What is the best 3D description of Gravitational waves?

[imatfaal]

Rob

 

1.That last section of the equation gives you a nice oscillation (it is a pretty standard expression in waves) this is what it looks like

 

http://www.wolframalpha.com/input/?i=cos%5B2*(sqrt((6.67e-11)*(1.3e32)%2F(100000%5E3)))(t-4e16)%5D

 

[latex] \cos\left[2\omega(t - R/c)\right] [/latex]

 

Bear in mind - It is a cos or sin function so the answer will be between -1 and 1 whatever the inputs

 

2. This bit deals with the inclination above the orbital plane - actually the angle between the perpendicular (ie angular momentum direction) and the viewer

 

[latex] (1+\cos^2\theta) [/latex]

 

3. This is to do the overall strength [latex] \frac{G^2}{c^4}\, \frac{2 m_1 m_2}{r} [/latex]

 

4. And this is the fall off with distance [latex] -\frac{1}{R}[/latex] - you see it is one over distance not the one of distance squared of gravitation attraction

[mp][/mp]

 

You can also see that at no point in the equations is there a [latex]\phi[/latex] (with all due respect to our Chief Executive Offworlder) - so there is no variance in the amplitude in the plane of the orbit ( if we assume [latex]\theta[/latex] the polar does "latitude" and [latex]\phi[/latex] the azimuthal does "longitude"). You would need avariation in both [latex]\theta[/latex] and [latex]\phi[/latex] for a spiral

[Robittybob1]

 

 

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.

Yes I understand that now thanks. The period of a wave at that separation was 7.003 millisecs so within that short period of time the wave will make a full oscillation.

 

( does this seem right 1000 millisec = 1 sec.

 

1000 / wave frequency

 

=> 1000 / (142.7965894) = 7.002968378 millisecs are there enough waves to fill in the 0.2 sec (200 millisecs) chirp time?

Edited March 29, 2016 by Robittybob1

[Strange]

Yes I understand that now thanks. The period of a wave at that separation was 7.003 millisecs so within that short period of time the wave will make a full oscillation.

 

Note that the period of the wave at any distance is the same.

[swansont]

 

Note that the period of the wave at any distance is the same.

 

 

If the frequency is constant, which it isn't.

[Strange]

 

 

If the frequency is constant, which it isn't.

 

ah yes, very good point.

[Robittybob1]

Rob

 

1.That last section of the equation gives you a nice oscillation (it is a pretty standard expression in waves) this is what it looks like

 

http://www.wolframalpha.com/input/?i=cos%5B2*(sqrt((6.67e-11)*(1.3e32)%2F(100000%5E3)))(t-4e16)%5D

 

[latex] \cos\left[2\omega(t - R/c)\right] [/latex]

 

Bear in mind - It is a cos or sin function so the answer will be between -1 and 1 whatever the inputs

 

2. This bit deals with the inclination above the orbital plane - actually the angle between the perpendicular (ie angular momentum direction) and the viewer

 

[latex] (1+\cos^2\theta) [/latex]

 

3. This is to do the overall strength [latex] \frac{G^2}{c^4}\, \frac{2 m_1 m_2}{r} [/latex]

 

4. And this is the fall off with distance [latex] -\frac{1}{R}[/latex] - you see it is one over distance not the one of distance squared of gravitation attraction

[mp][/mp]

 

You can also see that at no point in the equations is there a [latex]\phi[/latex] (with all due respect to our Chief Executive Offworlder) - so there is no variance in the amplitude in the plane of the orbit ( if we assume [latex]\theta[/latex] the polar does "latitude" and [latex]\phi[/latex] the azimuthal does "longitude"). You would need avariation in both [latex]\theta[/latex] and [latex]\phi[/latex] for a spiral

I see at the bottom there is a value for "t" which I can't copy from the Wolfram page.. Imatfaal did you use a different separation than I was using? Did you use 100 km?

 

All the rest is understood but I'd like to mention that this is the simplified linear version. There was a mention before about another formula that takes it all into account.

This simplified formula was based on the barycenter staying fixed. With the mass loss occurring the barycenter will be changing. I covered that in another thread http://www.scienceforums.net/topic/93875-warped-spacetime-around-bh-and-the-barycenter/. The whole question was when does the mass loss occur and was it from both orbiting masses equally http://www.scienceforums.net/topic/93913-each-member-of-the-binary-produces-g-rad-at-an-equal-rate-why/?

