1 "Harald" 
Perihelion of Mercury with classical mechanics ?  donderdag 27 januari 2005 14:45 
2 "Greg Neill" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 28 januari 2005 2:35 
3 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 28 januari 2005 14:30 
4 "Greg Neill" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 28 januari 2005 14:59 
5 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 28 januari 2005 15:41 
6 "Bjoern Feuerbacher" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 28 januari 2005 16:50 
7 "Paul Schlyter" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 28 januari 2005 19:36 
8 "Harald" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 29 januari 2005 3:46 
9 "Harald" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 29 januari 2005 3:49 
10 "Harald" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 29 januari 2005 3:56 
11 "Greg Neill" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 29 januari 2005 6:10 
12 "Greg Neill" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 29 januari 2005 6:22 
13 "Greg Neill" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 29 januari 2005 22:58 
14 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 30 januari 2005 0:34 
15 "Harald" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 30 januari 2005 21:49 
16 "Harald" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 30 januari 2005 21:54 
17 "greywolf42" 
Re: Perihelion of Mercury with classical mechanics ?  maandag 31 januari 2005 20:45 
18 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  dinsdag 1 februari 2005 16:17 
19 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  dinsdag 1 februari 2005 16:17 
20 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  dinsdag 1 februari 2005 16:17 
21 "Greg Neill" 
Re: Perihelion of Mercury with classical mechanics ?  woensdag 2 februari 2005 4:22 
22 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  donderdag 3 februari 2005 20:14 
23 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  donderdag 3 februari 2005 20:50 
24 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 4 februari 2005 16:51 
25 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 4 februari 2005 17:20 
26 "Paul Stowe" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 5 februari 2005 6:13 
27 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 6 februari 2005 0:55 
28 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 6 februari 2005 1:07 
29 "greywolf42" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 6 februari 2005 2:46 
30 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  zondag 6 februari 2005 22:32 
31 "greywolf42" 
Re: Perihelion of Mercury with classical mechanics ?  maandag 7 februari 2005 2:18 
32 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  vrijdag 11 februari 2005 14:49 
33 "greywolf42" 
Re: Perihelion of Mercury with classical mechanics ?  zaterdag 12 februari 2005 19:50 
34 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  maandag 14 februari 2005 15:43 
35 "Steve Carlip" 
Re: Perihelion of Mercury with classical mechanics ?  maandag 14 februari 2005 23:02 
36 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  dinsdag 16 februari 2005 14:52 
37 "N:dlzc D:aol T:com (dlzc)" 
Re: Perihelion of Mercury with classical mechanics ?  woensdag 16 februari 2005 14:58 
38 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  dinsdag 22 februari 2005 21:44 
39 "Nicolaas Vroom" 
Re: Perihelion of Mercury with classical mechanics ?  donderdag 24 februari 2005 19:54 
1. Does anyone know if the classical calculation of the perihelion of Mercury accounts for the attraction of the sun by Mercury?
I stumbled on an interesting article that discusses flaws in Newton's theory, and which suggests that the sun's motion is neglected even for the most accurate calculations. It would be surprising if that were true!
2. It also suggests that when taking it into account, the correct answer may be found. Any ideas if htat could be right? http://www.journaloftheoretics.com/Links/Papers/gravi.pdf Regretfully I know nothing about astronomy.
Harald
"Harald"
> 
1. Does anyone know if the classical calculation of the perihelion of
Mercury accounts for the attraction of the sun by Mercury?
I stumbled on an interesting article that discusses flaws in Newton's theory, and which suggests that the sun's motion is neglected even for the most accurate calculations. It would be surprising if that were true! 2. It also suggests that when taking it into account, the correct answer may be found. Any ideas if htat could be right? http://www.journaloftheoretics.com/Links/Papers/gravi.pdf Regretfully I know nothing about astronomy. Harald 
Attempts to calculate the precession of the perihelion of Mercury by purely classical means have taken into account any number of influences, including of course the motion of the Sun. Newton's theory is completely symmetric when expressed as differential equations to solve the problem.
Amongst the other factors included in "heavy duty" analyses include the influences of other solar system bodies, and the oblateness of the Sun due to its rotation.
No amount of fiddling around with classical mechanics can produce the correct result.
"Greg Neill"
> 

> > 
1. Does anyone know if the classical calculation of the perihelion of
Mercury accounts for the attraction of the sun by Mercury?
I stumbled on an interesting article that discusses flaws in Newton's theory, and which suggests that the sun's motion is neglected even for the most accurate calculations. It would be surprising if that were true! 2. It also suggests that when taking it into account, the correct answer may be found. Any ideas if htat could be right? http://www.journaloftheoretics.com/Links/Papers/gravi.pdf Regretfully I know nothing about astronomy. Harald 
> 
Attempts to calculate the precession of the perihelion of Mercury by purely classical means have taken into account any number of influences, including of course the motion of the Sun. Newton's theory is completely symmetric when expressed as differential equations to solve the problem. 
100 % Correct.
Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. For details go to my home page: https://www.nicvroom.be/ and study the ebook: The Reality Now and Understanding. https://www.nicvroom.be/now.htm
> 
Amongst the other factors included in "heavy duty" analyses
include the influences of other solar system bodies, and
the oblateness of the Sun due to its rotation.
No amount of fiddling around with classical mechanics can produce the correct result. 
There is no fiddling involved. To be even more precise: Within our solair system there is no dark matter. You do not need MOND.
Nicolaas Vroom
"Nicolaas Vroom"
> 
"Greg Neill" 
> > 

> > > 
1. Does anyone know if the classical calculation of the perihelion of
Mercury accounts for the attraction of the sun by Mercury?
I stumbled on an interesting article that discusses flaws in Newton's theory, and which suggests that the sun's motion is neglected even for the most accurate calculations. It would be surprising if that were true! 2. It also suggests that when taking it into account, the correct answer may be found. Any ideas if htat could be right? http://www.journaloftheoretics.com/Links/Papers/gravi.pdf Regretfully I know nothing about astronomy. Harald 
> > 
Attempts to calculate the precession of the perihelion of Mercury by purely classical means have taken into account any number of influences, including of course the motion of the Sun. Newton's theory is completely symmetric when expressed as differential equations to solve the problem. 
> 
100 % Correct. Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. 
Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. The magnitude of the effect is such that it would lead to obvious changes in the semimajor axes of the planets over relatively short periods of time.
>  For details go to my home page: https://www.nicvroom.be/ and study the ebook: The Reality Now and Understanding. https://www.nicvroom.be/now.htm 
> > 
Amongst the other factors included in "heavy duty" analyses include the influences of other solar system bodies, and the oblateness of the Sun due to its rotation. No amount of fiddling around with classical mechanics can produce the correct result. 
> 
There is no fiddling involved. To be even more precise: Within our solair system there is no dark matter. You do not need MOND. Nicolaas Vroom 
"Greg Neill"
> 
"Nicolaas Vroom" 
> > 
"Greg Neill" 
> > > 

> > > > 
1. Does anyone know if the classical calculation of the perihelion of
Mercury accounts for the attraction of the sun by Mercury?
I stumbled on an interesting article that discusses flaws in Newton's theory, and which suggests that the sun's motion is neglected even for the most accurate calculations. It would be surprising if that were true! 2. It also suggests that when taking it into account, the correct answer may be found. Any ideas if htat could be right? http://www.journaloftheoretics.com/Links/Papers/gravi.pdf Regretfully I know nothing about astronomy. Harald 
> > > 
Attempts to calculate the precession of the perihelion of Mercury by purely classical means have taken into account any number of influences, including of course the motion of the Sun. Newton's theory is completely symmetric when expressed as differential equations to solve the problem. 
