T. Bunn, J Baez, D.R. Lunsford and I have reached an
impasse in the thread,
Subject: The Meaning of the Einstein equation, redux
Newsgroups: sci.physics.research
Date: 2004-08-19 10:36:13 PST
concerning the invariance of Newton's g-constant "G".
Einstein in his classic GR assumed G is be a constant, and
therefore an implicit invariant. Schwarzschild in his famous
solution (SS), (ref Tolman's Relativity, Thermo... bottom of
pg. 203 followed by calling "2m" a constant, also implicitly
invariant.
Lunsford and Tucker have detailed reasons why G is not an
invariant. The ramifications impact the "exactness" of the SS,
and therefore the foundation of Black hole theory. It may also
help understand why G_uv = (- 8*pi*G/c^2)*T_uv does not precisely
define energy conservation as it was intended to do.
Some thoughts...
Each value allowed into GR must first pass the less stringent
test of SR. For example, the SS implies "m" to be invariant, but
how does that hold true relatively to a rapidly moving reference
K', that transforms the inertial and gravitational mass to,
m' = m * gamma , gamma = 1/sqr(1 - v^2/c^2)
likewise, radius is transformed in SR by
r' = r / gamma and as a result m/r is not an invariant in SR.
So how can we reasonably expect that to be invariant in GR?
In weak circumstances GR provides for Mass units to be expressed
in length units using L = (G/c^2)*M , however this relation can't
hold in SR. Instead a relation like,
L^u = K^uv E_v is necessary
where E_v is the Mass/Energy vector components, L^u is it's
expression in length units and K^uv is the "tensorfied"
(-8*pi*G/c^2) to yield an invariant,
K^uv T_uv in place of the usual T_uv in Einstein's Law.
For notation clarity set KT = K^uv T_uv, ((I hesitate to
use too many more G's in notation)), as a revision to Einstein's
Law, where the invariant KT == g^uv G_uv.
Regards
Ken S. Tucker
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