Gravity at the speed of light - Help needed.

[Robittybob1]

I'm just having a mental block how to understand this real simple diagram.

http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GravitationalWaves.pdf

 

So it looks like an oblique view of a binary orbiting system.

There is an observer O at r1 and r2 from the bodies.

 

Gravity propagates from the bodies to the observer at the speed of light c.

 

Since gravity takes a different length of time to reach O does he feel the pull of gravity as is drawn? I'm thinking the pull of gravity from the shorter arm will arrive earlier than the longer side.

 

OK that is as far as I got, but what position are the bodies at so that the gravity arrives at O at the same time?

I can think r1 > r2 so at any time t0 the propagation will be proportional to distance, "tpropagate = d/c"

Can you tell me what O feels at any moment in time for I thinking he can't experience 2 gravity "pulls" arriving at different times

So what exactly does he feel?

 

I just can't work out why I'm struggling with this one!

 

 

 

I'm sitting in my room surrounded by mass, so every mass is gravitationally attracted to me but and the time of propagation is all different but they all arrive at the same time. So if one object is moving I will be feeling the attraction to wherever it was but where is it when I feel it?

R1 is the distance to M1

R2 is the distance to M2

 

If R1 > R2 I will feel M2 first but I keep on feeling M2. Will therefore M2 have moved to a new position by the time I feel M1 so when I feel M1 and M2 at the same time am I not feeling them at the opposite sides of their binary orbit but in some other arrangement?

 

How would I work this out?

 

Is this true? Is the diagram is drawn wrong?

 

 

 

• Imagine observing a distant binary star and trying to measure the gravitational field at
your location. It is the sum of the field from the two individual components of the binary,
located at distances r1 and r2 from you.

.Can you sum the two components if they arrive at your position at different times?

Edited March 16, 2016 by Robittybob1

[Mordred]

I'm not sure why your having trouble.

 

Both objects have sufficient time for their gravity to reach you so you can sum their values. The influence you feel is the position of where it was.

 

We feel the effects of gravity from our sun 8 minutes ago, any change in potential would take 8 minutes to arrive.

 

Same goes for object a and b. The gravitational changes will arrive from each object at different times. When you receive the change the value of the sum changes.

Edited March 16, 2016 by Mordred

[Robittybob1]

I'm not sure why your having trouble.

 

Both objects have sufficient time for their gravity to reach you so you can sum their values.

They aren't in the same place at the time it takes for the gravity to reach O. Can you add the gravity at T1 and T2 when they are different times? I'm not sure that you are right. Thanks for trying all the same. Have you seen a problem like this before?

Edited March 16, 2016 by Robittybob1

[Mordred]

It's simple logic, the sum is the strength of the force of gravity by the signals you've received.

This type of problem is common. Two body problem for example for orbits.

 

Let's say you have a binary star system and one orbitting planet.

 

The orbit is determined by the sum of gravity from both suns. The planet will react to changes with the appropriate delay for the gravitational force to arrive.

 

 

For example if one star blew up, the planet will follow the path of the two star system until the loss of gravity reaches the planet then it will alter orbit.

Edited March 16, 2016 by Mordred

[Robittybob1]

It's simple logic, the sum is the strength of the force of gravity by the signals you've received.

This type of problem is common. Two body problem for example for orbits.

 

Let's say you have a binary star system and one orbitting planet.

 

The orbit is determined by the sum of gravity from both suns. The planet will react to changes with the appropriate delay for the gravitational force to arrive.

 

....

there would be a center of mass (CoM) and that CoM will stay in the same location even though the binary masses moved around in their orbit. So the combined strength will stay the same for the net effect will appear to be acting through the center of mass.

Now that would be a Newtonian approach but would the same hold true when the orbital speeds increased? Will the GR effects change this relationship?

I have read through this letter by S Carlip 2000 http://arxiv.org/pdf/gr-qc/9909087.pdf

The answer to the problem is in there.

 

Can anyone use this paper to explain the solution?

 

 

Thus up to terms of order v^2/c2, the force points not toward the retarded position of the

source, but toward the “linearly extrapolated” retarded direction

(1.9); the extra velocity

dependence in r eliminates aberration at order v/c.

