Robittybob1

What does "t" measure? does anyone know please? [latex]h_{\times} = -\frac{1}{R}\, \frac{G^2}{c^4}\, \frac{4 m_1 m_2}{r}\, (\cos{\theta})\sin \left[2\omega(t-R/c)\right][/latex] With GW150914 is it the period of the orbit?

We are told how to calculate [latex] \omega [/latex] and [latex] R/c [/latex] doesn't cause difficulty, but what was [latex] "t" [/latex]? Linear approximation https://en.wikipedia.org/wiki/Gravitational_wave#Linear_approximation Notice it is a linear approximation so unless we can get the equations that deal with near the black hole we are still stuck. In the part [latex] 2\omega(t-R/c) [/latex] variable "R" is a distance and "c" is the speed of light so "R/c" must be a time so "t" should also be a time measure (time - time). Is it the period of the binary orbit. It doesn't say what "t" refers to in the Wikipedia article. T is the stress energy momentum tensor but would that make sense in that equation (stress energy momentum tensor - time)?

Of all the variables in the equations I'm not certain about [latex]"t"[/latex] [latex]h_{\times} = -\frac{1}{R}\, \frac{G^2}{c^4}\, \frac{4 m_1 m_2}{r}\, (\cos{\theta})\sin \left[2\omega(t-R/c)\right][/latex] [latex]t[/latex] in the part [latex] 2\omega(t-R/c) [/latex] We are told how to calculate [latex] \omega [/latex] and [latex] R/c [/latex] doesn't cause difficulty, but what was [latex] "t" [/latex]? Linear approximation https://en.wikipedia.org/wiki/Gravitational_wave#Linear_approximation Notice it is a linear approximation so unless we can get the equations that deal with near the black hole we are still stuck.

Why not? Free moving test masses even if they were free falling toward the Earth they would all be staying together as a ring, there would be no strain on them.

They wouldn't be able to use pendulums. It will have to be a totally different method used AFAIK. Three satellites around the Sun. Would they just be measuring the distances between the satellites? If the apparent distances between the satellites fluctuate for no known reason it could be a gravitational wave.

it is a good question. One point I can think of is that the photons in the laser light will have the same energy in the two arms, the stretching of spacetime will not change their energy content so their frequency should stay the same, so do they get out of phase because the distance the laser has to travel in one arm changes compared to the other?

I had heard about the effect you are studying. It seems to be in the public domain already. https://en.wikipedia.org/wiki/Hygiene_hypothesis

#59 I displayed formulas based on EFE to calculate the amplitude of the waves. They are not that difficult we could test out what will happen when two waves meet.

That is true they are suspended as pendulums, so they need the earth's gravitational pull to keep them hanging steady. Would it be more like another type of wave? A water wave? what about sound waves? What type of waves could we compare them too? If you refer to #68 and had 2 water sprinklers operating at once would that give you a model of how 2 sets of gravitational waves would add together? You can only get more water never less water with the summation of the two waves. There will never be destructive interference but constructive interference seems possible.

Swansont - did my answers help? #66 and #68 Can you see why a 3D sphere makes sense?

"sockpuppets" would be the biggest issue from memory.

A more easy to understand question would be what if there were two BBHs (a total of 4 BHs) producing GWs at the same time. Do their vectors add?

The forum has a thread tracking why people were banned. Do you analyse that thread?

Gravitational waves are waves of gravitons (gravitational effect) and the effective amplitudes (or strength of effect) is a pattern like a 3D spiral (like water out of the garden sprinkler each one of the droplets is going straight but there is the overall image of a spiral of water. Stand close to it and you will get waves of water at intervals 2 per revolution).

