Re: variation of appropriate degrees of freedom of metric

On Feb 13, 12:23 pm, babalu...@xxxxxxxxx wrote:
In Carroll's text

Provide a link.

Einestein's equations are obtained from varying the
Hilbert action with respect to the full degrees of freedom of the
metric, i.e. the metric variation is not restricted to preserving
constant signature.

What's a "signature" ?

Assuming that constant signature underlies the
Bianchi identity, requiring that the energy-momentum tensor be
conserved preserves the Bianchi identity in the final result and maybe
saves us from having to vary with respect to the appropriate degrees
of freedom of the metric. I would like to know if the above thoughts
are correct.

IMHO: It depends on the "domain of
applicability" that is intended.
For pragmatic reasons specialized
simplifications can be done.
However, doing so will likely make the
interface of GR and Wave-Mechanics
impossible.
((That might be a question better posted
in sci.phy.research where expert math
physicist's could be more clarifying)).

Turning the problem inside-out, I'm in
favor of harmonic solutions, since all
physical measurement requires the use
of EMR frequency's, to survey effects
of gravitational fields.

Particularly, if I try to explore what happens if the matter term in
the
action is ignored and vary with the full (inappropriate) degrees of
freedom of
the metric, I obtain solutions that violate the Bianchi identity.
Imagine a
boundary condition on G_mu_nu such that it is non-zero in some regions
of a
spacelike hypersurface then the equation resulting from this
variation: G_mu_nu=0 outside the boundary would mean the G is not
conserved. This violation seems to me more a result of varying with
respect to inappropriate degrees of freedom of metric than of ignoring
the matter term in the action.
Is this correct?

Let me express AE's Law simply by
G=T
and it's covariant derivative as
(G=T);w =0.

Focus a bit on T;w=0.
Look carefully at that and see how
energy is exchanged via quanta, (not
continuously).
The differential variation of a quantum
increment is zero, we can prove that.
Take the derivative of the following
series, assuming h is a constant,

T = h + h + h......

dT = 0, but T is NOT a constant.

Thank you for any guidance on this.

Thank you for clarifying a question.
I'm intersted in the 5th rank Bianchi's
too.
Regards
Ken S.Tucker
PS Nice post.
.

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