Pioneer Anomaly Anomalous No More.

Max, I'm going trim almost all your reply so we can
focus on the question of signed velocities. We can
come back to the other bits after that.

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

.. Take Mercury's maximum
radial velocity relative to the Sun on its journey to and from
its perihelion, for example. v of course cannot be negative,
regardless of direction.
....
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:

What on earth is a negative velocity? A car moving toward me at
60km/h is not moving at -60km/h toward me. Nor can it be moving
at -60km/h away from me if it's going the other way. Wherever
it's going it is traveling at +60km/h.

That is the speed, not the velocity.

If I'm following along at
a constant distance behind the car, the car's velocity is zero
relative to me. If I increase my speed by 10km/h, is the car
traveling at -10km/h ?

Relative to you yes. The distance between you is
reducing so to get a negative distance change in
a positive time you must have a negative relative
velocity. Of course both cars are still moving forward
so the first might be moving at 80km/h while the
second is doing 90km/h and the relative velocity
is 80-90 = -10.

Of course it's not, it's moving toward
me at 10km/h. And it's moving away at 10km/h if I decrease my
speed at that rate.

If you decrease your speed to 70km/h then the same
calculation now gives you 80-70 = +10 as you expect.

The + - switch should be applied to direction, not velocity.

Velocity is a vector and includes the direction so if
we think in terms of a car then we have a 2D situation
and we can define the velocity as north-south and
east-west components. For a car moving at 80km/h
going east, the velocity would be (0, 80). If it was going
west, it would be (0, -80). North is (80,0) while west is
(-80, 0). The velocity of a car moving with a speed of
80km/h in a north-easterly direction is (56.6, 56.6).

Speed is a scalar and is the magnitude of the velocity
is if an object is moving at (vn, ve) then its speed is
sqrt(vn^2 + vs^2).

The location of a car can similarly be defined by
distance east and distance north of some reference
location. Velocity is the derivative of location so if
the distance east decreases, then the easterly
component of the velocity is negative.

What you are doing is invalid.

What I am doing is basic standard physics and
entirely correct. What you are doing only works if
you know the direction in advance, but near
perihelon and aphelion, you may not be able to
know the direction until you work th acceleration
to find out whether the object has changed from
outward to inward, so you don't know which
equation to apply until you have applied it. The
standard scientific method I am using always
works since the direction is encoded in the sign.

George

Negative values for velocity were introduced early in my
11th-grade physics course (Chapter 5, "Motion along a
straight-line path"). They follow directly from the math
we learned in 7th grade.

If you want to measure positions and velocities of things,
you probably start doing so without even thinking about
the positive and negative directions. If I'm driving a
car, the forward direction naturally gets positive values,
and if I later need to drive backwards, it naturally gets
negative values. If my language is written left-to-right,
I will naturally count up from zero from left to right,
and will naturally assign higher positive values to greater
distances and speeds to the right. If I then want to find
distances or speeds to the left, they will naturally get
negative values. If I am launching rockets, I would
naturally count my progress in altitude or vertical speed
as larger positive values, and start running when I see
large negative values.

On the other hand, if I were drilling into the Earth or
diving under water, I might assign larger positive values
to greater depths and speeds of descent, and later, as the
need arises, assign negative values to heights above the
waterline and speeds of ascent.

Using a single continuum for positive and negative values
of any physical quantity is much easier to deal with
mathematically and numerically than using two separate
equations to deal with motion in opposite directions, and
seems obviously to make much more sense.

-- Jeff, in Minneapolis

"George Dishman" wrote in message
ups.com...
Max, I'm going trim almost all your reply so we can
focus on the question of signed velocities. We can
come back to the other bits after that.

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

.. Take Mercury's maximum
radial velocity relative to the Sun on its journey to and from
its perihelion, for example. v of course cannot be negative,
regardless of direction.
....
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:

What on earth is a negative velocity? A car moving toward me at
60km/h is not moving at -60km/h toward me. Nor can it be moving
at -60km/h away from me if it's going the other way. Wherever
it's going it is traveling at +60km/h.

