Pioneer Anomaly Anomalous No More.

"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
Maths is a tool independent of the physics. Your
equation is supposed to tell me what effect your
anisotropy has and I have just applied that.

Don't be ridiculous. Did the maths invent the big bang universe,
or did that universe determine how the maths would be applied?
If theory predicts a CMBR, the maths is applied accordingly. If
a gravity anisotropy is predicted, the maths is applied
accordingly. ...

The maths must be designed
to incorporate such a thing.

And I assume that is what your equation does.

It certainly does in the zero origin universe.

Then when I use your equation, it will tell me what will
happen in the zero origin universe. That's what I do and
the result is a consequence you didn't anticipate and don't
like, but that's what it is.

You are still in the wrong universe. My equation doesn't design
the zero origin universe, the universe designs the equation.

Sure, and then the equation tells us what will happen to
the planets in your universe - they spiral into the Sun.
Theories often have consequences that weren't anticipated
by their originators. Planck could never have known the
consequences of his original lecture on quantisation as a
solution to black body radiation.

And that sent you all off on another wild goose chase. _Blackbody
radiation has never had anything to do with quantitization_. The
value for the photon has just been plucked out of the air. It
could have been anything at all. But you must surely know that?

Take this part of the formula for example: (2*pi*h*c^2).
Since every component is a constant, the whole thing can be
treated as a simple multiplier. That's all it is, isn't it! So
why isn't it stated as 3.747e-16. But 3.747e-16 what? It's
nothing more than a nominated surface area per unit radius from
the blackbody source on which to establish an energy quantity
according to wavelength. If the unit radius is 1 meter, the
designated surface area is 3.747e-16 square meters.

According to Planck, the energy received from a 4000 K radiator
at 7.25e-7 meter wavelength is
(2 * pi * h * c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
= 1.315e+13 (# units of energy, whatever they are)

Here's an equation that gives the emissive power per wavelength
received at unit radius from the same blackbody source for a 1
square meter surface area. w is wavelength, m is surface area.
m^2 / w^5 / (1.0145^(m / (w * t)) - 1) = 3.509e+28 (# units)

3.509e+28 * 3.747e-16 = 1.315e+13, as the Planck "photon" would
specify it for the tiny surface area.

Try any temperature-wavelength combination you like and you will
always get the right answer.

Planck's photon is entirely dependent on "e" which has no claim
to fame other than it never repeats. That is crap George.

You've done nothing else for weeks and it is based on
a fantasy world weher unlimited money is available for
launching spacecrfat without having any idea what
instruments to put on board to resolve a question in
the hope of blundering across an answer.

How hard would it be to send exactly the same Pioneer 10-11
configuration on a trip to Neptune and back?

Several million dollars hard.

How hard would it be
to detect the anomalous acceleration on the outward and inward
legs? How hard would it be to notice that one is pointing in the
opposite direction to the other, relative to the Sun? Whatever is
the cause, there should be some sort of anisotropy evident in the
result

Sure, but that gets you no farther forward in identifying
the _cause_ of the effect.

It would certainly prove that the effect is real though. Then we
can move on.
---

But GR's amazing feat is quite irrelevant anyway because
instantaneous action at a distance was never a requirement in
anybody's gravity.

It was in Newton's.

Only because he didn't understand gravity at all. The fact is
that something was missing in the way he analyzed planetary
motion. And that is clearly still the case today.
---

No Max, you are off into fantasy world again. The
focal point of elliptical orbits is the barycentre. The
slight effect of a delay would move them from there
so you get a diagram like this:

0-


+ . +


-0

I've fixed the diagram for you.

No, you have screwed it up again so I restored what
I wrote. Go and learn vectors, then read up on the
two-body problem

The following accompanies an animation at this address.
http://www.optusnet.com.au/~maxkeon/binstar.html

The animation depicts a binary star pair (blue) in a concentric
orbit about each other. The star masses are identical. This an
impossible scenario of course because the orbit velocity of each
star is faster than light speed. The geometry can obviously be
altered to overcome that problem, but my point is still made.

The action of gravity at one point around the orbit only is
shown, and that is obviously common throughout the entire orbit
cycle. The nearest red dot in the direction of motion is the
retarded position of each companion star from the viewpoint of
each star.

The '\' line travels the straight line distance to the intersect
point with its companion at the speed of light. The action of
gravity on each star is necessarily shifted to point toward the
retarded image of its companion, which is 90 degrees offset from
its true position around the orbit path. The apparent gravity
link with each companion star can obviously never be generated
where the companion currently resides.

The focal point of the orbit radius is perpendicular to the
natural tangent, which would cause the stars to fly apart. But
the distance from each star to where its companion appears to be
is .707 of the instantaneous distance. According to 1/r^2, the
pull of gravity toward each other is 1 / .707^2 which is double
that required to hold the stars at the instantaneous orbit
radius. The stars would be pulled inward to point in the
direction of the instantaneous orbit tangent.

Even if that's not enough to do the trick, the universe has no
rules which dictate that a binary star pair of specific mass can
only reside at a specific radius for a specific orbit velocity.
A stable orbit exists for every circumstance. Denying that is
absurd.

There is no reason whatever why the universe should comply with
theory. Nature sets the rules, not you, or me.
---

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.

In which case the anisotropic force pushes the body
in the same direction regardless of the direction of
motion, the opposite of what you told me.

Even an idiot could understand my meaning.
---

That wasn't the case at all. I'll try to explain each step of the
process in some detail (for my own benefit).

The conclusion was that no energy (or momentum) can be
immediately absorbed by the matter of the universe,

That's fine, but energy is being removed by the anisotropy
immediately, it is up to you to say where it goes.

so the force
remains like a spring ..

No it isn't. A spring always pulls in the same direction
like gravity but your anisotropy changes direction. A
spring removes and stores energy while being stretched
and returns it while being relaxed because the force is
in the same direction as the speed.

The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif.jpg

which is applying a constant restraining
force on Mercury's orbital motion. That action will of course
initially slow Mercury, which in turn will begin a slow
acceleration toward the Sun. As you say, that reaction cannot be
elastic, and momentum is lost.

As is energy.

But still no energy has been
transferred away from what can only be a locally closed system.

You could hypothesise that the energy and momentum
are stored somewhere locally but that is for you to
work out.

The momentum loss immediately converts to potential energy, which
in turn slowly converts to kinetic energy.

Wrong, momentum and energy are not interchangeable, they
are separate quantities, both of which must be separately
conserved (as must the three spatial components of momentum).

