Anisotropy In The Gravity Force Proven.

"Phineas T Puddleduck" wrote in message

More word salad from the Profoundly Dumb one.

More stupidity from Lil Gay Duckie!

HJ

"John "C"" wrote in message
et...

"Phineas T Puddleduck" wrote in message

More word salad from the Profoundly Dumb one.

More stupidity from Lil Gay Duckie!

HJ


There is no such thing as gravity ... the Earth sucks,
just like Fart DickLess and Penisless Puddlequack.

On 30 Mar, 01:52, "Max Keon" wrote:
"George Dishman" wrote in message
oups.com...
On 26 Mar, 14:52, "Max Keon" wrote:
"George Dishman" wrote in message
groups.com...
On 16 Mar, 03:10, "Max Keon" wrote:
I have also modified a standard two-body simulator
to investigate the effect of his extra term. It
turns out that the inclusion of the anisotropy makes
Keplerian orbits unstable. The effect causes any
slight eccentricity to increase with time though I
haven't done this to the point of determining likely
timescales. However, it does _not_ produce any
perihelion advance as Max has ocassionally claimed.

The perihelion advance of Mercury is not caused by motion
relative to the mass of the universe of course. It's to do with
Mercury's radial motion relative to the Sun.
Yes.
But you are wrong anyway because it's physically impossible for
Mercury's orbit orientation with the universe to remain constant
if gravity was to vary enroute to and from the perihelion, or at
perihelion or aphelion.
No, you are forgetting the effect is symmetrical
and it turns out the effect on prerhelion moving
inward is cancelled by the effect when it is moving
outwards.

You have apparently forgotten that it's not symmetrical. Which
has always been your main excuse for claiming an overall energy
loss in each completed orbit.

Max, YOU TOLD ME that the gravitational force is
REDUCED as a body approaches the Sun - that removes
kinetic energy from the body. YOU TOLD ME that the
gravitational force is INCREASED as a body moves
away from the Sun - that also removes kinetic energy
from the body.

The Sun-Mercury link is a closed
system of course,

No, not any more. With only the Newtonian force it
would be closed but that is no longer true because
your anisotropy removes energy.

so there's no avenue created that directs
energy out of that system. Because the pull of gravity is varied
by the anisotropy, Mercury and the Sun both retreat slightly more
from each other than normal on the inward leg of Mercury's orbit,
and advance together slightly more than normal on the outward leg,
as would be expected in a closed, and thus _elastic_ system.

Nope, your equation defines a force which is _inelastic_.
If you don't like that, change the equation.

The latter offers the most obvious
demonstration of my point.

If the pull of gravity is reduced at aphelion, so that "o" is
held in a circular orbit around "0", the aphelion will continue
to advance until the original pull of gravity is restored. That
will be the updated aphelion. The perihelion is obviously
likewise adjusted as a consequence. The same proportionally
applies if the pull of gravity is lessened anywhere enroute to
the perihelion.

o New aphelion
.
.
. o o o
o o
o 0 o
o o
o o o

I've pointed this out to you time and time again you know.
I know and I keep telling you it is wrong. At least
this time you drew something that shows an advance.

I'm not convinced that you genuinely believe what you write, but
if you do, I can only suggest that you put down your maths book
and have a good look at the physical processes involved. Remember
that your math does not support a gravity anisotropy and will
never give you the right answer, _ever_.

Max, when will you get it through your head that the
only maths I am using to calculate the effect of the
anisotropy is YOUR equation.

In the above diagram, I take it that you agree that the
aphelion will continue to advance while the pull of gravity
remains at the reduced level required to maintain a concentric
orbit trajectory around the Sun?

That is an assumption you are making with no evidence.
It is impossible to determine by inspection and unless
you show it by integrating along the path, it is nothing
more than a hand-waving guess on your part.

And you agree that when normal
gravity is restored, Mercury falls to an advanced perihelion?

