On May 14, 9:08*pm, Koobee Wublee wrote:
On May 14, 7:10 am, Mike wrote:

On May 13, 5:46 pm, Koobee Wublee wrote:
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. *You are welcome to
transform to another coordinate system with a different metric.

No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.

Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

Is there even one person who agrees with you on this?


I think you are pounding on a non-issue. The *real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.

Hmmm... *You are very confused.

Thus, the solution is empirically falsified.

shrug

On May 15, 1:08*am, Koobee Wublee wrote:
On May 14, 7:10 am, Mike wrote:

On May 13, 5:46 pm, Koobee Wublee wrote:
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. *You are welcome to
transform to another coordinate system with a different metric.

No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.

Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

I think you are pounding on a non-issue. The *real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.

Hmmm... *You are very confused.

About what?

(a) DO yopu think that the Schwarzschild metric gines Minskowski
spaceitme in the limit?

(b) do you think Minkowski space time alone can explain gravity?

(c) Do you think is pretty scary that everyone else is confused but
you?

Mike





Thus, the solution is empirically falsified.

shrug

On May 14, 4:10 pm, Mike wrote:

The real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical
reality
there is gravity is such limit.

Yes, General Relativity has not Newtonian limit and usual derivations
of Newtonian limits are either incorrect or not care about
mathematical consistency.

Absence of Newtonian limit is directly related to the impossibility to
quantize General Relativity. Notice you can quantize Newtonian
gravity, the resulting equation is the famous Schrödinger-Newton
equation

http://en.wikipedia.org/wiki/Schr%C3...wton_equations

Those equation are already under experimentalists' target!

Notice one finds problems *only* with the geometrical approach to
gravity. Those problems are *absent* in nongeometrical theories of
gravity such as FTG

http://www.amazon.com/Feynman-Lectur.../dp/0201627345

or R-AAAD

http://arxiv.org/pdf/physics/0612019v10

The usual general relativist motto:

(\blockquote
Mass grips space by telling it how to curve area, space grips mass
by telling it how to move.
)

was once believed to be true.

--
My newsserver continues down

On May 15, 2:15*am, "Juan R."
wrote:
On May 14, 4:10 pm, Mike wrote:

The *real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical
reality
there is gravity is such limit.

Yes, General Relativity has not Newtonian limit and usual derivations
of Newtonian limits are either incorrect or not care about
mathematical consistency.

It really is too bad that you can't just /accept/ that nobody cares.
It is well understood - there is a whole page of discussion in MTW
about this - that linearized GR is inconsistent with itself. It is
also well understood _why_ the inconsistency is there and _why_ it
doesn't matter.

[snip]

On May 15, 6:45*am, Eric Gisse wrote:
On May 15, 2:15*am, "Juan R."
wrote:

On May 14, 4:10 pm, Mike wrote:

The *real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical
reality
there is gravity is such limit.

Yes, General Relativity has not Newtonian limit and usual derivations
of Newtonian limits are either incorrect or not care about
mathematical consistency.

It really is too bad that you can't just /accept/ that nobody cares.
It is well understood - there is a whole page of discussion in MTW
about this - that linearized GR is inconsistent with itself. It is
also well understood _why_ the inconsistency is there and _why_ it
doesn't matter.

Yep, inconsistencies do not matter in your inconsistent world full of
inconsistent brains like you.

Have you got that ODE thing straight yet?

http://groups.google.gr/group/sci.ph...b?dmode=source

Do you think it is consistent thiikning that you call a = GM/r^2 an
Euler ODE?


Mike



[snip]

On May 15, 6:15*am, "Juan R."
wrote:
On May 14, 4:10 pm, Mike wrote:

The *real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical
reality
there is gravity is such limit.

Yes, General Relativity has not Newtonian limit and usual derivations
of Newtonian limits are either incorrect or not care about
mathematical consistency.

Absence of Newtonian limit is directly related to the impossibility to
quantize General Relativity. Notice you can quantize Newtonian
gravity, the resulting equation is the famous Schrödinger-Newton
equation

http://en.wikipedia.org/wiki/Schr%C3...wton_equations

Those equation are already under experimentalists' target!


