On May 7, 11:43*am, Koobee Wublee wrote:
[snip]

Once again, I have told you there is no such transformation. *shrug
Each solution to the field equations is independent of each other
using the same set of coordinate system.

Then prove it.


Do you not know how, or do you somehow think that proves nothing?

Can you dig that, multi-year super-senior?

On May 7, 2:40 pm, Eric Gisse wrote:
On May 7, 11:43 am, Koobee Wublee wrote:

Once again, I have told you there is no such transformation. shrug
Each solution to the field equations is independent of each other
using the same set of coordinate system.

Then prove it.

The troll sign comes out of you once again. As a multi-year super-
senior, is this not getting old?

I have already presented unchallengeable arguments to my claim. You
need to address my claim instead of resorting to these trollish
behaviors. shrug

On May 6, 9:55*pm, Koobee Wublee wrote:
On May 6, 8:28 pm, JanPB wrote:

On May 6, 2:24 pm, Koobee Wublee wrote:
Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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

It's the same solution as Schwarzschild's, you merely changed the
numbers labelling the spheres of symmetry.

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.

You did not specify whether you wrote both formulas with respect to
the same coordinate system or not.

Assuming you meant the same coordinates in both formulas, then you are
WRONG: the _first_ formula is NOT a solution. You can verify (it's
quite tedious) that in fact it doesn't satisfy Ricci=0. The second one
does.

On the other hand, if you intended the first formula to be obtained by
substituing r in the second, then you are also WRONG: in that case
both formulas describe the same dot product (i.e., the same solution).

It's exactly the same with dr in polar coordinates and x/
sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
covectors. Different formulas summing up to the same value. Same thing
when these covectors sit inside a formula for ds^2.

Well, both are solutions to the field equations using the spherically
symmetric polar coordinate system and nothing else.

So you say "the" spherically symmetric polar, so it looks like you
indeed intend both expressions for ds^2 to be written in the same
coordinate system. In this case your first formula is not a solution.

There are no
coordinate transformations involved. *They are just independent
solutions to the field equations. *shrug

Only the second one is a solution in this case.

As I said before, for a mere $99,999, I will design the universe of
your choice from the field equations.

You've just failed.

[...]

--
Jan Bielawski

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

These solutions are solved after the specific forms of
the field equations are laid out.

Not "after the specific forms of the field equations are laid out" but
after a coordinate system has been chosen. This choice is only a
computational convenience, not anything fundamental.

You are still wrong. shrug

These solutions are solved after the specific forms ofthe field
equations are laid out.

At this moment, the choice of
coordinate system should be well established, and no one can change
the coordinate. shrug

What you've just said is like: once the choice of inches as units has
been made during the calculation, no one can convert the result to
centimetres.

No, that is not what I said. Of course, you are free to convert it to
whatever coordinate system of your choice after the solutions are
derived. However, at these solutions only apply to one same
particular set of coordinate system. shrug

I beg your pardon. Is there a typo on my part?

It's not a solution if the (t,r,theta,phi) are the same as in the
second (Schwarzschild's) formula.

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).

Both should satisfy G_ij = 0, and thus R_ij = 0.

No, the first one doesn't if its (t,r,theta,phi) are the same as those
in the second formula.

shrug

On the other hand, if you intended the first formula to be obtained by
substituing r in the second, then you are also WRONG: in that case
both formulas describe the same dot product (i.e., the same solution).

There you go again worshipping that dot product as if a true God.
shrug

Saying "both formulas describe the same dot product" is worshipping
dot product? How?

Because you believe in nonsense. shrug

Both equations do not describe the same dot product.

It's exactly the same with dr in polar coordinates and x/
sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
covectors. Different formulas summing up to the same value. Same thing
when these covectors sit inside a formula for ds^2.

Yes?

Well, if they are equal, then what miracle suddenly makes them unequal
when they are inserted in the expression for ds^2?

But both equations do not specify the coordinate system being (t, x,
y, z). shrug

Accusation is very cheap. shrug

So are the two formulas written wrt to one coordinate system or two?
If two, what transformation did you use?

I am telling you both equations employ the same coordinate system.
Now, I see you are resorting back to your trollish behavior. There is
no point to continue any further, and in the future, please don’t
bring up that you have discussed this issue thoroughly with me because
it never did. You always resort to this kind of trollish behavior.
Now, go back to that fat castle in the air that you call GR and
fantasize about yourself being the queer of England, you majesty.

On May 8, 12:08*pm, Koobee Wublee wrote:

[snip]


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).

This is where your misunderstandings flare up. The coordinate labels
do not matter, but the definitions of the coordinates /do/ and if the /
definitions/ or (t,t,theta,phi) are the same as in Schwarzschild they
are different solutions. However if they are just /labels/, then they
are the same solution because I can [and have] found the isomorphism
between the two supposedly different solutions.

This relates to your inability to compute the area of a sphere - since
you blindingly assert '4pir^2' over and over rather than actually
compute the answer, you cannot see not only why the modern
Schwarzschild solution is preferred but what makes it different from
Hilbert's solution and other iterations.

Plus JanPB is _more_ than capable of computing elements of
differential geometry on his own - he does not need a software
package. The software just makes it crazy easy to dispute your
stupidities.

[snip]

On May 8, 9:34 pm, Eric Gisse wrote:
[snip crap]

This is where your misunderstandings flare up. The coordinate labels
do not matter, but the definitions of the coordinates /do/ and if the /
definitions/ or (t,t,theta,phi) are the same as in Schwarzschild they
are different solutions. However if they are just /labels/, then they
are the same solution because I can [and have] found the isomorphism
between the two supposedly different solutions.


You even don't know what a isomorphism is, ****head!

On May 8, 12:56*pm, Albertito wrote:
On May 8, 9:34 pm, Eric Gisse wrote:
[snip crap]

This is where your misunderstandings flare up. The coordinate labels
do not matter, but the definitions of the coordinates /do/ and if the /
definitions/ or (t,t,theta,phi) are the same as in Schwarzschild they
are different solutions. However if they are just /labels/, then they
are the same solution because I can [and have] found the isomorphism
between the two supposedly different solutions.

You even don't know what a isomorphism is, ****head!

And you do? Can you define an isomorphism in your own words?

On May 8, 3:03 am, Eric Gisse wrote:
On May 7, 10:48 pm, Koobee Wublee wrote:

The troll sign comes out of you once again. As a multi-year super-
senior, is this not getting old?

I have already presented unchallengeable arguments to my claim. You
need to address my claim instead of resorting to these trollish
behaviors. shrug

Your argument is nothing even close to "unchallengeable" - all you do
is assert some crap about "metric times coordinate" and therefore
state what the **** ever.

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

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’.

On May 8, 3:36*pm, Koobee Wublee wrote:
On May 8, 3:03 am, Eric Gisse wrote:

On May 7, 10:48 pm, Koobee Wublee wrote:
The troll sign comes out of you once again. *As a multi-year super-
senior, is this not getting old?

I have already presented unchallengeable arguments to my claim. *You
need to address my claim instead of resorting to these trollish
behaviors. *shrug

Your argument is nothing even close to "unchallengeable" - all you do
is assert some crap about "metric times coordinate" and therefore
state what the **** ever.

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.