wrote:
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.

Yes. And the metric cannot have any zero eigenvalues, as it must be
non-degenerate.


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).

But the consistency conditions on the boundaries require that such
non-zero values be accompanied by appropriate non-zero derivatives (and
possible second derivative, I'm not sure). You must also be sure your
"spacial portion" is sufficient to be a Cauchy surface....


This is a very general restriction
which I don't think violates any constraints on the Riemann tensor.

But it violates the consistency conditions for the boundary values of
the Einstein field equation.


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.

That should indicate your error -- such "infinite derivatives" cannot
possibly be valid. But the consistency conditions for the boundary
values ensure that the Riemann tensor can be continuous at the boundary.


Also I think it would mean non-conservation of G.

Cannot happen -- the Bianchi identities ensure that the covariant
derivative of G is identically zero. Attempting to violate the Bianchi
identities is like attempting to violate 1+1=2 -- such an attempt is
complete nonsense (they are inherent in the definition of Riemann).


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.

If the metric is a quantum field, ask what the corresponding vacuum is.
It does not make sense for there to be "no metric". So the usual
approach is to separate the metric into two pieces, a Minkowski part
(\eta) and the rest (h). Now quantize h. Note, however, this is known to
fail....


Tom Roberts