That is the speed, not the velocity.

Why shouldn't speed and velocity be expected to abide by the same
rules? The dictionary doesn't specify any unique properties for
velocity that can't be applied to speed.

If I'm following along at
a constant distance behind the car, the car's velocity is zero
relative to me. If I increase my speed by 10km/h, is the car
traveling at -10km/h ?

Relative to you yes. The distance between you is
reducing so to get a negative distance change in
a positive time you must have a negative relative
velocity. Of course both cars are still moving forward
so the first might be moving at 80km/h while the
second is doing 90km/h and the relative velocity
is 80-90 = -10.

I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though. How can anyone possibly have any confidence in
mathematics if it's going to give the conflicting result that
you're proposing?

This time I'll adopt your "toy value" system.
G = 1: M = 1: r = 1: c = 1.
# = (c+v) * -(G*M/r^2)+(G*M/r^2)
Note that -(G*M/r^2) = -1 and (G*M/r^2) = 1.

For v = .01 (c+v=1.01)
# = 1.01 * -1 + 1 = -.01

For v = -.01 (c+v=.99)
# = .99 * -1 + 1 = .01

Since you didn't include the + (G*M/r^2) in your analysis, while
the difference between the two results is still the same, they
are both negative. # = 1.01 * -1 = -1.01
# = .99 * -1 = -.99

If you don't get the same answer when you start shifting numbers
about, you've done something wrong. It's hardly likely that the
math will be at fault.

-----

Max Keon

"Max Keon" wrote in message
...

"George Dishman" wrote in message
ups.com...
Max, I'm going trim almost all your reply so we can
focus on the question of signed velocities. We can
come back to the other bits after that.

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

.. Take Mercury's maximum
radial velocity relative to the Sun on its journey to and from
its perihelion, for example. v of course cannot be negative,
regardless of direction.
...
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:

What on earth is a negative velocity? A car moving toward me at
60km/h is not moving at -60km/h toward me. Nor can it be moving
at -60km/h away from me if it's going the other way. Wherever
it's going it is traveling at +60km/h.

That is the speed, not the velocity.

Why shouldn't speed and velocity be expected to abide by the same
rules? The dictionary doesn't specify any unique properties for
velocity that can't be applied to speed.

Your dictionary should have told you that velocity
includes direction. I explained the difference but
you snipped it:

Velocity is a vector and includes the direction so if
we think in terms of a car then we have a 2D situation
and we can define the velocity as north-south and
east-west components. For a car moving at 80km/h
going east, the velocity would be (0, 80). If it was going
west, it would be (0, -80). North is (80,0) while west is
(-80, 0). The velocity of a car moving with a speed of
80km/h in a north-easterly direction is (56.6, 56.6).

Speed is a scalar and is the magnitude of the velocity
is if an object is moving at (vn, ve) then its speed is
sqrt(vn^2 + vs^2).

This is going to be critical for resolving our disagreements
so please make sure you understand it. Note the velocities
are all shown as number pairs. Try examples in all four
quadrants to see where the components are negative and
positive:

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=68

If you don't have Java installed you can get it he

http://www.java.com/en/

This doesn't use Java:

http://www.glenbrook.k12.il.us/GBSSC...ors/u3l1b.html

If I'm following along at
a constant distance behind the car, the car's velocity is zero
relative to me. If I increase my speed by 10km/h, is the car
traveling at -10km/h ?

Relative to you yes. The distance between you is
reducing so to get a negative distance change in
a positive time you must have a negative relative
velocity. Of course both cars are still moving forward
so the first might be moving at 80km/h while the
second is doing 90km/h and the relative velocity
is 80-90 = -10.

I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.

You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.

How can anyone possibly have any confidence in
mathematics if it's going to give the conflicting result that
you're proposing?

