# Do supernovae push neighboring stars outward?

I know that a supernova can mess up the heliosphere of nearby stars, but I'm wondering if it could physically push neighboring stars off their trajectories.

It's fun to imagine all the stars surrounding a supernova being propelled outward and tumbling out of the galactic arm!

I would expect that a really close star, such as a partner in a binary pair, would get really messed up. I'm thinking more about the neighbors a few light-years away.

I realize that a supernova involves both the initial EM burst and the mass ejection which arrives later. I'm open to the effects of any of these things.

• A silly but entertaining approach is to look at a universe simulation like tng-project.org/movies/tng/tng100_sb0_8fields_1080p.mp4. I think you can spot supernovas on the gas velocity map and then eyeball the stellar column density map to see if there is corresponding movement. Jan 22 '19 at 10:56

Consider a star of mass $$M$$ and radius $$R$$ at a distance $$r$$ from the supernova. For a back-of-the-envelope estimate, consider how much momentum would be transferred to the star by the supernova. From that, we can estimate the star's change in velocity and decide whether or not it would be significant.

First, for extra fun, here's a review of how a typical core-collapse supernova works :

Nuclear matter is highly incompressible. Hence once the central part of the core reaches nuclear density there is powerful resistance to further compression. That resistance is the primary source of the shock waves that turn a stellar collapse into a spectacular explosion. ... When the center of the core reaches nuclear density, it is brought to rest with a jolt. This gives rise to sound waves that propagate back through the medium of the core, rather like the vibrations in the handle of a hammer when it strikes an anvil. .. The compressibility of nuclear matter is low but not zero, and so momentum carries the collapse beyond the point of equilibrium, compressing the central core to a density even higher than that of an atomic nucleus. ... Most computer simulations suggest the highest density attained is some 50 percent greater than the equilibrium density of a nucleus. ...the sphere of nuclear matter bounces back, like a rubber ball that has been compressed.

That "bounce" is allegedly what creates the explosion. According to ,

Core colapse liberates $$\sim 3\times 10^{53}$$ erg ... of gravitational binding energy of the neutron star, 99% of which is radiated in neutrinos over tens of seconds. The supernova mechanism must revive the stalled shock and convert $$\sim 1$$% of the available energy into the energy of the explosion, which must happen within less than $$\sim 0.5$$-$$1$$ s of core bounce in order to produce a typical core-collapse supernova explosion...

According to , one "erg" is $$10^{-7}$$ Joules. To give the idea the best possible chance of working, suppose that all of the $$E=10^{53}\text{ ergs }= 10^{46}\text{ Joules}$$ of energy goes into the kinetic energy of the expanding shell. The momentum $$p$$ is maximized by assuming that the expanding shell is massless (because $$p=\sqrt{(E/c)^2-(mc)^2}$$), and while we're at it let's suppose that the collision of the shell with the star is perfectly elastic in order to maximize the effect on the motion of the star. Now suppose that the radius of the star is $$R=7\times 10^8$$ meters (like the sun) and has mass $$M=2\times 10^{30}$$ kg (like the sun), and suppose that its distance from the supernova is $$r=3\times 10^{16}$$ meters (about 3 light-years). If the total energy in the outgoing supernova shell is $$E$$, then fraction intercepted by the star is the area of the star's disk ($$\pi R^2$$) divided by the area of the outgoing spherical shell ($$4\pi r^2$$). So the intercepted energy $$E'$$ is $$E'=\frac{\pi R^2}{4\pi r^2}E\approx 10^{-16}E.$$ Using $$E=10^{46}$$ Joules gives $$E'\approx 10^{30}\text{ Joules}.$$ That's a lot of energy, but is it enough? Using $$c\approx 3\times 10^8$$ m/s for the speed of light, the corresponding momentum is $$p=E'/c\approx 3\times 10^{21}$$ kg$$\cdot$$m/s. Optimistically assuming an elastic collision that completely reverses the direction of that part of the shell's momentum (optimistically ignoring conservation of energy), the change in the star's momentum will be twice that much. Since the star has a mass of $$M=2\times 10^{30}$$ kg, its change in velocity (using a non-relativistic approximation, which is plenty good enough in this case) is $$2p/M\approx 3\times 10^{-9}$$ meters per second, which is about $$10$$ centimeters per year. That's probably not enough to eject the star from the galaxy. Sorry.

References:

 Page 43 in Bethe and Brown (1985), "How a Supernova Explodes," Scientific American 252: 40-48, http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/transparents/SN%20-%20Bethe%20e%20Brown.pdf

 Ott $$et al$$ (2011), "New Aspects and Boundary Conditions of Core-Collapse Supernova Theory," http://arxiv.org/abs/1111.6282

 Table 9 on page 128 in The International System of Units (SI), 8th edition, International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf

• That's.... actually a lot more than I'd have thought intuitively. I underestimated supernovae. Jan 21 '19 at 15:03
• @JaccovanDorp Yeah, it's hard to imagine. Do beware, though, that I made several "optimistic" simplifications here, which overestimate the effect to some degree. A more careful analysis would come up with a smaller effect -- but probably still impressively large. On the other hand, accounting for the asymmetry noted in PM2Ring's answer could make the effect bigger again... Jan 21 '19 at 15:07
• @jacco Everyone always does underestimate supernovae. Use the Randall Monroe rule: when asked "is the supernovae bigger/brighter/etc than X", answer yes. Classic example, which is brighter: a supernovae at the orbit of Pluto, or a full strength Tsar Bomba pressed against your eyeball? -- what-if.xkcd.com/73
– Yakk
Jan 21 '19 at 16:26
• Well, people also tend to underestimate how far a lightyear is, and we were thinking about 3 lightyears away. That's quite a bit further than pluto's orbit, or even Mars'. I was giving the benefit of the doubt to the wrong mind-boggling huge concept. Jan 22 '19 at 9:15

Probably not. Supernovae are powerful, but space is really big. ;)

Supernova energies are often measured in foe; one foe is $$10^{44}$$ joules. According to Wikipedia, a big supernova can release around 100 foe as kinetic energy of ejecta, plus 1 to 5 foe for the light & other EM energy released. (The energy of the released neutrinos is higher than the EM energy, but that's only an issue if you're really close to the supernova).

