4
$\begingroup$

Goodmorning everyone. I would like to share with you a question that has been gripping me for some time, but which I have never been able to give a convincing answer. When representing the ergosphere or the external event horizon of a black hole, it is often not taken into account that the coordinates used (if space-time is Kerr, the most usual are those of Boyer-Lindquist) have no physical meaning, in the sense that they do not allow us to "see" what the real form of such spatial hypersurfaces would be if they could be "spied" from the earth.

Now, I tried to formulate the embedding, so that the line element of the metric was the Euclidean one IE $$ds ^ 2 = dx ^ 2 + dy ^ 2 + dz ^ 2;$$ the problem (which I found also in the literature) is that this process is not always possible (for example if the spin of the black hole exceeds a certain critical value).

My question is: imagining a rotating black hole with very high angular velocity (t.c. angular momentum at = 0.99 in natural units), what should I see? And how do I understand analytically what geometric shape would have external horizon and ergosphere (spied from the earth) if I cannot embed them in a 3D space?

Improve this question
$\endgroup$
3
  • $\begingroup$ Thanks for the reply! But my question is not so much inherent in what I would see in reality (which is actually interesting), but it is more related to what it means to the fact that such a form cannot be embedded. In practice, imagining the ergosphere/horizon as a colored surface, since this cannot be embedded in a 3D space (beyond a certain critical value of angular momentum), how could I represent it through physical coordinates? And if this is not possible, how can a geometric shape not be representable? How can the horizon not have a proper 3D shape? Thank you again. $\endgroup$
    – Alessandro Rovetta
    Commented Dec 2, 2018 at 22:28
  • $\begingroup$ Hi and thank you again for the answer. I'm not talking about making all the space-time flat, but to embed the hypersurface "event horizon" (or ergosphere) in a 3D euclidean space and this can be done (under some conditions) and it makes sense to do it. In fact if we have $$g_{\mu \nu} dx^\mu dx^\nu\;|\;dt=0$$ So as to have a "photo" of the geometry. Then the remaining terms $$g_{\mu \nu} (r,\theta,\phi)$$ are spacelike. Now, by placing $$r=M+\sqrt{M^2-a^2cos^2\theta}$$ in the line element, it is a matter of making a coordinate change such that $$ds^2=dx^2+dy^2+dz^2$$ and this is possible $\endgroup$
    – Alessandro Rovetta
    Commented Dec 3, 2018 at 12:24
  • $\begingroup$ (here my calculations [link]ilrad1online.it/wp-content/uploads/2018/04/…). My problem is what it means that this visualization procedure is not possible due to too high angular velocities of the black hole. Thank you again for the interest. $\endgroup$
    – Alessandro Rovetta
    Commented Dec 3, 2018 at 12:25

2 Answers 2

Reset to default
1
$\begingroup$

I think I solved my own question: the embedding process is a distortion of space-time as it transforms a curved geometry into a flat one. Ergo, it is not correct to say "spied" FROM the earth but rather "observed" IN a flat space. Although it is still interesting to understand why embedding is not feasible under certain conditions, what I would see if I observed a black hole obviously depends on the curvature of the space-time i.e. on the way light reaches my measuring instrument [1].

Sources:

[1] https://arxiv.org/abs/1502.03808 p.27 - image (c).

Improve this answer
$\endgroup$
1
$\begingroup$

You might find this paper helpful: https://arxiv.org/abs/0809.2369

It explains how to go about constructing an embedding, and the restrictions that arise. In the case of the ergosphere, unlike the horizon, there is no intrinsic 2-dimensional geometry, since the 3d ergosurface in spacetime is timelike, not null, so the 2-geometry you get when slicing it by a spacelike cut depends on the cut.

Improve this answer
$\endgroup$

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.