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Can LIGO actually detect gravitational waves?


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#1 aramis720

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Posted 2 June 2017 - 08:55 PM

I have a basic question that I've never actually seen answered in discussions of LIGO and gravitational waves: if these waves are warping space itself (actually spacetime), then all matter occuping that space will be warped to exactly the same degree that space is warped, making such warps in principle undetectable. So if an interferometer like LIGO, with two perpendicular arms, is set up to measure such waves, what is it actually measuring? Any distortion of the arms in the direction of the waves will not be detected because that arm(s) will be distorted to exactly the same degree that space itself is distorted. Help? 


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#2 iNow

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Posted 2 June 2017 - 09:22 PM

We've already detected gravitational,waves with LIGO three separate times now. Here's a primer on how: https://www.ligo.cal.../page/ligos-ifo
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#3 Strange

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Posted 2 June 2017 - 10:26 PM

I have a basic question that I've never actually seen answered in discussions of LIGO and gravitational waves: if these waves are warping space itself (actually spacetime), then all matter occuping that space will be warped to exactly the same degree that space is warped, making such warps in principle undetectable. So if an interferometer like LIGO, with two perpendicular arms, is set up to measure such waves, what is it actually measuring? Any distortion of the arms in the direction of the waves will not be detected because that arm(s) will be distorted to exactly the same degree that space itself is distorted. Help? 

 

 

This is actually a really good question. What is being measured is the travel time of light along the two arms. The speed of light is not affected by the gravitational waves, so the minute difference in the time to travel along each arm can be measured (as a phase difference).

 

More here: http://stuver.blogsp...ssed-by-gw.html


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#4 aramis720

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Posted 2 June 2017 - 11:45 PM

We've already detected gravitational,waves with LIGO three separate times now. Here's a primer on how: https://www.ligo.cal.../page/ligos-ifo

Thanks, but it doesn't address the point I'm asking about: if space itself is waving/warping, how can any interferometer, no matter how long, detect it? It seems to me pretty clear that it can't, in principle, because any attempt to measure it will use tools that are distorted by exactly the same amount as space itself is warped. No? 


 

 

This is actually a really good question. What is being measured is the travel time of light along the two arms. The speed of light is not affected by the gravitational waves, so the minute difference in the time to travel along each arm can be measured (as a phase difference).

 

More here: http://stuver.blogsp...ssed-by-gw.html

Thanks, but I don't see how measuring the speed of light gets around the problem I'm referring to. Any speed of light refers to distance traveled divided by time, but if that distance is distorted by gravitational waves any measuring instrument will also be distorted by exactly the same amount, making such detection in principle impossible, or so it seems to me. Or am I missing something? 


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#5 Mordred

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Posted 3 June 2017 - 02:30 AM

The length of the detector arms must be of sufficient length to encompass a quarter gravitational wave. In order to understand this you must first be aware of the nature of a quadrupole wave.

Ligo can only detect a small range of gravitational waves just as an antennea can be tuned to pick up certain frequencies by increasing or decreasing the antenna length.

In the GW wave you have simultaneous x and y changes. As the x axis contracts the y axis expands and vise versa. This is specifically why the L shape is required. The deviations of the x and y axis of the GW wave induces strain which is what is being measured.

Thanks, but it doesn't address the point I'm asking about: if space itself is waving/warping, how can any interferometer, no matter how long, detect it? It seems to me pretty clear that it can't, in principle, because any attempt to measure it will use tools that are distorted by exactly the same amount as space itself is warped. No? 

Thanks, but I don't see how measuring the speed of light gets around the problem I'm referring to. Any speed of light refers to distance traveled divided by time, but if that distance is distorted by gravitational waves any measuring instrument will also be distorted by exactly the same amount, making such detection in principle impossible, or so it seems to me. Or am I missing something? 


