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The official limits for a neutron star is $1.4 - 3.2\;M_\odot$. But I read that the limit depends on the particular structure of a star to estimate which mass it must have. I also read that neutron stars with less than $1.4\;M_\odot$ were observed. Given this information, I wonder if we can be sure that our Sun has definitely not enough mass to become a neutron star. Are there absolut limits (without the need of further information) for a star to become a neutron star or a black hole ?

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Can you provide a link to observations of neutron stars with a mass of less than 1.4 times the solar mass. – John Rennie May 20 '14 at 11:25
John, see table 1 of The Nuclear Equation of State and Neutron Star Masses. There are several examples. – DavePhD May 20 '14 at 12:02
I received the following warning : Wait! Some of your past questions have not been well-received, and you're in danger of being blocked from asking any more. For help formulating a clear, useful question, see: How do I ask a good question? Also, edit your previous questions to improve formatting and clarity. Should I reopen the closed questions ? I do not think this makes sense because the questions do not seem to fulfill the criterias here. What shall I do not to be blocked ? – Peter May 26 '14 at 9:44
up vote 3 down vote accepted

There are two questions here, namely about the limits on neutron star masses, and about the possibility of our sun becoming one. I'll try to argue that they are different questions, viz. the first about the stability and the second about the formation of such objects.

1) DavePhD's reference in the comments (here, for completeness) answers it completely. There is a lot of room for neutron star masses, because it depends intrinsically on the equation of state of nuclear (and possibly sub-nuclear) matter. Since we don't know the correct equation of state is hard to give strict boundaries. Without an equation of state one could have a mass as large as desired, just by increasing radius. So qualitatively the best one can do depends on the interplay between mass and radius, or density if you will.

The strictest limit comes from Schwarzschild radius, that is if you make too dense a star it would generate an event horizon and collapse into a black hole. Next to this, one notes that the speed of sound escalates with density, so if you try to make too dense a star it will have speed of sound greater than the speed of light, violating causality. This gives a limitation in the different equations of state possible. The upper bounds of about 3.5 solar masses comes from this consideration. You'll find all this more deeply discussed in the aforementioned paper. The summary is in Figure 3, page 51. I am completely ignorant of an analogous argument for lower bounds on the masses that use only some physical principles (in spite of my first, incorrect, answer that related it to angular momentum and Rob Jeffries kindly corrected me on the comments) so I have deleted the incorrect previous part.

2)Somewhat independently of the previous discussion, we can be pretty sure that the sun will never become a neutron star, no matter what equation of state is correct. This is because gravitational collapse of a star is a highly non-linear process, that besides the different nuclear fusion cycles, will generate shock waves. Therefore it will not proceed adiabatically, on the contrary this processes will shed most of a star's mass. Therefore to produce a neutron star we need to start with a very heavy one, typically of the order of tens of solar masses. This is the reason we attribute neutron star formation to supernova events.

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The boundary for the production of a neutron star - which is the key point of the question - is around $8M_{\odot}$. The minimum mass of a neutron star has little to do with its rotation.… – Rob Jeffries Jun 2 '15 at 8:43
@RobJeffries, thanks for the 8 solar masses bound, but if you noticed I only mentioned rotation as an hypotheses. If you use the lowest angular momentum measured you can get a lower bound (which the papers describes), but it is clear that if you assume zero angular momentum this bound would not be applicable. I was just summarizing the paper. Without an equation of state it is only possible to get lower bounds with angular momentum. Hopefully the question you linked will complement this discussion with considerations from the equation of state – cesaruliana Jun 2 '15 at 22:40
don't know what you mean by "papers". I glanced at the Lattimer review. It (in section 2.1) does not discuss the minimum mass in terms of rotation. Low mass neutron stars, if they exist, would be large, not " tiny". The figure on p.51 has a curve representing a stability line for something rotating as fast as a millisecond pulsar. This is the maximum rotation ever observed, not the minimum. Rotation does not determine the minimum possible mass for a neutron star. – Rob Jeffries Jun 2 '15 at 23:11
@RobJeffries, you are entirely correct, of course, I'm very sorry about this. I wrote incorrectly "lowest" instead of "highest". According to my notes from a Friedman lecture the argument goes as this: if you try to make a low mass, small radius neutros star and put angular momentum on it then you get an instability. But low mass large radius are subjet to lots of non-equtilibrium processes. Therefore one does not expect neutron stars with arbitrarily low massess. It is not a bound but a heuristic guide. I'll rewrite later to reflect that, thank you – cesaruliana Jun 2 '15 at 23:36
For a given specific angular momentum, the ratio of centrifugal force to gravity scales as 1/r. Thus the effect of rotation on the structure of low mass neutron stars, which would have radii of ~200 km, will be smaller than standard neutron stars. – Rob Jeffries Jun 3 '15 at 6:08

Observed neutron stars range from $1.0 \pm 0.1 M_{\odot}$ to $2.7 \pm 0.2 M_{\odot}$ according to table 1 of The Nuclear Equation of State and Neutron Star Masses, which lists dozens of examples. Keep in mind that the mass of the neutron star is typically substantially smaller than the mass of its progenitor star; late in the stellar life cycle a lot of mass is blown away, for instance a star that goes though an AGB phase may lose >50% of its mass. So our $1M_\odot$ Sun is likely to end up as a stellar remnant with $M < 1M_\odot$, probably a white dwarf.

