<p>Black holes are created because the core of the star becomes very <strong>dense</strong>, not just because the star is massive. Prior to the creation of the black hole, the core is able to create enough outward pressure to prevent the core from gravitationally collapsing to the density needing to create a black hole.</p>

CC BY-SA 3.0
2014-09-24T21:18:06.617Z
2014-09-24T21:18:06.617Z
Aaron
<p>A supernova <em>may</em> actually be necessary in the creation of a stellar black hole.</p>

<p>At the ends of their lives the cores of massive stars are made mostly of iron-peak nuclei from which you cannot extract more fusion energy. To support their weight, these stars rely on electron degeneracy pressure - the pressure caused by the Pauli exclusion principle allowing no more than one electron to share the same quantum state.</p>

<p>In principle a star might be supported by degeneracy pressure forever as it gradually cools - this is the fate of most white dwarfs.</p>

<p>However, the core of a massive star is just too big for that to work. The density increases until all the electron are moving at close-to the speed of light and that's as high as the degeneracy pressure can get. If the core exceeds the Chandrasekhar mass, it will collapse and as it does so, the rest of the star collapses with it (a little more slowly).</p>

<p>The collapse is triggered by the removal of electrons by electron capture into nuclei to form neutrons. At some point enough neutrons are produced for neutron degeneracy pressure to halt or at least slow the collapse. This and the release of a lot of gravitational potential energy are ultimately what power a supernova explosion. But if the collapse is not halted then even neutron degeneracy pressure will not support the star and collapse to a black hole becomes inevitable. A black hole status is reached once a proportion of its mass is compressed inside its Schwarzschild radius $r_s = 2GM/c^2$. i.e. once its density achieves
$$ \rho &gt; \frac{3M}{4\pi r_s^{3}}$$
i.e. when a central mass $M$ has a density that exceeds
$$ \rho &gt; \frac{3}{32\pi} \frac{c^6}{G^3 M^2} = 1.8\times10^{19} \left(\frac{M}{M_{\odot}}\right)^{-2}\ {\rm kg/m}^3$$
This is a ball park figure and assumes spherical symmetry and neglects any detailed GR treatment, but is more or less correct - a few times higher than typical neutron star densities. </p>

<p>In other words it is the <em>density</em> of the material that largely determines whether something becomes a black hole. The mass is only an indirect parameter.</p>

CC BY-SA 3.0
2014-12-11T15:35:38.523Z
2015-01-15T21:43:08.383Z
2015-01-15T21:43:08.383Z
ProfRob
<p>The star before it turns into a black hole has what is known as radiation pressure, i.e. the fusing of elements that creates nuclear explosions. This outward radiation pressure counter balances the inward force of gravity but when the star finally runs out of fuel that radiation pressure stops.  Thus the only remaining force is gravity.  Gravity then attracts what is left of the star after the super nova inwards into a deep dense core which then forms the black hole.</p>

CC BY-SA 3.0
2014-09-25T06:56:19.273Z
2014-09-25T06:56:19.273Z
James Willcox
<p>Whether an object is a black hole is not just determined by its mass.  It's determined by whether that mass is entirely within its <a href="http://en.wikipedia.org/wiki/Schwarzschild_radius" rel="nofollow">Schwarzschild radius</a>.</p>

<p>In principle, any object can be a black hole <em>if</em> all its mass is concentrated into a sufficiently small volume.</p>

<p>For example, the Sun's Schwarzschild radius is about 3.0 km -- but its actual radius is about 700,000 km.  It could become a black hole only if it were compressed down to a radius of 3.0 km.</p>

CC BY-SA 3.0
2014-12-12T02:15:49.037Z
2014-12-12T02:15:49.037Z
Keith Thompson
<p>You are correct in saying that a star loses a lot of its mass in a supernova. However, there <em>is</em> a reason why the star still becomes a black hole. Actually, I suppose the question here is "Why doesn't a star become a black hole before it even undergoes a supernova?"</p>

<p>There is a reason for a supernova (I'll assume you're talking about type II supernovae, which result from incredibly massive stars). Stars undergo nuclear fusion, and this leads to "thermal pressure", which counteracts the force of gravity. Without this pressure, gravity would indeed make a sufficiently large star collapse upon itself. Gravitational collapse occurs when there is not enough pressure to counteract gravity; the result is a spectacular supernova. So stars only become black holes (or other compact objects, such as neutron stars) when they cannot produce enough energy to counteract the force of gravity due to their own mass.</p>

<p>As for the first part of your question (sorry for answering in reverse), light in the area of a black hole cannot escape if it is inside its event horizon or on a trajectory towards it. The radius of the event horizon for a non-rotating black hole is its Schwarzschild radius, which is proportional to the mass of the black hole. The reason this is non-applicable in stars is because the Schwarzschild radius in stars is deep inside its interior, and there is not a strong enough gravitational field to produce an event horizon to trap any light near it.</p>

