George, you have removed alt.astronomy again, perhaps by
accident. Cross posting certainly should not be encouraged, but
they have come along for the ride so far and it seems a bit
pointless to kick them off the bus at this late stage.
Don't you think?

"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
---
That means that whether we are looking at Mercury or
Pioneer, we only need to consider the radial part of the
velocity when working out the anisotropic force.

Both the Sun generated radial component and the universe
generated component are active simultaneously, but each component
needs to be analyzed separately.

Again I agree, what I am trying to do is get you to
confirm which way the force acts so that we can do
a basic analysis of a two-body problem. Once we have
that, we can extend it to more complexe situations
such as including the "rest of the universe".

In the case of Pioneer, the
universe generated anisotropy is the anomaly.

I understand that part of your ideas, I just want to
make sure I have correctly understood what you are
saying for the direction of the anisotropic effect
when a body is moving towards the Sun. Why can't
you just say yes or no to whether I have that part
right?

These two, part paragraphs from the web page describe my meaning
perfectly.

----------
According to the laws of that universe, the entire
dimension surrounding every bit of matter in the universe is
shifting inward into its own gravity well at the rate of
(G*M/r^2)*2 meters in each second and is updated at the speed of
light. Meaning that its acceleration capability diminishes to
zero for anything moving at light speed toward its center of
mass.

The equation representing an upward moving mass relative to a
gravity source is ((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2), while
((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) represents a downward moving
mass.
----------

Velocity is added to c for the up moving mass, and it's
subtracted from c for the down moving mass. The reason for that
should be very clear. This next paragraph, which followed the
above, replaced the always positive velocity with a signed
velocity, _on your insistence_.

----------
According to the conventional method of identifying gravity force
direction, and the conventional method of identifying velocity
direction relative to a gravity source, just the one equation is
all that's required. But what it attempts to describe is not as
clear. ((c+v)^2/c^2)^.5*(-G*M/r^2)+(G*M/r^2)
---------

Perhaps I should have described how velocity is signed as well.
v is negative for velocity away from a gravity source, and is
positive for velocity toward a gravity source. But it was you
who told me that George, so how can you be so confused?

...
If Pioneer, or Mercury due to its eccentric orbit, is moving
away from the Sun, there will be a gravitational effect which
will slow that outward motion. There is only one effect but
we can split it into two parts, the conventional effect given
by GR or, to a reasonable approximation, by Newton's Law,
and your extra anisotropic part. You have agreed that the
extra part slows the object more, thus it shows up in Pioneer
as an excess slowing of the craft, it isn't speeding up. I
believe you don't dispute that, we are in agreeement. It is
an indisputable fact that Pioneer is currently losing energy
due to the effect of the anomaly and your equation says
the same.

If Pioneer never returns to this place in the universe, energy
will be permanently lost in time (a long time) ..

Pioneer is losing energy due to the anomaly today, and
that is what your equation says for a body moving away
from the Sun. That part is not in dispute, I just need
to know what your theory says if Pioneer was moving
towards the Sun, then I can do the analysis.

As Pioneer moves away from the Sun in a normal system, it loses
kinetic energy and gains potential energy. If the Sun's mass was
to suddenly double, Pioneer would lose twice the kinetic energy
per time, but that loss would be offset by its position of higher
gravitational potential. Its potential energy is doubled for
that same radius. So nothing has really changed.

The anisotropy causes Pioneer to lose additional kinetic energy,
which is converted to an equivalent in potential energy because
the pull of gravity is increased by the anisotropy. _Nothing is
lost_.

The negative of that scenario must occur when Pioneer is in
freefall directly back toward the Sun. If the Sun's mass was
to again suddenly double, Pioneer's potential energy would
double and kinetic energy must halve to accommodate that change.
Which would seem to be wrong because kinetic energy now
increases more rapidly relative to the Sun. But it's not
increasing relative to the base of dimension that is shifting
past Pioneer into the Sun's gravity well at twice its freefall
rate. It is in fact still decreasing. When the Sun's mass
doubles, the inflow rate of dimension also doubles. Pioneer is
now twice as far from that base.

Pioneer's motion toward the Sun generates a gravity anisotropy
which reduces the pull of gravity, so potential energy reduces
and kinetic energy increases by falling at a lesser rate toward
the base of the inflowing dimension. Again, _nothing is lost_.

If Pioneer was in orbit around the Sun, the anisotropy would
reduce to zero at the orbit aphelion and perihelion, but its
consequences enroute to each point still must be considered.
The Sun-Mercury gravity link is exactly the same.
---

I'll snip the rest until we get this sorted out because what
your theory predicts depends critically on your answer.

I think the critical test is for you to prove I'm wrong in this
part that you snipped.

Sure, but as I have said several times now, I can't
reply sensibly until you decide which direction the
force is pointing.

This is how it all works for the Sun-Mercury system.

It would also apply for Pioneer if it was
in orbit around the Sun.

----------
I'll shift course to Mercury's eccentric
orbit in the Sun-Mercury relationship.

No energy can be immediately absorbed by the Sun either, and
the Sun-Mercury link is a closed system. The orbit velocity
slowing caused by the anisotropy on the trip to the aphelion
immediately converts to potential energy which in turn converts
to kinetic energy as the orbit trajectory curves more to the Sun
than would normally be expected. Momentum can only be conserved
by the Sun being pulled slightly toward Mercury.

Orbit velocity increases as Mercury falls, until centrifugal
forces build to counteract the effect of the anisotropy. When it
reaches what should be the aphelion, its orbit velocity is
greater than that required to maintain the orbit radius because
the anisotropy would be zero if it followed that radius. So it
continues on its outward trajectory a little further.

On the trip to the perihelion, the pull to the Sun is lessened.
Mercury's trajectory then falls away from the Sun as it slows
and centrifugal forces decrease. Momentum can only be conserved
by the Sun moving slightly away from Mercury.

When Mercury reaches the 180 degree mark from its last aphelion
position, it will be at a greater radius from the Sun than would
normally be expected and it will be traveling slower. So it will
continue to fall to the Sun until its velocity is such that
centrifugal forces again counteract the fall.
----------

I don't see any problems with that.

-----

Max Keon