<p>The blind spots are caused by the way the detectors work. They are sensitive to a gravitational wave (GW) changing the <em>relative</em> path length along interferometer arms at right angles to each other.</p>

<p>Gravitational waves come in two polarisations (plus and cross). These polarisations cause alternate perpendicular expansions and contractions in space, but are rotated by 45 degrees with respect to each other (plot from <a href="http://inspirehep.net/record/818927/plots" rel="nofollow noreferrer">Kalmus 2009</a>).</p>

<p><a href="https://i.sstatic.net/hUYfQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hUYfQ.png" alt="Plus and cross polarisation"></a></p>

<p>A GW source would normally be a mixture of both. The sensitivity to each polarisation depends on the orientation of the interferometer arms with respect to the source direction. For example, a source that is directly "overhead" will only be detected in the polarisation state that lines up with the arms and not in the other, because it would make the arm lengths change by the same amount. e.g. Imagine your detector is lined up with the x and y axis indicators in the picture above, then only the plus (<span class="math-container">$+$</span>) polarised waves would be detected.</p>

<p>However, if the waves come from a source <em>in the plane</em> of the interferometer, then neither polarisation causes a relative difference in the arm lengths if the source lies along the bisector of the two arms or along an equivalent line at right angles to this. Thus there are 4 blind spots on the sky (see picture from <a href="https://arxiv.org/abs/1208.4596" rel="nofollow noreferrer">Hayama &amp; Nishizawa 2012</a>, which shows the sensitivity as a function of position for the Hanford interferometer). Note that if a source is perfectly polarised (e.g. an edge-on merging binary system), then there will be additional blind spots.</p>

<p><a href="https://i.sstatic.net/K0zc4.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/K0zc4.jpg" alt="Hanford Antenna pattern"></a></p>

<p>This discussion applies to each interferometer separately. The Washington and Louisiana instruments are clearly at different locations on the Earth's surface, so don't operate in <em>exactly</em> the same plane and so don't have identical blind spots (but they are close and the addition of VIRGO in Italy is hugely important). Nevertheless, if one detector doesn't see the source then that makes it very difficult to put any constraints on the position, since it relies heavily on measuring time differences between detections at the different instruments.</p>

CC BY-SA 4.0
2020-02-08T16:04:42.390Z
2020-02-08T16:34:30.770Z
2020-02-08T16:34:30.770Z
ProfRob
<p><a href="http://arxiv.org/pdf/1304.0670.pdf" rel="nofollow noreferrer">This paper</a> should tell you all you need to know about how LIGO observes and localizes gravitational waves, and in particular its blind spots. The map below is taken from the paper and shows the observability of the system when in the  HLV configuration (Hanford-LIGO, Livingston-LIGO &amp; Virgo). To triangulate a source by definition you need at least 3 measurements so that is why Virgo is included, although the two LIGO detectors can make very broad estimates for source localisation on their own (GW150914 was assigned a 600 square degree error box). </p>

<p>The (x) denotes a blind spot.</p>

<p><a href="https://i.sstatic.net/UzMbR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UzMbR.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2016-02-18T12:41:54.840Z
I don't understand how this answers does LIGO have a blind spot? You imply that any blind spots are limited to some crosses (how big are hese spots) drawn on the surface of the *Earth* rather than the sky. I imagine that the blind "spots" are in fact significant areas on the sky that depend on the sidereal time of the observation. An answer may also need to discuss GW polarisation?
ProfRob
CC BY-SA 3.0
2016-02-18T13:42:33.950Z
Sorry I forgot to mention that this figure is for the localization accuracy of a face-on binary neutron-star systems. And the ellipses show 90% confidence localization areas based upon timing triangulation alone, and the red crosses show regions of the sky where the signal would not be confidently detected (i.e. blind spots), this map is projected onto the Earth for simplification.
Dean
CC BY-SA 3.0
2016-02-21T08:03:32.147Z
@RobJeffries - The gravitational wave detected by the LIGO installations in Washington state and Louisiana lasted but 0.2 seconds. The signal had to be observed by both sites to qualify as a detection. Since the signal was of such a short duration and since the detectors are fixed with respect to the rotating Earth, the blind spots are necessarily Earth-based rather than sky-based.
David Hammen
CC BY-SA 4.0
2020-02-08T15:06:31.083Z
The referred to paper may contain an explanation, but this answer does not explain anything.
ProfRob
CC BY-SA 4.0
2020-02-08T15:14:24.113Z
And the picture shown is not in that paper and the paper doesn't mention "blind spots".
ProfRob
CC BY-SA 3.0
2016-02-18T11:53:14.477Z
2016-02-18T11:53:14.477Z
Dean
<p>If I understand the principle of operation of operation of LIGO, it detects relative distortions of the two perpendicular arms. So if both arms are distorted the same way, nothing would be spotted.</p>

