Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Best Regrads,

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland

On Sunday, October 15, 2023 at 1:35:22 AM UTC-7, Hannu Poropudas wrote:
> Spherically symmetric metrics which satisfies
> Einstein's vacuum field equations.
>
> (c=1,G=1 units)
>
> matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
>
> (c=1,G=1 units)
>
> ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
>
> (m -> m*G/c^2 , if SI-units are used.)
>
> I don't know that would this solution have any astrophysical applications?
>
> There exist a book called something like
> "Exact Solutions of the Einstein Field Equations",
> which have about 740 pages and
> I don't know if this solution is among them?
>
> Three singularity points of the metrics are the following:
>
> r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
>
>
> I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
>
> Reference:
> Tolman R. C., 1934.
> Effect of inhomogeneity on cosmological models.
> Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
>
> Best Regrads,
>
> Hannu Poropudas
>
> Kolamäentie 9E
> 90900 Kiiminki / Oulu
> Finland

There is only one solution to the spherically symmetric vacuum equations
(see e.g. Hawking & Ellis).

So you either made a mistake somewhere or your solution is the same
as Schwarzschild's, only written differently.

--
Jan

On Monday, October 16, 2023 at 5:21:43 PM UTC-7, JanPB wrote:
> On Sunday, October 15, 2023 at 1:35:22 AM UTC-7, Hannu Poropudas wrote:
> > Spherically symmetric metrics which satisfies
> > Einstein's vacuum field equations.
> >
> > (c=1,G=1 units)
> >
> > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> >
> > (c=1,G=1 units)
> >
> > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> >
> > (m -> m*G/c^2 , if SI-units are used.)
> >
> > I don't know that would this solution have any astrophysical applications?
> >
> > There exist a book called something like
> > "Exact Solutions of the Einstein Field Equations",
> > which have about 740 pages and
> > I don't know if this solution is among them?
> >
> > Three singularity points of the metrics are the following:
> >
> > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> >
> >
> > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> >
> > Reference:
> > Tolman R. C., 1934.
> > Effect of inhomogeneity on cosmological models.
> > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> >
> > Best Regrads,
> >
> > Hannu Poropudas
> >
> > Kolamäentie 9E
> > 90900 Kiiminki / Oulu
> > Finland
> There is only one solution to the spherically symmetric vacuum equations
> (see e.g. Hawking & Ellis).
>
> So you either made a mistake somewhere or your solution is the same
> as Schwarzschild's, only written differently.
>
> --
> Jan

Ha, spherically linearly, spherically rotationally, symmetric, ....

"Only one", ....

It's usually Schwarzschild and Chandrasekhar on the inside,
and Kruszkeles and Kerr, or so, on the outside, one stationary black hole, spherical.

Outside the forces are like inverse cube outside but inside linear,
the inverse square singularity, cube wall, here that the inertial
system is usualy considred to be inside and stationary, also,
the stationary, non-rotating well of a gravitational black hole.

Mathematically, ....

It seems you are to post it to "arxiv" then have others include it.

On Sunday, October 15, 2023 at 1:35:22 AM UTC-7, Hannu Poropudas wrote:
> Spherically symmetric metrics which satisfies
> Einstein's vacuum field equations.
>
> (c=1,G=1 units)
>
> matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
>
> (c=1,G=1 units)
>
> ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
>
> (m -> m*G/c^2 , if SI-units are used.)
>
> I don't know that would this solution have any astrophysical applications?
>
> There exist a book called something like
> "Exact Solutions of the Einstein Field Equations",
> which have about 740 pages and
> I don't know if this solution is among them?
>
> Three singularity points of the metrics are the following:
>
> r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
>
>
> I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
>
> Reference:
> Tolman R. C., 1934.
> Effect of inhomogeneity on cosmological models.
> Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
>
> Best Regrads,
>
> Hannu Poropudas
>
> Kolamäentie 9E
> 90900 Kiiminki / Oulu
> Finland

I congratulate you, H. Poropudas. Is it Doctor Poropudas by chance? In any event, I would like to request that you apply your new metric to a) a photon grazing the surface of the sun, and b) the apogee of planet Mercury. If you decide to do this, then will you please report back your results to this forum at your earliest convenience. Thank you.

PS--What have you heard about a new gadanken experiment that is reputed to have originated in the Northwestern region of the United States which appears to be taking the world of relativity by storm. I believe they call it the Big Ben Paradox.

sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> Spherically symmetric metrics which satisfies
> Einstein's vacuum field equations.
>
> (c=1,G=1 units)
>
> matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
>
> (c=1,G=1 units)
>
> ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
>
> (m -> m*G/c^2 , if SI-units are used.)
>
> I don't know that would this solution have any astrophysical applications?
>
> There exist a book called something like
> "Exact Solutions of the Einstein Field Equations",
> which have about 740 pages and
> I don't know if this solution is among them?
>
> Three singularity points of the metrics are the following:
>
> r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
>
>
> I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
>
> Reference:
> Tolman R. C., 1934.
> Effect of inhomogeneity on cosmological models.
> Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
>
> Best Regrads,
>
> Hannu Poropudas
>
> Kolamäentie 9E
> 90900 Kiiminki / Oulu
> Finland

I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral
phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > Spherically symmetric metrics which satisfies
> > Einstein's vacuum field equations.
> >
> > (c=1,G=1 units)
> >
> > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> >
> > (c=1,G=1 units)
> >
> > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> >
> > (m -> m*G/c^2 , if SI-units are used.)
> >
> > I don't know that would this solution have any astrophysical applications?
> >
> > There exist a book called something like
> > "Exact Solutions of the Einstein Field Equations",
> > which have about 740 pages and
> > I don't know if this solution is among them?
> >
> > Three singularity points of the metrics are the following:
> >
> > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> >
> >
> > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> >
> > Reference:
> > Tolman R. C., 1934.
> > Effect of inhomogeneity on cosmological models.
> > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> >
> > Best Regrads,
> >
> > Hannu Poropudas
> >
> > Kolamäentie 9E
> > 90900 Kiiminki / Oulu
> > Finland
> I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> due three integration constants from Euler-Lagrange equations does not have
> same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
>
> MG = 6.292090968*10^11,
> 2*MG=1.258418194*10^12.
> I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
>
> 2.720522631*10^11<=r<=8.306841627*10^11
> +,- sign for integral
> phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
>
> and
>
> -1.103327381*10^12<=rr<=0
> +,- sign for integral
> phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
>
> I calculated also these integrals but their formulae are too long to copy here.
> Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> Real parts = 0 in these integrals.
> How to interpret pure imaginary phi and phiphi angles?
> How to interpret these Imaginary angle plots?
>
> Best Regards,
> Hannu Poropudas

Your solution is either:

(a) incorrect, or:

(b) isometric to Schwarzschild's.

Don't waste your time.

--
Jan

JanPB wrote:

> On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas
>> I calculated also these integrals but their formulae are too long to
>> copy here. Both sign can be taken into account when plotting Imaginary
>> parts 0<=P<=Pi. Real parts = 0 in these integrals. How to interpret
>> pure imaginary phi and phiphi angles? How to interpret these Imaginary
>> angle plots? Best Regards, Hannu Poropudas
>
> Your solution is either: (a) incorrect, or: (b) isometric to
> Schwarzschild's. Don't waste your time.

anyone that's not constantly on the verge of tears over what kikes have
done to ruin reality on this planet, simply doesn't understand what kikes
have done to ruin reality on this planet ··· Faggot kike channel

torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > Spherically symmetric metrics which satisfies
> > > Einstein's vacuum field equations.
> > >
> > > (c=1,G=1 units)
> > >
> > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > >
> > > (c=1,G=1 units)
> > >
> > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > >
> > > (m -> m*G/c^2 , if SI-units are used.)
> > >
> > > I don't know that would this solution have any astrophysical applications?
> > >
> > > There exist a book called something like
> > > "Exact Solutions of the Einstein Field Equations",
> > > which have about 740 pages and
> > > I don't know if this solution is among them?
> > >
> > > Three singularity points of the metrics are the following:
> > >
> > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > >
> > >
> > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > >
> > > Reference:
> > > Tolman R. C., 1934.
> > > Effect of inhomogeneity on cosmological models.
> > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > >
> > > Best Regrads,
> > >
> > > Hannu Poropudas
> > >
> > > Kolamäentie 9E
> > > 90900 Kiiminki / Oulu
> > > Finland
> > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > due three integration constants from Euler-Lagrange equations does not have
> > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> >
> > MG = 6.292090968*10^11,
> > 2*MG=1.258418194*10^12.
> > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> >
> > 2.720522631*10^11<=r<=8.306841627*10^11
> > +,- sign for integral
> > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> >
> > and
> >
> > -1.103327381*10^12<=rr<=0
> > +,- sign for integral
> > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> >
> > I calculated also these integrals but their formulae are too long to copy here.
> > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > Real parts = 0 in these integrals.
> > How to interpret pure imaginary phi and phiphi angles?
> > How to interpret these Imaginary angle plots?
> >
> > Best Regards,
> > Hannu Poropudas
> Your solution is either:
>
> (a) incorrect, or:
>
> (b) isometric to Schwarzschild's.
>
> Don't waste your time.
>
> --
> Jan

Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu

On Thursday, October 19, 2023 at 11:54:12 PM UTC-7, Hannu Poropudas wrote:
> torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > Spherically symmetric metrics which satisfies
> > > > Einstein's vacuum field equations.
> > > >
> > > > (c=1,G=1 units)
> > > >
> > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > >
> > > > (c=1,G=1 units)
> > > >
> > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > >
> > > > (m -> m*G/c^2 , if SI-units are used.)
> > > >
> > > > I don't know that would this solution have any astrophysical applications?
> > > >
> > > > There exist a book called something like
> > > > "Exact Solutions of the Einstein Field Equations",
> > > > which have about 740 pages and
> > > > I don't know if this solution is among them?
> > > >
> > > > Three singularity points of the metrics are the following:
> > > >
> > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > >
> > > >
> > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > >
> > > > Reference:
> > > > Tolman R. C., 1934.
> > > > Effect of inhomogeneity on cosmological models.
> > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > >
> > > > Best Regrads,
> > > >
> > > > Hannu Poropudas
> > > >
> > > > Kolamäentie 9E
> > > > 90900 Kiiminki / Oulu
> > > > Finland
> > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > due three integration constants from Euler-Lagrange equations does not have
> > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > >
> > > MG = 6.292090968*10^11,
> > > 2*MG=1.258418194*10^12.
> > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > >
> > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > +,- sign for integral
> > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > >
> > > and
> > >
> > > -1.103327381*10^12<=rr<=0
> > > +,- sign for integral
> > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > >
> > > I calculated also these integrals but their formulae are too long to copy here.
> > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > Real parts = 0 in these integrals.
> > > How to interpret pure imaginary phi and phiphi angles?
> > > How to interpret these Imaginary angle plots?
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > Your solution is either:
> >
> > (a) incorrect, or:
> >
> > (b) isometric to Schwarzschild's.
> >
> > Don't waste your time.
> >
> > --
> > Jan
> Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
>
> Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
>
> Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> but you are correct in point of view that it may not be physically acceptable solution
> of these equations at our present orthodoxic physical knowledge.
> This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
>
> There exist also few other integration constants from Euler-Largrange equations,
> but I have selected randomly only one couple of them in this example calculation.
>
> Hannu

It's either "more symmetries" or "less symmetries", or same.

