<p>The gravitational wave strain is not isotropic. More power is emitted (per unit solid angle) along the orbital axis of the binary than in the orbital plane.</p>
<p>For a circular orbit (and they tend to be circular when the black holes get close), the angular distribution is azimuthally symmetric around the axis of the binary, but varies with inclination <span class="math-container">$i$</span> as <span class="math-container">$(1 + \cos^2 i)/2$</span> for <span class="math-container">$h_+$</span> polarised waves and as <span class="math-container">$\cos i$</span> for <span class="math-container">$h_\times$</span> polarised waves.</p>
<p>When the binary is observed &quot;edge-on&quot;, <span class="math-container">$i=90$</span> degrees and we only see <span class="math-container">$h_+$</span> polarised waves, with a relative strain amplitude of 0.5. For a &quot;face-on&quot; binary with <span class="math-container">$i=0$</span> we are looking down the orbital axis and see an equal mixture of <span class="math-container">$h_+$</span> and <span class="math-container">$h_\times$</span> polarisations, both with a relative amplitude of 1.</p>
<p>The ratio of gravitational wave <em>power</em> per unit solid angle is proportional to the square of the strain and is 8 times bigger along the orbital axis than in the orbital plane.</p>

CC BY-SA 4.0
2023-10-12T08:53:32.267Z
It might be worth noting that this is based on the quadrupole approximation. There will be contributions from higher order modes, which change the distribution with inclination, although it will be always true (for spin aligned binaries) that edge-on the signal will be only $h_{+}$.
TimRias
CC BY-SA 4.0
2023-10-12T09:50:34.767Z
@TimRias thanks. I'm unfamiliar with those details. Do higher order modes propagate equally to large distances? How far from the factor of 8 does this change things?
ProfRob
CC BY-SA 4.0
2023-10-13T20:35:16.740Z
Thank you for the answer. I understood some of it. But I would guess the biggest waves were emitted in the orbital plane. This is also the way all illustrations, I so far has found, describes it. 
The waves are spreading along the “fabric of space” in the orbital plane. This makes also sense in my mind as it corresponds to the theory that the spiral arms in the Milky Way are a result of gravity waves emitted from two rotating black holes in the center of the galaxy.
Peter
CC BY-SA 4.0
2023-10-13T21:15:11.390Z
@Peter No, the opposite, as I have explained in my answer. The spiral arms have nothing to do with gravitational waves.
ProfRob
CC BY-SA 4.0
2023-10-24T13:41:58.440Z
@Peter Two black holes? There are numerous black holes in our galaxy, but there's only one large BH in the centre of the galaxy (Sagittarius A*), and although it *is* massive, it only makes a relatively small contribution to the galaxy's gravitational field.
PM 2Ring
CC BY-SA 4.0
2023-10-11T23:49:30.227Z
2023-10-11T23:49:30.227Z
ProfRob
<p>As a supplement to @ProfRob's <a href="https://astronomy.stackexchange.com/a/55002/29199">answer</a>.</p>
<p>The answer there only takes into account the leading order contribution coming from <span class="math-container">$\ell=m=2$</span> quadrupole mode. For binaries that are far from merger, and merging equal mass (non-spinning) black holes this is by far to most dominant mode, but for unequal mass binaries in the strong field the situation can be quite different.</p>
<p>In the general, the distribution of the gravitational wave strain along the celestial sphere can be written:</p>
<p><span class="math-container">$$ h(\Theta_{\rm obs},\Phi_{\rm obs}) = \sum_{\ell=2}^\infty\sum_{m=-\ell}^m h_{\ell m} {_{-2}}Y_{\ell m}(\Theta_{\rm obs},\Phi_{\rm obs}),$$</span>
where the <span class="math-container">$ {_{-2}}Y_{\ell m}$</span>, are so called spin-weighted spherical harmonics, the <span class="math-container">$h_{\ell m}$</span> are the referred to as the multipole modes.</p>
<p>As mentioned above, typically the <span class="math-container">$h_{22}$</span> is by far the largest mode, but all modes are emitted in any particular gravitational event, unless there is additional symmetry (such as in a head on collision). However, you can deviated quite far from the standard picture. Below is a graphic representation of the GW strain distribution caused by a small-mass orbit a Kerr black hole with near extremal spin (<span class="math-container">$a=0.995GM/c^2$</span>) on the innermost stable circular orbit (ISCO), obtained as an approximation linear in the mass of the smaller object,
<a href="https://i.sstatic.net/m96WE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/m96WE.png" alt="Strain distribution" /></a></p>
<p>The <span class="math-container">$\ell=m=2$</span> mode is still the largest contribution, but the <span class="math-container">$\ell=m=3$</span> and other <span class="math-container">$\ell=m=4$</span> modes are very close. The consequence is that the maximal strain is not obtain &quot;face-on&quot; (<span class="math-container">$\Theta_{\rm obs}=0$</span>), but at viewing angle of <span class="math-container">$\Theta_{\rm obs} \approx 42.9$</span> degrees.</p>

CC BY-SA 4.0
2023-10-24T12:42:38.840Z
2023-10-24T19:14:03.083Z
2023-10-24T19:14:03.083Z
TimRias
<p>I have a question regarding gravitational waves that I can't figure out. Hope some wise minds here can help a simple amateur without technical or scientific education.</p>
<p>Two black holes rotating around each other generate and emit gravitational waves. But, in what pattern do they spread?</p>
<p>Do they spread out as an expanding sphere or as a disk? What happens to the direction of propagation as the holes merge into a single hole?</p>
<p>Answers explained in a simple way are appreciated.</p>

CC BY-SA 4.0
2023-10-11T19:34:08.093Z
Hello Peter,  I count 6 different questions here.  Some may be related, but a basic "rule" of this site is "one question at a time".  The first question is good (though I think the answer is contained within https://physics.stackexchange.com/questions/256527/do-gravitational-waves-radiate-isotropically) Perhaps an answer that uses less maths and describes the situation qualitatively could usefully be written.  However applying the one question per post rule, I've hidden the other questions (they are still there if you [edit], and you can ask them in separate questions later)
James K
<black-hole><gravitational-waves><ligo>
Are gravitational waves emitted equally in all directions?
Peter
# Are gravitational waves emitted equally in all directions?

