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Author Topic: The implications of zero gravitational acceleration  (Read 1484 times)

Offline jeffreyH

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For GM/r^2 = 0 we must assume that the mass in the equation equals zero. We cannot have a well defined value for r=0. However we can have a well defined value for m=0 AND r=0 so that at the centre of gravity of a massive object the mass term must be at zero as well as the radius. This shows that only in a cavity at the centre of gravity can the equation be applied where the centre of mass is evacuated. Since zero mass equals zero gravity this is obvious since all other terms contributing to the gravitational field cancel at the COG.

What equation we write for a photon passing through the COG?


 

Offline GoC

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Re: The implications of zero gravitational acceleration
« Reply #1 on: 04/01/2016 15:13:29 »
Light created in the center of mass is red shifted so time and distance due to dilation in the COA changes. The photon distance increases along with the electron distance being equal. The result is the speed of light is measured to be the same at every geodesic position. your measuring stick increases length due to dilation. Math is the same. What the math means is not the same except for the ratio of distance to how we measure time.

Basically time slows down to make up for the extra distance dilation causes in space.
 

Offline jeffreyH

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Re: The implications of zero gravitational acceleration
« Reply #2 on: 04/01/2016 18:18:48 »
Unfortunately, however interesting that may or may not be, it isn't an equation.
 

Offline Thebox

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Re: The implications of zero gravitational acceleration
« Reply #3 on: 04/01/2016 23:03:37 »
Unfortunately, however interesting that may or may not be, it isn't an equation.

zero
 

Offline jeffreyH

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Re: The implications of zero gravitational acceleration
« Reply #4 on: 04/01/2016 23:38:26 »
Unfortunately, however interesting that may or may not be, it isn't an equation.

zero

All the photons energy is kinetic. Therefore E is all kinetic energy. Since E = hf and all the gravitational forces cancel there is no shift applied to the frequency f at the centre of gravity. This indicates that the frequency is highest at this point and thus has a maximum energy. The shift of the wavelength is to the red end of the spectrum in any direction away from the centre of gravity. If we apply this to a particle with mass we cannot say the same as the relationship between wavelength and energy is now velocity dependent. This does imply that the highest energy for any particular photon is at the centre of gravity. Dependent upon the initial energy of the photon when emitted. The centre of gravity of large bodies can therefore be a source of energy if photons are passed through it.
 

Offline GoC

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Re: The implications of zero gravitational acceleration
« Reply #5 on: 05/01/2016 03:12:42 »
All the photons energy is kinetic. Therefore E is all kinetic energy. Since E = hf and all the gravitational forces cancel there is no shift applied to the frequency f at the centre of gravity. This indicates that the frequency is highest at this point and thus has a maximum energy. The shift of the wavelength is to the red end of the spectrum in any direction away from the centre of gravity. If we apply this to a particle with mass we cannot say the same as the relationship between wavelength and energy is now velocity dependent. This does imply that the highest energy for any particular photon is at the centre of gravity. Dependent upon the initial energy of the photon when emitted. The centre of gravity of large bodies can therefore be a source of energy if photons are passed through it

Interesting point if the relationship of energy is greatest in the center of mass. What we generally consider energy is a difference in potential so gravity is something we relate to energy. light created in the center of mass is red shifted due to dilation when detected in a less dilated energy position.

its your perspective that challenges your understanding. I could be incorrect and energy is greatest at the COA. But if you follow equivalence between GR and SR You start at 32 feet/s/s and linear reduce acceleration for 4000 miles and become inertial. The speed you obtain would cause a clock to tick at the same rate as the COA of the Earth. That is equivalence of SR and GR red shift. A lower clock speed than in space away from as much gravitational influence. All equations can confuse the maze to understanding the cause of relativity.
 

