Anisotropy and Mercury (2)

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

Forgive me for starting a new thread, but the old one has become
fairly redundant due to the huge bulk of posts that seem to serve
no purpose other than to bury the thread under a layer of wader
depth bull****.

much snipped as there is a major probleme here

That means that any time, the acceleration adds an
amount to the velocity which is equal to the product
of the time and the acceleration. It doesn't add or
subtract from the radius, I don't know where you got
that idea.

For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.

I'm inclined to think you actually believe that George, which is
a bit disconcerting.

I hadn't realised you had a problem with that. Of
course it is fundamental, the word acceleration
is defined as the rate of change of velocity and
is unarguable and the other equations you have
tried to use are all derived form this under
various conditions.

I'll snip the rest and concentrate on this major error, which is
obviously the root of all of the confusion.

What you say is true for Mercury while in its stable eccentric
orbit around the Sun, so long as the anisotropy isn't included.

What I said above is true for all objects changing
speed for any reason whatsoever under any circumstances.

Certainly _not_ under any circumstances George. Your calculations
apply for any normal trajectory taken by an object naturally
moving to or from a gravity source. The curved trajectory forms
a natural part of the orbit shape, regardless of whether or not
a complete orbit will eventually form. The only reason for it not
forming is that there is other matter in the universe. But in all
cases, the average radius per orbit, or potential orbit, remains
unchanged.

You of course agree that an object in a sustainable concentric
orbit around the Sun will not shorten the radius between it and
the Sun? You also agree that the radius will shorten at the full
gravity rate only if its orbital speed is zero. AND ONLY THEN?
The same of course applies for an eccentric orbit.

Do you reject any of that so far?

Centrifigal forces change at the rate of orbital speed squared,
so if Mercury was traveling at an average of 24000 m/sec instead
of the average 48000 m/sec, it would be restrained from falling
at the full rate by 24000^2 / 48000^2 = .25 of the .0395 m/sec^2
gravity rate. The fall rate is .25 * .0395 = 9.875e-3 m/sec^2.
I hope you can see that now.

If you weren't aware of that then obviously nothing
else I said will have made any sense to you so I can
see why the conversation has been so difficult.

The acceleration variations throughout Mercury's orbit cycle are
not really changing anything. Mercury is not permanently shifting
from its stable orbit path, and orbit velocity doesn't vary from
what is an integral part of the stable orbit structure.

Sorry Max, that isn't true. You said the anisotropy
causes an extra acceleration, it acts to displace
Mercury and since it is not directly along the path
but rather points towards the Sun, it changes the
direction as well as the speed.

You are claiming that the anisotropy will cause Mercury's orbital
speed to automatically change the moment that it's applied, which
is impossible.

It is what you have been telling me. "Acceleration" is
the rate at which speed changes and normally is given by

a = -GM / r^2

in Newtonian gravitation. You said that was changed to

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

where v is the radial component of the velocity. That is
what my program models and you have seen the consequences.

For example, if the pull of gravity is doubled,
Mercury's orbital speed doesn't magically increase to comply with
the change.

Right, but it's _acceleration_ changes immediately
and the speed is then the integral of that. The
speed at any instant is changing by the value of
the acceleration _at_that_time_.

There's no point in replying to the rest of your post until
this has all been cleared up. You are repeating the same old
mistakes over and over again, again.

The gravity force is pointing directly at the Sun, so unless
Mercury falls closer to the Sun on average its orbital speed
cannot be increased. Adding a new force does not change the pull
direction, so orbital speed cannot change from the normal unless
the average radial length changes. Can you now see that?

-----

Max Keon

Max Keon wrote:

[...]

Centrifigal forces change at the rate of orbital speed squared,
so if Mercury was traveling at an average of 24000 m/sec instead
of the average 48000 m/sec, it would be restrained from falling
at the full rate by 24000^2 / 48000^2 = .25 of the .0395 m/sec^2
gravity rate. The fall rate is .25 * .0395 = 9.875e-3 m/sec^2.

[...]

The gravity force is pointing directly at the Sun, so unless
Mercury falls closer to the Sun on average its orbital speed
cannot be increased. Adding a new force does not change the pull
direction, so orbital speed cannot change from the normal unless
the average radial length changes. Can you now see that?

Have you asked yourself why the centrifugal force is (m v^2 / r)?

In spacetime with the Schwarzschild metric, the Euler-Lagrange
equation associated with r can simply be derived as the following
where the orbital motion is confined to the equatorial plane of the
gravitating body.

d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r

Where

** O = Longitude
** s = Spacetime
** U = G M / c^2 / r

Say the orbit is circular. That mean the following.

d^2r/ds^2 = 0

We have the following.

(1 - 3 U) r (dO/ds)^2 = U / r

Or

(1 - 3 U) r^2 (dO/ds)^2 = U

Or

v^2 / c^2 = U / (1 - 3 U)

Where (through BS but another chapter of discussion)

** v / c = r dO/ds

It actually takes a little bit higher speed than Newtonian result to
complete an orbit.

Thus back to the Euler-Lagrange equation associated with r, we must
have the centrifugal force identified as the following.

m c^2 (1 - 3 U) r (dO/ds)^2

The curvature in Schwarzschild spacetime actually causes a weaker
centrifugal force. In order to keep an equilibrium orbit, the speed
must be a little higher than the Newtonian result. If not, it will
fly spirally away.

Thus, the advance of Mercury's perihelion must have two components
identified as follows.

