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From: Ken S. Tucker on 24 Sep 2008 16:19
Harald, Juan and friends.

On Sep 23, 6:07 am, "harry" <harald.vanlintelButNotT...(a)epfl.ch>
wrote:
> "Juan R. González-Álvarez" <juanREM...(a)canonicalscience.com> wrote in
> messagenews:pan.2008.09.22.10.08.57(a)canonicalscience.com...

> > harry wrote on Fri, 19 Sep 2008 06:12:17 -0600:
>
> > Contrary to some posters in this ng, (F = ma) does not give us THE
> > definition of force but a second order form of the one-body equation of
> > motion.
>
> > The more general definition of force is (F == -@V/@x) with V the physical
> > potential.
>
> > Using the equation of motion
>
> > dp/dt = -@V/@x
>
> > One may obtain a second (less general) definition of force as
>
> > dp/dt == F
>
> > "as the time-derivative of momentum"
>
> >http://en.wikipedia.org/wiki/Force#Newtonian_mechanics
>
> > Sometimes one is interested in change of *kinetic* momentum. Then one may
> > split the above equation (by simplicity I consider only small velocities
> > and EM interaction) using (p = mv + eA)

> It's unclear to me what eA is; and see further.

I *think* Juan means the time component of momentum,
aka (also known as) energy of position due to potential.

> > d(mv)/dt + e dA/dt = F

The idea is e dA/dt=F is unreal, that equation is
a macroscopic idea, but in the quantum realm,
electronic charges in the realm of the nucleus
behave by photon exchange , please review,
http://en.wikipedia.org/wiki/Quantum_chemistry
The electron orbitals change by emission or
absorption of quanta (photons).
The e dA/dt vanishes in GR too, because there
is no such thing as a continuous variation of the
potential "A", and it has never been measured.
Regards
Ken S. Tucker
....


From: Tom Roberts on 24 Sep 2008 19:56
Ken S. Tucker wrote:
> The fact of light deflection in a g-field proves the
> vectors ExB=C are NOT orthogonal, because
> they aren't perpendicular.

No, it doesn't "prove" that they are not orthogonal. It is perfectly
possible that E.B=0 at every point along the light ray's trajectory, and
yet the light ray is deflected by the gravitation of a massive object,
because of the curvature of spacetime. Indeed, this is precisely how GR
models this: the E and B fields of a light ray ARE perpendicular (==
orthogonal, == E.B=0), at EVERY point along the trajectory; and yet that
trajectory is "deflected" by gravity.

Hint: the "deflection" has nothing whatsoever to do with E or B or E.B
or ExB. Hint2: ExB lies right along the trajectory at every point.
Hint3: both "." and "x" involve the metric (here x = cross-product of
3-vectors).

[There are subtle issues related to defining E and B
that I ignore; they do not affect the conclusion. There
are further subtleties related to 3-vectors and cross-
products which also do not affect the conclusion.]


Tom Roberts
From: Ken S. Tucker on 25 Sep 2008 11:38
Hi Fred.
On Sep 24, 10:17 pm, "FrediFizzx" <fredifi...(a)hotmail.com> wrote:
> http://press.web.cern.ch/press/PressReleases/Releases2008/PR10.08E.html
> Sorry folks, no collisions or quantum black holes until then. ;-)
> Best,
> Fred Diether

IIRC you and Andy and also Jay have made some
predictions based on possible LHC results, if so
would you guy's mind providing links...TIA.

Those delays in obtaining empirical evidence can
be irritating, I can empathize.
Back in the mid 90's I formulated an alternative
solution to AE's EFE's, that nulled "frame dragging"
(GP-b) and g-wave detection (LIGO). Well the wife
rightly asked me for evidence and I advised that the
measurements should be available by 2000.
I suppose that my predictions are true...so far, but
she keeps an eye on that science so she can give
me more chores because she'll claim my science
stinks, if GP-b OR LIGO works. It's 2008 now, and
I'm still avoiding chores :-).
Regards
Ken S. Tucker
From: Ken S. Tucker on 25 Sep 2008 11:52
On Sep 24, 4:56 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote:
> Ken S. Tucker wrote:
> > The fact of light deflection in a g-field proves the
> > vectors ExB=C are NOT orthogonal, because
> > they aren't perpendicular.
>
> No, it doesn't "prove" that they are not orthogonal. It is perfectly
> possible that E.B=0 at every point along the light ray's trajectory, and
> yet the light ray is deflected by the gravitation of a massive object,
> because of the curvature of spacetime. Indeed, this is precisely how GR
> models this: the E and B fields of a light ray ARE perpendicular (==
> orthogonal, == E.B=0), at EVERY point along the trajectory; and yet that
> trajectory is "deflected" by gravity.
>
> Hint: the "deflection" has nothing whatsoever to do with E or B or E.B
> or ExB. Hint2: ExB lies right along the trajectory at every point.
> Hint3: both "." and "x" involve the metric (here x = cross-product of
> 3-vectors).
>
> [There are subtle issues related to defining E and B
> that I ignore; they do not affect the conclusion. There
> are further subtleties related to 3-vectors and cross-
> products which also do not affect the conclusion.]
>
> Tom Roberts

