<p>Yes, it has to do something with metallicity.</p>
<p>Brown dwarf SDSS J0104+1535 <span class="math-container">$\to 0.086\rm\, M_\odot\to[Fe/H]=-2.4$</span></p>
<p>Red dwarf EBLM J0555-57Ab <span class="math-container">$\to 0.081\rm\, M_\odot\to [Fe/H]=-0.24$</span></p>
<p>There is <a href="https://web.archive.org/web/20061114221813/http://space.newscientist.com/article/dn9771-mass-cutoff-between-stars-and-brown-dwarfs-revealed.html#:%7E:text=Theory%20predicts%20that%20the%20mass%20cut%2Doff%20for%20what%20constitutes%20a%20star%20is%20different%20for%20objects%20of%20different%20metallicity%2C%20which%20refers%20to%20the%20proportion%20of%20elements%20heavier%20than%20hydrogen%20the%20object%20contains." rel="noreferrer">an article</a> which answers just what you have asked. Let me quote the important part (I striked the wrong statement, helium is non-metal as well in astronomical sense):</p>
<blockquote>
<p>Theory predicts that the mass cut-off for what constitutes a star is different for objects of different metallicity, which refers to the proportion of elements heavier than <strike>hydrogen</strike> the object contains.</p>
<p>For objects with a metallicity similar to that of the Sun, theory suggests that anything with less than 7.5% the mass of the Sun – or about 75 Jupiters – will be a brown dwarf.</p>
<p>In the case of NGC 6397, which has a metallicity 100 times lower than that of the Sun, the dividing line is expected to be at 8.3% the mass of the Sun, or about 83 Jupiters.</p>
</blockquote>
<p>If we take a look at the red dwarf, it has approximately <span class="math-container">$10^{\text{[Fe/H]}}=10^{-0.24}=0.58$</span> times the metallicity of the Sun, so its metallicity is pretty Sun-like. Therefore, according to the article, anything above <span class="math-container">$0.075\rm\, M_\odot$</span> will be a star. Our red dwarf therefore lies in the star region.</p>
<p>However, if we take a look at the brown dwarf, in <a href="https://arxiv.org/pdf/1702.02001.pdf" rel="noreferrer">another article</a>, it is stated that the calculated mass limit for a star with such metallicity is <span class="math-container">$0.088\rm\, M_\odot$</span>, which is just slightly more than <span class="math-container">$0.086\rm\, M_\odot$</span> which is the real mass of the brown dwarf, so it still lies in the brown dwarf region.</p>
<hr />
<p>But how does the metallicity influence the minimum mass for a star? <a href="https://astronomy.stackexchange.com/users/2531/profrob">ProfRob</a> wraps this neatly in <a href="https://physics.stackexchange.com/questions/168437/metal-rich-star-formation/485679#485679">his answer</a>. In short, the minimum mass is approximately given by
<span class="math-container">$$ M_{\rm min} \simeq 0.08 \left( \frac{\mu}{0.5} \right)^{-3/2} \left(\frac{\mu_e}{1.2}\right)^{-1/2}$$</span>
where <span class="math-container">$\mu$</span> is the number of mass units per particle in core and <span class="math-container">$\mu_e$</span> is the number of mass units per electron in core which depends on the metallicity.</p>