> > 
100 % Correct. Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. 
> 
Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. 
Can you be more spefic what you mean. Does this mean that the distance (to the Sun) increases ? What is magtitude of this effect ? Any way what is wrong with the assumption that for example the distance of Mars is not constant ? I have seen studies that if you assume that the distance of our Earth is not constant you can explain the ice ages.
(Anyway how do you compare your reply with an expanding Universe ?)
>  The magnitude of the effect is such that it would lead to obvious changes in the semimajor axes of the planets over relatively short periods of time. 
I expect you mean the semimajor axis of the orbits of the planets ? What is wrong with that ? What are the time periods involved ?
Any way what I have also done is to simulate the perihelion advance for one complete revolution. The results are quite interesting and ofcourse I would like to compare them with observations. Have you done, such a simulation, using GR ?
> > 
For details go to my home page:
https://www.nicvroom.be/
and study the ebook:
The Reality Now and Understanding.
https://www.nicvroom.be/now.htm
Nicolaas Vroom 
Nicolaas Vroom posting 5 wrote: [snip]
>  I have seen studies that if you assume that the distance of our Earth is not constant you can explain the ice ages. 
Reference, please.
>  (Anyway how do you compare your reply with an expanding Universe ?) 
Why do you think that is in any way relevant?
[snip]
Bye, Bjoern
Nicolaas Vroom posting 5 wrote:
> 
"Greg Neill" 
>> 
"Nicolaas Vroom" 
>>>  Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. 
>> 
Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. 
> 
Can you be more spefic what you mean. 
In Newtonian physics, any force with a finite propagation speed would be subjected to aberration as seen from a moving object. In the case of a planet orbiting the Sun, that would imply a small component of the force in the direction of motion which would slowly increase the speed of the planet, moving it into a larger orbit. And that would violate the principle of conservation of energy.
>  Does this mean that the distance (to the Sun) increases ? 
It would  and quite measurably so. But that does not happen.
>  What is magtitude of this effect ? 
For your proposed speed of gravity of 300*c, Mercury's energy would increase by some 10% in less than a year, which in turn would increase the mean distance of Mercury from the Sun by a comparable amount. Now, we've carefully observed the positions of the planets during several centuries, and less carefully over a few millennia. And these observations are very clear about this: such a rapid change of Mercury's distance to the Sun just does not happen. And it does not happen for the Earth, or any other planet, either.
>  Any way what is wrong with the assumption that for example the distance of Mars is not constant ? 
This distance has periodic variations, sure. But the long term average is, as far as we can measure, constant over time. And we can measure this to some 810 digits of accuracy.
>  I have seen studies that if you assume that the distance of our Earth is not constant you can explain the ice ages. 
Are you referring to the Milakovitch theory? Well, Milankovitch talks about periodic variations in the eccentricity of the Earth's orbit, and of the inclination and orientation of the axis of the Earth. But he does not assume a long term change of the mean distance EarthSun.
>  (Anyway how do you compare your reply with an expanding Universe ?) 
That's a completely different subject.
>>  The magnitude of the effect is such that it would lead to obvious changes in the semimajor axes of the planets over relatively short periods of time. 
> 
I expect you mean the semimajor axis of the orbits of the planets ? What is wrong with that ? 
It is contradicted by observations.
>  What are the time periods involved ? 
For your proposed speed of gravity = 300*c, the mean distance would increase some 10% over a time scale of just a few years. Observations show very clearly that it does not happen.
>  Any way what I have also done is to simulate the perihelion advance for one complete revolution. The results are quite interesting and ofcourse I would like to compare them with observations. Have you done, such a simulation, using GR ? 
GR also shows the perihelion advance  AND it predicts that the mean distance of the planets from the Sun does not change over the long term. A good fundamental textbook in celestial mechanics will show you how.
>>> 
For details go to my home page:
https://www.nicvroom.be/
and study the ebook:
The Reality Now and Understanding.
https://www.nicvroom.be/now.htm
Nicolaas Vroom 

Paul Schlyter, Grev Turegatan 40, SE114 38 Stockholm, SWEDEN
email: pausch at stockholm dot bostream dot se
WWW: http://www.stjarnhimlen.se/
http://home.tiscali.se/pausch/
>  No amount of fiddling around with classical mechanics can produce the correct result. 
Now that's for sure at least an exaggeration, as I know for fact that in 1898 Paul Gerber derived from Newtonian mechanics the same equation as Einstein for the perihelion, and that he calculated c from it with high accuracy. But I don't have his paper so I don't know the details of how he did it. I only have a copy of the original end part as printed in the book Einstein plus two, and Gerber found c = 305500 km/sec.
Harald
Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. For details go to my home page: https://www.nicvroom.be/
See my reply to Greg Neil; thus I must assume that you made an error somewhere. But I doubt that I am qualified to find it. If I get the paper of Gerber then I can send the PDF on to you.
Harald
Paul Schlyter posting 7 wrote:
>>  Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. 
I suppose an upgrade to SRT would take care of that, right? Thanks, Harald
"Paul Schlyter"
[snip]
Thanks, Paul, for saving me the trouble of replying with all the details. You hit the nail on the head.
Cheers.
"Harald"
>  Greg Neil posting 2 wrote: 
> >  No amount of fiddling around with classical mechanics can produce the correct result. 
> 
Now that's for sure at least an exaggeration, as I know for fact that in 1898 Paul Gerber derived from Newtonian mechanics the same equation as Einstein for the perihelion, and that he calculated c from it with high accuracy. But I don't have his paper so I don't know the details of how he did it. I only have a copy of the original end part as printed in the book Einstein plus two, and Gerber found c = 305500 km/sec. 
I believe that Gerber's derivation was shown to be incorrect, even though he arrived at the correct answer.
"Harald"
As your belief led to your strong statement above, I would like to know
what his error was  especially as Petr Beckman repeated his derivation
in a slightly different way with the same result.
>>>
Greg Neill posting 2 wrote:
>>> >
No amount of fiddling around with classical mechanics can
produce the correct result.
>
SNIP
>>
I believe that Gerber's derivation was shown to be incorrect,
even though he arrived at the correct answer.
>
Here's a link that may help:
http://www.mathpages.com/home/kmath527/kmath527.htm
Nicolaas Vroom
> 
"Greg Neill" 
>> 
"Nicolaas Vroom" 
[...]
>> >  Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. 
>>  Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. 
>  Can you be more spefic what you mean. Does this mean that the distance (to the Sun) increases ? What is magtitude of this effect ? 
See problem 12.4 of Lightman et al., _Problem book in relativity and gravitation_, for a simple derivation. For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would have been at the edge of the Sun about 120,000 years ago.
Laplace considered the effect of a finite speed of gravity in Newtonian mechanics in 1805, and showed that observations of the orbit of the Moon required a speed of at least 7x10^6 c.
Steve Carlip
>>>  Greg Neill posting 2 wrote: 
>>>>  No amount of fiddling around with classical mechanics can produce the correct result. 
>  SNIP 
>>  I believe that Gerber's derivation was shown to be incorrect, even though he arrived at the correct answer. 
>  Here's a link that may help: http://www.mathpages.com/home/kmath527/kmath527.htm 
Thanks, I had not thought of looking there! Note however that that site is less sure than you are; and for the moment I have not yet seen a strong argument against the validity of the first paper that I cited. Cheers, Harald
Nicolaas Vroom posting 3 wrote:
> 
Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. For details go to my home page: https://www.nicvroom.be/ and study the ebook: The Reality Now and Understanding. https://www.nicvroom.be/now.htm 
Hey Nicolaas, that's an interesting site! And I look forward to use and (if I can manae) to adapt your programs so as to include the attraction of the sun by Mercury.