You have to look at the type of strain spin 2 quadrupole waves induces. The animation on this page shows the simultaneous movements.

https://en.m.wikiped...vitational_wave

Using a L shape detector allows you to measure both the x and y axis changes.
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#6 aramis720

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Posted 3 June 2017 - 02:50 AM

The length of the detector arms must be of sufficient length to encompass a quarter gravitational wave. In order to understand this you must first be aware of the nature of a quadrupole wave.

Ligo can only detect a small range of gravitational waves just as an antennea can be tuned to pick up certain frequencies by increasing or decreasing the antenna length.

In the GW wave you have simultaneous x and y changes. As the x axis contracts the y axis expands and vise versa. This is specifically why the L shape is required. The deviations of the x and y axis of the GW wave induces strain which is what is being measured.

You have to look at the type of strain spin 2 quadrupole waves induces. The animation on this page shows the simultaneous movements.

https://en.m.wikiped...vitational_wave

Using a L shape detector allows you to measure both the x and y axis changes.

I guess I'm still not seeing how the interferometer arm is detecting anything at all b/c if a gravitational wave is defined (as it is) as a ripple in spacetime itself, then anything in that ripple of spacetime will be distorted exactly the same degree to which spacetime is distorted. So how can an interferometer arm(s) occupying that spacetime detect the wave in any way? 


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#7 Mordred

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Posted 3 June 2017 - 02:55 AM

The ripples are in different directions. Hence the required length to frequency ratio and shape of the arms.

Picture a ball squeeze it, the sides not being squeezed expand. Then next cycle this reverses. This action induces strain.

Did you read the wiki section on "effects of passing?" In essence you are measuring changes in the detector arms lengths.

\delta L(t)=\delta L_x-\delta L_y=h L(t) where h is the measured strain.

Edited by Mordred, 3 June 2017 - 03:14 AM.

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#8 StringJunky

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Posted 3 June 2017 - 06:51 AM

Time varies, relatively, in different parts of a varying gravitational gradient, which is what a gravitational wave is, so,  the arms will alternately shrink and stretch slightly as the wave passes through; as measured by the time it takes for the laser to traverse the length of each arm. Relative to each other, there will be a small difference in the time between them.


Edited by StringJunky, 3 June 2017 - 06:59 AM.

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#9 geordief

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Posted 3 June 2017 - 07:28 AM

I can see why  the OP  may still  think his question has not been answered. I too cannot understand why the outcome is as it is (and that all deviations in Spacetime should  not be somehow "mirror-imaged out" )

 

However  the GW clearly is detectable and has been detected  exactly as predicted.

 

I imagine that my lack of understanding is very simply down to  a lack of understanding and familiarity with the theory behind and the intricacies of this   experimental test.

 

In fact the word  "imagine" does too little justice to my own failure to appreciate   what is clearly ,to this point an extremely successful prediction and (doubtless) interpretation.


Edited by geordief, 3 June 2017 - 07:30 AM.

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#10 Mordred

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Posted 3 June 2017 - 07:35 AM

what is the simplist possible explanation. hrrm k you get length contraction on the x axis at the same time length expansion on the y axis. Then next cycle reverse the two.

Measure the contraction/expansion with a modified Michelson interferometer.

Edited by Mordred, 3 June 2017 - 07:36 AM.

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#11 Strange

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Posted 3 June 2017 - 07:51 AM

Thanks, but I don't see how measuring the speed of light gets around the problem I'm referring to. Any speed of light refers to distance traveled divided by time, but if that distance is distorted by gravitational waves any measuring instrument will also be distorted by exactly the same amount, making such detection in principle impossible, or so it seems to me. Or am I missing something?

 
 
It is the change in distance that is being measured. Because the distance changes, the time taken by the light changes. From the article I linked:

Since LIGO and detectors like it effectively measure the length of its arms and compares them to each other, how can we rely on light to measure any length changes from a passing gravitational wave?

The solution begins to become clear when you start thinking of the laser light as a clock instead of a ruler.


