Max Keon's "anisotropic gravity".

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

Since Pioneer's trajectory is only the outward leg of its
orbit around the Sun, I suppose it would still fall toward the
Sun. But how on earth could a 60 watt energy source not be very
very obvious?

The RTGs give off about 2kW in waste heat which in
theory is emitted isotropically, but some is reflected from
the back of the dish. If 1030W ended up going into deep
space and 970W went back towards the inner solar
system, that would explain the anomaly. It really is a
_very_ small anomaly.

That's not very small at all, it's enormous. It should be *very*
easy to determine the cause. The whole thing can be reproduced
and tested here on Earth anyway. The effect appeared in both
Pioneer configurations, so it should appear again. Such an
enormous variation in global emissions wouldn't be too hard to
detect surely.

You might think that but it is actually quite hard. Have
a skim over Louis Scheffer's analysis to get an idea of
the complexity:

I could detect a 60 watt temperature variation according to the
bend rate of a stick of plastisine.

http://www.lscheffer.com/pioneer/PioneerRTG.html

Mercury's maximum trajectory angle to the Sun is around 11
degrees. The average for the entire orbit is then 7.8 degrees,

No, I think you will find the average is zero. Remember it
is approaching the Sun for half the orbit.

In that sense it is zero. Positive and negative anisotropies are
generated during the respective half cycles and the result is
zero relative to the Sun's center of mass. But relative to the
Sun's inertial frame and the universe, the orientation of the
orbit ellipse has changed.

No, the change in perihelion is an advance on one half
of the orbit but a retardation on the other half so they
cancel to first order. You can see it in your own diagram

http://members.optusnet.com.au/~maxkeon/peri2.jpg

The green line is retarded on the positive half cycle but
advanced on the negative half.

That image was intended only as a comparison between using a
fixed or variable orbit velocity. The gravity anisotropy was
inadvertently excluded from the original graph anyway (and the
other graph).

I've replaced the second (apparently confusing) graph with
http://www.optusnet.com.au/~maxkeon/peri3.jpg which compares
Mercury's natural orbit path around the Sun (blue) and the
anisotropy affected path (red). The perihelion and aphelion radii
are exactly proportional to the real scenario, but the circular
path is of course not. The comparison between the two is still
quite valid though. The gravity anisotropy is magnified 5E+6
times.

As you can no doubt see, Mercury's trajectory toward the natural
perihelion is at an angle to the natural orbit path and that
retards the perihelion by around 18 degrees. The trajectory
across the retarded perihelion is pointing away from the natural
orbit path, toward the Sun, and that trajectory will advance more
toward the Sun because Mercury has entered that path at a slower
rate than if it was traveling the natural path. Have a good look
at the diagram. The negative anisotropy is falling to zero so the
pull from the Sun is increasing. The perihelion will be much
further retarded before Mercury begins its climb away from the
Sun.

Mercury's trajectory to the aphelion is closer to the Sun than
the natural path so its orbit velocity will be faster. When it
reaches the natural aphelion (which is now retarded) it's moving
outward at an angle across the natural orbit path. It's also
traveling at a higher velocity than would be normally. That
trajectory will also progress far past the marked aphelion before
it begins the return trip toward the Sun.

For this scenario to comply with observation, the perihelion
would need to advance by around 130 degrees per complete orbit.
Which is not too far from what that picture is demonstrating.

I've updated the link just in case you find it more user friendly
that way. http://www.optusnet.com.au/~maxkeon/mercury.html

------
------

Your "gravity anisotropy" is an extra force towards the sun
(on Pioneer for example) when the object is moving away
from the sun but an extra force away from the sun when
moving towards it as you explained it. Thhose are the
characteristics of a drag.

It is an extra force, but like any other entirely elastic force,
no energy is transferred. The positive and negative anisotropies
all add up to a simple stretch and release of the elastic link
between Mercury and the Sun, and the same will apply for Pioneer
when it finally returns from its excursion over the orbit
aphelion.

Your equation doesn't describe an elastic force. Think
of a simple spring. The restoring force is proportional
to the displacement from the rest position. You must
have done that at school. If you move the end of the
spring to the right, the force is to the left so slows
a mass on the end. As the mass stops and then moves
back to the left, the force continues to be to the left
since it depends on the displacement so now it
accelerates the mass returning the stored energy. A mass
on a spring on a low-friction table will oscillate.

Your equation on the other hand describes a force which
is proportional to the speed and always opposes it.
Think of a rough block on a bench moving to the right.
The frictional force is to the left and slows the block.
this time if the block stop and tries to move to the
left the frictional force immediately point to the right
and again reduces the energy. A mass on a spring with
friction still oscillates but that oscillation dies due
to the friction.

Your anisotropy equation relates the force to the speed,
not the displacement so acts like a frictional force, it
is _inelastic_. That may not have been your intention but
it is what the equation says.

You should be able to understand it now?
------
------

If you could explain how this drag works I may be able to follow
your reasoning.

Easy, as the velocity changes from outwards to inwards
at aphelion, your anisotropic gravity changes from inwards
to outwards as you said. The force is proportional to the
radial speed which is a drag term. Think of the first order
term in the simple harmonic oscillator.

That has nothing to do with gravity George.

I know that but you insist it does, you keep saying
there is speed-dependent anisotropy in the gravitational
force.

E/M radiation is THE
means of transferring energy through space. What makes you think
they are comparable? When did gravity ever transfer energy from
one place to another?

I said nothing about EM. I am discussing _your_
gravity equation, nothing else.

No, you didn't. But it's the only kind of harmonic oscillator
that can transfer energy through space and eventually come to
rest. Gravitational oscillations can't do that can they! There
is no tranfer of energy, so there is no drag.

That puts me back to square 1, where it all began. The Pioneer
anomaly is caused by Pioneer's motion relative to the Sun, in a
direct line away from the Sun. But it's not an anomaly in nature,
it's an anomaly in the math that is incapable of correctly
describing the universe. That's what seems to be conveniently
overlooked. Depending on the structure of the math flaw, the
anomalous acceleration curve can take on any shape at all.