Cheers, Harald
PS I have feedback on your Twin problem, but I suppose that has been sorted out by now
"Steve Carlip"
> 
Nicolaas Vroom 
> > 
"Greg Neill" 
> >> 
"Nicolaas Vroom" 
> 
[...] 
> >> >  Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. 
> 
> >> 
Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. 
> 
> > 
Can you be more spefic what you mean. Does this mean that the distance (to the Sun) increases ? What is magtitude of this effect ? 
> 
See problem 12.4 of Lightman et al., _Problem book in relativity and gravitation_, for a simple derivation. For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would 
have had to
>  have been at the edge of the Sun about 120,000 years ago. 
to be at it's present position, today.
{I don't have a copy of Lightman at hand. But I presume that Lightman is competent in wielding the aberration argument  #3, below.}
>  Laplace considered the effect of a finite speed of gravity in Newtonian mechanics in 1805, and showed that observations of the orbit of the Moon required a speed of at least 7x10^6 c. 
Steve is being deliberately dishonest, here. He is attempting to "motivate" you, so that you don't "waste your time" with theories that Steve does not support. (This is not inadverntent. He has done it before, and been called on it, several times.) In this immediate response, Steve has mixed two counteracting forces (aberration: Lightman, and drag: Laplace) in such a way as to make you think that they are addressing the same force.
There are five components to this deliberate distortion.
1) Steve is not telling you the name or type of the gravitational theory that Laplace was addressing. The theory is called Le Sagian gravity, and was proffered by Georges Louis Le Sage, in 1782. This theory derives Newton's gravitational law (actually it derives the weakfield limit of GR) from the partial absorption of 'ultramundane corpuscles' by mass. {A search on Le Sage or Lesage will bring up quite a few recent discussions on the theory.}
2) The 'drag' effect mentioned by Steve is based on the drag of a matter body as it moves through a *medium.* It is not the speed of gravity  per se  that would cause the Earth to shrink its orbit; it is the impact of those 'ultramundane corpuscles.'
3) The effect that arises in *any* gravitational theory with a finite speed of gravity (including GR) is gravitational aberration. And gravitational aberration will tend to *increase* the radius of an orbit. Steve did a paper on just this effect  to try to save GR from the issue.
4) Laplace (and just about everyone since, including Feynman and Poincare) determined their "requirement" for high speed on the basis of drag, alone. And never considered the potential balancing of the two forces. In fact, Steve will tell you that the aberration term will *always* overpower the drag term (for the Earth).
Steve will likely tell you that such is done simply to avoid "confusion."

greywolf42
ubi dubium ibi libertas
{remove planet for return email}
"Greg Neill"
> 
"Paul Schlyter" [snip] Thanks, Paul, for saving me the trouble of replying with all the details. You hit the nail on the head. Cheers. 
Because Paul did all this good work as a token of appreciation why don't you give your opinion about my latest posting in the news groups sci.astro.research or sci.physics.research: "How important is GR in order to calculate the precession of Mercury" I would really appreciate that.
Cheers
Nicolaas Vroom.
"Paul Schlyter"
>  Nicolaas Vroom posting 5 wrote: 
> > 
>  In Newtonian physics, any force with a finite propagation speed would be subjected to aberration as seen from a moving object. In the case of a planet orbiting the Sun, that would imply a small component of the force in the direction of motion which would slowly increase the speed of the planet, moving it into a larger orbit. 
I fully agree with you. The issue is how much.
>  And that would violate the principle of conservation of energy. 
> > 
What is magtitude of this effect ? 
> 
For your proposed speed of gravity of 300*c, Mercury's energy would increase by some 10% in less than a year, which in turn would increase the mean distance of Mercury from the Sun by a comparable amount. Now, we've carefully observed the positions of the planets during several centuries, and less carefully over a few millennia. And these observations are very clear about this: such a rapid change of Mercury's distance to the Sun just does not happen. And it does not happen for the Earth, or any other planet, either. 
If you go to my homepage and you select my free ebook
The Reality Now and Understanding
and specific the chapter (program) about Mercury.
This program allows you to simulate all the planets
around the Sun and to study the stability of each.
(by means of the parameter subtests)
See also: https://www.nicvroom.be/now_mercury.htm
Specific read what is written in paragraph 4.1 Test 2A
This test not only discusses the stability of the Earth around the Sun (for v=0) but also the planets Venus, Mars Jupiter Saturn and Uranus.
The results show that Jupiter is the most unstable and increases after each revolution of roughly 12 years with a distance for speed of gravity equal to c with 810 km out of a distance of 778300000 km. However for a speed of gravity equal to 100c this increase is reduced to 8 km and if sog = 300c this increase is equal to 3km for each revolution of jupiter around the Sun.
> >  Any way what is wrong with the assumption that for example the distance of Mars is not constant ? 
> 
This distance has periodic variations, sure. But the long term average is, as far as we can measure, constant over time. And we can measure this to some 810 digits of accuracy. 
For Mars for sog = c the increase is almost equal to zero.
> >  What are the time periods involved ? 
> 
For your proposed speed of gravity = 300*c, the mean distance would increase some 10% over a time scale of just a few years. Observations show very clearly that it does not happen. 
My simulations are not in agreement with this statement.
> >  Any way what I have also done is to simulate the perihelion advance for one complete revolution. The results are quite interesting and ofcourse I would like to compare them with observations. Have you done, such a simulation, using GR ? 
> 
GR also shows the perihelion advance  AND it predicts that the mean distance of the planets from the Sun does not change over the long term. A good fundamental textbook in celestial mechanics will show you how. 
What is the prediction of the precession of the perihelion
angle (currently 43 seconds of arc per century)
of mercury for one complete revolution of the precession
angle ? Is this angle constant ?
"My" prediction is that it is not.
> >>> 
For details go to my home page:
https://www.nicvroom.be/
and study the ebook:
The Reality Now and Understanding.
https://www.nicvroom.be/now.htm
Nicolaas Vroom 
"Steve Carlip"
> 
Nicolaas Vroom 
> > 
"Greg Neill" 
> >> 
"Nicolaas Vroom" 
> 
[...] 
> >> >  Newton's theory assumes that gravity act instantaneous. However if you take into acount that the speed of gravity is not infinite but equal to 300*c you can correctly simulate the perihelion precession of Mercury. 
> 
> >> 
Unfortunately, this leads to the problem of the energy of the orbit changing due to the noncentral nature of the resulting force. 
> 
> > 
Can you be more spefic what you mean. Does this mean that the distance (to the Sun) increases ? What is magtitude of this effect ? 
> 
See problem 12.4 of Lightman et al., _Problem book in relativity and gravitation_, for a simple derivation. For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would have been at the edge of the Sun about 120,000 years ago. 
My simulations of the stability of the Earth show that for a speed of gravity equal c the distance of the Earth increases with 1 km out of a distance of 149600000 km for each revolution (1 year) For a speed of gravity equal to 300*c this increase is equal to 1/300 km.
For more details see my reply to Paul Sluyter.
> 
Laplace considered the effect of a finite speed of gravity in
Newtonian mechanics in 1805, and showed that observations of the
orbit of the Moon required a speed of at least 7x10^6 c.
Steve Carlip 
Nicolaas Vroom
"Nicolaas Vroom"
> 
"Greg Neill" 
> > 
"Paul Schlyter" [snip] Thanks, Paul, for saving me the trouble of replying with all the details. You hit the nail on the head. Cheers. 