** Orbital geometric anomaly (dwelled on by physicists)
** Orbital speed anomaly (eluded physicists sigh)

GR prediction based on orbital geometric anomaly alone somewhat
satisfies the observation. However, including the orbital speed
anomaly, it falls far short. This of course is based on the
speculation that the geodesic motion follows the path of maximum
accumulated spacetime. If the model of geodesics is to follow the
path of least accumulated time or the principle of least time, you
will find the orbital geometric anomaly describes almost the result of
observation while the orbital speed anomaly is null. The net result
fits the observation almost perfectly. But don't celebrate the
achievement of GR yet, because under this model of geodesics it
predicts photons would travel beyond the speed of light due to the
Euler-Lagrange equation associated with r. Dealing with second order
effects, everything must fit in place before betting on a new
hypothesis. GR does not.

On May 29, 11:41 pm, Koobee Wublee wrote:
Max Keon wrote:
[...]

Centrifigal forces change at the rate of orbital speed squared,
so if Mercury was traveling at an average of 24000 m/sec instead
of the average 48000 m/sec, it would be restrained from falling
at the full rate by 24000^2 / 48000^2 = .25 of the .0395 m/sec^2
gravity rate. The fall rate is .25 * .0395 = 9.875e-3 m/sec^2.

[...]

The gravity force is pointing directly at the Sun, so unless
Mercury falls closer to the Sun on average its orbital speed
cannot be increased. Adding a new force does not change the pull
direction, so orbital speed cannot change from the normal unless
the average radial length changes. Can you now see that?

Have you asked yourself why the centrifugal force is (m v^2 / r)?

I wonder if you have ever studied Newton's second law in a rotating
coordinate system....


In spacetime with the Schwarzschild metric, the Euler-Lagrange
equation associated with r can simply be derived as the following
where the orbital motion is confined to the equatorial plane of the
gravitating body.

It isn't "spacetime", chuckles. It is an affine parameter.


d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r

Where

** O = Longitude
** s = Spacetime
** U = G M / c^2 / r

[snip all]

Wrong.

The actual radial equation of motion is:

d^2r/dl^2 + GM/r^3 * (r - 2GM) * (dt/dl)^2 - GM/r(r-2GM) *(dr/dl)^2 -
r(r-2GM)[(d\theta/dl)^2 + sin(\theta)^2*(d\phi/dl)^2] = 0, where l is
an arbitrary affine parameter [none of this 'spacetime' bull****].

Notice this isn't anywhere even close to your idiotic equation. As
usual you get even the simplest mathematics totally and completely
wrong.

On May 30, 1:24 pm, Eric Gisse wrote:
On May 29, 11:41 pm, Koobee Wublee wrote:

In spacetime with the Schwarzschild metric, the Euler-Lagrange
equation associated with r can simply be derived as the following
where the orbital motion is confined to the equatorial plane of the
gravitating body.

It isn't "spacetime", chuckles. It is an affine parameter.

d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r

Where

** O = Longitude
** s = Spacetime
** U = G M / c^2 / r

The actual radial equation of motion is:

d^2r/dl^2 + GM/r^3 * (r - 2GM) * (dt/dl)^2 - GM/r(r-2GM) *(dr/dl)^2 -
r(r-2GM)[(d\theta/dl)^2 + sin(\theta)^2*(d\phi/dl)^2] = 0, where l is
an arbitrary affine parameter [none of this 'spacetime' bull****].

This is wrong. It should be corrected as follows using your symbols
and notations.

d^2r/dl^2 + GM/r^3 * (r - 2GM) * (dt/dl)^2 - (GM/r)/(r-2GM) *(dr/dl)^2
- (r-2GM)[(d\theta/dl)^2 + sin(\theta)^2*(d\phi/dl)^2] = 0

For simplicity without sacrificing the accuracy of the mathematical
model, we can easily establish the following.

** \theta = 0
** \phi = O
** ds = dl
** U = G M [/ c^2] / r

In doing so, your equation after my correction leads to the following.

dr^2/ds^2 - (1 - 2 U) r (dO/ds)^2 = - [c^2] (1 - 2 U) (dt/ds)^2 U / r
+ (dr/ds)^2 (U / r) / (1 - 2 U)

Furthermore, we can also easily establish the following from the
spacetime equation with the Schwarzschild metric.

** [c^2] (1 - 2 U) (dt/ds)^2 = 1 + (dr/ds)^2 / (1 - 2 U) + r^2 (dO/
ds)^2

Thus finally, your equation after my correction eventually leads to a
much simpler and elegant equation as I have already presented. Here
it is once again.

d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r

Centrifugal force is not a farce. It is very real in direct contrast
from denials from Dr. Roberts.

Notice this isn't anywhere even close to your idiotic equation. As
usual you get even the simplest mathematics totally and completely
wrong.

You know how to copy equations without understanding them. shrug
You have no capability to copy equations correctly. sigh
You are hopelessly beyond any help. get lost

He knows not and not knows that he knows not is a fool. Shun Gisse
and his pathetic cronies.

On May 30, 2:54 pm, Koobee Wublee wrote:
On May 30, 1:24 pm, Eric Gisse wrote:



On May 29, 11:41 pm, Koobee Wublee wrote:
In spacetime with the Schwarzschild metric, the Euler-Lagrange
equation associated with r can simply be derived as the following
where the orbital motion is confined to the equatorial plane of the
gravitating body.

