<p>Using density is invalid.  As the radius of the event horizon for a given mass increases linearly, the volume of that radius increases as the cube and the density therefore decreases.  Looking at it the other way, the density increases as the event horizon decreases.</p>

<p>You can calculate the size of the <a href="https://en.wikipedia.org/wiki/Event_horizon" rel="noreferrer">event horizon</a> for any given mass.  You just need to find the point at which the <a href="https://en.wikipedia.org/wiki/Escape_velocity" rel="noreferrer">escape velocity</a> exceeds the speed of light. We can use the speed of light in the formula for escape velocity and solve for the radius</p>

<p>Escape velocity formula <img src="https://i.sstatic.net/xbdbH.png" alt="enter image description here"> solving for r gives 
<img src="https://i.sstatic.net/QsssV.gif" alt="enter image description here"></p>

<p>I put together a <a href="https://docs.google.com/spreadsheet/ccc?key=0Akd7u7y3k3jVdEZVQ0F1emliSGZreG1hekdmVWlfUmc&amp;usp=sharing" rel="noreferrer">spreadsheet</a> with the numbers. I calculate that a 3.2 solar mass black hole would have a radius of 4.752km, meaning that a neutron star of 3.2 solar masses were to become a black hole it would have to shrink to 9.504km and have a density of 7.13E18 kg/m^3.    Conversely the super-massive black hole at the center of our galaxy has an event horizon radius of about 6 billion km and a density of only 4.34E6 kg/m^3.  A black hole the size of a proton would need 350 million metric tons and have a density of 1.5E56 kg/m^3.</p>

<p>I think you are probably off on some of your numbers.  Specifically you are using ranges of numbers at the top end of a spectrum and an 'about' figure for the radius of a neutron star as though 12km is a single constant radius for all neutron stars.  In fact a 1.4 solar mass neutron star would nave a radius somewhere between 10.4 and 12.9 km (<a href="http://arxiv.org/abs/1205.6871" rel="noreferrer">source</a>)</p>

<p><a href="https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_about.html" rel="noreferrer">https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_about.html</a>
<img src="https://i.sstatic.net/lYH3Z.jpg" alt="enter image description here"></p>

CC BY-SA 3.0
2013-12-05T22:17:12.133Z
2013-12-05T22:48:49.463Z
2013-12-05T22:48:49.463Z
Jason Goemaat
<p>Let's just go back to the time when a red supergiant goes supernova. When it goes supernova, its outer shells are blown off because of the explosion. What happens next depends on the mass of the remnant. If the mass is 1.4 to 3 times the mass of the sun, it becomes a neutron star. If it is 3 times the mass or greater it becomes a black hole. Neutron stars cannot have the event horizons of black holes because their supernova remnant was simply not massive enough.</p>

CC BY-SA 4.0
2019-04-01T06:24:32.020Z
2019-04-01T06:24:32.020Z
Samvart Upadhyay
<p>It is said that neutron stars bend space/time so strongly that parts of the back are visible from the front! Of course a neutron star is essentially one very <em>very</em> large ball of neutrons with all the light elements on the surface. Some scientists now believe that simple neutron star collisions do not generate all heavy elements but the existence of elements heavier than iron is due to black hole-neutron star collisions. If so then they don't have an event horizon despite their enormous gravity because the matter is too spread out, whereas for a true black hole its all concentrated in one place.
In fact it is believed that the escape velocity for a typical neutron star is around 1/3 to 1/2 the speed of light, still a large number and incidentally life <em>may</em> be possible on a planet orbiting a neutron star with radiation tolerance sufficient even in a bacterium like Deinococcus radiodurans as long as the planet's orbit kept it well away from the jets.
A variant of this concept is when a neutron star hits a red supergiant briefly igniting helium fusion if the whole thing doesen't blow up first.<br>
<a href="https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/" rel="nofollow noreferrer">https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/</a></p>

CC BY-SA 4.0
2019-04-10T22:49:23.730Z
Heavier elements are coming from supernovas, NS-BH collision is very rare.
peterh
CC BY-SA 3.0
2018-03-23T05:49:25.870Z
2018-03-25T07:53:21.547Z
2018-03-25T07:53:21.547Z
Conundrum
<p>As Francesco Montesano points out, using the wrong mass leads to the wrong answer.  Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.</p>

<p>Since the density scales as &rho; ~ M/R^3 and the Schwarzschild radius as R<sub>s</sub>~M, the density of BHs scales as &rho;~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.</p>