As Van Flandern has stressed, though, astronomical observations require a more complete

cancellation: aberration terms of order v^3/c3 must be eliminated as well. To understand such

a cancellation, we can stand the argument of Ref. [1] on its head. As that paper emphasized,

a retarded purely central force with no velocity-dependent terms inevitably leads to the

drastic nonconservation of orbital (“mechanical”) angular momentum and energy in a binary

system. But by Noether’s theorem, any theory derived from a Lagrangian invariant under

rotations and time translations must conserve total angular momentum and energy. For an

isolated, bound system, this is only possible if changes in mechanical angular momentum

and energy are balanced by changes in the angular momentum and energy of radiation.

For electromagnetism, conservation of charge implies that there can be no monopole

radiation, and the power radiated in dipole radiation is proportional to |d2d/dt2|2

, where d

is the electric dipole moment of the source. Since the first derivative dd/dt is proportional

to the velocity, a charge moving at a constant velocity can radiate no angular momentum or

energy. Hence at least to first order in velocity, any nonconservation of mechanical angular

momentum and energy due to finite propagation speed must be compensated by additional

(velocity-dependent) terms in the interaction.

cancellation of velocity dependent terms so the force of gravity acts through the extrapolated position and not from where it actually is. Something like that. Does anyone understand it so they could give us simpler description?

 

There was another abstract "How to Measure Gravitational Aberration?" https://www.researchgate.net/publication/248627550_How_to_Measure_Gravitational_Aberration

 

In 2000, Carlip showed that in general relativity gravitational aberration is almost cancelled out by velocity–dependent interactions. We show how the actual value of the gravitational aberration can be obtained by measurement of a single angle at a suitable time t∗ corresponding to the perihelion of an elliptic orbit.

.

Edited March 16, 2016 by Robittybob1

[Mordred]

The CoM is the sum of vector forces. It's the same thing. I've already given a simpler example.

 

However if you want it worded in CoM terms each object will react to change when that object receives the information of a change.

 

So the sum of forces (vectoral) will point to the retarded CoM position until the change in field strength reaches each object.

Edited March 16, 2016 by Mordred

[Strange]

How would I work this out?

 

You have mentioned using Excel before. This looks like an example where it could be useful. Have a column for angle, a column for each distance and then a column for the corresponding force from each body. And finally, a column for the vector sum of the two forces.

 

Then plot a graph of the forces against angle and see what it shows.

[MigL]

Consider two massive objects moving uniformly and parallel to each other at an appreciable fraction of the speed of light.

Since each object feels gravitational attraction from the other to a point behind its 'current' position, i.e. not orthogonally opposed because of the finite speed of propagation, then there must be a vector component of the effective gravity acting 'backwards'. This would tend to slow down both masses ( equally if equal masses ).

Where is this energy going ( its gotta be conserved ) ?

 

( had this discussion before on another forum, but it was never addressed to my satisfaction )

( and although this may seem like a hijack of RobbityBob's OP, it is relevant to his questions )

[J.C.MacSwell]

Keep in mind that using the CoM is based on a spherical distribution of mass. Other distributions won't always sum to the same thing.

 

The gravitational vectors for Newtonian Gravity point directly to current position (no lag), and for GR they tend to the equivalent of that for lower speeds and where no abrupt changes take place, even though their is a lag in GR.

 

For, say, two masses some distance apart stationary in the same inertial frame, the gravitational effect would be directly toward each other. It would not be any different in other frames of reference...(measurements of what is considered "current" or simultaneous aside)...otherwise the two mass system could have a net self drag effect in a frame measuring them to have velocities parallel to each other due to the lag, and of course this does not happen.

 

Just saw MigL's post...same example

Edited March 16, 2016 by J.C.MacSwell

[swansont]

Consider two massive objects moving uniformly and parallel to each other at an appreciable fraction of the speed of light.

Since each object feels gravitational attraction from the other to a point behind its 'current' position, i.e. not orthogonally opposed because of the finite speed of propagation, then there must be a vector component of the effective gravity acting 'backwards'. This would tend to slow down both masses ( equally if equal masses ).

Where is this energy going ( its gotta be conserved ) ?

 

( had this discussion before on another forum, but it was never addressed to my satisfaction )

( and although this may seem like a hijack of RobbityBob's OP, it is relevant to his questions )

 

 

There is no problem here. They are at rest with respect to each other. There is no retardation issue, since there is no relative motion. We can always analyze an interaction in the rest frame, or any other inertial frame, at our convenience. As JCM states, the effect is directly at the other object.

