Warped spacetime around BH and the barycenter

[Robittybob1]

....

Analyzing graphs doesn't sound like a particularly useful way to gain insight into the science.

Today I listened to the 1 hour long announcement by LIGO team on the discovery of the G-waves ('We have detected gravitational waves!', breakthrough discovery confirmed (FULL PRESSER)) Two things were confirmed: The LIGO team had to compare the results of computer simulations to the chirp trace. This is how they were able to put masses to the BHs i.e. from the analysis of graphs.

Their simulations were made using Einstein's equations i.e. relativity and I never heard anyone use the word "barycenter", hence I will try and communicate with the LIGO team and let them comment on my speculation.

If my speculation is correct the orbital plane and the Earth were somewhat aligned. How else would the effects of the delayed and the advanced waves be picked up by LIGO unless the BHs were orbiting in near alignment with the Earth?

So that is a prediction that can be tested/verified.

Has anyone got their address? An email address will do.

Edited March 13, 2016 by Robittybob1

[swansont]

Today I listened to the 1 hour long announcement by LIGO team on the discovery of the G-waves ('We have detected gravitational waves!', breakthrough discovery confirmed (FULL PRESSER)) Two things were confirmed: The LIGO team had to compare the results of computer simulations to the chirp trace. This is how they were able to put masses to the BHs i.e. from the analysis of graphs.

Yes, they analytically compared their mathematical model (the simulation) to the data. This is not the same as analyzing a graph.

 

Their simulations were made using Einstein's equations i.e. relativity and I never heard anyone use the word "barycenter", hence I will try and communicate with the LIGO team and let them comment on my speculation.

 

Prepare to be ignored.

 

If my speculation is correct the orbital plane and the Earth were somewhat aligned. How else would the effects of the delayed and the advanced waves be picked up by LIGO unless the BHs were orbiting in near alignment with the Earth?

So that is a prediction that can be tested/verified.

 

Define "somewhat aligned."

[Strange]

Other than the results from the LIGO graphs what have they to go on. All the information is in the results recorded by these two machines and they make their calculations from those recordings.

 

You have that backwards. They have been studying these sort of situations for decades. The LIGO results nicely confirm that the theory is correct.

 

How do you know that is not what is happening within their programs already?

 

Because (several years ago) I read a lot about the subject.

If my speculation is correct the orbital plane and the Earth were somewhat aligned.

 

The alignment was calculated in a paper that someone posted in another (or this?) thread.

This paper: http://arxiv.org/abs/1602.03840

[Robittybob1]

 

You have that backwards. They have been studying these sort of situations for decades. The LIGO results nicely confirm that the theory is correct.

 

Because (several years ago) I read a lot about the subject.

 

The alignment was calculated in a paper that someone posted in another (or this?) thread.

This paper: http://arxiv.org/abs/1602.03840

What does it say about the orbital plane?

 

 

The luminosity distance is strongly correlated to the inclination

of the orbital plane with respect to the line of

sight [17]. For precessing systems, the orientation of the

orbital plane is time-dependent. We therefore describe the

source inclination by θJN , the angle between the total angular

momentum (which typically is approximately constant

throughout the inspiral) and the line of sight, and we

quote its value at a reference gravitational-wave frequency

fref = 20 Hz The posterior PDF shows that an orientation of the total orbital angular momentum of the BBH strongly misaligned to the line of sight is disfavoured; the probability that 45◦ < θJN < 135◦ is 0.35.

So does that mean they are saying the orbital plane "probably" aligns with the line of sight? (end of page 6 beginning 7) My prediction was just for the last 3 orbits before the merger. During that time I predict it was aligned with the line of site.

 

 

You have that backwards. They have been studying these sort of situations for decades. The LIGO results nicely confirm that the theory is correct.

OK that is what I meant, they compared the results of animations and simulations to the Interferometer strain readings to predict what sort of BBH situation they were dealing with. I think we are saying the same thing in fact. Some animations could be completed after the recording of the chirp was made too. That doesn't matter in reality but the point is they are comparing computer analyses of hypothetical situations to an actual recording.

