Gravitational Waves and the Role of Photons in LIGO

[KlowD9x]

This observatory is designed to detect gravitational waves, however, I feel that it won't work. I am sure I am wrong in my assumption, but I would like someone to explain to me why I am wrong.

I feel that this machine will not work because the gravitational waves passing over the machine will not limit themselves to distorting just the shape of the tunnels, but will distort the photons from the laser at the same time, always giving a measurement of 0 in the end.

If not, why are photons excluded from this space-time distortion?

 

[Holtman]

I think that's exactly how and why the detectors work, by distorting the photons. The interferometer detects any of those distortions in its field. Taken straight from http://www.ligo.caltech.edu/advLIGO/"

The effect of a propagating gravitational wave is to deform space in a quadrupolar form. The effect alternately elongates space in one direction while compressing space in an orthogonal direction and vice versa, with the frequency of the gravitational wave

My $0.02

 

[ZapperZ]

This observatory is designed to detect gravitational waves, however, I feel that it won't work. I am sure I am wrong in my assumption, but I would like someone to explain to me why I am wrong.

I feel that this machine will not work because the gravitational waves passing over the machine will not limit themselves to distorting just the shape of the tunnels, but will distort the photons from the laser at the same time, always giving a measurement of 0 in the end.

If not, why are photons excluded from this space-time distortion?

Have you, for example, read a review of the physics of LIGO? For example have you read

B.P. Abbott et al Rep. Prog. Phys. v.72, p.076901 (2009)?

Zz.

 

[FredMadison]

Hi!

I have the same thoughts as the original poster on this subject. I don't know how to get access to the article you mentioned, ZapperZ.
But doesn't the change in the metric influence the wavelength of the light as well as the length of the interferometer arm, thus cancelling the expected phase shift?

 

[cesiumfrog]

I feel that this machine will not work because the gravitational waves passing over the machine will not limit themselves to distorting just the shape of the tunnels, but will distort the photons from the laser at the same time

But doesn't the change in the metric influence the wavelength of the light as well as the length of the interferometer arm, thus cancelling the expected phase shift?

Do you agree the speed of light will stay constant?

 

[FredMadison]

I guess I would have to say "yes" to that... Please tell me more.

 

[cesiumfrog]

Forgetting the wavelength, if the light travels at a constant speed but the distance it has to travel changes then wouldn't you expect a change in the time period it takes to return? (That is, a comparative phase shift between the light beams returning from different arms of the interferometer?)

 

[bcrowell]

Suspicions about the nonobservability of gravitational waves go all the way back to Einstein. Some discussion of this:

http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.2 (subsection 8.2.4)

http://en.wikipedia.org/wiki/Sticky_bead_argument

I think you need to consider that the frequency of the gravitational waves being searched for is typically below 1 Hz. Therefore, a light wave's frequency is not modified significantly while it's in flight. I think this is different from, e.g., cosmological redshifts, where the photon's wavelength stretches while it's in flight for billions of years.

 

[heldervelez]

the speed of light will NOT stay constant under gravitational field

Do you agree the speed of light will stay constant?

No,
from Einstein (Ann. d. Phys. 35 1911) :
eq. (3) c=c0(1+phy/c^2)

the rate will be (1+phy/c^2) slower than at co-ordinate origin.

 

[cesiumfrog]

from Einstein (Ann. d. Phys. 35 1911)

You want to disagree based on citations predating general relativity theory?

 

[bcrowell]

Another thing to point out about the original argument made by KlowD9x is that it sounds sort of like a (correct) argument that you can't use one measuring rod to measure a change in the length of another measuring rod. If this was what LIGO was doing, then it would indeed be destined to fail. One way of seeing that you can't use one rod to measure a change in the length of another is that there is no such thing as intrinsic curvature in one dimension. Any real experiment to detect spacetime curvature essentially has to act out something resembling the definition of the Riemann tensor, which essentially involves seeing the path-dependence of parallel transport. In one dimension, you can't have two different paths from A to B that enclose a finite area.

 

[cesiumfrog]

you can't use one measuring rod to measure a change in the length of another measuring rod.

Sort of sounds like a bar detector (or sticky bead experiment)?

 

[heldervelez]

It reminds me the MM experiment (Michelson-Moreley) and the way Lorentz show how it failed.
(It is in line with the OP and cesiumfrog last post)
Another possible explanation: the expected effect does not exist. (http://en.wikipedia.org/wiki/Sticky...nts_on_the_properties_of_gravitational_waves")

 

[bcrowell]

Sort of sounds like a bar detector (or sticky bead experiment)?

In the sticky bead method, you have two dimensions: one spatial dimension (along the length of the rod) and one time dimension (because you detect motion of the bead).

 

[heldervelez]

You want to disagree based on citations predating general relativity theory?

Sorry, I've made a mistake, unintentional. The above equation was quoted from a paper of Einstein that was written before the formal presentation of GR. That equation gave a wrong value about the light deflection.

But I was not in absolute error... as c can vary:
"http://en.wikipedia.org/wiki/Schwar....29_formulations_of_the_Schwarzschild_metric""
quoting
...
the metric then becomes
. . .
In the terms of these coordinates, the velocity of light at any point is the same in all directions, but it varies with radial distance r1 (from the point mass at the origin of coordinates), where it has the value:

(see there pls, but the 3 at expoent in not ok, I think)

 

[cesiumfrog]

In the terms of these coordinates, the velocity of light

That's the coordinate speed, not the physical speed.