Max Keon wrote:
"George Dishman" wrote in message
...
"Max Keon" wrote in message
...

Max, there are two major difficulties in what you are
saying. I'll trim the rest until we resolve those.

"George Dishman" wrote in message
oups.com...
Max Keon wrote:
...
... The
moving matter will be slowed in the direction of motion according
to a combination of the two equations.

The equations describe slowing for an inward moving
mass but increasing speed for an outward moving
mass.

I understand what you are saying George, but your description
is invalid in this case.

What I wrote is simply applying your equations and that
was the point, one of your equations must be invalid.
Let's look at the details.

((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) ...

--- an aside ---

Try applying the laws on this page

http://www.mathsisfun.com/associativ...tributive.html

To start, look at the very last example and see if you
can replace their "16" with "(G*M/r^2)" in your equation.
Then using the other laws, see if you can reach:

(v/c) * (G*M/r^2) if you take the positive root, or

-(2 + v/c) * (G*M/r^2) if you take the negative root.

This is the question you refer to.
What is 6 * 16 + 4 * 16?
6 * 16 + 4 * 16 = (6+4) * 16 = 10 * 16 = 160

I fail to see the point of the exercise.

First, if you do the excercise, you will discover
why my equation and yours are identical, second,
you will be in a position to follow my arguments
when I use maths to prove a point, and third it
will make it much easier to see how the sign
changes with the direction of motion so you are
less likely to make the sort of mistake we are
discussing below.

Perhaps we should
just stick to the equations as they were initially written.

Perhaps not.

Note that the latter result, when added to the normal
Newtonian acceleration would produce triple the usual
value hence I assume you intended the positive root.

--- end aside ---

.. applies for motion away from
a gravity source, which naturally increases velocity relative to
the in-moving dimension, and consequently increases the pull
toward the gravity source.

Newton's acceleration due to gravity is -GM/r^2.
Note the negative sign which indicates the force is inward,

My terminology has apparently been the root of much confusion.
I can't see why though.

It is not your terminology in this case, the problem is
that you insist on using two equations with a change
of sign which means the acceleration is alwys in the
same direction regardless of the directon of motion.

or more accurately in the direction of reducing r, that
is towards the mass M. Note that well Max, it will also
be important later, the acceleration acts towards the
body of mass M which is producing it.

To align with your terminology the equations would be
# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# is now negative for outward motion, which satisfies your
description of a force pointing inward, and positive for inward
motion.

Let's check the first which is for outward motion hence
v is positive. Let's take a toy value of 0.01c:

# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)

v = +0.01

(c+v) = 1.01c

(c+v)^2 = 1.0201 c^2

((c+v)^2/c^2) = 1.0201

((c+v)^2/c^2)^.5 = 1.01

I won't bother putting number in for GM/r^2:

((c+v)^2/c^2)^.5* ( -(G*M/r^2) )

= 1.01 * (- G*M/r^2)

= -1.01 * (G*M/r^2)

so

# = [ -1.01 * (G*M/r^2) ] + (G*M/r^2)

and using the example on the web page above that is

# = [ -1.01 + 1] * (G*M/r^2)

and finally

# = -0.01 * (GM/r^2)

That is negative so the force is inward.

Now try your second equation which applies for inward
motion and hence v is negative. Let's take a similar toy
value of -0.01c:

# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)

v = -0.01

(c-v) = 1.01c { does that look familiar ;-) }

(c+v)^2 = 1.0201 c^2

((c+v)^2/c^2) = 1.0201

((c+v)^2/c^2)^.5 = 1.01

I won't bother putting number in for GM/r^2:

((c+v)^2/c^2)^.5* ( -(G*M/r^2) )

= 1.01 * (- G*M/r^2)

= -1.01 * (G*M/r^2)

so

# = [ -1.01 * (G*M/r^2) ] + (G*M/r^2)

and using the example on the web page above that is

# = [ -1.01 + 1] * (G*M/r^2)

and finally

# = -0.01 * (GM/r^2)

That is again negative so the force is again inward.

