Questions about the relationship between the metric tensor and the gravitational field

[Jay R. Yablon]

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)

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