<p>You cannot have a large mass of &quot;normal&quot; matter that is both cold <em>and</em> in equilibrium. Cold matter will collapse towards its minimum energy density configuration. For masses below the &quot;Chandrasekhar mass&quot;, of about <span class="math-container">$1.4 M_\odot$</span> for most common compositions, that will be a white dwarf supported by electron degeneracy pressure. For bigger masses that will be a higher density neutron star.</p>
<p>Worth noting that a hydrogen white dwarf could theoretically be supported by electron degeneracy pressure up to much higher masses. However, there will <em>inevitably</em> be nuclear fusion, even in cold material, because at high densities you get &quot;pycnonuclear&quot; reactions that will fuse the hydrogen into helium.</p>
<p>The limit to the neutron star mass is <em>at least</em> <span class="math-container">$2M_{\odot}$</span>, since at least two have measured masses as large as this (e.g. <a href="http://www.nature.com/nature/journal/v467/n7319/full/nature09466.html" rel="nofollow noreferrer">Demorest et al. 2010</a>).</p>
<p>The answer to your question is the same as the answer to what is the maximum mass of a neutron/quark star, since if you compress matter of any sort, this is ultimately what it will become.</p>
<p>The answer to this question is also <em>unknown</em> and depends on the uncertain equation of state (the relationship between pressure, density and composition) at ultra-high densities, but must be somewhere between the <span class="math-container">$2M_{\odot}$</span> I mentioned above and an upper limit of around <span class="math-container">$3.5M_{\odot}$</span>, which is imposed by General Relativity and an equation of state where the speed of sound equals the speed of light (the hardest possible equation of state).</p>
<p>Whether these highly compressed objects are &quot;solid&quot; is also a matter of debate and research. The conventional picture of an old-ish white dwarf is that it is indeed a crystalline solid. The bulk of the interior of neutron stars of moderate mass is almost certainly a fluid of neutrons along with some protons and electrons. The neutrons <em>may</em> solidify at very high densities and this is thought to be one of the &quot;harder&quot; options for the equation of state.</p>
<p>To answer a comment by @zephyr - cold in this context means that the Fermi energies of degenerate fermion species are very much higher than <span class="math-container">$kT$</span>. White dwarfs and neutron stars (older than a few seconds) are &quot;cold&quot;, despite having interior temperatures, in the case of the latter, of a billion degrees. Making them even colder does not change the size/density of the object but does enable the interior of a white dwarf to crystallise.</p>

CC BY-SA 3.0
2016-07-08T20:51:00.003Z
So if you pile up normal matter, you get a white (or black) dwarf necessarily?  You can't have a lower density gradient other than the way a star does it with generated heat?
JDługosz
CC BY-SA 3.0
2016-07-08T21:20:32.670Z
@JDługos White dwarfs and neutron stars are made of "normal matter".
ProfRob
CC BY-SA 4.0
2021-11-03T22:58:15.780Z
@JDługosz [Why don't question-askers accept answers?](https://astronomy.meta.stackexchange.com/q/773/7982)
uhoh
CC BY-SA 4.0
2021-11-04T14:24:17.983Z
I've Accepted about 1/3 of the questions I've posted here, but the total number I've asked is too small for meaningful statistics.  In this instance, neutron star material is not what I call "normal matter" as stated clearly in the OP.  Whether such exotic matter is "solid" or not has nothing to do with what I was asking.
JDługosz
CC BY-SA 4.0
2021-11-04T14:51:44.527Z
@JDługosz if you make a large cold object it will necessarily collapse to its minimum energy density configuration. Which for material $>1.4 M_{\odot}$ *is* a neutron star, and for a mass greater than somewhere between 2 and 3.5 solar masses is a black hole. You cannot have a large mass of "normal matter" that is cold *and* in equlibrium.
ProfRob
CC BY-SA 4.0
2016-07-08T17:03:07.057Z
2021-11-04T14:56:03.777Z
2021-11-04T14:56:03.777Z
ProfRob
<p>I understand that a neutron star is dense enough that adding more matter will increase the amount of degenerate matter directly, and the limit to its size is about 1.4 solar masses.</p>
<p>But if an object were formed <em>not</em> by an explosion that crushes it down to neutron matter, is it necessarily so?</p>
<p>Given normal atoms of any desired species, and it is piled up carefully so it does not get hot. When it gets around one solar mass, will it necessarily collapse into a neutron star, or can it get more massive?  Will such an object eventually collapse catastrophically or can it have a degenerate matter core and a substantial thickness of less-compressed matter on top and normal matter on top of that?</p>
<p>What is the largest possible mass of a cold solid object?  Can it be several solar masses?</p>
<p>By <em>cold</em> I mean that it’s not a star, puffed up by ongoing consumption of energy. Whatever it's made out of is not being “used up”.</p>
<h1>Edit</h1>
<p>I did not mean to side-track people about what is meant by <em>cold</em>.  I just mean that the object was not <em>crushed</em> down to degenerate matter in a separate step.  <em>Cold</em> here simply means it did not start undergoing fusion due to the heat of formation.</p>

