A black hole is defined as a part of spacetime where gravity is so strong, that spacetime curvature reaches extreme levels. Not even light can escape.


Now as spacetime is curved inside the black hole, is matter curved too inside the black hole?

Can anything straigt (3D) exist inside the black hole? I mean except the direction towards the singularity.

If a straight (3D) rod (not lying in the direction towards the singularity) enters the gravitational zone of the black hole, will it bend?

enter image description here

Gravity bends spacetime around the black hole as per GR, and even the photon's path gets bent (actually it is a straight line in 4D spacetime).


  1. Now will the rod bend next to the black hole because of spacetime curvature or will it stay straight, and which way will it bend?
  • 2
    $\begingroup$ It depends on how you define "straight". Normally, we can define "straight" as the path of a light ray in a vacuum. In the vicinity of a black hole, would we use the same definition? $\endgroup$ – S. McGrew Sep 26 '19 at 22:39
  • $\begingroup$ @safesphere According to Einstein equivalence principle, a local free fall is inertial with no tidal forces This is wrong. You only expected no tidal forces if the stick is infinitely small. For any finite length of stick, it is impossible for the entire stick to be in freefall at the same time, leading to tidal forces. $\endgroup$ – mmeent Sep 27 '19 at 10:03
  • $\begingroup$ @S.McGrew correct, if we use the definition with the path of the photon, will the rod be bent the same way? $\endgroup$ – Árpád Szendrei Sep 27 '19 at 15:32
  • $\begingroup$ @safesphere I believe my question is 2. Will the rod be bent like the path of the photon? $\endgroup$ – Árpád Szendrei Sep 27 '19 at 15:33
  • $\begingroup$ "Normally" means under ordinary conditions -- in flat space. "Straight" in a curved space needs to be a generalization of "straight" in flat space. $\endgroup$ – S. McGrew Sep 27 '19 at 17:17

That depends on the rigidity of the stick, but since all the tidal forces are finite in the case of a nonrotating black hole in theory it should be possible to have a straight stick around it (in principle just like around the earth, where light is also slightly curved, but it is still possible to compensate for that on a stick).

One thing you'll have to choose is the geometric definition of a straight line, if you define it with a light ruler a straight line can also be a closed circle (if it goes exactly around the photon sphere), or you can define it as locally bent but straight in some arbitrary coordinate system, for example that of a stationary or free falling observer.

This is also possible inside the black hole, although the stick would have to be in free fall since a stick with one part outside of the black hole and the other part already inside would break if the part outside of the black hole was held stationary.

Closer to the singularity the differential of the local free fall velocity between the various parts of the stick will eventually exceed c (when exactly that happens depends on the lenght of the stick and the mass of the black hole), then the stick will have to bend (or break, if it is rigid).

With a rotating black hole it's a different story, since the differential of the frame dragging velocity from one part of the stick to another may exceed the speed of light even outside of the horizon (but inside of the ergosphere), in which case even a maximally rigid stick would have to break.

| cite | improve this answer | |
  • $\begingroup$ All wrong, the misconception is yours because you seem to use external bookeeper coordinates for the inside instead of Finkelstein or Gullstrand 🤡 Of course there can be a valid reference frame for that, for example in the frame of an observer sitting on the lower part of the infalling stick it is very well possible that the higher part of the stick is still outside of the horizon. In the frame of an external bookeeper all the parts of the stick are always outside of the horizon due to gravitational time dilation and shapiro delay, but not in the frame of an observer who already fell in. $\endgroup$ – Gendergaga Sep 27 '19 at 4:29
  • $\begingroup$ Ah I remember you are that one guy that has his own interpretation of relativity, which is in conflict with the general consensus. Of course you are entitled to your own private theory, but the general consensus is that you can cross the horizon in finite proper time and notice nothing special when crossing it. If you don't believe me see MTW, § 25.5: i.imgur.com/MhUdfy4.png $\endgroup$ – Gendergaga Sep 27 '19 at 4:48
  • $\begingroup$ But if you think I got it wrong and you got it right don't hesitate to write a better answear, I promise I won't downvote it but if you really think that no frame can reach the horizon I'm afraid others will. $\endgroup$ – Gendergaga Sep 27 '19 at 4:53
  • $\begingroup$ @safesphere - How many percent of the users on Stackexchange agree with your interpretation? Your links are only arxiv and wordpress and I've never heard of those authors before. I think MTW is the more authorative source than some no name arxiv and wordpress articles, but if you doubt my words why not write an own answear instead of only comments and go for the reputation points? As said, I won't downvote it and if you are really right others will upvote your answear. $\endgroup$ – Gendergaga Sep 27 '19 at 6:08
  • 1
    $\begingroup$ @safesphere your logic is deeply flawed. $\endgroup$ – mmeent Sep 27 '19 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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