View Full Version : Invisible light
Would it be possible to cause enough gravity within a small enough region to bend the light enough, then equalizing the gravity enough where the light would circle around the gravity source in a continous orbit?
If so would this light be perceived as invisible?
That's the situation at the photospher of a black hole, where light may orbit the hole continuously.
Light is always invisible until it hits an observer, irrelevant of where it going straight or orbiting.
That was my fault, your right i wasn't thinking. But is it possible for us to control light enough where it would circle around our gravitational force
No. It's more a matter of controlling gravity, increasing the space-time curvature to black hole proportions.
curioucity
03-05-04, 04:58 AM
But wait.... if a wave is forced to circle a massive object constantly, is there any chance that it will lose energy and end up being absorbed?
>> is there any chance that it will lose energy and end up being absorbed?
Via Cromptom scattering it would be eventually red shifted out of detection range....
This is basically what a cosmological red shift is caused by.
But why exactly can't we see light? Is it just a property of light or is it the speed of light that's causing it to appear invisible? If it is the speed which is causing to to appear invisible then by circle it around a massive object wouldn't it appear visible over a duration of time?
curioucity
03-05-04, 11:56 AM
>> is there any chance that it will lose energy and end up being absorbed?
Via Cromptom scattering it would be eventually red shifted out of detection range....
This is basically what a cosmological red shift is caused by.
You mean, the wave is still there but with 0 frequency or what?
But why exactly can't we see light? Is it just a property of light or is it the speed of light that's causing it to appear invisible?
The only thing we ever see is light of appropriate wavelength and sufficient intensity that enters our eyes. If something is not visible, it means that no suitable light from that something is entering your eyes and being absorbed by your retina.
Now if you have a perpetually orbiting beam of light, then if you could see it, it means that the light isn't orbiting any more because it's being absorbed by your retina.
Once you see some light, then that particular light you saw is now gone.
But wait.... if a wave is forced to circle a massive object constantly, is there any chance that it will lose energy and end up being absorbed?
Yes, if it ever interacts with anything, it will be absorbed, reflected, or refracted, and probably not be circling any more.
You mean, the wave is still there but with 0 frequency or what?
He means the wave is still there but with frequency too low to detect.
Note that there are big question marks around Crompton Scattering... it's mainly endorsed by people who have a vested interest in maintaining that standard Cosmology is fundamentally misguided (Zarkov, for example. It's very much a minority point of view... not that that means it's necessarily wrong).
You mean, the wave is still there but with 0 frequency or what?
Yes the wavelength becomes so long that it barely causes a ripple in the ether of magnetism... or its energy becomes so low that our detectors can not detect any disturbance any longer.
>> Crompton Scattering... it's mainly endorsed by people who have a vested interest in maintaining that standard Cosmology is fundamentally misguided
Cromptom scattering is an observed effect, no interpretation. If a theory takes no notice of observed results of experiments then it is doomed,
Pete, these are almost your sentiments. If the theory ignores such evidence I can certainly say that that theory is fundamentally misguided....
:)
blackholesun
03-06-04, 12:08 AM
Don't you guys mean Compton scattering? Try to get it right if it's suppose to help your theory.
Compton Scattering - The scattering of photons from charged particles is called Compton scattering after Arthur Compton who was the first to measure photon-electron scattering in 1922.
I have no idea what "Cromptom Scattering" is
:o
Oh dear...
I need to do a little more research before I post.
What I should have said is that the use of Compton scattering theory to produce a tired light model of Cosmology has big question marks around it.
See here for more info:
http://www.astro.ucla.edu/~wright/tiredlit.htm
http://www.astro.ucla.edu/~wright/kierein.html
To expand:
Via Comptom scattering it would be eventually red shifted out of detection range....
This is basically what a cosmological red shift is caused by.
Zarkov appears to be supporting a theory proposed by John Kierein (http://www.angelfire.com/az/BIGBANGisWRONG/index.html). Unfortunately, to make his model work Kierein adds some extra bits to Compton scattering which are not observed, as described here (http://www.astro.ucla.edu/~wright/kierein.html).
Sorry for the spelling error, spelling never was a good point
Here are the objections to Tired Light Theories, from
http://www.astro.ucla.edu/~wright/kierein.html
>> * increase the electron scattering cross-section by a wavelength dependent factor that is as large as 1 trillion for VHF radio waves,>>>
1. The cross sectional area is increased due to the electromagnetic potential gradient within very very large spinning systems
>> * imagine that Compton scattering can change the wavelength of radiation without changing its direction, even though the both the standard formula and Kierein's modified formula give zero wavelength change for zero scattering angle,>>>
2. The direction is changed and most times like light passing through glass comes out coherent again. Light passing through spinning electrodynamic systems, ring currents of positive ions, or fixed harmonic electrodynamic standing wave structures, would re-emerge coherent. For light to pass through a medium there are two scatterings, and they are both equal but opposite if the medium is symetrical/harmonic in shape.