[Robittybob1]

 

You have mentioned using Excel before. This looks like an example where it could be useful. Have a column for angle, a column for each distance and then a column for the corresponding force from each body. And finally, a column for the vector sum of the two forces.

 

Then plot a graph of the forces against angle and see what it shows.

I had started doing that but I couldn't work out the "corresponding force from each body" unless I used a type of Newtonian gravity which was instantaneous, but as it appears from Carlip even though Cg (speed of gravity) = C because some velocity dependent terms terms drop out ..... not sure entirely sure what Carlip means sorry.

 

So do I use Newtonian gravitational strength in spreadsheet?

Consider two massive objects moving uniformly and parallel to each other at an appreciable fraction of the speed of light.

Since each object feels gravitational attraction from the other to a point behind its 'current' position, i.e. not orthogonally opposed because of the finite speed of propagation, then there must be a vector component of the effective gravity acting 'backwards'. This would tend to slow down both masses ( equally if equal masses ).

Where is this energy going ( its gotta be conserved ) ?

 

( had this discussion before on another forum, but it was never addressed to my satisfaction )

( and although this may seem like a hijack of RobbityBob's OP, it is relevant to his questions )

That is along the right line of enquiry I'm wanting to know how velocity will affect gravity. I'm not saying you are right though, but that is right on the topic.

Keep in mind that using the CoM is based on a spherical distribution of mass. Other distributions won't always sum to the same thing.

 

The gravitational vectors for Newtonian Gravity point directly to current position (no lag), and for GR they tend to the equivalent of that for lower speeds and where no abrupt changes take place, even though their is a lag in GR.

 

For, say, two masses some distance apart stationary in the same inertial frame, the gravitational effect would be directly toward each other. It would not be any different in other frames of reference...(measurements of what is considered "current" or simultaneous aside)...otherwise the two mass system could have a net self drag effect in a frame measuring them to have velocities parallel to each other due to the lag, and of course this does not happen.

 

Just saw MigL's post...same example

Is that cross drag or something? If one is being dragged backward isn't the other being dragged forward, net effect is just attraction between two masses traveling parallel to each other.

Please explain what you meant by "and of course this does not happen". Is this in real life situations?

 

 

There is no problem here. They are at rest with respect to each other. There is no retardation issue, since there is no relative motion. We can always analyze an interaction in the rest frame, or any other inertial frame, at our convenience. As JCM states, the effect is directly at the other object.

Could you explain what they were talking about when they thought there would be the momentum loss? (Was this momentum loss the beginnings of the prediction of Gravitational radiation???)

The quote below seems to describe the effect I was seeing if I tried to work out the force felt by O from two orbiting bodies "at the one time" for I couldn't be sure of the position of the orbiting bodies when that force was being felt.

 

[both footnote references [3,4] could not be found in a form I could read. Does anyone have the book "Problem Book in Relativity and Gravitation"?]

 

Page 2 0f http://arxiv.org/pdf/gr-qc/9909087.pdf

 

 

1 Aberration in Electromagnetism

It is certainly true, although perhaps not widely enough appreciated, that observations

are incompatible with Newtonian gravity with a light-speed propagation delay added in [3,4].

If one begins with a purely central force and puts in a finite propagation speed by hand, the

forces in a two-body system no longer point toward the center of mass, and the resulting

tangential accelerations make orbits drastically unstable. A simple derivation is given in

problem 12.4 of Ref. [4], where it is shown that Solar System orbits would shift substantially

on a time scale on the order of a hundred years.

Edited March 16, 2016 by Robittybob1

[J.C.MacSwell]

Is that cross drag or something? If one is being dragged backward isn't the other being dragged forward, net effect is just attraction between two masses traveling parallel to each other.

Please explain what you meant by "and of course this does not happen". Is this in real life situations?

It doesn't happen (in reality, GR, or Newtonian) "otherwise" there would be different results in different frames

[Robittybob1]

It doesn't happen (in reality, GR, or Newtonian) "otherwise" there would be different results in different frames

But using the different equations gave different results. e.g. precession of Mercury could not be accounted for in Newtonian gravity.In real life if we had two planets orbiting they are not inertial frames etc BBH lose energy, and they are accelerated. They keep on giving us examples of the Sun being taken away and we would feel the effect for another 8 mins. With orbiting binaries the direction of the force of gravity is always changing too. What effect does this have when the spacetime curvature can only be updated at the SoL (speed of light) c.