<p>Thermal pressure reference: <a href="https://en.wikipedia.org/wiki/Gravitational_collapse">https://en.wikipedia.org/wiki/Gravitational_collapse</a></p>

<p>I hope this helps.</p>

CC BY-SA 3.0
2016-09-30T15:25:09.257Z
"My question was rather why a star that has MORE mass than the black hole doesn't trap light as well" Couldn't be simpler: "blackhole-ness depends on density, not mass".
Fattie
CC BY-SA 3.0
2014-09-18T23:17:07.680Z
Why would the gravitational field be any different for a black hole than for a massive star?  If the star has more mass than the black hole, shouldn't the field be stronger?  Or is there some aspect of density that affects gravity as well?
Scottie
CC BY-SA 3.0
2014-09-18T23:18:38.323Z
You're right; I meant that there are forces countering gravitational collapse, and the net force is not enough to produce a black hole.
HDE 226868
CC BY-SA 3.0
2014-09-19T08:36:28.903Z
@Scottie The strength of gravity depends upon the distance to the center of mass. When inside the star you only experience gravity from material closer to the center than you are. Black holes etc. are much more compact than a main sequence star, so you can get much closer to the center before being inside, and so can experience stronger gravity.
zibadawa timmy
CC BY-SA 3.0
2014-09-19T10:21:51.437Z
I think @zibadawatimmy's comment is what the OP is actually asking.
Warrick
CC BY-SA 3.0
2014-09-21T00:27:02.000Z
A valid point. @Scottie - Were you indeed asking about what zibadawatimmy mentioned?
HDE 226868
CC BY-SA 3.0
2014-09-21T20:53:14.483Z
Yes, my question wasn't so much about why a star collapses into a black hole.  I understand how the processes of outward pressure is in equalibrium with gravity.   My question was rather why a star that has MORE mass than the black hole doesn't trap light as well.  It would seem to me that more mass=more gravity=traps light.
Scottie
CC BY-SA 3.0
2014-09-21T20:56:56.920Z
@Scottie I tried to address that by saying that light cannot escape from within the event horizon; however, the event horizon is much smaller than the progenitor star.
HDE 226868
CC BY-SA 3.0
2014-09-21T20:58:15.010Z
Right.  It was answered well.  I was just clarifying my question since you asked.  Thank you for the answer!
Scottie
CC BY-SA 3.0
2014-09-21T21:00:03.967Z
I just wanted to know if you wanted another answer from zibadawatimmy, because I thought it would be interesting if he approached it from a different prespective. You're welcome!
HDE 226868
CC BY-SA 3.0
2014-09-18T23:13:00.780Z
2014-09-18T23:13:00.780Z
HDE 226868
<p>If the mass of a black hole is creating so much gravity that light cannot escape, why isn't the mass of the star that created the black hole (before it went supernova) trapping light as well?</p>

<p>By all accounts, that pre-supernova star should have boatloads more mass than the black hole post supernova, right?  Doesn't the star lose most of it's mass when going supernova?</p>

<gravity><black-hole>
Why isn't the star that created the black hole a black hole?
Scottie
# Why isn't the star that created the black hole a black hole?

(Asked by: username_0 on 2014-09-18)

If the mass of a black hole is creating so much gravity that light cannot escape, why isn't the mass of the star that created the black hole (before it went supernova) trapping light as well?

By all accounts, that pre-supernova star should have boatloads more mass than the black hole post supernova, right? Doesn't the star lose most of it's mass when going supernova?

**Tags:** gravity, black-hole


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## 5 answers

### Answer by username_5 on 2014-09-18. Score: 7 | Accepted answer

You are correct in saying that a star loses a lot of its mass in a supernova. However, there *is* a reason why the star still becomes a black hole. Actually, I suppose the question here is "Why doesn't a star become a black hole before it even undergoes a supernova?"

There is a reason for a supernova (I'll assume you're talking about type II supernovae, which result from incredibly massive stars). Stars undergo nuclear fusion, and this leads to "thermal pressure", which counteracts the force of gravity. Without this pressure, gravity would indeed make a sufficiently large star collapse upon itself. Gravitational collapse occurs when there is not enough pressure to counteract gravity; the result is a spectacular supernova. So stars only become black holes (or other compact objects, such as neutron stars) when they cannot produce enough energy to counteract the force of gravity due to their own mass.

As for the first part of your question (sorry for answering in reverse), light in the area of a black hole cannot escape if it is inside its event horizon or on a trajectory towards it. The radius of the event horizon for a non-rotating black hole is its Schwarzschild radius, which is proportional to the mass of the black hole. The reason this is non-applicable in stars is because the Schwarzschild radius in stars is deep inside its interior, and there is not a strong enough gravitational field to produce an event horizon to trap any light near it.