>>* increase the density of ordinary matter in the Universe by a factor of 1 thousand,>>
3. The concept of the density of the central matter, or using mass as a criteria for their criticism (that is only what you detect is all there is) that rest mass is the only cause for light to lose energy, it only supplies an 'stationary' inertial field density so the objection is invalid.
Curent astronomy practices use orbital velocity to determine the rest mass of the orbitiing body, however the calculations for electromagnetic waves must be carried out using the potential of the dynamic field, which is effectively doing all the work. The potential of the field is the result of the field spin velocity and the inertial field density caused by rest mass,.
>> * and require that only 0.01% of this matter can be in the form of stars.>>
4. Certainly spinning fields are not in the form of stars, but these can be quite dense to light. As an example see gravitational lensing.
>> * increase the electron scattering cross-section by a wavelength dependent factor that is as large as 1 trillion for VHF radio waves,>>>
1. The cross sectional area is increased due to the electromagnetic potential gradient within very very large spinning systems
Do you have calculations that show how electromagnetic potential gradients change the compton cross section?
Compton scattering cross section is dependant upon the number of particles as given by the Klein-Nishina formula.
'Gravity' as you call it is dependant upon the electric charge and the density of the magnetic charge and the velocity of the field spin, that an orbiting body finds itself in. So around any geomagnetic body there is a concentration of charged ions and rings of Lorentz gas, each flowing in opposite directions... The gradient of the potential change defines the compton scattering cross section, because this defines the number of particles that can interact with EMR.
It is easy to see that newtonian and GR considerations do not account for the space plasma physics, so any traditional criticism of Red Shift and Compton scattering is totally misleading.
Compton scattering cross section is dependant upon the number of particles as given by the Klein-Nishina formula.
'Gravity' as you call it is dependant upon the electric charge and the density of the magnetic charge and the velocity of the field spin, that an orbiting body finds itself in. So around any geomagnetic body there is a concentration of charged ions and rings of Lorentz gas, each flowing in opposite directions... The gradient of the potential change defines the compton scattering cross section, because this defines the number of particles that can interact with EMR.
It is easy to see that newtonian and GR considerations do not account for the space plasma physics, so any traditional criticism of Red Shift and Compton scattering is totally misleading.
Do you have calculations that show how electromagnetic potential gradients change the compton cross section such that the shift in wavelength will be proportional to the wavelength?
>> Do you have calculations that show how electromagnetic potential gradients change the compton cross section
No
>> Do you have calculations that show how electromagnetic potential gradients change the compton cross section
No
Let me now understand something.
On one side, there is a theory that is based on very simple assumptions: that on large scale the universe is isotropic, and that the matter distribution is homogenous. From these assumptions we get quite simple cosmological models that predict the expansion of the universe. They predict the Hubble law and the cosmic background radiation. These predictions are backed up by observation.
On the other side there are assumptions that say that it is all wrong. That the redshift is a consequence of compton effect but with no calculation that match the observations. I remind you that the wavelength shift in the compton effect is independent of the wavelength of the incident photon, but only on the mass of the scaterer, so that in order to have redshifts that match the observation observation, the number of scatering should be linear with wavelength (the Klein-Nishina formula might also be wrong in order to match the observation using the compton effect as explanation for redshift).
Hi 1100f
>> Let me now understand something.
Both the assertions above are resaonable.
Now which is more feasable, the Big Bang and expansion, when astronomers can see back almost to the BB, (philosophically impossible, by the way ) and red shift due to this, .... in truth this should be the other way around -----> red shift ----> BB
OR
Spinning fields causing the red shift, due to some process that we may not fully understand, with a steady macro state of the Universe. We observe a red shift in light projected against the Earth's gravitational field
You realise the cosmological model all depends upon the interpretation of the red shift.
We see a cosmic background radiation at 2.7o K ... if this is left over from the BB, then why is that still measurable ? is the Universe a merry go around where light circles and never stops.
Facts are that in theory light is supposedly the only effect that is energy independant, ( energy is not lost by movement) and gravity is the only force that is unidirectional.... sorry but both these assertions are contrary to the logic of the rest of the Universe. Both are obviously poorly understood... not a good foundation to build theory upon.