In an inelastic, head on collision between two equal masses
the two masses are brought to a halt relative to each other.
Heat energy is obviously generated during the process. So where
does that come from if not from lost momentum?
e=pc^2 . That's what I think.

According to your maths, that process continues until Mercury
hits the Sun in a million years or so.

No, according to _your_ equation, that's what happens.
I just applied it for you without making your mistake
of confusing energy and momentum.

The eccentricity would not be affected by the universe generated
gravity anisotropy at all when Mercury arrives at a stable orbit.

You won't be able to work out what it does until you
learn how to handle vectors, differential equations and
the difference between energy and momentum.

The need for instantaneous action at a distance to overcome the
problem of the planets spiraling into the Sun should have
sounded alarm bells.

It did and people were trying to resolve it for 200
years until Einstein found the solution.

You know as well as I do that the "solution" fails the Pound and
Rebka test. I suggest you start looking for another answer.

-----

Max Keon

Max Keon wrote:
"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
Maths is a tool independent of the physics. Your
equation is supposed to tell me what effect your
anisotropy has and I have just applied that.

Don't be ridiculous. Did the maths invent the big bang universe,
or did that universe determine how the maths would be applied?
If theory predicts a CMBR, the maths is applied accordingly. If
a gravity anisotropy is predicted, the maths is applied
accordingly. ...

The maths must be designed
to incorporate such a thing.

And I assume that is what your equation does.

It certainly does in the zero origin universe.

Then when I use your equation, it will tell me what will
happen in the zero origin universe. That's what I do and
the result is a consequence you didn't anticipate and don't
like, but that's what it is.

You are still in the wrong universe. My equation doesn't design
the zero origin universe, the universe designs the equation.

Sure, and then the equation tells us what will happen to
the planets in your universe - they spiral into the Sun.
Theories often have consequences that weren't anticipated
by their originators. Planck could never have known the
consequences of his original lecture on quantisation as a
solution to black body radiation.

And that sent you all off on another wild goose chase. _Blackbody
radiation has never had anything to do with quantitization_. The
value for the photon has just been plucked out of the air. It
could have been anything at all. But you must surely know that?

Take this part of the formula for example: (2*pi*h*c^2).
Since every component is a constant, the whole thing can be
treated as a simple multiplier. That's all it is, isn't it! So
why isn't it stated as 3.747e-16. But 3.747e-16 what? It's
nothing more than a nominated surface area per unit radius from
the blackbody source on which to establish an energy quantity
according to wavelength. If the unit radius is 1 meter, the
designated surface area is 3.747e-16 square meters.

According to Planck, the energy received from a 4000 K radiator
at 7.25e-7 meter wavelength is
(2 * pi * h * c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
= 1.315e+13 (# units of energy, whatever they are)

Here's an equation that gives the emissive power per wavelength
received at unit radius from the same blackbody source for a 1
square meter surface area. w is wavelength, m is surface area.
m^2 / w^5 / (1.0145^(m / (w * t)) - 1) = 3.509e+28 (# units)

3.509e+28 * 3.747e-16 = 1.315e+13, as the Planck "photon" would
specify it for the tiny surface area.

Try any temperature-wavelength combination you like and you will
always get the right answer.

Planck's photon is entirely dependent on "e" which has no claim
to fame other than it never repeats. That is crap George.

Nice rant Max, but it doesn't even touch on the reason
why quantisation is required, the shape of the spectrum.


How hard would it be
to detect the anomalous acceleration on the outward and inward
legs? How hard would it be to notice that one is pointing in the
opposite direction to the other, relative to the Sun? Whatever is
the cause, there should be some sort of anisotropy evident in the
result

Sure, but that gets you no farther forward in identifying
the _cause_ of the effect.

It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

But GR's amazing feat is quite irrelevant anyway because
instantaneous action at a distance was never a requirement in
anybody's gravity.

It was in Newton's.

Only because he didn't understand gravity at all. The fact is
that something was missing in the way he analyzed planetary
motion. And that is clearly still the case today.

No, GR resolved that problem 90 years ago. Science
has moved on a long way since then.

---

No Max, you are off into fantasy world again. The
focal point of elliptical orbits is the barycentre. The
slight effect of a delay would move them from there
so you get a diagram like this:

0-


+ . +


-0

I've fixed the diagram for you.

No, you have screwed it up again so I restored what
I wrote. Go and learn vectors, then read up on the
two-body problem

The following accompanies an animation at this address.
http://www.optusnet.com.au/~maxkeon/binstar.html

The animation depicts a binary star pair (blue) in a concentric
orbit about each other. The star masses are identical. This an
impossible scenario of course because the orbit velocity of each
star is faster than light speed.

I assume you mean your illustration shows impossible
speeds.

The geometry can obviously be
altered to overcome that problem, but my point is still made.

The geometry is not a problem. There are many such
binary systems in reality but obviously the orbital speed
is much slower. Something around 300 km/s is the top
speed for a "grazing" binary IIRC but binaries of nearly
equal mass are not unusual.

The action of gravity at one point around the orbit only is
shown, and that is obviously common throughout the entire orbit
cycle. The nearest red dot in the direction of motion is the
retarded position of each companion star from the viewpoint of
each star.

The '\' line travels the straight line distance to the intersect
point with its companion at the speed of light. The action of
gravity on each star is necessarily shifted to point toward the
retarded image of its companion, which is 90 degrees offset from
its true position around the orbit path. The apparent gravity
link with each companion star can obviously never be generated
where the companion currently resides.

Sure, but only because you have assumed there is abberation
which isn't the case in Newtonian gravity or GR, you are only
describing a strawman so I don't see the point.

The focal point of the orbit radius is perpendicular to the
natural tangent,

No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.

which would cause the stars to fly apart. But
the distance from each star to where its companion appears to be
is .707 of the instantaneous distance. According to 1/r^2, the
pull of gravity toward each other is 1 / .707^2 which is double
that required to hold the stars at the instantaneous orbit
radius. The stars would be pulled inward to point in the
direction of the instantaneous orbit tangent.

Even if that's not enough to do the trick, the universe has no
rules which dictate that a binary star pair of specific mass can
only reside at a specific radius for a specific orbit velocity.
A stable orbit exists for every circumstance. Denying that is
absurd.

Crap. For any radius, if the speed is too fast for a circular
orbit the object will move farther away and if it is too slow
it moves closer. Both give stable elliptical orbits because
energy and momentum are conserved, but as sonn as you
start to remove energy the orbit decays. That's what happens
in nature so that's what our maths has to copy.

