"Max Keon" wrote in message
u...

"George Dishman" wrote in message
...
"Max Keon" wrote in message
u...
George Dishman wrote:
Max Keon wrote:
I've updated the web page accordingly.
http://www.optusnet.com.au/~maxkeon/pionomor.html

You still have the incorrect equations show, they should
be removed.

"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)"

To anyone familiar with maths at high school level or
beyond, your versions are both incorrect and very confusing.
The Lorentz equations include ((c^2+v^2)/c^2)^.5 and it is
very easy to misread your version as that, I did for some
time before noticing what you were really saying. By far
the clearest way is to write this way:

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

Well why didn't Newton do that?

He didn't have your (1+v/c) term because it doesn't
exist.

But you said his equations included ((c^2+v^2)/c^2)^.5

I said Lorentz's equations included the term, not
Newton's. The Lorentz transform are a completely
different thing.

Other than that, we now take calculus for
granted as something we all learnt at school while
he had to invent it, so in explaining the physics
he had to write in a much more detailed way. I can
write "a" above and you know I mean acceleration
which is the second derivative.

And why do you think that a = (1 + v/c) * (-G*M/r^2) + (G*M/r^2)
is not as confusing as a = ((c+v)^2/c^2)^.5*(-G*M/r^2)+(G*M/r^2)

It would be, why have you added an extra term? The
total acceleration in your theory is

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

of which the Newtonian part is

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

and the anisotropy is

a = (-G*M/r^2) * (v/c)

Add the two together to get the total.
------

--
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. Even matter in a fixed position relative to a gravity
source is traveling outward through dimension because dimension
is traveling inward through it, hence the action of gravity.
--

It's obvious that velocity is added to or subtracted from the
speed of light. And that should be obvious to anyone at all.

Sure, but if you add a positive number or subtract a
negative number, the result has the same sign. Your
equations may look pretty to you but they give the
wrong answer.

You seem to be quite adamant that you can't understand my
meaning unless velocity to and from a gravity source is signed
differently.

No, I understand your equation perfectly but it tells
me the acceleration is in the opposite direction from
what you say in words.

If I use the single equation, it's very confusing
when applied for the universe generated anisotropy. The two
equations are simultaneously active, as a = (-G*M/r^2) * (v/c)
and a = (-G*M/r^2) * (v/c). v is of course negative in one of
them, but I'm not permitted to show which one. Using just the one
equation, v is both positive and negative and thus cancels to
become zero.

Suppose 'A' and 'B' are two galaxies and 'p' is Pioneer:

A p B
--

Pioneer is moving from left to right. The distance
from A in increasing so the derivative (which is
the 'v' in your equation) is positive. The extra
acceleration from A is:

a_A = (-G*M/r^2) * (v/c)

which is negative and points _towards_ A or from
right to left.

The distance from B in decreasing so the derivative
(which is the 'v' in your equation this time) is
negative. The extra acceleration from A is:

a_A = (-G*M/r^2) * (v/c)

which is positive and points _away_from_ B or again
from right to left. Using my single version, the
effects add to produce a larger effect.

With your two equations, you would swap the sign on
one of the results and the contributions would cancel.

Three cheers for mathematics, the universe is once again saved.
You can call them one equation if you like, but they are no such
thing.

You need to do some serious remedial work on your
maths.

------

snip stuff on vectors, it is too advanced at the moment

Your vectors would need to include instantaneous conservation of
momentum at a distance. And that is obviously wrong.

Vectors are a branch of maths, just a tool you will
need to know and be able to use if you are ever to
do any physics.

The gravitational effect on the Sun
does that even though the craft is more than ten light hours
away. If you abandon GR for Newton then you have what he
called "instantaneous action at a distance" and whether the
distance is ten light hours or ten light years makes no
difference. Your equation is based on -GM/r^2 which applies
instantaneously, there is no delay term in the equation.

That has always been a totally absurd statement. Of course
there's no delay term built into the equation. Why should there
be? Nature provides the delay, not mathematics.

Don't be stupid Max, you know that in order to work,
the maths must be a model of nature. If nature has a
delay, that must be reflected in the maths.

Then why isn't it? The delay obviously exists in nature.

The delay does not exist in nature.

That statement
implies that math dictates how nature must behave.

No, it says the maths must be written to reflect nature.

????

Maths is a tool we use to _model_ the way the universe
works. If the maths is to be useful, it must be an
accurate model.

You can't
simply gesture hypnotically and brush the obvious truth aside,
that a time delay in the transfer of momentum in the physical
world is very clearly a part of nature.

But this has nothing whatever to do with Newton anyway. The zero
origin universe has its own very specific rules, which most
certainly don't include instantaneous action at a distance. Light
speed is the absolute limit, **for a very good reason**. Such a
time delay is certainly expected in that universe.

Then change your maths to show that delay.

The universe generated gravity anisotropy depends on there being
a delay _in nature_. The maths assumes that the delay exists.