Offline Space Flow

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Re: The implications of zero gravitational acceleration
« Reply #6 on: 05/01/2016 05:36:15 »
This does imply that the highest energy for any particular photon is at the centre of gravity. Dependent upon the initial energy of the photon when emitted. The centre of gravity of large bodies can therefore be a source of energy if photons are passed through it.
That doesn't quite compute to me. However it is something easily tested. The Centre of Mass of the Pluto/Charon binary is outside the radius of both bodies.If amplification of EMR is possible by photons passing through the centre of Mass then this should be detectable.
 

Offline jeffreyH

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Re: The implications of zero gravitational acceleration
« Reply #7 on: 05/01/2016 13:19:45 »
This does imply that the highest energy for any particular photon is at the centre of gravity. Dependent upon the initial energy of the photon when emitted. The centre of gravity of large bodies can therefore be a source of energy if photons are passed through it.
That doesn't quite compute to me. However it is something easily tested. The Centre of Mass of the Pluto/Charon binary is outside the radius of both bodies.If amplification of EMR is possible by photons passing through the centre of Mass then this should be detectable.

Lookup UV spectrum of the pluto-charon system.
 

Offline GoC

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Re: The implications of zero gravitational acceleration
« Reply #8 on: 05/01/2016 18:32:27 »
That doesn't quite compute to me. However it is something easily tested. The Centre of Mass of the Pluto/Charon binary is outside the radius of both bodies.If amplification of EMR is possible by photons passing through the centre of Mass then this should be detectable

That is not the center of mass. That is the focal balanced attraction point between the two masses. The boundary point of equal attraction to each COA. That position is also related to the mass of each body. Its the equal dilation point between the two bodies.
 

Offline Space Flow

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Re: The implications of zero gravitational acceleration
« Reply #9 on: 05/01/2016 21:23:52 »
That is not the center of mass. That is the focal balanced attraction point between the two masses. The boundary point of equal attraction to each COA. That position is also related to the mass of each body. Its the equal dilation point between the two bodies.
GoC I have to strongly disagree. That is the centre of mass of the two body system in anyway we can measure it from outside both bodies.
If there was such a thing as a Gravitational force or field, all the observations we are capable of making from outside the orbit of those two bodies would indicate that it is emanating from this empty point in space.
If Gravity was propagating outward by the mediation of Gravitons, then they would also be emanating from this empty point in space.
If we tried to orbit the system we like the existing 4 moons would also have to orbit this empty point in space.
How many more examples would you like to prove that that empty point in space is the COG?

Lookup UV spectrum of the pluto-charon system.
Jeffrey, you have obviously found something in support of your claim.
I'm about to leave for an operation on my back later but will certainly look into it on my return.
 

Offline jeffreyH

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Re: The implications of zero gravitational acceleration
« Reply #10 on: 05/01/2016 23:39:05 »
Quote from: jeffreyH on Today at 13:19:45Lookup UV spectrum of the pluto-charon system.Jeffrey, you have obviously found something in support of your claim.I'm about to leave for an operation on my back later but will certainly look into it on my return.

No it was something of interest I found on the binary planet system. A COG outside an object won't do as the metric is entirely different and the strength of gravity is too low. It is like comparing apples and pears and saying they are identical.
 

Offline jeffreyH

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Re: The implications of zero gravitational acceleration
« Reply #11 on: 05/01/2016 23:47:00 »
Imagine a mass that is a perfect sphere with a perfectly spherical cavity at the centre. If you slice it into two hemispheres then the slice can be said to be coincident with the x/y plane with a z-axis perpendicular to that plane and with the COG at the origin. If motion is constrained to move along the z-axis only, then the only point of equilibrium is at the COG. If the point remains at the centre of gravity the sphere is rotationally symmetric. You cannot say this about a centre of gravity that lies outside the body of a system. So the Pluto-Charon system is a very bad choice of laboratory.
 

