| Topic: |
Science > Physics |
| User: |
"Jack Sarfatti" |
| Date: |
06 Dec 2004 10:55:19 AM |
| Object: |
Common confusion about General Relativity |
On Dec 5, 2004, at 10:25 PM, wrote:
Jack Sarfatti wrote:
On Dec 5, 2004, at 6:41 PM, wrote:
The metric for an x-y vacuum domain wall can be written as
ds^2 = -(1 - g|z|/c^2)^2 . d(ct)^2 + (1 - g|z|/c^2)^2 .e^-(2kappa
ct).(dx^2 + dy^2) + dz^2
So what?
1. That is ONLY the linear weak field approximation I suspect.
You "suspect". You *are* a suspect!
Loopy polemics. Look I am too busy to read the paper in detail. From my
quick scan he only does a weak field case.
This is the exact solution for the vacuum domain wall. It is not a
weak-field approximation.
What equation in the paper? Even if it is exact solution so what? This
or that solution, exact or not has nothing, whatever to do with the real
problem here, which is your clueless confusion, worthy of Mrs Malaprop,
over the meaning of "inertial motion". Indeed YOU VIOLATE NEWTON'S FIRST
LAW as well as Einstein's GR.
"Inertial motion" means FORCE-FREE MOTION.
There is NO INERTIAL COMPENSATION of forces in inertial motion.
Let's take your really stupid idea in its simplest application to
Newton's mechanics in Galilean relativity.
Inertial motion there means motion of a point test particle in a
straight line in Euclidean space at constant speed.
There is ZERO FORCE there. "Zero" means zero. Your kakamany idea is that
even in that case you have INERTIAL COMPENSATION i.e.
F(real force) + F(inertial force) = 0
This is obviously stupid.
Your equation for curved space-time that in inertial motion (i.e.
timelike geodesic in curved space-time) along which the torsion-free
plain vanilla Levi-Civita connection {LC} can be set to zero in a
sequence of LIFs (geodesic coordinates) is
0 = {LC} = T + N
where you say T is a GCT 3rd rank tensor =/= 0 but that you have
"inertial compensation" i.e.
T + N = 0
T = - N =/= 0
But this is the same as saying that in Newtonian inertial motion
F(real force) + F(inertial force) = 0
where
T = F(real force)
N = F(inertial force)
Now, why have you made such an incompetent stupid mistake?
I have seen a lot of stupid mistakes in physics but really Paul, you
take the cake on this one. This is the stupidest mistake I have seen in
a long time and you have wasted years of futile effort on it.
You are suffering cognitive dissonance in the switch in the meaning of
"inertial motion" in the paradigm shift from Newton's mechanics of
forces acting at a distance to Einstein's LOCAL geometrodynamics. Now
this cognitive dissonance is epidemic and accounts for a lot of crackpot
psychoceramic attacks on Einstein and general relativity. So it is worth
explicating it. Traces of it underly Hal Puthoff's PV theory.
Start with Chapter 10 of Landau & Lifshitz's The Classical Theory of
Fields for Newtonian mechanics in Galilean relativity with absolute
simultaneity.
Consider the APPROXIMATE Lagrangian L for a uniform constant gravity
field g in the reference frame of the surface of the Earth, which in
Newtonian 17th century mechanics is considered an INERTIAL FRAME
(neglecting relatively small Coriolis inertial forces from the fact that
the Earth rotates once every 24 hours etc.
L = (1/2)mv^2 - mgz
z = altitude from surface of Earth where z = 0
In Newton's paradigm GRAVITY IS AN EXTERNAL FORCE in the INERTIAL FRAME
of the Earth. Gravity's potential energy in this approximation is
V(gravity) ~ mgz
The REAL FORCE of gravity in Newton's picture is
F(real force of gravity) = -dV(gravity)/dz = - mg pointing downward to
surface of Earth.
Consider the test particle m. Since it is in a real external gravity
force field, no different ontologically from an external electric field
in Newton's world picture, the REST FRAME of the test particle is
NON-INERTIAL. Therefore, you can and do use INERTIAL COMPENSATION in
that case. INERTIAL COMPENSATION, i.e. the cancellation of a real with a
"fictitious" inertial force only has meaning in NON-INERTIAL FRAMES OF
REFERENCE.
The Newtonian NON-INERTIAL REST FRAME of the test particle executing a
parabolic path in general in this uniform gravity force field, trivially
obeys
g - g = 0
where the first g is from the inertial "fictitious" force of the
accelerating rest frame of the test particle, and the second g is from
the Newtonian external field. But since m(gravity) = m(inertial) the
common m cancels out.
To make this more clear, change the problem for a moment. Suppose the
test particle has charge e in electrostatic field E without gravity.
Then the inertial compensation in the non-inertial rest frame of the
charge of mass m is
ma - eE = 0
That is
a = eE/m
Obviously Paul, your fancy words "inertial compensation" really mean
Newton's Second Law of Motion
F(external real force) = ma
seen from the POV of the test particle's non-inertial rest frame.
Now Paul your fundamental blunder at the heart of your darkness is
trying to apply Newton's distinction between inertial and non-inertial
motion to Einstein's theory!