You mention integration constants and so we call that "quantities",
when quantities aren't atomic algebraically to their symbolic notation.
(Analytically.)

The differential quantities, here is for an ideal, "the field equations
of a singularity", "the orbit equations of a singularity", as above
this "cube wall" reflects geometrically, the large and small,
inside/outside, that the horizon is local everywhere, makes
for not necessarily centralized moments, why outside it is
inverse cube and inside flat, or that ideally that's zero instead
of infinity, in or out of a singularity.

Then it's figured that the inertial _systems_, are in their potential
and actual, for example having "singularities that vanish"
vis-a-vis "singularities that as infinite ideally always grow
while growing less than their growing Scharzschild radius",
within which matter is so dense as singular, affecting inverse cube.

It's theories of stellar pulsation. Accretion, this and that,
it's like, when twenty or more years ago,
science announced "science now makes black holes on Earth",
that, there was "of course by definition that would eat the Earth",
what "definitions" result, singularity theories in ideals,
that "maintaining the inertial center" of a black hole, is arbitrary.

Here it's about inverse cube or weight to power, "gravity",
then in coordinates "the dynamics of the singularity, in
effect". Then these are linear broadly and orthogonal
directly, "cube wall" and "oncoming cube" and "incoming cube",
inverse cube and flat.

Good luck Hannu, here then it's as about "constants and quantities",
differential.

On Saturday, October 21, 2023 at 11:02:52 AM UTC-7, Ross Finlayson wrote:
> On Thursday, October 19, 2023 at 11:54:12 PM UTC-7, Hannu Poropudas wrote:
> > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > Spherically symmetric metrics which satisfies
> > > > > Einstein's vacuum field equations.
> > > > >
> > > > > (c=1,G=1 units)
> > > > >
> > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > >
> > > > > (c=1,G=1 units)
> > > > >
> > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > >
> > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > >
> > > > > I don't know that would this solution have any astrophysical applications?
> > > > >
> > > > > There exist a book called something like
> > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > which have about 740 pages and
> > > > > I don't know if this solution is among them?
> > > > >
> > > > > Three singularity points of the metrics are the following:
> > > > >
> > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > >
> > > > >
> > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > >
> > > > > Reference:
> > > > > Tolman R. C., 1934.
> > > > > Effect of inhomogeneity on cosmological models.
> > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > >
> > > > > Best Regrads,
> > > > >
> > > > > Hannu Poropudas
> > > > >
> > > > > Kolamäentie 9E
> > > > > 90900 Kiiminki / Oulu
> > > > > Finland
> > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > due three integration constants from Euler-Lagrange equations does not have
> > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > >
> > > > MG = 6.292090968*10^11,
> > > > 2*MG=1.258418194*10^12.
> > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > >
> > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > +,- sign for integral
> > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > >
> > > > and
> > > >
> > > > -1.103327381*10^12<=rr<=0
> > > > +,- sign for integral
> > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > >
> > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > Real parts = 0 in these integrals.
> > > > How to interpret pure imaginary phi and phiphi angles?
> > > > How to interpret these Imaginary angle plots?
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > Your solution is either:
> > >
> > > (a) incorrect, or:
> > >
> > > (b) isometric to Schwarzschild's.
> > >
> > > Don't waste your time.
> > >
> > > --
> > > Jan
> > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> >
> > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> >
> > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > but you are correct in point of view that it may not be physically acceptable solution
> > of these equations at our present orthodoxic physical knowledge.
> > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> >
> > There exist also few other integration constants from Euler-Largrange equations,
> > but I have selected randomly only one couple of them in this example calculation.
> >
> > Hannu
> It's either "more symmetries" or "less symmetries", or same.
>
> You mention integration constants and so we call that "quantities",
> when quantities aren't atomic algebraically to their symbolic notation.
> (Analytically.)
>
> The differential quantities, here is for an ideal, "the field equations
> of a singularity", "the orbit equations of a singularity", as above
> this "cube wall" reflects geometrically, the large and small,
> inside/outside, that the horizon is local everywhere, makes
> for not necessarily centralized moments, why outside it is
> inverse cube and inside flat, or that ideally that's zero instead
> of infinity, in or out of a singularity.
>
> Then it's figured that the inertial _systems_, are in their potential
> and actual, for example having "singularities that vanish"
> vis-a-vis "singularities that as infinite ideally always grow
> while growing less than their growing Scharzschild radius",
> within which matter is so dense as singular, affecting inverse cube.
>
> It's theories of stellar pulsation. Accretion, this and that,
> it's like, when twenty or more years ago,
> science announced "science now makes black holes on Earth",
> that, there was "of course by definition that would eat the Earth",
> what "definitions" result, singularity theories in ideals,
> that "maintaining the inertial center" of a black hole, is arbitrary.
>
>
> Here it's about inverse cube or weight to power, "gravity",
> then in coordinates "the dynamics of the singularity, in
> effect". Then these are linear broadly and orthogonal
> directly, "cube wall" and "oncoming cube" and "incoming cube",
> inverse cube and flat.
>
>
> Good luck Hannu, here then it's as about "constants and quantities",
> differential.

How is a vacuum round?
The closed universe is flat geometry and in the hypersphere boundary

perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > Spherically symmetric metrics which satisfies
> > > > Einstein's vacuum field equations.
> > > >
> > > > (c=1,G=1 units)
> > > >
> > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > >
> > > > (c=1,G=1 units)
> > > >
> > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > >
> > > > (m -> m*G/c^2 , if SI-units are used.)
> > > >
> > > > I don't know that would this solution have any astrophysical applications?
> > > >
> > > > There exist a book called something like
> > > > "Exact Solutions of the Einstein Field Equations",
> > > > which have about 740 pages and
> > > > I don't know if this solution is among them?
> > > >
> > > > Three singularity points of the metrics are the following:
> > > >
> > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > >
> > > >
> > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > >
> > > > Reference:
> > > > Tolman R. C., 1934.
> > > > Effect of inhomogeneity on cosmological models.
> > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > >
> > > > Best Regrads,
> > > >
> > > > Hannu Poropudas
> > > >
> > > > Kolamäentie 9E
> > > > 90900 Kiiminki / Oulu
> > > > Finland
> > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > due three integration constants from Euler-Lagrange equations does not have
> > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > >
> > > MG = 6.292090968*10^11,
> > > 2*MG=1.258418194*10^12.
> > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > >
> > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > +,- sign for integral
> > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > >
> > > and
> > >
> > > -1.103327381*10^12<=rr<=0
> > > +,- sign for integral
> > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > >
> > > I calculated also these integrals but their formulae are too long to copy here.
> > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > Real parts = 0 in these integrals.
> > > How to interpret pure imaginary phi and phiphi angles?
> > > How to interpret these Imaginary angle plots?
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > Your solution is either:
> >
> > (a) incorrect, or:
> >
> > (b) isometric to Schwarzschild's.
> >
> > Don't waste your time.
> >
> > --
> > Jan
> Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
>
> Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
>
> Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> but you are correct in point of view that it may not be physically acceptable solution
> of these equations at our present orthodoxic physical knowledge.
> This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
>
> There exist also few other integration constants from Euler-Largrange equations,
> but I have selected randomly only one couple of them in this example calculation.
>
> Hannu