No, of course not, the anisotropy isn't switched on
and off, it is a continuously varying factor which
smoothly reduces to precisely zero as the velocity
of Mercury becomes perpendicular to the Sun-Mercury
line at perihelion. There's no point in continuing
guessing Max, you know how to write Basic so try
writing a program to simulate the two-body problem.
Do it first using GM/r^2 only without any anisotropy
and check your code gives an ellipse. You need to be
careful because the steps in the code will cause a
slight drift away from an exact ellipse.

Then when that works, add in your equation for the
anisotropy and see what happens.

....
I have
applied your equation to a two-body simulator and
this is what actually happens as a consequence of
your anisotropy:

You cannot accommodate a gravity anisotropy using your maths.
Can't you see that?

When will you grasp that I am applying _YOUR_ equation.
This is the result:


. o o o
. o o
. o 0 o
/ o o
/ o o o
/
New aphelion

On each subsquent orbit the aphelion gets farther
from the Sun but it is always at the left hand side,
there is no advance or retardation.

BTW, compare those two drawings and you should see
why the one on your website that only shows half an
orbit doesn't prove anything either way. Run your
simulation program for two or three orbits and you'll
find I'm right (assuming your code is correct of
course).

You should now be able to understand the error in you reasoning.

There is no reasoning involved, I just blindly apply
_YOUR_ equation and that is what happens. Orbital
mechanics often gives unexpected results. I'll double
check my code to make sure I got your equation right
but I am not introducing any intepretation of my own.

... The star is quickly driven inward by the deflection force
applied by the anisotropy until centrifugal forces have increased
by an amount that's equal to the elastic force applied by the
anisotropy.
No, your equation is _inelastic_.

The equation is applied to gravity George. That makes it elastic.

No, your equation changes gravity into an inelastic force.

The head wind scenario in my previous reply was fairly redundant ..

Completely, your equation defines the effect of the
anisotropic force so you don't need to explain further.

---- orbit direction

- o - force direction

M o 0 o M

- o -

Two things that you really need to get firmly fixed in your mind
are that the relationship between the star and the mass of the
universe cannot react to accommodate any energy loss due to the
star's velocity slowing in the direction of motion, and that the
star-galaxy relationship does not exist in the star-universe
relationship at all. It can only react to variation, if any, that
occur in the star-universe relationship.

Whatever Max, your equation removes energy from the system
and does not return it. You can either suggest it is stored
somewhere or that it is simply lost but it makes no difference,
all you do is take the dot product of the anisotropy with the
velocity and that tells you the rate at which energy is being
lost.

A star in a concentric orbit around the galaxy center is cycling
around the galaxy within an all pervasive elastic web of gravity
from every bit of matter in the universe. It's fairly uniformly
drawn to everywhere. The anisotropy retards the web's point of
uniformity to always lag behind the star. That would seem to slow
the star. But if there is no avenue open to transfer energy, ..

Ther is an avenue though Max, your anisotrpopy takes it away.

the
star cannot slow at all. The force offset becomes just another
elastic component which must be overcome by centrifugal forces,
and that requires a greater orbit velocity.

Yes, that's what I showed you before. In fact the speed rises
and the radius falls which is precisely why the orbit decays.

http://www.optusnet.com.au/~maxkeon/merc-un.gif demonstrates
how the star would cycle around within the background universe.
I don't think the link will work though because nothing else on
my web page does. I should really find out what's going on there.

Yes, the site seems broken. I have to leave now but
I'll look if you get it fixed.

There's no point in creating a graph that can't be accessed,

I'll try to find the time to upload the simulation
with your anisotropy so you can see what happens
but I'm out tonight and tomorrow.

Fortunately, I have a great deal of faith in the integrity of
the folk of the physics community. Truth will prevail in the
end.

Truth in science means applying the equations accurately
which is what I have done. I am explaining the physics
to you honestly but you need to listen and understand the
mistake you are making.