CC BY-SA 3.0
2016-07-07T21:13:39.750Z
How cold does it have to be? Brown dwarfs might have a solid core, but they would have deuterium fusion.
called2voyage
CC BY-SA 3.0
2016-07-07T21:52:54.060Z
I'm flexible on that.  Hmm, what if it didn't have duterium? Would it collapse?
JDługosz
CC BY-SA 3.0
2016-07-07T21:58:36.207Z
I'm not sure about that. I do know that brown dwarfs eventually run out of deuterium, but I'm not familiar enough with them to know what happens then.
called2voyage
CC BY-SA 3.0
2016-07-07T22:02:22.950Z
But a brown dwarf would not be 2 or 4 solar masses!
JDługosz
CC BY-SA 3.0
2016-07-07T22:03:01.483Z
No, I have my doubts that something that large would be cold by any stretch of the term, but I am doing some digging.
called2voyage
CC BY-SA 3.0
2016-07-07T22:08:20.733Z
Ok, so it seems when brown dwarfs run out of deuterium they just get cooler, so no problems there. I don't think anything over 75 Jupiter masses could be considered cool though. I'll leave an answer to those more knowledgeable than I.
called2voyage
CC BY-SA 3.0
2016-07-07T22:08:32.820Z
@called2voyageAs I understand it, after deuterium fusion ends, lithium burning begins. After *that* ends, the brown dwarf begins contracting again until it is supported by electron degeneracy pressure in its core. At that point, it reaches about its final radius, but will continue cooling and radiating away thermal energy.
HDE 226868
CC BY-SA 3.0
2016-07-07T22:15:15.620Z
The 1.4$M_\odot$ [TOV limit](https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit) for neutron stars might be regarded as the limit because it accounts for neutron degeneracy pressure, not electron degeneracy pressure. As I understand it, [quark stars](https://en.wikipedia.org/wiki/Quark_star) or other [exotic stars](https://en.wikipedia.org/wiki/Exotic_star) *might* be more massive and supported by other types of degeneracy pressure, but that's an enormous "might".
HDE 226868
CC BY-SA 3.0
2016-07-08T02:32:35.557Z
I don't really understand the question here. Degenerate matter can form from either high densities or low temperatures. So it can be extremely hot if it is formed from density, or extremely cool if it is formed from low temperatures.
Sir Cumference
CC BY-SA 3.0
2016-07-08T13:54:05.613Z
Just a small point, the $1.4\:M_\odot$ appears to be a reference to the [Chandrasekhar Limit](https://en.wikipedia.org/wiki/Chandrasekhar_limit) for white dwarfs supported by electron degeneracy pressure. Neutron stars are of course supported by neutron degeneracy pressure and they have an upper mass limit described by the TOV Limit which ranges from $1.5\:M_\odot$ to $3.0\:M_\odot$, depending on the equation of state you choose (which EOS is correct is not currently known).
zephyr
CC BY-SA 3.0
2016-07-08T13:56:32.520Z
I think there also needs to be a good definition of what is meant by "cold" and more specifically, how you're defining "temperature". There are many ways to consider the concept of temperature in large planets and stars and they're not all equivalent.
zephyr
CC BY-SA 3.0
2016-07-08T14:05:05.587Z
@zephyr You're right about the limit; that's a typo on my part.
HDE 226868
<astrophysics><temperature><degenerate-matter>
How massive can a cold solid object be?
JDługosz
# How massive can a cold solid object be?

(Asked by: username_0 on 2016-07-07)

I understand that a neutron star is dense enough that adding more matter will increase the amount of degenerate matter directly, and the limit to its size is about 1.4 solar masses.

But if an object were formed *not* by an explosion that crushes it down to neutron matter, is it necessarily so?

Given normal atoms of any desired species, and it is piled up carefully so it does not get hot. When it gets around one solar mass, will it necessarily collapse into a neutron star, or can it get more massive? Will such an object eventually collapse catastrophically or can it have a degenerate matter core and a substantial thickness of less-compressed matter on top and normal matter on top of that?

What is the largest possible mass of a cold solid object? Can it be several solar masses?

By *cold* I mean that it’s not a star, puffed up by ongoing consumption of energy. Whatever it's made out of is not being “used up”.

# Edit

I did not mean to side-track people about what is meant by *cold*. I just mean that the object was not *crushed* down to degenerate matter in a separate step. *Cold* here simply means it did not start undergoing fusion due to the heat of formation.

