<p>Stars that have a mass lower than about $0.5 M_{\odot}$ will not ignite helium in their cores, in an analogous fashion to the way that stars with $M&lt;8M_{\odot}$ have insufficiently massive cores that never reach high enough temperatures to ignite carbon.</p>

<p>The cause in both cases is the onset of electron degeneracy pressure, which is independent of temperature and allows the core to cool at constant pressure and radius. [A normal gas would contract and become hotter as it loses energy!]</p>

<p>The end result for a $0.5M_{\odot}$ star will be a helium white dwarf with a mass (depending on uncertain details of the mass-loss process) of around $0.2M_{\odot}$. Such things do exist in nature now, but only because they have undergone some kind of mass transfer event in a binary system that has accelerated their evolution. The collapse to a degenerate state would be <strong>inevitable</strong> even for the lowest mass stars (which would of course then be very low-mass white dwarfs). As an inert core contracts it loses heat and cools - a higher density and lower temperate eventually lead to degenerate conditions that allow the core to cool without losing pressure.</p>

<p>The lowest mass stars ($&lt;0.3 M_{\odot}$) do get there via a slightly different route - they are fully convective, so the "core" doesn't exist really, it is always mixed with the envelope. They do not develop into red giants and thus I guess will suffer much less mass loss.</p>

<p>The remnant <em>would</em> be a white dwarf in either case and is fundamentally different from a brown dwarf both in terms of size and structure, because it would be made of helium rather than (mostly) hydrogen. This should have an effect in two ways. For the same mass, the brown dwarf should end up bigger because the number of mass units per electron is smaller (1 vs 2) and also because the effects of a finite temperature are larger in material with fewer mass units per particle - i.e. its outer, non-degenerate layer would be more "puffed up". NB: The brown dwarfs we see today are Jupiter-sized, but are still cooling. They will get a bit smaller and more degenerate.</p>

<p>A simple size calculation could use the approximation of an ideal, cold, degenerate gas. A bit of simple physics using the virial theorem gives you
$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.013\left(\frac{\mu_e}{2}\right)^{-5/3} \left(\frac{M}{M_{\odot}}\right)^{-1/3},$$
where $\mu_e$ is the number of atomic mass units per electron.
Putting in appropriate numbers I get $0.32\ R_{Jup}$ for a $0.07M_{\odot}$ Helium white dwarf versus $1.01\ R_{Jup}$ for a $0.07M_{\odot}$ completely degenerate Hydrogen brown dwarf (in practice it would be a bit smaller because it isn't all hydrogen).</p>

<p>However, it would be interesting to see some realistic calculations of what happens to a $0.07M_{\odot}$ brown dwarf versus a $0.08M_{\odot}$ star in a trillion years or so. I will update the answer if I come across such a study.</p>

<p>EDIT: I knew I'd seen something on this. Check out <a href="http://adsabs.harvard.edu/abs/1997ApJ...482..420L" rel="nofollow">Laughlin et al. (1997)</a>, which studies the long-term evolution of very low-mass stars. Low-mass stars do not pass through a red giant phase, remain fully convective and can thus convert almost all their hydrogen into helium over the course of $10^{13}$ years and end up cooling as degenerate He white dwarfs.</p>

CC BY-SA 4.0
2023-12-04T04:30:23.067Z
I know this is a very late comment, but I have a question about μe, what numbers do you use for that? I used 0.00054 as, looking it up, that is the only value I can get on how much an electron weighs in amu. And I don't get how that would differ, as, as I understand it, electrons remain the same mass regardless of what atom they are a part of, the only thing that differs is the amount of electrons. I understand if this is too late a question and you forgot. Edit: I got 0.40 Jupiter Radii, assuming the result was in Km. Seems you used 0.00061738 to get 0.32 and 0.0003097 for 1.01. But how got?
DanceroftheStars
CC BY-SA 4.0
2023-12-04T04:33:40.947Z
I apologize for the bad grammar in the last sentence, I was running out of space.
DanceroftheStars
CC BY-SA 4.0
2023-12-04T15:49:52.570Z
@DanceoftheStars $\mu_e$ is the number of mass units per electron **in the gas**. For ionised helium, carbon or oxygen, $\mu_e = 2$. For ionised hydrogen $\mu_e=1$.
ProfRob
CC BY-SA 3.0
2015-04-02T11:50:09.030Z
2015-04-02T17:22:03.463Z
2015-04-02T17:22:03.463Z
ProfRob
<p>OK, first, I know there's a variety of sizes and types of red dwarf stars and the universe is too young for any of them to have reached the end of their main sequence phase yet, so it's all theoretical and/or modeling.</p>
<p><a href="http://en.wikipedia.org/wiki/Red_dwarf" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Red_dwarf</a></p>
<p>But what is the theoretical size needed for a star to undergo the electron degeneracy process which turns a small star from at least Jupiter sized, usually bigger, into to earth sized super dense object, where, as I understand it, the electrons are squeezed off the nuclei - electron degeneracy.</p>
<p>It would seem to me that a 7.5% solar mass star, which gradually burns hydrogen into helium but doesn't burn helium, might not have the mass to compact into a true white dwarf but might end it's life looking more like a brown dwarf / super Jupiter - well, talking appearance, not really, because super Jupiter &amp; Brown dwarfs are mostly hydrogen while and an end of life red dwarf should be mostly helium - which, in and of itself, might make the difference.</p>
<p>It's just my curiosity whether all red dwarfs turn into white dwarfs at the end of their burning phase or is there a theoretical mass that's needed for that level of shrinkage to occur?</p>
<p>Thanks.</p>

<stellar-evolution><stellar-astrophysics>
death of a red dwarf star / minimum mass needed for a white dwarf?
userLTK
# death of a red dwarf star / minimum mass needed for a white dwarf?