This time I'll adopt your "toy value" system.
G = 1: M = 1: r = 1: c = 1.
# = (c+v) * -(G*M/r^2)+(G*M/r^2)
Note that -(G*M/r^2) = -1 and (G*M/r^2) = 1.

For v = .01 (c+v=1.01)
# = 1.01 * -1 + 1 = -.01

That is negative so indicates the anisotropic force is
towards the Sun.

For v = -.01 (c+v=.99)
# = .99 * -1 + 1 = .01

That is positive so indicates the anisotropic force is
away from the Sun. I believe those answers are both
what you expected.

Since you didn't include the + (G*M/r^2) in your analysis, while
the difference between the two results is still the same, they
are both negative. # = 1.01 * -1 = -1.01
# = .99 * -1 = -.99

Yes, we tend to lay aside the normal gravity when
discussing your slight change. Overall for a positive
velocity indicating motion away from the Sun at 0.01c
the total pull towards the Sun is increased to 101% of
the usual Newtonian value while for an object moving
towards the Sun at 0.01c the pull is reduced to 99%.
Correct me if I'm wrong but I thought that was what
you expected.

If you don't get the same answer when you start shifting numbers
about, you've done something wrong. It's hardly likely that the
math will be at fault.

Indeed but my 'shifting about' always gave the same
answers as your equations.

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

always gives the same answers as

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

while

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

always gives the same answers as

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

The rules are mostly he

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

but also (x^a)^b = x^(a*b) so (x^2)^0.5 = x^1 = x or simply
the square root of the square of x is just x.

The problem with the direction was something quite different,
you didn't realise physics uses signed velocities and that
change of sign means you don't need to use two different
equations, a single equation does the job.

George

"George Dishman" wrote in message
...
"Max Keon" wrote in message
...
------
------
I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.

You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.

I've updated the web page accordingly.
http://www.optusnet.com.au/~maxkeon/pionomor.html
While I was there, I made another attempt at finding the words
to better describe how Mercury's apparent loss of orbit momentum
is conserved when it finally comes to rest in a stable orbit that
counteracts the influence from the universe. It's impossible to
justify any energy transfer between Mercury and the mass of the
surrounding universe. It must be accounted for locally.

The fact that the Pioneer anomaly exists demands that the
question be resolved, one way or another, and a dedicated mission
seems to be the only way to do that. Such a mission can tell us
**much** about the universe, and is absolutely vital to the
progress of physics.

Afterall, truth is the ultimate goal, whatever the cost. So why
waste time and money chasing rainbows if that experiment has the
potential to turn physics upside down? It needs to be resolved
before we go searching for dark matter, surely?

-----

Max Keon

Repeating post lost by ISP:

"Max Keon" wrote in message
...

"George Dishman" wrote in message
...
"Max Keon" wrote in message
...
------
------
I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.

You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.

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)

It is then obvious that your are taking the conventional
Newtonian equation (-G*M/r^2) which everyone recognises
and modifying it by the (1 + v/c) term. It also makes it
clear that the result is first order where anyone looking
at your versions would see the speed appearing only as
"(c+v)^2" and assume it is second order.

While I was there, I made another attempt at finding the words
to better describe how Mercury's apparent loss of orbit momentum
is conserved when it finally comes to rest in a stable orbit that
counteracts the influence from the universe.

It is still wrong, it does not conserve the momentum. Have
you learned stuff about vector addition yet? Momentum is a
vector so you _cannot_ conserve it without dealing with the
direction issue.

Assuming you are now familiar with the scalar laws, you
next need to revise the vectors. You snipped that part:

Velocity is a vector and includes the direction so if
we think in terms of a car then we have a 2D situation
and we can define the velocity as north-south and
east-west components. For a car moving at 80km/h
going east, the velocity would be (0, 80). If it was going
west, it would be (0, -80). North is (80,0) while west is
(-80, 0). The velocity of a car moving with a speed of
80km/h in a north-easterly direction is (56.6, 56.6).