I know what theories I pitch for !
>> On one side, there is a theory that is based on very simple assumptions:
Actually these theories are quite complex and invoke numerous conceptions of "new" physics...... thumbs down IMO, and in other researcher's opinions
Talk about ignoring observation, the BB and associated assertions are just dogma, conceived by deluded cosmologists....
You realise the cosmological model all depends upon the interpretation of the red shift.
We see a cosmic background radiation at 2.7o K ... if this is left over from the BB, then why is that still measurable ? is the Universe a merry go around where light circles and never stops.
No, All the accepted cosmological models depend on the assumption that the universe is isotropic and homogeneous. From these two assumptions you get the metric of the univerese. From these metrics you get the Hubble law.
>> No, All the accepted cosmological models depend on the assumption that the universe is isotropic and homogeneous. From these two assumptions you get the metric of the univerese. From these metrics you get the Hubble law.
Please expand on this point of view. The Big Bang model was coined to explain the red shift... I think you are mistaken.
:)
>> No, All the accepted cosmological models depend on the assumption that the universe is isotropic and homogeneous. From these two assumptions you get the metric of the univerese. From these metrics you get the Hubble law.
Please expand on this point of view. The Big Bang model was coined to explain the red shift... I think you are mistaken.
:)
Just take any introductory book on general relativity and/or cosmology, go to the Robertson-Walker metrics and Friedman models and read them.
Frieman model was done in the years 1922-1924. Hubble law was discovered in 1927. In the early 30's, it was shown that the Robertson-Walker metric was the only one which is consistant with space being isotropic and homogenous.
BTW, I strongly recommend as introductory book on GR the book "Gravitation and Spacetime" by Hans C Ohanian.
Thank you for your advice 1100f, but I have read relevant parts. I do not understand why you fail to acknowledge that the conceptual explanation to account for the red shift is integral to the modern model of the Universe. Hubble, BB, CMB and concepts such as isotrophic and homogeneous are all tied into the logic of the modern Universe... all based on red shift due to "expansion", which lead to the BB concept.
The "tired light" concept was investigated and was discarded in favour of the "expansion" concept... however the theory used to discount was flawed IMO, because the understanding of the nature of gravity was flawed.
A correct model of the Universe can not be formulated until the key element "gravity" is fully understood.
I have read several texts on SR, and I find that informative, basically similar to QM, just electrodynamics.
However GR is totally adhorrent for my rationality. I can not accept fudged metrics to account for poor understanding, nor can I allow time to exist in any real sense. I know too much to be gullible enough to accept the logic presented in GR........
SR basically is a compilation of other researcher's work, and not a good one at that.
Einstein is a fraud IMO and can offer nothing to the real world.... yep great for the SciFi freaks....
1100f, sorry we will remain on opposite sides of the fence...
:)
Thank you for your advice 1100f, but I have read relevant parts. I do not understand why you fail to acknowledge that the conceptual explanation to account for the red shift is integral to the modern model of the Universe. Hubble, BB, CMB and concepts such as isotrophic and homogeneous are all tied into the logic of the modern Universe...
I really do not understand why the concept of the universe being isotropic and homogenous is a wrong concept.
all based on red shift due to "expansion", which lead to the BB concept.
As I allready told you, the basic is that the universe is isotropic and homogeneous. BB and redshift are consequences of these assumptions
The "tired light" concept was investigated and was discarded in favour of the "expansion" concept...
As far as I understand, the tired light concept was discarded because it is wrong.
however the theory used to discount was flawed IMO, because the understanding of the nature of gravity was flawed.
IYO
A correct model of the Universe can not be formulated until the key element "gravity" is fully understood.
As I allready told you, go and learn GR, then maybe you will understand it.
I have read several texts on SR, and I find that informative, basically similar to QM, just electrodynamics.
Did you really understood what you read?
However GR is totally adhorrent for my rationality. I can not accept fudged metrics to account for poor understanding, nor can I allow time to exist in any real sense. I know too much to be gullible enough to accept the logic presented in GR........
I really don't understand you.
In another thread you proudly announce that you discovered that for all the planets of our solar system rv^2 = K. From the equations of motion this comes from the fact that inertial mass and gravitational mass are the same. This is called the principle of equivalence. This is the basis of GR.
You are just spitting into the well you drink of.
SR basically is a compilation of other researcher's work, and not a good one at that.