OK, you are saying it is a failing in GR to model the
cause of the anomaly. That's one possibility being
considered

But it will never be the reason, will it. The anomaly will
just remain an anomaly until it can be justified without
upsetting GR. That's how it's been for 30 years.

If you can publish an equation that matches Pioneer
_without_ failing on the planets then people will
listen.

Let's hope we have a huge audience then.

-----

Max Keon

"Max Keon" wrote in message
u...

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

snip

You might think that but it is actually quite hard. Have
a skim over Louis Scheffer's analysis to get an idea of
the complexity:

I could detect a 60 watt temperature variation according to the
bend rate of a stick of plastisine.

http://www.lscheffer.com/pioneer/PioneerRTG.html

I'll leave it to you to work out how to get Pioneer
back to do the measurement.

Mercury's maximum trajectory angle to the Sun is around 11
degrees. The average for the entire orbit is then 7.8 degrees,

No, I think you will find the average is zero. Remember it
is approaching the Sun for half the orbit.

In that sense it is zero. Positive and negative anisotropies are
generated during the respective half cycles and the result is
zero relative to the Sun's center of mass. But relative to the
Sun's inertial frame and the universe, the orientation of the
orbit ellipse has changed.

No, the change in perihelion is an advance on one half
of the orbit but a retardation on the other half so they
cancel to first order. You can see it in your own diagram

http://members.optusnet.com.au/~maxkeon/peri2.jpg

The green line is retarded on the positive half cycle but
advanced on the negative half.

That image was intended only as a comparison between using a
fixed or variable orbit velocity. The gravity anisotropy was
inadvertently excluded from the original graph anyway (and the
other graph).

Well something caused a deviation in those graphs and
they were pretty much as I expected.

I've replaced the second (apparently confusing) graph with
http://www.optusnet.com.au/~maxkeon/peri3.jpg which compares
Mercury's natural orbit path around the Sun (blue) and the
anisotropy affected path (red). The perihelion and aphelion radii
are exactly proportional to the real scenario, but the circular
path is of course not. The comparison between the two is still
quite valid though. The gravity anisotropy is magnified 5E+6
times.

As you can no doubt see, Mercury's trajectory toward the natural
perihelion is at an angle to the natural orbit path and that
retards the perihelion by around 18 degrees.

That is the "fixed phase shift" I mentioned a few posts
back. The end of the orbit joins up nicely to the start
so it will be 18 degrees on every orbit, there will be
no secular advance. Again your diagram confirms what I
have been saying.

The trajectory
across the retarded perihelion is pointing away from the natural
orbit path, toward the Sun, and that trajectory will advance more
toward the Sun because Mercury has entered that path at a slower
rate than if it was traveling the natural path. Have a good look
at the diagram. The negative anisotropy is falling to zero so the
pull from the Sun is increasing. The perihelion will be much
further retarded before Mercury begins its climb away from the
Sun.

Mercury's trajectory to the aphelion is closer to the Sun than
the natural path so its orbit velocity will be faster. When it
reaches the natural aphelion (which is now retarded) it's moving
outward at an angle across the natural orbit path. It's also
traveling at a higher velocity than would be normally. That
trajectory will also progress far past the marked aphelion before
it begins the return trip toward the Sun.

For this scenario to comply with observation, the perihelion
would need to advance by around 130 degrees per complete orbit.
Which is not too far from what that picture is demonstrating.

No, the diagram doesn't show any advance at all. It shows
an orbit that repeats with perihelion in the same place
each time. To match what is measured, your diagram needs
to look like this:

http://www.wordsmith.org/~anu/java/spirograph.html

I've updated the link just in case you find it more user friendly
that way. http://www.optusnet.com.au/~maxkeon/mercury.html

------
------

Your "gravity anisotropy" is an extra force towards the sun
(on Pioneer for example) when the object is moving away
from the sun but an extra force away from the sun when
moving towards it as you explained it. Thhose are the
characteristics of a drag.

It is an extra force, but like any other entirely elastic force,
no energy is transferred. The positive and negative anisotropies
all add up to a simple stretch and release of the elastic link
between Mercury and the Sun, and the same will apply for Pioneer
when it finally returns from its excursion over the orbit
aphelion.

Your equation doesn't describe an elastic force. Think
of a simple spring. The restoring force is proportional
to the displacement from the rest position. You must
have done that at school. If you move the end of the
spring to the right, the force is to the left so slows
a mass on the end. As the mass stops and then moves
back to the left, the force continues to be to the left
since it depends on the displacement so now it
accelerates the mass returning the stored energy. A mass
on a spring on a low-friction table will oscillate.

Your equation on the other hand describes a force which
is proportional to the speed and always opposes it.
Think of a rough block on a bench moving to the right.
The frictional force is to the left and slows the block.
this time if the block stop and tries to move to the
left the frictional force immediately point to the right
and again reduces the energy. A mass on a spring with
friction still oscillates but that oscillation dies due
to the friction.

Your anisotropy equation relates the force to the speed,
not the displacement so acts like a frictional force, it
is _inelastic_. That may not have been your intention but
it is what the equation says.

You should be able to understand it now?

You haven't addressed my comments at all. There are two
different aspects we are discussing here, the advance of
the perihelion which you are trying to demonstrate with
the diagrams and the decay of the radial oscillation about
the mean due to drag which you haven't tackled.

If you could explain how this drag works I may be able to follow
your reasoning.

Easy, as the velocity changes from outwards to inwards
at aphelion, your anisotropic gravity changes from inwards
to outwards as you said. The force is proportional to the
radial speed which is a drag term. Think of the first order
term in the simple harmonic oscillator.

That has nothing to do with gravity George.

I know that but you insist it does, you keep saying
there is speed-dependent anisotropy in the gravitational
force.

E/M radiation is THE
means of transferring energy through space. What makes you think
they are comparable? When did gravity ever transfer energy from
one place to another?