There is a link to another (very simple) explanation at the end of the article:
http://www.americans...4/11/wavy-gravy
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#12 aramis720

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Posted 3 June 2017 - 10:28 PM

The ripples are in different directions. Hence the required length to frequency ratio and shape of the arms.

Picture a ball squeeze it, the sides not being squeezed expand. Then next cycle this reverses. This action induces strain.

Did you read the wiki section on "effects of passing?" In essence you are measuring changes in the detector arms lengths.

\delta L(t)=\delta L_x-\delta L_y=h L(t) where h is the measured strain.

But again there is no detectable change in any dimension because the waves are literally waves of space, so anything occupying that space (whether it's falling masses or interferometer arms or a space gerbil) will be distorted by EXACTLY the same degree as space itself. So how is the oscillating squeeze of the ball you describe and as depicted on the wikipedia page taking place when there is no detectable change?  


 
 
It is the change in distance that is being measured. Because the distance changes, the time taken by the light changes. From the article I linked:

There is a link to another (very simple) explanation at the end of the article:
http://www.americans...4/11/wavy-gravy

Thanks for this additional link but I don't find it helpful b/c Shawhan is simply re-stating the conclusion as though it's an explanation. So the idea is this: the interferometer arm is stretched by the gravitational waves and therefore light takes a bit longer to travel that arm and therefore the phases of the light in the two arms doesn't match anymore, by a tiny amount (this is how all interferometers work). BUT, again, if the space in which the arm is located is stretching, the arm itself doesn't stretch in any detectable way. And therefore there is no additional distance for light to travel. It's not as though light continues to travel in a separate dimension at the "unstretched" speed. Light is occupying the same spacetime all physical things occupy. So we're back to the basic question of how LIGO is supposed to work. 


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#13 Mordred

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Posted 3 June 2017 - 10:29 PM

But again there is no detectable change in any dimension because the waves are literally waves of space  

Waves in spacetime is changes in length as per length contradiction. So the x dimension of length does contract. What you have above does apply. That's precisely is what were detecting Ripples in spacetime geometry dimensions (ct,x,y,z)

Edited by Mordred, 3 June 2017 - 10:32 PM.

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#14 aramis720

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Posted 3 June 2017 - 10:34 PM

what is the simplist possible explanation. hrrm k you get length contraction on the x axis at the same time length expansion on the y axis. Then next cycle reverse the two.

Measure the contraction/expansion with a modified Michelson interferometer.

But what I'm suggesting is that as these terms are defined in the theory itself there is no contraction or expansion of these arms that is detectable b/c it's space itself that is contracting and expanding, so anything in that space will contract and expand in exactly the same degree as space. 


Waves in spacetime is changes in length as per length contradiction. So the x dimension of length does contract. What you have above does apply. That's precisely is what were detecting Ripples in spacetime geometry dimensions (ct,x,y,z)

Changes in length of what? Space itself, right? So, if we agree that space itself is contacting, how can anything occupying that same space measure that contraction? 


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#15 Mordred

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Posted 3 June 2017 - 10:44 PM

So your stating GR length contraction does not occur ?

It is detectable within a range of wavelengths when you change the x axis lengths you change space. Thats what Ligo is designed to detect.

It's not just math.

But what I'm suggesting is that as these terms are defined in the theory itself there is no contraction or expansion of these arms that is detectable b/c it's space itself that is contracting and expanding, so anything in that space will contract and expand in exactly the same degree as space. 

Changes in length of what? Space itself, right? So, if we agree that space itself is contacting, how can anything occupying that same space measure that contraction? 


Because not all particles in the same space is changing in the same direction.

As we have stated numerous times. Why do you think the arms are 4 km in length if memory serves.? wave length polarities Ripples.

Edited by Mordred, 3 June 2017 - 10:41 PM.

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#16 aramis720

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Posted 3 June 2017 - 11:35 PM

So your stating GR length contraction does not occur ?

It is detectable within a range of wavelengths when you change the x axis lengths you change space. Thats what Ligo is designed to detect.

It's not just math.

