When computing the distance at which the escape velocity is equal to the speed of light, the result is equal to the Schwarzschild radius. Is this just a coincidence, or is there a physical reason why the Newtonian approximation give the correct result?

If this is not a coincidence, then why is it OK to use Newtonian physics here?

  • $\begingroup$ This question has previously come up on the Physics Stack Exchange: physics.stackexchange.com/questions/39176/… $\endgroup$ – antispinwards Jan 20 at 18:34
  • $\begingroup$ How odd that we can't close a question which is satisfactorily answered on another SE site. Something for the PowersThatBe to consider. Certainly would be simpler than forcing this question to migrate and then be closed. $\endgroup$ – Carl Witthoft Jan 20 at 18:59
  • 3
    $\begingroup$ @CarlWitthoft It's contrary to the design philosophy to close things simply because they can be found on another stack exchange. Questions can be on-topic on multiple stack exchanges, and each should (be allowed to) contain their own Q&A for it. I don't entirely see why visibility isn't identical for questions closed as duplicates with links to the other-site Q&A, but such is what it is right now. If it is off-topic here but on-topic elsewhere (say physics), then that's when a move can be warranted (though that's harder when the base site is in beta). $\endgroup$ – zibadawa timmy Jan 20 at 23:25
  • $\begingroup$ @zibadawatimmy Guess I'd be in favor of the equivalent of a "softlink" so we don't clutter things up with duplicate answers in different SE sites. The sites should work together instead of maintaining walls between each other. $\endgroup$ – Carl Witthoft Jan 21 at 15:54
  • 2
    $\begingroup$ Note, the result is the same in this case, other similar calculations give very different results. For example, the Sun bends starlight twice more in the GR as in NM (the GR value is observed). $\endgroup$ – peterh - Reinstate Monica Jan 21 at 19:38

As explained in this answer, dimensional analysis indicates that this radius depends on $GM$ and on a velocity squared. However, the fact that the two quantities are exactly equal, rather than there being a constant factor between the two is a pure coincidence.

| improve this answer | |

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.