A black hole swallows a star.

The star remains intact.

Then goes supernova.

What would happen to the star?

Could we detect this event?

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    $\begingroup$ How would the star remain intact? I don't think this can happen. Therefore the whole scenario is a bit constructed. $\endgroup$ Feb 22, 2019 at 12:38
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    $\begingroup$ The largest supermassive black holes have a Schwarzchild radius around $6\times 10^{13}$ meters. Tidal forces near that radius would be roughly $10^{-12} g/m$ or a few milli g across the diameter of a not too large star. So it is definitely possible for a star to fall into such a black hole without being immediately destroyed. $\endgroup$ Feb 22, 2019 at 12:49
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    $\begingroup$ I'm voting to close this question as off-topic because the premise is fatally flawed - the situation could never occur. $\endgroup$ Feb 22, 2019 at 16:35
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    $\begingroup$ @CarlWitthoft (also Max0815) please explain what makes it impossible. As Steve Linton notes, the event horizon of a large SMBH can easily be 200 AU in radius, and there are a number of plausible situations where a star, white dwarf or neutron star could go supernova at that distance. $\endgroup$ Feb 23, 2019 at 11:38
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    $\begingroup$ @Chappo right yeah the roche limit of supermassive black hole is inside event horizon. $\endgroup$
    – Max0815
    Feb 24, 2019 at 16:01

1 Answer 1


We could not-- whatever happens inside a black hole stays inside a black hole. One way this is often explained is that the time coordinate inside the "event horizon" of a black hole looks to us on the outside like a radial coordinate-- events that occur at sequentially different times inside the event horizon, such as your supernova, seem to us on the outside to happen at sequentially smaller radii, until the last event for your dying star is to reach the center. The same holds for all the ejecta and even the light and neutrinos emitted-- it all goes to the center in a finite time. So nothing escapes from the event horizon.

Another way to say this is that "black holes have no hair", meaning there is not information outside the event horizon stemming from anything happening on the inside. All that happens to the outside stems from what falls into the hole-- how much mass, charge, and angular momentum crosses the event horizon. What it does while it is in there has no effect on the outside, so could never be detected.

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    $\begingroup$ You might add that it can't be observed because we can't observe anything cross an event horizon. $\endgroup$
    – ProfRob
    Feb 23, 2019 at 23:06

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