Speed is a scalar and is the magnitude of the velocity
is if an object is moving at (vn, ve) then its speed is
sqrt(vn^2 + vs^2).

This is going to be critical for resolving our disagreements
so please make sure you understand it. Note the velocities
are all shown as number pairs. Try examples in all four
quadrants to see where the components are negative and
positive:

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=68

If you don't have Java installed you can get it he

http://www.java.com/en/

This doesn't use Java:

http://www.glenbrook.k12.il.us/GBSSC...ors/u3l1b.html

You are going to need to know that maths in order to understand
the next steps.

Just as a reminder of where your theory is at the moment,
applying the equation above, either your form (the one with
the correct sign) or mine, with a value of "the mass of the
rest of the universe" taken from the Pioneer anomaly produces
a simple decay:

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

That rules out your model though a smaller mass which is
insufficient to explain Pioneer may allow your idea to survive
that test.

It's impossible to
justify any energy transfer between Mercury and the mass of the
surrounding universe. It must be accounted for locally.

In a collision between two objects, momentum is redistributed
between them which can be considered local. The effect of
gravity is not local. Right now Pioneer is being slowed by
the Sun and losing momentum. If the total is to be conserved
then that has to be matched by some other equal and opposite
change at the same time. 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.

The fact that the Pioneer anomaly exists demands that the
question be resolved, one way or another, and a dedicated mission
seems to be the only way to do that. Such a mission can tell us
**much** about the universe, and is absolutely vital to the
progress of physics.

Afterall, truth is the ultimate goal, whatever the cost. So why
waste time and money chasing rainbows if that experiment has the
potential to turn physics upside down? It needs to be resolved
before we go searching for dark matter, surely?

Many people would like to see a mission perhaps as an adjunct
to an existing plan, but spending money to chase a gas leak
or whatever when major plans are being postponed or cancelled
to fund Bush's publicity stunts is unjustifiable.

George

"George Dishman" wrote in message
ups.com...
Repeating post lost by ISP:

"Max Keon" wrote in message
...

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

I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.

You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.

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?

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)
Your simplified equation is not describing the cause of the
gravity anisotropy at all. It's very confusing in my opinion.

In the original format, as two separate equations, each equation
described very clearly how an object moving toward or away from a
gravity source will be affected. In order to achieve that goal,
there was nothing wrong with the way they were written in this
extract from the web page.

--
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.

It is then obvious that your are taking the conventional
Newtonian equation (-G*M/r^2) which everyone recognises
and modifying it by the (1 + v/c) term. It also makes it
clear that the result is first order where anyone looking
at your versions would see the speed appearing only as
"(c+v)^2" and assume it is second order.

While I was there, I made another attempt at finding the words
to better describe how Mercury's apparent loss of orbit momentum
is conserved when it finally comes to rest in a stable orbit that
counteracts the influence from the universe.

It is still wrong, it does not conserve the momentum. Have
you learned stuff about vector addition yet? Momentum is a
vector so you _cannot_ conserve it without dealing with the
direction issue.

Assuming you are now familiar with the scalar laws, you
next need to revise the vectors. You snipped that part:

Velocity is a vector and includes the direction so if
we think in terms of a car then we have a 2D situation
and we can define the velocity as north-south and
east-west components. For a car moving at 80km/h
going east, the velocity would be (0, 80). If it was going
west, it would be (0, -80). North is (80,0) while west is
(-80, 0). The velocity of a car moving with a speed of
80km/h in a north-easterly direction is (56.6, 56.6).

Speed is a scalar and is the magnitude of the velocity
is if an object is moving at (vn, ve) then its speed is
sqrt(vn^2 + vs^2).

This is going to be critical for resolving our disagreements
so please make sure you understand it. Note the velocities
are all shown as number pairs. Try examples in all four
quadrants to see where the components are negative and
positive:

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=68

If you don't have Java installed you can get it he

http://www.java.com/en/

This doesn't use Java:

http://www.glenbrook.k12.il.us/GBSSC...ors/u3l1b.html

You are going to need to know that maths in order to understand
the next steps.

