View Single Post
  #50  
Old January 29th 07, 11:36 AM posted to sci.astro
George Dishman[_1_]
external usenet poster
 
Posts: 2,509
Default Pioneer Anomaly Anomalous No More.



On 29 Jan, 09:16, "Max Keon" wrote:
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?


My ISP doesn't carry the alt group and I think its inclusion
might be the cause of some of my messages being dropped.
I would prefer to leave it of but add it back if you feel it is
important. However, nobody from there has contributed
AFAICS.

"George Dishman" wrote in ...
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_.


Velocity is _always_ signed Max, it is a vector. If you
invent bizarre ideas like "always positive velocity",
nobody will know what you are talking about unless
you explain it and your new notation for dealing with
the directional information that you have discarded.

----------
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?


What I said was that v is the derivative of the radius r from
the Sun so v = dr/dt and that is of course positive if the
radius is increasing, i.e velocity is positive if it is _away_
from the gravity source and negative (because radius is
reducing) if the motion is toward it. Now you teel me you
are using the opposite convention.

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. ...


OK. (Don't worry about mass doubling, it just confuses
things.)

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_.


I have concerns about that but let me accept what you say
for the moment.

The negative of that scenario must occur when Pioneer is in
freefall directly back toward the Sun. ...


OK , again forget mass changes.

Pioneer's motion toward the Sun generates a gravity anisotropy
which reduces the pull of gravity,

^^^^^^^^^

That's what I wanted you to confirm, thanks.

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_.


The kinetic energy of Pioneer increases because the
force pulls it towards the Sun, but you just said the
anisotropy _reduces_ that pull, so the speed (and
kinetic energy) increases by _less_ than it would
without the anisotropy. The effect of the anisotropy is
to _slow_ the inward fall. Do you get it yet?

Without the anisotropy, Pioneer loses some amount X
moving outwards from radius R1 to radius R2 and would
regain the same amount if it returned from R2 to R1.

With the anisotropy, it loses _more_ than X moving outward
because the pull is increased and regains _less_ than X
moving inwards because the pull is decreased so there is
a net loss in going from radius R1 to R2 and back to R1.

Your description would imply the missing energy would
remain in the form of potential energy but potential energy is
a function of the radius and independent of the history so that
doesn't work either. If you want the energy to be returned to
Pioneer, the anisotropy would have to _increase_ the pull of
the Sun when the motion was inwards, not reduce it as you
said.

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.


Yes, a perfectly circular orbit would not decay in a two-body
situation but energy is lost through the anisotropy en route
between any two radii whether the motion is inwards or
outwards.

The Sun-Mercury gravity link is exactly the same.


Yes, and so is the effect of any third body such as Jupiter
or all the bits of "the rest of the universe".

George