Re: For everyone who thinks they can revolutionize science...

On Jun 20, 8:51 am, Koobee Wublee <koobee.wub...@xxxxxxxxx> wrote:
On Jun 20, 1:27 am, Eric Gisse <jowr...@xxxxxxxxx> wrote:

Who was it that claims the Minkowski metric does not have a
Lagrangian? Oh thats right...YOU

The metric of spacetime does not have a valid Lagrangian. Hilbert's
Lagrangian that gives birth to the field equations is not a Lagrangian
according to the principle of least action. <shrug>

Disproved by explicit construction. You are wrong.


Who was it that claims one can introduce curvature through a
coordinate transformation? YOU.

No, not I.

Liar.

http://groups.google.com/group/sci.physics.relativity/msg/3a57c809c15fa348?dmode=source


Geometry is invariant. In order to describe this invariant geometry
using a particular set of coordinate system, a proper metric must be
utilized.

For example, if you are near-sighted, you need a particular pair of
prescription glasses to see the world in perfection. The world is the
geometry, your unaided vision is the coordinate system, and your
glasses are the metric. If the world is not distorted, yes you can
say the glasses reflect the real world. However, if the world is
curved, you can never tell how much curved by examining the glasses
alone.

This means the metric cannot be unique. The metric alone cannot
possibly describe the geometry if the geometry can be curved. You
will also need to know the coordinate system to complete your quest.
<shrug>

Being an idiot is not half-way to being an idiot savant. Your faulty
line of arguing is worthless not just because it is, as usual,
mathematics free but because it is WRONG.

You are completely ignorant of differential geometry and its'
application to general relativity. For example - the [Riemann]
curvature tensor is defined only in terms of derivatives of the
metric. The metric _IS_ the geometry - it will look different in
different coordinate systems because the coordinate representation of
the metric is the result of projecting the metric components down
along a particular coordinate axis.


Who was it that claims tensors are not invariant? YOU.

I don't really give a damn about tensors. The metric tensor and the
Ricci tensor are nothing more than 4-by-4 matrices. The Riemann
curvature tensor is a 4-by-4-by-4by-4 matrix. <shrug>

Wrong again. You /still/ don't know the difference between a linear
operator [matrix] and a bilinear operator [tensor].

The Riemann curvature tensor is NOT a 4x4x4x4 matrix, it is
represented by a set of 16 4x4 matrices.


I have claimed the metric cannot be invariant. See above. And so is
the Riemann curvature tensor, the Ricci curvature tensor, and the
Ricci scalar.

That's right - you CLAIM. That is all you do - claim. You are unable
to PROVE, with mathematics, anything you have said.

All you have to do is do some simple calculations to prove yourself
right.


Who was it that claims special relativity cannot handle the twins
paradox? YOU.

Yes, yours truly has decided on that based on the mathematics of the
Lorentz transformation. <shrug>

That's your problem.

I gave you a solution to the twins paradox, and you simply reject it.
Not based upon any error I made, but because you think it is garbage.


The twin's paradox results because of the combination of both of the
following properties of the Lorentz transformation.

** Time dilation
** Principle of relatiivity

The only way to resolve this paradox is to prove either or both of the
above wrong. That means to abandon the Lorentz transformation.
<shrug>

Yet more claims without proof.

Why don't you USE special relativity and prove your statements?


You are unfit to judge.

If you don't understand it, you are unfit to judge. <shrug>


.

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