The action of gravity, only, has
increased. The acceleration is not applied in the direction of
the outward moving mass, it's applied inward.

For a body moving away from mass M, the value of v is
positive and your equation gives a positive value so it
_decreases_ the effect of gravity.

Only according to the description that follows from using
-(GM/r^2).

If you swap the sign in the equation when the sign
of v swaps then the answer will always be in the
same direction. You will only get what you want if
you stick with the same equation for both directions.
That way the direction of the anomaly reverses when
the motion reverses.

The opposite of
course applies for motion toward the gravity source. (c+v) is
replaced with (c-v).

For a body moving towards the mass, v is negative. Since
the sign of v has changed and you have replaced (c+v) with
(c-v), the anisotropic force acts in the same direction.
Two negatives make a positive.

It would be far less confusing if you would just go back to the
original equations. Even the updated version, above, is very easy
to understand. Either way, the two equations don't both point in
the same direction at all.

Yes they do, even your new ones.

Using the updated version, # is added to the result of -(GM/r^2),
for both the up and the down case. The result of -(GM/r^2) plus
the negative # value compared with the result of -(GM/r^2) plus
the positive # value is exactly as it should be. Both are now
negative of course, but the difference between them is still
correct.

"Both are now negative of course" as you say, so both point
in the same direction, only now they both point inward instead
of outward.

Pioneer

For Pioneer, the two directions are similar so we can look
at Max's numbers:

... The Universe alone is responsible for the
anomalous acceleration which appears to be directed toward the
Sun.

((c+v)^2/c^2)^.5 * (G*M/r^2) - (G*M/r^2) = 8.34E-10 m/sec^2.

However, the anomalous acceleration of Pioneer 10 is
-8.74E-10 m/s^2 hence in the opposite direction. This
has already been pointed out to Max.

You still have it all wrong. For Pioneer's 12500m/sec velocity
away from the Sun, relative to the Sun, at the radius of e.g.
Neptune, ((c+v)^2/c^2)^.5*(G*M/r^2)-(G*M/r^2) = 2.626E-10m/sec^2
added pull to the Sun. That's an acceleration toward the Sun,
not away from it.

It is a positive number so it is away from the Sun. If it
was towards the Sun it would be -2.626E-10m/s^2.

Well we should now be in agreement.

Yes, on Pioneer. You now have the force always inward.

.....
I'm pretty sure if you just say the Newtonian acceleration
is modified to be

a = -GM/r^2 * (1 + v/c)

then you will get all the numbers you are expecting.

I'll continue to use the original or the updated equations, if
that's OK. They tell the story as it should be told. And they
really don't lead to confusion.

The next paragraph addresses that but is full of
contradictions. I'll extract the parts relevant to the
direction:

Pioneer 10's trajectory is 11 degrees off a line through the Sun.

Yes, that is correct.

Its motion relative to the universe is generating a force
.. also along that line
and pointing back to the Sun.

You just noted that "along that line" differs from
"pointing back to the Sun" by 11 degrees.

Yes. I was paying more attention to the wording to avoid any
more confusion.

Which of course sets the anomalous
acceleration in the direction of the Sun along its trajectory
path.

Again "in the direction of the Sun" differs from
"along its trajectory path" by 11 degrees.

Pioneer is accelerating toward the Sun, anomalously. What's
wrong with that?

The 11 degree difference.

Which is it to be Max? All the stuff above says it is
a modification of GM/r^2 which is "in the direction of
the Sun", not "along its trajectory path" so please
decide which it is to be.

This is a *path* that Pioneer has scribed through space on its
travels. It's a plot of Pioneer's trajectory as it traveled
outward, and it's obvious where Pioneer "o" is currently
pointing. But surely I can specify whatever direction I like
along that *path*? I can state that Pioneer's motion away from
the Sun increases the pull of gravity from the Sun, *along the
line of that path*, if I so choose. There's nothing confusing
about that.