Offline Space Flow

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Re: The implications of zero gravitational acceleration
« Reply #12 on: 06/01/2016 03:28:36 »
Imagine a mass that is a perfect sphere with a perfectly spherical cavity at the centre. If you slice it into two hemispheres then the slice can be said to be coincident with the x/y plane with a z-axis perpendicular to that plane and with the COG at the origin. If motion is constrained to move along the z-axis only, then the only point of equilibrium is at the COG. If the point remains at the centre of gravity the sphere is rotationally symmetric. You cannot say this about a centre of gravity that lies outside the body of a system. So the Pluto-Charon system is a very bad choice of laboratory.
No worries.
I give up. You suggested a possibility for an Electromagnetic amplification effect at the "centre of gravity".
All I did  was point you towards an easily accessible "centre of gravity" where the hypothesis could be tested.
You are now putting so many restrictions on what constitutes a "centre of gravity", that you are effectually putting the original claim in the totally untestable bracket. There is a very clear scientific definition of what constitutes a "centre of gravity".
When the goal posts start to move to suit one's views, it is a signal for me to leave a thread.
Enjoy...
 

Offline Thebox

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Re: The implications of zero gravitational acceleration
« Reply #13 on: 07/01/2016 10:44:07 »
Unfortunately, however interesting that may or may not be, it isn't an equation.

zero

All the photons energy is kinetic. Therefore E is all kinetic energy. Since E = hf and all the gravitational forces cancel there is no shift applied to the frequency f at the centre of gravity. This indicates that the frequency is highest at this point and thus has a maximum energy. The shift of the wavelength is to the red end of the spectrum in any direction away from the centre of gravity. If we apply this to a particle with mass we cannot say the same as the relationship between wavelength and energy is now velocity dependent. This does imply that the highest energy for any particular photon is at the centre of gravity. Dependent upon the initial energy of the photon when emitted. The centre of gravity of large bodies can therefore be a source of energy if photons are passed through it.
It won't let me post the equation, you asked about light passing through the COG, you did not mention any substance that would cause the propagation to compress.
Imagine the center of gravity having no physical substance, i.e rock ,stone, particle.

Imagine a void , at the center of the void is the COG, space itself(dark energy) applying a centripetal pressure on itself, at this singular point of all of space, the pressure creates the first kE which in turn creates work that forces the dark energy to gain motion. 

The pressure gains magnitude by the dark energy (negative) being attracted to the first kE positive, then all of the infinite void (dark energy) rushes in to create a big bang.

I tried a while ago to explain this






« Last Edit: 07/01/2016 10:52:20 by Thebox »
 

Offline jeffreyH

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Re: The implications of zero gravitational acceleration
« Reply #14 on: 07/01/2016 11:29:17 »
Imagine a mass that is a perfect sphere with a perfectly spherical cavity at the centre. If you slice it into two hemispheres then the slice can be said to be coincident with the x/y plane with a z-axis perpendicular to that plane and with the COG at the origin. If motion is constrained to move along the z-axis only, then the only point of equilibrium is at the COG. If the point remains at the centre of gravity the sphere is rotationally symmetric. You cannot say this about a centre of gravity that lies outside the body of a system. So the Pluto-Charon system is a very bad choice of laboratory.
No worries.
I give up. You suggested a possibility for an Electromagnetic amplification effect at the "centre of gravity".
All I did  was point you towards an easily accessible "centre of gravity" where the hypothesis could be tested.
You are now putting so many restrictions on what constitutes a "centre of gravity", that you are effectually putting the original claim in the totally untestable bracket. There is a very clear scientific definition of what constitutes a "centre of gravity".
When the goal posts start to move to suit one's views, it is a signal for me to leave a thread.
Enjoy...

The metrics are not equivalent and therefore the results will not be equivalent. If you don't see that then you need to learn more before attempting to rewrite physics. In my initial post I did mention a cavity at the centre of gravity. Maybe you missed that point.
 

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Re: The implications of zero gravitational acceleration
« Reply #14 on: 07/01/2016 11:29:17 »

 

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