Is that your understanding too?

 

.

[swansont]

Define "somewhat aligned."

 

 

OK that is what I meant, they compared the results of animations and simulations to the Interferometer strain readings to predict what sort of BBH situation they were dealing with. I think we are saying the same thing in fact. Some animations could be completed after the recording of the chirp was made too. That doesn't matter in reality but the point is they are comparing computer analyses of hypothetical situations to an actual recording.

Is that your understanding too?

 

It's doubtful that they made comparisons to animations. That's just not how it's done.

[Robittybob1]

Yes, they analytically compared their mathematical model (the simulation) to the data. This is not the same as analyzing a graph.

 

 

Prepare to be ignored.

 

 

Define "somewhat aligned."

What form did "the data" have? It was a wave pattern which I'm calling a graph. Graph might not be the right word but it is close.

 

One never knows if you are being ignored or not?

 

"Define "somewhat aligned"." I predict it would have a line of sight that would have been within 30 degrees of the orbital plane (above or below).

 

Define "somewhat aligned."

 

 

It's doubtful that they made comparisons to animations. That's just not how it's done.

 

Animations, simulations, simulated strain graphs that type of data to compare actual data with. That is what they were saying in that recording of the conference at the announcement.

Edited March 13, 2016 by Robittybob1

[swansont]

What form did "the data" have? It was a wave pattern which I'm calling a graph. Graph might not be the right word but it is close.

Numbers, most likely. The analysis was probably something like a regression analysis when comparing the data to the prediction. It's possible that nobody ever looked at a graph for long stretches of the analysis.

 

"Define "somewhat aligned"." I predict it would have a line of sight that would have been within 30 degrees of the orbital plane (above or below).

And how did you arrive at this number?

[Robittybob1]

 

...

 

 

And how did you arrive at this number?

As I said before from looking at the delayed wave and advanced wave and thinking about how at a greater the angle these differences in distance would disappear. (Trigonometric considerations)

[swansont]

As I said before from looking at the delayed wave and advanced wave and thinking about how at a greater the angle these differences in distance would disappear. (Trigonometric considerations)

 

So let's see the calculation.

[Robittybob1]

So let's see the calculation.

We have the orbital plane (OP) that extends indefinitely. that could be like the x axis. and we have the Earth defining the line of sight back to the BBHs. The line of sight meets the orbital plane at an angle Theta. The distance along the line of sight becomes the hypotenuse of a right angle triangle.

If each of the BHs produce a wave and they are at opposite sides of the binary orbit and the BHs have a different mass their orbital radii will be related by the barycenter calculation, the length of the hypotenuse changes depending whether you are closest to the inner or outer BH. If the orbit is divided perpendicular to the line of sight,in each half orbit period this will change. In one half M1 is closest then the next half orbital period M2 will be closest

 

Let r be the distance that the BHs are apart

Let M1 be the mass of the lighter BH

Let M2 be the mass of the heavier BH

Distance to the barycenter for M1 is A1

Distance to the barycenter for M2 is A2

If the barycenter is (0,0) A1 is further from (0,0) than A2 so theta changes whether the wave originated from M1 (theta1 is greater) or M2 (theta2 is lesser).

 

If the wave originated from M1 theta1 is greater.

If the wave originated from M2 theta2 is lesser.

 

So the adjacent side (along the x axis) and the hypotenuse changes when you are observing the different BHs

Hypotenuse to M1 = H1

Hypotenuse to M2 = H2

 

From the side of the orbit where the waves that reach the Earth come from:

 

A1 = r * M2 / (M1 + M2) [A1 is longer hence M1 closer to Earth]

A2 = r * M1 / (M1 + M2) [A2 is shorter hence M2 further from Earth]

 

So if the length of the adjacent to M1 is X1 the length to M2 is X2 then X2 = X1 + ( A1 - A2 ) [Note: X2 > X1 because A1 > A2]

 

Tan Theta1 = H1 / X1

Tan Theta2 = H2 / X2

 

[Note: as theta increases the ratio of changes in H1 and H2 become less marked.]