Just as a reminder of where your theory is at the moment,
applying the equation above, either your form (the one with
the correct sign) or mine, with a value of "the mass of the
rest of the universe" taken from the Pioneer anomaly produces
a simple decay:

That is to be expected of course because the math was never
required to incorporate a velocity related gravity anisotropy.
You later claim that momentum must be immediately conserved, but
that is clearly an impossibility. There is always a time delay
involved.

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

That rules out your model though a smaller mass which is
insufficient to explain Pioneer may allow your idea to survive
that test.

It's impossible to
justify any energy transfer between Mercury and the mass of the
surrounding universe. It must be accounted for locally.

In a collision between two objects, momentum is redistributed
between them which can be considered local.

And a time delay is involved in the process if there is any
component separation at all, even within the colliding
components themselves. It takes time to redistribute momentum.

The effect of
gravity is not local. Right now Pioneer is being slowed by
the Sun and losing momentum. If the total is to be conserved
then that has to be matched by some other equal and opposite
change at the same time.

The Sun-Pioneer relationship is fairly constant, so even though
they are both being drawn toward each other in a way the appears
to immediately conserve momentum, the ten hour signal delay is
still present. Neither the Sun or Pioneer could react instantly
if the other suddenly ceased to exist.

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. That statement
implies that math dictates how nature must behave. 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.

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 fact that the Pioneer anomaly exists demands that the
question be resolved, one way or another, and a dedicated mission
seems to be the only way to do that. Such a mission can tell us
**much** about the universe, and is absolutely vital to the
progress of physics.

Afterall, truth is the ultimate goal, whatever the cost. So why
waste time and money chasing rainbows if that experiment has the
potential to turn physics upside down? It needs to be resolved
before we go searching for dark matter, surely?

Many people would like to see a mission perhaps as an adjunct
to an existing plan, but spending money to chase a gas leak
or whatever when major plans are being postponed or cancelled
to fund Bush's publicity stunts is unjustifiable.

Gas leaks or not, until the Pioneer anomaly has been properly
resolved, there is no point whatever searching for the pot of
dark matter at the end of the rainbow. Which is what it all boils
down to in the end, isn't it?

A dedicated mission would seem to be absolutely essential.

-----

Max Keon

"Max Keon" wrote in message
u...

"George Dishman" wrote in message
ups.com...
Repeating post lost by ISP:

"Max Keon" wrote in message
...

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

I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.

You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.

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. 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.

Your simplified equation is not describing the cause of the
gravity anisotropy at all. It's very confusing in my opinion.

I think you just need to learn these basic laws of
arithmetic well enough to be able to read equations:

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


In the original format, as two separate equations, each equation
described very clearly how an object moving toward or away from a
gravity source will be affected. In order to achieve that goal,
there was nothing wrong with the way they were written in this
extract from the web page.

Except that the have the acceleration pointing in the same
direction whether the body is moving towards or away from
the Sun.

--
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.

It is then obvious that your are taking the conventional
Newtonian equation (-G*M/r^2) which everyone recognises
and modifying it by the (1 + v/c) term. It also makes it
clear that the result is first order where anyone looking
at your versions would see the speed appearing only as
"(c+v)^2" and assume it is second order.

and that is why your version is confusing.

While I was there, I made another attempt at finding the words
to better describe how Mercury's apparent loss of orbit momentum
is conserved when it finally comes to rest in a stable orbit that
counteracts the influence from the universe.

It is still wrong, it does not conserve the momentum. Have
you learned stuff about vector addition yet? Momentum is a
vector so you _cannot_ conserve it without dealing with the
direction issue.

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

Just as a reminder of where your theory is at the moment,
applying the equation above, either your form (the one with
the correct sign) or mine, with a value of "the mass of the
rest of the universe" taken from the Pioneer anomaly produces
a simple decay:

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

That is to be expected of course because the math was never
required to incorporate a velocity related gravity anisotropy.