That equation hold the key to the future of physics when people
finally get to their senses and abandon SR and GR after realizing the
inconsistencies present in those theories/


Mike



Notice one finds problems *only* with the geometrical approach to
gravity. Those problems are *absent* in nongeometrical theories of
gravity such as FTG

http://www.amazon.com/Feynman-Lectur...tiers-Physics/...

or R-AAAD

http://arxiv.org/pdf/physics/0612019v10

The usual general relativist motto:

(\blockquote
*Mass grips space by telling it how to curve area, space grips mass
*by telling it how to move.
)

was once believed to be true.

--
My newsserver continues down

On May 15, 6:56 pm, Mike wrote:

That equation hold the key to the future of physics when people
finally get to their senses and abandon SR and GR after realizing
the
inconsistencies present in those theories/

I am not so optimistic.

A subset of 'believers' will continue to support and promote geometric
approach to gravity without caring about how many recent theoretical
or experimental evidence goes against.

On May 15, 12:42 am, Mike wrote:
On May 15, 1:08 am, Koobee Wublee wrote:

(a) DO yopu think that the Schwarzschild metric gines Minskowski
spaceitme in the limit?

Yes, of course. This should be a no brainer. shrug

(b) do you think Minkowski space time alone can explain gravity?

No. Gravity has nothing to do with a curvature in space. Only
gravitational time dilation manifests gravity according to spacetime.
Since Minkowski sapcetime does not manifest gravitational time
dilation, it cannot explain gravity. PERIOD.

(c) Do you think is pretty scary that everyone else is confused but
you?

No, the god everyone else believes in is very incompetent. shrug

On May 14, 10:16*pm, Koobee Wublee wrote:
On May 13, 11:51 pm, JanPB wrote:

On May 13, 2:48 pm, Koobee Wublee wrote:
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate.

Not _spacetime_, the _coordinate representation of the metric_. There
is no such thing as "spacetime" that can be "applied" to a coordinate
system. The concept doesn't exist.

You have grossly misunderstood what I said, and I cannot tell if it is
due to your incompetence or just being malicious. *There is no point
to go on.

You said, this is a direct quote: "The spacetime described by the
Schwarzschild metric only applied to the spherically symmetric polar
coordinate". Remember always to define your private terms. We cannot
guess what you mean by terms you invent for your own use. As it
stands, and without further definition of terms, the statement I
quoted is nonsensical.

If you use the words "applied to a coordinate system" in some other
manner that allows "spacetime" in this context, then define this
usage.

--
Jan Bielawski

On May 14, 10:08*pm, Koobee Wublee wrote:
On May 14, 7:10 am, Mike wrote:

On May 13, 5:46 pm, Koobee Wublee wrote:
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. *You are welcome to
transform to another coordinate system with a different metric.

No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.

Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

dx^2 + dy^2

and:

dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

For example, consider vector v defined in Cartesian coordinates like
this:

v is situated at point (1,1),
v = (1,2)

Then, applying ds^2 = dx^2 + dy^2:

ds^2(v) = v.v = dx(v) * dx(v) + dy(v) * dy(v) = 1^2 + 2^2 = 5

Now calculate using the polar representation of ds^2:

v is situated at (sqrt(2), pi/4),
v = (3/sqrt(2), -1/2)

Then, applying ds^2 = dr^2 + r^2 dtheta^2:

ds^2(v) = v.v = dr(v) * dr(v) + (sqrt(2))^2 * dtheta(v) *
dtheta(v) =

= (3/sqrt(2))^2 + 2 * (-1/2)^2 =

= 9/2 + 2 * 1/4 =

= 9/2 + 1/2 = 10/2 = 5

Same result! Yout hink that's an accident? :-) You'd obtain the
corresponding same results with any vector. It must be so because both
expressions for ds^2 are equal.

Same thing with Schwarzschild, etc.

The term "metric" as used by everyone refers to that mapping that
takes tangent vectors (like v) to their squared lengths (like v.v).
That mapping is coordinate-INdependent. OTOH its _coordinate
representation_ of course depends on coordinates (by definition). But
dx^2+dy^2 and dr^2 + r^2 dtheta^2 are two different coordinate
representations of the same metric (the same mapping taking v to v.v,
or equivalently, a mapping taking vector pairs v,w to v.w - that's
what metric _is_).

--
Jan Bielawski