I said nothing about EM. I am discussing _your_
gravity equation, nothing else.

No, you didn't.

Thank you.

But it's the only kind of harmonic oscillator
that can transfer energy through space and eventually come to
rest. Gravitational oscillations can't do that can they!

Yes they can. That effect s causing the Moon to recede
from the Earth and causes "tidal heating" of the moons
of the gas giants and was used by Hulse and Taylor to
confirmed gravitatonal radiation in a binary pulsar
system.

There
is no tranfer of energy, so there is no drag.

I'll say it again, your equations describe a drag. I
don't care how much you hand-wave about it, until you
change the equations there is no doubt about this
because you have the force depending on the radial
speed, not the displacement, and it is a first order
equation.

If you can publish an equation that matches Pioneer
_without_ failing on the planets then people will
listen.

Let's hope we have a huge audience then.

First you need to find your equations because
you don't have them at the moment.

George

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

You might think that but it is actually quite hard. Have
a skim over Louis Scheffer's analysis to get an idea of
the complexity:

I could detect a 60 watt temperature variation according to the
bend rate of a stick of plastisine.

http://www.lscheffer.com/pioneer/PioneerRTG.html

I'll leave it to you to work out how to get Pioneer
back to do the measurement.

As I said before, the Pioneer configuration can and no doubt has
been reproduced here on Earth, and everything possible that could
cause a 60 watt global temperature variation would have been well
and truly scrutinized because until a cause can be identified,
your precious GR fails. The Pioneer GR falsification has been
with us for 30 years, so you've not only had compelling
motivation to do it, but plenty of time as well.

The anomaly is as significant as the Pioneer mission itself and
it would be very strange physics practice if the local test
wasn't carried out. Very strange indeed.

Positive and negative anisotropies are
generated during the respective half cycles and the result is
zero relative to the Sun's center of mass. But relative to the
Sun's inertial frame and the universe, the orientation of the
orbit ellipse has changed.

No, the change in perihelion is an advance on one half
of the orbit but a retardation on the other half so they
cancel to first order. You can see it in your own diagram

http://members.optusnet.com.au/~maxkeon/peri2.jpg

The green line is retarded on the positive half cycle but
advanced on the negative half.

That image was intended only as a comparison between using a
fixed or variable orbit velocity. The gravity anisotropy was
inadvertently excluded from the original graph anyway (and the
other graph).

Well something caused a deviation in those graphs and
they were pretty much as I expected.

As I said, the graph you refer to demonstrated the difference
between using fixed and variable orbit velocities. The graph that
has been removed was so difficult to follow that I recently
updated it because I assumed that the gravity anisotropy was
excluded from that as well. But that wasn't the case at all, so
now it's gone altogether.

I've replaced the second (apparently confusing) graph with
http://www.optusnet.com.au/~maxkeon/peri3.jpg which compares
Mercury's natural orbit path around the Sun (blue) and the
anisotropy affected path (red). The perihelion and aphelion radii
are exactly proportional to the real scenario, but the circular
path is of course not. The comparison between the two is still
quite valid though. The gravity anisotropy is magnified 5E+6
times.

As you can no doubt see, Mercury's trajectory toward the natural
perihelion is at an angle to the natural orbit path and that
retards the perihelion by around 18 degrees.

That is the "fixed phase shift" I mentioned a few posts
back. The end of the orbit joins up nicely to the start
so it will be 18 degrees on every orbit,

Of course it does. That's what is was designed to do. But 18
degrees relative to what George? Relative to where it was last
time around compared with the universe, the Sun's frame, or what?
The orbit aphelion-perihelion line is of course not compelled to
align with anything at all, is it! It is in fact rotating around
in a kind of orbit cycle of its own. That rotating picture is
what I described, as a rotating picture.

http://www.optusnet.com.au/~maxkeon/peri3.jpg very clearly
demonstrates my point. Why it advances is **very** obvious.
------
------

Your equation doesn't describe an elastic force. Think
of a simple spring. The restoring force is proportional
to the displacement from the rest position. You must
have done that at school. If you move the end of the
spring to the right, the force is to the left so slows
a mass on the end. As the mass stops and then moves
back to the left, the force continues to be to the left
since it depends on the displacement so now it
accelerates the mass returning the stored energy. A mass
on a spring on a low-friction table will oscillate.

Your equation on the other hand describes a force which
is proportional to the speed and always opposes it.
Think of a rough block on a bench moving to the right.
The frictional force is to the left and slows the block.
this time if the block stop and tries to move to the
left the frictional force immediately point to the right
and again reduces the energy. A mass on a spring with
friction still oscillates but that oscillation dies due
to the friction.

Your anisotropy equation relates the force to the speed,
not the displacement so acts like a frictional force, it
is _inelastic_. That may not have been your intention but
it is what the equation says.

You should be able to understand it now?

You haven't addressed my comments at all. There are two
different aspects we are discussing here, the advance of
the perihelion which you are trying to demonstrate with
the diagrams and the decay of the radial oscillation about
the mean due to drag which you haven't tackled.

Any "drag" relating to the planets outside the Sun-Mercury
relationship has nothing to do with the gravity anisotropy which
causes the perihelion to advance. The effects of that kind of
drag have been well established. I hope you're not still going on
about some unidentifiable drag effect in the Sun-Mercury gravity
link?
------
------

E/M radiation is THE
means of transferring energy through space. What makes you think
they are comparable? When did gravity ever transfer energy from
one place to another?

I said nothing about EM. I am discussing _your_
gravity equation, nothing else.

No, you didn't.

Thank you.

But it's the only kind of harmonic oscillator
that can transfer energy through space and eventually come to
rest. Gravitational oscillations can't do that can they!

Yes they can. That effect s causing the Moon to recede
from the Earth

I thought it was the tidal effect of the Moon shifting the
water in Earth's oceans that caused it? Anyway, such an effect
would increase Mercury's orbit radius.

and causes "tidal heating" of the moons
of the gas giants

There's not going to be too much tidal heating caused by
Mercury's surface shifting around, is there.

and was used by Hulse and Taylor to
confirmed gravitatonal radiation in a binary pulsar
system.