Yeah, we heard that. Einstein have stollen the SR from other researcher. But since he was the most famouse patent office clerk, he was acknowledged by all the physics community to be the father of SR.
Einstein is a fraud IMO and can offer nothing to the real world.... yep great for the SciFi freaks....
Correction: Just one thing or two (like SR, GR the explanation of the photoelectric effect, the laser)
1100f, sorry we will remain on opposite sides of the fence...
:)
At least, I am not sorry to be on the correct side of the fence.
>> At least, I am not sorry to be on the correct side of the fence.
Oh then it is all relative.....
IMPOSSIBLE I SAY !!!!!!
nothing is relative, the whole Universe has only preferred frames of reference, without worrying about other minor details.
GR is so poorly conceived it is basically laughable that 'physicists' have been paying homage to it for so long.
Definately shows that the state of human's minds are very degraded.
GR is so poorly conceived it is basically laughable that 'physicists' have been paying homage to it for so long.
I don't know what is so laughable about the principle of equivalence.
The only laugh that I had about is that a few days ago, I found that someone made the discovery that rv^2 of all the planets around the sun is constant.
>> I found that someone made the discovery that rv^2 of all the planets around the sun is constant.
LOL
and you may add all the moons, all the satellites around their centre of spin... in fact all cosmic orbiting bodies reflect Gcentral spin back on their source of spin
BUT only in a specified spin systems is rv^2 invarient.
Now this is OBSERVED, nothing made up, Gcentral spin is constant in any specific spin system. Each spin system has its own specific Gcentral spin.
AND YET, physicists/astronomers use a contrived/derived constant Gnewton, and then say we can calculate a derved/ contrived mass value.
That is observed value= made up value / inferred value
I laught at this madness.
in plain speak ......Gcentral spin = Gnewton / mass
The left hand side is measured and observed
The right hand side is total conjecture. In the objection you raised, you have show the profound lack of scientific method in current science.
Gcentral spin could have been calculated and employed for at least the last 100 years, and yet it is completely overlooked.
Now you tell me where are the scientific minds????????????????????????????????
(definately not on this planet)
Now you tell me where are the scientific minds????????????????????????????????
I am sure that someone that does want to study what has been done during the last 400 years, takes the 3rd Kepler's law and rephrase it as if it was something that he discovered is surely not a scientific mind.
Kepler's observations.... no law there... any explanations??????, as to why planets orbit is a closed harmonic motion...
You know all about it, don't you, read it in some text book somewhere???
Kepler's observations.... no law there... any explanations??????, as to why planets orbit is a closed harmonic motion...
You know all about it, don't you, read it in some text book somewhere???
It is called Kepler's law even though it is an observation.
The orbit of all the planet is given by solving the differential equation known as Newton's second law with the force given by Newton's force of gravitation. The orbit is closed because the gravitational force goes as 1/r^2.
Sorry, explains nothing.... just another observation.
:)
Sorry, explains nothing.... just another observation.
:)
No, the orbits of the planets are solutions of Newton's laws. If you have instead of 1/r^2, another force, then the orbit are not closed.
>> then the orbit are not closed.
Interesting, can that be proved?
>> then the orbit are not closed.
Interesting, can that be proved?
Yes, if you add for example a perturbation of the form 1/r^3 to the 1/r^2 force, then the bound states orbits are precessing ellipses.
All the planets, stars etc are precessing and their orbits are ellipses.
All the planets, stars etc are precessing and their orbits are ellipses.
It is well known, that the precession is due to perturbations from the other planets and from GR effects. If you take only the 1/r^2 force, then the solution for the bound state is an ellipse (not precessing).
>> If you take only the 1/r^2 force, then the solution for the bound state is an ellipse (not precessing).
I would have thought an inertial circle. But you are right, what you interpret from the observations all depend upon your model. Newton did a great analysis.
The new orbiting object is in a highly elliptical orbit contrary to expectations via conventional wisdom that it is not in a circular orbit...
afterall most of the mass is in the Sun itself.
>> Sedna's path is highly elliptic. It ranges from 76 astronomical units (AU) when it is closest to the Sun to 1,000 AU when it is farthest. One AU is the distance from Earth to the Sun.
http://www.space.com/scienceastronomy/sedna_earth_040316.html
Pluto has a reasonably eccentric orbit.
I would expect this to be the case, as an object gets further out, the orbit becomes more eccentric due to warped interplanetary magnetic field lines.
Nick Stenson
03-17-04, 08:08 PM
i think this comes back to... if a tree falls in the woods... does it make a sound??? noone knows... as holds true with the invisable light theroy...is light even light... is sound even sound... there must b an observer for everything or nothing at all exhists...