It isn't "spacetime", chuckles. It is an affine parameter.

d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r

Where

** O = Longitude
** s = Spacetime
** U = G M / c^2 / r

The actual radial equation of motion is:

d^2r/dl^2 + GM/r^3 * (r - 2GM) * (dt/dl)^2 - GM/r(r-2GM) *(dr/dl)^2 -
r(r-2GM)[(d\theta/dl)^2 + sin(\theta)^2*(d\phi/dl)^2] = 0, where l is
an arbitrary affine parameter [none of this 'spacetime' bull****].

This is wrong. It should be corrected as follows using your symbols
and notations.

No, it isn't. This is the radial equation of motion for the
Schwarzschild metric. Refer to any relativity textbook, or derive the
damn thing yourself.


d^2r/dl^2 + GM/r^3 * (r - 2GM) * (dt/dl)^2 - (GM/r)/(r-2GM) *(dr/dl)^2
- (r-2GM)[(d\theta/dl)^2 + sin(\theta)^2*(d\phi/dl)^2] = 0

For simplicity without sacrificing the accuracy of the mathematical
model, we can easily establish the following.

** \theta = 0
** \phi = O
** ds = dl
** U = G M [/ c^2] / r

I was using geometrized units - there was no c, so you have no idea if
you are inserting it in the right spots.


In doing so, your equation after my correction leads to the following.

dr^2/ds^2 - (1 - 2 U) r (dO/ds)^2 = - [c^2] (1 - 2 U) (dt/ds)^2 U / r
+ (dr/ds)^2 (U / r) / (1 - 2 U)

Furthermore, we can also easily establish the following from the
spacetime equation with the Schwarzschild metric.

The phrase "spacetime equation" is nonsense. Quit expecting people to
understand your nonstandard and self-created vocabulary about a
technical subject which you routinely mangle.

Your inability to communicate in an understandable language says much
about your understanding of general relativity.


** [c^2] (1 - 2 U) (dt/ds)^2 = 1 + (dr/ds)^2 / (1 - 2 U) + r^2 (dO/
ds)^2

Thus finally, your equation after my correction eventually leads to a
much simpler and elegant equation as I have already presented. Here
it is once again.

d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r

Wrong, and wrong.

Where did dt/ds go? Where did c go? You insisted on putting it in, why
did you remove it? Where did the free floating 1 go? Where did the "3"
come from? There is no way for you to pick up an extra (1 - U)r(dO/
ds)^2 term.

The equations of motion /eventually/ fold into a single equation of
motion with an effective potential and effective energy _after_ the
application of 3 conserved quantities - none of which you have
mentioned or show any amount of understanding. The rigorous derivation
of said equation of motion takes about a page of math, _all_ of which
you have ignored - and isn't anywhere NEAR what you have written
down.


Centrifugal force is not a farce. It is very real in direct contrast
from denials from Dr. Roberts.

Stooooopid. That's classical mechanics, which you don't even
understand.

Centrifugal force follows directly from the expression of Newton's 2nd
law in a rotating coordinate system. Tom Roberts knows this, because
he has an actual education in physics instead of whatever you pulled
from a cracker jack box. Centrifugal force exists only in coordinate
systems that are rotating - it is a fake force that arises from your
coordinate choice. Tom Roberts also knows this, and has explained this
to you and others on numerous occasions, apparently to no effect.

[snip remaining whining]

On 30 May, 00:38, "Max Keon" wrote:
"Max Keon" wrote in message
u...
"George Dishman" wrote in message
...

Forgive me for starting a new thread, but the old one has become
fairly redundant due to the huge bulk of posts that seem to serve
no purpose other than to bury the thread under a layer of wader
depth bull****.

I totally agree.

much snipped as there is a major probleme here
That means that any time, the acceleration adds an
amount to the velocity which is equal to the product
of the time and the acceleration. It doesn't add or
subtract from the radius, I don't know where you got
that idea.

For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.

I'm inclined to think you actually believe that George, which is
a bit disconcerting.

I hadn't realised you had a problem with that. Of
course it is fundamental, the word acceleration
is defined as the rate of change of velocity and
is unarguable and the other equations you have
tried to use are all derived form this under
various conditions.

I'll snip the rest and concentrate on this major error, which is
obviously the root of all of the confusion.

What you say is true for Mercury while in its stable eccentric
orbit around the Sun, so long as the anisotropy isn't included.

What I said above is true for all objects changing
speed for any reason whatsoever under any circumstances.

Certainly _not_ under any circumstances George.

Yes, under _any_ circumstances Max. The word
"acceleration" is _defined_ as rate of change
of velocity so in a short time dt the velocity
will change from an initial value v_i to a final
value v_f given by

v_f = v_i + a * dt

where a is the acceleration. v_i, v_f and a are
all vectors and can be handled as x and y
components as I do in the code. (You need z too
in general of course but our orbits are two
dimensional).

Your calculations
apply for any normal trajectory taken by an object naturally
moving to or from a gravity source.

No, it applies to _all_ motion of _any_ nature
whatsoever. That is fundamental to the whole
of dynamics and the process of integrating
acceleration to find velocity is pure maths.

....
You of course agree that an object in a sustainable concentric
orbit around the Sun will not shorten the radius between it and
the Sun?

Obviously since concentric just means at constant
radius.

You also agree that the radius will shorten at the full
gravity rate only if its orbital speed is zero. AND ONLY THEN?