There is no reason whatever why the universe should comply with
theory. Nature sets the rules, not you, or me.

Yeah, Max. You see the thing is that we know nature is just
going to do its own thing and ignore us, so the rule for science
is that the maths we use is required to mimic nature

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.

In which case the anisotropic force pushes the body
in the same direction regardless of the direction of
motion, the opposite of what you told me.

Even an idiot could understand my meaning.

Probably, but your equation says the opposite of your meaning.

That wasn't the case at all. I'll try to explain each step of the
process in some detail (for my own benefit).

The conclusion was that no energy (or momentum) can be
immediately absorbed by the matter of the universe,

That's fine, but energy is being removed by the anisotropy
immediately, it is up to you to say where it goes.

so the force
remains like a spring ..

No it isn't. A spring always pulls in the same direction
like gravity but your anisotropy changes direction. A
spring removes and stores energy while being stretched
and returns it while being relaxed because the force is
in the same direction as the speed.

The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif.jpg

Wrong, you have no anisotropy in that.

which is applying a constant restraining
force on Mercury's orbital motion. That action will of course
initially slow Mercury, which in turn will begin a slow
acceleration toward the Sun. As you say, that reaction cannot be
elastic, and momentum is lost.

As is energy.

But still no energy has been
transferred away from what can only be a locally closed system.

You could hypothesise that the energy and momentum
are stored somewhere locally but that is for you to
work out.

The momentum loss immediately converts to potential energy, which
in turn slowly converts to kinetic energy.

Wrong, momentum and energy are not interchangeable, they
are separate quantities, both of which must be separately
conserved (as must the three spatial components of momentum).

In an inelastic, head on collision between two equal masses
the two masses are brought to a halt relative to each other.
Heat energy is obviously generated during the process. So where
does that come from if not from lost momentum?

From lost energy. There is no momentum lost in the crash.

e=pc^2 . That's what I think.

It is nonsense. Until you learn what a vector is, you won't
even understand this answer. Momentum is a set of three
numbers, one for each of the three spatial directions. Energy
is a single number, it doesn't have a direction so you cannot
equate them that way at all.

According to your maths, that process continues until Mercury
hits the Sun in a million years or so.

No, according to _your_ equation, that's what happens.
I just applied it for you without making your mistake
of confusing energy and momentum.

The eccentricity would not be affected by the universe generated
gravity anisotropy at all when Mercury arrives at a stable orbit.

You won't be able to work out what it does until you
learn how to handle vectors, differential equations and
the difference between energy and momentum.

The need for instantaneous action at a distance to overcome the
problem of the planets spiraling into the Sun should have
sounded alarm bells.

It did and people were trying to resolve it for 200
years until Einstein found the solution.

You know as well as I do that the "solution" fails the Pound and
Rebka test.

What are you talking about Max? Pound-Rebka is one of
the classic tests which _confirm_ GR ! There are no tests
whatsoever that GR fails, only known limitations in merging
it with QM, which could as easily indicate a failing of QM.

I suggest you start looking for another answer.

I suggest you learn how to handle negative numbers and
what a vector is because we cannot continue this
discussion beyond primary school level until you have
those basic maths tools at your disposal.

George

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

You are still in the wrong universe. My equation doesn't design
the zero origin universe, the universe designs the equation.

Sure, and then the equation tells us what will happen to
the planets in your universe - they spiral into the Sun.
Theories often have consequences that weren't anticipated
by their originators. Planck could never have known the
consequences of his original lecture on quantisation as a
solution to black body radiation.

And that sent you all off on another wild goose chase. _Blackbody
radiation has never had anything to do with quantitization_. The
value for the photon has just been plucked out of the air. It
could have been anything at all. But you must surely know that?

Take this part of the formula for example: (2*pi*h*c^2).
Since every component is a constant, the whole thing can be
treated as a simple multiplier. That's all it is, isn't it! So
why isn't it stated as 3.747e-16. But 3.747e-16 what? It's
nothing more than a nominated surface area per unit radius from
the blackbody source on which to establish an energy quantity
according to wavelength. If the unit radius is 1 meter, the
designated surface area is 3.747e-16 square meters.

According to Planck, the energy received from a 4000 K radiator
at 7.25e-7 meter wavelength is
(2 * pi * h * c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
= 1.315e+13 (# units of energy, whatever they are)

Here's an equation that gives the emissive power per wavelength
received at unit radius from the same blackbody source for a 1
square meter surface area. w is wavelength, m is surface area.
m^2 / w^5 / (1.0145^(m / (w * t)) - 1) = 3.509e+28 (# units)

3.509e+28 * 3.747e-16 = 1.315e+13, as the Planck "photon" would
specify it for the tiny surface area.

Try any temperature-wavelength combination you like and you will
always get the right answer.

Planck's photon is entirely dependent on "e" which has no claim
to fame other than it never repeats. That is crap George.

Nice rant Max, but it doesn't even touch on the reason
why quantisation is required, the shape of the spectrum.

It's a little off topic, but the easiest way to demonstrate my
point (to anyone) is to set up a simple program. Each equation
set can be extracted and compared outside the program. This
program of course runs as it is in Qbasic.

Notice the absence of Planck's constant in the alternative
equations. All of the magic in Planck's formula is performed by
the EXP function, which is the same as 2.71828^(h*f) ect.


CLS
pi = 3.14159
c = 3E+08
k = 1.3807E-23
h = 6.6262E-34
m = 1 'Meter unit.

INPUT " Radiator temperature (degrees K)"; t
IF t = 0 THEN END
PRINT " Note: Peak emission wave length is";(4000 / t) * 725; "nm"
PRINT "For spectral energy density the peak is";(4000/t)*1285;"nm"

ma: INPUT " Wave length in nanometers."; w
IF w = 0 THEN END
w = w / 1E+09 'wavelength is reset to meters.
cm = .01 / w
f = c / w
PRINT
'---------------------------------------
PRINT " Planck Alternative"
PRINT
PRINT " Emissive power per wavelength."

plnka = (2*pi*h* c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
alta = m ^ 2 / w ^ 5 / (1.0145016# ^ (m / (w * t)) - 1)

PRINT plnka, alta * 3.747E-16; "per meter units"
'The multiplier is constant.

plnkb = (2*pi*h*(c*10)^2)/((w*10)^5*((EXP((h* f) / (k * t))) - 1))
altb = (m*10)^2/(w*10)^5/(1.0145016#^((m* 10) / (w * 10 * t)) - 1)

