Re: GR1916 question about g=1?

On Nov 20, 5:45 pm, John C. Polasek <jpola...@xxxxxxxxxx> wrote:
On Tue, 20 Nov 2007 14:56:31 -0800 (PST), JanPB <film...@xxxxxxxxx>
wrote:



On Nov 19, 11:26 pm, Koobee Wublee <koobee.wub...@xxxxxxxxx> wrote:

The metric can only be a matrix. <shrug>

Metric is a map which to any manifold point assigns a symmetric
bilinear function on the tangent space at that point (namely, the dot
product of vectors tangent at that point).

As such it can be represented by a table (matrix) of coefficients
_given a basis_. This table is different for each coordinate system.
That's why one cannot think of tensor as a fixed matrix. One can think
of tensor as a family of matrices - one matrix per coordinate system -
connected by the formula:

B = XAX^T

...where A is the matrix of the metric in coordinate system (1), B is
the matrix of the metric in coordinate system (2), and X is the matrix
of basis change between (1) and (2).

XAXtr is a similarity transform which is clearly forbidden in the
-1111 domain.

This is all basic linear algebra.

But it's not basic physics. It's algebraic nonsense.
A tensor embodies the few parameters that define a physical object,
subject to transformations and rotations of that object that are
mathematically tenable and interesting and which cannot possibly alter
the physical object. All this mapping crap is so much sophistry. Get
real.
The metric tensor is a 4x4 matrix that fails any such test: g00 stands
transfixed, like a lion on the library steps, with its feet fixed in
concrete.
John Polasek

I tend to agree with Jan, a matrix is
usually an n=2 array of scalars, but a
tensor is a definition of all possible
components including covariant and contra-
variant.
I learned that the hard way.
I took a metric tensor, gave it some
specification then used it in general
tensor analysis to form new equations,
which we're not generally covariant.
(incidentally that's a common error).

Now I avoid specifying any background,
including the assumption of spacetime,
until the equations are derived. Then
freeze the equations and substitute in
whatever dimensional and CS specs.

I got so darn neurotic on that issue I
let the tensor indices depart from being
integers to enable fractional dimensions.

Here's an example,
Suppose one were to differentiate x^2
(an area) to get 2x (a length). Usually
in tensors, an area is defined by a rank
2 tensor and length by a rank 1 tensor.
In simple terms, we're used to implicitly
thinking of reducing dimensionality by
differentiation, and increasing dimensions
by integration.

However the process of fractional
differentiation is logical, see here,
http://en.wikipedia.org/wiki/Fractional_calculus

((Technical Note: Integrate "0", "n"
times where "n" is >0.

$.....$ 0 dX = q X^n / n!
n

where "q" is a constant, and n! is
the factorial function.

When that reasoning is appled to the
shifting orbit of Mercury it worked,
using the radial dimensionality,

Force = GMm/r^(2.000 000 18).

What is r^(2.000 000 18) ?

It's between an area and a volume.

Suppose we find a field that varies as
r^(2.5)? Perhaps in subatomics.

What is the dimensionality of r^(2.5) ?

Let's write a tensor r^e where "e" is
not an integer, but we'll spec {u,v}
as being integers, then set, e = u + du,
to produce the Fractional Interdimensional
Transformation (FIT), then,

r^e = r^u + r^du .

In orbital physics, Mercury's orbit displays
that FIT, with the residual (000 000 18)
accounted for by r^du, at some particular
radius, (Mercury's).

So the applicable integer indice {u}
that applies in Newton's universe is
modified by a FIT to be r^e in AE's
universe with the quantity desired
being the FIT "du".

In mathematical terms, the usual process
of differentiation and integration needs
to be subtly adjusted in non-orthogonal
geometry, and normally this is done using
the Christoffel symbol in the covariant
derivative, which provides the usual means
of calulating Mercurys orbital relativistic
anomally.
The defect of that procedure may be the
requirement of a fixed spacetime 4D integer
background, while the FIT moves us in the
direction of being background free by
enabling the indices to be variables.

We can look deeper and ask, can these
indices be continuosly variable, or are
they quantized?

Have a look at "du", can du->0 continuously
or does du ->0 by du'=du-Du where "Du" is
the minimum variation of dimensionality?

The larger question is: Is dimensionality
quantized?

How does one produce an Area from a length?

From the above integral, set n=1,
to obtain the "virtual" area qx that
implicitly assumes q is vector
perpendicular to x.

Only when q=/=0 can a universe of dimensions
be founded by the process of integration,
hence, the use of integration depends upon
q=/=0 in our universe.

With vector "q>" perpendicular to vector
"X>" we can form an new vector,

S> = X> + q>

where the scalar product X>.q> =0,
follows naturally as does,

S>.S> = S^2 = X^2 + h , h = q>.q> .

We are still in a universe formed by
the calculus using q=/=0, from which
we can derive,

S dS = X dX,

as the foundation of General Covariance
iff "h" is constant, as specified in the
originating calculus, from the integration
of "0" above.

Now to demonstrate the problem, I need
to presume each complete integation
generates perpendicular *orthogonal*
dimensions such that,

X^2 = d_uv X^u X^v = X^u X^u = X_u X_u,

with d_uv being the metric tensor in an
orthogonal space, equivalent to the
Kronecker delta, and in the geometry of
nonorthogonality,

S^2 = g_uv X^u X^v = X^2 + h

where the g_uv = d_uv + A_u B_v,

and

h = A_u B_v X^u X^v.

At this point the product "A_u B_v"
provides a maximum of opportunity,
to determine how the original calculus
(q=/=0) can be described by tensor
analysis, and without any background
assumptions.

In physics I can close the program by
assuming "h" is Planck's constant and
"A_u" and "B_v" are potentials, however
by doing so I'll prejudice the ongoing
calculation.
Regards
Ken S. Tucker
kxsxt7

.

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