Re: The Absurd Claim of the Metric as a Tensor
- From: "bergeron" <badd_xi2@xxxxxxxxx>
- Date: 15 Jan 2007 06:56:11 -0800
You would need 50 more IQ points to realize
that you're incompetent.
Koobee Wublee wrote:
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.
The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.
**** Pros' argument:
The geometry described is invariant under any transformation. Thus, it
can be represented by an operator operating on the same vector twice.
ds^2 = f(dq,dq)
Where
** ds^2 = the invariant geometry
** f() = the operator
** dq = the coordinate vector
Of course, we all know that
ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n
Where
** dq' = Different coordinate system
The question is the matrices ([g] = [g']).
**** Cons's argument:
According to the pros, although the elements of the matrices (g_ij =/=
g'_ij), the martrix ([g] = [g']). In doing so, they can never
describe what constitutes how these 2 matrices are identical. They
avoid it as if a plague in fact. They can only hand-wave it by saying
over and over again that they are indeed identical.
**** What is at stake?
Does it matter if the metric is a tensor or not for the sake of the
mathematics involved? Yes, it does. The interpretation to the
infinite number of solutions to the field equations is at stake. The
existence of the black holes is at stake.
Apparently, the pros have never followed through the derivation of the
solutions to the field equations. Each solution in terms of g_ij is
only valid to the choice of coordinate system where each solution very
different from the others must describe a different geometry using the
same coordinate system. This means the field equations do have an
infinite number of solutions in which the Schwarzschild metric,
Schwarzschild's original metric, or any other is just as valid as any
other where each describes a different geometry. This would shatter
the general theory of relativity.
The field equations in free space are
R_ij(q^0, q^1, q^2, q^3) = 0
Where
** R_ij(q) = Ricci tensor as a function of q
** g1_ij(q) = Solution as function of q
** g2_ij(q) = Solution as function of q
** g3_ij(q) = Solution as function of q
** ...
And all these different geometries described by each solution with the
same coordinate system.
** ds1^2 = g1_ij dq^i dq^j
** ds2^2 = g2_ij dq^i dq^j
** ds3^2 = g3_ij dq^i dq^j
** ...
The silly claim by the pros is that ([g1] = [g2] = [g3] = ...). Thus,
(ds1^2 = ds2^2 = ds^3 = ...). They are wrong. The simple mathematics
shows so. They cannot prove why ([g1] = [g2] = [g3] = ...). The
Schwarzschild metric is not unique. The existence of black holes is
based on a non-unique solution to the field equations. 100 years of
physics is totally BS based on this wrong concept of linear algebra.
.
- References:
- The Absurd Claim of the Metric as a Tensor
- From: Koobee Wublee
- The Absurd Claim of the Metric as a Tensor
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