Re: Curvature vs matter by KST

On Jul 23, 11:51 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
My general thinking is to begin with the 4th rank
Riemann Christoffel (RC) Curvature tensor I'll denote
R_abcd. If the RC tensor is zero then the field can
be transformed to a state where the metric "g_uv"
are constants, indicative of no perturbing fields,
either EM or gravitational, thus defining empty space,
given by,

R_abcd = 0 == empty space. (1)

(I invite comments here).

Trying to do GR again? This will be fun to watch. I can't wait to see
how you will abuse "constant".


Presuming (1) is true, then

R_abcd =/=0 == a field effect, (2)

causal by the presence of matter, (non empty).

....and, wrong. Trivial counterpoint: Schwarzschild geometry.

There are plenty of geometries that are vacuum solutions but retain
curvature components.

Should we experiment with using (2) to provide
us with a description of matter and fields?

Let me write the RHS of (2) as,

R_abcd = T_abcd (3).

Of course one can find an invariant in a non-empty
space by,

R_abcd - T_abcd = 0 (4),

proving a "non-empty" space inclusive of matter
and fields can be described by a 4th rank tensor,
(invariant) defined by (4).

You have defined nothing. You set something equal to itself then took
the difference.


To recap, the components in (3) are not all zero,
hence the "g_uv" are not constant,

Now would be a good time for you to explain "constant", because there
are at least two versions of constant.

Covariantly constant? The metric satisfies that by definition.

Constant in coefficients? Not a chance. That is something that is only
true locally.

[snip remaining]


.

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