According to the definition of the concept of barycenter, the question arose for me as a mental experiment that if there were an ultra-heavy binary system such as two neutron stars so that the center of mass was outside the surface of two stars (such as the Sun and Jupiter). Given the definition of a black hole, and assuming that the total mass of the two neutron stars has passed a critical value, can we say that a black hole is formed at the point where the center of mass of this system is located?


1 Answer 1


No. The black hole will form in the volume with the greatest average density.

$r_s \propto M$ where $M$ is the mass inside $r_s$. Barycenter is irrelevant. What matters is density, which is a local quantity. Mass is only indirectly relevant, in that the more mass you have, the less density you need.


$r_s = k_1 M$

and $M = |\rho| V$, where $|\rho|$ is average density within spherical volume $V$


$V = k_2 r^3$

we have

$|\rho| k_2 k_1^3 M^3 = M$

$|\rho| = 1/(k_2k_1^3M^2) = k_3/M^2$

The black hole forms when the density inside a given spherical volume V becomes equal to a proportionality constant divided by the square of the mass inside the volume.

Dimensional analysis for a sanity check:

$k_2 = 4\pi/3$

$k_1 = 2G/c^2$

$k_3 = 3c^6/(32 \pi G^3)$

$c^6$ units: $m^6/s^6$

$G^3$ units: $m^9/kg^3s^6$

$c^6/(G^3M^2)$ units: $kg/m^3$

Those are density units, so I'm reasonably confident I haven't messed up.

  • $\begingroup$ Calculations for a flat space are not applicable to black holes. The volume $V$ inside the Schwarzschild radius $r_s$ is not $\dfrac{4}{3}\pi r_s^3$, but zero. The Schwarzschild radius is not a spatial radial distance to the origin (which is also zero), but a reduced circumference of the horizon. Geometry of a curved space is very different from the Euclidean. $\endgroup$
    – safesphere
    Sep 26, 2021 at 15:50

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