I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas

tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > Spherically symmetric metrics which satisfies
> > > > > Einstein's vacuum field equations.
> > > > >
> > > > > (c=1,G=1 units)
> > > > >
> > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > >
> > > > > (c=1,G=1 units)
> > > > >
> > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > >
> > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > >
> > > > > I don't know that would this solution have any astrophysical applications?
> > > > >
> > > > > There exist a book called something like
> > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > which have about 740 pages and
> > > > > I don't know if this solution is among them?
> > > > >
> > > > > Three singularity points of the metrics are the following:
> > > > >
> > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > >
> > > > >
> > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > >
> > > > > Reference:
> > > > > Tolman R. C., 1934.
> > > > > Effect of inhomogeneity on cosmological models.
> > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > >
> > > > > Best Regrads,
> > > > >
> > > > > Hannu Poropudas
> > > > >
> > > > > Kolamäentie 9E
> > > > > 90900 Kiiminki / Oulu
> > > > > Finland
> > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > due three integration constants from Euler-Lagrange equations does not have
> > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > >
> > > > MG = 6.292090968*10^11,
> > > > 2*MG=1.258418194*10^12.
> > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > >
> > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > +,- sign for integral
> > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > >
> > > > and
> > > >
> > > > -1.103327381*10^12<=rr<=0
> > > > +,- sign for integral
> > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > >
> > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > Real parts = 0 in these integrals.
> > > > How to interpret pure imaginary phi and phiphi angles?
> > > > How to interpret these Imaginary angle plots?
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > Your solution is either:
> > >
> > > (a) incorrect, or:
> > >
> > > (b) isometric to Schwarzschild's.
> > >
> > > Don't waste your time.
> > >
> > > --
> > > Jan
> > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> >
> > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> >
> > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > but you are correct in point of view that it may not be physically acceptable solution
> > of these equations at our present orthodoxic physical knowledge.
> > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> >
> > There exist also few other integration constants from Euler-Largrange equations,
> > but I have selected randomly only one couple of them in this example calculation.
> >
> > Hannu
> I put here those strange (NO ordinary physical interpretation) formulae of integration
> constants from Euler-Largrange equations:
>
> I mark now for convenience T = coordinate time and t = proper time.
>
> (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
>
> I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> to calculate two integration constants K1 and K2 of Euler-Largange equations
> (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
>
> K1 = +,- 0.7072727132*I,
> K2 = +,- 0.5943942676 +,- 0.5943942676*I,
>
> And I selected here randomly as an example two constants of integration
> in this my two analytic solutions calculation:
>
> K1 = - 0.7072727132*I
> and
> K2 = 0.5943942676 - 0.5943942676*I
>
> This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> (Phi(P) is pure imaginary angle and r(P) is real distance.
> Phiphi(P) is pure imaginary angle and rr(P) is real distance).
>
> Plot ([Im(phi(P)),r(P),P=0..Pi]);
> Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
>
> Those both plots resemble somehow pendulum orbit ?
>
> I have NO physical interpretations of these solutions
> and I think that these have NO real physical applications.
>
> Hannu Poropudas

I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.

I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas

keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > Spherically symmetric metrics which satisfies
> > > > > > Einstein's vacuum field equations.
> > > > > >
> > > > > > (c=1,G=1 units)
> > > > > >
> > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > >
> > > > > > (c=1,G=1 units)
> > > > > >
> > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > >
> > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > >
> > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > >
> > > > > > There exist a book called something like
> > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > which have about 740 pages and
> > > > > > I don't know if this solution is among them?
> > > > > >
> > > > > > Three singularity points of the metrics are the following:
> > > > > >
> > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > >
> > > > > >
> > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > >
> > > > > > Reference:
> > > > > > Tolman R. C., 1934.
> > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > >
> > > > > > Best Regrads,
> > > > > >
> > > > > > Hannu Poropudas
> > > > > >
> > > > > > Kolamäentie 9E
> > > > > > 90900 Kiiminki / Oulu
> > > > > > Finland
> > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > >
> > > > > MG = 6.292090968*10^11,
> > > > > 2*MG=1.258418194*10^12.
> > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > >
> > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > +,- sign for integral
> > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > >
> > > > > and
> > > > >
> > > > > -1.103327381*10^12<=rr<=0
> > > > > +,- sign for integral
> > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > >
> > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > Real parts = 0 in these integrals.
> > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > How to interpret these Imaginary angle plots?
> > > > >
> > > > > Best Regards,
> > > > > Hannu Poropudas
> > > > Your solution is either:
> > > >
> > > > (a) incorrect, or:
> > > >
> > > > (b) isometric to Schwarzschild's.
> > > >
> > > > Don't waste your time.
> > > >
> > > > --
> > > > Jan
> > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > >
> > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > >
> > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > but you are correct in point of view that it may not be physically acceptable solution
> > > of these equations at our present orthodoxic physical knowledge.
> > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > >
> > > There exist also few other integration constants from Euler-Largrange equations,
> > > but I have selected randomly only one couple of them in this example calculation.
> > >
> > > Hannu
> > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > constants from Euler-Largrange equations:
> >
> > I mark now for convenience T = coordinate time and t = proper time.
> >
> > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> >
> > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> >
> > K1 = +,- 0.7072727132*I,
> > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> >
> > And I selected here randomly as an example two constants of integration
> > in this my two analytic solutions calculation:
> >
> > K1 = - 0.7072727132*I
> > and
> > K2 = 0.5943942676 - 0.5943942676*I
> >
> > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> >
> > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> >
> > Those both plots resemble somehow pendulum orbit ?
> >
> > I have NO physical interpretations of these solutions
> > and I think that these have NO real physical applications.
> >
> > Hannu Poropudas
> I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
>
> It seems to me that this integral is too complicated to calculate analytically, but it could be so
> with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> in this above case that the coordinate time T could be two dimensional complex number ?
>
> This also seems to support what I said above.
> I have NO physical interpretations of these solutions
> and I think at the moment that these have NO real physical applications.
>
> And we should study two dimensional complex mathematics of two dimensional
> coordinate time (T) in this complicated integral better,
> if we try to better understand this situation,
> if this would be sensible at all ?
>
> Best Regards,
> Hannu Poropudas

CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

On Wednesday, October 25, 2023 at 2:01:55 AM UTC-7, Hannu Poropudas wrote:
> tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > Spherically symmetric metrics which satisfies
> > > > > > Einstein's vacuum field equations.
> > > > > >
> > > > > > (c=1,G=1 units)
> > > > > >
> > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > >
> > > > > > (c=1,G=1 units)
> > > > > >
> > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > >
> > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > >
> > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > >
> > > > > > There exist a book called something like
> > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > which have about 740 pages and
> > > > > > I don't know if this solution is among them?
> > > > > >
> > > > > > Three singularity points of the metrics are the following:
> > > > > >
> > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > >
> > > > > >
> > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > >
> > > > > > Reference:
> > > > > > Tolman R. C., 1934.
> > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > >
> > > > > > Best Regrads,
> > > > > >
> > > > > > Hannu Poropudas
> > > > > >
> > > > > > Kolamäentie 9E
> > > > > > 90900 Kiiminki / Oulu
> > > > > > Finland
> > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > >
> > > > > MG = 6.292090968*10^11,
> > > > > 2*MG=1.258418194*10^12.
> > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > >
> > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > +,- sign for integral
> > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > >
> > > > > and
> > > > >
> > > > > -1.103327381*10^12<=rr<=0
> > > > > +,- sign for integral
> > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > >
> > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > Real parts = 0 in these integrals.
> > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > How to interpret these Imaginary angle plots?
> > > > >
> > > > > Best Regards,
> > > > > Hannu Poropudas
> > > > Your solution is either:
> > > >
> > > > (a) incorrect, or:
> > > >
> > > > (b) isometric to Schwarzschild's.
> > > >
> > > > Don't waste your time.
> > > >
> > > > --
> > > > Jan
> > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > >
> > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > >
> > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > but you are correct in point of view that it may not be physically acceptable solution
> > > of these equations at our present orthodoxic physical knowledge.
> > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > >
> > > There exist also few other integration constants from Euler-Largrange equations,
> > > but I have selected randomly only one couple of them in this example calculation.
> > >
> > > Hannu
> > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > constants from Euler-Largrange equations:
> >
> > I mark now for convenience T = coordinate time and t = proper time.
> >
> > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> >
> > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> >
> > K1 = +,- 0.7072727132*I,
> > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> >
> > And I selected here randomly as an example two constants of integration
> > in this my two analytic solutions calculation:
> >
> > K1 = - 0.7072727132*I
> > and
> > K2 = 0.5943942676 - 0.5943942676*I
> >
> > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> >
> > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> >
> > Those both plots resemble somehow pendulum orbit ?
> >
> > I have NO physical interpretations of these solutions
> > and I think that these have NO real physical applications.
> >
> > Hannu Poropudas
> I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
>
> It seems to me that this integral is too complicated to calculate analytically, but it could be so
> with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> in this above case that the coordinate time T could be two dimensional complex number ?
>
> This also seems to support what I said above.
> I have NO physical interpretations of these solutions
> and I think at the moment that these have NO real physical applications.
>
> And we should study two dimensional complex mathematics of two dimensional
> coordinate time (T) in this complicated integral better,
> if we try to better understand this situation,
> if this would be sensible at all ?
>
> Best Regards,
> Hannu Poropudas

Again: all vacuum spherically symmetric solutions are locally
isometric to the Schwarzschild metric.

JanPB wrote:

>
> Stop wasting your time.

and he just wasted more of your time.

--
The Starmaker -- To question the unquestionable, ask the unaskable,
to think the unthinkable, mention the unmentionable, say the unsayable,
and challenge the unchallengeable.

keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > Einstein's vacuum field equations.
> > > > > > >
> > > > > > > (c=1,G=1 units)
> > > > > > >
> > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > >
> > > > > > > (c=1,G=1 units)
> > > > > > >
> > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > >
> > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > >
> > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > >
> > > > > > > There exist a book called something like
> > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > which have about 740 pages and
> > > > > > > I don't know if this solution is among them?
> > > > > > >
> > > > > > > Three singularity points of the metrics are the following:
> > > > > > >
> > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > >
> > > > > > >
> > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > >
> > > > > > > Reference:
> > > > > > > Tolman R. C., 1934.
> > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > >
> > > > > > > Best Regrads,
> > > > > > >
> > > > > > > Hannu Poropudas
> > > > > > >
> > > > > > > Kolamäentie 9E
> > > > > > > 90900 Kiiminki / Oulu
> > > > > > > Finland
> > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > >
> > > > > > MG = 6.292090968*10^11,
> > > > > > 2*MG=1.258418194*10^12.
> > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > >
> > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > +,- sign for integral
> > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > >
> > > > > > and
> > > > > >
> > > > > > -1.103327381*10^12<=rr<=0
> > > > > > +,- sign for integral
> > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > >
> > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > Real parts = 0 in these integrals.
> > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > How to interpret these Imaginary angle plots?
> > > > > >
> > > > > > Best Regards,
> > > > > > Hannu Poropudas
> > > > > Your solution is either:
> > > > >
> > > > > (a) incorrect, or:
> > > > >
> > > > > (b) isometric to Schwarzschild's.
> > > > >
> > > > > Don't waste your time.
> > > > >
> > > > > --
> > > > > Jan
> > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > >
> > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > >
> > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > of these equations at our present orthodoxic physical knowledge.
> > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > >
> > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > but I have selected randomly only one couple of them in this example calculation.
> > > >
> > > > Hannu
> > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > constants from Euler-Largrange equations:
> > >
> > > I mark now for convenience T = coordinate time and t = proper time.
> > >
> > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > >
> > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > >
> > > K1 = +,- 0.7072727132*I,
> > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > >
> > > And I selected here randomly as an example two constants of integration
> > > in this my two analytic solutions calculation:
> > >
> > > K1 = - 0.7072727132*I
> > > and
> > > K2 = 0.5943942676 - 0.5943942676*I
> > >
> > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > >
> > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > >
> > > Those both plots resemble somehow pendulum orbit ?
> > >
> > > I have NO physical interpretations of these solutions
> > > and I think that these have NO real physical applications.
> > >
> > > Hannu Poropudas
> > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> >
> > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > in this above case that the coordinate time T could be two dimensional complex number ?
> >
> > This also seems to support what I said above.
> > I have NO physical interpretations of these solutions
> > and I think at the moment that these have NO real physical applications..
> >
> > And we should study two dimensional complex mathematics of two dimensional
> > coordinate time (T) in this complicated integral better,
> > if we try to better understand this situation,
> > if this would be sensible at all ?
> >
> > Best Regards,
> > Hannu Poropudas
> CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> Sorry that I confused these two letters.
>
> Hannu

torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > Einstein's vacuum field equations.
> > > > > > > >
> > > > > > > > (c=1,G=1 units)
> > > > > > > >
> > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > >
> > > > > > > > (c=1,G=1 units)
> > > > > > > >
> > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > >
> > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > >
> > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > >
> > > > > > > > There exist a book called something like
> > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > which have about 740 pages and
> > > > > > > > I don't know if this solution is among them?
> > > > > > > >
> > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > >
> > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > >
> > > > > > > >
> > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > >
> > > > > > > > Reference:
> > > > > > > > Tolman R. C., 1934.
> > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > >
> > > > > > > > Best Regrads,
> > > > > > > >
> > > > > > > > Hannu Poropudas
> > > > > > > >
> > > > > > > > Kolamäentie 9E
> > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > Finland
> > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > >
> > > > > > > MG = 6.292090968*10^11,
> > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > >
> > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > +,- sign for integral
> > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > >
> > > > > > > and
> > > > > > >
> > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > +,- sign for integral
> > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > >
> > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > Real parts = 0 in these integrals.
> > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > How to interpret these Imaginary angle plots?
> > > > > > >
> > > > > > > Best Regards,
> > > > > > > Hannu Poropudas
> > > > > > Your solution is either:
> > > > > >
> > > > > > (a) incorrect, or:
> > > > > >
> > > > > > (b) isometric to Schwarzschild's.
> > > > > >
> > > > > > Don't waste your time.
> > > > > >
> > > > > > --
> > > > > > Jan
> > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > >
> > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > >
> > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > of these equations at our present orthodoxic physical knowledge.
> > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > >
> > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > >
> > > > > Hannu
> > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > constants from Euler-Largrange equations:
> > > >
> > > > I mark now for convenience T = coordinate time and t = proper time.
> > > >
> > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > >
> > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > >
> > > > K1 = +,- 0.7072727132*I,
> > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > >
> > > > And I selected here randomly as an example two constants of integration
> > > > in this my two analytic solutions calculation:
> > > >
> > > > K1 = - 0.7072727132*I
> > > > and
> > > > K2 = 0.5943942676 - 0.5943942676*I
> > > >
> > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > >
> > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > >
> > > > Those both plots resemble somehow pendulum orbit ?
> > > >
> > > > I have NO physical interpretations of these solutions
> > > > and I think that these have NO real physical applications.
> > > >
> > > > Hannu Poropudas
> > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > >
> > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > in this above case that the coordinate time T could be two dimensional complex number ?
> > >
> > > This also seems to support what I said above.
> > > I have NO physical interpretations of these solutions
> > > and I think at the moment that these have NO real physical applications.
> > >
> > > And we should study two dimensional complex mathematics of two dimensional
> > > coordinate time (T) in this complicated integral better,
> > > if we try to better understand this situation,
> > > if this would be sensible at all ?
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > Sorry that I confused these two letters.
> >
> > Hannu
> I found one interesting reference, which show that there
> are really only few astrophysically significant exact solutions to Einstein's field equations.
>
> Ishak, M. 2015.
> Exact Solutions to Einstein's Equations in Astrophysics.
> Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> 33 pages.
> https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
>
> Please take a look.
>
> Best Regards,
> Hannu Poropudas

On Friday, October 27, 2023 at 12:46:55 AM UTC-7, Hannu Poropudas wrote:
> torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > >
> > > > > > > > > (c=1,G=1 units)
> > > > > > > > >
> > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > >
> > > > > > > > > (c=1,G=1 units)
> > > > > > > > >
> > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > >
> > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > >
> > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > >
> > > > > > > > > There exist a book called something like
> > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > which have about 740 pages and
> > > > > > > > > I don't know if this solution is among them?
> > > > > > > > >
> > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > >
> > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > >
> > > > > > > > > Reference:
> > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > >
> > > > > > > > > Best Regrads,
> > > > > > > > >
> > > > > > > > > Hannu Poropudas
> > > > > > > > >
> > > > > > > > > Kolamäentie 9E
> > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > Finland
> > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > >
> > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > >
> > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > +,- sign for integral
> > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > >
> > > > > > > > and
> > > > > > > >
> > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > +,- sign for integral
> > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > >
> > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > >
> > > > > > > > Best Regards,
> > > > > > > > Hannu Poropudas
> > > > > > > Your solution is either:
> > > > > > >
> > > > > > > (a) incorrect, or:
> > > > > > >
> > > > > > > (b) isometric to Schwarzschild's.
> > > > > > >
> > > > > > > Don't waste your time.
> > > > > > >
> > > > > > > --
> > > > > > > Jan
> > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > >
> > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > >
> > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > of these equations at our present orthodoxic physical knowledge..
> > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > >
> > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > >
> > > > > > Hannu
> > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > constants from Euler-Largrange equations:
> > > > >
> > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > >
> > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > >
> > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > >
> > > > > K1 = +,- 0.7072727132*I,
> > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > >
> > > > > And I selected here randomly as an example two constants of integration
> > > > > in this my two analytic solutions calculation:
> > > > >
> > > > > K1 = - 0.7072727132*I
> > > > > and
> > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > >
> > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > >
> > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > >
> > > > > Those both plots resemble somehow pendulum orbit ?
> > > > >
> > > > > I have NO physical interpretations of these solutions
> > > > > and I think that these have NO real physical applications.
> > > > >
> > > > > Hannu Poropudas
> > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > >
> > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > >
> > > > This also seems to support what I said above.
> > > > I have NO physical interpretations of these solutions
> > > > and I think at the moment that these have NO real physical applications.
> > > >
> > > > And we should study two dimensional complex mathematics of two dimensional
> > > > coordinate time (T) in this complicated integral better,
> > > > if we try to better understand this situation,
> > > > if this would be sensible at all ?
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > Sorry that I confused these two letters.
> > >
> > > Hannu
> > I found one interesting reference, which show that there
> > are really only few astrophysically significant exact solutions to Einstein's field equations.
> >
> > Ishak, M. 2015.
> > Exact Solutions to Einstein's Equations in Astrophysics.
> > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > 33 pages.
> > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> >
> > Please take a look.
> >
> > Best Regards,
> > Hannu Poropudas
> In order to me more mathematically complete I calculate also
> approximate proper time t integral (primitive function)
> and plotted both real part and imaginary part of it.
> I have NO interpretations of these.
>
> ># Approximate proper time t integral calculated HP 27.10.2023
> ># REMARK: My letter convenience t=proper time T=coordinate time
> ># Real part and Imaginary part plotted
> >#K3:=0;
> >#K1 := -0.7072727132*I;
> >#K2 := 0.5943942676-0.5943942676*I;
> >#m := MG;
> >#MG := 0.6292090968e12;
> >#2*MG := 0.1258418194e13;
> >#a2<=r<=a1, definition area
> >#a2=2.720522631*10^11, a1=8.306841627*10^11
>
> >#Real part of primitive function t approx.
> ># Series approx at r = MG up to 7 degree.
> >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0..6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
>
> >#Imaginary part of primitive function t approx.
> >># Series approx at r = MG up to 7 degree.
> >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0..6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
>
> > #a2<=r<=a1, definition area
> > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > #MG := 0.6292090968e12;
> > #2*MG := 0.1258418194e13;
>
> >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
>
> > #a2<=r<=a1, definition area
> > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > #MG := 0.6292090968e12;
> > #2*MG := 0.1258418194e13;
>
> >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
>
>
> Best Regards,
> Hannu Poropudas

perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > >
> > > > > > > > > (c=1,G=1 units)
> > > > > > > > >
> > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > >
> > > > > > > > > (c=1,G=1 units)
> > > > > > > > >
> > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > >
> > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > >
> > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > >
> > > > > > > > > There exist a book called something like
> > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > which have about 740 pages and
> > > > > > > > > I don't know if this solution is among them?
> > > > > > > > >
> > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > >
> > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > >
> > > > > > > > > Reference:
> > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > >
> > > > > > > > > Best Regrads,
> > > > > > > > >
> > > > > > > > > Hannu Poropudas
> > > > > > > > >
> > > > > > > > > Kolamäentie 9E
> > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > Finland
> > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > >
> > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > >
> > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > +,- sign for integral
> > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > >
> > > > > > > > and
> > > > > > > >
> > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > +,- sign for integral
> > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > >
> > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > >
> > > > > > > > Best Regards,
> > > > > > > > Hannu Poropudas
> > > > > > > Your solution is either:
> > > > > > >
> > > > > > > (a) incorrect, or:
> > > > > > >
> > > > > > > (b) isometric to Schwarzschild's.
> > > > > > >
> > > > > > > Don't waste your time.
> > > > > > >
> > > > > > > --
> > > > > > > Jan
> > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > >
> > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > >
> > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > of these equations at our present orthodoxic physical knowledge..
> > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > >
> > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > >
> > > > > > Hannu
> > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > constants from Euler-Largrange equations:
> > > > >
> > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > >
> > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > >
> > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > >
> > > > > K1 = +,- 0.7072727132*I,
> > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > >
> > > > > And I selected here randomly as an example two constants of integration
> > > > > in this my two analytic solutions calculation:
> > > > >
> > > > > K1 = - 0.7072727132*I
> > > > > and
> > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > >
> > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > >
> > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > >
> > > > > Those both plots resemble somehow pendulum orbit ?
> > > > >
> > > > > I have NO physical interpretations of these solutions
> > > > > and I think that these have NO real physical applications.
> > > > >
> > > > > Hannu Poropudas
> > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > >
> > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > >
> > > > This also seems to support what I said above.
> > > > I have NO physical interpretations of these solutions
> > > > and I think at the moment that these have NO real physical applications.
> > > >
> > > > And we should study two dimensional complex mathematics of two dimensional
> > > > coordinate time (T) in this complicated integral better,
> > > > if we try to better understand this situation,
> > > > if this would be sensible at all ?
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > Sorry that I confused these two letters.
> > >
> > > Hannu
> > I found one interesting reference, which show that there
> > are really only few astrophysically significant exact solutions to Einstein's field equations.
> >
> > Ishak, M. 2015.
> > Exact Solutions to Einstein's Equations in Astrophysics.
> > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > 33 pages.
> > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> >
> > Please take a look.
> >
> > Best Regards,
> > Hannu Poropudas
> In order to me more mathematically complete I calculate also
> approximate proper time t integral (primitive function)
> and plotted both real part and imaginary part of it.
> I have NO interpretations of these.
>
> ># Approximate proper time t integral calculated HP 27.10.2023
> ># REMARK: My letter convenience t=proper time T=coordinate time
> ># Real part and Imaginary part plotted
> >#K3:=0;
> >#K1 := -0.7072727132*I;
> >#K2 := 0.5943942676-0.5943942676*I;
> >#m := MG;
> >#MG := 0.6292090968e12;
> >#2*MG := 0.1258418194e13;
> >#a2<=r<=a1, definition area
> >#a2=2.720522631*10^11, a1=8.306841627*10^11
>
> >#Real part of primitive function t approx.
> ># Series approx at r = MG up to 7 degree.
> >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0..6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
>
> >#Imaginary part of primitive function t approx.
> >># Series approx at r = MG up to 7 degree.
> >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0..6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
>
> > #a2<=r<=a1, definition area
> > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > #MG := 0.6292090968e12;
> > #2*MG := 0.1258418194e13;
>
> >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
>
> > #a2<=r<=a1, definition area
> > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > #MG := 0.6292090968e12;
> > #2*MG := 0.1258418194e13;
>
> >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
>
>
> Best Regards,
> Hannu Poropudas

On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
> perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> > torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > > >
> > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > >
> > > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > > >
> > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > >
> > > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > > >
> > > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > > >
> > > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > > >
> > > > > > > > > > There exist a book called something like
> > > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > > which have about 740 pages and
> > > > > > > > > > I don't know if this solution is among them?
> > > > > > > > > >
> > > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > > >
> > > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > > >
> > > > > > > > > >
> > > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > > >
> > > > > > > > > > Reference:
> > > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > > >
> > > > > > > > > > Best Regrads,
> > > > > > > > > >
> > > > > > > > > > Hannu Poropudas
> > > > > > > > > >
> > > > > > > > > > Kolamäentie 9E
> > > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > > Finland
> > > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > > some time ago in this sci.physics.relativity Google Group.. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > > >
> > > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > > >
> > > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > > +,- sign for integral
> > > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > > >
> > > > > > > > > and
> > > > > > > > >
> > > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > > +,- sign for integral
> > > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > > >
> > > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > > >
> > > > > > > > > Best Regards,
> > > > > > > > > Hannu Poropudas
> > > > > > > > Your solution is either:
> > > > > > > >
> > > > > > > > (a) incorrect, or:
> > > > > > > >
> > > > > > > > (b) isometric to Schwarzschild's.
> > > > > > > >
> > > > > > > > Don't waste your time.
> > > > > > > >
> > > > > > > > --
> > > > > > > > Jan
> > > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > > >
> > > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > > >
> > > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > > of these equations at our present orthodoxic physical knowledge.
> > > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > > >
> > > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > > >
> > > > > > > Hannu
> > > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > > constants from Euler-Largrange equations:
> > > > > >
> > > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > > >
> > > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > > >
> > > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > > >
> > > > > > K1 = +,- 0.7072727132*I,
> > > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > > >
> > > > > > And I selected here randomly as an example two constants of integration
> > > > > > in this my two analytic solutions calculation:
> > > > > >
> > > > > > K1 = - 0.7072727132*I
> > > > > > and
> > > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > > >
> > > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > > >
> > > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > > >
> > > > > > Those both plots resemble somehow pendulum orbit ?
> > > > > >
> > > > > > I have NO physical interpretations of these solutions
> > > > > > and I think that these have NO real physical applications.
> > > > > >
> > > > > > Hannu Poropudas
> > > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > > >
> > > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > > >
> > > > > This also seems to support what I said above.
> > > > > I have NO physical interpretations of these solutions
> > > > > and I think at the moment that these have NO real physical applications.
> > > > >
> > > > > And we should study two dimensional complex mathematics of two dimensional
> > > > > coordinate time (T) in this complicated integral better,
> > > > > if we try to better understand this situation,
> > > > > if this would be sensible at all ?
> > > > >
> > > > > Best Regards,
> > > > > Hannu Poropudas
> > > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > > Sorry that I confused these two letters.
> > > >
> > > > Hannu
> > > I found one interesting reference, which show that there
> > > are really only few astrophysically significant exact solutions to Einstein's field equations.
> > >
> > > Ishak, M. 2015.
> > > Exact Solutions to Einstein's Equations in Astrophysics.
> > > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > > 33 pages.
> > > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> > >
> > > Please take a look.
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > In order to me more mathematically complete I calculate also
> > approximate proper time t integral (primitive function)
> > and plotted both real part and imaginary part of it.
> > I have NO interpretations of these.
> >
> > ># Approximate proper time t integral calculated HP 27.10.2023
> > ># REMARK: My letter convenience t=proper time T=coordinate time
> > ># Real part and Imaginary part plotted
> > >#K3:=0;
> > >#K1 := -0.7072727132*I;
> > >#K2 := 0.5943942676-0.5943942676*I;
> > >#m := MG;
> > >#MG := 0.6292090968e12;
> > >#2*MG := 0.1258418194e13;
> > >#a2<=r<=a1, definition area
> > >#a2=2.720522631*10^11, a1=8.306841627*10^11
> >
> > >#Real part of primitive function t approx.
> > ># Series approx at r = MG up to 7 degree.
> > >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
> >
> > >#Imaginary part of primitive function t approx.
> > >># Series approx at r = MG up to 7 degree.
> > >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
> >
> > > #a2<=r<=a1, definition area
> > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > #MG := 0.6292090968e12;
> > > #2*MG := 0.1258418194e13;
> >
> > >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
> >
> > > #a2<=r<=a1, definition area
> > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > #MG := 0.6292090968e12;
> > > #2*MG := 0.1258418194e13;
> >
> > >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
> >
> >
> > Best Regards,
> > Hannu Poropudas
> Error estimations of these two series approximations (function - (series approx function), not integrated here):
>
> For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
> Max negative error about -2.15 near 2.72*10^11,
> Max negative error about -0.2 near 8.307*10^11.
>
> For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
> Max negative error about -1.09 near 2.72*10^11,
> Max negative error about -0.1 near 8.307*10^11.
>
> These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?
>
> Best Regards,
> Hannu Poropudas