> 
Because Paul did all this good work as a token of appreciation why don't you give your opinion about my latest posting in the news groups sci.astro.research or sci.physics.research: "How important is GR in order to calculate the precession of Mercury" I would really appreciate that. 
I'm afraid I haven't seen that posting. Perhaps it's my newsreader acting up.
However, in answer to your question, "How important is GR in order to calculate the precession of Mercury", I would say that it is essential if one wishes to consider a theoretical approach that agrees with all available data.
greywolf42
> 
"Steve Carlip" 
[...]
>>  Laplace considered the effect of a finite speed of gravity in Newtonian mechanics in 1805, and showed that observations of the orbit of the Moon required a speed of at least 7x10^6 c. 
>  Steve is being deliberately dishonest, here. He is attempting to "motivate" you, so that you don't "waste your time" with theories that Steve does not support. 
That is not true.
>  [...] In this immediate response, Steve has mixed two counteracting forces (aberration: Lightman, and drag: Laplace) in such a way as to make you think that they are addressing the same force. 
This is simply wrong. Go back and read Laplace, _Celestial Mechanics_, section X.VII.22. It's true that elsewhere in X.VII, Laplace deals with drag. But this section, which contains the limit that I quoted, deals *explicitly* with aberration, *not* drag.
>  There are five components to this deliberate distortion. 
Of which you list four?
>  1) Steve is not telling you the name or type of the gravitational theory that Laplace was addressing. The theory is called Le Sagian gravity, and was proffered by Georges Louis Le Sage, in 1782. This theory derives Newton's gravitational law (actually it derives the weakfield limit of GR) from the partial absorption of 'ultramundane corpuscles' by mass. {A search on Le Sage or Lesage will bring up quite a few recent discussions on the theory.} 
It may be that Laplace had LeSage in mind. I don't know. In particular, I have been unable to find any reference to LeSage in section X.VII of Laplace's _Celestial Mechanics_. Perhaps it's elsewhere  I haven't read the whole book. Would you care to provide a specific citation?
>  2) The 'drag' effect mentioned by Steve is based on the drag of a matter body as it moves through a *medium.* It is not the speed of gravity  per se  that would cause the Earth to shrink its orbit; it is the impact of those 'ultramundane corpuscles.' 
That is incorrect. The issue in this thread has been the effect of finite propagation speed in Newtonian gravity, and that's what I addressed. I did not say, or imply, anything about "drag." Contrary to your claim, the limit I quoted from Laplace also had nothing to do with drag, but came from the effect of putting a finite propagation speed into Newtonian gravity.
>  3) The effect that arises in *any* gravitational theory with a finite speed of gravity (including GR) is gravitational aberration. And gravitational aberration will tend to *increase* the radius of an orbit. [...] 
Right. That's what I said. "For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would have been at the edge of the Sun about 120,000 years ago." That's an increase in the radius of the orbit, right?
>  4) Laplace (and just about everyone since, including Feynman and Poincare) determined their "requirement" for high speed on the basis of drag, alone. And never considered the potential balancing of the two forces. In fact, Steve will tell you that the aberration term will *always* overpower the drag term (for the Earth). 
Once again: Laplace, _Celestial Mechanics_, section X.VII.22, is about finite propagation speed, not drag.
>  Steve will likely tell you that such is done simply to avoid "confusion." 
No, I will say that greywolf wrote a fictional account that had nothing to do with what I said. Unlike him, I will not charge "deliberate distortion" or accuse him of "deliberate dishonesty." He may have misremembered Laplace, or only read someone else's description, and leapt to conclusions without actually paying much attention to the post he was responding to.
Steve Carlip
Nicolaas Vroom
> 
"Steve Carlip" 
>>  See problem 12.4 of Lightman et al., _Problem book in relativity and gravitation_, for a simple derivation. For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would have been at the edge of the Sun about 120,000 years ago. 
>  My simulations of the stability of the Earth show that for a speed of gravity equal c the distance of the Earth increases with 1 km out of a distance of 149600000 km for each revolution (1 year) 
If
(1) you're looking at Newtonian gravity in the "force" description
(F = GMm/r^2 = ma), but with the direction and magnitude of the
force depending on the retarded position of the gravitating mass;
and
(2) you're looking at a twobody problem,
then the problem can be analyzed analytically, and gives a result that is drastically different from your claim. If this is the case, then there's something wrong with your simulation.
If you are looking at Newtonian gravity in the "potential" description, with a potential that depends on the retarded position of the gravitating mass, then the effect is suppressed. Even then, I suspect that you will get in trouble with the Lunar orbit, and you will certainly run into contradictions with pulsar observations. For that model, Mercury's perihelion advance can also be computed analytically, and disagrees with observation.
Or are you doing neither of these things?
Steve CArlip
"Steve Carlip"
> 
Nicolaas Vroom 
> > 
"Steve Carlip" 
>  [...] 
> >> 
See problem 12.4 of Lightman et al., _Problem book in relativity and gravitation_, for a simple derivation. For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would have been at the edge of the Sun about 120,000 years ago. 
> 
> > 
My simulations of the stability of the Earth show that for a speed of gravity equal c the distance of the Earth increases with 1 km out of a distance of 149600000 km for each revolution (1 year) 
> 
If
(1) you're looking at Newtonian gravity in the "force" description
(F = GMm/r^2 = ma), but with the direction and magnitude of the
force depending on the retarded position of the gravitating mass;
and then the problem can be analyzed analytically, and gives a result that is drastically different from your claim. If this is the case, then there's something wrong with your simulation. 
That is what I have done. What should be the result (increase in distance) for Jupiter after one revolution with speed of gravity equal to c? The same but for 300*c ?
I will try to do the same simulation but with 3 bodies, but it will take me some days to prepare.
>  If you are looking at Newtonian gravity in the "potential" description, with a potential that depends on the retarded position of the gravitating mass, then the effect is suppressed. Even then, I suspect that you will get in trouble with the Lunar orbit, and you will certainly run into contradictions with pulsar observations. For that model, Mercury's perihelion advance can also be computed analytically, and disagrees with observation. 
I did not use any "potential" description.
I did not study Earth and Moon or Sun Earth and Moon.
I did not study twobody simulation with one being pulsar.
If you give me mass, initial speed and distance than I will try.
> 
Or are you doing neither of these things?
Steve CArlip 
"Greg Neill"
> 
"Nicolaas Vroom" 
> > 
Because Paul did all this good work as a token of appreciation why don't you give your opinion about my latest posting in the news groups sci.astro.research or sci.physics.research: "How important is GR in order to calculate the precession of Mercury" I would really appreciate that. 
> 
I'm afraid I haven't seen that posting. Perhaps it's my newsreader acting up. However, in answer to your question, "How important is GR in order to calculate the precession of Mercury", I would say that it is essential if one wishes to consider a theoretical approach that agrees with all available data. 
In fact what I have proposed is to do a transformation
of all the observations of a certain time and date
into a frame which contains synchronised clocks
at the grid points
and then I ask my self the question what are the rules
which describe the movement of the bodies in that frame.
(In fact I removed the visible/observational aspects of the bodies
and I made them invisible)
IMO you do not have to take SR into account in that frame.
If you do it using GR than how does "it look".
There are no moving clocks in that frame.
What is the function of c in that frame in order to describe
the behaviour of the bodies.
If you need more you can also see a summary of the
postings of this thread at my homepage:
https://www.nicvroom.be/grcalc.htm
or for more discussions:
https://www.nicvroom.be/usenet.htm
Nicolaas Vroom
On Thu, 3 Feb 2005 19:14:24 +0000 (UTC), carlipnospam@physics.ucdavis.edu posting 22 wrote:
> 
greywolf42 
>> 
"Steve Carlip" 
> 
A related question to aberration... If there IS an aberrative vector that tends to cause a Star & its Planet to soon be parted, where in the universe does the energy to do so come from?
