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What is the best 3D description of Gravitational waves?


Robittybob1

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to a layman it seems like they are closely related though. Is the connection that mass (et al) curves spacetime and that gravitational waves (as an analogy) "pass a current" through that curvature?

 

 

 

But you could have this effect in flat spacetime as well. AFAICT it doesn't require that another source of gravity be present. The LIGO test masses don't oscillate in the direction of the earth's gravitational pull. It's a bystander in this experiment.

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But you could have this effect in flat spacetime as well. AFAICT it doesn't require that another source of gravity be present. The LIGO test masses don't oscillate in the direction of the earth's gravitational pull. It's a bystander in this experiment.

That is true they are suspended as pendulums, so they need the earth's gravitational pull to keep them hanging steady.

 

That is the question I thought was being asked. They do combine, but not in a simple way like light waves.

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.

Edited by Robittybob1
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Would it be more like another type of wave?

 

No. It would require the application of the Einstein Field Equations. I think you do get constructive and destructive interference but not in the simple way that other waves can be summed.

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No. It would require the application of the Einstein Field Equations. I think you do get constructive and destructive interference but not in the simple way that other waves can be summed.

#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.

Edited by Robittybob1
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You could do that without gravity, though. You can do this in space.

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.

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#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.

 

You cannot simply apply that to the combination of two waves.

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But you could have this effect in flat spacetime as well. AFAICT it doesn't require that another source of gravity be present. The LIGO test masses don't oscillate in the direction of the earth's gravitational pull. It's a bystander in this experiment.

I only introduced the extra black hole for purposes of demonstration. I concede that it doesn't add anything essential to the situation since spacetime seems to be always curved to one degree or another .

 

Perhaps there is too much about black holes in the daily news these days and it has got to me. :embarass:

 

The test masses you bring up do not oscillate in the direction of the earth but the presence of the Earth does affect their motion does it not?

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You cannot simply apply that to the combination of two waves.

Why not?

....

 

The test masses you bring up do not oscillate in the direction of the earth but the presence of the Earth does affect their motion does it 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.

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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.

 

 

No, not pendulums. But you can do this in space.

http://lisa.nasa.gov

The test masses you bring up do not oscillate in the direction of the earth but the presence of the Earth does affect their motion does it not?

 

Yes. They refer to it as noise, and it's a limit on the sensitivity of the device.

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Nice your finally posting latex. Well done now post your understanding of those equations.

 

PS +1

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]?

 

You must link to sources that you cite. Further, animations may only be conceptual, and those would not be to scale. Especially in Anstronomy/Astrophysics, where you couldn't even put a picture of our solar system up and have everything be portrayed to scale.

 

And we are not talking about the same model. My model is General Relativity. You are proposing something else, so your model can't be the same as mine.

 

 

It's been mentioned multiple times that the GWs emanate from the system rather than the individual BHs. Unless you can point to some reference says there is interference or provide a model to us, you need to drop it from the discussion.

 

 

The part that says linear approximation shows the wave equation. The generic function A contains the three spatial coordinates, so it works in 3D.

 

Pretty ugly, no? That's the math you are trying to reverse-engineer from simulations.

 

Linear approximation https://en.wikipedia.org/wiki/Gravitational_wave#Linear_approximation

 

 

 

Linear approximation

The equations above are valid everywhere — near a black hole, for instance. However, because of the complicated source term, the solution is generally too difficult to find analytically. We can often assume that space is nearly flat, so the metric is nearly equal to the [latex]\eta^{\alpha \beta} \[/latex] tensor. In this case, we can neglect terms quadratic in [latex] \bar{h}^{\alpha \beta} \[/latex], which means that the [latex]\tau^{\alpha \beta} \[/latex] field reduces to the usual stress–energy tensor [latex]T^{\alpha \beta} \[/latex]. That is, Einstein's equations become

[latex]\Box \bar{h}^{\alpha \beta} = -16\pi T^{\alpha \beta} \[/latex], .

If we are interested in the field far from a source, however, we can treat the source as a point source; everywhere else, the stress–energy tensor would be zero, so [latex]\Box \bar{h}^{\alpha \beta} = 0 \[/latex], .

Notice it is a linear approximation so unless we can get the equations that deal with near the black hole we are still stuck.