No! You are missing the whole point. Compare two
similar but slightly different motions for Mercury.
In the first it is in a perfectly circular orbit at
some radius with an orbital speed of 48km/s:


^
|
Sun M
|

In the second we have a snapshot of part of some
more complex path when the planet is also moving
at 48km/s:

^
\
Sun M
\

The difference is in the direction of motion, not
the speed. If the angle between the two paths is
just 1 degree the Mercury will move 48000*sin(1)
or 837.7m closer to the Sun in 1 second and that
is without even considering additional acceleration
effects.

The same of course applies for an eccentric orbit.

Do you reject any of that so far?

It is virtually all wrong, you don't seem to know the
definition of acceleration and you have completely
failed to grasp the importance of the direction of
motion.

There's no point in replying to the rest of your post until
this has all been cleared up. You are repeating the same old
mistakes over and over again, again.

I'm not making any mistakes Max, everything I
said follows directly from the definition of
velocity and acceleration. The problem is that
you have forgotten about the effect of the
direction of motion and discarded fundamentals
to try to get the answer you want. Velocity is
the integral of acceleration, _always_.

The gravity force is pointing directly at the Sun, ...

Yes, and Mercury is moving almost at right angles to
that, so the major influence is to change the direction
of motion, not the speed.

so unless
Mercury falls closer to the Sun on average its orbital speed
cannot be increased. Adding a new force does not change the pull
direction, so orbital speed cannot change from the normal unless
the average radial length changes. Can you now see that?

Change of orbital speed is a secondary effect due to
the slow reduction of radius (Mercury moves faster
than the outer planets), what you are missing is that
increasing or decreasing the pull towards the Sun
changes the rate at which the direction of motion
alters. My code includes that effect and the results
follow.

George

"George Dishman" wrote in message
ps.com...
On 30 May, 00:38, "Max Keon" wrote:
George Dishman wrote:
"Max Keon" wrote in message
...

I'll snip the rest and concentrate on this major error, which is
obviously the root of all of the confusion.

What you say is true for Mercury while in its stable eccentric
orbit around the Sun, so long as the anisotropy isn't included.

What I said above is true for all objects changing
speed for any reason whatsoever under any circumstances.

Certainly _not_ under any circumstances George.

Yes, under _any_ circumstances Max. The word
"acceleration" is _defined_ as rate of change
of velocity so in a short time dt the velocity
will change from an initial value v_i to a final
value v_f given by

v_f = v_i + a * dt

where a is the acceleration. v_i, v_f and a are
all vectors and can be handled as x and y
components as I do in the code. (You need z too
in general of course but our orbits are two
dimensional).

Your calculations
apply for any normal trajectory taken by an object naturally
moving to or from a gravity source.

No, it applies to _all_ motion of _any_ nature
whatsoever. That is fundamental to the whole
of dynamics and the process of integrating
acceleration to find velocity is pure maths.

You of course agree that an object in a sustainable concentric
orbit around the Sun will not shorten the radius between it and
the Sun?

Obviously since concentric just means at constant
radius.

You also agree that the radius will shorten at the full
gravity rate only if its orbital speed is zero. AND ONLY THEN?

No! You are missing the whole point. Compare two
similar but slightly different motions for Mercury.
In the first it is in a perfectly circular orbit at
some radius with an orbital speed of 48km/s:


^
|
Sun M
|

In the second we have a snapshot of part of some
more complex path when the planet is also moving
at 48km/s:

^
\
Sun M
\

The difference is in the direction of motion, not
the speed. If the angle between the two paths is
just 1 degree the Mercury will move 48000*sin(1)
or 837.7m closer to the Sun in 1 second and that
is without even considering additional acceleration
effects.

The same of course applies for an eccentric orbit.

Do you reject any of that so far?

It is virtually all wrong, you don't seem to know the
definition of acceleration and you have completely
failed to grasp the importance of the direction of
motion.

There's no point in replying to the rest of your post until
this has all been cleared up. You are repeating the same old
mistakes over and over again, again.

I'm not making any mistakes Max, everything I
said follows directly from the definition of
velocity and acceleration. The problem is that
you have forgotten about the effect of the
direction of motion and discarded fundamentals
to try to get the answer you want. Velocity is
the integral of acceleration, _always_.

The gravity force is pointing directly at the Sun, ...

Yes, and Mercury is moving almost at right angles to
that, so the major influence is to change the direction
of motion, not the speed.

so unless
Mercury falls closer to the Sun on average its orbital speed
cannot be increased. Adding a new force does not change the pull
direction, so orbital speed cannot change from the normal unless
the average radial length changes. Can you now see that?

Change of orbital speed is a secondary effect due to
the slow reduction of radius (Mercury moves faster
than the outer planets), what you are missing is that
increasing or decreasing the pull towards the Sun
changes the rate at which the direction of motion
alters. My code includes that effect and the results
follow.

I don't know how to get it through to George, but you are wrong
and I am right. Go and study it properly.
http://members.optusnet.com.au/maxkeon/peri.html

It seems to me that you may have a problem understanding what is
actually going on here. A gravity anisotropy is something you've
never encountered before and you are trying to explain it using
reasoning that you are accustomed to. That doesn't work.

Perhaps it might help if you realize that you are in no better
a position to see the truth than anyone else who has evolved
through any other society where "truth" has been indoctrinated
into them over many years. I use the word "indoctrinated" because
that's exactly what it is in every form of learning. One cannot
progress if they don't, at least temporarily, blindly accept
certain elements of the learning program.