To gain radii doesn't REAL energy have to be expended???
What about if we stop thinking so linearly. Say there are two bodies (the Star & its Planet) whos real position is like so,
[S] (P)Thus, for JUST a center point vector these would be > <
Who are actually aberrated so as to look to each other as,
(P) [S]and INSTEAD of a leading vector, the arrow does point to the baricenter as a curve. Thus the pointing vectors at each body points so that the remains a central radial acceleration only. The energy of the system remains conserved, AND, at each differential point along the curved path, the vector always remains center pointing?
Why is this not possible?
Nicolaas Vroom
> 
"Steve Carlip" 
[...]
>> 
If
(1) you're looking at Newtonian gravity in the "force" description
(F = GMm/r^2 = ma), but with the direction and magnitude of the
force depending on the retarded position of the gravitating mass;
and then the problem can be analyzed analytically, and gives a result that is drastically different from your claim. If this is the case, then there's something wrong with your simulation. 
> 
That is what I have done. What should be the result (increase in distance) for Jupiter after one revolution with speed of gravity equal to c? The same but for 300*c ? 
To a very good approximation, for nearly circular orbits the radius at time t will satisfy
r^2  (r_0)^2 = (4GM/c_g)(tt_0)
where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years.
Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy.
Steve Carlip
Paul Stowe
>  A related question to aberration... If there IS an aberrative vector that tends to cause a Star & its Planet to soon be parted, where in the universe does the energy to do so come from? 
If your theory is just Newtonian gravity with time delay stuck in, then energy isn't conserved. The energy doesn't come from anywhere; it just appears.
If you want to look at a model in which energy is conserved, the answer will depend on the details of the model. In models having only a gravitational field, the field itself can carry energy in the form of gravitational radiation, and a consistent theory has to automatically balance field energy and orbital kinetic energy. You can use this to get estimates of the effect of aberration by assuming selfconsistency; typically, you find that there must be other interactions (velocity dependent forces) that at least partially counteract the effect of finitevelocity propagation. Of course, this argument doesn't tell you what those interactions are  that will again depend on the specific model. But this is no different than most arguments appealing to conservation, which typically tell you that something must happen but don't in themselves tell you exactly what.
In a theory with more than just gravitational fields  a LeSage model, for instance, or a model describing gravity in terms of fluid flows  the extra stuff in the theory (LeSagean particles, fluid,...) can carry energy as well. You can still appeal to energy conservation, if you've checked that your model really does conserve energy. But to draw any real conclusions, you also need a fairly detailed understanding of the rate of energy transfer between the gravitating objects and whatever else is in the model.
Steve Carlip
"Steve Carlip"
> 
greywolf42 
> > 
"Steve Carlip" 
> 
[...] 
> >>  Laplace considered the effect of a finite speed of gravity in Newtonian mechanics in 1805, and showed that observations of the orbit of the Moon required a speed of at least 7x10^6 c. 
> 
> > 
Steve is being deliberately dishonest, here. He is attempting to "motivate" you, so that you don't "waste your time" with theories that Steve does not support. 
> 
That is not true. 
Well, that was been your stated purpose for this very same deliberate distortion in the past. I see you continue your deliberate distortion in your parallel post to Nicolaas Vroom.
> >  [...] 
Steve attempts to avoid the fact that he has made this same deception many
times. Replacing the snip:
======================
> >  (This is not inadverntent. He has done it before, and been called on it, several times.) 
> >  In this immediate response, Steve has mixed two counteracting forces (aberration: Lightman, and drag: Laplace) in such a way as to make you think that they are addressing the same force. 
> 
This is simply wrong. Go back and read Laplace, _Celestial Mechanics_, section X.VII.22. It's true that elsewhere in X.VII, Laplace deals with drag. But this section, which contains the limit that I quoted, deals *explicitly* with aberration, *not* drag. 
My apologies for not checking the section number.
So, instead of deceiving Nicolaas about drag *and* aberration, you are simply deceiving Nicolaas about the very existence of drag.
> >  There are five components to this deliberate distortion. 
> 
Of which you list four? 
Bad numbering system. The first was addressed immediately above.
> >  1) Steve is not telling you the name or type of the gravitational theory that Laplace was addressing. The theory is called Le Sagian gravity, and was proffered by Georges Louis Le Sage, in 1782. This theory derives Newton's gravitational law (actually it derives the weakfield limit of GR) from the partial absorption of 'ultramundane corpuscles' by mass. {A search on Le Sage or Lesage will bring up quite a few recent discussions on the theory.} 
> 
It may be that Laplace had LeSage in mind. I don't know. 
And it is irrelevant. For the point is not whether Laplace had Le Sage specifically in mind. But that Laplace was (and you are) addressing Le Sagetype theories.
>  In particular, I have been unable to find any reference to LeSage in section X.VII of Laplace's _Celestial Mechanics_. Perhaps it's elsewhere  I haven't read the whole book. Would you care to provide a specific citation? 
I haven't read the whole thing for years now, either. Nor do I have a copy. Nor is it relevant to the issue of your deception  keeping Nicolaas ignorant of the type of theory being discussed. At issue is not simply a single section of Laplace, but the essence of the argument of Laplace, Lightman, and yourself.
> >  2) The 'drag' effect mentioned by Steve is based on the drag of a matter body as it moves through a *medium.* It is not the speed of gravity  per se  that would cause the Earth to shrink its orbit; it is the impact of those 'ultramundane corpuscles.' 
> 
That is incorrect. The issue in this thread has been the effect of finite propagation speed in Newtonian gravity, and that's what I addressed. 
Which is why I accurately described your action as deliberate deception, and not an outright lie. Your statements are quite literally true  and also deliberately deceptive.
>  I did not say, or imply, anything about "drag." 
That *is* the deception on your part. You are well aware that theories of the sort that you (and Laplace and Lightman) were addressing *also* have a drag component. But knowing this  and knowing the possibility exists of a balance  you did not tell Nicolaas about this.
>  Contrary to your claim, the limit I quoted from Laplace also had nothing to do with drag, but came from the effect of putting a finite propagation speed into Newtonian gravity. 
But Laplace *does* have a section on drag. From which, you are attempting to divert.
> >  3) The effect that arises in *any* gravitational theory with a finite speed of gravity (including GR) is gravitational aberration. And gravitational aberration will tend to *increase* the radius of an orbit. 
>  Right. That's what I said. "For a speed of gravity of 300c within Newtonian gravity, the Earth's orbit is unstable enough that it would have been at the edge of the Sun about 120,000 years ago." That's an increase in the radius of the orbit, right? 
Yep. As I noted. Your statement about orbital increase is specifically true  and deliberately deceptive. Because you are assuming zero drag effect. (I believe the value in your calculation may be in error by about a factor of 1 million. What aberration factor did you use for the Earth?) With a drag effect, you can't make the above claim. That is the deception.
> >  [...] 
Here is what Steve snipped:
> >  Steve did a paper on just this effect  to try to save GR from the issue. 
And this is unavoidable. Steve knows that GR suffers from *precisely* the same "problem" of aberration. But aberration never acts alone. That is Steve's deception. For GR, Steve discussed "miraculous" (and nonspecific) backaction.
> >  4) Laplace (and just about everyone since, including Feynman and Poincare) determined their "requirement" for high speed on the basis of drag, alone. And never considered the potential balancing of the two forces. In fact, Steve will tell you that the aberration term will *always* overpower the drag term (for the Earth). 
