On May 14, 2005, at 3:35 PM, wrote:
Jack Sarfatti wrote:
On May 13, 2005, at 1:44 AM, wrote:
Jack Sarfatti wrote: That is not valid reasoning Paul. You have to
look at the math.
[Z] Of course you have to look at the math, but the math has already
been looked at very carefully, and these technical questions about
Einstein's t_uv were pretty much settled in 1918.
False. You often misinterpret the math with your words.
I meant *these* technical points (Schroedinger's and Bauer's) about
Einstein t_uv were settled. No one disputes them.
Only because it's all been forgotten years ago!
Einstein's t_uv can be zero where there is G_uv =/= 0, and it can be
made non-zero even in empty spacetime just by choosing polar coordinates.
Even Einstein accepted this.
Prove that. The conclusion would be, in such a case, that such an object
is not physical, i.e. cannot be directly measured unless one believes in
Magick! However, until you actually do the calculation it's not clear
what is being claimed here.
Start with flat
ds^2 = (cdt)^2 - dx^2 - dy^2 - dz^2
Go to spherical polar coordinates.
ds^2 = (cdt)^2 - dr^2 - r^2(dtheta^2 + sin^2thetadphi^2)
OK standard text book (forgetting making dimensions consistently
commensurate for now) gives for only only non-zero connection coefficients
(LC)^122 = -r
(LC)^233 = -sinthetacostheta
(LC)^133 = -rsin^2theta
(LC)^313 = 1/r
(LC)^212 = 1/r
(LC)^323 = cottheta
Note that there are no (LC) =/= 0 that mix in the time component 0.
Those are the physical ones that correspond to inertial forces in
accelerating local frames.
0 = ct
1 = r
2 = theta
3 = phi
This will give
Ruv = 0
So what tuv are you talking about? There, sure ain't no tuv here!
BTW even if you put in SSS distortion you will still get
Ruv = 0
You never get a tuv(Gravity-Matter) in any purely vacuum problem.
So what are you talking about. What metric do you start from to see what
you are alluding to?
As far as I know no one in in the literature on this subject disputes
Schroedinger's or Bauer's technical assertions about Einstein t_uv. I
know Einstein didn't.
That means nothing. Everyone was quite naive mathematically back then in
1921. It was all very new. One can now see in hindsight that Einstein's
identification of tuv with gravity energy was a big blunder, completely
unnecessary making a bogus problem.
Including, no one in the current literature on the subject.
There isn't any current literature on the subject I am aware of. Such as?
That means nothing to you?
That's right, until you can define what the problem is. I mean show me
an example worked out where you get a tuv in a complete vacuum Ruv = 0.
I say that's mathematically impossible. Prove me wrong.
Tuv(Matter) = 0 there. So how do you get tuv(Gravity-Matter) =/= 0?
Just show us the math in a simple example.
I was just trying to make it intuitively plausible by pointing to the
natural dependence of Einstein t_uv on the metric gradients g_uv, w,
which in GR determine the gravitational field strength.
Paul, I already wrote explicitly
tuv(Matter-Gravity)^,v =
-{LC(Gravity)}u^l^vTlv(Matter) - {LC(Gravity)}l^l^vTuv(Matter)
This is a set of INHOMOGENEOUS "externally forced" first order partial
differential equations for the desired
Yes, "externally forced" by the GAMMA coefficients.
Right but trite - stating the obvious, which I stated many times already.
So in Einstein's model there is a coupling of moving matter T_uv with
the metric gradients, expressed by this split of covariant T_uv; v into
an ordinary derivative (of a tensor density) and the "coupling term"
You are repeating what I said several times already.
{LC(Gravity)}l^l^vTuv(Matter)
(where Tuv(Matter) is also a tensor density).
Obviously this law governs the exchange of energy between moving sources
T_uv and the Einstein field (represented locally by the GAMMAs) that
results in gravitational radiation and thus transport of energy-momentum
through the vacuum.
For all practical purposes you can completely neglect the gravity waves
as a really tiny perturbation and they have nothing to do directly with
the terms we are talking about here . The overwhelming effect is NEAR
FIELD, so you are distorting the proper perspective because you have no
sense here of the numbers - what is big and what is small. The above
formulae are valid even if there are no gravity waves. In fact they have
very little to do with gravity waves because
Guv(Gravity)^;v = 0
all by themselves in 1916 GR and obviously, gravity waves are RIPPLES in
the Ruvwl field!
Ruvwl(Gravity) = Ruvwl(Near Field) + Ruvwl(Gravity Waves)
Ruvwl(Near Field) >> Ruvwl(Gravity Waves)
In vacuum
Ruv(Total Gravity) = 0
Therefore
Ruv(Near Field)^;v = - Ruv(Gravity Waves)^;v
So what's your problem Paul?
The question is, consistent with this "conservation law", exactly where
does the energy lost by the material sources go, and exactly how does it
get there?
