Re: GRT and higher order effects?



Joe Fischer wrote:
> On Thu, Tom Roberts <tjroberts@xxxxxxxxxx> wrote:
>
> >Harry wrote:
> >> for a situation that r does not change much, we have a nearly
> >> homogeneous field, in which case the factor that is only a first order
> >> approximation cannot be the exact solution if the equivalence principle is
> >> exactly true!
> >
> >The equivalence principle is not exact. It is merely a local
> >approximation (i.e. exact only at a single point in an appropriate limit).
> >Tom Roberts tjroberts@xxxxxxxxxx
>
> What is this "exact (solution)" supposed to be? Does
> _anybody_ really think that there can be a "formula" that gives
> the acceleration of gravity at any point in the vicinity of any
> massive object?
>
> The Principle of Equivalence is what it is, the first
> insight into the complicated world of gravity, using a very
> simple technique.
>
> Joe Fischer

Yeah Joe...
The PoE seems to me to be an explanation to
justify the covariant derivative g_uv;w=0 that
is used everywhere in GR. It states - more or
less - if the metric components are found to be
constant in one CS, then g_uv;w=0 in that CS,
and g_uv;w=0 being a tensor is true in all CS's.

Every ordinary tensor analysis book shows how

g_uv;w=0 ,(Principle of Equivalence, PoE)

is the basis for the Christoffel Symbol, which in
turn is the basis for the g-field in GR because it
provides the geodesics, the so-called "paths".

In advanced research I'm informed the g_uv;w=0
is not necessarily true, as H. Weyl tried to demon-
strate, in his failed unified field theory attempt.

Thinking more deeply, if a field required
g_uv;w =/= 0 then an absolute acceleration
would exist in that field and no CS could be found
to be an origin, relatively to which all else moves.
That would screw up relativity.

Since an origin is arbituary, a CS can always be
found where g_uv;w=0 and absolute acceleration
vanishes, i.e. can't exist.

So I would say, g_uv;w=0 unifies the PoE and the
requirement of the relativity of accelerated motion,
to be indistinguishable.

I'd be interested in other peoples intepretation of
g_uv;w=0.

Ken S. Tucker

.

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