Steve apparently has confirmed this last sentence. Both by not contradicting it, and by trying to remove all consideration of balancing drag forces from this post.
>  Once again: Laplace, _Celestial Mechanics_, section X.VII.22, is about finite propagation speed, not drag. 
But the *other* sections contain drag calculations.
> >  Steve will likely tell you that such is done simply to avoid "confusion." 
> 
No, I will say that greywolf wrote a fictional account that had nothing to do with what I said. 
But that is simply a false statement, Steve. Unlike your prior distortion (which was explicitly true, but deceptive), this statement is demonstrably false. My statements have everything to do with what you've said  both on this post and on prior exchanges.
>  Unlike him, I will not charge "deliberate distortion" 
But Steve will continue to deliberately distort the physical situation. ;)
>  or accuse him of "deliberate dishonesty." 
Because Steve knows that nothing in my post is either a distortion, or dishonest.
>  He may have misremembered Laplace, or only read someone else's description, and leapt to conclusions without actually paying much attention to the post he was responding to. 
So, Steve will continue to try to deliberately deceive Nicolaas about the *fact* that all orbital dynamical calculations contain both drag and aberration terms. But Steve will pretend to be noble and professional.
Steve, all you have to do to be honest and professional is to mention that there are two competing forces in real, physical, causative theories. Drag and aberration. And that *IF* one of these two forces overpowers the other, then the planet will either spiral in or out. But you can't honestly continue to claim that either approach  alone  demonstrates that physical theories of gravity don't work.
Of course, that acknowledges the issue that you wish to avoid.

greywolf42
ubi dubium ibi libertas
{remove planet for email}
greywolf42
> 
"Steve Carlip" 
[A good deal of ad hominem snipped...]
>>  I did not say, or imply, anything about "drag." 
>  That *is* the deception on your part. You are well aware that theories of the sort that you (and Laplace and Lightman) were addressing *also* have a drag component. But knowing this  and knowing the possibility exists of a balance  you did not tell Nicolaas about this. 
The question in this thread was *explicitly* about Newtonian gravity with a propagation delay. Period. If you need clarification on this, see http://groupsbeta.google.com/group/sci.astro/msg/15c024f0ce180044.
Since this was the question, this is what I responded to.
[more ad hominem snipped...]
>  So, Steve will continue to try to deliberately deceive Nicolaas about the *fact* that all orbital dynamical calculations contain both drag and aberration terms. But Steve will pretend to be noble and professional. 
Nicolaas knows perfectly well that there are other models in which additional forces act. You should not insult him by assuming such ignorance. In this thread, those other models were not at issue. Go back and read a little!
Steve Carlip
"Steve Carlip"
> 
Nicolaas Vroom 
> > 
"Steve Carlip" 
> 
[...] 
> >> 
If
(1) you're looking at Newtonian gravity 
But, Steve, at issue is not simply "Newtonian gravity." Nicolaas was responding to Greg's claims about "classical" theories of gravity:
"Attempts to calculate the precession of the perihelion of Mercury by purely classical means have taken into account any number of influences, including of course the motion of the Sun."
Why do you continue to try to divert solely into Newton's empirical formula ... then change the formula?
> >> 
in the "force" description
(F = GMm/r^2 = ma), but with the direction and magnitude of the
force depending on the retarded position of the gravitating mass;
and (2) you're looking at a twobody problem, then the problem can be analyzed analytically, and gives a result that is drastically different from your claim. If this is the case, then there's something wrong with your simulation. 
> 
> > 
That is what I have done. 
> 
To a very good approximation, for nearly circular orbits the radius at time t will satisfy r^2  (r_0)^2 = (4GM/c_g)(tt_0) where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years. Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy. 
But this is simply trying to hack an empirical formula (Newton's). Equation mining is not generally useful. A real theory includes a cause, that gives rise to a finite speed of gravity. Not simply trying to slap a new term into an equation.
The fact that the orbit is not stable simply indicates that your crude approach has failed to accurately (or completely) model the process. Not that gravity does not have a finite speed.

greywolf42
ubi dubium ibi libertas
{remove planet for email}
"Steve Carlip"
> 
Nicolaas Vroom To a very good approximation, for nearly circular orbits the radius at time t will satisfy r^2  (r_0)^2 = (4GM/c_g)(tt_0) where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years. Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy. 
I have studied the book from a library but still I have a couple of unsettled questions.
In the above equation the stability is a function of the mass
of the Sun.
The results of my simulation show a different result.
For a summary see chapter 4.1 in
https://www.nicvroom.be/now_mercury.htm
more specific paragraph 4.1.6 mathematics.
What my simulations show is that
1. if you increase the mass of the planet with a factor 2
the stability decreases with a factor of 2.
Delta R becomes 2 * Delta R per Year
2. if you increase the distance of the planet with a factor 2
the stability increases with a factor of 2.
Delta R becomes Delta R / 2 per Year
The reason IMO comes from the equation at page 350
that the earth's energie increases at a function of
v * theta = v * v / c.
IMO that is wrong.
IMO the sun's energie is a function of angle v / c
i.e. v+/c with v+ being the speed of the earth.
IMO the earth's energie is a function of angle v0/c
with v0 being the speed of the sun.
In total earth's energie is a function of v+*v0/c
(which is a factor M+/M0 smaller)
As such I expect that the above equation should be: r^2  (r_0)^2 = (4GM+/c_g)(tt_0) (ie not a function of M0 but of M+ = mass planet)
In my simulation the force/acceleration of the Earth points to (is influenced by) the retarded position of the Sun which is a function of v0/c (Speed Sun) And the force/accelaration of the Sun points to the retarded position of the Earth which is a function of v+/c (speed of the Earth).
For a good description of aberration see paragraph
1.363 page 23 in the book "Explanatory supplement
to the astronomical almanac" by K. Seidelmann.
In that paragraph they discuss correction for lighttime
(independent motion of observer) and stellar aberration
(independent of motion and distance of object).
The sum of the two is called planetary abberration.
Nicolaas Vroom
>  Steve Carlip 
carlipnospam@physics.ucdavis.edu posting 27 wrote:
{snip}
> 
To a very good approximation, for nearly circular orbits the radius
at time t will satisfy
r^2  (r_0)^2 = (4GM/c_g)(tt_0) where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years. Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy. 
The effect of aberration delay in Newtonian physics is a tangential acceleration roughly proportional to (m / M) (v / v_g). You have left out the first term. Quite simply, you (and Lightman) forgot that the planet and the Sun orbit the centerofmass  not the midpoint of the orbit. It is only when the masses are equal that the aberration angle is proportional to v / v_g, alone.
Hence, your equation that resulted in 400 years for aberration, alone (if you did the calculation correctly), would result in a 400 million year time when v_g = c. (Since m / M is about 1 / 1,000,000 for the Earth and Sun.)
greywolf42
ubi dubium ibi libertas
In order to test the stability of the Earth and Jupiter I have
done a new simulation with includes 3 objects:
Sun Earth and Jupiter.
This simulation takes eccentricity into account and is performed
in 2D (x and y, z=0)
The largest distance for the Earth are as follows:
(first value is total initial distance, all others last 4 digits)
with speed of gravity equal to zero.
152096309, 6449, 6560, 5986, 4349, 2281, 1392, 2540, 4634
6131, 6570, 6418, 6314, 6479, 6537, 5823, 4058, 2050, 1436,
2815,
The 13th value 6314 is the value after 12 revolution which
is almost identical as the first value because at that instance
Jupiter has made one revolution.
What this means is that the average distance of the Earth
is constant.
For c = 300000 the 13th value is equal to 152096330.
This means that the average distance of the Earth increases
per year with a value equal to 16/12.
