Questions about the relationship between the metric tensor and the gravitational field
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From: Jay R. Yablon on 14 Nov 2009 20:50 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: jyablon(a)nycap.rr.com co-moderator: sci.physics.foundations Weblog: http://jayryablon.wordpress.com/ Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
From: BURT on 14 Nov 2009 23:03 On Nov 14, 5:50 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote: > 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: jyab...(a)nycap.rr.com > co-moderator: sci.physics.foundations > Weblog:http://jayryablon.wordpress.com/ > Web Site:http://home.roadrunner.com/~jry/FermionMass.htm The curve could be round geometry a sphere radiating outward from the center of mass. Mitch Raemsch
From: eric gisse on 15 Nov 2009 02:05 Jay R. Yablon wrote: > In the linear approximation, the metric tensor g^uv is related to > the gravitational field h^uv according to (k=sqrt(16 pi G)): The metric tensor g{^,_)uv _is_ the gravitational field. > > 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) It contains the exact same information as the metric, just expressed in an irrelevant but slightly different way which adds nothing to the discourse. > > I would like to know what (1) and (2) become, exactly, when the > gravitational fields become very strong. They become irrelevant. The linearized limit presumes weak gravitation. The full answer lies back with the full nonlinear field equations. > 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) No. > > and that the relationship (2) stays intact. Is this so? No. > If not, what > are the correct relationships for gravitational field of any strength? G_uv = R_uv - 1/2 R g_uv = 8piG T_uv > > 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) NO! Stop the idiocy. You are backporting a meaningless quantity from the weak field limit back to the full nonlinear theory and are getting complete bullshit. [snip meaningless rest]
From: BURT on 15 Nov 2009 03:29 On Nov 14, 11:05 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > Jay R. Yablon wrote: > > In the linear approximation, the metric tensor g^uv is related to > > the gravitational field h^uv according to (k=sqrt(16 pi G)): > > The metric tensor g{^,_)uv _is_ the gravitational field. > > > > > 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) > > It contains the exact same information as the metric, just expressed in an > irrelevant but slightly different way which adds nothing to the discourse.. > > > > > I would like to know what (1) and (2) become, exactly, when the > > gravitational fields become very strong. > > They become irrelevant. The linearized limit presumes weak gravitation. The > full answer lies back with the full nonlinear field equations. > > > 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) > > No. > > > > > and that the relationship (2) stays intact. Is this so? > > No. > > > If not, what > > are the correct relationships for gravitational field of any strength? > > G_uv = R_uv - 1/2 R g_uv = 8piG T_uv > > > > > 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) > > NO! Stop the idiocy. > > You are backporting a meaningless quantity from the weak field limit back to > the full nonlinear theory and are getting complete bullshit. > > [snip meaningless rest] If you pick a point in Schwarzschild geometry how can it be parabolic? How can a parabolic metric make elliptical orbits? Mitch Raemsch
From: Jay R. Yablon on 15 Nov 2009 10:07
"eric gisse" <jowr.pi.nospam(a)gmail.com> wrote in message news:hdo998$ij5$1(a)news.eternal-september.org... > Jay R. Yablon wrote: > >> In the linear approximation, the metric tensor g^uv is related to >> the gravitational field h^uv according to (k=sqrt(16 pi G)): > > The metric tensor g{^,_)uv _is_ the gravitational field. > >> >> 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) > > It contains the exact same information as the metric, just expressed > in an > irrelevant but slightly different way which adds nothing to the > discourse. Yes, they have the same information, but in a different form. Relevance depends on what you are trying to do and cannot be prejudged. >> >> I would like to know what (1) and (2) become, exactly, when the >> gravitational fields become very strong. > > They become irrelevant. The linearized limit presumes weak > gravitation. The > full answer lies back with the full nonlinear field equations. Obviously one uses the non-linear field equations. But that is not my question. The entity psi^uv is taken in quantum theory to represent the graviton. Yes, it has the same information as g^uv but is a function of g^uv. I am simply asking what that actual function becomes. If you know how to write it with R^uv, T^uv, etc., that is fine. What this is relevant to, is that I am trying to determine what would be the field of integration "field" in a path integral $Dfield, for gravitation. If you know that for sure and can back it up, we can skip all this other stuff. There was no reason for you to start pissing. {snipped pissing} Jay. |