On May 8, 3:37*pm, Koobee Wublee wrote:
On May 8, 1:34 pm, Eric Gisse wrote:

On May 8, 12:08 pm, Koobee Wublee wrote:
Have you personally plug it into the field equations and verify that.
Eric Gisse has the software to do so. *I am sure if that is not a
solution, someone would have complained long ago. *So, the first
equation is still a valid solution to the field equations using the
same coordinate system of (t, r, theta, phi) as the second equation
(the Schwarzschild metric).

[crap snipped]

I say ‘get lost’.

cry?

On May 8, 4:54 pm, Eric Gisse wrote:
On May 8, 3:36 pm, Koobee Wublee wrote:

Go back to read my older posts, and don’t come back without
understanding them. Now, get lost.

I have a better idea - I'll continue to respond to your stupidities
and call them out as such, then laugh as you ineffectively defend
yourself with your shield of stupidity.

Well, I have even a better idea. You can choose to ignore you stupid
and trollish comments. Next time, whenever you try to bring up how I
did not address your stupid and trollish comments, you will know why.
That is assuming you possess some rudimentary reasoning. If not, it
is no sweat. That is why you remain a multi-year super-senior at the
University of Alaska majoring in basket weaving. shrug

On May 8, 1:08 pm, Koobee Wublee wrote:
On May 8, 1:23 am, JanPB wrote:

On May 8, 12:21 am, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– r^4 dO^2 / K^2

The Schwarzschild metric has the following form.

ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2

For the queer of England to continue living in that fat castle in the
air, he/she/it needs to talk himself/herself/itself into believing
both of the above equations are exactly the same.

Both coordinate systems are exactly the same. shrug

Then the first formula is not a solution. Calculate its Ricci tensor
and you'll see it's nonzero.

(I'll skip the rest.)

--
Jan Bielawski

On May 9, 1:11 am, JanPB wrote:
On May 8, 1:08 pm, Koobee Wublee wrote:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– r^4 dO^2 / K^2

The Schwarzschild metric has the following form.

ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2

Then the first formula is not a solution. Calculate its Ricci tensor
and you'll see it's nonzero.

Oh, you are correct. It is my mistake --- a typo. Please allow me to
correct the first equation.

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– K^2 (1 + r^4 / K^4) dO^2

The following is also a solution where the gravitational force follows
the inverse cubed law.

ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
– r^4 dO^2 / K^2

On May 8, 4:55 pm, Eric Gisse wrote:

cry?

Yes, crying won’t do you any good. You remain a multi-year super-
senior. shrug

On May 9, 1:11 am, JanPB wrote:
On May 8, 1:08 pm, Koobee Wublee wrote:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– r^4 dO^2 / K^2

The Schwarzschild metric has the following form.

ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2

Then the first formula is not a solution. Calculate its Ricci tensor
and you'll see it's nonzero.

Oh, you are correct. It is my mistake --- a typo. Please allow me to
correct the first equation.

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

The following is also a solution where the gravitational force follows
the inverse cubed law.

ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
– r^4 dO^2 / K^2

On May 9, 12:32 pm, Koobee Wublee wrote:
On May 9, 1:11 am, JanPB wrote:

On May 8, 1:08 pm, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– r^4 dO^2 / K^2

The Schwarzschild metric has the following form.

ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2

Then the first formula is not a solution. Calculate its Ricci tensor
and you'll see it's nonzero.

Oh, you are correct. It is my mistake --- a typo. Please allow me to
correct the first equation.

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci curvature is still nonzero if r denotes the same coordinate
as in Schwarzschild. Try it. I recommend Cartan moving frame method as
it's much faster than Christoffel symbols.

The following is also a solution where the gravitational force follows
the inverse cubed law.

ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
– r^4 dO^2 / K^2

Same thing: if r is the same then Ricci of the above is nonzero. OTOH
if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
0 but in that case it's the same solution as Schwarzschild - you
simply change the labels you attach to the spheres from "r" to "r^2/
K".

--
Jan Bielawski

On May 10, 3:09 am, JanPB wrote:
On May 9, 12:32 pm, Koobee Wublee wrote:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci curvature is still nonzero if r denotes the same coordinate
as in Schwarzschild. Try it.

