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. 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.

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?

Thank you for any guidance on this.

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

On Feb 13, 12:23 pm, 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.

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

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

Eric Gisse wrote:
On Feb 15, 7:11 am, Tom Roberts wrote:
the technique is valid only for SMALL
variations, and the definition of "small" must include not changing the
metric signature.

So is it correct to say that one cannot continuously change the
signature of a metric?
Is a signature change a discontinuous process like a parity inversion?

Certainly. The signature of a metric is an integer, and an integer
cannot possibly be changed "continuously".

[Some authors call "+++-" the "signature"; others use
the sum +1+1+1-1=2; I mean the latter sense here.]


Tom Roberts

On Feb 15, 6:08 pm, Tom Roberts wrote:
Eric Gisse wrote:
On Feb 15, 7:11 am, Tom Roberts wrote:
the technique is valid only for SMALL
variations, and the definition of "small" must include not changing the
metric signature.

So is it correct to say that one cannot continuously change the
signature of a metric?
Is a signature change a discontinuous process like a parity inversion?

Certainly. The signature of a metric is an integer, and an integer
cannot possibly be changed "continuously".

Sure, but its' a sum. I see your point though, this just isn't
something I have thought much about.


[Some authors call "+++-" the "signature"; others use
the sum +1+1+1-1=2; I mean the latter sense here.]

Tom Roberts

Eric you're ****ed up.
Buy a slide-rule, use 1 dimension,
explain what g11,g_11,g^11 is.

I think you're saying that my boundary condition conflicts with
Einstein's vacuum equation (through the "consistency" conditions).
Since I'm saying the Einstein vacuum equation conflicts with my
boundary condition then we would be saying the same thing in different
word ordering! Maybe that's the miscommunication here.

Thank you much anyways for spending your time to check this. I will do
some more reading before coming back to it.