It was probably one or more stellar black holes, most of which are born perhaps twice the mass of our sun. Update: Corrected to account for visible supernovae ejecting 75% of their mass If the implosion of the supernova were completely symmetrical, no gravitational waves would be emitted outside the star - However, computer simulations (and some recent observations) suggest that a supernova implosion is a very chaotic process, and often very asymmetrical, meaning that gravitational waves may be detectable from a nearby supernova (and neutrinos too - but astronomers have been waiting since 1987). Changes in the shape of spacetime/Gravitational Field would propagate within this zone, but this is typically within a radius of a few light-seconds. There are major changes in the gravitational field between (the original surface of the star) and (the surface of the new black hole). So the gravitational field outside of the star does not really change before and after the supernova, so the gravitational field does not need to "travel" for light-years. This point is now at the center of the newly-formed black hole. Before the supernova, the distortion of spacetime outside the surface of the star emulates the distortion as if all the mass of the star existed at a single point at its center (thanks to Newton's shell theorem) - After the supernova, the distortion of spacetime outside the (original) surface of the star shell of ejected material emulates the distortion as if all the mass of the star existed at a single point at its center (thanks to Newton's shell theorem). This star bends spacetime around the star - At the end of its life, after it has burnt all the fuel in the core to iron, the star explodes/implodes as a supernova, forming a black hole, with almost the same the same order of magnitude as the mass of the star before it imploded. Stellar-mass black holes typically start off as a massive star. one that does not change the curvature of spacetime itself.Ī Gravitational Field can be viewed as a distortion in spacetime (thanks to Einstein). I've always wondered if that is only true for a "test mass" dropped in toward a black hole, i.e. Obviously that narrative is very different from the usual version of what happens when something falls towards a black hole and is said to never reach the event horizon - as far a distant observer is concerned. That may all happen in a finite amount of time for an observer well away from the black hole. Having engulfed it, spacetime settles back down to a typical Schwarzschild solution again but now with a slightly greater mass parameter. One reasonable idea is that there is a significantly different, if temporary, solution to the EFE when the rock is near the (former) event horizon of the black hole, the event horizon extends outward to meet the rock falling in and does engulf it. The PopSci explanations usually gloss over this issue completely. Overall, I can't really say what happens - which is why I was asking. However, the geometry of spacetime isn't going to be exactly Schwarzschild with a rock close to the event horizon. The rock never reaches the black hole event horizon IF the space in that region retains its Schwarzschild geometry. Anyway, I am inclined to agree with the later comments you made. It's the mass parameter of the Black Hole that I was more interested in. However, your statements explain what happens there well. Hi and thanks It wasn't really the mass parameter of the rock I was interested in.
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