conditions under which precession is a maximum or a minimum. The Minkowski force equation is the rigorously correct relativistic Newton equation, as is well known. It is given as a problem in Marion and Thornton chapter fifteen. The really important result is that we have found precession from the well known Minkowski force equation when applied ot planar orbital theory. In UFT228 ff this was not clear.

To: EMyrone

Sent: 05/05/2017 09:03:58 GMT Daylight Time

Subj: Re: Precessions from ECE2 RelativityI did the calculations with eq.(8) of the note and only made a trial with the modified equation (inverted gamma factor). Obviously a factor <1 in multiplied to the Newtonian force gives a negative precession and a factor >1 gives a positive precession.

Horst

Am 05.05.2017 um 08:35 schrieb EMyrone:

Many thanks, precession from the Minkowski equation is an important result sign. Your equation below has the right sign, but it should be the same as Eq. (8) of the note:

r double dot bold = – MG r bold / (gamma fourth r cubed)

The relativistic linear momentum relevant to the Minkowski force equation is the well known

p bold = d r bold / d tau = gamma d r bold / dt = gamma p0

where tau is the proper time. The relativistic angular momentum is therefore

L bold = r bold x p bold

In the non relativistic limit this becomes the Newtonian angular momentum for a planar orbit which is constant, so your results seem to be right. The relativistic angular momentum depends on gamma:

L bold = gamma r bold x p0 bold

The relativistic total energy is E = gamma m c squared and this is a constant of motion for a free particle. For orbital motion the constant of motion is the relativistic hamiltonian:

H = gamma m c squared + U

I will work out the relativistic Euler Lagrange equation in plane polar coordinates to calculate the relativistic angular momentum and see whether it is a constant analytically.

To: EMyrone

Sent: 04/05/2017 20:25:26 GMT Daylight Time

Subj: Re: Precessions from ECE2 RelativityI did some calculations with model parameters so that numerical precision is not an issue. It comes out that the precession angle is always negative for the Minkowski equation. Only when the gamma factor is inverted artificially, i.e.

bold r dotdot = – gamma^4 G M bold r / r^3,

precession is positive. Another issue is the relativistic angular momentum. This is not constant but the non-relativistic angular momentum is constant, quite strange. I will check Marion and Thornton tomorrow.

Horst

Am 04.05.2017 um 13:41 schrieb EMyrone:

This is very important in my opinion, this is the final form of the orbital equation in ECE2 relativity. In future papers this theory can be tied up with fluid gravitation.

Negative Precession from ECE2 Relativity

I will do some more calculations with test data. It would be interesting to see under which conditions a negative precession comes out.

Horst

Am 04.05.2017 um 12:51 schrieb EMyrone:

This is a very important result! Einsteinian general relativity (EGR) is not capable of giving negative precession. I should think that the Minkowski type equation will give many interesting results, especially with choice of spin connection.

To: EMyrone

Sent: 04/05/2017 11:45:19 GMT Daylight Time

Subj: Re: 376(6): Comparison of Precession EquationsI calculated the orbit of the S2 star with the Minkowski equation. The resulting orbit time interval, max. radius and epsilon are nearly identical to the previous calculation. However the precession seems to be affected significantly and is negative now. I will further check this. The value is -0.002 rad/orbit while the exp. value is > -0.017 rad/orbit so we are in the experimental range as before.

Horst

Am 04.05.2017 um 10:31 schrieb EMyrone:

This note shows that the rigorously correct ECE2 orbital precession is produced by the Minkowski type orbital equation (8) and the relativistic Euler Lagrange equation (9). These are both newly discovered equations. The previously used precession equation (10) is produced by the Euler Lagrange equation (11), in which t appears. The correct and newly inferred equation (9) has the proper time tau on the right hand side. Both Eqs. (9) and (11) produce orbital precessions, but the rigorously correct one is Eq. (9). it must be solved simultaneously with the field equations of ECE2 gravitostatics. So the set of equations for numerical solution is Eq. (21) to (23). I will write up UFT376, Sections 1 and 2, reporting the discovery of these new equations. Section 3 can be dedicated to showing that they produce precession, so EGR becomes obsolete and unnecessary. There is also a relativistic Hamilton Principle of Least Action of ECE2 relativity. It will be interesting to compare these new precessions with those in the S2 star system.

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