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Anisotropy and Mercury (2)



 
 
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  #1  
Old May 30th 07, 12:38 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Max Keon
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Posts: 262
Default 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



  #2  
Old May 30th 07, 07:41 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Koobee Wublee
external usenet poster
 
Posts: 815
Default Anisotropy and Mercury (2)

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.

  #3  
Old May 30th 07, 08:26 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
George Dishman[_1_]
external usenet poster
 
Posts: 2,509
Default Anisotropy and Mercury (2)

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

  #4  
Old May 30th 07, 09:24 PM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Eric Gisse
external usenet poster
 
Posts: 1,465
Default Anisotropy and Mercury (2)

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.




  #5  
Old May 30th 07, 10:54 PM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Koobee Wublee
external usenet poster
 
Posts: 815
Default Anisotropy and Mercury (2)

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.

  #6  
Old May 31st 07, 12:20 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Eric Gisse
external usenet poster
 
Posts: 1,465
Default Anisotropy and Mercury (2)

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]

  #7  
Old June 2nd 07, 10:36 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Max Keon
external usenet poster
 
Posts: 262
Default Anisotropy and Mercury (2)


"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,



  #8  
Old June 2nd 07, 12:01 PM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
George Dishman[_1_]
external usenet poster
 
Posts: 2,509
Default Anisotropy and Mercury (2)


"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



  #9  
Old June 3rd 07, 01:46 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Max Keon
external usenet poster
 
Posts: 262
Default Anisotropy and Mercury (2)

During the process of a quick edit just prior to posting, I
turned one paragraph into what can probably be describes as
illegible babble. I can't leave it that way.
----------------------

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

---

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.

---

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 of the 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.

In the aforementioned program, Richard Dawkins made it very clear
that _nobody_ has the right to impose their own personal morality
on anybody else. What he didn't mention is the fact that that
imposition eventually extends to one's entire version of reality.
Professor Dawkins would no doubt point the critical finger at me
for trying to impose my reality on the world. But my reality
cannot be scientifically faulted. And that reality demands
absolute compliance with the laws of nature on this planet, if
we are to survive with any purpose.

That is all an impossible dream unless we learn to live together
as one united community. Otherwise 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,



  #10  
Old June 3rd 07, 01:49 AM posted to sci.physics.relativity,sci.physics,sci.astro,alt.astronomy
Phineas T Puddleduck[_2_]
external usenet poster
 
Posts: 1,121
Default Anisotropy and Mercury (2)

In article ,
"Max Keon" wrote:

In the aforementioned program, Richard Dawkins made it very clear
that _nobody_ has the right to impose their own personal morality
on anybody else. What he didn't mention is the fact that that
imposition eventually extends to one's entire version of reality.
Professor Dawkins would no doubt point the critical finger at me
for trying to impose my reality on the world. But my reality
cannot be scientifically faulted.



Yawn. Another loon with the sole track to truth.




--
COOSN-174-07-82116: Official Science Team mascot and alt.astronomy's favourite
poster (from a survey taken of the saucerhead high command).

Official maintainer of the supra-cosmic space fluid pump (Mon and Tues only).
 




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