I've drawn a sketch of two equal mass BBH in circular orbit. If the distance across between the BHs is 3 lms (light milliseconds) and the tangential velocity is 0.333c they would advance in their orbit by about a third of their diameter. Each BH would run into the gravity left there by the opposite BH. The gravity appears stronger and advanced, so would it accelerate even more? But if one gains momentum the other must lose it.

So they both seem to be losing and gaining momentum. It would be interesting to see if the effects talked about above by Carlip could ever result in more being lost that what is gained.

[Strange]

I've drawn a sketch of two equal mass BBH in circular orbit. If the distance across between the BHs is 3 lms (light milliseconds) and the tangential velocity is 0.333c they would advance in their orbit by about a third of their diameter. Each BH would run into the gravity left there by the opposite BH. The gravity appears stronger and advanced, so would it accelerate even more?

 

<cough> numerical methods <cough> supercomputer <cough>

 

You are looking for simple answers where there are none.

[Robittybob1]

 

<cough> numerical methods <cough> supercomputer <cough>

 

You are looking for simple answers where there are none.

No it is just like the example Mordred gave of the Sun disappearing. It is only a mental exercise to see what effects gravity at the SoL has. Yesterday when I was trying to figure it out my brain didn't feel like any sort of computer, it just froze.

[J.C.MacSwell]

But using the different equations gave different results. e.g. precession of Mercury could not be accounted for in Newtonian gravity.In real life if we had two planets orbiting they are not inertial frames etc BBH lose energy, and they are accelerated. They keep on giving us examples of the Sun being taken away and we would feel the effect for another 8 mins. With orbiting binaries the direction of the force of gravity is always changing too. What effect does this have when the spacetime curvature can only be updated at the SoL (speed of light) c.

I am not suggesting Newtonian never differs from reality or GR. I am saying Newtonian results are always consistent, regardless of what frame you choose. They give the same result. This would also be true in GR compared to the calculations from different frames in GR (not that I could do them), and, I would like to think, in reality as well.

 

If you could somehow quickly shift the Sun to where Mercury is (or have it disappear) you would get different results in Newtonian or in GR, but in each case you would get consistent results for each, independent of choice of frame. Newtonian would instantly shift the gravitational vector to the new position of the Sun (or force would disappear). GR would do the equivalent (approximately) but only after a lag...the effect would remain in the direction toward where the Sun was going to be until after the lag. This is I think approximately right, and assuming an equivalent set of assumptions for each frames calculations. In reality the Sun shift would have to have some cause, and of course a disappearance impossible.

[Robittybob1]

I am not suggesting Newtonian never differs from reality or GR. I am saying Newtonian results are always consistent, regardless of what frame you choose. They give the same result. This would also be true in GR compared to the calculations from different frames in GR (not that I could do them), and, I would like to think, in reality as well.

 

If you could somehow quickly shift the Sun to where Mercury is (or have it disappear) you would get different results in Newtonian or in GR, but in each case you would get consistent results for each, independent of choice of frame. Newtonian would instantly shift the gravitational vector to the new position of the Sun (or force would disappear). GR would do the equivalent (approximately) but only after a lag...the effect would remain in the direction toward where the Sun was going to be until after the lag. This is I think approximately right, and assuming an equivalent set of assumptions for each frames calculations. In reality the Sun shift would have to have some cause, and of course a disappearance impossible.

Did you understand that letter from S Carlip?

http://arxiv.org/pdf/gr-qc/9909087.pdf

Your analysis "Newtonian would instantly shift the gravitational vector to the new position of the Sun (or force would disappear). GR would do the equivalent (approximately) but only after a lag...the effect would remain in the direction toward where the Sun was going to be until after the lag." would be different if I have any true understanding of what he says. No one has mentioned his work as yet. I think he is saying there is no or little lag in GR either but I'd like someone to confirm that.

Edited March 16, 2016 by Robittybob1

[Mordred]

No he's saying there is some lag as defined by the speed of light. In this case the speed of gravity. The lag is what he's describing by retarded position

[J.C.MacSwell]

No he's saying there is some lag as defined by the speed of light. In this case the speed of gravity. The lag is what he's describing by retarded position

Reading the abstract I think that is correct:

 

The observed absence of gravitational aberration requires that “Newtonian” gravity propagate at a speed c g > 2 × 1010 c. By evaluating the gravitational effect of an accelerating mass, I show that aberration in general relativity is almost exactly canceled by velocity-dependent interactions, permitting c g = c.