PRINT plnkb, altb * 3.747E-16; "per cm units."
PRINT

PRINT " Spectral energy density."
PRINT " Frequency ="; f; "~/sec. 0r"; cm; "~/cm."
plnkc = (8*pi* h * f ^ 3) / (c ^ 2 * (EXP((h * f) / (k * t)) - 1))
altc = f ^ 3 / (c ^ 2 * (1.0145016# ^ (m / ((c / f) * t)) - 1))

PRINT plnkc, altc * 1.6654E-32; "per meter units"
'The multiplier is again constant.

plnkd = (8*pi*h*f^3)/((c * 10) ^ 2 * (EXP((h * f) / (k * t)) - 1))
altd = f^3/((c*10)^2*(1.0145016#^(m*10/(((c * 10) / f) * t)) - 1))

PRINT plnkd, altd * 1.6654E-32; "per cm units"

GOTO ma
------

These are two sets of randomly chosen results for a 4000 k
radiator.
------
Radiator temperature (degrees K)? 4000
Note: Peak emission wave length is 725 nm
For spectral energy density the peak is 1285 nm
(which can obviously be converted to frequency).

Wave length in nanometers.? 400
Planck Alternative
Emissive power per wavelength.
4.523494E+12 4.523528E+12 per meter units
4.523494E+09 4.523528E+09 per cm units.
Spectral energy density.
Frequency = 7.5E+14 ~/sec. 0r 25000 ~/cm.
9.650121E-09 9.650578E-09 per meter units
9.650121E-11 9.650578E-11 per cm units

Wave length in nanometers.? 800
Planck Alternative
Emissive power per wavelength.
1.285609E+13 1.28561E+13 per meter units
1.285609E+10 1.28561E+10 per cm units.
Spectral energy density.
Frequency = 3.75E+14 ~/sec. 0r 12500 ~/cm.
1.097053E-07 1.097097E-07 per meter units
1.097053E-09 1.097097E-09 per cm units


And two results for a 2.73 K radiator.

Radiator temperature (degrees K)? 2.73
Note: Peak emission wave length is 1062271 nm
For spectral energy density the peak is 1882784 nm

Wave length in nanometers.? 1e6
Planck Alternative
Emissive power per wavelength.
1.929899E-03 1.929902E-03 per meter units
1.929899E-06 1.929902E-06 per cm units.
Spectral energy density.
Frequency = 3E+11 ~/sec. 0r 10 ~/cm.
2.573199E-17 2.573306E-17 per meter units
2.573199E-19 2.573306E-19 per cm units

Wave length in nanometers.? 2e6

Planck Alternative
Emissive power per wavelength.
9.028213E-04 9.028193E-04 per meter units
9.028213E-07 9.028193E-07 per cm units.
Spectral energy density.
Frequency = 1.5E+11 ~/sec. 0r 5 ~/cm.
4.815047E-17 4.815229E-17 per meter units
4.815047E-19 4.815229E-19 per cm units
------

Frequency is also given in cycles per cm, which is how the CMBR
monopole is normally depicted. A common multiplier is of course
required to align the power spectrum peak with 384 MJy/sr.
Power * 7.94e+18 does the trick.

How hard would it be
to detect the anomalous acceleration on the outward and inward
legs? How hard would it be to notice that one is pointing in the
opposite direction to the other, relative to the Sun? Whatever is
the cause, there should be some sort of anisotropy evident in the
result

Sure, but that gets you no farther forward in identifying
the _cause_ of the effect.

It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

Yes, but you still think it could be caused by systematic errors.
So you really have no idea what the actual cause might be.

But GR's amazing feat is quite irrelevant anyway because
instantaneous action at a distance was never a requirement in
anybody's gravity.

It was in Newton's.

Only because he didn't understand gravity at all. The fact is
that something was missing in the way he analyzed planetary
motion. And that is clearly still the case today.

No, GR resolved that problem 90 years ago. Science
has moved on a long way since then.

And how much farther it would have gone if it was on the right
path. Mankind might not be in the mess it's now in. The clock is
certainly ticking.

Go and learn vectors, then read up on the
two-body problem

The following accompanies an animation at this address.
http://www.optusnet.com.au/~maxkeon/binstar.html

The animation depicts a binary star pair (blue) in a concentric
orbit about each other. The star masses are identical. This an
impossible scenario of course because the orbit velocity of each
star is faster than light speed.

I assume you mean your illustration shows impossible
speeds.

The geometry can obviously be
altered to overcome that problem, but my point is still made.

The geometry is not a problem. There are many such
binary systems in reality but obviously the orbital speed
is much slower. Something around 300 km/s is the top
speed for a "grazing" binary IIRC but binaries of nearly
equal mass are not unusual.

The action of gravity at one point around the orbit only is
shown, and that is obviously common throughout the entire orbit
cycle. The nearest red dot in the direction of motion is the
retarded position of each companion star from the viewpoint of
each star.

The '\' line travels the straight line distance to the intersect
point with its companion at the speed of light. The action of
gravity on each star is necessarily shifted to point toward the
retarded image of its companion, which is 90 degrees offset from
its true position around the orbit path. The apparent gravity
link with each companion star can obviously never be generated
where the companion currently resides.

Sure, but only because you have assumed there is abberation
which isn't the case in Newtonian gravity or GR, you are only
describing a strawman so I don't see the point.

No. I'm merely describing how gravity works in the real universe.

The focal point of the orbit radius is perpendicular to the
natural tangent,

No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.

Take another look at the animation.
http://www.optusnet.com.au/~maxkeon/binstar.gif There is no
doubt that each star is drawing toward where its companion
appears to be. That's the direction of the pull of gravity at
the star even if that path curves back to the companion. It
doesn't make any difference if the gravity link is curved or
straight, but the distance to where the companion was is
shorter. Dimension was being drawn into the companion's gravity
well at that time, so even if the companion is no longer in the
same place, the full affect of its presence will be felt when
its image coincides with the star's position in space.

which would cause the stars to fly apart. But
the distance from each star to where its companion appears to be
is .707 of the instantaneous distance. According to 1/r^2, the
pull of gravity toward each other is 1 / .707^2 which is double
that required to hold the stars at the instantaneous orbit
radius. The stars would be pulled inward to point in the
direction of the instantaneous orbit tangent.

Even if that's not enough to do the trick, the universe has no
rules which dictate that a binary star pair of specific mass can
only reside at a specific radius for a specific orbit velocity.
A stable orbit exists for every circumstance. Denying that is
absurd.

Crap. For any radius, if the speed is too fast for a circular
orbit the object will move farther away and if it is too slow
it moves closer.