Think about it my friends...
The new orbiting object is in a highly elliptical orbit contrary to expectations via conventional wisdom
:rolleyes:
Where do you find your notions of 'conventional wisdom', Zarkov?
Halley: "What shape would the orbit of a planet around the Sun take if it was attracted by a force operating under an inverse-square law?"
Newton: "Elliptical"
if a tree falls in the woods... does it make a sound???
Why not just leave a few hundred tapes on "record" in the woods. Once a tree falls down we willl know if it makes a sound or not.
if a tree falls in the woods... does it make a sound???
If my wife doesn't hear me, am I still wrong? :D :D
>> Where do you find your notions of 'conventional wisdom', Zarkov?
Your logic re 1/r^2 is not sound, it is specific theory oriented.
It would appear that the closer objects get near to each other, the relationship of "attraction" is 1/r
Terrestrial gravity as per on Earth is 1/r^2, it is the result of a (1/r X 1/r^3)^.5 interaction.
Planetary "gravity" is more complex than terrestrial gravity.
The new solar object just found in the Solar System would be bound by 1/r^3
Spin gravity would predict this sequence of events.
Conventional explanations of gravity would predict that the new solar object should be in an inertial circular orbit....... however it has an exceptionally eccentric orbit.
>> Where do you find your notions of 'conventional wisdom', Zarkov?
Your logic re 1/r^2 is not sound, it is specific theory oriented.
It would appear that the closer objects get near to each other, the relationship of "attraction" is 1/r
Aren't the masses in Cavendish experiment close enough?
Anyway, in this experiment the force between the masses goes as 1/r^2.
Terrestrial gravity as per on Earth is 1/r^2, it is the result of a (1/r X 1/r^3)^.5 interaction.
Yes gravity as per on earth is 1/r^2.
Planetary "gravity" is more complex than terrestrial gravity.
Planetary gravity around the sun also goes as 1/r^2.
As a result of this, if you take rv^2 of each planet, it is a constant.
The new solar object just found in the Solar System would be bound by 1/r^3
Spin gravity would predict this sequence of events.
Conventional explanations of gravity would predict that the new solar object should be in an inertial circular orbit....... however it has an exceptionally eccentric orbit.
Wrong. Conventionnal gravity predicts that the orbit of each body around the sun would be in an elliptic orbit. The excentricity of the orbit depends on the angular momentum of the planet.
>> Planetary gravity around the sun also goes as 1/r^2.
As a result of this, if you take rv^2 of each planet, it is a constant.
Can you show that relationship is direct mathematically, I am interested.
Hi Zarkov,
Do you want it for circular or elliptical orbits?
Circular orbits is trivial.
Acceleration of an object in uniform circular motion = v<sup>2</sup> / r
Acceleration due to gravity = k/r<sup>2</sup>
Equating the two:
v<sup>2</sup>/r = k/r<sup>2</sup>
r.v<sup>2</sup> = k
You might also be interested inNewton's derivations of Kepler's laws (http://www.shef.ac.uk/physics/people/vdhillon/teaching/phy105/phy105_derivation.html)
Your logic re 1/r<sup>2</sup> is not sound, it is specific theory oriented.
In the context, elliptical orbits were observed by Kepler. Newton then determined that an inverse square law would produce elliptical orbits.
Here's an exercise: Determine what sort of orbit would a 1/r<sup>3</sup> law produce. Understandable analysis is required.
It would appear that the closer objects get near to each other, the relationship of "attraction" is 1/r
Do you have any observational records to support that assertion?
Conventional explanations of gravity would predict that the new solar object should be in an inertial circular orbit.
Your understanding of conventional explanations of gravity appears to be flawed.
Can you explain why you think that conventional explanations of gravity would predict a circular orbit?
>> Can you explain why you think that conventional explanations of gravity would predict a circular orbit?
Hi Pete,
yep, as per the most recent discussion, the orbit of a body under 1/r^2 is an ellipse.... no really a circle.... as this would be the result of prolonged precession of the orbit if it was given enough time.... and we have time to spare.
So an object in Sun orbit at the average distance of Sedna would certainly be in an almost circular orbit
However this is not as per observation....a real problem for astronomers to explain....
(Sedna's path is highly elliptic. It ranges from 76 astronomical units (AU) when it is closest to the Sun to 1,000 AU when it is farthest.)
I am sure the progression of planets/moon is produced by core ejection (fission model) and as long as the Sun pumps out separated static electric charge each orbiting body will slowly move into a higher orbital path.
