On Jan 3, 3:02 am, Eric Gisse wrote:
On Jan 2, 10:42 pm, Koobee Wublee wrote:
On Jan 2, 12:03 am, Eric Gisse wrote:

ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2

They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.

Your notation is screwed up. The r of the 2nd is not the same as r of
the 3rd. The phi of the 2nd is the same as theta of the 3rd.
Otherwise, they can all represent the same geometry. shrug

See? You are a **** hair's distance away from understanding.

What is that again?

THE LABELS DO NOT MATTER.

Wrong! The labels do matter. shrug

The COORDINATES do not matter. They all
represent the _same_ geometry.

Wrong again. To describe a geometry, you need both the labels
(coordinate) and how the labels are connected (metric). shrug

This is not what I have been talking about. It looks like very few
have understood what I am saying. That is why I feel very lonely to
be the few who after Riemann and perhaps Christoffel that have
understood the curvature business very well. shrug

So where did you properly learn this "curvature business"? All I'm
asking is for the book you learned it from. Is that asking so much?

Just my intelligence which you lack. shrug

What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.

** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2

Where

** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

Except the coordinate labels are not the same in both coordinate
systems.

What do you mean they are not the same? I have chosen them to be the
same. shrug

Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.

I did. Just look up on my older post.

What coordinate transformation? u is any function of r. This is very
basic mathematics. shrug

Yes, and it is called a "coordinate transformation". You start out in
the (t,u,\theta,\phi) coordinate system and end up in the (t,r,\theta,
\phi) coordinate system.

I did not start out in anything. I merely presented these two
independent geometries of spacetime without showing any favor of one
to the other.

You are still dreaming and jacking off in your mathemagical Realm. In
doing so, probably wishing Einstein can join you. shrug

Let me write down the two very different spacetime in a more concise
way.

** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2

THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC
OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
BOTH ds1 and ds2 IS EXACTLY IDENTICAL.

If they are "exactly identical" they should give the same values for
the surface area of the sphere, as #1 is in Schwarzschild isotropic
coordinates which has spheres of constant (r,t) of surface area 4 pi
r^2.

They are not identical. shrug

Let us see you calculate the surface area of #2.

I would also like you calculate the Ricci scalars of both. You'll
notice something interesting.

Calculating the area does not make any difference. To any observer
observing the geometry of any curved spacetime, the answer is always
going to be 4 pi r^2. shrug

Can you dig this, punk? I don’t think you possess any minute
intelligence to get after all these years, and I really don’t care.
shrug

With any matheMagical tricks, show me how you can coordinate-transform
from one equation to the other and what justify one solution is more
unique than the others. Remember that the coordinate system is the
same with both equations.

Except the coordinate systems are different

Hey, I am the one presenting the problem. Both ds1 and ds2 share the
same coordinate system. They are all solved from the set of field
equations with the same coordinate system. shrug

- the second one is
displaced from the origin by an amount K, which you can see
immediately by looking at the coefficients of the angular components.
Or by calculating the surface area. Or by calculating the Ricci
scalar. Or by looking at the determinant of the metric.

Nonsense! There is no displacement of anything. There is no
coordinate transformation. shrug

I cannot claim one solution is more unique than the other, which is
the whole point of my explanation. I just prefer Schwarzschild in
isotropic coordinates. I can easily convert from Schwarzschild and
every /OTHER/ "different" solution to yours - just chain the
transformations.

Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in
isotropic coordinate system. Again, for the n’th time, these
solutions are static, spherically symmetric, and asymptotically flat.
I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by
example. shrug

Your ranting is because you refuse the God you worship is a nitwit, a
plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a
plagiarist, and a liar. Despite the nitwit, the plagiarist, and the
liar did not come up with all these nonsense, whatever is credited to
this nitwit, this plagiarist, and this liar is utter nonsense in the
first place. shrug