CC BY-SA 4.0
2022-05-29T22:35:55.043Z
"But how does the metallicity influence the minimum mass for a star? Unfortunately, I haven't been able to find that out [...]" As a simple rule of thumb, heavier elements require more heat/pressure to fuse, and release less energy when they do. For iron (and larger atoms) fusion is endothermic, so about the only time it happens is in the first stage of a supernova. So, a sun-massed ball of mostly hydrogen and some helium emits a lot of energy. With more metal (especially iron) present, you get less energy out. A sun-massed ball of pure iron would be basically inert.
Jerry Coffin
CC BY-SA 4.0
2022-05-30T03:57:34.680Z
IIRC metallicity decreases the radiation pressure, so I would think that it would mean you need more mass when you have less metallicity to balance out the increased pressure, meaning a more massive lower limit.
NathanOliver
CC BY-SA 4.0
2022-05-30T06:16:32.233Z
NathanOliver I think your explanation is the correct one. @JerryCoffin, I thought of that as well, but this explanation (lower metallicity implies low boundary mass) is exactly the opposite of what the researchers have found (lower metallicity implies high boundary mass). Please correct me if I misunderstood it.
User123
CC BY-SA 4.0
2022-05-30T08:14:44.737Z
@User123: I guess I didn't really address the area you mostly care about. Right at the lower limit, iron helps simply by having relatively high density, even without gravity. The denominator in Newton's equation for gravity is r^2, so starting with some material that's dense on its own means you get higher gravity for less total mass. With only hydrogen/helium, you get kind of a Catch-22, where you need gravity to get density, but you need density to get gravity.
Jerry Coffin
CC BY-SA 4.0
2022-05-30T11:04:37.777Z
Some details about why metallicity affects the star/BD threshold can be found here https://physics.stackexchange.com/a/485679/43351 . It has nothing to do with radiation pressure, it is to do with how many electrons you have per unit mass and how many particle you have per unit mass.
ProfRob
CC BY-SA 4.0
2022-05-29T12:39:03.937Z
2022-05-30T17:08:06.780Z
2022-05-30T17:08:06.780Z
User123
<p><a href="https://www.space.com/36242-most-massive-purest-brown-dwarf.html" rel="noreferrer">SDSS J0104+1535</a> is about 90 times more massive than Jupiter, making it the heaviest known brown dwarf. <a href="https://en.wikipedia.org/wiki/EBLM_J0555-57" rel="noreferrer">EBLM J0555-57Ab</a> has a mass of about 85.2±4 Jupiter masses, or 0.081 Solar masses.</p>
<p>I am confused. How is a brown dwarf more massive than a red dwarf? Why hasn't it ignited? Does lower metallicity has something to do with this?</p>

CC BY-SA 4.0
2022-05-29T15:08:01.893Z
See also https://astronomy.stackexchange.com/questions/48069/is-stellar-ignition-all-or-nothing
ProfRob
CC BY-SA 4.0
2022-05-30T11:05:20.027Z
and https://physics.stackexchange.com/a/485679/43351 for a brief explanation of why metallicity affects the star/brown dwarf boundary mass.
ProfRob
<star-formation><brown-dwarf><red-dwarf>
How can a brown dwarf be more massive than a star?
Binita Rimal
# How can a brown dwarf be more massive than a star?

(Asked by: username_0 on 2022-05-29)

[SDSS J0104+1535](https://www.space.com/36242-most-massive-purest-brown-dwarf.html) is about 90 times more massive than Jupiter, making it the heaviest known brown dwarf. [EBLM J0555-57Ab](https://en.wikipedia.org/wiki/EBLM_J0555-57) has a mass of about 85.2±4 Jupiter masses, or 0.081 Solar masses.

I am confused. How is a brown dwarf more massive than a red dwarf? Why hasn't it ignited? Does lower metallicity has something to do with this?

**Tags:** star-formation, brown-dwarf, red-dwarf
> **Comments:**
> **username_1**: See also https://astronomy.stackexchange.com/questions/48069/is-stellar-ignition-all-or-nothing  
> **username_1**: and https://physics.stackexchange.com/a/485679/43351 for a brief explanation of why metallicity affects the star/brown dwarf boundary mass.  

---

## 1 answers

### Answer by username_2 on 2022-05-29. Score: 16 | Accepted answer

Yes, it has to do something with metallicity.