<p>In particular, any wave coming in the direction 45 degrees between the arms (both ways), or from above or from below - or generally any direction in the symmetry plane 45 degrees to the arms - would be completely invisible, and anything close to it would be very muted as the relative distortion would be minimal.</p>

<p>Is that correct?</p>

CC BY-SA 3.0
2016-02-18T06:46:44.067Z
I always asked myself why LIGOs do not have three arms, the third being perpendicular on the other two. Maybe the computations are too difficult?
Florin Ghita
CC BY-SA 3.0
2016-02-18T07:06:47.187Z
@FlorinGhita: I believe it would about triple the price, three pairs of laser beams instead of one. It isn't just "adding a third beam" but "adding two new gravitometers".
SF.
CC BY-SA 3.0
2016-02-18T17:59:25.783Z
@FlorinGhita because building a 4km tower and keeping its top exactly steady and unaffected by wind and other disturbances to a accuracy of less than a proton diameter is technically challenging.
James K
CC BY-SA 3.0
2016-02-18T21:07:17.593Z
@JamesKilfiger: Building a 4km mine shaft is a routine work though.
SF.
CC BY-SA 3.0
2016-02-19T09:43:47.063Z
Not so routine. That would be nearly twice as deep as Homestake mine, the deepest in the US. I dount knoe anything of the geology, but I do know that a 4km hole nothing simple. And would probably have little scientific value
James K
CC BY-SA 3.0
2016-02-19T09:51:00.320Z
@JamesKilfiger The TauTona Mine or Western Deep No.3 Shaft, is a gold mine in South Africa. 3.9 kilometers. And the value would be in making it a self-contained full 3D gravitometer, but I guess it IS easier just to build another surface facility far enough for Earth's curvature to kick in to provide the change of plane of operation.
SF.
<gravitational-waves><ligo>
Does LIGO have a blind spot?
SF.
# Does LIGO have a blind spot?

(Asked by: username_0 on 2016-02-18)

If I understand the principle of operation of operation of LIGO, it detects relative distortions of the two perpendicular arms. So if both arms are distorted the same way, nothing would be spotted.

In particular, any wave coming in the direction 45 degrees between the arms (both ways), or from above or from below - or generally any direction in the symmetry plane 45 degrees to the arms - would be completely invisible, and anything close to it would be very muted as the relative distortion would be minimal.

Is that correct?

**Tags:** gravitational-waves, ligo
> **Comments:**
> **username_1**: I always asked myself why LIGOs do not have three arms, the third being perpendicular on the other two. Maybe the computations are too difficult?  
> **username_0**: @FlorinGhita: I believe it would about triple the price, three pairs of laser beams instead of one. It isn't just "adding a third beam" but "adding two new gravitometers".  
> **username_2**: @FlorinGhita because building a 4km tower and keeping its top exactly steady and unaffected by wind and other disturbances to a accuracy of less than a proton diameter is technically challenging.  
> **username_0**: @JamesKilfiger: Building a 4km mine shaft is a routine work though.  
> **username_2**: Not so routine. That would be nearly twice as deep as Homestake mine, the deepest in the US. I dount knoe anything of the geology, but I do know that a 4km hole nothing simple. And would probably have little scientific value  
> **username_0**: @JamesKilfiger The TauTona Mine or Western Deep No.3 Shaft, is a gold mine in South Africa. 3.9 kilometers. And the value would be in making it a self-contained full 3D gravitometer, but I guess it IS easier just to build another surface facility far enough for Earth's curvature to kick in to provide the change of plane of operation.  

