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I want to ask some very naive questions in general relativity. I have the background of PDE and few Riemannian geometry.

After Schwazchild gave a solution, people study its singularity and predict the black hole, but for some PDEs, people can get a solution with singularity by intuition but in fact this PDE has a smooth solution, for example $$\Delta u = e^u - (x^2+y^2)e^{-u}$$ on $B_1(0)$, $\frac{1}{2}log(x^2+y^2)$ is obviously a solution with singularity, but by prior estimate we can find that it has a smooth solution, my question is: are there smooth solutions to Einstein equation? In these universe there is no black hole？

Elio Li
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Remember that while a PDE might have whole families of solutions, in practice we often want to look at the solution(s) that satisfy certain initial and/or boundary conditions.

The Schwarzschild "black hole" is a specific case of a more general solution, that makes the following assumptions:

  1. The system is spherically symmetrical about the (spatial) origin.

  2. The system is not changing with respect to time.

  3. The system "flattens out" in the limit as the distance from the origin goes to infinity.

Under these assumptions, you get the classic "rubber sheet with a mass on it" kind of behaviour in the solution, and the amount of distortion at a given point is a function of (a) how far the point is from the origin, and (b) how much mass is closer to the origin than the point (i.e. if the point is at distance $R$, then how much mass is at distance $r \leq R$).

If you look at that distortion function, then in fact as long as the mass is sufficiently spread out then the entire thing is smooth, with no discontinuities. The discontinuities only occur when the mass is sufficiently compact - specifically, when an amount of mass is present in a volume contained within its own Schwarzschild radius. Even then, with an appropriate change of co-ordinates you will still have a smooth metric everywhere (the Schwarzschild metric assumes that all of the mass is contained in a point at the origin which is why it has an unremovable singularity there, whereas other solutions that assume a broader distribution of mass can avoid that to some extent).

ConMan
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  • Thanks for your answer! I also have another question, why all the existing solutions of Einstein equation (Reisner-Nordstrom, Kerr, Schwazschild and Kerr-Newman) has singularity? Is this a coincidence or. – Elio Li Nov 07 '23 at 06:04
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    Probably not - they're all coming from the same base set of equations, so it's not surprising that they have some features in common. – ConMan Nov 07 '23 at 08:51
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    The first answer addresses an essential point in understanding what solution of Einstein field equations in physics is - it means solving of initial value problem not just finding a solution function. In case of static perfect fluid sphere the equations reduce to the second order linear parametric linear ODE with two physical boundary conditions. One can show that in all corresponding solutions for some critical parameter the central pressure diverges what is commonly interpreted as 'singularity’. – JanG Nov 20 '23 at 17:04