According to Structure of Quark Stars, the mass is the only parameter to consider for neutron stars (but not hypothetical quark stars), although I would think rotation rate would be a factor.

This reference also states that neutron stars can be as small as $0.1 M_{\odot}$, but this does not imply that the sun will actually become a neutron star.

According to Possible ambiguities in the equation of state for neutron stars, it is the theory (equation of state) of neutron stars that is causing the current uncertainty about the limits of neutron stars.

Also, it is unknown whether or not neutron stars may become quark stars before becoming black holes. There is a term "quark nova" for such a hypothetical event.

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+1, and added a mention of the distinction between the stellar remnant mass and the stellar progenitor mass, which seems to be a point of confusion in the question. – Kyle Oman May 20 '14 at 19:32
Yes, it is often not appreciated that the smallest neutron stars are less massive than what many people think is the "Chandrasekhar limit". A new, precise measurement exists for a neutron star at $1.174 \pm 0.004 M_{\odot}$ This is still a little above the Chandrasekhar mass for degenerate iron under GR conditions. – Rob Jeffries Oct 14 '15 at 11:31

Yes, there are absolute limits (with some theoretical uncertainty) for the mass of a progenitor star that can become a neutron star or black hole and the Sun is well below that limit.

The other answers here talk about the range of masses of neutron stars, but do not directly answer the question you pose: the answer arises from considerations of what happens in the core of a star during the course of its evolution.

In a star of similar mass to the Sun, core hydrogen burning produces a helium ash. After about 10 billion years, the core is extinguished and hydrogen burning in a shell results in the production of a red giant. The red giant branch is terminated with the onset of core helium burning, leaving a core ash of carbon and oxygen via the triple alpha process. After the core is extinguished again, there is a complicated cycle of hydrogen and helium burning in shells around the core. During this phase, the star swells enormously to become an asymptotic red giant branch star (AGB). AGB stars are unstable to thermal pulsations and lose a large fraction of their envelopes via a massive wind. The Sun is expected to lose about $0.4-0.5M_{\odot}$ at this time.

Now we get to the crux of the answer. What is left behind is a core of carbon and oxygen, with maybe a thin layer of hydrogen/helium on top. With no nuclear reactions going on, this core contracts as far as it is able and cools. In a star governed by "normal" gas pressure, this process would continue until the centre was hot enough to ignite carbon and oxygen burning (a higher temperature is needed to overcome the greater Coulomb repulsion between more proton-rich nuclei). However, the cores of progenitor stars with masses $<8M_{\odot}$ are so dense that electron degeneracy pressure takes over. The electrons in the gas are compressed so much that the Pauli Exclusion Principle results in all the low energy states being filled completely, leaving many electrons with very high energies and momenta. It is this momentum that provides the pressure that supports the star. Crucially, this pressure is independent of temperature. This means that the core can continue to cool without contracting any further. As a result it does not get any hotter in the centre and fusion never restarts. The final fate of stars like the Sun, and anything with a main sequence mass of $<8M_{\odot}$ is to be a cooling white dwarf. The figure of $8M_{\odot}$ is uncertain by about $\pm 1M_{\odot}$, because the details of mass loss during the AGB phase are not completely solved theoretically and it is difficult to empirically estimate the progenitor masses of white dwarfs.

Stars more massive than this have cores which do contract sufficiently to begin further stages of fusion, resulting in the production of an iron/nickel core. Fusion cannot produce any more energy from these nuclei, which are at the peak of the binding energy per nucleon curve, and thus the star will ultimately collapse and has a core mass greater than can be supported by electron degeneracy pressure. It is this collapsing core which forms a neutron star or black hole.

An interesting caveat to my answer is that there may be an evolutionary route for a star like the Sun to become a neutron star if it were in a binary system. Accretion from a companion might increase the mass of the white dwarf star, pushing it above the Chandrasekhar mass - the maximum mass that can be supported by electron degeneracy pressure. Though in principle this might form a neutron star, it is considered that a more likely scenario is that the entire star will detonate as a Type Ia Supernova, leaving nothing behind.

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