Robittybob1

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 - when the two BBH were approaching the merger even at that probable 60 degrees inclination to the line of site the Shapiro Effect could have some effect. The size of the EHs at that stage are quite large in proportion to their separation. Would the size of the photosphere have to be considered as well?

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? 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. 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? 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

I read them as distances: OK position vector: Makes sense. I was not talking about that particular BBH but the possible limits of LIGO, seeing whether it was even possible to get the "Gravity" lines to overlap and superimpose as they did in that 3D animation. I came to the conclusion it is virtually impossible as gravity operates at the speed of light and if the orbit is large the orbital frequency is low so the G-waves are always really well separated.

Yes that how I understand it too now. Thanks.

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? 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 .

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?

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

The thought of going off on the tangent makes the idea of free falling not so bad. It was how I got it into my head but I'm always flying off.

Think about where it would be if it didn't fall. It would go straight ahead and would be nearly out of the Solar System by now.

http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GravitationalWaves.pdf Just look at that diagram do you think that is correct? I don't, for if the masses are moving the line representing gravity should be a spiral or a dotted line but not a continuous line. I know what you are saying and it has been playing on my mind, so I'm not saying you are wrong but I'm trying to understand what it means. So from the diagram if the BBH was directly above the LIGO detectors and the two distances were equal the detectors would get a signal compressing one way and after the BBH rotated through another 90 degrees and compressing in the other arm, another 90 degrees compressing in the first arm and so on , 2 cycles per orbit. I can see that but the point of maximum gravity comes from each body and the effect of this is a spiral through space heading toward the LIGO detectors. So are you going to get places where the gravity lessens or superimposes in space with spirals or pulses where the distances to the LIGO detectors are not equal? I have tried to work this out. When the BBH are the same distance apart as the two LIGO detectors and there was an absolute side on view to the orbital plane with a line of sight to the Earth of zero degrees the gravity from both black holes at two points in their orbit could reach one of each of the two facilities at the same time so they would be basically measuring parts of different orbits. If the LIGO facilities are 2000 km apart and the BBH binaries are 2000 km apart the gravity from the furthest BH will travel through space and hit the closest detector at the same time as the closer BH will send gravity to the furthest detector. Here they would be measuring parts of 2 different orbits. This could be the worst case scenario. Would the optimum case be a black hole binary, with the orbital plane parallel to the plane through the ligo facilities, directly overhead the mid point between the two facilities with a 4 km separation of the binary bodies? At 2000 km the orbital velocity would depend on their masses but the solar mass BHs will not being orbiting very fast even at this distance but if the LIGO team start picking up the lower frequency waves at greater separation of the BBHs they could end up with more issues with interference. Am I understanding this right?

http://www.scienceforums.net/topic/93472-gravitational-lens-and-gravitational-waves-question/page-3#entry909686

It was this statement that did not seem to fit into what was displayed in Fig 5. Do you still stand by "if the signal was from above, I don't think you'd register a signal"? From the discussion I thought the opposite was true. When I said "Above and below no matter from which direction." I'm only describing where the GWs could be coming from. Basically meaning from all directions but according to figure 5 a BBH directly above the LIGO will be equidistant and sensitive to detection.

Howabout you starting it and prove the null hypothesis, that you advocate? 1. You have convincing evidence that it is Not relevant to this situation (i.e. that there is NO independent "signal" from each black hole); and 2. You know how to calculate the Shapiro delay in an extreme case like that of a spinning black hole. It is not easy is it?

I thought you had accepted that there were two waves coming from the quadrupole. I was thinking if it was proven by DanMP it would be hidden within a thread on a different topic. It is not my baby but Dan's.

What I was trying to say in my layman way was that there is a sort of a plane created by the horizon and there is above and below that horizontal plane, but there is also the 360 degree rotation (that is the "no matter which direction" bit). From that I say "Above and below (the horizontal plane) and no matter from which direction (on the 360 degree rotation)" You might have a more concise way of saying that? I see they use the term "tangent to the Earth's surface" for my horizontal plane. https://dcc.ligo.org/public/0109/P1300187/023/ms.pdf Figure 5 explains it.

I'm looking for that physicist. Kip Thorne has tried to explain it "Kip S. Thorne-Gravitational Waves: A New Window onto the Universe" YT but...

Time delay at the LIGO depend on things like direction of line of sight. If you don't know where the signal is coming from you don't know how much of the delay was due to the position, The precession of the BBH affects the position of the nodes (where the GW signal will be strongest. So how will you get LIGO to pick it up, so it will only be theoretical. What happens to light may not happen to gravity, but you could look at the effects if it was. Is DanMP a crackpot? His estimate of the strength of the STD effect seemed rather high, but that in itself is not enough to reject the idea.

@Strange I was meaning right at the site where the two merged, not necessarily at the LIGO detectors. But no doubt we will need to consider both. Do you think we need to start another thread specifically on the Shapiro Time Delay Effect?

.Bit of a goose chase so far and I'm still unsure. From the above link I got: From that footnote link led me to http://arxiv.org/pdf/1602.03840v1.pdf .Looking at that 103, 104 footnote: Gave background work on using the time delay. . That reads a bit like a "treasure map" when reading "to two islands of probability that cannot be distinguished". I'm not sure whether GW on some lines of sight can't be detected. There will always be two islands for there are those waves coming in from above the plane of the detectors and below. Above and below no matter from which direction. Defining some terms: GW150914 I suppose that is Gravity Wave 2015, Sept 09 and 14th day ... GW 15/09/14 >>> GW150914 PyCBC

I find we are in the middle of this most exciting break through, the discovery of gravitational waves and all the information is available on the internet. It is truly amazing. In that other thread I wanted to understand what causes the loss of momentum leading to the decayed orbit, it would be hard to think that the Shapiro Time Delay (STD) would have anything to do with it but who knows? Does the STD affect the speed or direction of gravity at the level of the binary orbits too? (gravitational lensing effect and the STD effects combing somehow?)

So if I understand you if the line of site was directly above Hanford (H) and Livingston (L) the two signals would possibly arrive at the same time, but since H and L are separated by around 2000 km and L gets the signal first the signal came from somewhere on a line of sight looking from the general direction of H to L? 2 directions "two possible answers" because the waves will go through the Earth as well.

So is the "strong interaction" the same or connected to the "binding energy" in some way? OK I'll do some more homework.

What are the things needed for this calculation? @Dan are you thinking gravity is delayed as light was delayed?

Didn't someone use the word "above" so "beneath" is below that! What I was thinking was that these BBHs had been producing G-waves for the last million years or so how can you identify when the signal starts, but there was a definite maximum peak and that followed by the ringdown. If you see the paper that mentions how they established the timing please post it.