(Asked by: username_0 on 2015-04-02)

OK, first, I know there's a variety of sizes and types of red dwarf stars and the universe is too young for any of them to have reached the end of their main sequence phase yet, so it's all theoretical and/or modeling.

<http://en.wikipedia.org/wiki/Red_dwarf>

But what is the theoretical size needed for a star to undergo the electron degeneracy process which turns a small star from at least Jupiter sized, usually bigger, into to earth sized super dense object, where, as I understand it, the electrons are squeezed off the nuclei - electron degeneracy.

It would seem to me that a 7.5% solar mass star, which gradually burns hydrogen into helium but doesn't burn helium, might not have the mass to compact into a true white dwarf but might end it's life looking more like a brown dwarf / super Jupiter - well, talking appearance, not really, because super Jupiter & Brown dwarfs are mostly hydrogen while and an end of life red dwarf should be mostly helium - which, in and of itself, might make the difference.

It's just my curiosity whether all red dwarfs turn into white dwarfs at the end of their burning phase or is there a theoretical mass that's needed for that level of shrinkage to occur?

Thanks.

**Tags:** stellar-evolution, stellar-astrophysics


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## 1 answers

### Answer by username_1 on 2015-04-02. Score: 2 | Accepted answer

Stars that have a mass lower than about $0.5 M\_{\odot}$ will not ignite helium in their cores, in an analogous fashion to the way that stars with $M<8M\_{\odot}$ have insufficiently massive cores that never reach high enough temperatures to ignite carbon.

The cause in both cases is the onset of electron degeneracy pressure, which is independent of temperature and allows the core to cool at constant pressure and radius. [A normal gas would contract and become hotter as it loses energy!]

The end result for a $0.5M\_{\odot}$ star will be a helium white dwarf with a mass (depending on uncertain details of the mass-loss process) of around $0.2M\_{\odot}$. Such things do exist in nature now, but only because they have undergone some kind of mass transfer event in a binary system that has accelerated their evolution. The collapse to a degenerate state would be **inevitable** even for the lowest mass stars (which would of course then be very low-mass white dwarfs). As an inert core contracts it loses heat and cools - a higher density and lower temperate eventually lead to degenerate conditions that allow the core to cool without losing pressure.

The lowest mass stars ($<0.3 M\_{\odot}$) do get there via a slightly different route - they are fully convective, so the "core" doesn't exist really, it is always mixed with the envelope. They do not develop into red giants and thus I guess will suffer much less mass loss.

The remnant *would* be a white dwarf in either case and is fundamentally different from a brown dwarf both in terms of size and structure, because it would be made of helium rather than (mostly) hydrogen. This should have an effect in two ways. For the same mass, the brown dwarf should end up bigger because the number of mass units per electron is smaller (1 vs 2) and also because the effects of a finite temperature are larger in material with fewer mass units per particle - i.e. its outer, non-degenerate layer would be more "puffed up". NB: The brown dwarfs we see today are Jupiter-sized, but are still cooling. They will get a bit smaller and more degenerate.

A simple size calculation could use the approximation of an ideal, cold, degenerate gas. A bit of simple physics using the virial theorem gives you
$$ \left(\frac{R}{R\_{\odot}}\right) \simeq 0.013\left(\frac{\mu\_e}{2}\right)^{-5/3} \left(\frac{M}{M\_{\odot}}\right)^{-1/3},$$
where $\mu\_e$ is the number of atomic mass units per electron.
Putting in appropriate numbers I get $0.32\ R\_{Jup}$ for a $0.07M\_{\odot}$ Helium white dwarf versus $1.01\ R\_{Jup}$ for a $0.07M\_{\odot}$ completely degenerate Hydrogen brown dwarf (in practice it would be a bit smaller because it isn't all hydrogen).

However, it would be interesting to see some realistic calculations of what happens to a $0.07M\_{\odot}$ brown dwarf versus a $0.08M\_{\odot}$ star in a trillion years or so. I will update the answer if I come across such a study.

EDIT: I knew I'd seen something on this. Check out [Laughlin et al. (1997)](http://adsabs.harvard.edu/abs/1997ApJ...482..420L), which studies the long-term evolution of very low-mass stars. Low-mass stars do not pass through a red giant phase, remain fully convective and can thus convert almost all their hydrogen into helium over the course of $10^{13}$ years and end up cooling as degenerate He white dwarfs.

> **Comments:**
> **username_2**: I know this is a very late comment, but I have a question about μe, what numbers do you use for that? I used 0.00054 as, looking it up, that is the only value I can get on how much an electron weighs in amu. And I don't get how that would differ, as, as I understand it, electrons remain the same mass regardless of what atom they are a part of, the only thing that differs is the amount of electrons. I understand if this is too late a question and you forgot. Edit: I got 0.40 Jupiter Radii, assuming the result was in Km. Seems you used 0.00061738 to get 0.32 and 0.0003097 for 1.01. But how got?  
> **username_2**: I apologize for the bad grammar in the last sentence, I was running out of space.  
> **username_1**: @DanceoftheStars $\mu_e$ is the number of mass units per electron **in the gas**. For ionised helium, carbon or oxygen, $\mu_e = 2$. For ionised hydrogen $\mu_e=1$.  

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