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I read an article on LIGO, and I heard it mentioned that it is a nontrivial argument to say that the effect can be measured by interferometry. What happens to space as the wave passes? Does the light change wavelength due to the space that it is traveling through expanding? or does it keep its original wavelength according to a reference frame unaffected by the wave?

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The effect is in space and time. Conceptually, you send a beam around a rectangular loop, two sides of which are actually in time, and you carefully keep track of the phase when it gets back. In practice you use two beams since you can only go forwards in time. A passing wave stretches the space-like edges differently, which causes the phase to shift. – genneth Oct 20 '11 at 23:37
Related: Can LIGO measure anything? – Emilio Pisanty Feb 14 at 2:18

Let's assume the gravitational wave passes along the z-axis and has only the plus-polarized component. In this case the metric tensor of the rippled spacetime g(t) can be represented using linearized gravity as the sum of the flat spacetime metric tensor η and the perturbation caused by the gravitational wave h(t)

\begin{equation} g(t) = \eta + h(t) \end{equation}

where the perturbation is

\begin{equation} h(t) = \left[ \begin{array}{cccc} 0 & 0 & 0 & 0 \newline 0 & h_+(t-\frac{z}{c}) & 0 & 0 \newline 0 & 0 & -h_+(t-\frac{z}{c}) & 0 \newline 0 & 0 & 0 & 0 \end{array} \right] \end{equation}

The inertial coordinate frame in which this perturbation takes on this simple form is called transverse-traceless coordinate frame or TT-frame. It can be proven that for each plus-polarized wave such convenient frame can be found.

The corresponding line element in the TT-frame is

\begin{equation} ds^2 = -c^2dt^2 + [1+h_+(t-\frac{z}{c})]dx^2 + [1-h_+(t-\frac{z}{c})]dy^2 + dz^2 \end{equation}

This shows that the proper time of the interferometer is equal to the coordinate time and hence the frequency of the laser does not change.

Also, the proper length of the x-arm of the interferometer is

\begin{equation} L_x(t)=\int_0^{l}\sqrt{1+h_+(t)}dx \approx l[1+\frac{1}{2}h_+(t)] \end{equation}

Similarly, the proper length of the y-arm is

\begin{equation} L_y(t)=\int_0^{l}\sqrt{1-h_+(t)}dy \approx l[1-\frac{1}{2}h_+(t)] \end{equation}

where l is the length of each arm in the absence of the wave.

This shows that while the wave is passing the x-arm will shrink and at the same time the y-arm will extend, then they will both return to length l and then the x-arm will extend and the y-arm will shrink and so on until the wave passes. This also explains why two arms are used: the essence of the operation of the interferometer is to constantly compare Lx(t) with Ly(t), hence both arms are needed.

The important consequence of the above metric is that the ripple of gravitational wave affects the length in directions perpendicular to the direction of motion of the wave and does not affect time nor the length in the direction of motion of the wave.

Thus, since time remains unaffected (precisely: proper time of the interferometer is its coordinate time, i.e. just as in the case of flat spacetime) the frequency of the laser remains constant and hence (since the speed of light is constant) the wavelength also does not change while the distances from the splitter to the mirrors of the interferometer (Lx(t) and Ly(t)) do change.

In order to establish the direction from which the wave is coming, more than one interferometer is needed. In fact, LIGO includes interferometers in two locations: Livingstone, Luisiana and Hanford, Washington for this very purpose.

See also this wikipedia article for more details and visualization of the effects of a gravitational wave on spacetime. See this for more advanced details on gravitational waves including their detection.

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Can you explain why time remains constant? – Andrius Feb 12 at 10:21
There is a missing part of the argument isn't there? The wavelengths are stretched by the gravitational wave but only whilst they are in the interferometer. Because the travel time of the waves is much less than the timescale for the stretch to occur - i.e. the wave frequency is like 100 Hz, but the light travels through the interferometer in $\ll 0.01$ s. – Rob Jeffries Feb 15 at 15:03

When the laser pulses are generated they are in phase. When they arrive at the photodiode detector they destructively interfeer resulting in zero signal. A distortion that affects the length of either arm of the L shaped system will cause the beams to be out of phase and produce a signal. The sensitivity of the detector is on the order of the width of a proton, so virtually anything can cause a signal. So they're looking for a very particular pattern in the data that would match the gravity waves generated by events that Relativity describes well.

As the wave passes, the size of the one or both of the arms will shift. The reason why the detector is L shaped is so that they can determine the direction of the source of the wave. Best as I can tell, the wavelength of light is not affected, they become out of phase due to this change in shape of the detector.

You can read more at:

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