That will have been already accounted for in the Sun-Mercury
relationship. Why should I need to account for that as well when
my concern is only with the gravity anisotropy? Why do you keep
on throwing up smoke screens?

There
is no tranfer of energy, so there is no drag.

I'll say it again, your equations describe a drag. I
don't care how much you hand-wave about it,

The pot calling the kettle black again I think?

My equations describe Mercury's behavior very well. I may not have
the math to plot the rotation rate of the aphelion-perihelion line
but according to http://www.optusnet.com.au/~maxkeon/mercury.html
it's not going to be too far from what is actually observed. The
geometry is certainly in the ball park.

I'll get there 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:

You might think that but it is actually quite hard. Have
a skim over Louis Scheffer's analysis to get an idea of
the complexity:

I could detect a 60 watt temperature variation according to the
bend rate of a stick of plastisine.

http://www.lscheffer.com/pioneer/PioneerRTG.html

I'll leave it to you to work out how to get Pioneer
back to do the measurement.

As I said before, the Pioneer configuration can and no doubt has
been reproduced here on Earth,

It has not. There is a replica in the Smithsonian:

http://www.nasm.si.edu/exhibitions/gal100/pioneer.html

but it hasn't been measured to my knowledge and
it is probably only a shell and not representative.

and everything possible that could
cause a 60 watt global temperature variation would have been well
and truly scrutinized

The scrutiny has been through theoretical modelling.
Aspects such as the change of emissivity of the paint
due to radiation effects in space ahve been considered
but they can only be inferred for craft that have
returned to Earth.

because until a cause can be identified,
your precious GR fails.

Rubbish. The planets are not affected by the
acceleration therfore it isn't universal, almost
certainly it applies to small craft only and is
extremely unlikely to have anyhting to do with
gravity at all.

The Pioneer GR falsification has been
with us for 30 years, so you've not only had compelling
motivation to do it, but plenty of time as well.

The anomaly is as significant as the Pioneer mission itself and
it would be very strange physics practice if the local test
wasn't carried out. Very strange indeed.

Positive and negative anisotropies are
generated during the respective half cycles and the result is
zero relative to the Sun's center of mass. But relative to the
Sun's inertial frame and the universe, the orientation of the
orbit ellipse has changed.

No, the change in perihelion is an advance on one half
of the orbit but a retardation on the other half so they
cancel to first order. You can see it in your own diagram

http://members.optusnet.com.au/~maxkeon/peri2.jpg

The green line is retarded on the positive half cycle but
advanced on the negative half.

That image was intended only as a comparison between using a
fixed or variable orbit velocity. The gravity anisotropy was
inadvertently excluded from the original graph anyway (and the
other graph).

Well something caused a deviation in those graphs and
they were pretty much as I expected.

As I said, the graph you refer to demonstrated the difference
between using fixed and variable orbit velocities. The graph that
has been removed was so difficult to follow that I recently
updated it

I am referring to your new updated version. It still
confirms what I said.

because I assumed that the gravity anisotropy was
excluded from that as well. But that wasn't the case at all, so
now it's gone altogether.

I've replaced the second (apparently confusing) graph with
http://www.optusnet.com.au/~maxkeon/peri3.jpg which compares
Mercury's natural orbit path around the Sun (blue) and the
anisotropy affected path (red). The perihelion and aphelion radii
are exactly proportional to the real scenario, but the circular
path is of course not. The comparison between the two is still
quite valid though. The gravity anisotropy is magnified 5E+6
times.

As you can no doubt see, Mercury's trajectory toward the natural
perihelion is at an angle to the natural orbit path and that
retards the perihelion by around 18 degrees.

That is the "fixed phase shift" I mentioned a few posts
back. The end of the orbit joins up nicely to the start
so it will be 18 degrees on every orbit,

Of course it does. That's what is was designed to do. But 18
degrees relative to what George? Relative to where it was last
time around compared with the universe, the Sun's frame, or what?

Relative to an inertial (non-rotating) frame.

The orbit aphelion-perihelion line is of course not compelled to
align with anything at all, is it! It is in fact rotating around
in a kind of orbit cycle of its own. That rotating picture is
what I described, as a rotating picture.

http://www.optusnet.com.au/~maxkeon/peri3.jpg very clearly
demonstrates my point. Why it advances is **very** obvious.

It is impossible to tell from the picture, you need
to show multiple orbits and as I said it should then
look like a spirograph.

Your equation doesn't describe an elastic force. Think
of a simple spring. The restoring force is proportional
to the displacement from the rest position. You must
have done that at school. If you move the end of the
spring to the right, the force is to the left so slows
a mass on the end. As the mass stops and then moves
back to the left, the force continues to be to the left
since it depends on the displacement so now it
accelerates the mass returning the stored energy. A mass
on a spring on a low-friction table will oscillate.

Your equation on the other hand describes a force which
is proportional to the speed and always opposes it.
Think of a rough block on a bench moving to the right.
The frictional force is to the left and slows the block.
this time if the block stop and tries to move to the
left the frictional force immediately point to the right
and again reduces the energy. A mass on a spring with
friction still oscillates but that oscillation dies due
to the friction.

Your anisotropy equation relates the force to the speed,
not the displacement so acts like a frictional force, it
is _inelastic_. That may not have been your intention but
it is what the equation says.

You should be able to understand it now?

You haven't addressed my comments at all. There are two
different aspects we are discussing here, the advance of
the perihelion which you are trying to demonstrate with
the diagrams and the decay of the radial oscillation about
the mean due to drag which you haven't tackled.

Any "drag" relating to the planets outside the Sun-Mercury
relationship has nothing to do with the gravity anisotropy which
causes the perihelion to advance.

As I have pointed out repeatedly, your equation
describes a drag force, period.

The effects of that kind of
drag have been well established. I hope you're not still going on
about some unidentifiable drag effect in the Sun-Mercury gravity
link?