CC BY-SA 4.0
2019-04-10T15:34:40.127Z
"more massive BHs are less dense" And of course an interesting consequence of this is that assuming a flat and non-expanding space, takings a volume of any positive density and scaling up its size in three dimensions while keeping density within it constant will eventually result in a black hole.
Shufflepants
CC BY-SA 3.0
2013-12-22T00:33:28.243Z
+1, though there's another reason why this density is bad: volume is completely frame-dependent. Wiki's density figures use Euclidean volume where the geometry is strongly non-Euclidean. With the metric in the Tolman-Oppenheimer-Volkoff ansantz, a spherically symmetric simple neutron star would have volume
$$V_\text{TOV} = \int_0^R \frac{4\pi r^2\,\mathrm{d}r}{\sqrt{1-\frac{2GM(r)}{rc^2}}}\text{,}$$
which is never Euclidean. In another frame, it'd be something else still. We could still use the Euclidean "overall density" to compare neutron stars, but the figure itself doesn't mean much.
Stan Liou
CC BY-SA 3.0
2013-10-20T11:19:47.007Z
2013-10-20T11:19:47.007Z
AstroFloyd
<p>Trying to compare density of Black Holes and Neutron Stars I came up with the following:</p>
<blockquote>
<p>A typical neutron star has a mass between about 1.4 and 3.2 solar masses<a href="http://curious.astro.cornell.edu/question.php?number=219" rel="noreferrer">1</a>[3] (see Chandrasekhar Limit), with a corresponding radius of about 12 km. (...) Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m^3 <a href="http://en.wikipedia.org/wiki/Neutron_star" rel="noreferrer">[1]</a></p>
</blockquote>
<p>and</p>
<blockquote>
<p>You can use the Schwarzschild radius to calculate the &quot;density&quot; of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x10^16 g/cm^3) x (Msun / M)^2 (...)</p>
<p>The value of the Schwarzschild radius works out to be about (3x10^5 cm) x (M / Msun) <a href="http://curious.astro.cornell.edu/question.php?number=219" rel="noreferrer">[2]</a></p>
</blockquote>
<p>Let's take a neutron star from the top of the spectrum (3.2 Msun) and same mass black hole.</p>
<p>Converting units:</p>
<ul>
<li>Neutron star: 5.9×10^17 kg/m^3 = 5.9 × 10^14 g/cm^3</li>
<li>Black hole: 1.8x10^16 g/cm^3 x (1/5.9)^2 = 5.2 x10^14 g/cm^3</li>
</ul>
<p>The radius of the black hole would be (3x10^5 cm) x ( 5.2 ) = 15.6km</p>
<p>The 3.2Msun Neutron Star of this density would have volume of 1.08 x 10^13 m^3 which gives radius of 13.7 kilometers</p>
<p>According to Shell Theorem, spherical objects' gravity field strength at given distance is the same for spheres as for point masses so at the same distance from center of same mass (point - black hole, sphere - neutron star) the gravity will be the same.</p>
<p>That would put the surface of the neutron star below the surface of event horizon of equivalent black hole. Yet I never heard about even horizon of neutron stars.</p>
<p>Either I made a mistake in my calculations (and if I did, could you point it out?) or... well, why?</p>

CC BY-SA 4.0
2019-04-01T06:35:30.043Z
Why has this got so many upvotes. It contains a trivial error in the Schwarzschild radius. R_s is 2.96 km per solar mass.
ProfRob
CC BY-SA 3.0
2013-10-18T19:14:48.400Z
There is an error: where did you get the *5.9* in the equation for the black hole and the 5.2 in the radius of the black hole? You must use 3.2. In this way you get 1.7x10^15 g/cm^3 as density and 9.6km as radius
Francesco Montesano
<black-hole><neutron-star>
Why don't Neutron Stars form event horizon?
SF.
# Why don't Neutron Stars form event horizon?

(Asked by: username_0 on 2013-10-18)

Trying to compare density of Black Holes and Neutron Stars I came up with the following:

> A typical neutron star has a mass between about 1.4 and 3.2 solar masses[1](http://curious.astro.cornell.edu/question.php?number=219)[3] (see Chandrasekhar Limit), with a corresponding radius of about 12 km. (...) Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m^3 [[1]](http://en.wikipedia.org/wiki/Neutron_star)

and

> You can use the Schwarzschild radius to calculate the "density" of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x10^16 g/cm^3) x (Msun / M)^2 (...)
>
> The value of the Schwarzschild radius works out to be about (3x10^5 cm) x (M / Msun) [[2]](http://curious.astro.cornell.edu/question.php?number=219)

Let's take a neutron star from the top of the spectrum (3.2 Msun) and same mass black hole.

Converting units:

* Neutron star: 5.9×10^17 kg/m^3 = 5.9 × 10^14 g/cm^3
* Black hole: 1.8x10^16 g/cm^3 x (1/5.9)^2 = 5.2 x10^14 g/cm^3

The radius of the black hole would be (3x10^5 cm) x ( 5.2 ) = 15.6km

The 3.2Msun Neutron Star of this density would have volume of 1.08 x 10^13 m^3 which gives radius of 13.7 kilometers

According to Shell Theorem, spherical objects' gravity field strength at given distance is the same for spheres as for point masses so at the same distance from center of same mass (point - black hole, sphere - neutron star) the gravity will be the same.

That would put the surface of the neutron star below the surface of event horizon of equivalent black hole. Yet I never heard about even horizon of neutron stars.

Either I made a mistake in my calculations (and if I did, could you point it out?) or... well, why?