Well really you cannot specify whatever direction you like,
the direction should be dictated by the physics so when
you say the anisotropy is produced because "dimension"
is flowing towards the Sun, that flow should determine
the direction and leave you no choice. However, I don't have
a problem with you telling me what the direction is, I only
insist that you make that choice and stick with it. You
rae trying to produce a new law of gravitation and that law
_must_ be universal.

--------------------Trajectory path---------------o
Sun

Not quite, it is like this:


Sun


--------------------Trajectory path---------------o

There is about an 11 degree difference between the
trajectory and a line pointing to the Sun. I just want
you to choose of those two directions and stick to it.
We can't do a sensible analysis if you keep changing
your mind.

Then there's the reaction to the velocity slowing and
consequent loss of momentum, which emerges in the perpendicular
plane as a velocity increase.

That is complete rubbish, any motion other than in the
direction of the force violates conservation of momentum.

Overall, momentum is conserved because the Pioneer craft
is exerting an equal and opposite force on the Sun which
produces a tiny acceleration. That would have the value

a = -Gm/r^2 * (1 + v/c)

where m is the mass of Pioneer, about 241kg from memory.

If Pioneer isn't coming back, what you say is correct.

What I say is correct whether it comes back or not.

The Sun
will be permanently shifted in the direction of Pioneer's travels
and Pioneer's momentum loss is transferred to the Sun and is
easily accounted for. But if it was in Mercury's elliptical orbit
the momentum lost to the Sun on Pioneer's outward leg would be
reclaimed on the inward leg as the tension in the gravity link
between Pioneer and the Sun is reduced. The Sun would recoil back
to (almost) where it was at last perihelion, as would be expected
in a closed system.

Yes that is also true.

But all of that is quite irrelevant because it's the relationship
between Pioneer and the universe that generates the anomaly.
That's the one which can't be concealed. Considering that there
is no means of transferring energy between Pioneer and the
universe other than gravitational radiation, ..

Not at all, the 'rest of the universe' is ust a collection
of other bodies like the Sun so the laws must be the
same. The tiny acceleration of Pioneer due to the
gravity of Sirius chauses a change in momentum
which is balanced by a change in the momentum
of Sirius and so on for all the other stars.

that becomes a
localized system. Meaning that everything must be accounted for
locally.

Pioneer's lost momentum due to its 8.4E-10m/sec^2 velocity
reduction must go somewhere. It can't just disappear.

I assume you mean momentum is conserved in
your theory and that is fine.

The logical
answer is that it's shifted into the perpendicular plane relative
to the motion direction, in the direction of the Sun.

No, that doesn't work. Momentum is a vector quantity so
to conserve the total, if Pioneer is accelerated towards
the Sun like this:

S - P

then some other body must be accelerated in exactly the
opposite direction, and of course that is the Sun

S - - P

Gravity is like a spring stretched between the bodies and
the spring pulls on _both_ ends.

If you want an extra motion of Pioneer perpendicular to that
it looks like this:

^
|
S P

and to conserve momentum the Sun would have to accelerate
in the opposite direction like this:

^
|
S P
|
v

Put the two together and you get motion at 45 degrees to the
sun-craft line:

\
S P
\

snip
The natural fall from its trajectory should be perpendicular to
that *path*, not directly toward the Sun.

OK, there is an 11 degree difference which is too small for me
to show but the same comments apply, the sum of an inward
acceleration and an acceleration perpendicular to the trajectory
is not zero and must be compensated by a corresponding
"equal and opposite" change of momentum of the Sun.

The same applies to the

After all, the only way
the loss of momentum along its trajectory can be accounted for is
if it's tranferred into a totally different plane as a momentum
increase.

Nope, that doesn't work, momentum is a vector but you
are treating it like a scalar.

It's a momentum loss in one plane which becomes a
momentum gain in an entirely different plane. Momentum MUST shift
planes, otherwise nothing changes.

Momentum is a vector. If you want it to be conserved
overall then the components in _each_ direction have
to be conserved separately as well.

George