There is no delay and the maths you have written so
far correctly assumes there is none. We have been
writing equations like this

a = G * M / r^2

That's just an example, I've left out the anisotropy
bit just to keep it short. The point is that G and M
are constants but the acceleration and radial distance
vary with time. What we have been writing is a shortcut,
and both a and r should be shown as functions of time
like this:

a(t) = G * M / r(t)^2

If you want to add a delay, then you must change it to
be something like this:

a(t-r/c) = G * M / r(t)^2

which says the acceleration at some time in the future
depends on the value of the radius now, the time being
the current radius divided by the speed of light. Of
course you need to decide whether it is the radius now
or the radius at the future time or maybe a combination
of both that determines the acceleration.

You assume the existence of dark matter because it ties in with
the maths.

No, I assume the existence of dark matter in galaxies
because it is needed to fit the same model that
correctly models other observed gravitaional effects.
The maths is nothing more than a tool used to perform
the comparison of observations.

I assume the existence of a delay in action at a
distance because the maths requires it. It's also the only
logical conclusion, and is a direct prediction as well.

Your maths did not requre it. See above for a hint
on the change you need to make to the math to model
a delay.

What you
will find is something Newton knew, that it will
produce aberration of the gravitational force and
again cause the planets to spiral into the Sun. He
didn't like the instantaneous nature of forces but
he knew he had no choice if his maths was to work.

I think he should have put more thought into the physical side of
the problem instead of letting the maths confuse him.

This is a binary star pair. 0 are their instantaneous positions,
while + is where each appears to be. They will spiral away from
each other, losing momentum, and their orbit velocities will slow
until they reach a stable orbit radius. The only consequence is
that they would be orbiting a little slower than the maths would
predict. _But that couldn't be noticed because the masses of the
stars are determined by orbit velocity_.

Why do you think they would continue to lose momentum and spiral
together?

Because the same diagram applies on the next orbit and
the one after that. There is no stable configuration
for them to reach.

+ 0-


-0 +

If the next scenario was possible, the stars would gain
additional momentum as they are driven inward, and would thus
continuously spiral away from each other.

Indeed but this never happens since changing the
speed moves the plus sign to the other side.

+ 0-


-0 +

Is that how you see it?

Energy can be almost immediately removed from interacting
charges and stored in space in the form of E/M radiation. But
gravity is equivalent to only a single wave that extends to
infinity. So there is obviously nowhere to store the energy
equivalent of Pioneer's momentum loss due to its motion
relative to the mass of the universe. Momentum is by no means
immediately conserved, but it is conserved in time.

The word "conserved" means it has the same value
AT ALL TIMES.

Pioneer's velocity will continue to slow and that will cause it
to be drawn in the direction of the focal point of its trajectory
path radius, and the Sun. Its momentum is not (immediately)
conserved, but it isn't lost forever.

That doesn't matter, as you admit the momentum is
not conserved unless you postulate that the missing
momentum is somehow stored in the vacuum somehow to
be returned later.

When its fall rate
in the direction of the Sun and focal point is equal to the
slowing rate applied by the universe in its direction of motion,
all energy would be accounted for.

If Pioneer was in a circular orbit around the Sun it would
eventually arrive at a stable orbit radius, where it would be
orbiting faster than your maths would suggest.

No, your maths is broken, there is no stable orbit.

Dark matter does not explain the Pioneer anomaly,

No, but the Pioneer anomaly explains why your search for dark
matter is futile. It's essential that some effort be put in to
prove that the anomaly is in fact nothing more than a glitch in
the system. Until that is done, the search for dark matter is a
waste of time. If the Pioneer anomaly is real, then so is the
zero origin universe.

Sorry, Max, that's simply not true. You need to do a
lot of remedial revision of basic maths before you try
to work this out. If the Pioneer anomaly was due to
your anisotropy caused by the mass of the rest of the
universe, then Mercury would spiral into the Sun in a
million years. This diagram shows what your theory
predicts:

http://www.georgedishman.f2s.com/max/Mercury.png

and
your idea doesn't explain either that or galactic
rotation curves.

That's just hand waving George. If the anisotropy exists, then it
explains those things, and much more.

No it doesn't, it does not affect the velocity curves
at all since the acceleration is so slow that the
speed and radius are in equilibrium at all times, and
it is not hand-waving either, I have done the maths
and showed you that even for Mercury the effect is
many orders of magnitude too small to be detected.

However, you skills in maths are a
long way short of being able to manipulate the equations
to the point where you can follow the derivations. Either
you start learning stuff like this or you will be stuck
with taking my word for it (or that of others). It is
basic arithmetic that you should have learnt at least in
your first years of senior school so I don't know how you
could have missed out:

It has been a long time, but I know what I'm doing even if you
don't. You'll see the light eventually.

I know exactly what you are trying to do but you are
making serious mistakes in the maths and getting only
imaginary numbers that your theory doesn't actually
predict.

George