I was wondering how much you had to crush an object for it to become a black hole. Recently I learned that anything could become a black hole (even you) if it were crushed down small enough, for example our sun would have to be crushed to the size of Manhattan to become a black hole (According to a video I saw). I did a little research and found out that something would have to be crushed down to Schwarzchild radius meaning all of that objects mass would have to be crushed into that objects radius. But if you say crush the sun down to the size of Manhattan, that is much smaller than the sun's actual radius. So I haven't really gotten a clear answer and I'm a little confused, thanks for clarifying =)

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    $\begingroup$ But if you say crush the sun down to the size of Manhattan, that is much smaller than the sun's actual radius. - Sure, but that doesn't mean that an object can't be made smaller. $\endgroup$
    – HDE 226868
    Commented Jul 7, 2015 at 23:47
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    $\begingroup$ The Schwarzchild radius is the definitive answer. It depends on mass. Small mass = small radius; big mass = big radius. That's all. $\endgroup$ Commented Jul 8, 2015 at 0:08
  • $\begingroup$ @FlorinAndrei You should consider posting that as an answer so the OP can mark the question as solved. $\endgroup$
    – fantasia
    Commented Jul 8, 2015 at 1:37
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    $\begingroup$ If you have some assumptions about the matter and object, you can put more detailed limits. For example, if your matter cannot support superluminal sound waves, then a shell of it would only need to be smaller than $25/24$ of its Schwarzschild radius in order to necessarily become a black hole, and a uniformly-dense ball of it smaller than $4/3$ of its Schwarzschild radius. Outside toy problems like that, though, it becomes a tough question. ... I wonder how small, e.g., a neutron star can be given a fixed mass before it necessarily collapses, and how this depends on assumed EOS, etc. $\endgroup$
    – Stan Liou
    Commented Jul 8, 2015 at 8:42
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    $\begingroup$ @StanLiou tuns out it is actually about 1.4 Schwarzschild radii. See my answer. $\endgroup$
    – ProfRob
    Commented Jul 8, 2015 at 17:41

3 Answers 3


Well, I wasn't going to answer but the other two answers are wrong, or at least incomplete. If you wish to make a black hole from a stellar-sized object, then there is no need to compress it to as small as the Schwarzschild radius (though that would certainly work and would certainly be the answer for smaller objects with negligible self-gravity). Instead, you just need to compress it to a size at which it cannot be possibly supported by any plausible equation of state against further gravitational collapse. It turns out that this is somewhat bigger than the Schwarzschild radius and hence the density required is considerably lower. More details below.

There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse.

There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro & Teukolsky, pp.260-261, it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that $(GM/Rc^2)<0.405$.

The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R > 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$. A more accurate treatment in Lattimer (2013) suggests that a maximally compact neutron star has $R\geq 1.41R_s$.

If the equation of state is softer, then collapse will occur at smaller masses, and higher densities but at a similar multiple of $R_s$.

Thus it is not necessary to compress matter within $R_s$ to form a black hole.

The picture below (from Demorest et al. 2010) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled "causality" and which gives radii slightly larger than Shapiro & Teukolsky's approximate result) and then in the very top left, the border marked by "GR" coincides with the Schwarzschild radius. Neutron stars become unstable where their mass-radius curves peak, so stable neutron stars are always significantly larger than $R_s$ at all masses.

Neutron star mass-radius relations

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    $\begingroup$ Lord, relativistic kinetics! I'll happily accept the incomplete. My understanding of your post is that the equation I gave holds after everything settles down to a non-spinning steady state; not accounting for accretion or Hawking radiation. $\endgroup$ Commented Jul 9, 2015 at 2:14

$$r_s = \frac{2GM}{c^2}$$ --Here:

$r_s$ is the Schwarzschild radius;
$G$ is the gravitational constant;
$M$ is the mass of the object;
$c$ is the speed of light in vacuum.

The proportionality constant, $2G/c^2$, is approximately 1.48×1027 m/kg


Well, it has to be crushed (compressed would be the better word to use here) below the Schwarzschild radius.

The Schwarzschild radius is the radius of the object in which the escape velocity would be the speed of light from that object. When this radius becomes even smaller, even light cannot escape it and voilà, there's your black hole!


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