For a speed of gravity equal to 300*c that increase can
be neglected.
For the maximum distance of Jupiter we get the following values
with c=0:
816038505, 8349,7985, 7668, 7634, 7907, 8284, 8498, 8404,
8065, 7716, 7612, 7834
The 7th value is almost identical as the first one.
The 11 th value is after 118 revolutions of the Earth.
The 13 th value is after 142 revolutions of the Earth.
What the values show is that over a long period the
distance of Jupiter is constant but over a short distance not.
For cg = 300000 the first value is 810 higher,
the second 2 times 810, the third value 3 times 810 etc.
For cg = 300*c those values are not 810 but 810/300
What that means that it is very difficult based on actual observations if the distance of jupiter is truely constant or slightly in creases (over a period of 100 years) with a value of 810/300/12 per year.
I hope that there is someone else who also tries to perform the same simulation, based on Newton's Law and speed of gravity using the same model as I did.
My next chalenge is to perform the same simulation in 3D.
Nicolaas Vroom
Nicolaas Vroom
> 
"Steve Carlip" 
>> 
Nicolaas Vroom 
>> 
To a very good approximation, for nearly circular orbits the radius
at time t will satisfy
r^2  (r_0)^2 = (4GM/c_g)(tt_0) where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years. 
>>  Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy. 
>  I have studied the book from a library but still I have a couple of unsettled questions. 
>  In the above equation the stability is a function of the mass of the Sun. 
> 
The reason IMO comes from the equation at page 350
that the earth's energie increases at a function of
v * theta = v * v / c.
IMO that is wrong. IMO the sun's energie is a function of angle v / c i.e. v+/c with v+ being the speed of the earth. IMO the earth's energie is a function of angle v0/c with v0 being the speed of the sun. In total earth's energie is a function of v+*v0/c (which is a factor M+/M0 smaller) 
Ah... You're partly right. There is an additional ambiguity here in what one means by "Newtonian gravity with a finite propagation speed."
The most common model assumes, explicitly or implicitly, that something is traveling between the Earth and the Sun at a speed c_g, and then imparting a force in the direction of its motion. In that case, the relevant speed is the Earth's speed, and the tangential acceleration of the Earth is proportional to v+/c_g. (Think of the usual analogy of "walking in the rain"  it's your speed that determines the angle the rain hits you.) This is the assumption of, for example, Van Flandern, and is the starting point of Lightman et al.
But one could divorce oneself from this picture, and simply postulate that the Sun's gravitational force at time t points to the position of the Sun at time t  r/c_g. In that case, you would be right in saying the tangential acceleration of the Earth would be proportional to v0/c_g. This would give you your factor of M+/M0.
This still won't agree with observation, though. Probably the most clearcut case is the Moon (this is the example Laplace looked at). You should check my arithmetic, but even with a tangential acceleration proportional to the Moon's velocity, I get a change of about 2500 m/yr for gravity propagating at c. Thanks to Lunar laser ranging, we've known the position of the Moon to an accuracy of a centimeter or so for more than 30 years. An anomaly that size  or even one of 8 m/yr, for c_g = 300c  would be impossible to miss. If you assume, generously, that a steady increase of 1 cm/yr could have been missed for 35 years, you need c_g to be at least 250,000c.
Steve Carlip
"Steve Carlip"
> 
Nicolaas Vroom 
> > 
"Steve Carlip" 
> >> 
Nicolaas Vroom 
> 
> >> 
To a very good approximation, for nearly circular orbits the radius at time t will satisfy r^2  (r_0)^2 = (4GM/c_g)(tt_0) where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years. 
> 
> >> 
Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy. 
> 
> > 
I have studied the book from a library but still I have a couple of unsettled questions. 
> 
> > 
In the above equation the stability is a function of the mass of the Sun. 
>  [...] 
> > 
The reason IMO comes from the equation at page 350 that the earth's energie increases at a function of v * theta = v * v / c. IMO that is wrong. IMO the sun's energie is a function of angle v / c i.e. v+/c with v+ being the speed of the earth. IMO the earth's energie is a function of angle v0/c with v0 being the speed of the sun. In total earth's energie is a function of v+*v0/c (which is a factor M+/M0 smaller) 
> 
Ah... You're partly right. There is an additional ambiguity here in what one means by "Newtonian gravity with a finite propagation speed." The most common model assumes, explicitly or implicitly, that something is traveling between the Earth and the Sun at a speed c_g, and then imparting a force in the direction of its motion. In that case, the relevant speed is the Earth's speed, and the tangential acceleration of the Earth is proportional to v+/c_g. (Think of the usual analogy of "walking in the rain"  it's your speed that determines the angle the rain hits you.) This is the assumption of, for example, Van Flandern, and is the starting point of Lightman et al. But one could divorce oneself from this picture, and simply postulate that the Sun's gravitational force at time t points to the position of the Sun at time t  r/c_g. In that case, you would be right in saying the tangential acceleration of the Earth would be proportional to v0/c_g. This would give you your factor of M+/M0. This still won't agree with observation, though. Probably the most clearcut case is the Moon (this is the example Laplace looked at). You should check my arithmetic, but even with a tangential acceleration proportional to the Moon's velocity, I get a change of about 2500 m/yr for gravity propagating at c. Thanks to Lunar laser ranging, we've known the position of the Moon to an accuracy of a centimeter or so for more than 30 years. An anomaly that size  or even one of 8 m/yr, for c_g = 300c  would be impossible to miss. If you assume, generously, that a steady increase of 1 cm/yr could have been missed for 35 years, you need c_g to be at least 250,000c. Steve Carlip 
Accordingly to this url the increase is 3.8 cm / year. http://curious.astro.cornell.edu/question.php?number=124
For more info about McDonald Laser Ranging Station see: http://www.csr.utexas.edu/mlrs/
I have performed a simulation for an Earth Moon system with a speed of gravity cg equal to 3000km/sec The increases are in km per rev: 10 (0.32), 20(0.64) 30 (0.95) The values in brackets is the time in years.
That means there is an increase of 30 km / year. For cg = 300000 we get 30000/100 = 300 m / year For cg = 300*c we get 1 m /year = 100 cm / yr which is a factor 25 higher than observed.
For cg = 8000*c we get 3.8 cm / yr.
I doubt if that value of cg is small enough to simulate the precession of the perihelion of Mercury.
Nicolaas Vroom https://www.nicvroom.be
Dear Nicolaas Vroom:
"Nicolaas Vroom"
> 
"Steve Carlip" 
>> 
Nicolaas Vroom 
>> > 
"Steve Carlip" 
>> >> 
Nicolaas Vroom 
>> 
>> >> 
To a very good approximation, for nearly circular orbits the radius at time t will satisfy r^2  (r_0)^2 = (4GM/c_g)(tt_0) where r_0 is the radius at time t_0 , M is the mass of the Sun, and c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of the Sun and r the present radius of the Earth's orbit, and you can use this to compute tt_0, the time in the past that the Earth must have been at r_0. If c_g=c, this comes out to about 400 years. The time is directly proportional to c_g, so for c_g=300c, this becomes about 120,000 years. 
>> 
>> >> 
Again, the computation is fairly simple; see the Lightman reference I gave before. All you really have to do is to note that the effect of propagation delay in Newtonian gravity is to impart a tangential acceleration equal to v/c_g times the radial acceleration, and compute the change in energy. 
>> 
>> > 
I have studied the book from a library but still I have a couple of unsettled questions. 
>> 
>> > 
In the above equation the stability is a function of the mass of the Sun. 
>>  [...] 
