<p>I read that in principle, the formation of a double neutron star requires that at the time of the second supernova explosion the binary consists of a <span class="math-container">$\sim$</span> 1.4 solar mass neutron star, and a second star whose initial mass was &gt; 8 solar mass.</p>
<p>But it can be shown that a binary will become unbound should it lose more than 50% of its mass. This is equation 9.17 in <a href="https://www.astro.ru.nl/%7Efverbunt/binaries/compbin.pdf" rel="nofollow noreferrer">https://www.astro.ru.nl/~fverbunt/binaries/compbin.pdf</a></p>
<p>I know that neutron stars have masses of around 2.5 solar mass. If I take for example that a t the time of the second explosion the masses are 1.4 and 9 solar masses and later the neutron starts are 1.4 and 2.5 there is a mass loss of more than 50%.   So how can such a system ever remain bound?</p>

CC BY-SA 4.0
2024-02-18T19:22:02.870Z
I would recommend including the equations you're using to show that the binary would become unbound if it loses more than 50% mass. I'd suspect you're assuming the stars are in circular orbits, rather than elliptical ones, as the required mass loss for escape changes  depending on when it happens (less near periapsis, more near apopasis), and at any random time, the stars are more likely to be closer to their  apoapses.
notovny
CC BY-SA 4.0
2024-02-18T20:03:26.187Z
@notovny I added a link to the lecture notes with the relevant equation
some_math_guy
CC BY-SA 4.0
2024-02-19T10:34:10.870Z
That's a 137 page document.  For those wondering, the equation in question is at the bottom of page 117 of the pdf (page 114 of the document's internal numbering). I'll note that equation is exactly as notovny says: for circular orbits.  It is the following pages/section that deals with elliptical orbits, which are far more likely.
zibadawa timmy
CC BY-SA 4.0
2024-02-19T18:36:22.700Z
Section 9.1.15 onwards in the reference you quote is about how the problem you pose in your question is solved. In summary: mass transfer and binary evolution. Also note that the 8 solar mass lower limit is on the *initial* mass of a star that explodes.
ProfRob
CC BY-SA 4.0
2024-02-20T07:45:42.727Z
@I know it is a 137 document.That's why I provided the relevant eq number 9.17, which is the same as saying page 114 or 117 of the pdf. From that eq it follows that with a mass loss of ore than 50% the eccentricity is greater than 1, so the system unbounds. If it is an equation for the eccentricity, isn't it dealing with all orbits?
some_math_guy
CC BY-SA 4.0
2024-02-20T07:50:02.123Z
@ProfRob Would it be correct saying that the mass doesn't really leave the system but stays as a common envelope as a solution or possible scenario for my question?
some_math_guy
CC BY-SA 4.0
2024-02-20T10:25:55.803Z
@some_math_guy part of the solution can be that mass lost by one star does not leave the system, but is transferred to the other star. The notes you referenced are quite good in explaining various scenarios. There are a number of possibilities and indeed another is that the initial binary is quite eccentric.
ProfRob
<binary-star><neutron-star><star-formation>
Why do double neutron stars remain bounded?
some_math_guy
# Why do double neutron stars remain bounded?

(Asked by: username_0 on 2024-02-18)

I read that in principle, the formation of a double neutron star requires that at the time of the second supernova explosion the binary consists of a $\sim$ 1.4 solar mass neutron star, and a second star whose initial mass was > 8 solar mass.

But it can be shown that a binary will become unbound should it lose more than 50% of its mass. This is equation 9.17 in [https://www.astro.ru.nl/~fverbunt/binaries/compbin.pdf](https://www.astro.ru.nl/%7Efverbunt/binaries/compbin.pdf)

I know that neutron stars have masses of around 2.5 solar mass. If I take for example that a t the time of the second explosion the masses are 1.4 and 9 solar masses and later the neutron starts are 1.4 and 2.5 there is a mass loss of more than 50%. So how can such a system ever remain bound?

**Tags:** binary-star, neutron-star, star-formation
> **Comments:**
> **username_1**: I would recommend including the equations you're using to show that the binary would become unbound if it loses more than 50% mass. I'd suspect you're assuming the stars are in circular orbits, rather than elliptical ones, as the required mass loss for escape changes  depending on when it happens (less near periapsis, more near apopasis), and at any random time, the stars are more likely to be closer to their  apoapses.  
> **username_0**: @notovny I added a link to the lecture notes with the relevant equation  
> **username_2**: That's a 137 page document.  For those wondering, the equation in question is at the bottom of page 117 of the pdf (page 114 of the document's internal numbering). I'll note that equation is exactly as notovny says: for circular orbits.  It is the following pages/section that deals with elliptical orbits, which are far more likely.  
> **username_3**: Section 9.1.15 onwards in the reference you quote is about how the problem you pose in your question is solved. In summary: mass transfer and binary evolution. Also note that the 8 solar mass lower limit is on the *initial* mass of a star that explodes.  
> **username_0**: @I know it is a 137 document.That's why I provided the relevant eq number 9.17, which is the same as saying page 114 or 117 of the pdf. From that eq it follows that with a mass loss of ore than 50% the eccentricity is greater than 1, so the system unbounds. If it is an equation for the eccentricity, isn't it dealing with all orbits?  
> **username_0**: @ProfRob Would it be correct saying that the mass doesn't really leave the system but stays as a common envelope as a solution or possible scenario for my question?  
> **username_3**: @some_math_guy part of the solution can be that mass lost by one star does not leave the system, but is transferred to the other star. The notes you referenced are quite good in explaining various scenarios. There are a number of possibilities and indeed another is that the initial binary is quite eccentric.  

---

## 0 answers