Since Sedna is so far out, its electric charge would be degraded, much as is the case with Pluto,,,,,, so basically the force controlling the orbit is a magnetic 1/r^3 resulting in a highly elliptic orbit.
>> Acceleration of an object in uniform circular motion = v2 / r
Acceleration due to gravity = k/r2
Equating the two:
v2/r = k/r2
r.v2 = k >>>
LOL, planetary orbits are just a little more complicated in their mechanism :)
*****
Thanks Pete for the Newton link.... good stuff, I will see if I can apply it to Mercury.... but the logic presented fails to acknowledge precessional advance tending towards a circular orbit... all a little too simplistic, but I shall investigate ....
added later..
From that link Pete
>> Newton generalized Kepler's first law. He deduced that when one body moves under the gravitational influence of another,
I am not sure what is meant by this statement but in my view:-
"one body (the orbiting body, that is not causing separation of electric charge, eg the Earth)....
...moves under the gravitational influence of another ( the other being in the case of our Solar System, is the source of electric charge separation, eg the Sun)"
This problem is not a 'two body' problem IMO, the situation is not symetrical (relative)....
On body here is passive.. if the Sun ceased charge separation, the Earth would be drawn magnetically into the Sun... The Earth is only kept where it is because both the Sun and the Earth have a common and like charge.
The Earth is moved electrically... "planetary gravity" is not the same as "terrestrial gravity". Terrestrial gravity is due to the reaction (which is induced by a non-inertial acceleration of a mass in the spinning energy field of the planet).
This is a reaction to the created crossed (orthogonal) concentric electric and magnetic vectors set up around the planet, and this spin field is manifest by the differential velocities of the Sun's field spin and the size of the native field of the planet, which spins the field of the planet and moves it into orbit.
With no external preturbations the orbit will be circular.
Since the planet is accumulating a greater and greater electric charge, the planet is moving away from the sun, and slowing, so the field spin is slowing.. creating a centripetal resultant on any non inertial mass in the planet's field spin, which is directed to the centre of field spin of the planet.
Planetary gravity is due to the primary forces that set up the criteria for terrestrial gravity. These forces are raw electic drivers generating moving magnetic fields (the planet) in the static magnetic field of the Sun. This creates in a sense a magnetic electric jelly of standing waves.
Both forms of 'gravity' are subject to the same constants, however planetary gravity to be a 1/r^2 relation does require a set range in distance from the source of electric charge separation and so can become purely magnetic, given enough distance from the Sun.
Sedna demonstrates this aspect very well.
IMO drawn from the analysis of spin gravity theory.
from that link
>> due to a central force such as gravitation, the torque vanishes.
Newton never could explain why the planets are moving in the first place if gravity just pulls, first assumption. No torque no rotation in the first instance. But all objects spin in the Universe,,, ah but oh, thta is an observation that just doesn't fit. LOL
Cosmology overcomes this problem by adding a second assumption that of an accretion disk....all travelling in the same direction.
and the assumptions continue..... all special contrived logical extensions of the initial premise of "attraction" as the mechanism of action.
An the tower of bable is so built....until Sedna, that is a real black eye.
That link, Pete has only incestuous logic to offer, sophist logic... but it is a good exercise in one eyed logic.
There are more rational ways to treat the problem that only assume a limited and incestuous set of parameters and a fixed logic... that transends time.
Surely one is preferrential to the sophist other. As I said in his time Newton performed a miracle, however understanding has grown since then. Clearly the planets do not orbit in a relative manner either.
>> This tells us that the more massive body orbits closer to the centre of mass than the less massive body.
This is again negated by observational evidence.
Both the positions of Uranus and Neptune upset this conclusion.
>> Can you explain why you think that conventional explanations of gravity would predict a circular orbit?
Hi Pete,
yep, as per the most recent discussion, the orbit of a body under 1/r^2 is an ellipse.... no really a circle.... as this would be the result of prolonged precession of the orbit if it was given enough time.... and we have time to spare.
Conventional gravity predicts that the orbit is an ellipse (no pecession), how could this not existing precession, change the orbit from an ellipse to a circle?
So an object in Sun orbit at the average distance of Sedna would certainly be in an almost circular orbit
However this is not as per observation....a real problem for astronomers to explain....
(Sedna's path is highly elliptic. It ranges from 76 astronomical units (AU) when it is closest to the Sun to 1,000 AU when it is farthest.)
You have a lot of comets with an orbit as large as sedna's orbit which are far from being circular.