I am "going on" about _YOUR_ equation that
you wrote in this thread, nothing else. It
describes drag.

But it's the only kind of harmonic oscillator
that can transfer energy through space and eventually come to
rest. Gravitational oscillations can't do that can they!

Yes they can. That effect s causing the Moon to recede
from the Earth

I thought it was the tidal effect of the Moon shifting the
water in Earth's oceans that caused it? Anyway, such an effect
would increase Mercury's orbit radius.

Yes, these are just examples how orbital motion can
transfer energy. The point is you cannot make the
blanket statement that "Gravitational oscillations"
cannot transfer energy, you need to look at specifics.

and causes "tidal heating" of the moons
of the gas giants

There's not going to be too much tidal heating caused by
Mercury's surface shifting around, is there.

and was used by Hulse and Taylor to
confirmed gravitatonal radiation in a binary pulsar
system.

That will have been already accounted for in the Sun-Mercury
relationship. Why should I need to account for that as well when
my concern is only with the gravity anisotropy? Why do you keep
on throwing up smoke screens?

Just correcting an errant statement, you were
over-simplifying.

There
is no tranfer of energy, so there is no drag.

I'll say it again, your equations describe a drag. I
don't care how much you hand-wave about it,

The pot calling the kettle black again I think?

No, just stating a mathematical fact.

My equations describe Mercury's behavior very well.

No, you are guessing and your digram does not
actually show what would happen.

I may not have
the math to plot the rotation rate of the aphelion-perihelion line
but according to http://www.optusnet.com.au/~maxkeon/mercury.html
it's not going to be too far from what is actually observed. The
geometry is certainly in the ball park.

I'll get there eventually.

If I get some time later, I might post the maths
but I have a shed roof to re-cover and we are
going out tonight so I might not get the chance.
I can explain the approach if you want to try
yourself.

George

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

You might think that but it is actually quite hard. Have
a skim over Louis Scheffer's analysis to get an idea of
the complexity:

I could detect a 60 watt temperature variation according to the
bend rate of a stick of plastisine.

http://www.lscheffer.com/pioneer/PioneerRTG.html

I'll leave it to you to work out how to get Pioneer
back to do the measurement.

As I said before, the Pioneer configuration can and no doubt has
been reproduced here on Earth,

It has not. There is a replica in the Smithsonian:

http://www.nasm.si.edu/exhibitions/gal100/pioneer.html

but it hasn't been measured to my knowledge and
it is probably only a shell and not representative.

Perhaps someone should go and check it out?

and everything possible that could
cause a 60 watt global temperature variation would have been well
and truly scrutinized

The scrutiny has been through theoretical modelling.

That's very hard to believe. What kind of physics would not
investigate ever possible avenue available in the search for
evidence that the anomaly was not due to a failing of GR. There
would not be one stone left unturned by now, since the problem
has been unresolved for 30 years.

Simulating spacelike conditions here on Earth just to demonstrate
how paint would be affected doesn't seem to be all that difficult.
It would certainly be cheaper than launching a full scale Pioneer
replica. I think the result is already known though.

Aspects such as the change of emissivity of the paint
due to radiation effects in space ahve been considered
but they can only be inferred for craft that have
returned to Earth.

It seems to me more likely that the paint surface facing the Sun
would fade far more than the rest, and consequently emit more of
the residual heat energy in the direction of the Sun. The
anomalous acceleration would be away from the Sun. I would say
that that has been completely ruled out as a possible cause.

because until a cause can be identified,
your precious GR fails.

Rubbish. The planets are not affected by the
acceleration therfore it isn't universal, almost
certainly it applies to small craft only and is
extremely unlikely to have anyhting to do with
gravity at all.

But the planets **are** affected by the acceleration.
i.e. Mercury.
------
------

I've replaced the second (apparently confusing) graph with
http://www.optusnet.com.au/~maxkeon/peri3.jpg which compares
Mercury's natural orbit path around the Sun (blue) and the
anisotropy affected path (red). The perihelion and aphelion radii
are exactly proportional to the real scenario, but the circular
path is of course not. The comparison between the two is still
quite valid though. The gravity anisotropy is magnified 5E+6
times.

As you can no doubt see, Mercury's trajectory toward the natural
perihelion is at an angle to the natural orbit path and that
retards the perihelion by around 18 degrees.

That is the "fixed phase shift" I mentioned a few posts
back. The end of the orbit joins up nicely to the start
so it will be 18 degrees on every orbit,

Of course it does. That's what is was designed to do. But 18
degrees relative to what George? Relative to where it was last
time around compared with the universe, the Sun's frame, or what?

Relative to an inertial (non-rotating) frame.

The orbit according to the gravity link between the Sun and
Mercury is as indicated in the above image. The only modification
would be a shortened perihelion and an extended aphelion. Any
designated starting point will be exactly the finish point. From
the Sun's frame the designated point will advance in the orbit
direction because the orbit path length is longer in length, and
longer in time, than that of the unaffected orbit, which would
remain fixed with the Sun's frame. Whether or not the designated
point has shifted in the Sun's frame is completely irrelevant to
the orbit shape.
------
------

You haven't addressed my comments at all. There are two
different aspects we are discussing here, the advance of
the perihelion which you are trying to demonstrate with
the diagrams and the decay of the radial oscillation about
the mean due to drag which you haven't tackled.

Any "drag" relating to the planets outside the Sun-Mercury
relationship has nothing to do with the gravity anisotropy which
causes the perihelion to advance.

As I have pointed out repeatedly, your equation
describes a drag force, period.

And as I've asked repeatedly, what drag force? How does it work?
How is energy transferred in this case?

The effects of that kind of
drag have been well established. I hope you're not still going on
about some unidentifiable drag effect in the Sun-Mercury gravity
link?

I am "going on" about _YOUR_ equation that
you wrote in this thread, nothing else. It
describes drag.

But it's the only kind of harmonic oscillator
that can transfer energy through space and eventually come to
rest. Gravitational oscillations can't do that can they!

