OJ287 binary black hole system perihelion shifts

[hapor...@gmail.com]

I calculated OJ287 binary black hole system with Swarzshild General Relativity system and I got two different perihelion shift values (not rounded numbers): 27.73298398 degrees and 27.79877741 degrees per one orbit. I got two analytic solutions also but above was calculated with primary solution. Ref. Lankeswar Dey et al. 2018. The Astrophysical Journal. 866:11 (20pp), 2018 October 20. Table 2/page 7. Best Regards, Hannu Poropudas, Finland.

[hapor...@gmail.com]

I give here those two analytic solutions:First primitive function is following:e=2.117198112*EllipticF(sin(P),0.2601014318), r=(0.1627132736e34*sin(P)^2-0.2405124138e35)/(0.2726824096e18*sin(P)^2-0.3378949220e18), definition areas are 0<=P<=Pi/2 and 0.7117964733e17<=r<=0.3438620569e18. Second primitive function is following: ee=2.117198112*EllipticF(sin(P),0.9655812990), rr=(0.2051871297*sin(P)^2)/(0.5967134948e16*sin(P)^2+0.3378949220e18), definition areas are 0<=P<=Pi/2 and 0<=rr<=0.5967134948e16. Maple 9 used and cgs unit system with c=1 Weinberg's used units. Best Regards Hannu Poropudas, Finland

[hapor...@gmail.com]

I forget to mention that full orbits in both General Relativity analytic solutions are achieved when one takes + and - signs for first +-e and second +-ee elliptic integrals of second kind. Remark that Maple9 uses e-letter in those formulae as 10^ meaning also, not to confuse about this. Best Regards, Hannu Poropudas, Finland

[hapor...@gmail.com]

Elliptic integral of the first kind was correct name. Plotting is done in polar coordinates. Best Regards, Hannu Poropudas, Finland.

[hapor...@gmail.com]

I have used P^2=(4*Pi^2*a^3)/(G*M) to calculate semimajoraxis a, with P=12.062 yr and M=18348*10^6*M_SUN. But z=0.306 From 2014 Thesis of Pihajoki P. University of Turku, ser. A1 - 492. Annales Universitatis Turkuensis. If I would take this redshift into account should I recalculte all by P=12.062yr/(1+0.306)=9.235834609 yr? (I use not rounded numbers. Hannu

[hapor...@gmail.com]

I recalculated with accounting z=0.306:e=2.142521522*EllipticF(sin(P),0.2877094699), definition areas are 0<=P<=Pi/2 and 0.5957449715e17<=r<=0.2877986650e18. r=(0. 1389227423e34*sin(P)^2-0.1678282376e35)/(0.2282241678e18*sin(P)^2-0.2817115471e18). Second analytic slution: ee=2.142521522*EllipticF(sin(P),0.9577177353), rr=(0. 1751864413e34*sin(P)^2)/(0.6087117927e16*sin(P)^2+0.2817115471e18), definition areas are 0<=P<=Pi/2 and 0<=rr<=0.6087117927e16. Problem with calculation precession angle per orbit in first parametric formula which gives Imaginary number for P which corresponds aphelion. P=1.57079637-0.1503825754e-4*I,and P=-1.570796327+0.1503825754e-4*I,Maple 9 gives or these 34.01372910 degrees per orbit and 34.04443070 degrees per orbit. Best Regards, Hannu Poropudas, Finland.

[hapor...@gmail.com]

J= - 1.731669280*10^16 total angular momentum per unit mass and E= 1.015569493 total energy per unit mass. Weinberg gives in his book p. 187 formula for dP/dt = J*(1-2*M*G/r)/r^2, M=3.648591540*10^43 g. G=6.674083100*10^(-8), c=1 and cgs units. I give this due orbit seems to be strange pendulum orbit with no proper aphelion. Best Regards, Hannu Poropudas, Finland.

[hapor...@gmail.com]

Not dP/dt. Correct is de/dt. Sorry, Hannu

[hapor...@gmail.com]

Aphelion directions e=197.01454 degrees and e=-197.01454 degrees. Polar coordinates used. Hannu

[hapor...@gmail.com]

Perihelion directions (also two) are small: +, - 7.675394121*10^(-4) degrees. Hannu

[hapor...@gmail.com]

Remark: Calculations are made with primitive functions. This is function which does not account constant of integration or in other words constant of integration is taken as zero. Hannu

[Hannu Poropudas]

perjantai 15. toukokuuta 2020 klo 16.27.32 UTC+3 hapor...@gmail.com kirjoitti:
> Remark: Calculations are made with primitive functions. This is function which does not account constant of integration or in other words constant of integration is taken as zero. Hannu

Both analytic solutions has +,- sign branchs. These both are in both analytic solutions
most easily visible with Maple 9 plotting command range of the solution parameter

P = -Pi/2..Pi/2.

Result is strange pendulum orbit.

Hannu