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

I recommend Cartan moving frame method as
it's much faster than Christoffel symbols.

There is no short cut in such type of mathematics. shrug

The following is also a solution where the gravitational force follows
the inverse cubed law.

ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
– r^4 dO^2 / K^2

Same thing: if r is the same then Ricci of the above is nonzero.

But mathematically, its Ricci and Einstein tensors also vanish.
shrug

OTOH
if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
0 but in that case it's the same solution as Schwarzschild - you
simply change the labels you attach to the spheres from "r" to "r^2/
K".

Perhaps, there is a flaw in your Cartan whatever. As I said, anyone
who possesses mathematical software like Eric Gisse can easily verify
it, and I do not see any complaint from that multi-year super-senior.
shrug

On May 10, 8:42*pm, Koobee Wublee wrote:
On May 10, 3:09 am, JanPB wrote:

On May 9, 12:32 pm, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
* * *– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci curvature is still nonzero if r denotes the same coordinate
as in Schwarzschild. Try it.

Its Ricci and Einstein tensors both vanish. *In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Then evaluate surface area of a sphere in both coordinate systems.
You'll see something interesting if you actually do the computation.


I recommend Cartan moving frame method as
it's much faster than Christoffel symbols.

There is no short cut in such type of mathematics. *shrug

Based on your years of not studying the subject?


The following is also a solution where the gravitational force follows
the inverse cubed law.

ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
* * *– r^4 dO^2 / K^2

Same thing: if r is the same then Ricci of the above is nonzero.

But mathematically, its Ricci and Einstein tensors also vanish.
shrug

Not if they use the Schwarzschild coordinate system.


OTOH
if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
0 but in that case it's the same solution as Schwarzschild - you
simply change the labels you attach to the spheres from "r" to "r^2/
K".

Perhaps, there is a flaw in your Cartan whatever. *As I said, anyone
who possesses mathematical software like Eric Gisse can easily verify
it, and I do not see any complaint from that multi-year super-senior.
shrug

That's because we [JanPB, me] understand that the r you use is
different from the r in Schwarzschild even if you don't.

You appear to think that the coordinates determine the physics. You
claim otherwise and you can't explain why Minkowski space in Cartesian
coordinates is different from Minkowski space in spherical
coordinates, but when presented with the Schwarzschild metric in two
different coordinates your GR myopia sets in and you spray ink to
avoid discussing the subject.

On May 10, 9:42 pm, Koobee Wublee wrote:
On May 10, 3:09 am, JanPB wrote:

On May 9, 12:32 pm, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci curvature is still nonzero if r denotes the same coordinate
as in Schwarzschild. Try it.

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Calculate the Ricci tensor. It's not zero. The R_theta,theta is
probably the easiest nonzero component to compute.

I recommend Cartan moving frame method as
it's much faster than Christoffel symbols.

There is no short cut in such type of mathematics. shrug

Cartan's method utilizes the skew symmetry of certain combinations of
Christoffel symbols (rather than symmetry of the symbols themselves).
It allows one to refactor the computations in a more efficient way. In
this regard it's a bit like the Euler-Lagrange equation which also
cleverly refactors the rather ugly brute-force general-coordinate form
of "F=ma".

The following is also a solution where the gravitational force follows
the inverse cubed law.

ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
– r^4 dO^2 / K^2

Same thing: if r is the same then Ricci of the above is nonzero.

But mathematically, its Ricci and Einstein tensors also vanish.
shrug

Does "mathematically vanish" mean something different than "vanish"?

All I'm saying is that Ricci curvatures of the two metrics you wrote
are nonzero if you think of (t,r,theta,phi) as the coordinate system
of Schwarzschild.

OTOH
if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
0 but in that case it's the same solution as Schwarzschild - you
simply change the labels you attach to the spheres from "r" to "r^2/
K".

Perhaps, there is a flaw in your Cartan whatever.

Use Christoffel symbols then. You'll see the terms don't cancel.

As I said, anyone
who possesses mathematical software like Eric Gisse can easily verify
it, and I do not see any complaint from that multi-year super-senior.
shrug

By all means, use a computer if you prefer.

--
Jan Bielawski