Yes they can. That effect s causing the Moon to recede
from the Earth

I thought it was the tidal effect of the Moon shifting the
water in Earth's oceans that caused it? Anyway, such an effect
would increase Mercury's orbit radius.

Yes, these are just examples how orbital motion can
transfer energy. The point is you cannot make the
blanket statement that "Gravitational oscillations"
cannot transfer energy, you need to look at specifics.

Those specifics are already accounted for. They don't alter the
velocity related gravity anisotropy one bit.
------
------

There
is no tranfer of energy, so there is no drag.

I'll say it again, your equations describe a drag. I
don't care how much you hand-wave about it,

The pot calling the kettle black again I think?

No, just stating a mathematical fact.

My equations describe Mercury's behavior very well.

No, you are guessing and your digram does not
actually show what would happen.

I'm simply following Mercury's trajectory across the perihelion
and the aphelion, noting that it's not pointing in the same
direction as it would be if it followed the unaffected orbit
path. I also note Mercury's slower velocity when entering the
perihelion zone and its higher velocity entering the aphelion
zone. All of these very clear observations lead me to conclude
that the perihelion will advance. That's not exactly guessing
is it.

I may not have
the math to plot the rotation rate of the aphelion-perihelion line
but according to http://www.optusnet.com.au/~maxkeon/mercury.html
it's not going to be too far from what is actually observed. The
geometry is certainly in the ball park.

I'll get there eventually.

If I get some time later, I might post the maths
but I have a shed roof to re-cover and we are
going out tonight so I might not get the chance.
I can explain the approach if you want to try
yourself.

Your tentative offer is much appreciated.

-----

Max Keon

Max Keon wrote:

That's very hard to believe. What kind of physics would not
investigate ever possible avenue available in the search for
evidence that the anomaly was not due to a failing of GR. There
would not be one stone left unturned by now, since the problem
has been unresolved for 30 years.

It has received a lot of scrutiny - just check arxiv.org - there must
be hundreds of papers. The problem is that there is not enough hard
data to choose between the theories.

Simulating spacelike conditions here on Earth just to demonstrate
how paint would be affected doesn't seem to be all that difficult.
It would certainly be cheaper than launching a full scale Pioneer
replica. I think the result is already known though.

You might be able to simulate it, but only if you knew exactly what the
paint was.Small changes in IR emission were never considered as part of
the design, and no one thought to save samples of the actual paint.
Results from LDEF (the long duration space exposure test) show that
minor paint changes, like the exact composition of the binder, can have
big effects on the aging of the paint. However, these are not
carefully controlled during the paint manufacture since they are
normally not relevent, nor were they documented when Pioneer was built.

Aspects such as the change of emissivity of the paint
due to radiation effects in space ahve been considered
but they can only be inferred for craft that have
returned to Earth.

And the only ones that have been returned are those from LDEF, which
was in low-earth orbit and hence subject to other effects like atomic
oxygen that Pioneer did not see.

It seems to me more likely that the paint surface facing the Sun
would fade far more than the rest, and consequently emit more of
the residual heat energy in the direction of the Sun. The
anomalous acceleration would be away from the Sun. I would say
that that has been completely ruled out as a possible cause.

The LDEF data does not agree with this. Some paints became better
emitters upon exposure, and some worse.

because until a cause can be identified,
your precious GR fails.

Rubbish. The planets are not affected by the
acceleration therfore it isn't universal, almost
certainly it applies to small craft only and is
extremely unlikely to have anyhting to do with
gravity at all.

This is in simple terms the conclusion of Iorio, in a formal paper,
http://arxiv.org/abs/gr-qc/0610050,
Can the Pioneer anomaly be of gravitational origin?

Lou Scheffer

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

snip

As I said before, the Pioneer configuration can and no doubt has
been reproduced here on Earth,

It has not. There is a replica in the Smithsonian:

http://www.nasm.si.edu/exhibitions/gal100/pioneer.html

but it hasn't been measured to my knowledge and
it is probably only a shell and not representative.

Perhaps someone should go and check it out?

AFAIK it is just a replica. It wouldn't be representative
and wouldn't give any representative data. As I
explained there's no way to know what effect being
in space had on the system and in particular the
intense radiation encountered in the Jupiter flyby
may have caused significant changes. Exposing
an acurate model to that environment would probably
turn it into toxic waste!

and everything possible that could
cause a 60 watt global temperature variation would have been well
and truly scrutinized

The scrutiny has been through theoretical modelling.

That's very hard to believe. What kind of physics would not
investigate ever possible avenue available in the search for
evidence that the anomaly was not due to a failing of GR. There
would not be one stone left unturned by now, since the problem
has been unresolved for 30 years.

Why would anyone spend a fortune building a
duplicate of a 30 year old craft, putting it into
an expansive vacuum chamber and spending
weeks testing it just to prove that the heat of
the RTGs shining off the back could explain the
anomaly if a calculation like Lou's can do the
same job for only his (unpaid) time? You seem
to have a bizarre idea of the economics of the
situation.

Simulating spacelike conditions here on Earth just to demonstrate
how paint would be affected doesn't seem to be all that difficult.
It would certainly be cheaper than launching a full scale Pioneer
replica. I think the result is already known though.

The result isn't known, see Lou's reply for details.

snip stuff covered by Lou

Rubbish. The planets are not affected by the
acceleration therfore it isn't universal, almost
certainly it applies to small craft only and is
extremely unlikely to have anyhting to do with
gravity at all.

But the planets **are** affected by the acceleration.
i.e. Mercury.

Nope, your maths is wrong. See later.

... But 18
degrees relative to what George? Relative to where it was last
time around compared with the universe, the Sun's frame, or what?

Relative to an inertial (non-rotating) frame.