George

"George Dishman" wrote in message
oups.com...
On 26 Mar, 14:52, "Max Keon" wrote:
"George Dishman" wrote in message
oups.com...
On 16 Mar, 03:10, "Max Keon" wrote:
---

A velocity generated gravity anisotropy is a direct prediction
of the zero origin concept

http://www.optusnet.com.au/~maxkeon/the1-1a.html
...

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

I have also modified a standard two-body simulator
to investigate the effect of his extra term. It
turns out that the inclusion of the anisotropy makes
Keplerian orbits unstable. The effect causes any
slight eccentricity to increase with time though I
haven't done this to the point of determining likely
timescales. However, it does _not_ produce any
perihelion advance as Max has ocassionally claimed.

The perihelion advance of Mercury is not caused by motion
relative to the mass of the universe of course. It's to do with
Mercury's radial motion relative to the Sun.

Yes.

But you are wrong anyway because it's physically impossible for
Mercury's orbit orientation with the universe to remain constant
if gravity was to vary enroute to and from the perihelion, or at
perihelion or aphelion.

No, you are forgetting the effect is symmetrical
and it turns out the effect on prerhelion moving
inward is cancelled by the effect when it is moving
outwards.

You have apparently forgotten that it's not symmetrical. Which
has always been your main excuse for claiming an overall energy
loss in each completed orbit. The Sun-Mercury link is a closed
system of course, so there's no avenue created that directs
energy out of that system. Because the pull of gravity is varied
by the anisotropy, Mercury and the Sun both retreat slightly more
from each other than normal on the inward leg of Mercury's orbit,
and advance together slightly more than normal on the outward leg,
as would be expected in a closed, and thus _elastic_ system.

The latter offers the most obvious
demonstration of my point.

If the pull of gravity is reduced at aphelion, so that "o" is
held in a circular orbit around "0", the aphelion will continue
to advance until the original pull of gravity is restored. That
will be the updated aphelion. The perihelion is obviously
likewise adjusted as a consequence. The same proportionally
applies if the pull of gravity is lessened anywhere enroute to
the perihelion.

o New aphelion
.
.
. o o o
o o
o 0 o
o o
o o o

I've pointed this out to you time and time again you know.

I know and I keep telling you it is wrong. At least
this time you drew something that shows an advance.

I'm not convinced that you genuinely believe what you write, but
if you do, I can only suggest that you put down your maths book
and have a good look at the physical processes involved. Remember
that your math does not support a gravity anisotropy and will
never give you the right answer, _ever_.

In the above diagram, I take it that you agree that the
aphelion will continue to advance while the pull of gravity
remains at the reduced level required to maintain a concentric
orbit trajectory around the Sun? And you agree that when normal
gravity is restored, Mercury falls to an advanced perihelion?

If gravity then remains as normal from there on, Mercury rises
from the perihelion radius to the previously updated aphelion,
and it will continue to cycle in that same orbit orientation.
Notice that neither the aphelion or perihelion radial length has
been altered as a consequence of the change in the pull of
gravity.

Now do the same at perihelion. This time the pull of gravity is
increased to hold Mercury in a concentric orbit at the perihelion
radius. The perihelion continues to advance until normal gravity
is restored and Mercury swings out to the aphelion radius. When
it returns, the perihelion will still be in the advanced
position.

It should be blatantly obvious that if the pull of gravity is
reduced _anywhere_ during the fall from aphelion to perihelion
the orbit orientation with the universe will advance. If it's
increased _anywhere_ during the rise to the aphelion the orbit
orientation with the universe will advance.

I have
applied your equation to a two-body simulator and
this is what actually happens as a consequence of
your anisotropy:

You cannot accommodate a gravity anisotropy using your maths.
Can't you see that?

. o o o
. o o
. o 0 o
/ o o
/ o o o
/
New aphelion

On each subsquent orbit the aphelion gets farther
from the Sun but it is always at the left hand side,
there is no advance or retardation.