(Asked by: username_0 on 2023-10-11)

I have a question regarding gravitational waves that I can't figure out. Hope some wise minds here can help a simple amateur without technical or scientific education.

Two black holes rotating around each other generate and emit gravitational waves. But, in what pattern do they spread?

Do they spread out as an expanding sphere or as a disk? What happens to the direction of propagation as the holes merge into a single hole?

Answers explained in a simple way are appreciated.

**Tags:** black-hole, gravitational-waves, ligo
> **Comments:**
> **username_1**: Hello Peter,  I count 6 different questions here.  Some may be related, but a basic "rule" of this site is "one question at a time".  The first question is good (though I think the answer is contained within https://physics.stackexchange.com/questions/256527/do-gravitational-waves-radiate-isotropically) Perhaps an answer that uses less maths and describes the situation qualitatively could usefully be written.  However applying the one question per post rule, I've hidden the other questions (they are still there if you [edit], and you can ask them in separate questions later)  

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## 2 answers

### Answer by username_2 on 2023-10-11. Score: 12

The gravitational wave strain is not isotropic. More power is emitted (per unit solid angle) along the orbital axis of the binary than in the orbital plane.

For a circular orbit (and they tend to be circular when the black holes get close), the angular distribution is azimuthally symmetric around the axis of the binary, but varies with inclination $i$ as $(1 + \cos^2 i)/2$ for $h\_+$ polarised waves and as $\cos i$ for $h\_\times$ polarised waves.

When the binary is observed "edge-on", $i=90$ degrees and we only see $h\_+$ polarised waves, with a relative strain amplitude of 0.5. For a "face-on" binary with $i=0$ we are looking down the orbital axis and see an equal mixture of $h\_+$ and $h\_\times$ polarisations, both with a relative amplitude of 1.

The ratio of gravitational wave *power* per unit solid angle is proportional to the square of the strain and is 8 times bigger along the orbital axis than in the orbital plane.

> **Comments:**
> **username_3**: It might be worth noting that this is based on the quadrupole approximation. There will be contributions from higher order modes, which change the distribution with inclination, although it will be always true (for spin aligned binaries) that edge-on the signal will be only $h_{+}$.  
> **username_2**: @TimRias thanks. I'm unfamiliar with those details. Do higher order modes propagate equally to large distances? How far from the factor of 8 does this change things?  
> **username_0**: Thank you for the answer. I understood some of it. But I would guess the biggest waves were emitted in the orbital plane. This is also the way all illustrations, I so far has found, describes it.  The waves are spreading along the “fabric of space” in the orbital plane. This makes also sense in my mind as it corresponds to the theory that the spiral arms in the Milky Way are a result of gravity waves emitted from two rotating black holes in the center of the galaxy.  
> **username_2**: @Peter No, the opposite, as I have explained in my answer. The spiral arms have nothing to do with gravitational waves.  
> **username_4**: @Peter Two black holes? There are numerous black holes in our galaxy, but there's only one large BH in the centre of the galaxy (Sagittarius A*), and although it *is* massive, it only makes a relatively small contribution to the galaxy's gravitational field.  

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### Answer by username_3 on 2023-10-24. Score: 4

As a supplement to @ProfRob's [answer](https://astronomy.stackexchange.com/a/55002/29199).

The answer there only takes into account the leading order contribution coming from $\ell=m=2$ quadrupole mode. For binaries that are far from merger, and merging equal mass (non-spinning) black holes this is by far to most dominant mode, but for unequal mass binaries in the strong field the situation can be quite different.

In the general, the distribution of the gravitational wave strain along the celestial sphere can be written:

$$ h(\Theta\_{\rm obs},\Phi\_{\rm obs}) = \sum\_{\ell=2}^\infty\sum\_{m=-\ell}^m h\_{\ell m} {\_{-2}}Y\_{\ell m}(\Theta\_{\rm obs},\Phi\_{\rm obs}),$$
where the $ {\_{-2}}Y\_{\ell m}$, are so called spin-weighted spherical harmonics, the $h\_{\ell m}$ are the referred to as the multipole modes.

As mentioned above, typically the $h\_{22}$ is by far the largest mode, but all modes are emitted in any particular gravitational event, unless there is additional symmetry (such as in a head on collision). However, you can deviated quite far from the standard picture. Below is a graphic representation of the GW strain distribution caused by a small-mass orbit a Kerr black hole with near extremal spin ($a=0.995GM/c^2$) on the innermost stable circular orbit (ISCO), obtained as an approximation linear in the mass of the smaller object,
[![Strain distribution](https://i.sstatic.net/m96WE.png)](https://i.sstatic.net/m96WE.png)

The $\ell=m=2$ mode is still the largest contribution, but the $\ell=m=3$ and other $\ell=m=4$ modes are very close. The consequence is that the maximal strain is not obtain "face-on" ($\Theta\_{\rm obs}=0$), but at viewing angle of $\Theta\_{\rm obs} \approx 42.9$ degrees.


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