That graph uses _your_ math of course.

You later claim that momentum must be immediately conserved, but
that is clearly an impossibility. There is always a time delay
involved.

Sorry, a delay is impossible, the word "conserved" means
that the total is unchanging. If Pioneer loses some momentum
at some time and the Sun gets it after a delay, then for the
duation of the delay, the total is lower so the value isn't
conserved, it changes.

In a collision between two objects, momentum is redistributed
between them which can be considered local.

And a time delay is involved in the process if there is any
component separation at all, even within the colliding
components themselves. It takes time to redistribute momentum.

Nope, if that were true, it would not be conserved by
definition.

The effect of
gravity is not local. Right now Pioneer is being slowed by
the Sun and losing momentum. If the total is to be conserved
then that has to be matched by some other equal and opposite
change at the same time.

The Sun-Pioneer relationship is fairly constant, so even though
they are both being drawn toward each other in a way the appears
to immediately conserve momentum, the ten hour signal delay is
still present. Neither the Sun or Pioneer could react instantly
if the other suddenly ceased to exist.

That is true in GR but not in the Newtonian gravity
that you are modifying.

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.

That statement
implies that math dictates how nature must behave.

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

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. 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.

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.

The fact that the Pioneer anomaly exists demands that the
question be resolved, one way or another, and a dedicated mission
seems to be the only way to do that. Such a mission can tell us
**much** about the universe, and is absolutely vital to the
progress of physics.

Afterall, truth is the ultimate goal, whatever the cost. So why
waste time and money chasing rainbows if that experiment has the
potential to turn physics upside down? It needs to be resolved
before we go searching for dark matter, surely?

Many people would like to see a mission perhaps as an adjunct
to an existing plan, but spending money to chase a gas leak
or whatever when major plans are being postponed or cancelled
to fund Bush's publicity stunts is unjustifiable.

Gas leaks or not, until the Pioneer anomaly has been properly
resolved, there is no point whatever searching for the pot of
dark matter at the end of the rainbow. Which is what it all boils
down to in the end, isn't it?

Dark matter does not explain the Pioneer anomaly, and
your idea doesn't explain either that or galactic
rotation curves. 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:

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

George

"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

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. 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.

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.
------

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.
------

The effect of
gravity is not local. Right now Pioneer is being slowed by
the Sun and losing momentum. If the total is to be conserved
then that has to be matched by some other equal and opposite
change at the same time.

The Sun-Pioneer relationship is fairly constant, so even though
they are both being drawn toward each other in a way the appears
to immediately conserve momentum, the ten hour signal delay is
still present. Neither the Sun or Pioneer could react instantly
if the other suddenly ceased to exist.

That is true in GR but not in the Newtonian gravity
that you are modifying.

This is a whole new ball game based on gravity in the zero origin
universe. It has nothing to do with Newton.

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.

That statement
implies that math dictates how nature must behave.

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

????

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.

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

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?

+ 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.

+ 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. 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.

The fact that the Pioneer anomaly exists demands that the
question be resolved, one way or another, and a dedicated mission
seems to be the only way to do that. Such a mission can tell us
**much** about the universe, and is absolutely vital to the
progress of physics.

Afterall, truth is the ultimate goal, whatever the cost. So why
waste time and money chasing rainbows if that experiment has the
potential to turn physics upside down? It needs to be resolved
before we go searching for dark matter, surely?

Many people would like to see a mission perhaps as an adjunct
to an existing plan, but spending money to chase a gas leak
or whatever when major plans are being postponed or cancelled
to fund Bush's publicity stunts is unjustifiable.

Gas leaks or not, until the Pioneer anomaly has been properly
resolved, there is no point whatever searching for the pot of
dark matter at the end of the rainbow. Which is what it all boils
down to in the end, isn't it?

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.

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.

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.

-----

Max Keon

"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