<p>I understand your question and wrestled with it myself for some time.</p>

<p>The reason that the Michelson-type interferometer setup works is that the travel time of the photons in each arm is <em>much</em> shorter than the time period of the gravitational wave. Thus the photons are not really stretched by the gravitational wave at all - they are simply emitted into an arm that has changed its apparent length. The effect is different in the two perpendicular arms (one is stretched whilst the other is contracted) and this produces a phase difference in the signal when the beams in the two arms are recombined. The phase difference is modulated with the frequency of the gravitational wave (which is <em>much</em> lower than the frequency of the laser light).</p>

<p>For example, the LIGO arms are effectively 1120 km long, because the light is bounced backwards and forwards 280 times. Light spends 3.7 micro-seconds in the apparatus. The gravitational waves that are being detected have frequencies below a kHz, so the arm length only changes significantly on times longer than a milli-second. Thus the photons "see" arms of fixed length.</p>

CC BY-SA 3.0
2017-02-13T15:31:06.327Z
Ah-ha, and thus the fact that we only can see mergers of certain masses in LIGO! Thank you @Rob Jeffries, great answer
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-02-13T16:45:52.477Z
@AtmosphericPrisonEscape The frequency sensitivity of LIGO is bounded by seismic noise at the low end ($<10$ Hz) and shot noise at high frequencies. The central part of this frequency sensitivity is matched by the orbital frequencies of $\sim 10$ solar mass black holes just prior to merger. The upper bound is not due to the approximation above, which wouldn't be an issue until you got to above 100 kHz I think.
ProfRob
CC BY-SA 3.0
2017-02-13T12:31:45.830Z
2017-02-13T12:31:45.830Z
ProfRob
<p>Geometry of gravitational waves is such, that it stretches and shrinks spacetime differently in perpendicular directions. So the detector notices the difference in flight time between two photons going along different axes of the detector, as the velocity of light is constant. "Stretching" of a photon would mean, that it changes color (wavelength) but not flight time.</p>

<p>If you're interested in the math have a look at <a href="http://bcc.impan.pl/13Gravitational/uploads/Whelan_geometry.pdf" rel="nofollow noreferrer">The Geometry of Gravitational Wave Detection</a> by John T Wheelan.</p>

CC BY-SA 3.0
2017-02-13T11:35:16.080Z
2017-02-13T11:35:16.080Z
Grimaldi
<p>We know that gravitational waves stretch space-time so light travelling through that space should be stretched as well. So my question is how do we know anything is stretching if the effect is experienced by every object in the universe. </p>

CC BY-SA 3.0
2017-02-08T15:43:49.553Z
For one, it happens to every object which happens to be where the gravitational wave is, which is not everywhere in the universe at once. Second, what we experience is the change in stretching. That is what was [measured by LIGO](https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves).
zephyr
CC BY-SA 3.0
2017-02-09T23:33:38.473Z
@zephyr, you can make that an answer.
James K
CC BY-SA 3.0
2017-02-09T23:35:15.293Z
@JamesK I know, I was just lazy. Feel free to take it though. Maybe later if no one has put an official answer down I will write something.
zephyr
<gravity><astrophysics><space-time>
Gravitational waves
user32409
# Gravitational waves

(Asked by: username_0 on 2017-02-08)

We know that gravitational waves stretch space-time so light travelling through that space should be stretched as well. So my question is how do we know anything is stretching if the effect is experienced by every object in the universe.

**Tags:** gravity, astrophysics, space-time
> **Comments:**
> **username_1**: For one, it happens to every object which happens to be where the gravitational wave is, which is not everywhere in the universe at once. Second, what we experience is the change in stretching. That is what was [measured by LIGO](https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves).  
> **username_2**: @zephyr, you can make that an answer.  
> **username_1**: @JamesK I know, I was just lazy. Feel free to take it though. Maybe later if no one has put an official answer down I will write something.  

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## 2 answers

### Answer by username_3 on 2017-02-13. Score: 4 | Accepted answer

I understand your question and wrestled with it myself for some time.

The reason that the Michelson-type interferometer setup works is that the travel time of the photons in each arm is *much* shorter than the time period of the gravitational wave. Thus the photons are not really stretched by the gravitational wave at all - they are simply emitted into an arm that has changed its apparent length. The effect is different in the two perpendicular arms (one is stretched whilst the other is contracted) and this produces a phase difference in the signal when the beams in the two arms are recombined. The phase difference is modulated with the frequency of the gravitational wave (which is *much* lower than the frequency of the laser light).

For example, the LIGO arms are effectively 1120 km long, because the light is bounced backwards and forwards 280 times. Light spends 3.7 micro-seconds in the apparatus. The gravitational waves that are being detected have frequencies below a kHz, so the arm length only changes significantly on times longer than a milli-second. Thus the photons "see" arms of fixed length.

> **Comments:**
> **username_4**: Ah-ha, and thus the fact that we only can see mergers of certain masses in LIGO! Thank you @Rob Jeffries, great answer  
> **username_3**: @AtmosphericPrisonEscape The frequency sensitivity of LIGO is bounded by seismic noise at the low end ($<10$ Hz) and shot noise at high frequencies. The central part of this frequency sensitivity is matched by the orbital frequencies of $\sim 10$ solar mass black holes just prior to merger. The upper bound is not due to the approximation above, which wouldn't be an issue until you got to above 100 kHz I think.  

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### Answer by username_5 on 2017-02-13. Score: 0

Geometry of gravitational waves is such, that it stretches and shrinks spacetime differently in perpendicular directions. So the detector notices the difference in flight time between two photons going along different axes of the detector, as the velocity of light is constant. "Stretching" of a photon would mean, that it changes color (wavelength) but not flight time.

If you're interested in the math have a look at [The Geometry of Gravitational Wave Detection](http://bcc.impan.pl/13Gravitational/uploads/Whelan_geometry.pdf) by John T Wheelan.


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