The orbit according to the gravity link between the Sun and
Mercury is as indicated in the above image. The only modification
would be a shortened perihelion and an extended aphelion. Any
designated starting point will be exactly the finish point. From
the Sun's frame the designated point will advance in the orbit
direction because the orbit path length is longer in length, and
longer in time, than that of the unaffected orbit, which would
remain fixed with the Sun's frame. Whether or not the designated
point has shifted in the Sun's frame is completely irrelevant to
the orbit shape.

The orbit shape is close to an ellipse and not of any
interest, it is the secular change of location of the
perihelion point relative to a non-rotating frame that
is measured.

You haven't addressed my comments at all. There are two
different aspects we are discussing here, the advance of
the perihelion which you are trying to demonstrate with
the diagrams and the decay of the radial oscillation about
the mean due to drag which you haven't tackled.

Any "drag" relating to the planets outside the Sun-Mercury
relationship has nothing to do with the gravity anisotropy which
causes the perihelion to advance.

As I have pointed out repeatedly, your equation
describes a drag force, period.

And as I've asked repeatedly, what drag force? How does it work?
How is energy transferred in this case?

And as I have told over and over again YOU have to
answer that Max, not me. It is YOUR equation that
describes a drag force. I think it doesn't exist and
your theory is nonsense.

The effects of that kind of
drag have been well established. I hope you're not still going on
about some unidentifiable drag effect in the Sun-Mercury gravity
link?

I am "going on" about _YOUR_ equation that
you wrote in this thread, nothing else. It
describes drag.

But it's the only kind of harmonic oscillator
that can transfer energy through space and eventually come to
rest. Gravitational oscillations can't do that can they!

Yes they can. That effect s causing the Moon to recede
from the Earth

I thought it was the tidal effect of the Moon shifting the
water in Earth's oceans that caused it? Anyway, such an effect
would increase Mercury's orbit radius.

Yes, these are just examples how orbital motion can
transfer energy. The point is you cannot make the
blanket statement that "Gravitational oscillations"
cannot transfer energy, you need to look at specifics.

Those specifics are already accounted for.

Correct, so you need to find adifferent explanation
for the new drag term you are introducing.

They don't alter the
velocity related gravity anisotropy one bit.
------
------

There
is no tranfer of energy, so there is no drag.

I'll say it again, your equations describe a drag. I
don't care how much you hand-wave about it,

The pot calling the kettle black again I think?

No, just stating a mathematical fact.

My equations describe Mercury's behavior very well.

No, you are guessing and your digram does not
actually show what would happen.

I'm simply following Mercury's trajectory across the perihelion
and the aphelion, noting that it's not pointing in the same
direction as it would be if it followed the unaffected orbit
path. I also note Mercury's slower velocity when entering the
perihelion zone and its higher velocity entering the aphelion
zone. All of these very clear observations lead me to conclude
that the perihelion will advance. That's not exactly guessing
is it.

See below.

I may not have
the math to plot the rotation rate of the aphelion-perihelion line
but according to http://www.optusnet.com.au/~maxkeon/mercury.html
it's not going to be too far from what is actually observed. The
geometry is certainly in the ball park.

I'll get there eventually.

If I get some time later, I might post the maths
but I have a shed roof to re-cover and we are
going out tonight so I might not get the chance.
I can explain the approach if you want to try
yourself.

Your tentative offer is much appreciated.

I have to break off now to go to a meeting (lunch is
over) but I'll reply again with this when I get some
time. It might be tomorrow night as I'm out again
this evening.

George

wrote in message
oups.com...
Max Keon wrote:
That's very hard to believe. What kind of physics would not
investigate ever possible avenue available in the search for
evidence that the anomaly was not due to a failing of GR. There
would not be one stone left unturned by now, since the problem
has been unresolved for 30 years.

It has received a lot of scrutiny - just check arxiv.org - there must
be hundreds of papers. The problem is that there is not enough hard
data to choose between the theories.

Simulating spacelike conditions here on Earth just to demonstrate
how paint would be affected doesn't seem to be all that difficult.
It would certainly be cheaper than launching a full scale Pioneer
replica. I think the result is already known though.

You might be able to simulate it, but only if you knew exactly what the
paint was.Small changes in IR emission were never considered as part of
the design, and no one thought to save samples of the actual paint.
Results from LDEF (the long duration space exposure test) show that
minor paint changes, like the exact composition of the binder, can have
big effects on the aging of the paint. However, these are not
carefully controlled during the paint manufacture since they are
normally not relevent, nor were they documented when Pioneer was built.

Place 100 different paint surfaces in front of an electric
radiator. It would be extremely unlikely for any of them to turn
into reflectors, don't you think? If you ever do find one that
does, you'll need to continue the experiment until enough data is
available to determine the ratio of possibility that both
Pioneers were painted with one of these types of paints.

Aspects such as the change of emissivity of the paint
due to radiation effects in space ahve been considered
but they can only be inferred for craft that have
returned to Earth.

And the only ones that have been returned are those from LDEF, which
was in low-earth orbit and hence subject to other effects like atomic
oxygen that Pioneer did not see.

It seems to me more likely that the paint surface facing the Sun
would fade far more than the rest, and consequently emit more of
the residual heat energy in the direction of the Sun. The
anomalous acceleration would be away from the Sun. I would say
that that has been completely ruled out as a possible cause.

The LDEF data does not agree with this. Some paints became better
emitters upon exposure, and some worse.

What percentage does "some" represent? And where is the LDFE data
that led to that conclusion? Words are not very convincing you
know.

because until a cause can be identified,
your precious GR fails.

Rubbish. The planets are not affected by the
acceleration therfore it isn't universal, almost
certainly it applies to small craft only and is
extremely unlikely to have anyhting to do with
gravity at all.

This is in simple terms the conclusion of Iorio, in a formal paper,
http://arxiv.org/abs/gr-qc/0610050,
Can the Pioneer anomaly be of gravitational origin?

I was rather dismayed to note the incredible ease with which
postulates are thrown about without the slightest justification.
And bad data that doesn't fit is just shoved aside. To top it
all off, a conclusion was reached which was considered to be a
very positive demonstration that the Pioneer anomaly was not
gravity caused. I think I must be dreaming.