But if you do know that what I say is correct, that's a whole
new ball game.

I'm going to need to elaborate quite a bit, so don't nod off.

I just happened to catch the final episode of a series, titled
"The Root Of All Evil", within the parent program "Compass" which
is aired on the ABC (Australia). Richard Dawkins of Oxford
University (you perhaps know him) highlighted the consequences
of the religious indoctrination of the young. He described it as
a virus of faith which is transmitted from the older generation
to the new. It's a never ending cycle that divides a community
and leads to a religious intolerance that extends to everyone
who is not likewise indoctrinated.

He also mentioned that even though science is constantly
falsifying the basis for creationism, the message doesn't seem
to be getting through.

The problem is, no matter what evidence science may find, it
could have been created as an integral part of the universe
while the universe was being made in the designated time of 6
or 7 days, depending what one wants to believe. So, it really
doesn't matter a damn what evidence is found.

I don't want to seem overly critical but offering the big bang
theory as an alternative reality is absolutely useless. It does
not describe how the universe began, or how it will end. All it
does is attempt to explain why the universe is what it is.

How can that constitute reality to anyone? The door is wide open.

It's about time the zero origin universe assumed the role of
explaining reality, that's what I think. That universe naturally
has a beginning which can be seen when we look back in time. We
see how the universe has evolved up until now, but we can't see
into the future to the ultimate end of the universe.

If you understood the consequences of that origin and where we
are heading, you would probably realize that life is not about
the individual or the fulfillment of one's self indulgent
desires, including working toward claiming the pot of eternal
bliss at the end of life's rainbow. All life, even in a primitive
state of evolution, is unbounded in its potential for future
development. Every bit of life in this universe has infinitely
more chance of averting the ultimate fate of everything that
exists in this universe than no life at all.

I watch my cat loafing around with no apparent purpose in life.
What is he waiting for? What is the point to the life of a tiny
amoeba floating about at the edge of a backwater? What is the
point to my life, or your life? Life may not seem to offer much
hope in the grand scheme of the universe, BUT WITHOUT IT, THERE
IS NO HOPE AT ALL. No matter how mundane a life may appear to be,
it still can be of absolute importance. Who knows what the future
consequences of its existence will be?

But there is a catch22. The demand for the planet's dwindling
resources in the rapidly developing countries is rising
exponentially. And that doesn't really take into account the
enormous future impact that 220000 plus per day world population
growth. Even a blind man could see that it MUST eventually come
to a sticky end. Choosing to ignore that obvious fact and
blundering down the same old path will lead to the self
extinction of mankind and just about everything else on the
planet. So why did we even bother in the first place.

It would have been far better for us to stay in the backwater
with the amoebas and let something else have a go.

Even though Richard Dawkins would probably tell me I have no
right to impose my own personal morality on anyone, any more than
does anyone have the right to impose their own personal morality
on me, which is invariably not just a moral issue but always
extends to the imposition of one's entire "reality", one thing is
for certain, if we can't all learn to live together as one united
community, there is not one hope in hell of us ever getting out
of the mess that we have so stupidly created here on Earth.

I don't know what the fix will be, but I do know that I won't
like it any more than will anybody else. But failure here is
not an option.

Do you now understand the importance of knowing the truth?

-----

Max Keon

You know the question, just as I did.
What is the Matrix?
Power, control,

"Max Keon" wrote in message
u...
"George Dishman" wrote in message
ps.com...
On 30 May, 00:38, "Max Keon" wrote:
George Dishman wrote:
"Max Keon" wrote in message
...

I'll put back part you snipped as it is the
core of the disagreement:

For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.
....
What I said above is true for all objects changing
speed for any reason whatsoever under any circumstances.

Certainly _not_ under any circumstances George.

Yes, under _any_ circumstances Max. The word
"acceleration" is _defined_ as rate of change
of velocity so in a short time dt the velocity
will change from an initial value v_i to a final
value v_f given by

v_f = v_i + a * dt

where a is the acceleration. v_i, v_f and a are
all vectors and can be handled as x and y
components as I do in the code. (You need z too
in general of course but our orbits are two
dimensional).
....
.. it applies to _all_ motion of _any_ nature
whatsoever. That is fundamental to the whole
of dynamics and the process of integrating
acceleration to find velocity is pure maths.

snip application of above until we get
this agreed.

I don't know how to get it through to George, but you are wrong
and I am right. Go and study it properly.

These may help you:

http://www.staff.amu.edu.pl/~romango...eleration.html

http://hep.physics.indiana.edu/~rick...inematics.html

http://www.blurtit.com/q286865.html

You could also look at the "Force and Motion"
topic he

http://learningcenter.nsta.org/produ...e_objects.aspx

It concentrates mainly on position and velocity
and is a bit light on acceleration but it is
well presented and has lots of questions to let
you check your progress. Once you have gone
through that, look again at what I said at the
top:

For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.

Do you still deny that?

It seems to me that you may have a problem understanding what is
actually going on here. A gravity anisotropy is something you've
never encountered before and you are trying to explain it using
reasoning that you are accustomed to. That doesn't work.

I have encountered anisotropy in many areas but
actually that doesn't matter, you have done the
work on that side and all I need is your result.
You have told me that the effect is to change
the acceleration due to gravity from Newton's
equation:

a = -GM / r^2

to

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

where v is the radial speed, the rate of change
of the radius, and the direction is the same

as Newton's.