 

The minimum angle could be zero degrees where the difference in H1 and H2 will be most marked, hence the time to travel to Earth will cause the waves to bunch in groups of two. This bunching is also noted when the masses of the BH are markedly different.

 

The maximum angle to the orbital plane is 90 degrees. At that position the Earth would receive a constant flow of gravitational energy and hence it would not register on the LIGO. There could be angles where the two waves added together and appeared more powerful but then the frequency would be halved as the two waves were added together. (This effect is apparent on some of the simulations, particularly the one showing the 3D nature of the GWs.)

So 0 - 30 degrees is somewhat aligned to the orbital plane

60 - 90 degrees could be described as nearly orthogonal to the orbital plane.

30 - 60 degrees neither.

Edited March 13, 2016 by Robittybob1

[swansont]

 

The maximum angle to the orbital plane is 90 degrees. At that position the Earth would receive a constant flow of gravitational energy and hence it would not register on the LIGO.

 

 

But the signal is strongest in that direction.

http://www.scienceforums.net/topic/93472-gravitational-lens-and-gravitational-waves-question/page-2#entry909493

 

Can you cite anything that contradicts this, and supports you?

[Robittybob1]

 

 

But the signal is strongest in that direction.

http://www.scienceforums.net/topic/93472-gravitational-lens-and-gravitational-waves-question/page-2#entry909493

 

Can you cite anything that contradicts this, and supports you?

Are you willing to look at wave patterns on diagrams/animations? This was from my own analysis so it is another of the predictions.

I could see what @AstroKatie (Katie Mack, an astrophysicist) was talking about but the effect was sporadic and only stronger quite close to the merging BHs from studying the animation. This occurred where the two waves added together (superimposed) and the frequency in effect was halved. The waves were approaching each other at an angle (around 60 - 120 degrees) so they don't travel in that strong phase as they will pass through that point of superposition.

 

Do you want another YT video linked to the thread? I'll find it but it is complex in the extreme. You'll need to view it.

I brought the speed down to 0.25 and the best view was 10 secs into the animation. Maybe it's not the one I saw yesterday but it is close.

 

At about 8 - 10 secs into it the waves superimpose directly above the orbital plane.

 

The animation makes it look like that the gravitational waves are pulsations from the surface of the BH. The whole animation could be wrong.

I am tending to think it is more an artefact rather than real for if the GW is coming continuously of the body as it is accelerating continuously in orbit there is a spot which is equidistant from both of the orbiting BHs At these spots the GW superimpose and the intensity will be high but more or less continuous or at least slower in frequency and hence harder to pick up.

I'm in two minds over this one. It is much harder to think through.

Would I be right in thinking a strong low frequency GW is going to be more difficult to pick up. (Compared to the "chirp")

 

To have the waves as close as this animation the BHs would need to going faster than the speed of light. If the circumference of the orbit is 2 * pi() * r and they are going at 0.5c the radiation from each body should be

4 * pi() * r apart (the wave per each should be twice the orbital circumference away from the BH it emanates from) or every wave (that is from the 2 bodies in the binary) should be at about 2 * pi() * r apart radiating from each BH of an equal mass BBH going at 0.5c.

At 0.5c the waves should be around 12.6 times the orbital radius apart from its particular source. So you will get nodes where the waves are superimposed and the nodes will travel, but the frequency is halved and the intensity is doubled (2 waves merged into one) so the amount of GW is still the same over time. But because the orbital radius is changing rapidly and erratically in most cases where the masses are unequal these nodes also will be occurring in different positions every orbit. If you had a series of GW waves at these nodes and were being picked up by LIGO I wonder how they would distinguish them from the same frequency waves but coming from the orbital plane?

I would predict (from the above considerations) the chances of getting the chirp exactly dead center from equal mass BBHs is going to be very rare.

Edited March 14, 2016 by Robittybob1

[Strange]

What does it say about the orbital plane?

So does that mean they are saying the orbital plane "probably" aligns with the line of sight? (end of page 6 beginning 7)

 

No. It says that the line of sight is closely aligned with the "axis" not the orbital plane.

