Re: Aberation and the Speed of Gravity

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.htmlseems to 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.
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.

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" ?

--
bz

Good stuff "bz"
Ken S. Tucker

.

Relevant Pages

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