On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
> perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> > torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > > >
> > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > >
> > > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > > >
> > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > >
> > > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > > >
> > > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > > >
> > > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > > >
> > > > > > > > > > There exist a book called something like
> > > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > > which have about 740 pages and
> > > > > > > > > > I don't know if this solution is among them?
> > > > > > > > > >
> > > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > > >
> > > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > > >
> > > > > > > > > >
> > > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > > >
> > > > > > > > > > Reference:
> > > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > > >
> > > > > > > > > > Best Regrads,
> > > > > > > > > >
> > > > > > > > > > Hannu Poropudas
> > > > > > > > > >
> > > > > > > > > > Kolamäentie 9E
> > > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > > Finland
> > > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > > some time ago in this sci.physics.relativity Google Group.. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > > >
> > > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > > >
> > > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > > +,- sign for integral
> > > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > > >
> > > > > > > > > and
> > > > > > > > >
> > > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > > +,- sign for integral
> > > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > > >
> > > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > > >
> > > > > > > > > Best Regards,
> > > > > > > > > Hannu Poropudas
> > > > > > > > Your solution is either:
> > > > > > > >
> > > > > > > > (a) incorrect, or:
> > > > > > > >
> > > > > > > > (b) isometric to Schwarzschild's.
> > > > > > > >
> > > > > > > > Don't waste your time.
> > > > > > > >
> > > > > > > > --
> > > > > > > > Jan
> > > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > > >
> > > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > > >
> > > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > > of these equations at our present orthodoxic physical knowledge.
> > > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > > >
> > > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > > >
> > > > > > > Hannu
> > > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > > constants from Euler-Largrange equations:
> > > > > >
> > > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > > >
> > > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > > >
> > > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > > >
> > > > > > K1 = +,- 0.7072727132*I,
> > > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > > >
> > > > > > And I selected here randomly as an example two constants of integration
> > > > > > in this my two analytic solutions calculation:
> > > > > >
> > > > > > K1 = - 0.7072727132*I
> > > > > > and
> > > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > > >
> > > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > > >
> > > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > > >
> > > > > > Those both plots resemble somehow pendulum orbit ?
> > > > > >
> > > > > > I have NO physical interpretations of these solutions
> > > > > > and I think that these have NO real physical applications.
> > > > > >
> > > > > > Hannu Poropudas
> > > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > > >
> > > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > > >
> > > > > This also seems to support what I said above.
> > > > > I have NO physical interpretations of these solutions
> > > > > and I think at the moment that these have NO real physical applications.
> > > > >
> > > > > And we should study two dimensional complex mathematics of two dimensional
> > > > > coordinate time (T) in this complicated integral better,
> > > > > if we try to better understand this situation,
> > > > > if this would be sensible at all ?
> > > > >
> > > > > Best Regards,
> > > > > Hannu Poropudas
> > > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > > Sorry that I confused these two letters.
> > > >
> > > > Hannu
> > > I found one interesting reference, which show that there
> > > are really only few astrophysically significant exact solutions to Einstein's field equations.
> > >
> > > Ishak, M. 2015.
> > > Exact Solutions to Einstein's Equations in Astrophysics.
> > > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > > 33 pages.
> > > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> > >
> > > Please take a look.
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > In order to me more mathematically complete I calculate also
> > approximate proper time t integral (primitive function)
> > and plotted both real part and imaginary part of it.
> > I have NO interpretations of these.
> >
> > ># Approximate proper time t integral calculated HP 27.10.2023
> > ># REMARK: My letter convenience t=proper time T=coordinate time
> > ># Real part and Imaginary part plotted
> > >#K3:=0;
> > >#K1 := -0.7072727132*I;
> > >#K2 := 0.5943942676-0.5943942676*I;
> > >#m := MG;
> > >#MG := 0.6292090968e12;
> > >#2*MG := 0.1258418194e13;
> > >#a2<=r<=a1, definition area
> > >#a2=2.720522631*10^11, a1=8.306841627*10^11
> >
> > >#Real part of primitive function t approx.
> > ># Series approx at r = MG up to 7 degree.
> > >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
> >
> > >#Imaginary part of primitive function t approx.
> > >># Series approx at r = MG up to 7 degree.
> > >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
> >
> > > #a2<=r<=a1, definition area
> > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > #MG := 0.6292090968e12;
> > > #2*MG := 0.1258418194e13;
> >
> > >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
> >
> > > #a2<=r<=a1, definition area
> > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > #MG := 0.6292090968e12;
> > > #2*MG := 0.1258418194e13;
> >
> > >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
> >
> >
> > Best Regards,
> > Hannu Poropudas
> Error estimations of these two series approximations (function - (series approx function), not integrated here):
>
> For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
> Max negative error about -2.15 near 2.72*10^11,
> Max negative error about -0.2 near 8.307*10^11.
>
> For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
> Max negative error about -1.09 near 2.72*10^11,
> Max negative error about -0.1 near 8.307*10^11.
>
> These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?
>
> Best Regards,
> Hannu Poropudas

On Saturday, October 28, 2023 at 1:02:58 PM UTC-7, patdolan wrote:
> On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
> > perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> > > torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > > > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > > > >
> > > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > > >
> > > > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > > > >
> > > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > > >
> > > > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > > > >
> > > > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > > > >
> > > > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > > > >
> > > > > > > > > > > There exist a book called something like
> > > > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > > > which have about 740 pages and
> > > > > > > > > > > I don't know if this solution is among them?
> > > > > > > > > > >
> > > > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > > > >
> > > > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > > > >
> > > > > > > > > > >
> > > > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > > > >
> > > > > > > > > > > Reference:
> > > > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > > > >
> > > > > > > > > > > Best Regrads,
> > > > > > > > > > >
> > > > > > > > > > > Hannu Poropudas
> > > > > > > > > > >
> > > > > > > > > > > Kolamäentie 9E
> > > > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > > > Finland
> > > > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > > > >
> > > > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > > > >
> > > > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > > > +,- sign for integral
> > > > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > > > >
> > > > > > > > > > and
> > > > > > > > > >
> > > > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > > > +,- sign for integral
> > > > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1..103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > > > >
> > > > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > > > >
> > > > > > > > > > Best Regards,
> > > > > > > > > > Hannu Poropudas
> > > > > > > > > Your solution is either:
> > > > > > > > >
> > > > > > > > > (a) incorrect, or:
> > > > > > > > >
> > > > > > > > > (b) isometric to Schwarzschild's.
> > > > > > > > >
> > > > > > > > > Don't waste your time.
> > > > > > > > >
> > > > > > > > > --
> > > > > > > > > Jan
> > > > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > > > >
> > > > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > > > >
> > > > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > > > of these equations at our present orthodoxic physical knowledge.
> > > > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > > > >
> > > > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > > > >
> > > > > > > > Hannu
> > > > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > > > constants from Euler-Largrange equations:
> > > > > > >
> > > > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > > > >
> > > > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > > > >
> > > > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > > > >
> > > > > > > K1 = +,- 0.7072727132*I,
> > > > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > > > >
> > > > > > > And I selected here randomly as an example two constants of integration
> > > > > > > in this my two analytic solutions calculation:
> > > > > > >
> > > > > > > K1 = - 0.7072727132*I
> > > > > > > and
> > > > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > > > >
> > > > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance)..
> > > > > > >
> > > > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > > > >
> > > > > > > Those both plots resemble somehow pendulum orbit ?
> > > > > > >
> > > > > > > I have NO physical interpretations of these solutions
> > > > > > > and I think that these have NO real physical applications.
> > > > > > >
> > > > > > > Hannu Poropudas
> > > > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > > > >
> > > > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > > > >
> > > > > > This also seems to support what I said above.
> > > > > > I have NO physical interpretations of these solutions
> > > > > > and I think at the moment that these have NO real physical applications.
> > > > > >
> > > > > > And we should study two dimensional complex mathematics of two dimensional
> > > > > > coordinate time (T) in this complicated integral better,
> > > > > > if we try to better understand this situation,
> > > > > > if this would be sensible at all ?
> > > > > >
> > > > > > Best Regards,
> > > > > > Hannu Poropudas
> > > > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > > > Sorry that I confused these two letters.
> > > > >
> > > > > Hannu
> > > > I found one interesting reference, which show that there
> > > > are really only few astrophysically significant exact solutions to Einstein's field equations.
> > > >
> > > > Ishak, M. 2015.
> > > > Exact Solutions to Einstein's Equations in Astrophysics.
> > > > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > > > 33 pages.
> > > > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> > > >
> > > > Please take a look.
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > In order to me more mathematically complete I calculate also
> > > approximate proper time t integral (primitive function)
> > > and plotted both real part and imaginary part of it.
> > > I have NO interpretations of these.
> > >
> > > ># Approximate proper time t integral calculated HP 27.10.2023
> > > ># REMARK: My letter convenience t=proper time T=coordinate time
> > > ># Real part and Imaginary part plotted
> > > >#K3:=0;
> > > >#K1 := -0.7072727132*I;
> > > >#K2 := 0.5943942676-0.5943942676*I;
> > > >#m := MG;
> > > >#MG := 0.6292090968e12;
> > > >#2*MG := 0.1258418194e13;
> > > >#a2<=r<=a1, definition area
> > > >#a2=2.720522631*10^11, a1=8.306841627*10^11
> > >
> > > >#Real part of primitive function t approx.
> > > ># Series approx at r = MG up to 7 degree.
> > > >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
> > >
> > > >#Imaginary part of primitive function t approx.
> > > >># Series approx at r = MG up to 7 degree.
> > > >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
> > >
> > > > #a2<=r<=a1, definition area
> > > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > #MG := 0.6292090968e12;
> > > > #2*MG := 0.1258418194e13;
> > >
> > > >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
> > >
> > > > #a2<=r<=a1, definition area
> > > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > #MG := 0.6292090968e12;
> > > > #2*MG := 0.1258418194e13;
> > >
> > > >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
> > >
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > Error estimations of these two series approximations (function - (series approx function), not integrated here):
> >
> > For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
> > Max negative error about -2.15 near 2.72*10^11,
> > Max negative error about -0.2 near 8.307*10^11.
> >
> > For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
> > Max negative error about -1.09 near 2.72*10^11,
> > Max negative error about -0.1 near 8.307*10^11.
> >
> > These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?
> >
> > Best Regards,
> > Hannu Poropudas
> Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation. Will YOU, Hannu Poropudas? If not, why not?

