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Can LIGO actually detect gravitational waves?


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#221 swansont

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Posted 19 June 2017 - 12:53 PM

swansont,
 
I admit it is real handwavy but it goes to the consideration of how confident are we in the data.  That is, how "clear" is the data.  Can we see signals within the signal that we are not expecting, or can we only see what we are expecting to see?
 
My "math" to suggest one might be able to pickup a gw crossing another gw 1 ly away is a general figuring that a gw gradient stayed coherent after traveling 1.3billion lys and is "spread out" in terms of power in an inversed cubed type of way that results in the wave (initially with the force of the mass of multiple suns turned instantly to energy) only having the power to distort space the width of a partial proton over the distance leveraged by the experiment, yet we can read the effect that this distortion has on space through monitoring the interference of some EM waves that took two orthogonal routes to the detector...whereas the gw we sense today HAD to either have been crossed by or is soon to be crossed by another GW that we just saw or are about to see.  So the impulse, the wave of one gw should be embedded in the next.


Not in any measurable way, as has been discussed. The chirp is a light-second in length. 3 x 10^5 km. If there's any effect, you stretch or shrink that by a part in 10^-18. The pulse becomes ~10^-16 m shorter or longer. Its duration changes by nanoseconds. Undetectable.
 

And this fluctuation, this "interference" had to happen within a couple lys of here because the two waves crossed earth within a year of each other.  So space is curved within curved space, and the interaction should still have some coherence because relatively few other GWs have had a chance to add or subtract their impulse, and the interaction had to be a billion times closer than the original gradients.

 
Undetectable.
 

But again, I am trying to box in the answer to the thread question.  Either we can pick up GWs or we can't.  If we can, then we can make further study of them, and possibly use them as probes of space...able to in essence "feel" the universe around us.  And while having a fourth, or building  12, would be stupid if we cannot detect GWs, if we CAN detect GWs we should be able to learn a lot more about gravity than we know today, by studying in detail, GW waves.    If we can see a strong one from 1.3 billion lys, mathwise, we should be able to see a weak one from 1 ly.

 
There will be no GWs from a LY away since there isn't any appreciable mass at that distance, doing anything that would emit them. A little while ago we discussed scaling. Perhaps you could go back and review that. You need a huge amount of mass, and appreciable acceleration, to get anything we could detect. Only a handful of the binary pulsars we can observe emit enough gravitational radiation to potentially see their orbits degrade over the years, and that's not nearly enough to detect with LIGO.
 

I am having a hard time, mentally. figuring what the intersection of two GWs would look like in space.  It would not be one event, but an infinite number of events happening continually, at an infinite number of locations, within, at a particular instant, that area of space where the two expanding shells coexisted.  So while one GW passes through a spot on Earth in a sec, another portion of that expanding shell has to already be passing through another "close" GW...so depending on how precisely we can see one wave embedded in the prior, or the next,  we might, with 12 detectors, be able to map gravity waves that passed through already, before we had any detectors up. (by reading the record of other gws on the passing one)


Ripples on a pond might be a good starter for visualizing interference. The waves do not become embedded. You have no basis for continuing to make that claim.


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#222 tar

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Posted 19 June 2017 - 12:53 PM

so the strong wave would be in a plane less than 6 thousand miles thick, expanding out in that plane, the chances of that thin ring intersecting another thin ring of the same sort might be good, but it would only happen twice, for any two rings, max and this is definitely not enough interaction to be seen from 1ly which is the closest such an interaction could be and its imprint would NOT be on the part of the next ring that came through Earth from another direction...so SwansonT you are absolutely correct, there is no way to sense an intersection of two gws 1ly away.  My expanding shell model is not how it would be.  It is expanding rings.


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#223 swansont

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Posted 19 June 2017 - 12:58 PM

if the gradient energy stayed on the plane, mostly coherently for billions of lys, then the thickness of the plane would be max the diameter of the larger of the two BHs, and the diameter of the larger of the two had to be much less than the travel path of either mass during that last second which at .6C would be 118,000 miles

 

 

Where are you getting 118,000 miles?


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#224 tar

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Posted 19 June 2017 - 12:58 PM

.6c


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#225 swansont

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Posted 19 June 2017 - 01:17 PM

.6c

 

 

That's a speed, not a distance. How are you getting that number as a distance regarding two black holes orbiting each other?


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#226 tar

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Posted 19 June 2017 - 01:21 PM

I am sorry for the misdirection.   The validity of our gravity wave detection is in NO WAY challenged by an inability to read another GW embedded in the signal.  There is very little chance that such a signal is even present (even if two rings cross paths it is not likely they do so at the same time...that is one ring could pass were another was or will be, but in the expanse of space, the chance of meeting is near nil), much less that it would be strong enough to register on the equipment.  The use of gravity waves as plumbs of the space they transversed, is likewise lessened in likelihood. 


