Generally, when a solution to a system defined by an equation or set of generalized equations that is generally applicable to a broad range of possible initial conditions, or boundary constraints, is itself applicable only to a narrow subset of that broad range, the more specific constrained solution can be derived from the more general solution by fixing, or constraining, within the analysis, the values of the variable representing that constraint.

So, my question is: Can the Schwarzschild solution to GR be derived from the Kerr solution by simply setting the rotation rate to zero?

  • 4
    $\begingroup$ Have you just plugged $J=0$ into the standard form of the Kerr metric to check? $\endgroup$
    – J. Murray
    Commented Mar 10 at 15:30
  • $\begingroup$ No, Unfortunately, I am not mathematically literate enough to try that. But I will give it a try! B $\endgroup$ Commented Mar 11 at 2:28

1 Answer 1


Yes. The most general solution is the Kerr-Newman black hole solution, which has a mass $M$, electric charge $Q$, and angular momentum $J$.

  1. Set $Q=J=0$ to get the Schwarzschild black hole.
  2. Set $Q=0$ to get the Kerr black hole.
  3. Set $J=0$ to get the Reissner–Nordström black hole
  • 1
    $\begingroup$ The most general analytic black hole metric is Kerr Newman De Sitter, set Λ=0 to recover Kerr Newman. $\endgroup$
    – Yukterez
    Commented Mar 10 at 15:32
  • 1
    $\begingroup$ That black hole is not asymptotically flat and falls into a different class altogether. $\endgroup$
    – Prahar
    Commented Mar 10 at 15:33
  • $\begingroup$ This is a low quality answer, essentially just “yes” with no physical details. The Kerr solution has two horizons and a timelike singularity. The Schwarzschild solution has one horizon and a spacelike singularity. If we take a Schwarzschild solution and perturb it by an infinitesimal angular momentum, then what is the asymptotically smooth transition of the infinitely long spacelike singularity behind one horizon to a thin timelike ring singularity behind two horizons? This is a physics site - what is the physics behind this perturbation explaining the transition from Schwarzschild to Kerr? $\endgroup$
    – safesphere
    Commented Mar 11 at 4:01
  • 2
    $\begingroup$ @safesphere - I used to put answers like this in comments and was asked to create entire answers for them. That being said, the details that you suggest I put in are clearly will not be comprehensible to OP. I always gear my answers to the level that I perceive OP to be based on the question. It is clear that they do not understand what a "timelike singularity" even means (and that's fine of course!). There is no point giving these extra details at this point in their learning. $\endgroup$
    – Prahar
    Commented Mar 11 at 11:28
  • $\begingroup$ @Yukterez One might argue that the most general solution is the Plebanski-Demianski family of solutions (Which includes Kerr-Newman De Sitter). $\endgroup$
    – TimRias
    Commented Mar 11 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.