BTW, compare those two drawings and you should see
why the one on your website that only shows half an
orbit doesn't prove anything either way. Run your
simulation program for two or three orbits and you'll
find I'm right (assuming your code is correct of
course).

You should now be able to understand the error in you reasoning.

... The star is quickly driven inward by the deflection force
applied by the anisotropy until centrifugal forces have increased
by an amount that's equal to the elastic force applied by the
anisotropy.

No, your equation is _inelastic_.

The equation is applied to gravity George. That makes it elastic.

The head wind scenario in my previous reply was fairly redundant
because it was the reverse of what is actually happening. This
one is based on the real universe, but the anisotropy is
generated by the mass of the universe in a single plane only,
perpendicular to the plane of the orbit. Individually, the rest
of the universe will behave likewise. The "o" represents a star
in a concentric orbit around "0" which is the center of mass of
a galaxy.

---- orbit direction

- o - force direction

M o 0 o M

- o -

Two things that you really need to get firmly fixed in your mind
are that the relationship between the star and the mass of the
universe cannot react to accommodate any energy loss due to the
star's velocity slowing in the direction of motion, and that the
star-galaxy relationship does not exist in the star-universe
relationship at all. It can only react to variation, if any, that
occur in the star-universe relationship.

A star in a concentric orbit around the galaxy center is cycling
around the galaxy within an all pervasive elastic web of gravity
from every bit of matter in the universe. It's fairly uniformly
drawn to everywhere. The anisotropy retards the web's point of
uniformity to always lag behind the star. That would seem to slow
the star. But if there is no avenue open to transfer energy, the
star cannot slow at all. The force offset becomes just another
elastic component which must be overcome by centrifugal forces,
and that requires a greater orbit velocity.

http://www.optusnet.com.au/~maxkeon/merc-un.gif demonstrates
how the star would cycle around within the background universe.
I don't think the link will work though because nothing else on
my web page does. I should really find out what's going on there.

This link
http://www.astronomynotes.com/cosmolgy/s10.htm#A2.4.1
shows that the center of mass of the galaxy does not represent
the galaxy mass when calculating star orbit velocities. The
process is apparently enormously complex.

There's no point in creating a graph that can't be accessed,
so this is a short list of orbit velocities per orbit radius
that were generated using the same galaxy mass as previously
(2.388e+39 kg).

These are some of the orbit velocities per
radius required to counteract the anisotropy.
Sun radius * 2: v = 219837m/sec
* 1: v = 219845
* .5: v = 219878
* .25: v = 220010
* .125: v = 220534
* .0625: v = 222584
* .01: v = 284275

v goes to infinity for a zero radius at the center of mass of
the galaxy.

If you compare my figures with the graph, you will no doubt
consider them to be a poor match. But the margin for error in
determining what those velocities should be within the galaxy
environment is enormous. That graph can be tweaked all over the
place. And don't forget that my calculations were based on using
a point to represent the galaxy mass.
---

It also predicts the
anomalous galaxy rotation curves, which coincide exactly with
observation. Do you really believe that it's all just an amazing
coincidence?

No, what I really beieve is that you are incapable
of doing the maths needed to apply your equation,
which explains why your statements are wrong. Your
equation does _not_ predict any difference from
Newtonian curves other than that orbits will slowly
lose energy and quickly become unstable as the
eccentricity increases causing aphelion to increase
and perihelion to decrease.

Fortunately, I have a great deal of faith in the integrity of
the folk of the physics community. Truth will prevail in the
end.

-----

Max Keon

"George Dishman" wrote in message
oups.com...
On 30 Mar, 01:52, "Max Keon" wrote:
"George Dishman" wrote in message
oups.com...
On 26 Mar, 14:52, "Max Keon" wrote:
---

The latter offers the most obvious
demonstration of my point.

