<p>If a third body interacts with a binary system then it is possible for the third body to be ejected taking away energy from the binary system. The lowering of its (negative) total energy (the sum of kinetic and potential energy) causes the binary components to get closer and orbit each other faster.</p>
<p>We would say the &quot;binary has hardened&quot;.</p>
<p>What is meant by this is that if another star then interacts with that binary, it is likely to extract even more energy from the binary. Hard binaries are better at accelerating third bodies than are softer, wider binaries.</p>
<p>Hard and soft are qualitative terms. In the context you are dealing with, a &quot;hard binary&quot; is one capable of imparting enough additional speed to a third body that it will exceed the speed of most stars in the vicinity. That will happen when the orbital speed of the binary components reaches some threshold value - a value which depends on the mass ratio of its components as given in your question.</p>

CC BY-SA 4.0
2023-01-24T21:08:02.130Z
2023-01-24T21:08:02.130Z
ProfRob
<p>In <a href="https://arxiv.org/abs/2301.08640" rel="noreferrer">this paper</a>, the authors state in the Introduction that</p>
<blockquote>
<p>A hard SMBHB [SuperMassive Black Hole Binary] can eject surrounding stars to transfer their orbital energy and angular momentum, which may be efficient to drive two SMBHs coalesce quickly.</p>
</blockquote>
<p>In this context, a &quot;hard black hole binary&quot; is one which can eject stars out of the vicinity at high velocity.</p>
<p><a href="https://arxiv.org/pdf/astro-ph/9601092.pdf" rel="noreferrer">One of the references</a> states that</p>
<blockquote>
<p>a binary [SMBH] with masses <span class="math-container">$m_1 ≥ m_2$</span> should not be considered hard until its orbital velocity exceeds the background velocity dispersion by a factor that scales as <span class="math-container">$(1 + m_1/m_2)^{1/2}$</span>.</p>
</blockquote>
<p>What exactly does this scale factor mean?</p>

<orbital-mechanics><binary-star><supermassive-black-hole>
What's the meaning of "hard supermassive black hole binary"?
Jim421616
# What's the meaning of "hard supermassive black hole binary"?

(Asked by: username_0 on 2023-01-24)

In [this paper](https://arxiv.org/abs/2301.08640), the authors state in the Introduction that

> A hard SMBHB [SuperMassive Black Hole Binary] can eject surrounding stars to transfer their orbital energy and angular momentum, which may be efficient to drive two SMBHs coalesce quickly.

In this context, a "hard black hole binary" is one which can eject stars out of the vicinity at high velocity.

[One of the references](https://arxiv.org/pdf/astro-ph/9601092.pdf) states that

> a binary [SMBH] with masses $m\_1 ≥ m\_2$ should not be considered hard until its orbital velocity exceeds the background velocity dispersion by a factor that scales as $(1 + m\_1/m\_2)^{1/2}$.

What exactly does this scale factor mean?

**Tags:** orbital-mechanics, binary-star, supermassive-black-hole

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## 1 answers

### Answer by username_1 on 2023-01-24. Score: 6 | Accepted answer

If a third body interacts with a binary system then it is possible for the third body to be ejected taking away energy from the binary system. The lowering of its (negative) total energy (the sum of kinetic and potential energy) causes the binary components to get closer and orbit each other faster.

We would say the "binary has hardened".

What is meant by this is that if another star then interacts with that binary, it is likely to extract even more energy from the binary. Hard binaries are better at accelerating third bodies than are softer, wider binaries.

Hard and soft are qualitative terms. In the context you are dealing with, a "hard binary" is one capable of imparting enough additional speed to a third body that it will exceed the speed of most stars in the vicinity. That will happen when the orbital speed of the binary components reaches some threshold value - a value which depends on the mass ratio of its components as given in your question.


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