Ill-posed question.
In ordinary GR
Tuv(Matter)^;v = 0
Therefore
Tuv(Matter)^,v = -{LC(Gravity)}u^l^vTlv(Matter) -
{LC(Gravity)}l^l^vTuv(Matter)
Now Tuv(Matter)^,v is the FLAT T4 4-divergence of Minkowski space-time.
In a closed system in flat field theory
Tuv(Matter)^,v = 0
is local conservation of stress-energy current density, just like
ju^,u = 0 is local conservation of electric current or matter current
density etc.
Therefore, in the virtual non-physical FLAT POV
-{LC(Gravity)}u^l^vTlv(Matter) - {LC(Gravity)}l^l^vTuv(Matter)
is a SOURCE, but it is strange because the stress-energy nonlinearly
feeds back into itself in the minimal coupling to LC(Gravity) which of
course allows for far field gravity waves, but the adiabatic
non-propagating near field distortions dominate.
This would be analogous in EM theory to (static case)
Div.J = kJ
In a spherically symmetric problem this is
J,r = kJ
Therefore
J = e^kr
In general in the curved space-time POV
Tuv(Matter)^;v = 0
automatically includes the minimal coupling to the gravity field so
there is never any problem at all.
BTW note that when
Guv,^;v = 0
from the Bianchi identities
that is a WALL separating geometry from matter.
That is the pure gravity stress-energy current densities are locally
conserved by themselves as a CLOSED SYSTEM! That's what happens in 1916
GR where metric engineering is impossible. You need to break down the
Bianchi identities to get to where you want to go Ad Astra -- Through
Struggles To The Stars and Beyond!
That is, in 1916 GR, even when there is matter there is NO direct
mingling of the gravity stress-energy currents with the matter
stress-energy currents!
Note that the former Guv have second order derivatives of guv, whereas
the minimal coupling of gravity to matter only has first order
derivatives of guv to matter.
BTW, in my theory, guv is ~ first order derivative of argVacuumODLRO, so
(LC) is second order, and curvature is THIRD ORDER, i.e. "jerk" terms
like the radiation reaction in EM with nonlocality!
Dark energy destroys the Bianchi identities!
Forget Matter altogether. I mean it's a tiny perturbation only in the
usual scheme of things where Omega(Matter) ~ 0.04.
What we have is
Guv + /\zpfguv ~ 0
Where, now
Guv^;v =/= 0
Instead
Guv^;v = - /\zpf^,vguv
For the control system of a flying saucer
/\zpf^,v ~ |Vacuum ODLRO||Anyon ODLRO|sin[(argVacuumODLRO) -
(arganyonODLRO)][(argVacuumODLRO) - (arganyonODLRO)]^,v
Where /\zpf > 0 is the universally repulsive blue-shifting dark energy
density
/\zpf < 0 is the universally attractive red-shifting dark matter density
If you want to see what MAYBE this can do to metal look at the pictures in
http://www.svn.net/krscfs/EVOs%20and%20Hutchison%20Effect.pdf
My naive answer would be localized wave propagation at finite speed from
the matter sources into the vacuum, resulting in a definite objectively
describable localized flow and momentary distribution of energy-momentum
in and through the
gravitational vacuum.
Right but trite. That's a very small perturbation. Gravity waves really
weak on lab scale. Have you looked at the numbers for LISA & LIGO?
That's why those gravity wave papers at STAIF are ridiculous - except
perhaps for Ray Chiao's idea.
So in Einstein's theory, which quantity should properly represent the
density of energy transported through the vacuum?
Everything you say below is not even wrong. I showed you above how to
think about this.
My naive "redneck" response to this would be that this quantity should
take the form of an invariant integral of a tensor density over the
spacetime volume in question.
Einstein's answer was the energy pseudotensor, which is at least partly
in the mind of the "observer"; it is not an objective (i.e.,
observer-independent) physical quantity. It depends on the observer's
world line, and also on the choice of spacetime coordinates (which
should not in any sensible theory alter the observer's word line, but
only its mathematical description).
But then neither is Einstein's "gravitational-inertial field" an
objective physical quantity. It is not even an objective physical
entity. From a physical POV it is semi-fictitious, since it includes
frame-dependent inertial effects, according to the classic equivalence
hypothesis (NOT "EEP").
Spell out the difference explicitly otherwise no one knows what you are
talking about.
So why should anyone be surprised?
I have said all along that Einstein's 1916 theory of gravitation is
internally incoherent and even suffers from deep internalcontradictions,
logically, mathematically, and physically, and I continue to say this.
The chronic "energy problem" of GR is just one symptom of this
fundamental confusion.
Paul, I showed you why when
Guv(Gravity)^;v = 0
all by itself alone, that it has nothing much to do with
tuv(Gravity-Matter) in
Tuv(Matter)^;v = 0 all by itself alone as well.
Gravity vacuum waves are obviously in Guv!
.
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