Edited by Robittybob1
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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

 

T is the stress energy/momentum tensor.

https://en.m.wikipedia.org/wiki/Stress%E2%80%93energy_tensor

 

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)?

Edited by Robittybob1
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No it's the term that describes the energy/momentum of particles. It's not a time component.

 

It covers how particles behave within the field.

 

Actually I think I may have gotten that equation mixed up.

 

There is no coordinate on t. The stress tensor requires tensor coordinates.

 

T in this formula is time not the stress tensor. My apologies.

 

(Yeah the wiki page uses the Greek symbol for the stress tensor)

Edited by Mordred
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No it's the term that describes the energy/momentum of particles. It's not a time component.

 

It covers how particles behave within the field.

 

Actually I think I may have gotten that equation mixed up.

 

There is no coordinate on t. The stress tensor requires tensor coordinates.

 

T in this formula is time not the stress tensor. My apologies.

 

(Yeah the wiki page uses the Greek symbol for the stress tensor)

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?

Edited by Robittybob1
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No, the frequency is [latex]2\omega[/latex]

was "t" measured in units of time? You said before "It is how the strain changes with time" so what are the units of that?

 

I need some sort of value to put into the equation for GW150914. Have you any idea of what it would be? (even roughly)

 

[latex]\omega=\sqrt{G(m_1+m_2)/r^3}[/latex] I'll plug some values into that and see if the frequency is twice that.

Edited by Robittybob1
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was "t" measured in units of time? You said before "It is how the strain changes with time" so what are the units of that?

 

It is time. So if at 12:00 the strain is 0, then this equation tells you what it will be 1 second (or 1 nanosecond, or 1 minute) later.

 

This is the standard equation for describing a wave travelling through space [latex]\sin(f (t-x/v))[/latex]; where f = frequency, t = time, x = position and v = velocity (of the wave). You can see how this parallels (that part of) the equation above.

 

I need some sort of value to put into the equation for GW150914. Have you any idea of what it would be? (even roughly)

 

The frequency is changing all the time as the system evolves. So all you would have to choose some frequency. But all the other values are known, at least approximately ([latex]\theta[/latex] is probably the most uncertain).

 

[latex]\omega=\sqrt{G(m_1+m_2)/r^3}[/latex] I'll plug some values into that and see if the frequency is twice that.

 

That is the orbital frequency. Which is why there is [latex]2 \omega[/latex] in the wave equation.

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These seem to match the figures for GW150914.

 

Mass1 = 36 solar mass = 7.1604E+31 kg

Mass2 = 29 solar mass = 5.7681E+31 kg

Separation 350 km = 350000 meters

Omega = 448.6087

divided by 2 pi() = 71.3982947 Hz (orbital frequency at separation of 350 km)

wave frequency = 142.7965894 Hz

So knowing that what is "t" ?

 

 

I take omega to be the angular velocity not the orbital frequency.

Edited by Robittybob1
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So knowing that what is "t" ?

 

It is the time at which you want to know the strain (h). The strain changes over time with the waveform.

 

I take omega to be the angular velocity not the orbital frequency.

 

You are right. It is radians/sec not orbits/sec. So that is why the 2 is there.

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[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]R/c[/latex] is the distance to the BBH 1.3 billion ly in meters. and c the speed of light in m/sec so what will that give us? 1.3 billion years as a time? How can you add a millisecond to that?

 

Is t just 200 milliseconds greater than [latex]R/c[/latex] ?

 

Maybe I've got those the wrong way around? "t" might be the 1.3 billion years as a time, and R/c another time closer in towards the BBH. Does that sound right?


I think I'm understanding it now. At that frequency the amplitude at any one moment will depend on how far it has gone through a cycle.

 

From previous discussions the angle of sight was about 60 degrees wasn't it?

Edited by Robittybob1
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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?

 

 

No, it's not the period of the orbit, it's the time variable. It describes how the wave at any given position changes with time.

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I think I'm understanding it now. At that frequency the amplitude at any one moment will depend on how far it has gone through a cycle.

 

That's it. And how far through the cycle you are (the amplitude of h) depends on the time (because it is a constantly changing value) and where you are (because it is passing by you). The (t - R/c) bit captures both those aspects.

 

From previous discussions the angle of sight was about 60 degrees wasn't it?

 

Correct.

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