Whether the change in barycenter will prescribe a spiral I'll have to consider. (With increasing mass loss the barycenter will follow a spiral, well it isn't staying in the same location in space at least.)

 

 

If the frequency is constant, which it isn't.

What I thought Strange meant was that at any distance closer to the BBH the frequency would be constant yet the amplitude would be greater. You are both right depending on what you are meaning.

[imatfaal]

I see at the bottom there is a value for "t" which I can't copy from the Wolfram page.. Imatfaal did you use a different separation than I was using? Did you use 100 km?

 

 

t is the variable of the equation - it runs from one figure to another continuously - for each t there is one and only one h_cross and h_plus. And yes I used 100km just to put a number in there

[Robittybob1]

 

t is the variable of the equation - it runs from one figure to another continuously - for each t there is one and only one h_cross and h_plus. And yes I used 100km just to put a number in there

From the Wolfram page "cos[2*(sqrt((6.67e-11)*(1.3e32)/(100000^3)))(t-4e16)]" So from that I take it the R/c = 4e16 or as I'd write it in Excel 4.0E+16?

 

What has happened to the word "sin" in the original equation (below)? I see the word "cos" is still there in Wolfram equation.

[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]

 

Sorry for asking such silly questions.

Edited March 29, 2016 by Robittybob1

[imatfaal]

From the Wolfram page "cos[2*(sqrt((6.67e-11)*(1.3e32)/(100000^3)))(t-4e16)]" So from that I take it the R/c = 4e16 or as I'd write it in Excel 4.0E+16?

 

What has happened to the word "sin" in the original equation (below)? I see the word "cos" is still there in Wolfram equation.

[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]

 

Sorry for asking such silly questions.

 

4e16=4.0E+16=4*10¹⁶

 

h_plus is cos and h_cross is sin

 

cos and sine will be at a different phase from each other - so for the same t you will get different amounts of h_plus and h_cross

[Strange]

What I thought Strange meant was that at any distance closer to the BBH the frequency would be constant yet the amplitude would be greater. You are both right depending on what you are meaning.

 

Exactly. The simple wave equation you are looking at, at the moment has a constant frequency at all times and distance. But in reality, the source frequency is changing so the picture is a little more complex.

[Robittybob1]

 

Exactly. The simple wave equation you are looking at, at the moment has a constant frequency at all times and distance. But in reality, the source frequency is changing so the picture is a little more complex.

What I have been thinking is to find the max value of the last part of the equation and then fill the rest of the equation with values to see if we get the strain value they had in the paper on discovering GW150914

 

The signal sweeps upwards in frequency from 35 to 250 Hz with a peak gravitational-wave strain of 1.0 × 10^−21

I'll have to go back and see if Imatfaal has given this to us already it could just be 1, but it will probably be lesser for the line of sight was not direct.

What is the max value for this part of the equation?

[latex](\cos{\theta})\sin \left[2\omega(t-R/c)\right][/latex]

 

It has cos and sin in the same equation is that correct? for the formula in the Wolfram page seemed to only has cos not both or were both there?

"cos[2*(sqrt((6.67e-11)*(1.3e32)/(100000^3)))(t-4e16)]"

Edited March 29, 2016 by Robittybob1

[Strange]

What is the max value for this part of the equation?

[latex](\cos{\theta})\sin \left[2\omega(t-R/c)\right][/latex]

 

The maximum value of that is 1.

 

The amplitude is given by point 3 in imatfaal's earlier breakdown. So it increases as r (the orbital radius) decreases.

 

(As this is an approximation, I'm not sure how accurate it will be at the extremes.)

Edited March 29, 2016 by Strange

[Robittybob1]

 

The maximum value of that is 1.

 

The amplitude is given by point 3 in imatfaal's earlier breakdown. So it increases as r (the orbital radius) decreases.

 

(As this is an approximation, I'm not sure how accurate it will be at the extremes.)

it increases but it will never go over 1. That is a bit contradictory. The frequency of the wave "increases as r (the orbital radius) decreases". So if I just make that part = 1 we can progress to the strength of the amplitude. If the amplitude is too high we then could take the line of sight into consideration.