"That's a speed, not a distance."

 

but the whole signal, the 20 revolutions prior the merge happened in a second so the path either took to get around each other could not be longer than 118,000 miles   

 

divide that by pi and that by 20 and the diameter of the larger mass has to be less than that


Edited by tar, 19 June 2017 - 01:31 PM.

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#227 imatfaal

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Posted 19 June 2017 - 01:25 PM

 

 

No assumptions are made. The interaction is modelled to find the form that the gravitational waves take. The waves are strongest in the (equatorial) plane of the orbits and weakest in the direction of the axis of the orbits. (This may complicated slightly by the spin of the black holes as I think it depends on the total angular momentum of the system, not just the axis of the orbits.)

 

(OK. A few assumptions are made in the analysis: that GR is correct, is the main one. Also that the orbits are nearly circular.)

 

You sure about the intensity variation Strange?  Going back through my memory I seem to think that h_0 (on the axis of rotation) was strongest and that anywhere else you got a varying combo of h_+ and h_x  both of which were related to h_0 by being multiplied by sines and cosines of angles of inclination (ie necessairly less than 1)

 

Unfo I cannot find any docs on this at present.  My gut instinct is that you are correct - but memory is clear


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#228 Strange

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Posted 19 June 2017 - 01:54 PM

so the strong wave would be in a plane less than 6 thousand miles thick, expanding out in that plane

 

 

No.


divide that by pi and that by 20 and the diameter of the larger mass has to be less than that

 

The larger mass had a diameter of about 200 km. So you seem to be out by a factor of 1,000.


 

You sure about the intensity variation Strange? 

 

No!

 

I have just had another look at the parameter estimation paper but that just confused me. Maybe you can make more sense of it!

https://arxiv.org/abs/1602.03840


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#229 swansont

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Posted 19 June 2017 - 02:47 PM

I am sorry for the misdirection.   The validity of our gravity wave detection is in NO WAY challenged by an inability to read another GW embedded in the signal.  There is very little chance that such a signal is even present (even if two rings cross paths it is not likely they do so at the same time...that is one ring could pass were another was or will be, but in the expanse of space, the chance of meeting is near nil), much less that it would be strong enough to register on the equipment.  The use of gravity waves as plumbs of the space they transversed, is likewise lessened in likelihood. 


"That's a speed, not a distance."

 

but the whole signal, the 20 revolutions prior the merge happened in a second so the path either took to get around each other could not be longer than 118,000 miles   

 

divide that by pi and that by 20 and the diameter of the larger mass has to be less than that

 

 

0.6c * 1 sec / 20 = 9,000 km.   (0.6c is 111,600 miles/sec)

 

divide that by 2π and you get ~1400 km in radius

 

We went through this a few pages back. You agreed the orbit was, at largest, a few thousand miles in diameter.


 

You sure about the intensity variation Strange?  Going back through my memory I seem to think that h_0 (on the axis of rotation) was strongest and that anywhere else you got a varying combo of h_+ and h_x  both of which were related to h_0 by being multiplied by sines and cosines of angles of inclination (ie necessairly less than 1)

 

Unfo I cannot find any docs on this at present.  My gut instinct is that you are correct - but memory is clear

 

http://www.tapir.cal...ve_details.html

"The strongest radiation is along this rotation axis"

 

(This came up in another thread http://www.sciencefo...e-2#entry909493 )

 

 

Dipoles don't radiate along their axis, but quadrupoles do.


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#230 tar

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Posted 19 June 2017 - 04:06 PM

​"If we integrate this flux over a sphere around the source, we get the total luminosity, or energy emitted per unit time. The result cannot depend on the TT quadrupole moment, since "tranverse" can only refer to a specific direction of propagation. Instead it depends just on the traceless quadrupole moment IT, whose components are:"

 

SwansonT,

 

So what does that mean? Is the energy of the strain propagated only edge on, only in the direction of the axis or both in some combination where being 45 degrees from either position you would get some lesser but calculable energy?

 

That is the luminosity seems to be figured on a sphere, but the wave propagates in certain directions, not all directions.

 

Regards, TAR 


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#231 swansont

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Posted 19 June 2017 - 04:22 PM

​"If we integrate this flux over a sphere around the source, we get the total luminosity, or energy emitted per unit time. The result cannot depend on the TT quadrupole moment, since "tranverse" can only refer to a specific direction of propagation. Instead it depends just on the traceless quadrupole moment IT, whose components are:"

 

SwansonT,

 

So what does that mean? Is the energy of the strain propagated only edge on, only in the direction of the axis or both in some combination where being 45 degrees from either position you would get some lesser but calculable energy?