Because not all particles in the same space is changing in the same direction.

As we have stated numerous times. Why do you think the arms are 4 km in length if memory serves.? wave length polarities Ripples.

Let's zero in on the interferometer arm that is allegedly contracting. The arm is in the x direction of space, let's say. The grav wave comes in at the same x direction, waving up and down. So any physical object, whether it's an interferometer arm or a light wave, that occupies that space will wave to exactly the same degree as space itself. It won't matter how long or short the arm is because any distortion in that arm will be undetectable b/c it's distorting in exactly the same amount as space itself. Is that clear? 


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#17 KipIngram

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Posted 4 June 2017 - 02:31 AM

Is this really a relativistic effect?  I mean, I know gravity waves are.  But the response of the LIGO arms to gravity waves - isn't that just changes in the strength of gravity pulling on the material of the arms and wigging it around?  If the source was lined up with one of the arms, then it will "tug on" the near end first.  For a four kilometer length it will take 13.3 microseconds for the effect to reach the other end; that's plenty of time for modern electronics to respond.  When the wave first arrives (let's say it makes gravity a bit stronger) the near end will be pulled on and the arm will lengthen.  Laser light emitted from the far end and arriving at the near end will arrive a bit late due to that tug, observable as a phase shift.

 

Is there more to it than that?


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#18 Mordred

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Posted 4 June 2017 - 05:32 AM

To detect you essentially send a continous wavelength signal with a laser beam. You split the beam with mirrors via destructive interference. Then combine the two with constructive interference reforming the original wavelength.

This forms your baseline, when length contraction occurs on either arm the phases shift due to length contraction leads to destructive interference on recombination as one or both signals change which are detectable at the receivers

This is essentially how the Michelson interferometer works. Also keep in mind you cannot have length contraction without also having time dilation due to a GW wave.
If your interested I can post the mathematics specific to the spin 2 and quadrupole nature of a GW wave,as well as spin 1. Mechanical vibration and the electromagnetic waves are dipoler waves they do not have the same effects on phase shifts. There is no attenuation or dispersion in a GW wave. As such a lot of research went into filtering out those types of interference. Though there is a distinguishable difference between spin 1 and spin 2. (mechanical vibration is symmetric with spin 1 dipoler).

Though I'm less familiar with the detector end.

Let's zero in on the interferometer arm that is allegedly contracting. 

Thank you just wanted clarity we get all kinds on a forum lol

Edited by Mordred, 4 June 2017 - 06:14 AM.

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#19 Strange

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Posted 4 June 2017 - 06:25 AM

Is this really a relativistic effect?  I mean, I know gravity waves are.  But the response of the LIGO arms to gravity waves - isn't that just changes in the strength of gravity pulling on the material of the arms and wigging it around?  If the source was lined up with one of the arms, then it will "tug on" the near end first.  For a four kilometer length it will take 13.3 microseconds for the effect to reach the other end; that's plenty of time for modern electronics to respond.  When the wave first arrives (let's say it makes gravity a bit stronger) the near end will be pulled on and the arm will lengthen.  Laser light emitted from the far end and arriving at the near end will arrive a bit late due to that tug, observable as a phase shift.

 

 

The stretching takes place at right angles to the direction of travel. So if one of the arms was pointing directly at the source, then it wouldn't be affected at all.


There is another explanation here: https://www.ligo.caltech.edu/page/faq

 

(It basically says the same thing as the others, but sometimes i=different words can help...)

 

I have seen a very detailed explanation. I will see if I can find it again.


More here: https://www.quora.co...-being-affected


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#20 Mordred

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Posted 4 June 2017 - 07:03 AM

If you want the mathematical details here is how the interferometer works with GW waves

https://www.google.c...XmpQe1m0HWgMP9w

The essential formulas are included in particular photon shot noise. covered in the link as well as the unit polarization tensors derived via the linear Einstein equations for a quadrupole wave.

Edited by Mordred, 4 June 2017 - 07:08 AM.

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