Brown dwarf SDSS J0104+1535 $\to 0.086\rm\, M\_\odot\to[Fe/H]=-2.4$

Red dwarf EBLM J0555-57Ab $\to 0.081\rm\, M\_\odot\to [Fe/H]=-0.24$

There is [an article](https://web.archive.org/web/20061114221813/http://space.newscientist.com/article/dn9771-mass-cutoff-between-stars-and-brown-dwarfs-revealed.html#:%7E:text=Theory%20predicts%20that%20the%20mass%20cut%2Doff%20for%20what%20constitutes%20a%20star%20is%20different%20for%20objects%20of%20different%20metallicity%2C%20which%20refers%20to%20the%20proportion%20of%20elements%20heavier%20than%20hydrogen%20the%20object%20contains.) which answers just what you have asked. Let me quote the important part (I striked the wrong statement, helium is non-metal as well in astronomical sense):

> Theory predicts that the mass cut-off for what constitutes a star is different for objects of different metallicity, which refers to the proportion of elements heavier than hydrogen the object contains.
>
> For objects with a metallicity similar to that of the Sun, theory suggests that anything with less than 7.5% the mass of the Sun – or about 75 Jupiters – will be a brown dwarf.
>
> In the case of NGC 6397, which has a metallicity 100 times lower than that of the Sun, the dividing line is expected to be at 8.3% the mass of the Sun, or about 83 Jupiters.

If we take a look at the red dwarf, it has approximately $10^{\text{[Fe/H]}}=10^{-0.24}=0.58$ times the metallicity of the Sun, so its metallicity is pretty Sun-like. Therefore, according to the article, anything above $0.075\rm\, M\_\odot$ will be a star. Our red dwarf therefore lies in the star region.

However, if we take a look at the brown dwarf, in [another article](https://arxiv.org/pdf/1702.02001.pdf), it is stated that the calculated mass limit for a star with such metallicity is $0.088\rm\, M\_\odot$, which is just slightly more than $0.086\rm\, M\_\odot$ which is the real mass of the brown dwarf, so it still lies in the brown dwarf region.

---

But how does the metallicity influence the minimum mass for a star? [ProfRob](https://astronomy.stackexchange.com/users/2531/profrob) wraps this neatly in [his answer](https://physics.stackexchange.com/questions/168437/metal-rich-star-formation/485679#485679). In short, the minimum mass is approximately given by
$$ M\_{\rm min} \simeq 0.08 \left( \frac{\mu}{0.5} \right)^{-3/2} \left(\frac{\mu\_e}{1.2}\right)^{-1/2}$$
where $\mu$ is the number of mass units per particle in core and $\mu\_e$ is the number of mass units per electron in core which depends on the metallicity.

> **Comments:**
> **username_3**: "But how does the metallicity influence the minimum mass for a star? Unfortunately, I haven't been able to find that out [...]" As a simple rule of thumb, heavier elements require more heat/pressure to fuse, and release less energy when they do. For iron (and larger atoms) fusion is endothermic, so about the only time it happens is in the first stage of a supernova. So, a sun-massed ball of mostly hydrogen and some helium emits a lot of energy. With more metal (especially iron) present, you get less energy out. A sun-massed ball of pure iron would be basically inert.  
> **username_4**: IIRC metallicity decreases the radiation pressure, so I would think that it would mean you need more mass when you have less metallicity to balance out the increased pressure, meaning a more massive lower limit.  
> **username_2**: NathanOliver I think your explanation is the correct one. @JerryCoffin, I thought of that as well, but this explanation (lower metallicity implies low boundary mass) is exactly the opposite of what the researchers have found (lower metallicity implies high boundary mass). Please correct me if I misunderstood it.  
> **username_3**: @User123: I guess I didn't really address the area you mostly care about. Right at the lower limit, iron helps simply by having relatively high density, even without gravity. The denominator in Newton's equation for gravity is r^2, so starting with some material that's dense on its own means you get higher gravity for less total mass. With only hydrogen/helium, you get kind of a Catch-22, where you need gravity to get density, but you need density to get gravity.  
> **username_1**: Some details about why metallicity affects the star/BD threshold can be found here https://physics.stackexchange.com/a/485679/43351 . It has nothing to do with radiation pressure, it is to do with how many electrons you have per unit mass and how many particle you have per unit mass.  

---