It will find a stable orbit somewhere though, won't it.

Both give stable elliptical orbits because
energy and momentum are conserved, but as sonn as you
start to remove energy the orbit decays. That's what happens
in nature so that's what our maths has to copy.

What energy is removed in a simple orbit? A curved gravity link
isn't going to remove energy.
---

so the force
remains like a spring ..

No it isn't. A spring always pulls in the same direction
like gravity but your anisotropy changes direction. A
spring removes and stores energy while being stretched
and returns it while being relaxed because the force is
in the same direction as the speed.

The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif

Wrong, you have no anisotropy in that.

Of course not. The elastic force applied by the universe has
been counteracted by Mercury's increased centrifugal force as
it orbits the Sun at a faster rate than you would expect. That's
its natural orbit in the prevailing circumstances.
---

The momentum loss immediately converts to potential energy, which
in turn slowly converts to kinetic energy.

Wrong, momentum and energy are not interchangeable, they
are separate quantities, both of which must be separately
conserved (as must the three spatial components of momentum).

In an inelastic, head on collision between two equal masses
the two masses are brought to a halt relative to each other.
Heat energy is obviously generated during the process. So where
does that come from if not from lost momentum?

From lost energy. There is no momentum lost in the crash.

Momentum may not have been lost from the universe, but neither
mass has momentum relative to the other. They are stationary.
The expended energy would need to be put back to restore the
original relationship, and that could point them in any opposing
directions. Their momentum relative to each other, and relative
to the universe, can now be pointing anywhere.

So what are you on about?

Mercury's momentum is transferred to the "elastic sheet" provided
by the mass of the universe. Potential energy is an immediate
consequence of its orbit velocity slowing. That slowly converts
to kinetic energy as it pulls toward the Sun. Mercury's fall
halts when its orbit velocity has increased to the point where
centrifugal forces counteract the fall. It then just cycles
around within the elastic sheet.
---

The need for instantaneous action at a distance to overcome the
problem of the planets spiraling into the Sun should have
sounded alarm bells.

It did and people were trying to resolve it for 200
years until Einstein found the solution.

You know as well as I do that the "solution" fails the Pound and
Rebka test.

What are you talking about Max? Pound-Rebka is one of
the classic tests which _confirm_ GR !

The radioactive iron frequency generator located at the
Pound-Rebka tower base appears to cycle slower than a similar
piece located at the tower top. The logical conclusion is that
the generated frequencies are not the same. But if that was the
case, although GR could probably live with it, it would scuttle
the big bang theory. So the only available option is that the
frequencies generated in each piece must remain constant, that
they only appear to be different through some geometric quirk
of nature.

The problem is that the radioactive iron frequency generator is
much the same as the Caesium atom configuration that drives an
atomic clock. Atomic clocks can run for years, and all the while
the base clock would be losing ticks compared with the top clock.
Those lost ticks can't hide forever in the distance between the
tower base and the top.

When the clocks are brought together, there is no doubt that
the base clock will have lost time compared to the top clock.
_No doubt at all_. The time difference will depend almost
entirely on how long they have been separated over the tower
height.

If the cycle time of an atomic clock oscillator varies with
altitude, so did the radioactive iron frequency source used in
the Pound and Rebka experiment.

There are no tests
whatsoever that GR fails, only known limitations in merging
it with QM, which could as easily indicate a failing of QM.

I don't think that QM is going to be a problem.

-----

Max Keon

"Max Keon" wrote in message
u...

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

You are still in the wrong universe. My equation doesn't design
the zero origin universe, the universe designs the equation.

Sure, and then the equation tells us what will happen to
the planets in your universe - they spiral into the Sun.
Theories often have consequences that weren't anticipated
by their originators. Planck could never have known the
consequences of his original lecture on quantisation as a
solution to black body radiation.

And that sent you all off on another wild goose chase. _Blackbody
radiation has never had anything to do with quantitization_. The
value for the photon has just been plucked out of the air. It
could have been anything at all. But you must surely know that?

Take this part of the formula for example: (2*pi*h*c^2).
Since every component is a constant, the whole thing can be
treated as a simple multiplier. That's all it is, isn't it! So
why isn't it stated as 3.747e-16. But 3.747e-16 what? It's
nothing more than a nominated surface area per unit radius from
the blackbody source on which to establish an energy quantity
according to wavelength. If the unit radius is 1 meter, the
designated surface area is 3.747e-16 square meters.

According to Planck, the energy received from a 4000 K radiator
at 7.25e-7 meter wavelength is
(2 * pi * h * c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
= 1.315e+13 (# units of energy, whatever they are)

Here's an equation that gives the emissive power per wavelength
received at unit radius from the same blackbody source for a 1
square meter surface area. w is wavelength, m is surface area.
m^2 / w^5 / (1.0145^(m / (w * t)) - 1) = 3.509e+28 (# units)

3.509e+28 * 3.747e-16 = 1.315e+13, as the Planck "photon" would
specify it for the tiny surface area.

Try any temperature-wavelength combination you like and you will
always get the right answer.

Planck's photon is entirely dependent on "e" which has no claim
to fame other than it never repeats. That is crap George.

Nice rant Max, but it doesn't even touch on the reason
why quantisation is required, the shape of the spectrum.

It's a little off topic,

Completely, so I don't intend to continue much
beyond this response.

but the easiest way to demonstrate my
point (to anyone) is to set up a simple program. Each equation
set can be extracted and compared outside the program. This
program of course runs as it is in Qbasic.

Notice the absence of Planck's constant in the alternative
equations. All of the magic in Planck's formula is performed by
the EXP function, which is the same as 2.71828^(h*f) ect.

Yeah, Max, the maths only represents what happens in
nature so don't waste your time explaining trivial
maths identities, the question is why does nature
behave the way it does. In a random non-quantised
system you expect numbers of photons in equal frequency
bands so there should be higher amounts of energy at
higher frequencies. The observation is that it peaks
and then falls and the question is why. The answer is
that in nature energy must be emitted in discrete
chunks and if the energy in any chunk is proportional
to the frequency of that chunk then the total energy
will fall above some frequency because fewer chunks
can be produced from the limited energy available.
Analysing that then gives the maths.

snip examples

How hard would it be
to detect the anomalous acceleration on the outward and inward
legs? How hard would it be to notice that one is pointing in the
opposite direction to the other, relative to the Sun? Whatever is
the cause, there should be some sort of anisotropy evident in the
result

Sure, but that gets you no farther forward in identifying
the _cause_ of the effect.