---

## 2 answers

### Answer by username_4 on 2016-02-18. Score: 5 | Accepted answer

[This paper](http://arxiv.org/pdf/1304.0670.pdf) should tell you all you need to know about how LIGO observes and localizes gravitational waves, and in particular its blind spots. The map below is taken from the paper and shows the observability of the system when in the HLV configuration (Hanford-LIGO, Livingston-LIGO & Virgo). To triangulate a source by definition you need at least 3 measurements so that is why Virgo is included, although the two LIGO detectors can make very broad estimates for source localisation on their own (GW150914 was assigned a 600 square degree error box).

The (x) denotes a blind spot.

[![enter image description here](https://i.sstatic.net/UzMbR.png)](https://i.sstatic.net/UzMbR.png)

> **Comments:**
> **username_3**: I don't understand how this answers does LIGO have a blind spot? You imply that any blind spots are limited to some crosses (how big are hese spots) drawn on the surface of the *Earth* rather than the sky. I imagine that the blind "spots" are in fact significant areas on the sky that depend on the sidereal time of the observation. An answer may also need to discuss GW polarisation?  
> **username_4**: Sorry I forgot to mention that this figure is for the localization accuracy of a face-on binary neutron-star systems. And the ellipses show 90% confidence localization areas based upon timing triangulation alone, and the red crosses show regions of the sky where the signal would not be confidently detected (i.e. blind spots), this map is projected onto the Earth for simplification.  
> **username_5**: @RobJeffries - The gravitational wave detected by the LIGO installations in Washington state and Louisiana lasted but 0.2 seconds. The signal had to be observed by both sites to qualify as a detection. Since the signal was of such a short duration and since the detectors are fixed with respect to the rotating Earth, the blind spots are necessarily Earth-based rather than sky-based.  
> **username_3**: The referred to paper may contain an explanation, but this answer does not explain anything.  
> **username_3**: And the picture shown is not in that paper and the paper doesn't mention "blind spots".  

---

### Answer by username_3 on 2020-02-08. Score: 2

The blind spots are caused by the way the detectors work. They are sensitive to a gravitational wave (GW) changing the *relative* path length along interferometer arms at right angles to each other.

Gravitational waves come in two polarisations (plus and cross). These polarisations cause alternate perpendicular expansions and contractions in space, but are rotated by 45 degrees with respect to each other (plot from [Kalmus 2009](http://inspirehep.net/record/818927/plots)).

[![Plus and cross polarisation](https://i.sstatic.net/hUYfQ.png)](https://i.sstatic.net/hUYfQ.png)

A GW source would normally be a mixture of both. The sensitivity to each polarisation depends on the orientation of the interferometer arms with respect to the source direction. For example, a source that is directly "overhead" will only be detected in the polarisation state that lines up with the arms and not in the other, because it would make the arm lengths change by the same amount. e.g. Imagine your detector is lined up with the x and y axis indicators in the picture above, then only the plus ($+$) polarised waves would be detected.

However, if the waves come from a source *in the plane* of the interferometer, then neither polarisation causes a relative difference in the arm lengths if the source lies along the bisector of the two arms or along an equivalent line at right angles to this. Thus there are 4 blind spots on the sky (see picture from [Hayama & Nishizawa 2012](https://arxiv.org/abs/1208.4596), which shows the sensitivity as a function of position for the Hanford interferometer). Note that if a source is perfectly polarised (e.g. an edge-on merging binary system), then there will be additional blind spots.

[![Hanford Antenna pattern](https://i.sstatic.net/K0zc4.jpg)](https://i.sstatic.net/K0zc4.jpg)

This discussion applies to each interferometer separately. The Washington and Louisiana instruments are clearly at different locations on the Earth's surface, so don't operate in *exactly* the same plane and so don't have identical blind spots (but they are close and the addition of VIRGO in Italy is hugely important). Nevertheless, if one detector doesn't see the source then that makes it very difficult to put any constraints on the position, since it relies heavily on measuring time differences between detections at the different instruments.


---