By way of comparison, the Sun's gravitational binding energy (GBE) is around $$3.8 \times 10^{41}$$ joules. So if a sun-like star got hit by one thousandth of the light energy released by a supernova it would get seriously messed up.

But as I said at the start, space is really big. If you spread 1 foe of energy over the surface of a sphere of 1 light-year radius, an area on that sphere equal to the Sun's cross-section would get around $$1.35 \times 10^{29}$$ joules, which is a substantial quantity of energy, but it's around a trillionth of the Sun's GBE. So a supernova may do interesting things to the atmosphere of a star 1 light-year away, and the atmospheres of any planets in that system, but it won't disrupt the star, or cause a noticeable perturbation of its galactic orbit.

However, supernova explosions are notoriously asymmetrical. The energy and matter they release is not spread uniformly over a nice spherical surface. So there's a chance that the damage at 1 light-year is much worse than what I stated in the previous paragraph. In particular, the supernova remnant (the neutron star or black hole produced by the collapse) may be ejected at 500 km/s or faster. If you happen to be in the path of one of those, Bad Things are likely to occur. An extreme example is Pulsar B1508+55, a neutron star heading out of the galaxy at 1100 km/s.

• I love this answer because it helps explain why we observe binary neutron stars - I'd always wondered why the supernova that created each compact remnant didn't disrupt the other or at least kick it out of the binary orbit. Jan 22 '19 at 2:52
• @Chappo Thanks! I didn't really discuss what happens to any companion stars of a supernova, since the OP explicitly asked about neighbours that aren't that close. Normally, there is a big effect on companions: they tend to lose a fair bit of matter in the blast, although that varies a lot due to the asymmetry. And afterwards, the neutron star can siphon off more matter, accelerating the companion's evolution. See astronomynow.com/2018/10/12/… Jan 22 '19 at 3:33
• Interesting article, especially the confirmation of "super-stripped" SNE. The dynamics of binary stars is simply fascinating! Jan 22 '19 at 3:45
• @Chappo Indeed! Like many sci-fi fans, I'd love to see a binary system up close. OTOH, it may not be such a great place to live. ;) Jan 22 '19 at 3:49
• @uhoh :D Here's one I thought of a few weeks ago. According to the last paragraph in this section, the Milky Way has a radius of 129,000 light-years and a mass of 1.5 trillion solar masses, including dark matter. A circular disc of that diameter and mass, with the density of water (1g/cm³), would have a thickness of almost 0.64 millimetres, according to the Google Calculator: ((1.5*10^12 solar masses)/(1g/cm^3))/(pi*(129000 light-years)^2) Feb 23 '20 at 12:45

It actually can, through gravity. This is a mechanism some people use to explain why the dark matter density profile is not "cuspy" in dwarf galaxies. The idea is fairly simple: a super nova explodes, it pushes out gas. The gas is coupled through gravity with dark matter, so dark matter is also moved around. This changes the potential of the galaxy, which in turn makes the orbits of the stars change close to the center.

More formally, the potential energy required to assemble the dark halo is

$$W_{\rm DM} = -4\pi G \int_0^{R_{\rm vir}}{\rm d}r ~ \rho(r)M(r) r \tag{1}$$

with $$\rho$$ its density, and $$M$$ its cumulative mass profile. $$R_{\rm vir}$$ is the virtual radius of the galaxy, which for a dwarf spheroid is a few kpc. The energy released by SNe is

$$E_{\rm SN} = \frac{M_\star}{\langle m_\star\rangle}\xi(m_\star > 8 M_{\odot}) E_{\rm SN} \tag{2}$$

where $$M_{\rm star}$$ is the total stellar mass of the galaxy, $$\langle m_\star\rangle$$ is the average stellar mass, $$\xi$$ is the initial mass function of which only the upper tail is considered since the fraction producing stars with $$m_{\star} < 8 M_{\odot}$$ will not result in SNe, $$E_{\rm SN}$$ is the energy released in a SNII explosion.

Now, some of this released energy goes through transform the orbits of the DM particles, let's say a small fraction $$\eta$$. The important question is then: imagine you start with a galaxy with gravitational energy $$W_{\rm DM}^{(1)}$$, and want to transform it to a galaxy with gravitational energy $$W_{\rm DM}^{(2)}$$, can we do it with the amount of energy $$\eta E_{\rm SN}$$?

$$2\eta E_{\rm SN} > W_{\rm DM}^{(2)} - W_{\rm DM}^{(1)} \tag{3}$$

And the answer is a solid yes. You can see a couple of references here for more details.

It clearly depends on what the final product $$W_{\rm DM}^{(2)}$$ looks like, but it is definitely able to transform a galaxy with a density profile of the form $$\rho \sim r^{-1}$$ to galaxy with a density profile $$\rho \sim 1$$ for small $$r$$ in a reasonable time scale.

• Intetesting! I didn't consider the effect on dark matter. Jan 21 '19 at 23:21