 

[latex]h_{\times} = -\frac{1}{R}\, \frac{G^2}{c^4}\, \frac{4 m_1 m_2}{r}\,[/latex]

 

We have some values to put into all those terms.

G, c, M1, M2, and r = (350 km) R = "Luminosity distance 410 (+160 −180) Mpc" so there is a large error bar on the distance.

1 megaparsec = 3.086e+22 meters

300 2.80E-21

310 2.71E-21

320 2.63E-21

330 2.55E-21

340 2.47E-21

350 2.40E-21

360 2.34E-21

370 2.27E-21

380 2.21E-21

390 2.16E-21

400 2.10E-21

410 2.05E-21

420 2.00E-21

430 1.96E-21

440 1.91E-21

450 1.87E-21

460 1.83E-21

470 1.79E-21

480 1.75E-21

490 1.72E-21

500 1.68E-21

510 1.65E-21

520 1.62E-21

530 1.59E-21

 

First column is in megaparsecs and second is the hx value. They all seem too high. They reported "peak gravitational-wave strain of 1.0 × 10−21" but it is the right order.

I know what it is they calculated the max at the minimum separation. But doesn't make sense either as then these figures will really blow out.

 

Something really wrong here 1.06E-21 when r = 680 km

7.18E-21 when r = 100 km

strain, Km, meters

1.44E-20, 50, 50000

1.20E-20, 60, 60000

1.03E-20, 70, 70000

8.98E-21, 80, 80000

7.98E-21, 90, 90000

7.18E-21, 100, 100000

 

calculated strain seems to be 10 times to high.

Even if I take the distance to the max error limit and allow a 60 degree angle line of sight we are still 5 times too high.

 

This makes me think we have to adjust the mass as the energy is lost. Using the "RB Law" the mass loss as they fall is twice the final difference. So we can take 3 solar masses off each of the BHs as they fall. I wonder what happens then?

 

It helps but it doesn't fix the problem. Lower the angle of sight? There are so many variables that could be altered.

 

Can anyone see where I went wrong earlier?

I did get sensible values when the reduced masses were 15 and 8 so a final mass of 23 solar masses. Could they ever be that far out on their mass estimates? With those masses and a 60 degree line of sight at normal distance of 410 Megaparsecs we get the right amount of strain at LIGO.

1. What I haven't checked yet is whether those masses would orbit at the required rates, the frequencies that were recorded by LIGO.

M1 m2 M1+M2 r=((m1+m2)*G/w^2)^1/3 Omega

15 8 23 247555.666 448.6087162

 

By changing the orbital radius it is possible for lighter mass BHs to produce the same amplitude GW and the same frequency wave.

Edited March 30, 2016 by Robittybob1

[Robittybob1]

The maximum frequency was recorded at 250 Hz, that is the signal frequency, halve that and multiply it by 2*pi() to get the orbital velocity that gives us 785.4 radians/sec. Can we get lighter mass BHs to produce that sort of velocity?

For the above was not so good for when the lighter mass BHs were producing the right amplitude they weren't at the minimum separation. Can we get these other solutions?

 

The minimum separation is dependent on the Schwarzschild radius of each BH. So the smaller BHs had to get closer to merge.

[latex]r_\mathrm{s} < \frac{2 M G}{c^2},[/latex]

 

It is not so easy!

 

As the mass goes up the allowable separation increases due to the radius of the BH Schwarzschild increasing. You need a greater masses than what I found above to get these extreme orbital velocities.

I need to go back to the original paper and see why they chose the values of mass that they did.

Edited March 30, 2016 by Robittybob1

[imatfaal]

I'll have to go back and see if Imatfaal has given this to us already it could just be 1, but it will probably be lesser for the line of sight was not direct.

What is the max value for this part of the equation?

[latex](\cos{\theta})\sin \left[2\omega(t-R/c)\right][/latex]

 

It has cos and sin in the same equation is that correct? for the formula in the Wolfram page seemed to only has cos not both or were both there?

"cos[2*(sqrt((6.67e-11)*(1.3e32)/(100000^3)))(t-4e16)]"

 

Rob this is why these discussions are so frustrating. You are talking about your speculation being unchallenged in one thread - yet it becomes clear that you do not understand that cos or sine of anything can only vary between 1 and minus 1. This is like saying you should be playing for the NZ 2020 team in the Semi-final and then asking what this heavy lump of willow actually does.

