<p>Your question inspired me to have a look at the data that is used in plots produced by the stellar evolution code <a href="http://mesa.sourceforge.net" rel="nofollow noreferrer">MESA</a>. Here's a $\rho$-$T$ diagram showing boundaries of roughly where various nuclear reaction chains kick in as well as a solar model (black dashed line) and the a line indicating the onset of electron degeneracy (red line).
<a href="https://i.sstatic.net/suVDn.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/suVDn.png" alt="enter image description here"></a>
To answer the question, the point at which the "He burn" line intersects $10^8\,\mathrm{K}$ is around $4000\,\mathrm{g.cm}^{-3}$ but you can also see how the boundary varies somewhat as a function of temperature.</p>

<p>As a disclaimer, I don't actually know where these numbers come from but I'm asking around about it. If I find out, I'll update this answer.</p>

CC BY-SA 4.0
2019-05-23T19:49:27.197Z
I'd like to talk with you about this plot on the chat. Would you join me? https://chat.stackoverflow.com/rooms/193851/room-for-jasmeru-and-warrick
Jasmeru
CC BY-SA 3.0
2016-11-28T07:44:16.860Z
2016-11-28T07:44:16.860Z
Warrick
<p>But note that high-mass stars begin helium fusion when the electrons are still an ideal gas, but lower-mass stars like the Sun begin helium fusion when the electrons are degenerate.  So that produces a "helium flash," which the question is not about, but after the flash stabilizes (when the electrons are once again acting like an ideal gas), you get normal helium fusion in low-mass stars.  Typically, the density will be much higher in a low-mass star, perhaps an order of magnitude higher than what is quoted for the high-mass stars.  So the stellar mass will matter in answering the question, but you get the basic idea.</p>

CC BY-SA 3.0
2016-11-28T04:58:16.577Z
2016-11-28T04:58:16.577Z
Ken G
<p>The short answer: $\sim10^4$ grams per cubic centimeter.</p>

<p>From <a href="http://www.astro.cornell.edu/academics/courses/astro201/highmass.htm" rel="nofollow noreferrer">this webpage</a>, I have a few statistics regarding required mean density and during each phase of fusion (the fusion of lighter elements may happen in the "shells" outside the core).
$$\begin{array}{|c|c|}
\hline \text{Fusion phase}&amp;\text{Mean density (g/cm}^3)\\
\hline \text{Hydrogen} &amp; 5\\
\hline \text{Helium} &amp; 700\\
\hline \text{Carbon} &amp; 200,000\\
\hline \text{Neon} &amp; 4,000,000\\
\hline \text{Oxygen} &amp; 10,000,000\\
\hline \text{Silicon} &amp; 30,000,000\\
\hline
\end{array}$$
This is data for a star of $25M_{\odot}$.</p>

<p>Note, however, that this is <em>mean</em> density in the star, <em>not</em> the density in the core, where fusion is taking place. That is on the order of $\sim10^4$ grams per cubic centimeter (see e.g. <a href="https://astro.uni-bonn.de/~nlanger/siu_web/ssescript/new/chapter9.pdf" rel="nofollow noreferrer">here</a>). There may be a sharp density dropoff starting from the center of the star, showing that only the hot, dense inner regions can fuse the main element in each stage.</p>

CC BY-SA 3.0
2014-11-06T15:35:50.073Z
2016-11-27T19:13:49.047Z
2016-11-27T19:13:49.047Z
HDE 226868
<p>I have seen several references that say that helium burning begins in a star once the core temperature reaches $10^8$K (<a href="http://en.wikipedia.org/wiki/Triple-alpha_process" rel="nofollow">such as here</a>) but nowhere that says what density that corresponds to.  Does anybody know a reference that has this value?  And if it makes a difference, I would like to know the density for quiescent helium burning, not helium flash helium burning.</p>

CC BY-SA 3.0
2014-11-06T15:22:47.697Z
Order of $10^4$ g/cc I will elaborate when I get to my desktop
Cheeku
<star><helium>
At what density does helium burning start in a star?
NeutronStar
# At what density does helium burning start in a star?

(Asked by: username_0 on 2014-11-06)

I have seen several references that say that helium burning begins in a star once the core temperature reaches $10^8$K ([such as here](http://en.wikipedia.org/wiki/Triple-alpha_process)) but nowhere that says what density that corresponds to. Does anybody know a reference that has this value? And if it makes a difference, I would like to know the density for quiescent helium burning, not helium flash helium burning.

**Tags:** star, helium

> **Comments:**
> **username_1**: Order of $10^4$ g/cc I will elaborate when I get to my desktop  

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## 3 answers

### Answer by username_5 on 2014-11-06. Score: 6 | Accepted answer

The short answer: $\sim10^4$ grams per cubic centimeter.

From [this webpage](http://www.astro.cornell.edu/academics/courses/astro201/highmass.htm), I have a few statistics regarding required mean density and during each phase of fusion (the fusion of lighter elements may happen in the "shells" outside the core).
$$\begin{array}{|c|c|}
\hline \text{Fusion phase}&\text{Mean density (g/cm}^3)\\
\hline \text{Hydrogen} & 5\\
\hline \text{Helium} & 700\\
\hline \text{Carbon} & 200,000\\
\hline \text{Neon} & 4,000,000\\
\hline \text{Oxygen} & 10,000,000\\
\hline \text{Silicon} & 30,000,000\\
\hline
\end{array}$$
This is data for a star of $25M\_{\odot}$.

Note, however, that this is *mean* density in the star, *not* the density in the core, where fusion is taking place. That is on the order of $\sim10^4$ grams per cubic centimeter (see e.g. [here](https://astro.uni-bonn.de/~nlanger/siu_web/ssescript/new/chapter9.pdf)). There may be a sharp density dropoff starting from the center of the star, showing that only the hot, dense inner regions can fuse the main element in each stage.


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### Answer by username_2 on 2016-11-28. Score: 3

Your question inspired me to have a look at the data that is used in plots produced by the stellar evolution code [MESA](http://mesa.sourceforge.net). Here's a $\rho$-$T$ diagram showing boundaries of roughly where various nuclear reaction chains kick in as well as a solar model (black dashed line) and the a line indicating the onset of electron degeneracy (red line).
[![enter image description here](https://i.sstatic.net/suVDn.png)](https://i.sstatic.net/suVDn.png)
To answer the question, the point at which the "He burn" line intersects $10^8\,\mathrm{K}$ is around $4000\,\mathrm{g.cm}^{-3}$ but you can also see how the boundary varies somewhat as a function of temperature.

As a disclaimer, I don't actually know where these numbers come from but I'm asking around about it. If I find out, I'll update this answer.

> **Comments:**
> **username_3**: I'd like to talk with you about this plot on the chat. Would you join me? https://chat.stackoverflow.com/rooms/193851/room-for-username_3-and-username_2  

---

### Answer by username_4 on 2016-11-28. Score: 2

But note that high-mass stars begin helium fusion when the electrons are still an ideal gas, but lower-mass stars like the Sun begin helium fusion when the electrons are degenerate. So that produces a "helium flash," which the question is not about, but after the flash stabilizes (when the electrons are once again acting like an ideal gas), you get normal helium fusion in low-mass stars. Typically, the density will be much higher in a low-mass star, perhaps an order of magnitude higher than what is quoted for the high-mass stars. So the stellar mass will matter in answering the question, but you get the basic idea.


---