I am sure the progression of planets/moon is produced by core ejection (fission model) and as long as the Sun pumps out separated static electric charge each orbiting body will slowly move into a higher orbital path.
Since Sedna is so far out, its electric charge would be degraded, much as is the case with Pluto,,,,,, so basically the force controlling the orbit is a magnetic 1/r^3 resulting in a highly elliptic orbit.
How do you know that the 1/r^3 force gives an elliptical orbit with no preceesion (this is the meaning of elliptical). One solution at least of a central force that goes like 1/r^3 is a circular orbit, so how can you say that the highly elliptic orbit is due to a 1/r^3 force?
>> Acceleration of an object in uniform circular motion = v2 / r
Acceleration due to gravity = k/r2
Equating the two:
v2/r = k/r2
r.v2 = k >>>
LOL, planetary orbits are just a little more complicated in their mechanism :)
Bigger LOL. We all know that planetary motion is a little bit more complicated. For example when it is said that rv^2 of the planet is a constant, since r is not constant and v is not constant, you cannot say that rv^2 is constant all the time for the planets.
Thanks Pete for the Newton link.... good stuff, I will see if I can apply it to Mercury.... but the logic presented fails to acknowledge precessional advance tending towards a circular orbit... all a little too simplistic, but I shall investigate ....
added later..
From that link Pete
>> Newton generalized Kepler's first law. He deduced that when one body moves under the gravitational influence of another,
I am not sure what is meant by this statement but in my view:-
Instead of looking at internet sites who try to explain physics, why don't you just take a book in classical physics and study classical mechanics. Learn Calculus and differential equations. Learn what are Newton's law and how they are used to solve problems in classical mechanics.
"one body (the orbiting body, that is not causing separation of electric charge, eg the Earth)....
...moves under the gravitational influence of another ( the other being in the case of our Solar System, is the source of electric charge separation, eg the Sun)"
This problem is not a 'two body' problem IMO, the situation is not symetrical (relative)....
On body here is passive.. if the Sun ceased charge separation, the Earth would be drawn magnetically into the Sun... The Earth is only kept where it is because both the Sun and the Earth have a common and like charge.
The Earth is moved electrically... "planetary gravity" is not the same as "terrestrial gravity". Terrestrial gravity is due to the reaction (which is induced by a non-inertial acceleration of a mass in the spinning energy field of the planet).
This is a reaction to the created crossed (orthogonal) concentric electric and magnetic vectors set up around the planet, and this spin field is manifest by the differential velocities of the Sun's field spin and the size of the native field of the planet, which spins the field of the planet and moves it into orbit.
With no external preturbations the orbit will be circular.
Since the planet is accumulating a greater and greater electric charge, the planet is moving away from the sun, and slowing, so the field spin is slowing.. creating a centripetal resultant on any non inertial mass in the planet's field spin, which is directed to the centre of field spin of the planet.
Planetary gravity is due to the primary forces that set up the criteria for terrestrial gravity. These forces are raw electic drivers generating moving magnetic fields (the planet) in the static magnetic field of the Sun. This creates in a sense a magnetic electric jelly of standing waves.
Both forms of 'gravity' are subject to the same constants, however planetary gravity to be a 1/r^2 relation does require a set range in distance from the source of electric charge separation and so can become purely magnetic, given enough distance from the Sun.
Sedna demonstrates this aspect very well.
IMO drawn from the analysis of spin gravity theory.
You are saying long sentences by using words that you just don't understand the meaning.
from that link
>> due to a central force such as gravitation, the torque vanishes.
As I told you, take any book in classical mechanics and learn mechanics. Maybe then you will understand why a central force gives conservation of angular momentum.
>> For example when it is said that rv^2 of the planet is a constant, since r is not constant and v is not constant, you cannot say that rv^2 is constant all the time for the planets.
That is not correct
rv^2 for any system is a constant, Gcentral spin is unique to the spinning system.
All the planets give the same Gcentral spin, that of the Sun's system.
The Earth and satellites give the same Gcentral spin, that of the Earth's system.
Gcentral spin or rv^2 is invarient for inertial objects in a specified spinning system.
Observational evidence.
>> For example when it is said that rv^2 of the planet is a constant, since r is not constant and v is not constant, you cannot say that rv^2 is constant all the time for the planets.
That is not correct
rv^2 for any system is a constant, Gcentral spin is unique to the spinning system.
All the planets give the same Gcentral spin, that of the Sun's system.
The Earth and satellites give the same Gcentral spin, that of the Earth's system.