If the pull of gravity is reduced at aphelion, so that "o" is
held in a circular orbit around "0", the aphelion will continue
to advance until the original pull of gravity is restored. That
will be the updated aphelion. The perihelion is obviously
likewise adjusted as a consequence. The same proportionally
applies if the pull of gravity is lessened anywhere enroute to
the perihelion.

o New aphelion
.
.
. o o o
o o
o 0 o
o o
o o o

I've pointed this out to you time and time again you know.

I know and I keep telling you it is wrong. At least
this time you drew something that shows an advance.

I'm not convinced that you genuinely believe what you write, but
if you do, I can only suggest that you put down your maths book
and have a good look at the physical processes involved. Remember
that your math does not support a gravity anisotropy and will
never give you the right answer, _ever_.

Max, when will you get it through your head that the
only maths I am using to calculate the effect of the
anisotropy is YOUR equation.

But you are not applying it correctly.

In the above diagram, I take it that you agree that the
aphelion will continue to advance while the pull of gravity
remains at the reduced level required to maintain a concentric
orbit trajectory around the Sun?

That is an assumption you are making with no evidence.
It is impossible to determine by inspection and unless
you show it by integrating along the path, it is nothing
more than a hand-waving guess on your part.

Impossible to determine by inspection??? The extreme case where
the anisotropy completely halts the fall from the aphelion is
proof beyond doubt that the aphelion advances while the
anisotropy is active. The obvious consequence is that even the
slightest degree of anisotropy applied at the aphelion radius
will cause the aphelion to advance.

I'm sure you can understand why that would also be the case
wherever the anisotropy is applied during the fall to the
perihelion.

And you agree that when normal
gravity is restored, Mercury falls to an advanced perihelion?

No, of course not, the anisotropy isn't switched on
and off,

No, it's not. But that would be the extreme case, and still no
energy is lost from the Sun-Mercury system. And it demonstrates
very clearly why the perihelion advances.

it is a continuously varying factor which
smoothly reduces to precisely zero as the velocity
of Mercury becomes perpendicular to the Sun-Mercury
line at perihelion. There's no point in continuing
guessing Max, you know how to write Basic so try
writing a program to simulate the two-body problem.
Do it first using GM/r^2 only without any anisotropy
and check your code gives an ellipse. You need to be
careful because the steps in the code will cause a
slight drift away from an exact ellipse.

Then when that works, add in your equation for the
anisotropy and see what happens.

"understand what happens" would seem to be more appropriate.

Perhaps if I write the equation thus,
anisotropy = (v/c*M) * -G / r ^ 2
The anisotropy does in fact act to vary the apparent mass.

Even though the anisotropy isn't switched on and off at the
aphelion radius in the manner described in the above scenario,
exactly the same consequences would result. Mercury is drawn less
than normal to the Sun as it falls toward it, so it lags behind
at a radius of higher gravitational potential, just as it did in
the extreme case. Because the slower orbit speed is combined with
a position of higher potential, all energy is accounted for at
ever step along the way. Nothing is lost.

The extreme case advances the perihelion by 90 degrees if the
anisotropy is switched off at "x". Regardless of how minute the
true effect of the anisotropy may be, it's still directly
proportional to the extreme case.

"o" is the normal orbit path.

, x New aphelion for the extreme case.
, ' , , , (180 degrees to perihelion)
, , ' ' ,
,, o o o ',
o o ,
o ,etc 0 o,
o , # new perihelion for
o o' 'o (v/c*M) * -G / r ^ 2

On the journey away from the Sun, Mercury's orbit velocity
increases above the normal as it's drawn more to the Sun, and
it's in a position of lesser gravitational potential. Again,
nothing is lost or gained.

Perhaps this little analogy will help.
In a frictionless environment, picture a ball that's held by an
elastic rope in a concentric orbit around a much larger mass. If
the tension in the elastic rope is released slightly, the ball
spirals outward and finally arrives at a greater stable orbit
radius. While that condition remains, the time for each orbit
has increased. The original rope tension is reinstated and the
ball is drawn back to the original radius. Its orbit velocity has
not been changed by the excursion, has it!