You have specified the acceleration Max, all
that is left is to integrate it once to get
velocity and a second time to get position and
the results fall out of that calculation. I
don't need to know anything more about your
ideas at all. Now if that equation doesn't
represent your idea then obviously my comments
will similarly be misplaced but that is what
you have provided so I am taking it at face
value.

snip diatribe

Do you now understand the importance of knowing the truth?

Of course, so go and look up the definition of
"acceleration" and you will find I have been
telling you the truth throughout. Your analysis
is wrong for numerous reasons that I listed
before and which you have ignored.

I have only made one significant error during
the course of the discussion which was when I
did a very quick simulation of your equation
and accidentally got the sign wrong, I used
(1 - v/c) instead of (1 + v/c) so the
eccentricity increased instead of decreasing
but that was some time ago and the Basic code
has the correct sign for your theory. It is
available for anyone to peer review [1] and
I'll repeat the part relevant to the physics
here since we have changed thread since it was
posted:

r2 = x * x + y * y
radius = SQR(r2)

vr = (radius - lastradius) / dt
lastradius = radius

Newton = GM / r2 ' The constant GM is negative

anisotropy = Newton * (vr / c)
acceleration = Newton + anisotropy

ax = acceleration * (x / radius)
ay = acceleration * (y / radius)

vx = vx + dt * ax
vy = vy + dt * ay

x = x + dt * (vx + 0.5 * dt * ax)
y = y + dt * (vy + 0.5 * dt * ay)

time = time + dt

George

[1] If anyone actually peer reviews that code,
be aware that in the actual version I run,
the acceleration for the next time step
uses a linear prediction from the current
and previous acceleration values to get:
forward_mean = (3 * current - previous) / 2

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

I'll put back part you snipped as it is the
core of the disagreement:

I'll snip that and reply where it reappears.
---

I don't know how to get it through to George, but you are wrong
and I am right. Go and study it properly.

These may help you:

http://www.staff.amu.edu.pl/~romango...eleration.html

http://hep.physics.indiana.edu/~rick...inematics.html

http://www.blurtit.com/q286865.html

You could also look at the "Force and Motion"
topic he

http://learningcenter.nsta.org/produ...e_objects.aspx

It concentrates mainly on position and velocity
and is a bit light on acceleration but it is
well presented and has lots of questions to let
you check your progress. Once you have gone
through that, look again at what I said at the
top:

For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.

Do you still deny that?

I don't think I ever denied it in the sense that you've written
it. But I certainly disagree that velocity will change as you
propose if the force is that of gravity acting on an object which
is in a stable orbit. Radial velocity will vary according to the
degree of eccentricity, but the average orbit radius is not going
to change at all. Neither is average orbital speed.

If the gravity force was to double, accelerating the object at
4 m/sec^2, the average orbit radius is not going to shrink by
the added rate of 2 m/sec^2, but would fall inward at a much
lesser rate.

127645 m/sec orbital speed had previously counteracted the
2 m/sec^2 fall pointing directly at the Sun while the object was
held in a stable orbit at a radius of 8146563693 meters, but now
the required orbital speed to hold it in a stable orbit at that
radius around the double Sun mass is 180517 m/sec.

It should be obvious that if the object falls directly inward at
4 m/sec^2 when its orbital speed is zero and zero distance when
its orbital speed is 180517 m/sec, the direct fall rate is going
to reduce proportionally as orbital speed increases.

So what is the true direct fall rate of the object now that it's
traveling to slowly to counteract the direct fall? That rate is
obviously not linear, in the realm of gravity.
http://members.optusnet.com.au/maxkeon/falrate.jpg demonstrates
what that rate would be.

From the graph:

l ----------va------------------- l
l l -----vb------------- l
l----------l------------------------l
l --vc-- l

vc^2 / va^2 * gravity = fall rate

vc = va - vb
180517 - 127645 = 52872

52872^2 / 180517^2 * 4 = .343 m/sec^2 inward fall rate.

It seems to me that you may have a problem understanding what is
actually going on here. A gravity anisotropy is something you've
never encountered before and you are trying to explain it using
reasoning that you are accustomed to. That doesn't work.

I have encountered anisotropy in many areas but
actually that doesn't matter, you have done the
work on that side and all I need is your result.
You have told me that the effect is to change
the acceleration due to gravity from Newton's
equation:

a = -GM / r^2

to

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

where v is the radial speed, the rate of change
of the radius, and the direction is the same

as Newton's.

You have specified the acceleration Max, all
that is left is to integrate it once to get
velocity and a second time to get position and
the results fall out of that calculation. I
don't need to know anything more about your
ideas at all. Now if that equation doesn't
represent your idea then obviously my comments
will similarly be misplaced but that is what
you have provided so I am taking it at face
value.

snip diatribe

Let's hope my message has been noted and understood, however it
was presented.

Do you now understand the importance of knowing the truth?

Of course, so go and look up the definition of
"acceleration" and you will find I have been
telling you the truth throughout. Your analysis
is wrong for numerous reasons that I listed
before and which you have ignored.

I have only made one significant error during
the course of the discussion which was when I
did a very quick simulation of your equation
and accidentally got the sign wrong, I used
(1 - v/c) instead of (1 + v/c) so the
eccentricity increased instead of decreasing
but that was some time ago and the Basic code
has the correct sign for your theory. It is
available for anyone to peer review [1] and
I'll repeat the part relevant to the physics
here since we have changed thread since it was
posted:

I've added to your program extract so that it can be run and
tested. The resident text explains what I've done.