Ok, perhaps Tom you might provide us with your
understanding of how E.B=0 , E.C=0 , B.C=0 and
ExB=C, CxE=B , BxC=E in a GR g-field, in view
of the Shapiro Effect, the Deflection of light
and the "red-shift".
I'll caveat that those 6 equations hold in an
Orthogonal space and time, simultaneously, but
I maintain those 6 cannot hold simultaneously
in a NonOrthogonal SpaceTime.
Regards
Ken S. Tucker
From: Tom Roberts on 26 Sep 2008 22:08
Ken S. Tucker wrote:
> On Sep 24, 4:56 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote:
>> Ken S. Tucker wrote:
>>> The fact of light deflection in a g-field proves the
>>> vectors ExB=C are NOT orthogonal, because
>>> they aren't perpendicular.
>> No, it doesn't "prove" that they are not orthogonal. It is perfectly
>> possible that E.B=0 at every point along the light ray's trajectory, and
>> yet the light ray is deflected by the gravitation of a massive object,
>> because of the curvature of spacetime. Indeed, this is precisely how GR
>> models this: the E and B fields of a light ray ARE perpendicular (==
>> orthogonal, == E.B=0), at EVERY point along the trajectory; and yet that
>> trajectory is "deflected" by gravity.
>>
>> Hint: the "deflection" has nothing whatsoever to do with E or B or E.B
>> or ExB. Hint2: ExB lies right along the trajectory at every point.
>> Hint3: both "." and "x" involve the metric (here x = cross-product of
>> 3-vectors).
>>
>> [There are subtle issues related to defining E and B
>> that I ignore; they do not affect the conclusion. There
>> are further subtleties related to 3-vectors and cross-
>> products which also do not affect the conclusion.]
>
> Ok, perhaps Tom you might provide us with your
> understanding of how E.B=0 , E.C=0 , B.C=0 and
> ExB=C, CxE=B , BxC=E in a GR g-field, in view
> of the Shapiro Effect, the Deflection of light
> and the "red-shift".

[To make sense of this, I must assume C to be a UNIT 3-vector
along the trajectory, and E and B to be UNIT 3-vectors along
the corresponding field directions. In no other way can I
reconcile units consistently. That's OK.]

It's quite simple: E.B, E.C, B.C, ExB, CxE, BxC all involve the fields,
tangent, and metric at a single point along the trajectory. They all
have values as you say, at each and every point along the trajectory.
The Shapiro effect, the deflection of light by a massive object, and the
gravitational redshift of light are all NON-local phenomena -- the
curvature of the manifold induces them all. But the curvature of the
manifold does not affect its properties at a given point, which are what
matters for those dot- and cross-products.

Simple example: Consider cylindrical coordinates {r,\phi,z}
on E^3, and the circle r=R,z=0. Let E=d/dz, B=d/dr, and
C=(1/R)*d/d\phi (all 3 are unit vectors). At each and every
point of the circle, E.B=0, E.C=0, B.C=0, ExB=C, CxE=B,
BxC=E. Yet the circle is "curved". Yes, this is not a
geodesic path, but it illustrates the point that these 3
vectors can be orthogonal at each point along a curved path.

Exercise for the reader: construct a similar example on
the surface of a sphere S^2 (omit E and keep B and C on
the surface); use a circle of constant latitude. This
shows that a flat manifold is not required.


Note that all those dot- and cross-products hold independent of
coordinate system (given the proper definition of the cross-product in
terms of the underlying 2-form).

This also holds for my example, even though E,B,C are
defined in terms of a particular coordinate system.
Because they are vectors, and once defined they are
independent of coordinates.


> I'll caveat that those 6 equations hold in an
> Orthogonal space and time, simultaneously, but
> I maintain those 6 cannot hold simultaneously
> in a NonOrthogonal SpaceTime.

I don't know what you mean by "orthogonal space and time" or
"NonOrthogonal SpaceTime". AFAIK the adjective "orthogonal" is
incommensurate with the noun "manifold". Google gives some opaque usages
of the phrase that don't appear relevant to me.


Tom Roberts
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