On Saturday, October 28, 2023 at 1:19:28 PM UTC-7, Ross Finlayson wrote:
> On Saturday, October 28, 2023 at 1:02:58 PM UTC-7, patdolan wrote:
> > On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
> > > perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > > > > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > > > > >
> > > > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > > > >
> > > > > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > > > > >
> > > > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > > > >
> > > > > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > > > > >
> > > > > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > > > > >
> > > > > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > > > > >
> > > > > > > > > > > > There exist a book called something like
> > > > > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > > > > which have about 740 pages and
> > > > > > > > > > > > I don't know if this solution is among them?
> > > > > > > > > > > >
> > > > > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > > > > >
> > > > > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > > > > >
> > > > > > > > > > > >
> > > > > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > > > > >
> > > > > > > > > > > > Reference:
> > > > > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > > > > >
> > > > > > > > > > > > Best Regrads,
> > > > > > > > > > > >
> > > > > > > > > > > > Hannu Poropudas
> > > > > > > > > > > >
> > > > > > > > > > > > Kolamäentie 9E
> > > > > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > > > > Finland
> > > > > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > > > > >
> > > > > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > > > > >
> > > > > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > > > > +,- sign for integral
> > > > > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > > > > >
> > > > > > > > > > > and
> > > > > > > > > > >
> > > > > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > > > > +,- sign for integral
> > > > > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > > > > >
> > > > > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > > > > >
> > > > > > > > > > > Best Regards,
> > > > > > > > > > > Hannu Poropudas
> > > > > > > > > > Your solution is either:
> > > > > > > > > >
> > > > > > > > > > (a) incorrect, or:
> > > > > > > > > >
> > > > > > > > > > (b) isometric to Schwarzschild's.
> > > > > > > > > >
> > > > > > > > > > Don't waste your time.
> > > > > > > > > >
> > > > > > > > > > --
> > > > > > > > > > Jan
> > > > > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > > > > >
> > > > > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > > > > >
> > > > > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > > > > of these equations at our present orthodoxic physical knowledge.
> > > > > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > > > > >
> > > > > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > > > > >
> > > > > > > > > Hannu
> > > > > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > > > > constants from Euler-Largrange equations:
> > > > > > > >
> > > > > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > > > > >
> > > > > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > > > > >
> > > > > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > > > > >
> > > > > > > > K1 = +,- 0.7072727132*I,
> > > > > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > > > > >
> > > > > > > > And I selected here randomly as an example two constants of integration
> > > > > > > > in this my two analytic solutions calculation:
> > > > > > > >
> > > > > > > > K1 = - 0.7072727132*I
> > > > > > > > and
> > > > > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > > > > >
> > > > > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > > > > >
> > > > > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > > > > >
> > > > > > > > Those both plots resemble somehow pendulum orbit ?
> > > > > > > >
> > > > > > > > I have NO physical interpretations of these solutions
> > > > > > > > and I think that these have NO real physical applications.
> > > > > > > >
> > > > > > > > Hannu Poropudas
> > > > > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > > > > >
> > > > > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > > > > >
> > > > > > > This also seems to support what I said above.
> > > > > > > I have NO physical interpretations of these solutions
> > > > > > > and I think at the moment that these have NO real physical applications.
> > > > > > >
> > > > > > > And we should study two dimensional complex mathematics of two dimensional
> > > > > > > coordinate time (T) in this complicated integral better,
> > > > > > > if we try to better understand this situation,
> > > > > > > if this would be sensible at all ?
> > > > > > >
> > > > > > > Best Regards,
> > > > > > > Hannu Poropudas
> > > > > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > > > > Sorry that I confused these two letters.
> > > > > >
> > > > > > Hannu
> > > > > I found one interesting reference, which show that there
> > > > > are really only few astrophysically significant exact solutions to Einstein's field equations.
> > > > >
> > > > > Ishak, M. 2015.
> > > > > Exact Solutions to Einstein's Equations in Astrophysics.
> > > > > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > > > > 33 pages.
> > > > > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> > > > >
> > > > > Please take a look.
> > > > >
> > > > > Best Regards,
> > > > > Hannu Poropudas
> > > > In order to me more mathematically complete I calculate also
> > > > approximate proper time t integral (primitive function)
> > > > and plotted both real part and imaginary part of it.
> > > > I have NO interpretations of these.
> > > >
> > > > ># Approximate proper time t integral calculated HP 27.10.2023
> > > > ># REMARK: My letter convenience t=proper time T=coordinate time
> > > > ># Real part and Imaginary part plotted
> > > > >#K3:=0;
> > > > >#K1 := -0.7072727132*I;
> > > > >#K2 := 0.5943942676-0.5943942676*I;
> > > > >#m := MG;
> > > > >#MG := 0.6292090968e12;
> > > > >#2*MG := 0.1258418194e13;
> > > > >#a2<=r<=a1, definition area
> > > > >#a2=2.720522631*10^11, a1=8.306841627*10^11
> > > >
> > > > >#Real part of primitive function t approx.
> > > > ># Series approx at r = MG up to 7 degree.
> > > > >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
> > > >
> > > > >#Imaginary part of primitive function t approx.
> > > > >># Series approx at r = MG up to 7 degree.
> > > > >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
> > > >
> > > > > #a2<=r<=a1, definition area
> > > > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > > #MG := 0.6292090968e12;
> > > > > #2*MG := 0.1258418194e13;
> > > >
> > > > >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
> > > >
> > > > > #a2<=r<=a1, definition area
> > > > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > > #MG := 0.6292090968e12;
> > > > > #2*MG := 0.1258418194e13;
> > > >
> > > > >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
> > > >
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > Error estimations of these two series approximations (function - (series approx function), not integrated here):
> > >
> > > For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
> > > Max negative error about -2.15 near 2.72*10^11,
> > > Max negative error about -0.2 near 8.307*10^11.
> > >
> > > For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
> > > Max negative error about -1.09 near 2.72*10^11,
> > > Max negative error about -0.1 near 8.307*10^11.
> > >
> > > These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?
> > >
> > > Best Regards,
> > > Hannu Poropudas
> > Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation. Will YOU, Hannu Poropudas? If not, why not?
> Paddy what you do is go down to a flea market and find an old scale according to
> spring pendulum or the off-weight, toward a balance scale, and maybe a known
> fixed mass weight for a balance, then also one of those new densitometers of
> the sorts of bathroom scales, or whatever's considered dead-weight, then you
> yourself can confirm small differences among the two configurations of experiment,
> then explain how clocks drifted last year back around to being on time.

On Saturday, October 28, 2023 at 7:33:23 PM UTC-7, patdolan wrote:
> On Saturday, October 28, 2023 at 1:19:28 PM UTC-7, Ross Finlayson wrote:
> > On Saturday, October 28, 2023 at 1:02:58 PM UTC-7, patdolan wrote:
> > > On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
> > > > perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > > torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > > > > > >
> > > > > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > > > > >
> > > > > > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > > > > > >
> > > > > > > > > > > > > (c=1,G=1 units)
> > > > > > > > > > > > >
> > > > > > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > > > > > >
> > > > > > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > > > > > >
> > > > > > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > > > > > >
> > > > > > > > > > > > > There exist a book called something like
> > > > > > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > > > > > which have about 740 pages and
> > > > > > > > > > > > > I don't know if this solution is among them?
> > > > > > > > > > > > >
> > > > > > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > > > > > >
> > > > > > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > > > > > >
> > > > > > > > > > > > >
> > > > > > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Reference:
> > > > > > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Best Regrads,
> > > > > > > > > > > > >
> > > > > > > > > > > > > Hannu Poropudas
> > > > > > > > > > > > >
> > > > > > > > > > > > > Kolamäentie 9E
> > > > > > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > > > > > Finland
> > > > > > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > > > > > >
> > > > > > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > > > > > >
> > > > > > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > > > > > +,- sign for integral
> > > > > > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > > > > > >
> > > > > > > > > > > > and
> > > > > > > > > > > >
> > > > > > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > > > > > +,- sign for integral
> > > > > > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > > > > > >
> > > > > > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > > > > > >
> > > > > > > > > > > > Best Regards,
> > > > > > > > > > > > Hannu Poropudas
> > > > > > > > > > > Your solution is either:
> > > > > > > > > > >
> > > > > > > > > > > (a) incorrect, or:
> > > > > > > > > > >
> > > > > > > > > > > (b) isometric to Schwarzschild's.
> > > > > > > > > > >
> > > > > > > > > > > Don't waste your time.
> > > > > > > > > > >
> > > > > > > > > > > --
> > > > > > > > > > > Jan
> > > > > > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > > > > > >
> > > > > > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > > > > > >
> > > > > > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > > > > > of these equations at our present orthodoxic physical knowledge.
> > > > > > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > > > > > >
> > > > > > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > > > > > >
> > > > > > > > > > Hannu
> > > > > > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > > > > > constants from Euler-Largrange equations:
> > > > > > > > >
> > > > > > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > > > > > >
> > > > > > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > > > > > >
> > > > > > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > > > > > >
> > > > > > > > > K1 = +,- 0.7072727132*I,
> > > > > > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > > > > > >
> > > > > > > > > And I selected here randomly as an example two constants of integration
> > > > > > > > > in this my two analytic solutions calculation:
> > > > > > > > >
> > > > > > > > > K1 = - 0.7072727132*I
> > > > > > > > > and
> > > > > > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > > > > > >
> > > > > > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > > > > > (Phi(P) is pure imaginary angle and r(P) is real distance..
> > > > > > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > > > > > >
> > > > > > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > > > > > >
> > > > > > > > > Those both plots resemble somehow pendulum orbit ?
> > > > > > > > >
> > > > > > > > > I have NO physical interpretations of these solutions
> > > > > > > > > and I think that these have NO real physical applications..
> > > > > > > > >
> > > > > > > > > Hannu Poropudas
> > > > > > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > > > > > >
> > > > > > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > > > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > > > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > > > > > >
> > > > > > > > This also seems to support what I said above.
> > > > > > > > I have NO physical interpretations of these solutions
> > > > > > > > and I think at the moment that these have NO real physical applications.
> > > > > > > >
> > > > > > > > And we should study two dimensional complex mathematics of two dimensional
> > > > > > > > coordinate time (T) in this complicated integral better,
> > > > > > > > if we try to better understand this situation,
> > > > > > > > if this would be sensible at all ?
> > > > > > > >
> > > > > > > > Best Regards,
> > > > > > > > Hannu Poropudas
> > > > > > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > > > > > Sorry that I confused these two letters.
> > > > > > >
> > > > > > > Hannu
> > > > > > I found one interesting reference, which show that there
> > > > > > are really only few astrophysically significant exact solutions to Einstein's field equations.
> > > > > >
> > > > > > Ishak, M. 2015.
> > > > > > Exact Solutions to Einstein's Equations in Astrophysics.
> > > > > > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > > > > > 33 pages.
> > > > > > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> > > > > >
> > > > > > Please take a look.
> > > > > >
> > > > > > Best Regards,
> > > > > > Hannu Poropudas
> > > > > In order to me more mathematically complete I calculate also
> > > > > approximate proper time t integral (primitive function)
> > > > > and plotted both real part and imaginary part of it.
> > > > > I have NO interpretations of these.
> > > > >
> > > > > ># Approximate proper time t integral calculated HP 27.10.2023
> > > > > ># REMARK: My letter convenience t=proper time T=coordinate time
> > > > > ># Real part and Imaginary part plotted
> > > > > >#K3:=0;
> > > > > >#K1 := -0.7072727132*I;
> > > > > >#K2 := 0.5943942676-0.5943942676*I;
> > > > > >#m := MG;
> > > > > >#MG := 0.6292090968e12;
> > > > > >#2*MG := 0.1258418194e13;
> > > > > >#a2<=r<=a1, definition area
> > > > > >#a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > >
> > > > > >#Real part of primitive function t approx.
> > > > > ># Series approx at r = MG up to 7 degree.
> > > > > >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;
> > > > >
> > > > > >#Imaginary part of primitive function t approx.
> > > > > >># Series approx at r = MG up to 7 degree.
> > > > > >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;
> > > > >
> > > > > > #a2<=r<=a1, definition area
> > > > > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > > > #MG := 0.6292090968e12;
> > > > > > #2*MG := 0.1258418194e13;
> > > > >
> > > > > >plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);
> > > > >
> > > > > > #a2<=r<=a1, definition area
> > > > > > #a2=2.720522631*10^11, a1=8.306841627*10^11
> > > > > > #MG := 0.6292090968e12;
> > > > > > #2*MG := 0.1258418194e13;
> > > > >
> > > > > >plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
> > > > >
> > > > >
> > > > > Best Regards,
> > > > > Hannu Poropudas
> > > > Error estimations of these two series approximations (function - (series approx function), not integrated here):
> > > >
> > > > For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
> > > > Max negative error about -2.15 near 2.72*10^11,
> > > > Max negative error about -0.2 near 8.307*10^11.
> > > >
> > > > For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
> > > > Max negative error about -1.09 near 2.72*10^11,
> > > > Max negative error about -0.1 near 8.307*10^11.
> > > >
> > > > These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation. Will YOU, Hannu Poropudas? If not, why not?
> > Paddy what you do is go down to a flea market and find an old scale according to
> > spring pendulum or the off-weight, toward a balance scale, and maybe a known
> > fixed mass weight for a balance, then also one of those new densitometers of
> > the sorts of bathroom scales, or whatever's considered dead-weight, then you
> > yourself can confirm small differences among the two configurations of experiment,
> > then explain how clocks drifted last year back around to being on time.
> Ross, yes, that's how you do it. And it is easy to write out the equation(s) representing that experiment.
>
> I now request that any of the aforementioned write out the equations of the new gravitation, aka general relativity, so that we may check the results of the new theory against the old. I confidently reiterate that to do so is impossibly complex. And I give a my proof the fact that not a single denizen of this forum can or will do it; nor have they even deigned to respond why they won't do it. Hannu Poropudas and the rest just look at me as though I am crazy for even asking. Even though Hannu is in possession of a brand new metric contrived for this exact purpose.

perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
> torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
> > keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
> > > keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
> > > > tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
> > > > > perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
> > > > > > > On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
> > > > > > > > sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
> > > > > > > > > Spherically symmetric metrics which satisfies
> > > > > > > > > Einstein's vacuum field equations.
> > > > > > > > >
> > > > > > > > > (c=1,G=1 units)
> > > > > > > > >
> > > > > > > > > matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])
> > > > > > > > >
> > > > > > > > > (c=1,G=1 units)
> > > > > > > > >
> > > > > > > > > ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2
> > > > > > > > >
> > > > > > > > > (m -> m*G/c^2 , if SI-units are used.)
> > > > > > > > >
> > > > > > > > > I don't know that would this solution have any astrophysical applications?
> > > > > > > > >
> > > > > > > > > There exist a book called something like
> > > > > > > > > "Exact Solutions of the Einstein Field Equations",
> > > > > > > > > which have about 740 pages and
> > > > > > > > > I don't know if this solution is among them?
> > > > > > > > >
> > > > > > > > > Three singularity points of the metrics are the following:
> > > > > > > > >
> > > > > > > > > r = 0, r = m*G/c^2 and r = 2*m*G/c^2.
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.
> > > > > > > > >
> > > > > > > > > Reference:
> > > > > > > > > Tolman R. C., 1934.
> > > > > > > > > Effect of inhomogeneity on cosmological models.
> > > > > > > > > Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.
> > > > > > > > >
> > > > > > > > > Best Regrads,
> > > > > > > > >
> > > > > > > > > Hannu Poropudas
> > > > > > > > >
> > > > > > > > > Kolamäentie 9E
> > > > > > > > > 90900 Kiiminki / Oulu
> > > > > > > > > Finland
> > > > > > > > I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
> > > > > > > > due three integration constants from Euler-Lagrange equations does not have
> > > > > > > > same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
> > > > > > > > I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
> > > > > > > > some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)
> > > > > > > >
> > > > > > > > MG = 6.292090968*10^11,
> > > > > > > > 2*MG=1.258418194*10^12.
> > > > > > > > I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):
> > > > > > > >
> > > > > > > > 2.720522631*10^11<=r<=8.306841627*10^11
> > > > > > > > +,- sign for integral
> > > > > > > > phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
> > > > > > > > r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)
> > > > > > > >
> > > > > > > > and
> > > > > > > >
> > > > > > > > -1.103327381*10^12<=rr<=0
> > > > > > > > +,- sign for integral
> > > > > > > > phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
> > > > > > > > rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)
> > > > > > > >
> > > > > > > > I calculated also these integrals but their formulae are too long to copy here.
> > > > > > > > Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
> > > > > > > > Real parts = 0 in these integrals.
> > > > > > > > How to interpret pure imaginary phi and phiphi angles?
> > > > > > > > How to interpret these Imaginary angle plots?
> > > > > > > >
> > > > > > > > Best Regards,
> > > > > > > > Hannu Poropudas
> > > > > > > Your solution is either:
> > > > > > >
> > > > > > > (a) incorrect, or:
> > > > > > >
> > > > > > > (b) isometric to Schwarzschild's.
> > > > > > >
> > > > > > > Don't waste your time.
> > > > > > >
> > > > > > > --
> > > > > > > Jan
> > > > > > Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
> > > > > >
> > > > > > Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.
> > > > > >
> > > > > > Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
> > > > > > but you are correct in point of view that it may not be physically acceptable solution
> > > > > > of these equations at our present orthodoxic physical knowledge..
> > > > > > This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.
> > > > > >
> > > > > > There exist also few other integration constants from Euler-Largrange equations,
> > > > > > but I have selected randomly only one couple of them in this example calculation.
> > > > > >
> > > > > > Hannu
> > > > > I put here those strange (NO ordinary physical interpretation) formulae of integration
> > > > > constants from Euler-Largrange equations:
> > > > >
> > > > > I mark now for convenience T = coordinate time and t = proper time.
> > > > >
> > > > > (dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
> > > > > (1-2*m/r)*(dT/dt) = K2 (constant of integration)
> > > > > (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.
> > > > >
> > > > > I calculated for randomly selected numerical values of S2-star aphelion and perhelion
> > > > > distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
> > > > > for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
> > > > > to calculate two integration constants K1 and K2 of Euler-Largange equations
> > > > > (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):
> > > > >
> > > > > K1 = +,- 0.7072727132*I,
> > > > > K2 = +,- 0.5943942676 +,- 0.5943942676*I,
> > > > >
> > > > > And I selected here randomly as an example two constants of integration
> > > > > in this my two analytic solutions calculation:
> > > > >
> > > > > K1 = - 0.7072727132*I
> > > > > and
> > > > > K2 = 0.5943942676 - 0.5943942676*I
> > > > >
> > > > > This selection gave those two pure imaginary analytic solutions which I gave here earlier.
> > > > > (Phi(P) is pure imaginary angle and r(P) is real distance.
> > > > > Phiphi(P) is pure imaginary angle and rr(P) is real distance).
> > > > >
> > > > > Plot ([Im(phi(P)),r(P),P=0..Pi]);
> > > > > Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
> > > > > gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)
> > > > >
> > > > > Those both plots resemble somehow pendulum orbit ?
> > > > >
> > > > > I have NO physical interpretations of these solutions
> > > > > and I think that these have NO real physical applications.
> > > > >
> > > > > Hannu Poropudas
> > > > I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?
> > > >
> > > > It seems to me that this integral is too complicated to calculate analytically, but it could be so
> > > > with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
> > > > in this above case that the coordinate time T could be two dimensional complex number ?
> > > >
> > > > This also seems to support what I said above.
> > > > I have NO physical interpretations of these solutions
> > > > and I think at the moment that these have NO real physical applications.
> > > >
> > > > And we should study two dimensional complex mathematics of two dimensional
> > > > coordinate time (T) in this complicated integral better,
> > > > if we try to better understand this situation,
> > > > if this would be sensible at all ?
> > > >
> > > > Best Regards,
> > > > Hannu Poropudas
> > > CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
> > > Sorry that I confused these two letters.
> > >
> > > Hannu
> > I found one interesting reference, which show that there
> > are really only few astrophysically significant exact solutions to Einstein's field equations.
> >
> > Ishak, M. 2015.
> > Exact Solutions to Einstein's Equations in Astrophysics.
> > Texas Symposium on Relativistic Astrophysics, Geneva 2015.
> > 33 pages.
> > https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf
> >
> > Please take a look.
> >
> > Best Regards,
> > Hannu Poropudas
> In order to me more mathematically complete I calculate also
> approximate proper time t integral (primitive function)
> and plotted both real part and imaginary part of it.
> I have NO interpretations of these.
>
> ># Approximate proper time t integral calculated HP 27.10.2023
> ># REMARK: My letter convenience t=proper time T=coordinate time
> ># Real part and Imaginary part plotted
> >#K3:=0;
> >#K1 := -0.7072727132*I;
> >#K2 := 0.5943942676-0.5943942676*I;
> >#m := MG;
> >#MG := 0.6292090968e12;
> >#2*MG := 0.1258418194e13;
> >#a2<=r<=a1, definition area
> >#a2=2.720522631*10^11, a1=8.306841627*10^11
>
> >#Real part of primitive function t approx.
> ># Series approx at r = MG up to 7 degree.