The radial component of the Schwarzschild metric shows a dramatic metric stretching as the event horizon is approached from the outside and it shoots to infinity at the horizon.

Does this extreme metric stretching show up in other coordinate systems like the Eddington–Finkelstein or the Kruskal–Szekeres coordinates or is it just a property of the S-child metric?

  • $\begingroup$ It seems that by dramatic/extreme metric stretching, you essentially mean a coordinate singularity. $\endgroup$
    – Qmechanic
    Jul 3, 2018 at 17:14
  • 1
    $\begingroup$ The metric "stretching" to infinity simply means that time stops at the horizon as observed from afar. Other coordinate systems are deceiving, because instead of time they use combinations of time and space. So "time" defined in these systems doesn't stop at the horizon, but it is not the physical time. The physical time does not depend on which system you use and stops at the horizon regardless of such misleading mathematical tricks. While using these systems, one should always be careful and keep in mind the actual physical effects, not just abstract coordinates. $\endgroup$
    – safesphere
    Jul 4, 2018 at 4:44

1 Answer 1


As you suggest, it's a peculiarity of the Schwarzschild metric. The same point in other coordinate systems (Eddington–Finkelstein or the Kruskal–Szekeres are good examples) does not show any pathology and all the curvature invariants are regular.

A nice reference on the topic is Spacetime and Geometry by Carroll.

  • $\begingroup$ So, just to make sure, are you saying that there is zero metric stretching in the other coordinate systems or that it just doesn't go to infinity and become a coordinate singularity, as Qmechanic mentions above. Sorry to be picky but I'm having a hard time differentiating between metric stretching and regular gravitational effects. Is metric stretching just the special situation of a highly non-uniform gravitational field or is it a completely different critter? $\endgroup$
    – dcgeorge
    Jul 5, 2018 at 17:25
  • $\begingroup$ As you can directly verify looking at the "good" metrics, there is really no coordinate singularity. Also, I'm not sure about what you call "metric stretching", but tidal forces, related to the Riemann tensor, can be arbitrarily small (given a big enough black hole), therefore nothing physically relevant happen for a local observer. $\endgroup$
    – Rexcirus
    Jul 5, 2018 at 19:45

Your Answer

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

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