 

That is the luminosity seems to be figured on a sphere, but the wave propagates in certain directions, not all directions.

 

Regards, TAR 

 

 

The equations that follow the quoted bit tell you how the amplitude varies with angle.


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#232 tar

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Posted 19 June 2017 - 04:52 PM

no doubt

 

I was not able to see it, and was hoping for an English translation.


Stange had said one of the GWs we saw was 30 degrees from edge on 


which is also 60 degrees from axial

 

what percentage of the energy is going axially and what percentage is going edge on, in the case of GW150914?


I am trying to think in terms of radiation count, as if gravity was quantized into gravitons.

 

Some distant stars we "see" we piece together photon by photon over time.   We have no such luxury in this case, as all the gravitons were released in particular directions within about a second.  


Edited by tar, 19 June 2017 - 04:42 PM.

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#233 imatfaal

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Posted 19 June 2017 - 06:45 PM

No!

 

I have just had another look at the parameter estimation paper but that just confused me. Maybe you can make more sense of it!

https://arxiv.org/abs/1602.03840

 

Gotta luv an honest answer. 


...

http://www.tapir.cal...ve_details.html

"The strongest radiation is along this rotation axis"

 

(This came up in another thread http://www.sciencefo...e-2#entry909493 )

 

 

Dipoles don't radiate along their axis, but quadrupoles do.

 

Yes - that brings it all back.  Robittybob questioning whether AstroKatie knew what she was talking about - seeing the female bit before the astrophysicist bit.

 

And I was looking at that very Caltech page a few days ago - must have sunk in without me realising.  Thanks


 

 

I have just had another look at the parameter estimation paper but that just confused me. Maybe you can make more sense of it!

https://arxiv.org/abs/1602.03840

 

Using papers nomenclature

 

If we call  A_{GW} the maximum possible amplitude, the max amplitude at any time t  we denote as  A_{GW}(t)

 

 and we call the angle that the observers line of sight makes with the axis of rotation  \iota

 

then we can say (ignoring the phase) that

 

 h_+ =  A_{GW}(t) \cdot (1+cos^2 \iota) \cdot (phase\  angle)

 

 h_x = A_{GW}(t) \cdot (-2 \cos{\iota}) \cdot (phase\ angle)

 

 (1+cos^2 \iota)\ and\ (-2 \cos{\iota}) are the important bits which show how the amplitude varies with elevation.  Both will be at a maximum when  \cos{\iota} is equal to 1 or -1 and that will be the case when \iota = 0 or \pi - ie when you are on the axis

 

Because of the way the phase and magnitude are set when  \iota equals zero (ie along axis of rotation)  the gravitational waves are not only most intense they are circularly polarized (along the line of propagation the peaks of the waves would trace a helical path - I guess a double helix).

 

When  \iota equals  \pi/2   (ie in the plane of rotation) the gravitations waves are linearly polarized ( along the line of propagation the peaks would trace straight lines

 

When  \iota is between the two values the waves would be elliptically polarized (which is kinda mixture of both)


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#234 tar

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Posted 19 June 2017 - 07:49 PM

Well thank you Imatfaal.  I do comprehend English words better than Greek letters standing for whole bunches of English words.  But I still have the percentage question.  The highest amplitude waves are going out, in a equatorial disk, the thickness initially of  the diameter of the larger BH, whereas the axial power is going out at a max amplitude in exactly polar directions which I suppose is in two directions something like a barbershop pole coming out top and bottom, with two spirals on it, but the diameter of the orbit that put out the wave.  Drawing a hypothetical sphere around the event, most of the power would hit the equator, not the poles.    I am visualizing a couple degree size moon being the area of half the power going polar and a degree wide band on the horizon going 360 degrees around, putting maybe 100 times more power out on the equator than on the axis.


If you were under water, had a stick and you put it straight above your head and turned around in a little circle it seems that would take a lot less energy than holding the stick out making a big circle... but maybe I am thinking about conservation of angular momentum which probably does not apply here.


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#235 imatfaal

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Posted 19 June 2017 - 09:03 PM

Well thank you Imatfaal.  I do comprehend English words better than Greek letters standing for whole bunches of English words. ...

 
Obviously not that well - because your conclusion was this "The highest amplitude waves are going out, in a equatorial disk,"  which is the exact opposite of what I was explaining.  The highest amplitude are axial - ie at the poles not on the equator. 
 

the thickness initially of the diameter of the larger BH


Do you have any conception of the scales we are talking about? And we are dealing in mathematically ideals - the lowest amplitude linearly polarized radiation is on the plane of orbit; in reality you get what you get and any observations will have errors so high that plane or band are immaterial. 