The same applies for Mercury's orbit velocities at perihelion
and aphelion, where the gravity anisotropy is reduced to zero.
Everything is put back the way it was.

...
I have
applied your equation to a two-body simulator and
this is what actually happens as a consequence of
your anisotropy:

You cannot accommodate a gravity anisotropy using your maths.
Can't you see that?

When will you grasp that I am applying _YOUR_ equation.
This is the result:

. o o o
. o o
. o 0 o
/ o o
/ o o o
/
New aphelion

On each subsquent orbit the aphelion gets farther
from the Sun but it is always at the left hand side,
there is no advance or retardation.

BTW, compare those two drawings and you should see
why the one on your website that only shows half an
orbit doesn't prove anything either way. Run your
simulation program for two or three orbits and you'll
find I'm right (assuming your code is correct of
course).

You should now be able to understand the error in you reasoning.

There is no reasoning involved, I just blindly apply
_YOUR_ equation and that is what happens. Orbital
mechanics often gives unexpected results. I'll double
check my code to make sure I got your equation right
but I am not introducing any intepretation of my own.

... The star is quickly driven inward by the deflection force
applied by the anisotropy until centrifugal forces have increased
by an amount that's equal to the elastic force applied by the
anisotropy.
No, your equation is _inelastic_.

The equation is applied to gravity George. That makes it elastic.

No, your equation changes gravity into an inelastic force.

The head wind scenario in my previous reply was fairly redundant ..

Completely, your equation defines the effect of the
anisotropic force so you don't need to explain further.

---- orbit direction

- o - force direction

M o 0 o M

- o -

Two things that you really need to get firmly fixed in your mind
are that the relationship between the star and the mass of the
universe cannot react to accommodate any energy loss due to the
star's velocity slowing in the direction of motion, and that the
star-galaxy relationship does not exist in the star-universe
relationship at all. It can only react to variation, if any, that
occur in the star-universe relationship.

Whatever Max, your equation removes energy from the system
and does not return it. You can either suggest it is stored
somewhere or that it is simply lost but it makes no difference,
all you do is take the dot product of the anisotropy with the
velocity and that tells you the rate at which energy is being
lost.

Energy isn't lost through variations in gravity George. That's
something you are going to have to get used to.

The mass "M" is close by in this case.

-
o
M o 0 o
o o
o
-

The relationship between "M" and "o" is really no different
to the relationship between "0" and "o" in an eccentric orbit.
When "o" is moving to and from "M" the apparent mass of "M" is
varied by the anisotropy, but so what? Everything is back to
normal when radial motion ceases at the closest and farthest
points to "M" in the orbit. "o" will fall more to "0" while it
approaches "M" because it's drawn less to "M" and it will also
fall closer to "0" as it retreats from "M" because it's drawn
more to "M". But that's of no consequence in the "M"-"o" system.

-----

Max Keon

"Max Keon" wrote in message
u...
"George Dishman" wrote in message
oups.com...
....
Max, when will you get it through your head that the
only maths I am using to calculate the effect of the
anisotropy is YOUR equation.

But you are not applying it correctly.

That is possible, I asked for weeks but you were always vague
about the direction but eventually we seemed to agree so let's
check if I understand correctly.

Consider two bodies of masses M and m. In Newtonian gravity
there is a force f exerted on each by the other which produces
a change in their motion - an acceleration. I'll only show the
effect on the smaller mass m to keep the drawing uncluttered:

M --a-- m

| -- r -- |

The force on m points directly towards M and has a value given
by the equation:

f = -GMm/r^2

where r is the radial distance of m from M. That force causes
mass m to accelerate in the direction of the force at a rate
given by:

a = -GM/r^2

Now you have told me that the effect of your anisotropy is to
reduce that force if m is moving towards M by a fraction v/c
while if it is moving away, the anisotropy increases the force,
and you said the tangential speed doesn't count. So that means
the v you are using is the derivative of the radius:

v = dr/dt

and the acceleration is changed to:

a = -GM/r^2 * (1 - dr/dt)

Now please confirm, is that what you mean or if not what are
you saying?