 

so to answer your questions

 

[latex]\cos \theta[/latex] max value is 1 min value is minus 1.

 

[latex]\sin \left[2\omega(t-R/c)\right][/latex] max value is 1 min value is minus 1

 

obviously the max and min value of their product is the same.

 

I thought I had made it abundantly clear in my previous post that the [latex]\cos \theta[/latex] section deals with the change in the Amplitude of the certain polarisation with the change in the angle from the perpendicular whilst the [latex]\sin \left[2\omega(t-R/c)\right][/latex] section deals with the evolution of the function over time. You were claiming you did not see how such a small change in t could make any difference to the overall function - I hoped that the graph on wolfram alpha would show you that this formulation was the basis of a wave equation and in fact was very important in determining the sort of signal we expect

They all seem too high. They reported "peak gravitational-wave strain of 1.0 × 10−21" but it is the right order.

 

You know what theta is and you know you are taking either the sine or the 1+cos² so you know the value of the expression which deals with the change in Amplitude due to angle of observation. Take this into account - or have you already.

[Robittybob1]

 

Rob this is why these discussions are so frustrating. You are talking about your speculation being unchallenged in one thread - yet it becomes clear that you do not understand that cos or sine of anything can only vary between 1 and minus 1. This is like saying you should be playing for the NZ 2020 team in the Semi-final and then asking what this heavy lump of willow actually does.

 

so to answer your questions

 

[latex]\cos \theta[/latex] max value is 1 min value is minus 1.

 

[latex]\sin \left[2\omega(t-R/c)\right][/latex] max value is 1 min value is minus 1

 

obviously the max and min value of their product is the same.

 

I thought I had made it abundantly clear in my previous post that the [latex]\cos \theta[/latex] section deals with the change in the Amplitude of the certain polarisation with the change in the angle from the perpendicular whilst the [latex]\sin \left[2\omega(t-R/c)\right][/latex] section deals with the evolution of the function over time. You were claiming you did not see how such a small change in t could make any difference to the overall function - I hoped that the graph on wolfram alpha would show you that this formulation was the basis of a wave equation and in fact was very important in determining the sort of signal we expect

 

You know what theta is and you know you are taking either the sine or the 1+cos² so you know the value of the expression which deals with the change in Amplitude due to angle of observation. Take this into account

Thanks. I'm using Excel and I think I have understood how it works now. I have been playing around with the angle of observation to see what effects that has. I've never worked with this type of function before where you have to leave one value "t" as a variable.

So I am just looking at the maximum value of 1 and varying theta only.

 

Do you think in the calculation of omega (the angular velocity) that any part needs to be treated relativistically?

[swansont]

 

Do you think in the calculation of omega (the angular velocity) that any part needs to be treated relativistically?

 

 

It's possible. IIRC I've seen the frequency equation derived from Newtonian gravity, and assumes a circular orbit. So there may be additional corrections needed.

[Strange]

And it may be that those and other (presumably non-trivial) corrections are part of the reason this approximation does not give accurate results - especially as r gets smaller.

[Robittybob1]

And it may be that those and other (presumably non-trivial) corrections are part of the reason this approximation does not give accurate results - especially as r gets smaller.

The frequency they detected at LIGO had a max of 250 Hz. I converted that to angular velocity in radians per second. I then made up a simultaneous calculation adding the two Schwarzschild radii (Sr) of M1 and M2 and used M1 and M2 in the formula for omega solving for r.

The mass (M1 + M2) that gave the same result to both screens was 91.55 solar masses. The r value when the two Sr meet was around 270 km. It doesn't matter how you apportion those masses as long as the combined mass stays at 91.55 solar mass.

 

Initially I looked at an angle of sight of 60 degrees and varied the masses (proportions of 91.55) to see what proportion gave a measured strain of 1.0E-21 and the ratio was 10 and 81.55. Different degrees of line of sight resulted in these mass proportions needing to be adjusted. The greater the angle the more even split the two masses needed to be.

All these were at 410 Mps.

 

I now want to investigate the cosmological redshift factor of 0.09. Any suggestions as to how to handle that?