[swansont]

 

No. It says that the line of sight is closely aligned with the "axis" not the orbital plane.

 

 

And the axis of rotation is perpendicular to the orbital plane (if there's no precession)

Are you willing to look at wave patterns on diagrams/animations?

Do you want another YT video linked to the thread?

 

No.

 

I want scientific references.

[Robittybob1]

 

No. It says that the line of sight is closely aligned with the "axis" not the orbital plane.

Show me where it says that please? Define orbital plane as you use it please?

Here is my definition from Wikipedia "In the Solar System, the inclination of the orbit of a planet is defined as the angle between the plane of the orbit of the planet and the ecliptic.[2] Therefore Earth's inclination is, by definition, zero."

 

 

And the axis of rotation is perpendicular to the orbital plane (if there's no precession)

 

No.

 

I want scientific references.

Would references to the way waves superimpose help then? For to me the same physics would apply. Wave tank experiments showing where the waves increase in intensity? Its been awhile since I've seen those experiments.

This experiment would be similar to how I predict GWaves would spread out through space "Physics Lab Demo 14: Ripple Tank" that is a YT demonstration done by a teacher.

What we'd really need is a ripple tank with the two points rotating (constantly in the water).

Even that will only produces waves on the plane of orbit (Orbital plane)

Can you think how you could set up a wave tank that would show the waves above and below the orbital plane.

If you had a large deep tank and had 2 masses orbiting slowly deep under the water I wonder if the waves would appear on the surface? This is not the usual wave tank setup.

 

We would not want any reflection from the sides of the tank for we are not going to get reflection of GWaves.

 

In fact that would start the water rotating along with the masses causing the waves. so after a while you would get no waves or at least less waves being generated.

I can't think how you could physically demonstrate 3D waves being generated by a quadrupole.

Edited March 14, 2016 by Robittybob1

[imatfaal]

!

Moderator Note

 

Robbitybob

 

Please take this warning seriously - you have to start reading and understanding responses before replying. It is most disheartening to see questions answered in one of your threads only to be asked again in others of your threads, or the exact opposite of the correct answer being affirmed as a follow-up; what makes it worse is when your subsequent posting makes it clear you do not even understand the basic physics and terminology of the discourse. You are probably posting more than any other member at the moment and unfortunately the quality of your posts is declining with the volume.

 

Take time to read and digest posts - and if you do not understand them then perhaps a bit of reading around the subject is the way forward rather than your present parallel attack of asserting your arguments and asking very basic remedial questions.

 

Do not respond to this moderation within the thread - you can report this post if you feel it is unfair.

 

[Strange]

Show me where it says that please?

 

In the bit you quoted and asked about.

[Robittybob1]

 

In the bit you quoted and asked about.

Please, how do you define orbital plane? Define the angle of inclination of the orbital plane to the line of sight.

Edited March 14, 2016 by Robittybob1

[swansont]

Would references to the way waves superimpose help then?

 

 

No. But not because I already understand how waves superimpose.

 

The point here is that your claim that there are no waves along the axis is contrary to what an actual astrophysicist said. In this particular case, I don't want the explanation of your prediction. I want corroboration of what the actual physics says.

 

The reason for this is that if the accepted physics says one thing and you say another, then I don't care what your model is. It's wrong.

[Robittybob1]

 

 

No. But not because I already understand how waves superimpose.

 

The point here is that your claim that there are no waves along the axis is contrary to what an actual astrophysicist said. In this particular case, I don't want the explanation of your prediction. I want corroboration of what the actual physics says.

 

The reason for this is that if the accepted physics says one thing and you say another, then I don't care what your model is. It's wrong.

My model could be wrong and that statement I made earlier could be wrong (contained in #61) but I have thought the problem through since and on the basis of Gwaves being produced continuously and the way waves would superimpose. So my best shot at understanding how the physics is predicted to work was in #62 http://www.scienceforums.net/topic/93875-warped-spacetime-around-bh-and-the-barycenter/page-4#entry910990 You can either agree or disagree.