In the Paradigm Shift from Newton to Einstein the very idea of an
objective external gravity force is completely eliminated! Back to the
problem
L = (1/2)mv^2 - mgz
The test particle m is now in INERTIAL MOTION, i.e. on a timelike
geodesic in curved spacetime. Therefore "inertial compensation" makes no
sense other than to say
F = ma = 0
i.e. F = 0
i.e. your T = 0
A timelike geodesic is the straightest possible line with the most
uniform possible motion in the curved spacetime. It is by definition
FREE OF OBJECTIVE (e.g. GCT tensor forces). In Einstein's theory it is
the surface of the Earth that is in non-inertial motion and the test
particle on the parabola is in inertial motion. We feel weight on the
surface of the Earth because of inertial compensation of the inertial
g-force {LC}^z't't' against the electrical reaction real force of the
rock pushing us off timelike geodesics.
Now, Paul you have confused all this! BTW I do not deny that one can
have approximately uniform g-fields for some non-inertial observers from
a Tuv(matter) source like an unstable short-lived vacuum wall. I never
denied that. You keep raising this Red Herring. Indeed my vacuum defect
hedgehog explanation of the observed anomaly in NASA Pioneer 10&11
motion of cH ~ 10^-7 cm/sec^2 pointing back to the Sun is very similar
to Vilenkin's basic idea, but with a different topology for the vacuum
coherence!
.
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| User: "Ken S. Tucker" |
|
| Title: Re: Common confusion about General Relativity |
08 Dec 2004 05:47:49 AM |
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At the outset, Jack is in accord with classical GR., see Tolman's
Eq.103.1
or Weinberg's Eq. (5.1.11)...(referring to the main body of Jack's
argument).
It's Tucker who argues for a strict intrepretation of GR, much more
than
the classical presentations.
Note in each of those ref's, the usual geodesic described by the
absolute
derivative of the 4-velocity, U^u = dx^u/ds given by,
DU^u =0 (Einstein/Tucker's GR)
uses
DU^u = f^u where f_u = q*F_uv U^v (Classical GR)
the later being Lorentz's force equation.
Tucker argues for Einstein, ref GR1916, following Eq.(65a),
where Einstein argues the geodesic kappa_sigma =0, for
Lorentz force, ie.
f_u = 0, hence DU^u = 0 is always true in GR.
Tucker finds that is exactly what the quantum theory finds,
otherwise a non-zero force would predict a spiral of an
electron into the nucleus, that Planck's QT dismissed.
It's also obvious that,
f_u U^u =0 , due to asymmetry,
hence finding the force invariant "f" = 0 is true by GR prediction
and subsequently verified by QT and the results of that theory.
As a result, the usual "classical" geodisic
DU^u =0
that applies to bodies in "freefall", applies equally well to
bodies subject to EM/quantum variations, by Lorentz's
f_u =0 and f =0.
these latter I call "quantum geodesics" to formally distinguish
them from the usual "classical free-fall" geodesics.
Where GR is concerned, there are 6 matrix elements within
the metric g_uv that describes how spacetime projects, these
are non-symmetrics contained within, g01 ... g23, corresponding
to q*F01...q*F23, that describe the spacetime field, when EM
fields are apparent.
To get a good sense of reality, one only needs to take a pair
of bar magnets and play with the variables of attraction and
repulsion through various orientations, and satisfy themselves,
by experiment, that the only medium of communication is the
spacetime field, (assuming of course the effects are true in a
vacuum).
Tucker departs from Einstein on a very important point.
True Einstein introduced the nonsymmetrical metric, and
Tucker agrees, however Einstein (in his Non-Symmetric Field)
pushed forth an asymmetry of the Christoffels's themselves,
whereas Tucker maintains symmetrical connections, even
when the metrics are not.
I mention this, because Dr. John Moffat and I discussed this
issue (over dinner, Dr. Moffat is currently the worlds foremost
mathematician regarding asymmetrical connections) and I
departed with his view about the need for a non-symmetrical
connection.
Instead Tucker subsumes the antisymetrical metrics within the
Christoffel connection this way, set the metric in two parts.
guv = Suv + Auv == Symmetrical + Antisymmetrical
Using the useful rule for antisymmetric's
Auv,w + Awu,v + Avw,u =0
that coresponds to Maxwell's Seconds set of Equations
(see AE's GR1916, Eq. (60, 61)
By setting my connection trivially to
[ guv,w + gwu,v - gwv,u ]
means Tucker implants Maxwell's equations within the connection,
using the Einstein/Tucker antisymmetrical metric.
(A_uv == q*F_uv)
Tucker very definitely respects Einstein for understanding the need
for the antisymmetrical extension to the metric., but regret his want
to extend that to the connection.
Moving forthright to physics, we find within the connection
Maxwell's equations, that are the essense of measure, and
the change of measurement., by employing light waves, using
DU^u =0,
in accord with QT.
Tucker maintains DU^u =0, and the nonsymmetric metric
in accord with Einstein, but differs with Einstein and Moffat
by maintaining a symmetrical Christoffel, even when the
metric is nonsymmetrical.
Regards
Ken S. Tucker
.
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