Take note while it's running, that Mercury falls closer to the
Sun on both the up and the down leg of the orbit. Regardless of
what the math says, that is wrong. A fundamental consequence of
the anisotropy is that it will be drawn more to the Sun on the
way up and less on the way down. The trajectory on the down leg
should move toward being further from the Sun than normal, and
that doesn't happen.

The other thing that you still don't seem to grasp is the _fact_
that the whole process is entirely elastic. I've demonstrated why
many times. Should I do it again?

-----

Max Keon

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

'In the duplicated program following "ac:", the anisotropy is excluded.
'The program is now set up to test your updated method against your
'previous method. I know you can do it yourself because you wrote the
'program, but I've included some examples to make it easier for anyone
'else who wants to run it.

'The purple curve only, includes the anisotropy.

'Control-break halts the program. It's the only way out. Any other
'method just bogs the program down. F4 brings back the result.

DEFDBL A-Z
SCREEN 12
CLS
LOCATE 16, 66: PRINT "Aphelion"

c = 300000000#
GM = -1.327D+20
x = 70000000000#
multi = .000000005# ' Multiplier for the graphics.

mx = x ' Aligns program 2 with program 1.
' Program 2 has no anisotropy.

'---------------------------------
vy = 39000#: dt = 1000#
' vy = 39000#: dt = 100#
' vy = 8000#: dt = 50#
' vy = 8000#: dt = 10#
' The current program setup is for Mercury according to your latest
' program update. dt is 1000 second chunks. Swap the switch to run line
' 2 of the set and note the difference. Mercury now falls at a different
' rate. dt must therefore be set to 1 for it to run properly. But even
' then, Mercury is falling almost as fast without an anisotropy. I don't
' think that was intended. Why did you change the program anyway? It
' better represented your view before. Am I doing something wrong?
' I am very impressed with what your program does though.
'---------------------------------

mvy = vy ' Aligns program 2 with program 1.

aa: ' END ' Remove this switch to end the program.
r2 = x * x + y * y
radius = SQR(r2)

vr = (radius - lastradius) / dt
lastradius = radius

Newton = GM / r2 ' The constant GM is negative

anisotropy = Newton * (vr / c)
acceleration = Newton + anisotropy

ax = acceleration * (x / radius)
ay = acceleration * (y / radius)

vx = vx + dt * ax
vy = vy + dt * ay

'-------------------------------------
x = x + dt * (vx + .5 * dt * ax) ' Updated method.
y = y + dt * (vy + .5 * dt * ay)
' x = x + dt * vx ' Previous method.
' y = y + dt * vy ' Swap the switches in both programs.
'-------------------------------------

time = time + dt

IF time 1000 THEN GOSUB ab

GOSUB ac
GOTO aa

ab:
time = 0
CIRCLE (200 + x * multi, 250 + y * multi), 0, 13
CIRCLE (200 + mx * multi, 250 + my * multi), 0, 11
RETURN

ac:
mr2 = mx * mx + my * my
mradius = SQR(mr2)

mvr = (mradius - mlastradius) / dt
mlastradius = mradius

mNewton = GM / mr2 ' The constant GM is negative

' manisotropy = mNewton * (mvr / c)
macceleration = mNewton ' + manisotropy

max = macceleration * (mx / mradius)
may = macceleration * (my / mradius)

mvx = mvx + dt * max
mvy = mvy + dt * may

'----------------------------------------
mx = mx + dt * (mvx + .5 * dt * max) ' Updated method.
my = my + dt * (mvy + .5 * dt * may)
' mx = mx + dt * mvx ' Previous method.
' my = my + dt * mvy ' Swap the switches.
'----------------------------------------

RETURN

"Max Keon" wrote in message
u...

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

I'll put back part you snipped as it is the
core of the disagreement:

I'll snip that and reply where it reappears.
---

I don't know how to get it through to George, but you are wrong
and I am right. Go and study it properly.

These may help you:

http://www.staff.amu.edu.pl/~romango...eleration.html

http://hep.physics.indiana.edu/~rick...inematics.html

http://www.blurtit.com/q286865.html

You could also look at the "Force and Motion"
topic he

http://learningcenter.nsta.org/produ...e_objects.aspx

It concentrates mainly on position and velocity
and is a bit light on acceleration but it is
well presented and has lots of questions to let
you check your progress. Once you have gone
through that, look again at what I said at the
top:

For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.

Do you still deny that?

I don't think I ever denied it in the sense that you've written
it.

Here is the quote, it sounds as though you objected:

[GD:]
For example, if an acceleration of 2m/s^2 acts for
50s then the velocity will change by 100m/s.
[MK:]
I'm inclined to think you actually believe that George, which is
a bit disconcerting.

Anyway, you seem to say the same again:

But I certainly disagree that velocity will change as you
propose if the force is that of gravity acting on an object which
is in a stable orbit.

Acceleration is defined as being the rate at which the
velocity is changing. No matter what the circumstances,
that means that the velocity is found by integrating the
acceleration. If a constant acceleration of 2m/s^2, or
"2m/s per second" applies for 50 seconds then the change
of speed will be 50 * 2 or 100 m/s. The fact that it is
gravity that produces the acceleration is not important,
if the velocity didn't change by 100 m/s in the 50s then
the value of the acceleration would be something else,
it would be dv/50 where dv is the change in velocity.

