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My question is a follow up to this one: Does the collision of a neutron and anti-neutron produce energy?.

Quoting from an answer:

"The collision of a neutron and antineutron star would initiate a terrible strong annihillation. The result would be similar to a supernova with an extreme gamma photon flash. It is hard to say, what would be the result. In the Universe, no significant amount of antimatter exists."

Let's say this is the smallest possible neutron star and smallest possible anti-neutron star. Sure, there would be an extreme gamma photon flash. But I want to know how devastating this will be. Will it destroy the entire galaxy with material spewed out into inter-galactic space? Or will it be like any other supernova explosion? And will this result in a black hole or just a big explosion followed by nothing?

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    $\begingroup$ Great question to send to What-If . $\endgroup$ Apr 11, 2022 at 14:34

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Probably not even close. A back of the envelope calculation indicates that a neutron and an anti-neutron star colliding and completely annihilating would release energy of about $10^{47}~\rm J$. This is a staggering amount of energy, but it's also only a few hundred times larger than the largest supernovas we've seen.

Even if you add a couple orders of magnitude to this (to compensate for the fact that the vast majority of a supernova's energy is lost to neutrinos), when you include the $1/r^2$ factor for the explosion, the blast radius is "only" ~100 times larger than a supernova. This would be dramatic, no doubt, but it doesn't come close to the ~50,000 light years of the Milky Way.

Besides which it's very likely that the annihilation would blow the stars apart prematurely, and relatively little of their mass would actually annihilate.

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  • $\begingroup$ Great answer. But I didn't get the part about neutrinos. Are neutrinos particles that can be very energetic and yet not cause much damage? $\endgroup$ Apr 11, 2022 at 21:21
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    $\begingroup$ @RohitPandey Except at very, very high energies neutrinos don't interact much at all with anything. Every second, trillions of neutrinos pass through your body. Most of them go on to pass all the way through the earth like it isn't there. When a supernova goes off, we can detect it just from the neutrinos, before the light even arrives, and even if the detector is on the "wrong" side of the earth. $\endgroup$
    – Chris
    Apr 12, 2022 at 1:42
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Let us assume two minimal mass, $1.2M_\odot$ neutron and anti-neutron stars collide, totally annihilate, and produce a flash of GeV gamma rays.

The energy released will be $2.4M_\odot c^2 = 4\times 10^{47}$ J.

The gravitational binding energy of the inner part of the galaxy (excluding the dark matter halo that can't absorb gamma rays) is about $GM^2/R$, where $M\sim 10^{11}M_\odot$ and $R\sim 10$ kpc. This amounts to $\sim 10^{52}$ J.

Thus, even if the gamma rays were all absorbed (they wouldn't be and also some fraction of the annihilation energy is lost in the form of weakly interacting neutrinos), this falls short by around 5 orders of magnitude from injecting enough energy to unbind the galaxy.

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  • $\begingroup$ We can probably bump that energy up a bit. Using a crude Newtonian freefall calculation, two bodies of mass $1.2\,M_\odot$ and radius $\rm11\,km$ collide at a speed of ~$0.57c$, $\gamma\approx1.2$. Of course, they probably won't have a head-on collision, but will spiral in, releasing gravitational waves. $\endgroup$
    – PM 2Ring
    Feb 5, 2023 at 1:17
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    $\begingroup$ @PM2Ring The gravitational redshift of the outgoing radiation should cancel that exactly, no? Put another way, the system has energy $2M$ when the stars are distantly separated (and gravity can be neglected), where $M$ is the mass of each star. The stars accelerate toward each other, colliding and (in the idealized scenario) converting fully into radiation. In the limit that the radiation has traversed a large distance (so gravity can be again neglected), it should carry the same energy $2M$. $\endgroup$
    – Sten
    Feb 5, 2023 at 13:43
  • $\begingroup$ @Sten That sounds reasonable. $\endgroup$
    – PM 2Ring
    Feb 5, 2023 at 13:45

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