I know it's not your problem, but the claim in this part caption
which was attached to the graph depicting Pioneer's anomalous
acceleration up to 24 AU, "The measured anomalous acceleration
experienced by Jupiter, Saturn and Uranus according to
observationally determined perihelion rates by Pitjeva, are
shown." has left me totally bewildered. How the hell did Pitjeva
arrive at such enormous perihelion shift rate differences between
Jupiter Saturn and Uranus when the orbit eccentricities are
.05, .06 and .05 respectively?

Lou Scheffer

I had a look at your analysis at
http://www.lscheffer.com/pioneer/PioneerRTG.html
and every one of the things you mentioned can be tested here on
Earth. It shouldn't be too hard to simulate spacelike conditions
and determine why **both** Pioneers gave the same result.

-----

Max Keon

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

big snip covered by other replies

I may not have
the math to plot the rotation rate of the aphelion-perihelion line ..

I'll get there eventually.

If I get some time later, I might post the maths
but I have a shed roof to re-cover and we are
going out tonight so I might not get the chance.
I can explain the approach if you want to try
yourself.

Your tentative offer is much appreciated.

OK, the trick is that we are looking for very small
deviations from the classic elliptical orbit and
other effects which will cause some precession so the
approach needs to be one that will not be limited
by the resolution and accuracy of the calculation.

Start with a small body in a perfectly circular orbit
of radius R and period P. Now consider if the body
is given a small nudge nearly perpendicular to the
orbital velocity so that the kinetic energy is
unchanged but it is on a slightly outward orbit (part
of an ellipse where it crosses the circular orbit).

What we want to do is find an equation of the form

r" = -k r

where r is the deviation of the radius from R, r" is the
second derivative of r wrt time and k is some constant.
That provides a restoring force which means the body
will reach some maximum value of r and then return
and overshoot to some minimum value. In other words
if the maximum of r is x then perihelion is at radius
R+x and aphelion is at R-x.

If you only consider the inverse square force then you
won't get a solution, what you need to do is also take
into account the increase in gravitational potential
energy as the particle moves away from the orbited
body. That energy is taken from the kinetic energy
hence the speed falls and the centrifugal force is
reduced. The combination of that and the inverse
square gravitational change gives you the constant k.

Solve r" = -k * r to get a sine wave and we should
be able to show that the period is the same as that of
the basic circular orbit so that the path repeats.

Next you want to know the effect of your anisotropic
force. It adds a speed-dependent term so the equation
becomes this:

r" = -k r - b r'

where r' is the derivative of r, dr/dt (which is the
radial component of the speed of course) and b is again
a constant.

The solution to that is the classical damped harmonic
oscillator and the values of the constants will produce
a very under-damped system:

http://www.hep.vanderbilt.edu/~webst.../moscillat.pdf

See equations (8), (9) and (10).

Equation (10) is what you actually want because it
tells you the difference in frequency of the radial
oscillation versus that of the circular orbit.

What you will be less happy about is equation (8)
which is what I have been pointing out to you. The
first exponential term indicates that the amplitude of
the sine wave radial variation decays which is the
standard result for an under-damped oscillator. That
amplitude is related to the eccentricity and the fact
is that your theory requires elliptical orbits to
slowly decay to become circular.

Now the nicest thing about this approach is that you
don't need to bother doing most of the work :-) The
lambda factor determines both the frequency shift
_and_ the decay of the eccentricity so since you
know the rate of Mercury's perihelion shift, you can
directly calculate the time constant of the exponential
decay of the eccentricity. If you are lucky, you might
find it is a very long time but then you need to find
some observational confirmation of your prediction.

George

Max Keon wrote:
wrote in message
You might be able to simulate it, but only if you knew exactly what the
paint was. [...]

Place 100 different paint surfaces in front of an electric
radiator. It would be extremely unlikely for any of them to turn
into reflectors, don't you think? If you ever do find one that
does, you'll need to continue the experiment until enough data is
available to determine the ratio of possibility that both
Pioneers were painted with one of these types of paints.

See below - it's pretty clear all 100 paints will not behave the same
way.

The LDEF data does not agree with this. Some paints became better
emitters upon exposure, and some worse.

What percentage does "some" represent? And where is the LDFE data
that led to that conclusion? Words are not very convincing you
know.
The report can be found he
http://ntrs.nasa.gov/archive/nasa/ca...1999020109.pdf
(Warning - 15 MB pdf file)
Pages 45-55 describe the white paint samples.
See for example, table 12, samples 14-17. Note that YB71 paint became
slightly worse (0.901-0.880) but Z93 paint became slightly better
(0.915-0.918). YB71 over Z93 became considerably better
(0.849-0.880) but S13G became considerably worse (0.900-0.883).
Basically they tried 3 paints and 1 combination - 2 became better and 2
worse. There is lots more data in the report - spectra, photos, etc.
However even this data is hard to apply to Pioneer, since none of these
exact paints was used, and LDEF also had atomic oxygen effects that
Pioneer did not.

I had a look at your analysis at
http://www.lscheffer.com/pioneer/PioneerRTG.html
and every one of the things you mentioned can be tested here on
Earth. It shouldn't be too hard to simulate spacelike conditions
and determine why **both** Pioneers gave the same result.

Both Pioneers were built at the same time, from almost identical parts.
So if the cause is IR reflecting from the spacecraft, or directional
thermal emission from the instrument compartment, or paint fading on
the RTGs, you would expect it to be very similar on both spacecraft, to
well within the experimental errors. In my mind this does argue
against gas leaks as a cause, but does not distinguish between
"gravity" and "thermal" causes.

Also, note that Cassini, when the measured it, had a 3x bigger effect.
Cassini had the RTGs closer in, which makes the IR even harder to
predict to the accuracy needed. Of course you can argue that 1x of
this is new physics, and 2x is IR, so this does not prove there is no
effect. But another possible conclusion is that IR is hard to analyze,
and is perhaps wholly responsible for both experimental results.

Lou Scheffer