 

Primary black hole mass 36þ5 −4M⊙ Secondary black hole mass 29þ4 −4M⊙ Final black hole mass 62þ4 −4M⊙ Final black hole spin 0.67þ0.05 −0.07 Luminosity distance 410þ160 −180 Mpc Source redshift z 0.09 þ0.03 −0.04

Redshift means we are recording the frequencies slower than they were generated, is that right? Higher frequencies are going to demand even more mass in the BBH.

Do we have to look at both strain equations at the same time? Adding the strains together? The biggest determining factor of the mass is the omega calculation. If the Schwarzschild radii were not so critical and we were allowed to encroach on the BH Schwarzschild radius the masses could be reduced.

 

If the mass of a BH is at the singularity what difference does it make if the Schwarzschild radii touch?

(New thread http://www.scienceforums.net/topic/94222-what-difference-does-it-make-if-the-schwarzschild-radii-touch/)

Edited March 30, 2016 by Robittybob1

[swansont]

The frequency they detected at LIGO had a max of 250 Hz. I converted that to angular velocity in radians per second. I then made up a simultaneous calculation adding the two Schwarzschild radii (Sr) of M1 and M2 and used M1 and M2 in the formula for omega solving for r.

The mass (M1 + M2) that gave the same result to both screens was 91.55 solar masses. The r value when the two Sr meet was around 270 km. It doesn't matter how you apportion those masses as long as the combined mass stays at 91.55 solar mass.

 

 

 

But we know that mass is incorrect. If the formula is Newtonian and the speeds make the Newtonian equations inaccurate, it means the system has the kinetic energy of the Newtonian calculation at a lower frequency.

The r value when the two Sr meet was around 270 km. It doesn't matter how you apportion those masses as long as the combined mass stays at 91.55 solar mass.

 

 

No. The radius for the 36 solar-mass BH is about 106 km, so there is no way the sum of the two radii is as large as 250 km.

[Robittybob1]

 

 

But we know that mass is incorrect. If the formula is Newtonian and the speeds make the Newtonian equations inaccurate, it means the system has the kinetic energy of the Newtonian calculation at a lower frequency.

 

No. The radius for the 36 solar-mass BH is about 106 km, so there is no way the sum of the two radii is as large as 250 km.

I was just working with the formulas as presented. They may need corrections for the effects of relativity, but what factors will relativity affect?

I was working out the Sr for the 91.55 solar combined mass black hole = 270067 meters.

36 solar-mass Sr = 106198 meters

91.55 solar-mass Sr = 270067 meters

for GW150914 there were two BH of 36 and 29 solar-mass so the combined Sr = 191746 meters. so they couldn't get any closer than 191 km without their Schwarzschild radii touching. (Everything is idealised here, and there is no allowance for the Schwarzschild radii changing shape as the two BHs get closer to each other.)

 

Could you explain this again please?

"If the formula is Newtonian and the speeds make the Newtonian equations inaccurate, it means the system has the kinetic energy of the Newtonian calculation at a lower frequency."

 

Would the dimensions of the Schwarzschild radius be affected by relativity?

 

[ @moderators - do we have to start a new thread every time we get a new question? I don't mind how many threads there are but you might.]

Edited March 30, 2016 by Robittybob1

[swansont]

I was just working with the formulas as presented. They may need corrections for the effects of relativity, but what factors will relativity affect?

I was working out the Sr for the 91.55 solar combined mass black hole = 270067 meters.

 

 

 

What system has a combined 91 solar masses? This hasn't come up before.

[Strange]

Quite. 36+29=65, not 91.

[Robittybob1]

 

 

What system has a combined 91 solar masses? This hasn't come up before.

If you use the omega value for the GW150914 maximum value of 785.4 radians/sec. You can only get that orbital velocity if the masses add to 91 solar masses and at an r value of when their Sr are just touching. I worked out this value from using the formulas as presented. I'm not saying that is the mass involved in any system but the value that comes out of the maths.

Now we are trying to explain why it is different to the values given to GW150914.

Quite. 36+29=65, not 91.

I had already told you that #123

 

for GW150914 there were two BH of 36 and 29 solar-mass so the combined Sr = 191746 meters.

Edited March 30, 2016 by Robittybob1

[Strange]

Now we are trying to explain why it is different to the values given to GW150914.

 

I assume because you are using an approximation. These tend to get less accurate at the extremes.