**Tags:** black-hole, neutron-star

> **Comments:**
> **username_2**: There is an error: where did you get the *5.9* in the equation for the black hole and the 5.2 in the radius of the black hole? You must use 3.2. In this way you get 1.7x10^15 g/cm^3 as density and 9.6km as radius  
> **username_1**: Why has this got so many upvotes. It contains a trivial error in the Schwarzschild radius. R_s is 2.96 km per solar mass.  

---

## 4 answers

### Answer by username_7 on 2013-10-20. Score: 15 | Accepted answer

As username_2 points out, using the wrong mass leads to the wrong answer. Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.

Since the density scales as ρ ~ M/R^3 and the Schwarzschild radius as Rs~M, the density of BHs scales as ρ~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.

> **Comments:**
> **username_9**: +1, though there's another reason why this density is bad: volume is completely frame-dependent. Wiki's density figures use Euclidean volume where the geometry is strongly non-Euclidean. With the metric in the Tolman-Oppenheimer-Volkoff ansantz, a spherically symmetric simple neutron star would have volume $$V_\text{TOV} = \int_0^R \frac{4\pi r^2\,\mathrm{d}r}{\sqrt{1-\frac{2GM(r)}{rc^2}}}\text{,}$$ which is never Euclidean. In another frame, it'd be something else still. We could still use the Euclidean "overall density" to compare neutron stars, but the figure itself doesn't mean much.  
> **username_8**: "more massive BHs are less dense" And of course an interesting consequence of this is that assuming a flat and non-expanding space, takings a volume of any positive density and scaling up its size in three dimensions while keeping density within it constant will eventually result in a black hole.  

---

### Answer by username_3 on 2013-12-05. Score: 8

Using density is invalid. As the radius of the event horizon for a given mass increases linearly, the volume of that radius increases as the cube and the density therefore decreases. Looking at it the other way, the density increases as the event horizon decreases.

You can calculate the size of the [event horizon](https://en.wikipedia.org/wiki/Event_horizon) for any given mass. You just need to find the point at which the [escape velocity](https://en.wikipedia.org/wiki/Escape_velocity) exceeds the speed of light. We can use the speed of light in the formula for escape velocity and solve for the radius

Escape velocity formula ![enter image description here](https://i.sstatic.net/xbdbH.png) solving for r gives
![enter image description here](https://i.sstatic.net/QsssV.gif)

I put together a [spreadsheet](https://docs.google.com/spreadsheet/ccc?key=0Akd7u7y3k3jVdEZVQ0F1emliSGZreG1hekdmVWlfUmc&usp=sharing) with the numbers. I calculate that a 3.2 solar mass black hole would have a radius of 4.752km, meaning that a neutron star of 3.2 solar masses were to become a black hole it would have to shrink to 9.504km and have a density of 7.13E18 kg/m^3. Conversely the super-massive black hole at the center of our galaxy has an event horizon radius of about 6 billion km and a density of only 4.34E6 kg/m^3. A black hole the size of a proton would need 350 million metric tons and have a density of 1.5E56 kg/m^3.

I think you are probably off on some of your numbers. Specifically you are using ranges of numbers at the top end of a spectrum and an 'about' figure for the radius of a neutron star as though 12km is a single constant radius for all neutron stars. In fact a 1.4 solar mass neutron star would nave a radius somewhere between 10.4 and 12.9 km ([source](http://arxiv.org/abs/1205.6871))

<https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_about.html>
![enter image description here](https://i.sstatic.net/lYH3Z.jpg)


---

### Answer by username_4 on 2019-04-01. Score: 3

Let's just go back to the time when a red supergiant goes supernova. When it goes supernova, its outer shells are blown off because of the explosion. What happens next depends on the mass of the remnant. If the mass is 1.4 to 3 times the mass of the sun, it becomes a neutron star. If it is 3 times the mass or greater it becomes a black hole. Neutron stars cannot have the event horizons of black holes because their supernova remnant was simply not massive enough.


---

### Answer by username_5 on 2018-03-23. Score: -2

It is said that neutron stars bend space/time so strongly that parts of the back are visible from the front! Of course a neutron star is essentially one very *very* large ball of neutrons with all the light elements on the surface. Some scientists now believe that simple neutron star collisions do not generate all heavy elements but the existence of elements heavier than iron is due to black hole-neutron star collisions. If so then they don't have an event horizon despite their enormous gravity because the matter is too spread out, whereas for a true black hole its all concentrated in one place.
In fact it is believed that the escape velocity for a typical neutron star is around 1/3 to 1/2 the speed of light, still a large number and incidentally life *may* be possible on a planet orbiting a neutron star with radiation tolerance sufficient even in a bacterium like Deinococcus radiodurans as long as the planet's orbit kept it well away from the jets.
A variant of this concept is when a neutron star hits a red supergiant briefly igniting helium fusion if the whole thing doesen't blow up first.  
<https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/>

> **Comments:**
> **username_6**: Heavier elements are coming from supernovas, NS-BH collision is very rare.  

---