Thermal pressure reference: <https://en.wikipedia.org/wiki/Gravitational_collapse>

I hope this helps.

> **Comments:**
> **username_0**: Why would the gravitational field be any different for a black hole than for a massive star?  If the star has more mass than the black hole, shouldn't the field be stronger?  Or is there some aspect of density that affects gravity as well?  
> **username_5**: You're right; I meant that there are forces countering gravitational collapse, and the net force is not enough to produce a black hole.  
> **username_7**: @username_0 The strength of gravity depends upon the distance to the center of mass. When inside the star you only experience gravity from material closer to the center than you are. Black holes etc. are much more compact than a main sequence star, so you can get much closer to the center before being inside, and so can experience stronger gravity.  
> **username_8**: I think @zibadawatimmy's comment is what the OP is actually asking.  
> **username_5**: A valid point. @username_0 - Were you indeed asking about what zibadawatimmy mentioned?  
> **username_0**: Yes, my question wasn't so much about why a star collapses into a black hole.  I understand how the processes of outward pressure is in equalibrium with gravity.   My question was rather why a star that has MORE mass than the black hole doesn't trap light as well.  It would seem to me that more mass=more gravity=traps light.  
> **username_5**: @username_0 I tried to address that by saying that light cannot escape from within the event horizon; however, the event horizon is much smaller than the progenitor star.  
> **username_0**: Right.  It was answered well.  I was just clarifying my question since you asked.  Thank you for the answer!  
> **username_5**: I just wanted to know if you wanted another answer from zibadawatimmy, because I thought it would be interesting if he approached it from a different prespective. You're welcome!  
> **username_6**: "My question was rather why a star that has MORE mass than the black hole doesn't trap light as well" Couldn't be simpler: "blackhole-ness depends on density, not mass".  

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### Answer by username_1 on 2014-09-24. Score: 2

Black holes are created because the core of the star becomes very **dense**, not just because the star is massive. Prior to the creation of the black hole, the core is able to create enough outward pressure to prevent the core from gravitationally collapsing to the density needing to create a black hole.


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### Answer by username_2 on 2014-12-11. Score: 2

A supernova *may* actually be necessary in the creation of a stellar black hole.

At the ends of their lives the cores of massive stars are made mostly of iron-peak nuclei from which you cannot extract more fusion energy. To support their weight, these stars rely on electron degeneracy pressure - the pressure caused by the Pauli exclusion principle allowing no more than one electron to share the same quantum state.

In principle a star might be supported by degeneracy pressure forever as it gradually cools - this is the fate of most white dwarfs.

However, the core of a massive star is just too big for that to work. The density increases until all the electron are moving at close-to the speed of light and that's as high as the degeneracy pressure can get. If the core exceeds the Chandrasekhar mass, it will collapse and as it does so, the rest of the star collapses with it (a little more slowly).

The collapse is triggered by the removal of electrons by electron capture into nuclei to form neutrons. At some point enough neutrons are produced for neutron degeneracy pressure to halt or at least slow the collapse. This and the release of a lot of gravitational potential energy are ultimately what power a supernova explosion. But if the collapse is not halted then even neutron degeneracy pressure will not support the star and collapse to a black hole becomes inevitable. A black hole status is reached once a proportion of its mass is compressed inside its Schwarzschild radius $r\_s = 2GM/c^2$. i.e. once its density achieves
$$ \rho > \frac{3M}{4\pi r\_s^{3}}$$
i.e. when a central mass $M$ has a density that exceeds
$$ \rho > \frac{3}{32\pi} \frac{c^6}{G^3 M^2} = 1.8\times10^{19} \left(\frac{M}{M\_{\odot}}\right)^{-2}\ {\rm kg/m}^3$$
This is a ball park figure and assumes spherical symmetry and neglects any detailed GR treatment, but is more or less correct - a few times higher than typical neutron star densities.

In other words it is the *density* of the material that largely determines whether something becomes a black hole. The mass is only an indirect parameter.


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### Answer by username_3 on 2014-09-25. Score: 0

The star before it turns into a black hole has what is known as radiation pressure, i.e. the fusing of elements that creates nuclear explosions. This outward radiation pressure counter balances the inward force of gravity but when the star finally runs out of fuel that radiation pressure stops. Thus the only remaining force is gravity. Gravity then attracts what is left of the star after the super nova inwards into a deep dense core which then forms the black hole.


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### Answer by username_4 on 2014-12-12. Score: 0

Whether an object is a black hole is not just determined by its mass. It's determined by whether that mass is entirely within its [Schwarzschild radius](http://en.wikipedia.org/wiki/Schwarzschild_radius).

In principle, any object can be a black hole *if* all its mass is concentrated into a sufficiently small volume.

For example, the Sun's Schwarzschild radius is about 3.0 km -- but its actual radius is about 700,000 km. It could become a black hole only if it were compressed down to a radius of 3.0 km.


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