Thank you Tom, I admit I wasn't recognizing that a SMALL variation of
the metric is not likely to change signature since the sign of an
eigenvalue is not changed by small variation unless it equals zero...
silly of me but I feel better now.

The boundary condition I make on the metric is simply that G_mu_nu be
non-zero at any region of the spacial portion of the boundary (no
conditions on derivatives of G). This is a very general restriction
which I don't think violates any constraints on the Riemann tensor.
Minimizing the \integral R sqrt(-g) d^4x as you express it, should
then not lead to Einstein's vacuum equation, R_mu,nu=0 (equivalently
G_mu_nu=0) inside the boundary since that would mean an infinite
derivative of G_mu_nu at the boundary going suddenly from non-zero to
zero. Also I think it would mean non-conservation of G. I agree of
course that minimizing the above integral does lead to Einstein's
vacuum equations when G is assumed to be zero at the boundary, and I
do agree that when it does not equal zero a matter term must
conventionally be included. This is what I mean by exploring what
happens if the matter term is ignored: ignored when G is non zero at
the boundary and wondering if we then do not get Einstein's vaccum
equation but something else.

The reason why this is interesting to me is that particles/mass in QFT
are nothing more than excitations of a field, and in Einstein's old
view of a resolution of the quantum problem through GR the metric
would play the role of this field. If the metric is sufficient to
describe the particles/mass then fundamentally, there is no need for
the T tensor to do so, explainig why I'm exploring what happens when
it's ignored.

Thanks again for shooting down a misconception and hope you shoot more
of them down.
-Tarek Halabi

On Feb 15, 8:11*am, Tom Roberts wrote:
wrote:
In Carroll's text 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.

Maintaining constant signature is absolutely essential. But note the
variational technique is a CONTINUOUS technique, and both the manifold
and the metric are continuous as well. For any given metric, there is a
neighborhood in which the signature remains unchanged, and the variation
must occur within that neighborhood -- this is implicit in using such
variational techniques. IOW: the technique is valid only for SMALL
variations, and the definition of "small" must include not changing the
metric signature.

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.

I'm not sure what you mean. If one ignores the matter terms, then the
action is just \integral R sqrt(-g) d^4x, and varying that gives the
(vacuum) Einstein field equation, which clearly satisfies the Bianchi
identities, identically. I suspect you made a mistake.

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.

You cannot just dictate such a condition on a boundary. There are
constraint equations which must be satisfied on any Cauchy surface or
boundary.

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?

I doubt it. I suspect you tried to use unphysical initial or boundary
conditions (i.e. ones which do not satisfy the relevant constraints).

Tom Roberts