It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

Yes, but you still think it could be caused by systematic errors.
So you really have no idea what the actual cause might be.

Right on both counts, and your proposal to simply repeat
the mission would not shed any light on that.

But GR's amazing feat is quite irrelevant anyway because
instantaneous action at a distance was never a requirement in
anybody's gravity.

It was in Newton's.

Only because he didn't understand gravity at all. The fact is
that something was missing in the way he analyzed planetary
motion. And that is clearly still the case today.

No, GR resolved that problem 90 years ago. Science
has moved on a long way since then.

And how much farther it would have gone if it was on the right
path.

Given that it has never failed a test, it was the "right
path". Scientific theories are nothing more than tools
for predicting how nature will behave that allow us to
manipulate our world. GR achieves that goal. Some people
seem to think there is some "search for ultimate truth"
in science but that's just philosophical bull****.

Mankind might not be in the mess it's now in. The clock is
certainly ticking.

Go and learn vectors, then read up on the
two-body problem

The following accompanies an animation at this address.
http://www.optusnet.com.au/~maxkeon/binstar.html

The animation depicts a binary star pair (blue) in a concentric
orbit about each other. The star masses are identical. This an
impossible scenario of course because the orbit velocity of each
star is faster than light speed.

I assume you mean your illustration shows impossible
speeds.

The geometry can obviously be
altered to overcome that problem, but my point is still made.

The geometry is not a problem. There are many such
binary systems in reality but obviously the orbital speed
is much slower. Something around 300 km/s is the top
speed for a "grazing" binary IIRC but binaries of nearly
equal mass are not unusual.

The action of gravity at one point around the orbit only is
shown, and that is obviously common throughout the entire orbit
cycle. The nearest red dot in the direction of motion is the
retarded position of each companion star from the viewpoint of
each star.

The '\' line travels the straight line distance to the intersect
point with its companion at the speed of light. The action of
gravity on each star is necessarily shifted to point toward the
retarded image of its companion, which is 90 degrees offset from
its true position around the orbit path. The apparent gravity
link with each companion star can obviously never be generated
where the companion currently resides.

Sure, but only because you have assumed there is abberation
which isn't the case in Newtonian gravity or GR, you are only
describing a strawman so I don't see the point.

No. I'm merely describing how gravity works in the real universe.

Pardon? It thought your diagram was showing aberration,
which we know doesn't happen, so youwere describing the
flaw you perceived in Newtonian gravity. You would be
right if it wasn't instantaneous in his formulation.

The focal point of the orbit radius is perpendicular to the
natural tangent,

No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.

Take another look at the animation.
http://www.optusnet.com.au/~maxkeon/binstar.gif There is no
doubt that each star is drawing toward where its companion
appears to be.

That is not what is meant by the focus of an ellipse, maybe
we are at cross purpose. Check the definition:

http://en.wikipedia.org/wiki/Ellipse

That's the direction of the pull of gravity at
the star even if that path curves back to the companion. It
doesn't make any difference if the gravity link is curved or
straight, but the distance to where the companion was is
shorter. Dimension was being drawn into the companion's gravity
well at that time, so even if the companion is no longer in the
same place, the full affect of its presence will be felt when
its image coincides with the star's position in space.

which would cause the stars to fly apart. But
the distance from each star to where its companion appears to be
is .707 of the instantaneous distance. According to 1/r^2, the
pull of gravity toward each other is 1 / .707^2 which is double
that required to hold the stars at the instantaneous orbit
radius. The stars would be pulled inward to point in the
direction of the instantaneous orbit tangent.

Even if that's not enough to do the trick, the universe has no
rules which dictate that a binary star pair of specific mass can
only reside at a specific radius for a specific orbit velocity.
A stable orbit exists for every circumstance. Denying that is
absurd.

Crap. For any radius, if the speed is too fast for a circular
orbit the object will move farther away and if it is too slow
it moves closer.

It will find a stable orbit somewhere though, won't it.

That depends on whether it is moving faster than "escape
velocity". If so, then no, it just keeps moving away
forever.

Both give stable elliptical orbits because
energy and momentum are conserved, but as sonn as you
start to remove energy the orbit decays. That's what happens
in nature so that's what our maths has to copy.

What energy is removed in a simple orbit? A curved gravity link
isn't going to remove energy.

None, I was referring to your anisotropy from the
which constantly removes energy except in the
special case of a two-body problem and a circular
orbit.

so the force
remains like a spring ..

No it isn't. A spring always pulls in the same direction
like gravity but your anisotropy changes direction. A
spring removes and stores energy while being stretched
and returns it while being relaxed because the force is
in the same direction as the speed.

The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif

Wrong, you have no anisotropy in that.

Of course not. The elastic force applied by the universe has

How many times do I have to point out that your equation
does not permit the anisotropy to be elastic? You say that
the force is slowing Pioneer as it leaves the Sun but it
would push Pioneer away if it was approaching the Sun. Both
those reduce the energy of the craft so it nevers recovers
what is lost.

been counteracted by Mercury's increased centrifugal force as
it orbits the Sun at a faster rate than you would expect. That's
its natural orbit in the prevailing circumstances.
---

The momentum loss immediately converts to potential energy, which
in turn slowly converts to kinetic energy.

Wrong, momentum and energy are not interchangeable, they
are separate quantities, both of which must be separately
conserved (as must the three spatial components of momentum).

In an inelastic, head on collision between two equal masses
the two masses are brought to a halt relative to each other.
Heat energy is obviously generated during the process. So where
does that come from if not from lost momentum?

From lost energy. There is no momentum lost in the crash.

Momentum may not have been lost from the universe, but neither
mass has momentum relative to the other. They are stationary.

Before the impact, one had positive momentum while the
other had a negative value (it was moving in the opposite
direction) and the total was zero. After the impact, both
have zero so the total hasn't changed, the overall amount
is conserved.

The expended energy would need to be put back to restore the
original relationship,

Before the impact, both had positive values of kinetic
energy. After the impact both have zero and the energy
has been converted into heat in the materials or perhaps
sound and other forms.

and that could point them in any opposing
directions. Their momentum relative to each other, and relative
to the universe, can now be pointing anywhere.

So what are you on about?

Energy and momentum and distinct, separate quantities
which cannot be converted from one to the other. Each
is separately conserved. The same is true of the three
quantities that go to make up momentum. In familiar
terms, the amount of momentum in the east-west direction
must be conserved, so must that in the north-south
direction and also in the up-down direction. Again all
three must be individually conserved, they are not
interchangeable.