The exact answer to the question you've asked may not have yet been resolved. I don't know of any papers covering that particular question.

Katie Mack needs to give her reasons then for saying what she did in that tweet.

 

This is my best shot:

"So you will get nodes (above and below the orbital plane) where the waves are superimposed (nodes) and the nodes will travel, but the frequency (at the nodes) is halved and the intensity is doubled (2 waves merged into one) so the amount of GE is still the same (passing through a detector in line of sight) over time. But because the orbital radius is changing rapidly and erratically in most cases where the masses are unequal these nodes also will be occurring in different positions (in space) every orbit."

Above and below the orbital plane is it true there will never be superposition of the G-wave from the one BH as it orbits? I'm thinking the waves will always be far enough apart that no position of line of sight will allow a wave from a previous orbit to be superimposed on a subsequent one, no matter from which part of the orbit it originated. Does this logic hold true?

Edited March 14, 2016 by Robittybob1

[swansont]

My model could be wrong and that statement I made earlier could be wrong (contained in #61) but I have thought the problem through since and on the basis of Gwaves being produced continuously and the way waves would superimpose.

 

But we have a working model already. General relativity.

 

So my best shot at understanding how the physics is predicted to work was in #62 http://www.scienceforums.net/topic/93875-warped-spacetime-around-bh-and-the-barycenter/page-4#entry910990 You can either agree or disagree.

The exact answer to the question you've asked may not have yet been resolved. I don't know of any papers covering that particular question.

If your mental model is wrong, it will not lead to understanding anything.

 

Katie Mack needs to give her reasons then for saying what she did in that tweet.

 

That wasn't what was asked of her. The answer probably wouldn't fit into a tweet, anyway.

 

This is my best shot:

Exactly what I didn't ask for.

[Robittybob1]

 

But we have a working model already. General relativity.

 

If your mental model is wrong, it will not lead to understanding anything.

 

 

That wasn't what was asked of her. The answer probably wouldn't fit into a tweet, anyway.

 

Exactly what I didn't ask for.

General relativity is used, I agree, but once the waves are in space traveling at the speed of light why can't we revert to general physics after that?

If for some reason the spacial arrangement of the Gwaves were not isotropic around the BH then my model would definitely fail.

If it was possible to measure GE directly would any part of the BH be producing more energy at the level of the event horizon than at any other point? Or would it be like polar jets of GE? This would be an extreme example image https://upload.wikimedia.org/wikipedia/commons/thumb/f/f8/Galaxies-AGN-Inner-Structure.svg/270px-Galaxies-AGN-Inner-Structure.svg.png

Edited March 14, 2016 by Robittybob1

[imatfaal]

...

If for some reason the spacial arrangement of the Gwaves were not isotropic around the BH then my model would definitely fail. ..

For a binary viewed face-on, GWs are circularly polarized, whereas for a binary observed edge on, GWs are linearly polarized.

From a paper you claim to have read and has been discussed above. The waves are not isotropic in that their polarisation changes.

 

Others can confirm / deny my gut instinct heuristic ; if the waves were completely isotropic I fail to see how any Angular momentum could leave the two body system - and all reports I read state that the angular momentum decreases

Sorry - I have just realised that this is the thread I have moderated in; I thought that was the other one. Apologies

Edited March 14, 2016 by imatfaal
Apology

[Robittybob1]

 

From a paper you claim to have read and has been discussed above. The waves are not isotropic in that their polarisation changes.

 

Others can confirm / deny my gut instinct heuristic ; if the waves were completely isotropic I fail to see how any Angular momentum could leave the two body system - and all reports I read state that the angular momentum decreases

 

Very good observations thanks Imatfaal. Can anyone explain what polarisation means in this context? Is that the degrees of motion (directions) to which the test particles would move when the wave passed?

 

"circularly polarized"

"Linearly polarized"

[imatfaal]

Sorry - I have just realised that this is the thread I have moderated in; I thought that was the other one. I can participate or moderate - not both; I should step away from this thread. Apologies