Strange

I don't think DanMP is interested in proving (or even providing any support for) any of his ideas. He appears to be of the "I thought of it, so it makes sense and must therefore be right" school of thought.

If you do feel the need to start one then, before you do, make sure that: 1. You have convincing evidence that it is relevant to this situation (i.e. that there is an 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.

No, because an aircraft is not falling. The wings lift it. Did you look at the Newton Cannonball?

No, I think they are all in freefall. There is nothing stopping or slowing their fall.

Surely a time delay is a time delay; it will be the same wherever you measure it. I can't imagine why. It is just another crackpot idea from another random crackpot.

Free fall means there is no force preventing your fall under the influence of gravity. A sideways motion doesn't do that; for example, a ball thrown horizontally will fall to the ground at the same time as one that is just dropped. Newton came up with the cannonball analogy to explain orbits as an example of free fall: https://en.wikipedia.org/wiki/Newton's_cannonball

It may do, in principle, but my guess (until evidence is presented to the contrary) is that the effect is immeasurably small (i.e. less than the noise/errors in the measurements).

I have no idea. The Wikipedia page gives a simple approximation but notes that it is not applicable to situations involving black holes. Exactly. Gravitational waves travel through space-time and so any curvature of space-time will affect them in the same way that light is affected. This paper might give you some ideas: http://iopscience.iop.org/article/10.1086/310835/fulltext/5332.text.html "Our calculations demonstrate that, for reasonably compact and eccentric orbits, rotation of a sufficiently massive BH companion to the orbiting pulsar leads to approximately microsecond-order departures from pulse arrival times"

Or it may not. I'm going with "not" as I haven't seen anything to suggest otherwise. I have read several papers on black hole mergers, including skimming the papers on the recent detection and none of them have ever mentioned this as a possible factor. At that angle neither black hole is even close to being behind the other, so your speculation seems to be irrelevant.

Can you show us the calculations that support this being a significant factor that needs to be taken into account? From your description, it sounds as if this effect (if it exists) would only be relevant for gravitational waves that are detected "edge-on". But that is not the orientation in the LIGO detection. (And gravitational waves are strongest at right angles to that direction anyway.)

You could try emailing some researchers who look like they might have the data you need, with details of the analysis that you want to do to see if they have the relevant information in their data sets. They may be reluctant to make the data available (for privacy reasons - depending what is in the database) but may be able to extract the relevant subset for you or do some analysis (if they think your hypothesis is interesting enough).

It is here: http://arxiv.org/abs/1602.03839 You can find a helpful index and overview of the papers here: http://cplberry.com/2016/02/23/gw150914-the-papers/ (which is how I found that one)

More accurately, after nearly one month you still haven't understood your error.

You tell us. For example, have you worked out how much force will be generated? And at what cost?

The probability of some orientations is pretty much zero. What does "underneath" mean? I'm sure these questions are answered in one of the other papers describing the details of the detection and analysis. My guess would be that they used some sort of correlation function to determine the time difference. Looking at a single peak would not be very accurate.

Yes. The effect is relative to each observer (hence the theory of "relativity") so each observer (qaurk in this case) will see the proton flattened in a different direction. (note: flattened, not flat - i.e. slightly flatter, not completely flat) But each only sees it flattened in one direction. This is not different from the Earth which is also flatted by different amounts in different directions, depending which observer you consider. No. For one thing, protons are not flattened in all directions; only in the direction of motion of the observer. Also flattened in the sense of slightly squashed or foreshortened is not the same as the universe being geometrically flat (i.e. not curved) BTW: Matt Strassler's site is a really good source.

I am not defining anything (although I'm sure the meaning of orbital plane is pretty obvious: the plane that they are orbiting in). Note that (as you have seen in some other simulations) if the black holes are spinning then there can be precession and the orbital plane will change orientation. For this reason, the paper defines the angle very carefully - I don't have it to haand at the moment, but I seem to remember that they took the value at a particular orbital rate (20Hz?). The graph in the paper shows that the most likely angle between the normal to the plane (which they define as the vector of angular momentum) is about 150 degrees. So 30 degrees off perpendicular to the plane. Because the waves only exist in GR.

In the bit you quoted and asked about.

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

So what evidence do you have for your idea? How would we test it to see if it is correct? Actually, experiments have just confirmed that the current theory is correct about this. Can you show your calculations for the merger of two black holes?

Don't worry. There is no chance of you being in shock. Sometimes a stupid idea is just wrong.

He is referring to the relativistic Doppler effect which does work in this situation. However as there is no relative motion on his example, there is no relativistic Doppler either.

Yes, I know that. Polar jets are not black bodies. Are related t the poster who goes by the name of mpc775 (or similar)? Because he just posted exactly the same thing repeatedly with random, barely relevant quotes. It is like you are twins.

I assume it is from here: https://pl.wikipedia.org/wiki/Efekt_Dopplera

The CMB is a black body spectrum. Polar jets are not. Your theory fails. Go on then.