I'm just going to answer this in an off the cuff as I can comprehend the situation. If there was just 1 BH and if there were gravitons producing the gravity around the BH the effective graviton field would be centered on the center of mass (COM) of the one BH. If there were 2 BHs (non orbiting) and there were gravitons producing the gravity around the BHs the effective graviton field would be the summation centered on the center of mass (COM) of EACH of the BHs which effectively is the COM of both (the barycenter). If there were 2 BHs (orbiting each other) they would be binary black holes (BBH) and if there were gravitons producing the gravity around the BHs the effective graviton field would be the summation centered on the center of mass (COM) of EACH of the BHs. But now the center of mass (COM) of EACH of the BHs is moving in space relative to an observer so the observer will be getting waves of gravitons depending on the geometry of the BBH and the speed of the graviton. So from a very long way away it might appear for ease of calculation that the effect is from the COM (R in the above formulas #59 http://www.scienceforums.net/topic/94060-what-is-the-best-3d-description-of-gravitational-waves/page-3#entry913453)and the waves are spherical around this R). For the formulas can't be used to describe the waves closer in because they don't originate from the barycenter but the COM each mass forming the barycenter, and the distances to these COMs varies, depending on separation and orbital speed, (calculated by finding omega). So we would have a BBH geometry, waves (of gravitons) and the effect of gravitons on mass (yet to be discovered). I haven't mentioned path of a graviton - maybe we will work that one out in the thread on the Shapiro effect of BBHs. I would tend to think a graviton is not influenced by mass so the graviton always travel "straight" unlike light. No matter how massive a BH gets the gravitons still get out so the rules about gravitational time dilation can't apply to gravitons. BHs stop EM getting out but don't stop gravitons. Am I saying light and gravitons travel at different speeds? They seem too when coming out of BHs at least.

3 equations: [latex]\omega=\sqrt{G(m_1+m_2)/r^3}[/latex] [latex]h_{+} = -\frac{1}{R}\, \frac{G^2}{c^4}\, \frac{2 m_1 m_2}{r} (1+\cos^2\theta) \cos\left[2\omega(t - R/c)\right][/latex] [latex]h_{\times} = -\frac{1}{R}\, \frac{G^2}{c^4}\, \frac{4 m_1 m_2}{r}\, (\cos{\theta})\sin \left[2\omega(t-R/c)\right][/latex]

That's pretty amazing. Did he stay on as a teacher? You should have used your smartphone to report him.

There are rules against preaching. No rules against facts.

Try and answer the questions Phi for all posed to you. #34 http://www.scienceforums.net/topic/93659-the-theory-of-god/page-2#entry913404

Have you discussed this as a separate thread?

GR is a very complex calculation where for ease of finding solutions they make approximations, so from a large distance they would treat a spiral as a circle, or in this case a 3D spiral approximates a sphere. Now I would agree to that if the GW was from a point source. It is more than likely the source is the binary masses operating in unison so the source could likened to the two ends of a dumbbell, and hence a quadrupolar wave comes out, you won't get that quadrupolar wave from a point source. It is definitely not a point source. I was hoping from the analysis of those equations we would see if GR was treating it as a point source. Let me have some time to do the math please. I think they would but they can be neglected as they are even across the surface i.e both ends of the LIGO would have the same g due to the Earth. Strange has said the opposite because "they are nonlinear". The test masses in LIGO are always in the Earth's gravitational field, which is perpendicular to the LIGO arms so those parts are non linear. But the GW could come in from any angle so at times it won't be all nonlinear. If you were alongside a black hole the test masses would be affected by the gravity of the BH but since they would all accelerate toward the BH at the same rate. So the GW would still move the test particles in the same pattern. Thought for the day: if the 3D structure of a GW was a sphere or a circle from a point source all symmetrical rotating bodies would be able to produce GWs.

Is it a type of conservation of objects?

They seem to be predominantly those "one hit wonders", where they post a speculation but never come back to defend it. I regret not doing more maths at school and subsequent, for I do like it, but it doesn't come naturally. I am sure I think in terms of physics even without the maths. I am learning maths behind the scenes.

So you have to be willing to make mistakes, take the chance to be proven wrong. It is that inventive mind, you are looking for that variation that will make the difference, but you have to be able to express it, like it isn't a technical skill. To me it is a thought process, but not just trial and error. Good on you for being able to that Klaynos. Sounds like you are working on cutting edge stuff.

But that doesn't happen if the GW is a 3D spiral. One side of the point (the reduced ring of test particles) will feel a strain before the other. It will wobble. I haven't had to think it through too much, but if the plane wavefront approached along the z axis but the wavefront at an angle to the z axis one side of the point would be touched before the opposite side. Can you see that?