Gcentral spin or rv^2 is invarient for inertial objects in a specified spinning system.
Observational evidence.Show me where did you take the data that rv^2 is allways constant
>> This tells us that the more massive body orbits closer to the centre of mass than the less massive body.
This is again negated by observational evidence.
Both the positions of Uranus and Neptune upset this conclusion.
You didn't read that link very well, did you? The "more massive body" referred to in that sentence is the Sun in this context.
I never cease to be amazed at your astounding ability to maintain your confidence in your ability as a scientist, in the face of such gaffs.
Read it again... don't forget it's talking about the two body case.
(Yes, planetary motion is more complicated than that, because it's a many-body problem. That's why the orbits aren't perfect ellipses - they precess due to the gravitational influence of other planets.)
Let's be perfectly clear here Zarkov - is it a predicition of your ESGT that for all planets, rv<sup>2</sup> is a constant not just for the average figures, but for the particular radius and velocity at all points on all orbits?
To be specific, is it your prediction that rv<sup>2</sup> for Mars is the same at perihelion as it is at aphelion?
It is fair to warn you that the conventional wisdom says differently - specifically, that v<sup>2</sup> - 2.65 x 10<sup>5</sup>/r is constant for a planet, at all points in its orbit. (go on, ask me!)
>> v2 - 2.65 x 10^5/r = constant (go on, ask me!)
I ask, LOL :) That is a great concocted formula, true or not, Heeheee
All the natural orbiting objects in a spinning system behave as in a vortex.
>> Let's be perfectly clear here Zarkov - is it a predicition of your ESGT that for all planets, rv2 is a constant not just for the average figures,
For the inertial situation eg a circular orbit, the above is true.
>> but for the particular radius and velocity at all points on all orbits?
At perihelion the electric field is intensified overpowering the magnetic field... at aphelion the opposite occurs. This varying harmonic situation causes a precessional advance of the orbit (perihelion).
For the inertial situation eg a circular orbit, the above is true.
LOL, planetary orbits are just a little more complicated in their mechanism :)
>> For example when it is said that rv<sup>2</sup> of the planet is a constant, since r is not constant and v is not constant, you cannot say that rv<sup>2</sup> is constant all the time for the planets.
That is not correct
rv<sup>2</sup> for any system is a constant, Gcentral spin is unique to the spinning system.
All the planets give the same Gcentral spin, that of the Sun's system.
To be specific, is it your prediction that rv<sup>2</sup> for Mars (and other planets) is the same at perihelion as it is at aphelion?
Silverback
03-20-04, 04:01 AM
*sigh*
Yet one more thread hijacked by Zarkov to push his spin-gravity theory. :rolleyes:
Ok let the thread die...
Talk to yourselves
BYE
Ok let the thread die...
Talk to yourselves
BYE
You still didn't answer my question.
Where did you get the data showing tha rv^2 is a constant all the time?
BTW, do you agree with Newton's second law (F = ma)?
Because if you agree with Newton's second law, no matter what is the force that attract the planets to the sun, if the force is a central force, according to Newton's second law, angular momentum is conserved, and for elliptical orbit, you cannot have rv^2 = const all the time.
For the planets, since their orbit is an ellipse, rv^2 is not a constant.
What is a constant? It is r^3/T^2 which is a constant, where r is the axis of the ellipse. in the case where the ellipse degenerates to a circle, you can write that rv^2 = const.
For the planets, since the orbits are close to a circle, the radius is almost constant and it is <r><v>^2 which is a constant (or something like that) where <> stands for average.
Man! What are you guys? Scientists? Professors? Extremely educated hobbyists? This seems to be an 'exclusive club' chat. But interesting just the same. :)
>> You still didn't answer my question.
I would post the relevant sections of my book.... but there is only a one way discussion going on here.... I am telling you the way it is, and you guys can only either throw shit or established out of date logic.
OR tripe such as
>> *sigh*
Yet one more thread hijacked by Zarkov to push his spin-gravity theory. >>>
To hell with any theory, I am talking observations.... and it so happens thet ESGT is an analysis of observations.....
One way trip to hell trying to get a coherent discussion going on.
I am not posting in this thread anymore.... see new thread
"Cosmological no topic"
>> To hell with any theory, I am talking observations.... and it so happens thet ESGT is an analysis of observations.....
Where did you get the observation that rv^2 is constant all the time?
(third time that I am asking)
vBulletin® v3.8.1, Copyright ©2000-2009, Jelsoft Enterprises Ltd.