Problem on space-time curvature, black holes and planetary orbits

If the sun were to be replaced by a black hole of the same mass, how would the curvature of space-time change around the sun's initial location and around the region of Earth's orbit? What effect would the change have on Earth's orbit?

So I think that the Earth's orbit would not change because of the same mass, but I am not entirely sure about the curvature of space-time in each of the above locations. I am inclined to believe that, since mass seems to be a determining factor for the curvature of space-time, that it too would not change because the mass will be the same. However, can anyone here confirm or deny what I have written above?

2 Answers

Your question is related to Schwarzschild's spacetimes, which are two spherically symmetric solutions of the equations of General Relativity.

The first one is the exterior Schwarzschild solution (eSS) which describes the curvature (gravity) produced by a spherical object out of it. It is a solution of the vacuum, so that it does not explain the gravitational field inside the sphere. The latter could be modelled with the interior Schwarzschild solution (iSS), if we assume that the sun is a sphere of a perfect fluid with a radial distribution of density and pressure.

Now, let us consider the two important distances in the situation: the radius of the initial star, $r=R$, and the critical radius for the collapse, $r=R_{Sch}$, the so-called Schwarzschild radius that corresponds to the horizon of the black hole (for the real sun it is around 3 km!).

Obviously, $R_{Sch} < R$ (in other case we would have had a black hole from the beginning), so the universe can be separated into three regions:

I ($0 \leq r < R_{Sch}$),

II ($R_{Sch} \leq r < R$) and

III ($r \geq R$).

And you will have the following distribution:

• Before the collapse: iSS (I, II) and eSS (III).
• After the collapse: eSS (I , II, III). There is vacuum everywhere except at the singularity, and now the region I becomes exotic (it is the no-return zone of the black hole).

(Answer) As you can see, in the exterior of the initial star (the region III) the spacetime remains exactly the same, as well as the orbits of the planets.

Note. Both solutions (iSS and eSS) can be "paste" together at the point $r=R$ in a continuous and regular way.

If we assume that the sun is a perfectly spherical body, then nothing would change!

Birkhoff's theorem guarantees that the gravitational field in the exterior of a spherically symmetric body (in our case the Earth lives outside the sun) must be described by the 1-parameter (this parameter being the mass) family of solutions given by the Schwarzschild metric. Since you assume the black hole to have the same mass as the Sun, the two gravitational fields would be indistinguishable.

(Of course, the Sun is not a perfect sphere and moreover there could be other effects due to electromagnetic interactions, so the statement only really holds at leading order...)