4

This is some kind of a follow up of this nicely to the point answer to a provocative (but nevertheless upvoted!) question, about the legitimacy of black hole physics. The answer mentions, that the interior and exterior of a black hole are not completely decoupled, and about this exact relationship I'd like to learn some more not just hand waving details.

Of course I know the complementarity principle, which says that the perspective of what happens near the event horizon of a black hole an infalling observer who does not note anything special when crossing the event horizon and the description of the physics near the horizon by a staying outside observer who sees the infalling observer getting frozen on the horizon, are both equally legitimate descriptions of the same physics.

Is there some kind of a map, or exact dictionary by which one can transform back and forth between these two descriptions of the same physics, that describes the complementary principle for black holes mathematically?

An aside: I always thought that the holographic principle is soemething different from the complementarity principle in this context, so this should not be the trivial answer to my question (?) ... Maybe I have managed to confuse myself now :-)

Clarification after some comments

I am not just up to a coordinate transformation between spacetime inside and outside the even horizon, but I'd like to learn about how the degrees of freedom used by an infalling observer and the ones used by a staying outside observer to describe the same physics transform into each other.

Community
  • 1
Dilaton
  • 9,503
  • 2
    I think what you're looking for are what are called the Novikov coordinates of a black hole--you surrender manifest staticity of the spacetime and the coordinates don't cover the whole Schwarzschild patch, but in exchange, you have a coordinate set that is static with respect to infalling radial observers.. – Zo the Relativist Jun 05 '13 at 00:25
  • It's a fundamental principle of GR that it is formulated in a co-ordinate independant way. That automatically means that the description in all frames of reference describe the same physics. Transforming to and fro between any two co-ordinate systems is just a matter of straightforward if tedious algebra. If you're interested in linking the behaviour inside and outside the event horizon you'd use co-ordinates that are continuous across the horizon e.g. Kruskal-Szekeres co-ordinates (http://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates). – John Rennie Jun 05 '13 at 06:21
  • 2
    @JohnRennie : I agree with the general GR philosophy, but, practically, the transformations of freedom degrees of fields seen by the free-falling observer and a outside fixed observer are highly non-trivial. – Trimok Jun 05 '13 at 09:38
  • @Trimok yep exactly, I am not just up to a coordinate transformation between spacetime inside and outside the even horizon, but I'd like to learn about how the degrees of freedom used by an infalling observer and the ones used by a staying outside observer to describe the same physics transform into each other. I'll clarify the question. – Dilaton Jun 05 '13 at 09:44
  • I'm still not sure exactly what you're asking. As long as the free falling observer stays outside the event horizon the co-ordinate transformations are pretty straightforward, and once they've crossed the event horizon they are causally disconnected and there is no transformation between the co-ordinate systems. – John Rennie Jun 05 '13 at 13:19
  • @JohnRennie yes, I understand that there is no transformation between the coordinate systems. But nevertheless, the fallen in observer who still notices nothing special until he hits the real singularity at r=0 and the outside observer who sees the other one getting "frozen" on the horizon are talking about the same physics. I am interested in seeing how this works technically. – Dilaton Jun 05 '13 at 13:44
  • @Dilaton: the exterior Schwartzschild co-ordinates constitute a patch that does not cover the whole of the manifold. How this works technically is that once inside the event horizon you're outside the Schwarzschild patch. That's why there is no transformation between the two. In the Schwarzschild co-ordinates the inside of the black hole is technically covered by a separate patch that is disconnected from the exterior patch. However the interior patch is not physical so can't usefully be used to describe what happens. – John Rennie Jun 05 '13 at 14:17
  • 2
    Dear @Dilaton, that's a very good - and very hard - question that hasn't been fully answered, as the continued stream of new papers attempting to answer the "code" - such as the Maldacena+Susskind paper a few days ago - demonstrates. So while it seems clear that the degrees of freedom inside can't be treated quite as independent ones from those outside - i.e. that classical GR's causality is misleading in principle - the precise dependence hasn't been clarified. – Luboš Motl Jun 12 '13 at 17:47
  • Hi Lumo yep, when reading your additional thoughts concerning the Maldaena+Susskind paper in this article, it made me immediately think of this could point towards a possible solution of the issue, as I said in my comment on TRF :-). – Dilaton Jun 12 '13 at 18:30

1 Answers1

3

I share the same big interest about your question and I have not the answer, but I have one idea:

The idea is to study a simpler example, that is a non-uniform accelerating observer in special relativity. For instance, imagine a 100-meter race in a stadium. Typically, it takes 10 seconds for the runners to finish the race. The people in the stadium (as observers) see the runner crossing the finish line. We may call $z$ and $t$ the coordinates as seen by a stadium observer (S), so as the runner crosses the finish line $z = 0$ at $t = 0$. Now, we can choose an moving observer (M), which has a variable speed $v(t)$ relatively to the stadium observer. I think that one can choose $v(t)$ such as the observer M never sees the runner crossing the finish line. You need simply: $$\int_{-10}^0 \frac{dt}{\sqrt{1 - \frac{v^2(t)}{c^2}}} = + \infty$$ (I suppose here that we consider that the speed of the runner is neglectible relatively to the speed of light) So, it is a kind of horizon, it is a planar horizon at $z = 0$, instead of a spherical horizon in the Schwartzschild GR problem, but the logic is the same. Now, by knowing $v(t)$, you know the transformation law between $dz'$, $dt'$ (M coordinates) and $dz$, $dt$ (S coordinates)

From this, it would be possible, in principle, to get the transformation of fields, for instance, for a scalar field, we would have (with $x'=x$ and $y'=y$):

$$\phi'(x', y', z', t') = \phi(x, y, z, t)$$

For instance, it would be possible to modelize the runner as a cylindric or cubic wave-packet $\phi$, and we should be able to prove that we have : $$\phi'(x', y', z', t') = 0 ~ for ~ z > 0$$

Trimok
  • 17,567