# What could break up a neutron star?

Beside the obvious answer black hole, is there anything else? Could a nearby supernova blow it away/apart? Or any sort of (theoretically intense enough) field? I guess an electric field is out of the question no matter how strong.

Sorry if this a lazy/amateurish question, but I'd probably have to study degenerate matter a lot before I can answer this myself comprehensively. I suppose the question asking What is the binding energy of a neutron star? is related, but I don't know how to interpret it myself in layman's terms.

Do you mean anything in the real universe or just theoretically? If the latter, then I can think of a few phenomena:

• Heat: Just heat it up until the thermal velocity at the surface is greater than the escape volcity. Then neutrons will just fly off and it will evaporate (sublimate?).
• Spin: Wind it up until the tangential velocity at the equator reaches escape velocity. This might be self-limiting as the spun-off mass takes away angular momentum. You'd have to keep winding it.
• Tides: Drive it past a particularly dense Black Hole and let the tidal forces pull it apart.

In all these cases, I'm not sure how the nuclear Strong Force might modify the calculations (it might not be only gravity holding the neutrons down).

To expand on those answers, I did a few BoE calculations assuming only gravity, to see what kind of values we'd need for each of these cases. We start with a typical neutron star:

mass, $M = 1.4M_\odot$ (so, $2.8 \times 10^{30}kg$)

density, $\rho = 5 \times 10^{17}kg/m^3$. This gives:

radius, $r = \sqrt[3]{{3M}/{4\pi\rho}} = 11km$.

escape velocity, $v_e = \sqrt{{2GM}/r} = 1.8 \times 10^8 ms^{-1}$ (or $0.6c$)

• Heat: $v= \sqrt {{3kT}\over m}$, so $T = {{mv^2}\over{3k}}$, where $m$ is the neutron mass and $k$ is Boltzmann's constant. This comes out at around $10^{12}K$, which is rather hot (interestingly, it's within a factor of 10 of the core temperature in a supernova, where neutron stars are made).
• Spin: $v_e=r\omega$ and $w=2\pi f$ so $f = {v_e \over {2\pi r}}$. Plugging in the numbers gives a rotational frequency of 2.6 kHz or 150,000 rpm. Quite a rate for a thing the size of a mountain.
• Tides: The equation for the Roche Limit is $d = 1.26 R_M (M_M / M_m)$. If we find a black hole of $3M_{\odot}$ and a radius of 9km, we just have to steer our neutron star to within 24Km and it'll just fall apart.
• I like your answer, BUT: The typical mass of a neutron star would be $1.4 M_{\odot}$. Part (i), they're hard to evaporate, not melt. Part (ii) Why have you used $v=0.5c$? The criterion for rotational break up is roughly $\omega^2 r > GM^2/r$ (Newtonian). Part (iii) The Schwarzschild radius of a $3M_{\odot}$ BH is 9km. I get $d=16.2$ km. – Rob Jeffries Nov 5 '15 at 14:27
• @Rob: I got 0.5c from googling "escape velocity of neutron star". Ditto for mass and radius <blush/>. To be frank, I should've picked a mass (e.g., $1.4M_{\odot}$) then calculated $v_e$ and radius. For the BH, I used the photon sphere radius - jst trying to get a feel. I'm a bit surprised the RL scale is about the size of the thing, I thought it would be much less. For the spin, you got the $^2$ in the wrong place - should be $\omega = \sqrt {GM/r^2}$. That gives $f$ of the order of $10^6 s^{-1}$. Wonder why it doesn't match up... Hmmm. – Oscar Bravo Nov 5 '15 at 15:09
• Whoops. Should have been $\omega^2 r > GM/r^2$ sorry. So a rotation period of about 0.5 ms. Note that the fastest pulsar is about 1.4 ms. – Rob Jeffries Nov 5 '15 at 15:59
• Re: "Do you mean anything in the real universe or just theoretically?" I meant both and you've answered it quite nicely (as I was hoping) by going from what theoretical conditions are needed to what could cause them to happen in our universe. – Fizz Nov 5 '15 at 19:53
• I like the way you think. Spin, if you get it spinning fast enough, should work. Heat is harder as heat tends to get radiated away, though in theory you could heat up it's outer surface enough to radiate away a few molecules at a time, but you might end up putting more energy (mass) into it than you take away. Diminishing returns. – userLTK Nov 6 '15 at 4:13

I'm going to add another way to break up a neutron star. Shoot antimatter at it.