Of course the acceleration isn't constant and in a stable
orbit, integrating the acceleration over a whole orbit
must get the speed back to its original value - that's
what stable means.

The point here though is that you have changed the formula
for the acceleration by adding the anisotropy and we must
integrate that to find out whether the orbit is stable or
not.

Radial velocity will vary according to the
degree of eccentricity, but the average orbit radius is not going
to change at all. Neither is average orbital speed.

You don't know that until you do the calculation and
that's what the program is about. However, your own
description below matches mine so I'm not sure why
you are saying this.

If the gravity force was to double, accelerating the object at
4 m/sec^2, the average orbit radius is not going to shrink by
the added rate of 2 m/sec^2, but would fall inward at a much
lesser rate.

If the force doubled for just 1 second at the middle
of the inward leg and then went back to normal, it
would cause a slight change in the direction of the
motion towards the sun but the change of energy would
be negligible. That would decrease the perihelion and
since the planet is back in a stable elliptical orbit
with the same energy, it would also increase the
aphelion half an orbit later, hence it would change
the eccentricity.

In fact you said the anisotropy _decreases_ the force
on the inward leg so that increases the perihelion
distance and decreases the aphelion reducing the
eccentricity. Suppose the planet would move from
A to B to C to D without the anisotropy, it goes
from A to B to E to F if the anisotropy is switched
on between B and E then off again:

A
B
E
C
F

D







Sun

The same happens on the outward leg, the slight
_increase_ in gravity as the planet is moving away
pulls it round to point more towards the Sun again
reducing the aphelion.

Sun






A


B

E
C F

D


Because that happens on every orbit, the eccentricity
falls on both the outward and inward legs of every
orbit and it ends up circular, which is the only
stable two-body configuration when your anisotropy is
included.

snip diatribe

Let's hope my message has been noted and understood, however it
was presented.

We are both aiming to understand accurately what your
modification to the formula predicts so:

Do you now understand the importance of knowing the truth?

was unnecessary.

Of course, so go and look up the definition of
"acceleration" and you will find I have been
telling you the truth throughout. Your analysis
is wrong for numerous reasons that I listed
before and which you have ignored.

I have only made one significant error during
the course of the discussion which was when I
did a very quick simulation of your equation
and accidentally got the sign wrong, I used
(1 - v/c) instead of (1 + v/c) so the
eccentricity increased instead of decreasing
but that was some time ago and the Basic code
has the correct sign for your theory. It is
available for anyone to peer review [1] and
I'll repeat the part relevant to the physics
here since we have changed thread since it was
posted:

I've added to your program extract so that it can be run and
tested. The resident text explains what I've done.

Take note while it's running, that Mercury falls closer to the
Sun on both the up and the down leg of the orbit. Regardless of
what the math says, that is wrong.

I agree, that would be wrong, but it is farther
away when I run my original program. I'll recheck
and maybe post some numbers later in case you
changed something by mistake when editing but I
have other pressing (boring) stuff to do first.

A fundamental consequence of
the anisotropy is that it will be drawn more to the Sun on the
way up and less on the way down. The trajectory on the down leg
should move toward being further from the Sun than normal, and
that doesn't happen.

You are absolutely correct, that is what I said
above, and when I run my program that is what
happens. I'll run your version later and see if
I can find why you think it is the opposite.

Can you not see that moving the planet away from
the Sun on the inward leg increases perihelion
while drawing it toward the Sun on the up leg
decreases aphelion so both reduce the eccentricity
as I have been saying?

The other thing that you still don't seem to grasp is the _fact_
that the whole process is entirely elastic. I've demonstrated why
many times. Should I do it again?

Max, you have never demonstrated that at all and
you cannot do so. The requirement for it to be
elastic is that the force must be a function of
radial distance _only_. Since you include speed
in the formula, it isn't elastic.

Let me just pull out one other point:

' ... Why did you change the program anyway? It
' better represented your view before. Am I doing something wrong?

I assume you mean this bit:

vx = vx + dt * ax
vy = vy + dt * ay

'-------------------------------------
x = x + dt * (vx + .5 * dt * ax) ' Updated method.
y = y + dt * (vy + .5 * dt * ay)
' x = x + dt * vx ' Previous method.
' y = y + dt * vy ' Swap the switches in both programs.
'-------------------------------------

The equations apply to a single small time step
lasting dt seconds. Take some simple numbers for
the x values as an example, suppose the step is
dt = 3 seconds, the speed at the start is
vx = 100 m/s and the acceleration is ax = 6 m/s^2.

vx = vx + dt * ax

says that speed at the end of the step will have
risen from 100 m/s to 118 m/s.

The simpler equation

x = x + dt * vx

says the location has changed by 3 * 100 = 300 m
during that time, but the speed isn't a constant
100m/s, it changed from 100 m/s to 118 m/s. The
_average_ speed during the period is 109 m/s so
the accurate calculation of distance moved is
3 * 109 = 327 m.

The formula

x = x + dt * (vx + .5 * dt * ax)

gives a shift of

dt * (vx + .5 * dt * ax)

= 3 * (100 + 0.5 * 3 * 6)

= 3 * (100 + 9)

= 327 m

which is correct. The change just improves the
accuracy.

In the program I'm running at the moment, I have
another small change that does the same sort of
thing to account for the slight change of the
acceleration during the time step.

The aim of all this is that better accuracy of
the basic equations means we can use a larger
step duration so get the program to run faster.

George