Re: Aberation and the Speed of Gravity

On Sep 13, 8:24 pm, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote innews:1189710896.870202.285150@xxxxxxxxxxxxxxxxxxxxxxxxxxx:



Hi "bz", good of you to examine the Speed
of Gravity "paradox", from the standpoint
of aberration.

On Sep 13, 1:30 am, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote
innews:1189530341.575644.177940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

du1/ds = 0 = -1/2*g11*(g10,0 + g01,0 - g00,1)*u0*u0

Let g11~1 and u0~1, which provides,

du1/ds = 0 = 1/2*g00,1 - g10,0
(g10 is considered symmetrical herein.)

where

1/2*g00,1 = gravitational acceleration

g10,0 = inertial (centrifugal) acceleration.

If I remember my physics classes correctly, there is no such thing as
'centrifugal
force'.http://phun.physics.virginia.edu/topics/centrifugal.htmlseemsto
agree with me.

There is only the fact that mass has inertia and want to continue to
travel in a straight line (and would do so if there were nothing
exerting a centripetal force[pushing or pulling toward the center]).

I'm worried about semantical discussion, for
instance, a conventional "Celestrial Mechanic"
would be ok with using Centrifugal Force in
opposition to Gravitational Force to explain
circular orbital stability.

True, but centrifugal forces, when applied to things 'closer to home'
produces semantic and mathematically absurd conclusions.

Centrifugal force is based on treating a Rotating Frame of Reference as if
it were an inertial Frame of Reference. This can only be valid when the
region of interest is small. Extending the 'frame of reference' results in
absurd conclusions.

For instance, consider that happens when I stand in the middle of a
parking lot at night and spin in a circle.

The 'centrifugal forces' exerted on distant star in the 'frame of
reference' co-moving with me, those forces (and velocities) would be
absurd.

This, among other reasons, is why physics teaches us to avoid using the
term 'centrifugal force'.



The *names* that I applied to the geodesical terms
"g_00,1" and "g_01,0" are merely guidelines from a
more primitive foundation, for the purposes of
clarity and security of theoretical developement.

The 'centrifugal force' is a fiction invented to explain why someone
riding in a car 'feels like' they are being forced toward to outside of
the turn when really, the car is preventing them from traveling in a
straight line along the old path the car was taking.

Thus there is NO outward directed force. The earth is attempting to
continue in a straight line from its current position, TANGENTIAL to
the orbit, at this instant in time.

Does this change your picture or formula in any way?

No, because I used a math trick. I specified
a rotating CS, such that all 3-velocities
of the Sun relative to the Earth or vis-versa
are zero. That's reasonable to do because the
radius is considered to be constant, so relative
velocity is not detectable, (I imposed a circular
orbit for simplicity).

It would seem to, because in your pictures, with the forces slanted
[either way], the forces are NOT perpendicular to the direction of
travel, but they MUST be, because inertia can ONLY act to keep the
earth moving in a straight line and gravity can ONLY act perpendicular
to the current direction of travel (for a stable circular orbit)
otherwise the earth would be changing speed in addition to changing
direction of travel and if it changes speed, it can't be in a circular
orbit.

You've used the words "MUST" and "ONLY",
I think you do that from a Newtonian PoV.
GR uses the "apparent" position of the Sun
relative to the Earth for the gravitational
effects on Earth, which makes people crabby,
(made me crabby too).

So to eliminate the crabbiness we required
a geodesic solution in accord with GR, that
I posted.

I think that GR rules out 'centrifugal' forces, just as do any 'modern'
college or even high school level physics courses' taught over that last
100 years.

That's fine, what educators teach is their
business, I'm easy :-), it's just semantics.

From du1/ds = 0 = 1/2*g00,1 - g10,0,
the geodesic for a circular orbit becomes,

du1/ds = 0 = &(PHI)/&r + &V/&t

(PHI=GM/r, & is partial diff, u1=constant).

The &(PHI)/&r is fairly well known as
the analog of Newtonian Gravitational
acceleration.
The quantity &V/&t I regard as the analog
to Centrifugal acceleration, but I'm open
to improvements and semantic suggestions.

The bonus from that solution is a confirmation
of the requirement of asymmetric metrics such
as g12 = -g21 in rotating CS's such as the
mutual revolution of the Earth and Sun as
demo'd. That was suspected by Einstein and
has been explored more recently by my friend
Prof. John Moffat who has posted to ArXiv's
about "nonsymmetrical" field theory.

I think the solution itself is a mathematical
physics discovery of the 1st rank, because it
confirms the physical validity of asymmetrical
metrics, in application to a real problem in a
straight-forward way.

On the other hand, if the centripetal force pointed anywhere other than
to where the mass being orbited is CURRENTLY located, if it pointed to
where the visible image of the sun is seen to be rather than where the
sun actually is, then there could never be a stable circular orbit.

How do you define "CURRENTLY located" ?

The position as indicated by reference to an inertial frame of reference
co-moving with the center of gravity of the earth at any particular
instant of time, oriented with the 'x-y plane' coincident with the plane
of the earths circular orbit and the x axis aligned along a vector
pointing along a tangent pointing along the velocity vector at that
instant of time.

At that instant, the centripetal force vector 'must' be perpendicular to
that velocity vector in order for circular motion to continue around the
orbit, if I understand what I remember of my college physics and math
courses and what I have studied since then.

So the barycenter of the orbit 'must' coincide 'at that instant' with
where the centripetal force vector is pointing, even though the force
vector is being 'exerted' 9 light minutes away from the barycenter.

Let's try a crude gedanken. Suppose we string X-mas
lights from the Sun to the Earth, pulled as tight as
possible, and call that the "ABSOLUTE" direction
of the Sun as strung from Earth.
That string of lights will appear curved, compared to
the light from the Sun, because at the position of
Earth they will have different angles, because of
aberration. Then they will converge to the same
point on the Sun. What do you want to do?

I respect GR in that it requires we use the apparent
position of the Sun, and all gravitating bodies, to
compute reality, as we take measures on Earth.
Otherwise, we'll need to figure out where things
may have been.
Regards
Ken S. Tucker

.

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