Mercury's momentum is transferred to the "elastic sheet" provided
by the mass of the universe. Potential energy is an immediate
consequence of its orbit velocity slowing. That slowly converts
to kinetic energy as it pulls toward the Sun. Mercury's fall
halts when its orbit velocity has increased to the point where
centrifugal forces counteract the fall. It then just cycles
around within the elastic sheet.

That's a reasonable description of the ordinary view
of gravity but note that for the force to increase
the energy, the body has to be pulled towards the
Sun while it is moving towards it. You said your
anisotropy acts the opposite way, pushing outwards
when the body is moving inwards and that's what is
causing the energy loss. If you mis-stated it and
the anisotropy actually increases the gravitational
force pulling it towards the Sun faster than Newton's
equation then it could be elastic but that's not what
you told me (it is what your pair of equations say
though).

The need for instantaneous action at a distance to overcome the
problem of the planets spiraling into the Sun should have
sounded alarm bells.

It did and people were trying to resolve it for 200
years until Einstein found the solution.

You know as well as I do that the "solution" fails the Pound and
Rebka test.

What are you talking about Max? Pound-Rebka is one of
the classic tests which _confirm_ GR !

The radioactive iron frequency generator located at the
Pound-Rebka tower base appears to cycle slower than a similar
piece located at the tower top. The logical conclusion is that
the generated frequencies are not the same.

Indeed, that's one way of looking at it, and the amount
by which they are not the same is exactly what is
predicted by GR.

But if that was the
case, although GR could probably live with it, it would scuttle
the big bang theory.

Rubbish, it is the same thing that produces the Hubble
redshift! That is more often described as a velocity
but technically it is a "stretching of space" and the
same stretching is happening between the top and bottom
of the tower due to the Earth's gravity.

So the only available option is that the
frequencies generated in each piece must remain constant, that
they only appear to be different through some geometric quirk
of nature.

Gr is actually taught in modern courses based on a branch
of mathematics called "differential geometry" and "geometric
quirk of nature" is a pretty good way to summarise it :-)

The problem is that the radioactive iron frequency generator is
much the same as the Caesium atom configuration that drives an
atomic clock. Atomic clocks can run for years, and all the while
the base clock would be losing ticks compared with the top clock.
Those lost ticks can't hide forever in the distance between the
tower base and the top.

When the clocks are brought together, there is no doubt that
the base clock will have lost time compared to the top clock.
_No doubt at all_. The time difference will depend almost
entirely on how long they have been separated over the tower
height.

Yes, it is the GR equivalent of the "twins paradox".
In fact atomic clocks have regularly been moved
between different altitudes and proved what you say
to be true, HP had two factories where they made the
clocks and the effect made calibration more difficult.
Of course GR told them exactly how much the effect
would be so they were able to compensate precisely,
which again is why GR was the "right track".

If the cycle time of an atomic clock oscillator varies with
altitude, so did the radioactive iron frequency source used in
the Pound and Rebka experiment.

Yes, GR requires that and tells us the amount to expect.
If the frequency _didn't_ vary then it would be a problem
for GR but the experiment showed GR was exactly right.

There are no tests
whatsoever that GR fails, only known limitations in merging
it with QM, which could as easily indicate a failing of QM.

I don't think that QM is going to be a problem.

The maths of QM assumes that time is universal in a
sense so is incompatible with the idea of atomic
clocks running at different rates. The two can be
merged using approximations suitable for weak
gravity and the result, QED, is one of the most
successful and accurate theories ever developed,
but problems arise when gravity gets really strong
and the approximations reach their limit.

George

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

It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

Yes, but you still think it could be caused by systematic errors.
So you really have no idea what the actual cause might be.

Right on both counts, and your proposal to simply repeat
the mission would not shed any light on that.

If the anomaly points in opposite directions on the outward and
inward legs, that would rule out systematic errors as a possible
cause. But it would be pointless taking measurements anywhere
near the radius where the Sun becomes the major influence.
Somewhere between the Uranus orbit radius and Neptune would be
fine.
---

The focal point of the orbit radius is perpendicular to the
natural tangent,

No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.

Take another look at the animation.
http://www.optusnet.com.au/~maxkeon/binstar.gif There is no
doubt that each star is drawing toward where its companion
appears to be.

That is not what is meant by the focus of an ellipse,
maybe we are at cross purpose.

Probably so. The animation shows two focus points _to which each
star is drawn_, but the only thing that coincides at the center
is the line drawn between the two.

Check the definition:

http://en.wikipedia.org/wiki/Ellipse

That's the direction of the pull of gravity at
the star even if that path curves back to the companion. It
doesn't make any difference if the gravity link is curved or
straight, but the distance to where the companion was is
shorter. Dimension was being drawn into the companion's gravity
well at that time, so even if the companion is no longer in the
same place, the full affect of its presence will be felt when
its image coincides with the star's position in space.
---

The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif

Wrong, you have no anisotropy in that.

Of course not. The elastic force applied by the universe has

How many times do I have to point out that your equation
does not permit the anisotropy to be elastic? You say that
the force is slowing Pioneer as it leaves the Sun but it
would push Pioneer away if it was approaching the Sun. Both
those reduce the energy of the craft so it nevers recovers
what is lost.

You've switched from the universe generated anisotropy for
Mercury to the local anisotropy generated in the Sun-Pioneer
relationship. That's hardly relevant to the "elastic sheet"
comparison in the universe generated anisotropy for Mercury.

Pioneer is not in orbit around the Sun, so that argument becomes
totally disjointed. I'll shift course to Mercury's eccentric
orbit in the Sun-Mercury relationship. Much the same applies for
that relationship as for the Mercury-universe 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.

_The only energy loss is in the advance of the perihelion._

-----

Max Keon

On 23 Jan, 23:29, "Max Keon" wrote:
"George Dishman" wrote in ...
Max Keon wrote:
George Dishman wrote:
Max Keon wrote:---

It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

Yes, but you still think it could be caused by systematic errors.
So you really have no idea what the actual cause might be.

Right on both counts, and your proposal to simply repeat
the mission would not shed any light on that.

If the anomaly points in opposite directions on the outward and
inward legs, that would rule out systematic errors as a possible
cause.

It could rule out some but not others, but we are not
at the point of ruling out ideas that might fit to narrow
down the cause, we don't have _any_ explanations
that work (including yours).

....
How many times do I have to point out that your equation
does not permit the anisotropy to be elastic? You say that
the force is slowing Pioneer as it leaves the Sun but it
would push Pioneer away if it was approaching the Sun. Both
those reduce the energy of the craft so it nevers recovers
what is lost.