 

whereas the axial power is going out at a max amplitude in exactly polar directions which I suppose is in two directions something like a barbershop pole coming out top and bottom, with two spirals on it, but the diameter of the orbit that put out the wave.

What the hell is axial power?  Could you go away and read a primer on Gravitational waves?  And again with the scale thing - who cares?  Mathematically it is the axial direction that matters and observationally everything is too blurred to be important one way or another.

 

Drawing a hypothetical sphere around the event, most of the power would hit the equator, not the poles.    I am visualizing a couple degree size moon being the area of half the power going polar and a degree wide band on the horizon going 360 degrees around, putting maybe 100 times more power out on the equator than on the axis.
  I can dig out the equation which gives you the power for a given area -ie dE / dA dt but it is complex .  But no - for any given area the closer the the poles the more the energy transferred per second.  IF you take a big area near the plane and a small area near the poles then yes more power at the plane - but for equal areas no.

 

If you were under water, had a stick and you put it straight above your head and turned around in a little circle it seems that would take a lot less energy than holding the stick out making a big circle... but maybe I am thinking about conservation of angular momentum which probably does not apply here.

 

Angular momentum does apply here - gravitational waves rob the system of angular momentum as well as energy.  this is what first allowed us an indirect confirmation of their existence - the taylor hulse binary . 

 

But I am not getting into discussing hokey analogies - this is heavy duty physics; Einstein can get away with homespun thought experiments because he understood the real theory - but you cannot really work in the opposite direction


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#236 tar

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Posted 19 June 2017 - 09:30 PM

imatfaal,

 

"Obviously not that well - because your conclusion was this "The highest amplitude waves are going out, in a equatorial disk,"  which is the exact opposite of what I was explaining.  The highest amplitude are axial - ie at the poles not on the equator."

 

 

I thought your equations meant the highest amplitude H+ would be at the equator and the highest amplitude Hx would be at the poles, and my question was simply in GW150194 what percentage of the energy went out the top and bottom and what percentage went out on or near the equatorial disk.

 

Regards, TAR
 


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#237 Mordred

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Posted 19 June 2017 - 09:43 PM

No I posted previously h+ is positive polarity and h× is cross polarity.
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#238 tar

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Posted 19 June 2017 - 10:06 PM

OK I really do have to reread a bunch of stuff, I am completely confused.  I thought, in the LIGO chart, the amplitude up on the black lines was considered H+ and the amplitude down was considered Hx.  I was trying to visualize what that meant, in regards, to the merger and am now completely clueless.

 

What was that about space squishing along one arm of the LIGO while stretching on the other?  Are not H+ and Hx inversely related on the 90 degrees? The amplitude of the one high while the other low?


better yet, I will bow out...this stuff is obviously way beyond me


Edited by tar, 19 June 2017 - 10:02 PM.

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#239 Mordred

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Posted 19 June 2017 - 11:04 PM

Ok with electric field polarity is the vectors of charge.

With magnets polarity is vectors of poles (north, south)

Both above cases are dipolar.

GW is quadrupole instead of two vectors you have 4 (per wave cycle).

These occur regardless of propogation direction. They do not describe the direction the wave is travelling in. Though using the above one can calculate direction of travel.

Lets try this. Take a telegraph machine. Each bip on your morse code is a chirp. Each chirp will be dipolar (neg to positive and positive to negative). The signal still radiates outward depending on the emiiter antennae. Omnidirectional vs directional)

Now apply that to a GW wave each chirp corresponds to a change in angular momentum of the two BH's. A consequence of the conservation of angular momentum. The loss momentum is your GW wave. The chirp rate depends on the orbit changes.

Each chirp will have a quardupole polarity.
Here is the chirp mass formulas for a binary system (specifically).

https://en.m.wikiped...wiki/Chirp_mass

See image 5 for how a GW transverse polarity wave propogates in 3d. ( best image I could find) though still misleading somewhat all images tend to be in GR lol
http://www.thephysic...tational-waves/

Edited by Mordred, 19 June 2017 - 11:04 PM.

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#240 swansont

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Posted 20 June 2017 - 09:54 AM

OK I really do have to reread a bunch of stuff, I am completely confused.  I thought, in the LIGO chart, the amplitude up on the black lines was considered H+ and the amplitude down was considered Hx.  I was trying to visualize what that meant, in regards, to the merger and am now completely clueless.

 

What was that about space squishing along one arm of the LIGO while stretching on the other?  Are not H+ and Hx inversely related on the 90 degrees? The amplitude of the one high while the other low?

 

 

Amplitude is how big the excursion is. The direction is irrelevant.  Stretching vs compression is not big vs small; the amplitude is how much you have of each as measured from some equilibrium point.


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