0
$\begingroup$

If I do not interpret equivalence principle wrongly, it says that we could always free fall ourselves to achieve local inertial frame. The metric would be $ds^2=-dt^2+dx^2+dy^2+dz^2$.

Suppose an astronaut am nearby an object described by Schwarzschild metric, say a black hole. He free falls himself, subject to gravity only and follow the geodesic. What metric would he see? Is it still the ordinary Lorentz frame's metric or Schwarzschild metric?

And as an observer observing the tragedy, I am using Schwarzschild metric to view the event. Am I understanding it correctly?

$\endgroup$
4
  • $\begingroup$ My understanding of how equivalence comes about in relativity is that at every point on your space-time manifold you can go into the local Lorentz frame, i.e. you have the Minkowski metric at a single point. Physically speaking, this means that the astronaut sees the basic Minkowski metric in his local Lorentz frame. The Schwarzschild metric is defined at every point in space time, while you can only create a consistent flat metric at a single point, rather than on any reasonably large scale. $\endgroup$
    – J.V.Gaiter
    Apr 22 at 17:39
  • $\begingroup$ So you as an observer see things very near to you as if you are in a flat space-time, while looking far away you have to consider the whole Schwarzschild metric. $\endgroup$
    – J.V.Gaiter
    Apr 22 at 17:40
  • $\begingroup$ If I understand your statement correctly, that is to say if I am the astronaut, to my point of view, the way I am falling into the black hole has no difference from the way I am falling on the Earth. Am I correct? $\endgroup$
    – Simon219
    Apr 22 at 17:52
  • $\begingroup$ In principle, yes. As the astronaut falls into the black hole, at least a suitable distance from the event horizon, the cause of him falling towards the black hole is the same as the cause of someone else falling towards earth in free-fall. Of course, these two cases differ quite a bit in the magnitude of the "acceleration" that the observers see, due to differences in mass in the two circumstances. $\endgroup$
    – J.V.Gaiter
    Apr 22 at 18:01

Your Answer

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

Browse other questions tagged or ask your own question.