Questions about the relationship between the metric tensor and the gravitational field
Jay R. Yablon
the gravitational field h^uv according to (k=sqrt(16 pi G)):
g^uv = eta^uv + k h^uv (1)
Further, the "graviton" field psi^uv is related to h^uv according to
(what is the best thing to call psi^uv, in contrast to h^uv?):
psi^uv = h^uv - .5 g^uv h (2)
I would like to know what (1) and (2) become, exactly, when the
gravitational fields become very strong. I believe what happens is the
the sqrt(-g) factor kicks in, so that (1) now becomes:
sqrt(-g) g^uv = eta^uv + k sqrt(-g) h^uv (3)
and that the relationship (2) stays intact. Is this so? If not, what
are the correct relationships for gravitational field of any strength?
If the above is so, then combining (2) and (3), we obtain:
k sqrt(-g) psi^uv = (2-sqrt(-g)) g^uv - eta^uv, (4)
which subtracts off the constant flat background eta^uv. Thus, if we
take a variation (delta) of each side, which removes out the Minkowski
eta^uv=constant background, (4) leads to:
k delta(sqrt(-g) psi^uv) = 2 delta (g^uv) - delta (sqrt(-g) g^uv) (5)
which in the linear sqrt(-g)=1 approximation in rectilinear coordinates
leads to:
k delta(psi^uv) = delta (g^uv) (6)
and so up to the constant k, psi^uv and g^uv vary together. But, in the
non-linear case, (5) seems to suggest that sqrt(-g) psi^uv has two terms
contributing to the variation, one in relation to sqrt(-g) g^uv as in
the linear theory, and the other in relation to delta (g^uv).
Again, is this correct, and if not, what are the correct, exact
relationships in a field of unlimited strength?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Juan R.
> In the linear approximation, the metric tensor g^uv is related to the
> gravitational field h^uv according to (k=sqrt(16 pi G)):
>
> g^uv = eta^uv + k h^uv (1)
Very odd to put outside h a 'coupling' constant with G inside a square
root. A more sensible choice is the standard
g^uv = eta^uv + h^uv
> Further, the "graviton" field psi^uv is related to h^uv according to
> (what is the best thing to call psi^uv, in contrast to h^uv?):
>
> psi^uv = h^uv - .5 g^uv h (2)
Usually h^uv represent the graviton 'field'. Weinberg in the Chapter 10 of
his book on gravitation and Cosmology gives the expression for the
classical field, the source, the field equation, the formal solution to
the field equation (as an integration over retarded sources), plane wave
solutions and even *formally* discusses the quantum field \hat{h^uv} in
terms of creation and anihilation operators for graviton. Of course,
Weinberg warns about the some difficulties of rigorously defining quantum
gravity.
Evidently psi^uv is another way to write it, just as Hilbert-Einstein
equations can be written in two ways using traces:
(R^ab - 1/2 g^ab R) = kappa T^ab
R^ab = kappa (T^ab - 1/2 g^ab T).
> I would like to know what (1) and (2) become, exactly, when the
> gravitational fields become very strong.
As explained in textbooks, both are exact in strong regimes.
> I believe what happens is the
> the sqrt(-g) factor kicks in, so that (1) now becomes:
>
> sqrt(-g) g^uv = eta^uv + k sqrt(-g) h^uv (3)
What?
(snip rest)
A moderator said to you in this nws:
If you are interested in any of these issues, you really do need to read
through at least some of the many tomes written on QFT.
Add also GR to QFT :-D
--
http://www.canonicalscience.org/
BLOG:
http://www.canonicalscience.org/en/publicationzone/
canonicalsciencetoday/canonicalsciencetoday.html
carlip...@physics.ucdavis.edu
> In the linear approximation, the metric tensor g^uv is related to
> the gravitational field h^uv according to (k=sqrt(16 pi G)):
> g^uv = eta^uv + k h^uv (1)
This is a rather idiosynchratic definition of "gravitational field."
> Further, the "graviton" field psi^uv is related to h^uv according to
> (what is the best thing to call psi^uv, in contrast to h^uv?):
> psi^uv = h^uv - .5 g^uv h (2)
> I would like to know what (1) and (2) become, exactly, when the
> gravitational fields become very strong.
You'll have to explain what you mean. You can *always* write eqn. (1)
-- there's nothing to prevent you from taking the metric and subtracting
a flat metric -- and you can always define a new quantity psi as in your
eqn. (2). What do you mean by "exact"? Exactly what?
Steve Carlip
Jay R. Yablon
news:he456s$h5l$2...@skeeter.ucdavis.edu...
Fair question.
The underlying problem is this:
For electrodynamics, if one applies path integral quantization to the
classical Maxwell Lagrangian, the path integral is taken over the
integration field A^u, the vector potential. That is, one uses a field
element for the integration, of:
DA^u. (1)
Suppose one were to try to path integrate the Einstein-Hilbert action.
(I am not saying that this integration is easy or convergent, I am just
trying to set up the problem correctly.)
One would need in this situation to specify the field of integration,
and in this case, the field of integration would have to be related to
the metric tensor g_uv "in some way." I would like to know "exactly" in
what way.
Specifically, I am trying to figure out what would be the appropriate
field of integration for the E-H action, where the matter Lagrangian is
explicitly tied to the trace of the stress energy tensor, i.e., where
L_matter=-.5T, such that the E-H action, when subject to classical
variation, identically yields the Einstein field equation.
It seems clear to me that one would employ for the integration variable
some function of the contravariant g^uv, and that there would need to be
a sqrt(-g) to provide a proper field density given spacetime curvature.
So, it is my preliminary sense that the proper field of integration
should be:
D(sqrt(-g)g^uv) (2)
But then, I have in the back of my mind that, as you point out, one can
always subtract out a flat metric out, and so define:
h^uv == sqrt(-g)g^uv - eta^uv (3)
With this in mind, it seems to me that perhaps (3) is the proper
variable of integration, that is, that rather than (2), one should use:
Dh^uv (4)
in a path integration of the E-H action, because the eta^uv is just
background structure, not part of the gravitational field, and that
eta^uv should not therefore enter into the path integral as part of the
field variable being integrated.
So, where I am a bit stuck right now, is on this question: would one use
(2), or (4), or something else, as the field integration variable to do
a path integral of the E-H action, and if so, why? My guess is that (4)
is the correct construct to use, though if I had a choice in the matter
(i.e., if you said "you can use either one, it doesn't matter"), I would
pick (2) because the downstream math calculation is (a little bit)
simpler.
Again, I am trying to "set up" the problem correctly, recognizing that
it is still a formidable challenging to calculate the integral once it
is set up properly.
Thanks,
Jay