I'll answer the rest later but there's no point if I have
misunderstood what you are saying.

George

On Apr 3, 3:59 pm, "George Dishman" wrote:
"Max Keon" wrote in message

u...

"George Dishman" wrote in message
roups.com...
...
Max, when will you get it through your head that the
only maths I am using to calculate the effect of the
anisotropy is YOUR equation.

But you are not applying it correctly.

That is possible, I asked for weeks but you were always vague
about the direction but eventually we seemed to agree so let's
check if I understand correctly.

Consider two bodies of masses M and m. In Newtonian gravity
there is a force f exerted on each by the other which produces
a change in their motion - an acceleration. I'll only show the
effect on the smaller mass m to keep the drawing uncluttered:

M --a-- m

| -- r -- |

The force on m points directly towards M and has a value given
by the equation:

f = -GMm/r^2

where r is the radial distance of m from M. That force causes
mass m to accelerate in the direction of the force at a rate
given by:

a = -GM/r^2

Now you have told me that the effect of your anisotropy is to
reduce that force if m is moving towards M by a fraction v/c
while if it is moving away, the anisotropy increases the force,
and you said the tangential speed doesn't count. So that means
the v you are using is the derivative of the radius:

v = dr/dt

and the acceleration is changed to:

a = -GM/r^2 * (1 - dr/dt)

Would it be premature to point out how badly this represents reality?


Now please confirm, is that what you mean or if not what are
you saying?

I'll answer the rest later but there's no point if I have
misunderstood what you are saying.

George

On 4 Apr, 03:09, "Eric Gisse" wrote:
....
and the acceleration is changed to:

a = -GM/r^2 * (1 - dr/dt)

Ooops! I missed the 1/c term:

a = -GM/r^2 * (1 - 1/c * dr/dt)


Would it be premature to point out how badly this represents reality?

Bear in mind that Max's version of that equation looked
like this:

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.

It took some time to convince him that ((c+v)^2/c^2)^.5
is the same as (1 + v/c) so by all means point it out but
please use no more than the level of maths that he can
handle.

George

In article .com,
"Eric Gisse" wrote:

On Apr 3, 3:59 pm, "George Dishman" wrote:

snip

Now you have told me that the effect of your anisotropy is to
reduce that force if m is moving towards M by a fraction v/c
while if it is moving away, the anisotropy increases the force,
and you said the tangential speed doesn't count. So that means
the v you are using is the derivative of the radius:

v = dr/dt

and the acceleration is changed to:

a = -GM/r^2 * (1 - dr/dt)

Would it be premature to point out how badly this represents reality?

Not at all, but when that position was put forward, much earlier in the
discussion, it did not receive universal assent. The present exercise
appears to be an attempt to identify the precise point where the
participants' views diverge (from each other or from reality), through a
recapitulation in detail.

--
Odysseus

On Apr 3, 10:47 pm, "George Dishman" wrote:
On 4 Apr, 03:09, "Eric Gisse" wrote:
...

and the acceleration is changed to:

a = -GM/r^2 * (1 - dr/dt)

Ooops! I missed the 1/c term:

a = -GM/r^2 * (1 - 1/c * dr/dt)

Would it be premature to point out how badly this represents reality?

Bear in mind that Max's version of that equation looked
like this:

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.

It took some time to convince him that ((c+v)^2/c^2)^.5
is the same as (1 + v/c) so by all means point it out but
please use no more than the level of maths that he can
handle.

Why should I concern myself with the opinions of those who would not
understand simple arguments? Max has no entertainment value.


George