The difficulty with breaking up a neutron stars is that, once they undergo the compression to become Neutron stars, their gravitation tends to keep them there. The minimum size for a Neutron star to form is about 1.2-1.5 solar masses, but once it's shrunk down, the mass it needs to maintain the Neutron compression is a lot less than that, perhaps around 0.1 solar masses. See here for much more detailed answer on minimum Neutron star size.

The outside of a Neutron star is highly compact heavy ions and flowing electrons.

Source

So, if we shoot it with positrons, the positron hits an Electron and evaporates into gamma rays and you're left with a positively charged Neutron star, shoot it enough and the positive charges will gradually build a force that resists the gravity (granted, it also becomes more resistant to continued firing of Positrons), but if you manage to shoot it enough times and you might just be able to cause the Neutron star to expand and overcome the gravitational force that keeps it in it's Neutron state.

Now, maybe you'd need to hit it with anti protons / anti neutrons to reduce more mass as positrons alone might not get the job done, but I like the positron idea cause it preserves the baryonic matter and if you get the Neutron star to expand and when that happens, the Neutrons begin to decay into protons, so you're left with basically a hydrogen star. Granted one with an unnaturally high positive charge.

• This is a good idea. I'm wondering if the excess charge would distribute evenly or cling to the surface due to the skin effect? Anyway, if you could convert enough neutrons into protons, you'd end up with a massive nucleus - that would definitely fission! – Oscar Bravo Nov 6 '15 at 8:44
• Afaik shooting with particles with a non-neutral net charge will make the NS charged. After its charge steps over a potential, it will radiate away the excess in the form of well-formed proton (or ion) lightnings. Thus, I suspect beaming it with positrons would be a much effectiver way to destruct it as with antiprotons, although beaming the positron beam into it will be harder as its charge grows. It will also heat the star, and this heating would induce evaporation. – peterh says reinstate Monica Jul 19 '16 at 23:44

The neutron star is still "regular enough mater" for that it would react to anything a normal object would react.

To me the point is more "since its center is not far to collapsing to blackhole, is it possible to shake (or breakup) a neutron star without making it collapse".

• Yes, this is indeed what I was trying to ask, although I did not phrase it that well. – Fizz Nov 5 '15 at 19:47

The only likely mechanism by which a neutron star can break up is through a collision with another neutron star, in particular in binary neutron star mergers.

• Binary Neutron Star mergers eject small amounts of material due to tidal forces. The vast majority of material, however, is believed to collapse into a BH --- not be broken up. – DilithiumMatrix Nov 5 '15 at 15:12
• @DilithiumMatrix I suspect in the case of a direct collision (and not a merge), the result would significantly differ (large part of the stars would be ejected). Although they have only a very small chance, the Universe seem to expand until infinity, but the galaxies seem to be gravitationally bound (despite the expansion). And the only effect which could limit the lifetime of the neutron stars is the proton decay (and the merging to black holes). – peterh says reinstate Monica Jul 19 '16 at 23:31
• @peterh it actually doesn't really matter if its an orbital merger or direct collision. The way to see this is simply that the characteristic velocity will be basically identical (orbital at the NS radius, vs. escape/free-fall velocity) --- and both will be marginally relativistic. – DilithiumMatrix Jul 20 '16 at 17:27
• @DilithiumMatrix Well, that seems true. What is the case in a BH-NS direct collision? – peterh says reinstate Monica Jul 20 '16 at 17:45
• @peterh It depends on the mass-ratio, but in general it isn't much better than NS-NS (and in fact, usually worse) --- because there's no solid surface to collide with. – DilithiumMatrix Jul 20 '16 at 18:43