**Tags:** astrophysics, temperature, degenerate-matter

> **Comments:**
> **username_1**: How cold does it have to be? Brown dwarfs might have a solid core, but they would have deuterium fusion.  
> **username_0**: I'm flexible on that.  Hmm, what if it didn't have duterium? Would it collapse?  
> **username_1**: I'm not sure about that. I do know that brown dwarfs eventually run out of deuterium, but I'm not familiar enough with them to know what happens then.  
> **username_0**: But a brown dwarf would not be 2 or 4 solar masses!  
> **username_1**: No, I have my doubts that something that large would be cold by any stretch of the term, but I am doing some digging.  
> **username_1**: Ok, so it seems when brown dwarfs run out of deuterium they just get cooler, so no problems there. I don't think anything over 75 Jupiter masses could be considered cool though. I'll leave an answer to those more knowledgeable than I.  
> **username_2**: @username_1As I understand it, after deuterium fusion ends, lithium burning begins. After *that* ends, the brown dwarf begins contracting again until it is supported by electron degeneracy pressure in its core. At that point, it reaches about its final radius, but will continue cooling and radiating away thermal energy.  
> **username_2**: The 1.4$M_\odot$ [TOV limit](https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit) for neutron stars might be regarded as the limit because it accounts for neutron degeneracy pressure, not electron degeneracy pressure. As I understand it, [quark stars](https://en.wikipedia.org/wiki/Quark_star) or other [exotic stars](https://en.wikipedia.org/wiki/Exotic_star) *might* be more massive and supported by other types of degeneracy pressure, but that's an enormous "might".  
> **username_3**: I don't really understand the question here. Degenerate matter can form from either high densities or low temperatures. So it can be extremely hot if it is formed from density, or extremely cool if it is formed from low temperatures.  
> **username_4**: Just a small point, the $1.4\:M_\odot$ appears to be a reference to the [Chandrasekhar Limit](https://en.wikipedia.org/wiki/Chandrasekhar_limit) for white dwarfs supported by electron degeneracy pressure. Neutron stars are of course supported by neutron degeneracy pressure and they have an upper mass limit described by the TOV Limit which ranges from $1.5\:M_\odot$ to $3.0\:M_\odot$, depending on the equation of state you choose (which EOS is correct is not currently known).  
> **username_4**: I think there also needs to be a good definition of what is meant by "cold" and more specifically, how you're defining "temperature". There are many ways to consider the concept of temperature in large planets and stars and they're not all equivalent.  
> **username_2**: @username_4 You're right about the limit; that's a typo on my part.  

---

## 1 answers

### Answer by username_5 on 2016-07-08. Score: 6

You cannot have a large mass of "normal" matter that is both cold *and* in equilibrium. Cold matter will collapse towards its minimum energy density configuration. For masses below the "Chandrasekhar mass", of about $1.4 M\_\odot$ for most common compositions, that will be a white dwarf supported by electron degeneracy pressure. For bigger masses that will be a higher density neutron star.

Worth noting that a hydrogen white dwarf could theoretically be supported by electron degeneracy pressure up to much higher masses. However, there will *inevitably* be nuclear fusion, even in cold material, because at high densities you get "pycnonuclear" reactions that will fuse the hydrogen into helium.

The limit to the neutron star mass is *at least* $2M\_{\odot}$, since at least two have measured masses as large as this (e.g. [Demorest et al. 2010](http://www.nature.com/nature/journal/v467/n7319/full/nature09466.html)).

The answer to your question is the same as the answer to what is the maximum mass of a neutron/quark star, since if you compress matter of any sort, this is ultimately what it will become.

The answer to this question is also *unknown* and depends on the uncertain equation of state (the relationship between pressure, density and composition) at ultra-high densities, but must be somewhere between the $2M\_{\odot}$ I mentioned above and an upper limit of around $3.5M\_{\odot}$, which is imposed by General Relativity and an equation of state where the speed of sound equals the speed of light (the hardest possible equation of state).

Whether these highly compressed objects are "solid" is also a matter of debate and research. The conventional picture of an old-ish white dwarf is that it is indeed a crystalline solid. The bulk of the interior of neutron stars of moderate mass is almost certainly a fluid of neutrons along with some protons and electrons. The neutrons *may* solidify at very high densities and this is thought to be one of the "harder" options for the equation of state.

To answer a comment by @username_4 - cold in this context means that the Fermi energies of degenerate fermion species are very much higher than $kT$. White dwarfs and neutron stars (older than a few seconds) are "cold", despite having interior temperatures, in the case of the latter, of a billion degrees. Making them even colder does not change the size/density of the object but does enable the interior of a white dwarf to crystallise.

> **Comments:**
> **username_0**: So if you pile up normal matter, you get a white (or black) dwarf necessarily?  You can't have a lower density gradient other than the way a star does it with generated heat?  
> **username_5**: @JDługos White dwarfs and neutron stars are made of "normal matter".  
> **username_6**: @username_0 [Why don't question-askers accept answers?](https://astronomy.meta.stackexchange.com/q/773/7982)  
> **username_0**: I've Accepted about 1/3 of the questions I've posted here, but the total number I've asked is too small for meaningful statistics.  In this instance, neutron star material is not what I call "normal matter" as stated clearly in the OP.  Whether such exotic matter is "solid" or not has nothing to do with what I was asking.  
> **username_5**: @username_0 if you make a large cold object it will necessarily collapse to its minimum energy density configuration. Which for material $>1.4 M_{\odot}$ *is* a neutron star, and for a mass greater than somewhere between 2 and 3.5 solar masses is a black hole. You cannot have a large mass of "normal matter" that is cold *and* in equlibrium.  

---