You've switched from the universe generated anisotropy for
Mercury to the local anisotropy generated in the Sun-Pioneer
relationship. That's hardly relevant to the "elastic sheet"
comparison in the universe generated anisotropy for Mercury.

The "elastic sheet" model is a way of thinking about the
metric of GR. Your equation(s) would be like friction
between body and the sheet.

Pioneer is not in orbit around the Sun, so that argument becomes
totally disjointed.

The anisotropy must obey your equation regardless. You
said there was no effect from the tangential component
so a planet in a circular orbit would not feel any anisotropic
force. Please correct that if you have changed your mind.

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.

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 was moving towards the Sun, I think you have also
made it clear that the anisotropic force would point the other
way and what you said above matches that:

If the anomaly points in opposite directions on the outward and
inward legs, that would rule out systematic errors as a possible
cause.

What that means is that the conventional part of gravity
pulls towards the Sun so causes the object to increase
its speed towards the Sun as it approaches. Your words
tell me that the effect of the anisotropic force would be
to reduce that effect, the object would speed up slightly
less than expected. Is that right, or do you think it would
speed up _more_ than the conventional theory?

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

_The only energy loss is in the advance of the perihelion._

The advance of the perihelion doesn't change the energy,
it only causes the ellipse to slowly move round to point
in a different direction.

George

"George Dishman" wrote in message
ups.com...
Max Keon wrote:
George Dishman wrote:
---
How many times do I have to point out that your equation
does not permit the anisotropy to be elastic? You say that
the force is slowing Pioneer as it leaves the Sun but it
would push Pioneer away if it was approaching the Sun. Both
those reduce the energy of the craft so it nevers recovers
what is lost.

You've switched from the universe generated anisotropy for
Mercury to the local anisotropy generated in the Sun-Pioneer
relationship. That's hardly relevant to the "elastic sheet"
comparison in the universe generated anisotropy for Mercury.

The "elastic sheet" model is a way of thinking about the
metric of GR. Your equation(s) would be like friction
between body and the sheet.

That's completely unrelated. GR describes your universe George,
it plays no role in mine.

Pioneer is not in orbit around the Sun, so that argument becomes
totally disjointed.

The anisotropy must obey your equation regardless. You
said there was no effect from the tangential component
so a planet in a circular orbit would not feel any anisotropic
force. Please correct that if you have changed your mind.

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. In the case of Pioneer, the
universe generated anisotropy is the anomaly. The Sun generated
radial component is concealed by assuming that the Sun is more
massive than it actually is.
http://www.optusnet.com.au/~maxkeon/pionomor.html
The radial component would also be concealed in the Sun-Mercury
relationship, but because the orbit cycle is complete, its
consequences are not concealed, i.e. the advance of the orbit
perihelion.

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) as distant stars
shift to accommodate its changing position within their
gravitational fields, which of course extend to the edge of the
visible universe.

If Pioneer was moving towards the Sun, I think you have also
made it clear that the anisotropic force would point the other
way and what you said above matches that:

If the anomaly points in opposite directions on the outward and
inward legs, that would rule out systematic errors as a possible
cause.

What that means is that the conventional part of gravity
pulls towards the Sun so causes the object to increase
its speed towards the Sun as it approaches. Your words
tell me that the effect of the anisotropic force would be
to reduce that effect, the object would speed up slightly
less than expected. Is that right, or do you think it would
speed up _more_ than the conventional theory?

The description at the above link tells the story.

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. 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.
----------

_The only energy loss is in the advance of the perihelion._

The advance of the perihelion doesn't change the energy,
it only causes the ellipse to slowly move round to point
in a different direction.

There is a direction change in the orientation of the ellipse
relative to the universe. That is a change in momentum.

-----

Max Keon

"Max Keon" wrote in message
u...

"George Dishman" wrote in message
ups.com...
Max Keon wrote:
George Dishman wrote:
---
How many times do I have to point out that your equation
does not permit the anisotropy to be elastic? You say that
the force is slowing Pioneer as it leaves the Sun but it
would push Pioneer away if it was approaching the Sun. Both
those reduce the energy of the craft so it nevers recovers
what is lost.

You've switched from the universe generated anisotropy for
Mercury to the local anisotropy generated in the Sun-Pioneer
relationship. That's hardly relevant to the "elastic sheet"
comparison in the universe generated anisotropy for Mercury.

The "elastic sheet" model is a way of thinking about the
metric of GR. Your equation(s) would be like friction
between body and the sheet.

That's completely unrelated. GR describes your universe George,
it plays no role in mine.

I agree so let's lay it aside.

Pioneer is not in orbit around the Sun, so that argument becomes
totally disjointed.

The anisotropy must obey your equation regardless. You
said there was no effect from the tangential component
so a planet in a circular orbit would not feel any anisotropic
force. Please correct that if you have changed your mind.

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?

....
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.

...
If Pioneer was moving towards the Sun, I think you have also
made it clear that the anisotropic force would point the other
way and what you said above matches that:

If the anomaly points in opposite directions on the outward and
inward legs, that would rule out systematic errors as a possible
cause.

What that means is that the conventional part of gravity
pulls towards the Sun so causes the object to increase
its speed towards the Sun as it approaches. Your words
tell me that the effect of the anisotropic force would be
to reduce that effect, the object would speed up slightly
less than expected. Is that right, or do you think it would
speed up _more_ than the conventional theory?

The description at the above link tells the story.

No, it is still contradictory. Your two equations say
it pushes the body away from the Sun but then you give
the single version I wrote which says it pulls the body
towards the Sun. I just want you to tell me which way
the force acts, towards or away? It's a simple question
Max, why can't you clear up the confusion?

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.

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.
----------

_The only energy loss is in the advance of the perihelion._

The advance of the perihelion doesn't change the energy,
it only causes the ellipse to slowly move round to point
in a different direction.

There is a direction change in the orientation of the ellipse
relative to the universe. That is a change in momentum.

The direction of motion of the planet (hence of its
momentum) changes by 360 degrees in every orbit. At
any instant, the momentum points in the direction of
the tangent to the ellipse. However, we were talking
about energy, not momentum. You still seem to have
trouble keeping these ideas separate. You snipped this:

Energy and momentum and distinct, separate quantities
which cannot be converted from one to the other. Each
is separately conserved. The same is true of the three
quantities that go to make up momentum. In familiar
terms, the amount of momentum in the east-west direction
must be conserved, so must that in the north-south
direction and also in the up-down direction. Again all
three must be individually conserved, they are not
interchangeable.

George

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

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