 

He seems to be saying that whereas adding any lag at all to Newtonian gravity would have it differ from observation, with GR it can match observation even as slow as the speed of light.

[Robittybob1]

No he's saying there is some lag as defined by the speed of light. In this case the speed of gravity. The lag is what he's describing by retarded position

This is the paragraphs that seems to highlight the issue

 

 

. Although gravity propagates at the speed of light in general relativity, the expected aberration is almost exactly canceled by velocity-dependent terms in the interaction. While at first this cancellation seems to be “miraculous,” it can be explained from first principles by turning Van Flandern’s argument on its head: conservation of energy and angular momentum, together with the quadrupole nature of gravitational radiation, require that any causal theory have such a cancellation.

So does that mean we expect aberration due to gravity propagating at light speed. But this expected aberration is cancelled by velocity dependent terms in the interaction.

 

What does that actually mean?

Edited March 17, 2016 by Robittybob1

[Mordred]

To understand that you would need to understand the term stellar aberration.

 

https://en.m.wikipedia.org/wiki/Stellar_aberration_(derivation_from_Lorentz_transformation)

 

In this case he is discussing aberration of gravity. The first link gives the general idea though.

 

essentailly the paper is comparing the Lorentz invariance of light to gravity being variant or invariant.

 

If the graviton is truly massless, just as light is then gravity would be invariant. The emitters velocity would have no influence upon the propogation of gravity. (Velocity dependant cancellations).

 

If however gravity is not invariant then conservation of momentum can cause changes in the speed of the graviton.

 

Most papers discuss alternative competing theories for comparison. This is essentially what this paper is doing.

 

The comparison is Lorentz invariant or not invariant. For gravity.

The paper includes electromagnetic Metrics to help explain the variance and invariant theories.

 

The conclusion the paper presents is gravity is invariant.

Essentially stating that the speed of gravity will be the same to all observers regardless of the observer or emitter velocities.

 

 

I found it kind of humorous that the paper mentions Van Flandern... well you decide lol

 

https://en.m.wikipedia.org/wiki/Tom_Van_Flandern

[Robittybob1]

To understand that you would need to understand the term stellar aberration.

 

https://en.m.wikipedia.org/wiki/Stellar_aberration_(derivation_from_Lorentz_transformation)

 

In this case he is discussing aberration of gravity. The first link gives the general idea though.

 

essentailly the paper is comparing the Lorentz invariance of light to gravity being variant or invariant.

 

If the graviton is truly massless, just as light is then gravity would be invariant. The emitters velocity would have no influence upon the propogation of gravity. (Velocity dependant cancellations).

 

If however gravity is not invariant then conservation of momentum can cause changes in the speed of the graviton.

 

Most papers discuss alternative competing theories for comparison. This is essentially what this paper is doing.

 

The comparison is Lorentz invariant or not invariant. For gravity.

The paper includes electromagnetic Metrics to help explain the variance and invariant theories.

 

The conclusion the paper presents is gravity is invariant.

Essentially stating that the speed of gravity will be the same to all observers regardless of the observer or emitter velocities.

 

 

I found it kind of humorous that the paper mentions Van Flandern... well you decide lol

 

https://en.m.wikipedia.org/wiki/Tom_Van_Flandern

Did you link to the right aberration topic?

There were 501 articles in Google Scholar mentioning "T van flandern" Carlip is not agreeing with van Flandern and I suppose you don't either, so that can't be a bad thing.

What exactly were you implying about S Carlip?

Edited March 17, 2016 by Robittybob1

[Mordred]

Just the choice of counter model. Usually the counter model is more competitive. Ie LCDM vs MOND back when MOND was in stronger contention.

 

Though I suppose Van's model may have been more supported at one time.

Edited March 17, 2016 by Mordred

[Robittybob1]

Just the choice of counter model. Usually the counter model is more competitive. Ie LCDM vs MOND back when MOND was in stronger contention.

 

Though I suppose Van's model may have been more supported at one time.

Some of van's Flandern's ideas are a bit weird I agree.

[swansont]

This is the paragraphs that seems to highlight the issue

So does that mean we expect aberration due to gravity propagating at light speed. But this expected aberration is cancelled by velocity dependent terms in the interaction.

 

What does that actually mean?

 

 

It means we don't experimentally observe any (or only a small amount) of aberration. Not because it's absent, but because there are other effects which cancel it.