>> > 
The reason IMO comes from the equation at page 350 that the earth's energie increases at a function of v * theta = v * v / c. IMO that is wrong. IMO the sun's energie is a function of angle v / c i.e. v+/c with v+ being the speed of the earth. IMO the earth's energie is a function of angle v0/c with v0 being the speed of the sun. In total earth's energie is a function of v+*v0/c (which is a factor M+/M0 smaller) 
>> 
Ah... You're partly right. There is an additional ambiguity here in what one means by "Newtonian gravity with a finite propagation speed." The most common model assumes, explicitly or implicitly, that something is traveling between the Earth and the Sun at a speed c_g, and then imparting a force in the direction of its motion. In that case, the relevant speed is the Earth's speed, and the tangential acceleration of the Earth is proportional to v+/c_g. (Think of the usual analogy of "walking in the rain"  it's your speed that determines the angle the rain hits you.) This is the assumption of, for example, Van Flandern, and is the starting point of Lightman et al. But one could divorce oneself from this picture, and simply postulate that the Sun's gravitational force at time t points to the position of the Sun at time t  r/c_g. In that case, you would be right in saying the tangential acceleration of the Earth would be proportional to v0/c_g. This would give you your factor of M+/M0. This still won't agree with observation, though. Probably the most clearcut case is the Moon (this is the example Laplace looked at). You should check my arithmetic, but even with a tangential acceleration proportional to the Moon's velocity, I get a change of about 2500 m/yr for gravity propagating at c. Thanks to Lunar laser ranging, we've known the position of the Moon to an accuracy of a centimeter or so for more than 30 years. An anomaly that size  or even one of 8 m/yr, for c_g = 300c  would be impossible to miss. If you assume, generously, that a steady increase of 1 cm/yr could have been missed for 35 years, you need c_g to be at least 250,000c. Steve Carlip 
> 
Accordingly to this url the increase is 3.8 cm / year. http://curious.astro.cornell.edu/question.php?number=124 For more info about McDonald Laser Ranging Station see: http://www.csr.utexas.edu/mlrs/ I have performed a simulation for an Earth Moon system with a speed of gravity cg equal to 3000km/sec The increases are in km per rev: 10 (0.32), 20(0.64) 30 (0.95) The values in brackets is the time in years.
That means there is an increase of 30 km / year. For cg = 8000*c we get 3.8 cm / yr. I doubt if that value of cg is small enough to simulate the precession of the perihelion of Mercury. 
What if about half that value (3.8 cm/year) is due to actual angular momentum transfer (via tides) between the Earth and the Moon? What will that do to cg?
David A. Smith
"Nicolaas Vroom"
> 
Accordingly to this url the increase is 3.8 cm / year. http://curious.astro.cornell.edu/question.php?number=124 For more info about McDonald Laser Ranging Station see: http://www.csr.utexas.edu/mlrs/ 
Is there any one who can give me more informartion
how based on observations:
first the observed distance d1 Earth Moon,
(Using McDonald Laser Ranging Station ?)
secondly the actual distance
and finally this increase in distance of 3.8 cm / year
are calculated ?
How many observatories are involved as part of
these observations / calculations ?
Is the observed distance d1 at t1 calculated as:
(t2  t0) * c / 2
or is a more complex algorithm used ?
With t0 = moment of emission of light from laser
With t2 = moment of receiving of light from laser
With t1 = (t0+t2) / 2
Nicolaas Vroom https://www.nicvroom.be/
"Nicolaas Vroom"
> 
Is there any one who can give me more informartion
how based on observations:
first the observed distance d1 Earth Moon,
(Using McDonald Laser Ranging Station ?)
secondly the actual distance
and finally this increase in distance of 3.8 cm / year
are calculated ?
How many observatories are involved as part of
these observations / calculations ?
Is the observed distance d1 at t1 calculated as:
(t2  t0) * c / 2 or is a more complex algorithm used ? With t0 = moment of emission of light from laser With t2 = moment of receiving of light from laser With t1 = (t0+t2) / 2 Nicolaas Vroom https://www.nicvroom.be/ 
In order to get some idea about the Earth Moon distance I have studied the program MOON supplied as part of the book "Astronomy with your personal computer" by Peter DuffettSmith. This program calculates the EarthMoon distance at a certain time and place, but with some modifications you can also use that to calculate the maximum distances for a certain period.
Following is the result for 1984: Column 1 = distance in km Column 24 = date Column 57 = time
405608.3849943659 7 1 1984 10 1 46.99 406430.5054588925 3 2 1984 16 25 17.56 406714.4260363030 1 3 1984 17 25 55.86 406350.3714502475 28 3 1984 19 57 7.43 405395.6816477149 25 4 1984 15 20 49.34 404468.1197446492 24 5 1984 0 30 48.53 404243.9745394128 19 6 1984 21 49 21.88 404852.1254151030 17 7 1984 18 47 23.51 405781.0821973901 14 8 1984 14 35 26.28 406359.6689450809 10 9 1984 18 47 0.53 406322.6275803399 7 10 1984 19 16 32.25 405690.9634341389 3 11 1984 23 25 22.60 404813.0296634666 1 12 1984 19 46 35.07 404338.7797164128 29 12 1984 18 10 7.13 404628.8869445491 26 1 1985 16 46 53.47What this tells you is that the maximum distance highly flexible over a period of one year but what is more important that it is very difficult to explain that suppose that all the distance value are measured values that the value of 29 12 1984 should not be 404338 km and 77972 cm but 77968 cm if this 3.8 cm increase is not taken into account.
I expect that one important object that influences
these distances is the planet Jupiter.
That means you have to remove the influence of
Jupiter out of these observations.
In order to do that you have to know the positions
of the Earth and Jupiter.
Accurately ?
And if you do not know them accurately I expect
that makes any claim about an increase of the
EarthMoon distance rather controversial.
(in the near future I will give more information about a simulation with the four objects: Sun, Earth, Moon and Jupiter)
With the same program I also calculated the maximum distances over a much longer period:
406703.9617799576 28 12 1902 21 25 47.68 406707.6002670664 8 1 1921 16 35 35.06 406672.4259527441 27 1 1930 20 2 30.48 406696.1499911513 8 2 1948 15 8 22.16 406687.4182780707 30 1 1957 18 51 42.11 406710.1184460392 18 2 1966 22 16 29.22 406669.1009584760 8 10 1980 19 32 19.80 406714.4260363030 1 3 1984 17 25 55.86 406672.3202105450 8 11 2007 18 18 29.29 406656.9669761913 1 4 2011 16 28 15.11 406694.5184556800 23 3 2020 19 44 22.79 406705.0145049790 30 11 2043 20 37 8.66 406671.6101895833 23 4 2047 18 37 35.73 406707.5434456542 11 12 2061 15 46 49.81 406675.2184529630 30 12 2070 19 7 47.45 406698.4571414733 10 1 2089 14 13 52.00 406684.7218982943 2 1 2098 18 1 32.62 406712.3839595916 22 1 2107 21 21 57.43 406717.0200725148 2 2 2125 16 31 6.44 406667.1142930359 11 10 2148 17 21 43.40 406683.4550249315 4 3 2152 15 17 20.52 406698.3902277122 24 2 2161 18 48 20.10 406696.3255528670 15 3 2170 22 23 7.87 406701.2674920212 2 11 2184 19 39 35.86 406699.5379324314 26 3 2188 17 30 37.53 406704.4021879547 15 11 2202 14 48 44.01
Accordingly to my simulation the distance R increases with 100 cm / year = 1 m / year. That means with 1 km per 1000 years.
Does the above information, again assuming that they are actual measured/observed values invalidates such a claim ?
Nicolaas Vroom https://www.nicvroom.be/
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