I have been following a course on GR that at one point discusses the metric derived for the outside of a physical, non rotating uncharged massive object with spherical symmetry. For this situation I have seen the Schwarzschild metric derived, in Scharzschild coordinates.

From the form of the metric, it is observed that bad things may happen at $r = 2GM$ and $r = 0$, but this comes with the warning that since the metric is coordinate dependent, one should check the scalar contractions of the Riemann curvature tensor to see if any physical bad things happen. The course proceeds to show that this is the case for $r=0$ but not for $r=2GM$.

Then, light cones are studied by finding out what happens with the geodesic equation for massless particles, in Schwarzschild coordinates, and they appear to close up as $r$ nears $2GM$. This suggests that in these coordinates, light can reach the horizon, but not cross it.

A coordinate transformation to Eddington-Finkelstein coordinates is then used to show that the metric in these coordinates is benign at $r=2GM$ and an analysis of the light cones in these coordinates show that one edge of the light cone, the one that is oriented radially inward to the center of the geometry, appears to be unchanged from flat spacetime while light oriented radially outward has its edge of the light cone tipping over until at the horizon, this is aligned with the horizon.

This analysis shows that light can reach into the horizon, but cannot escape the region.

With this introduction and context, my question is how interpret the different analysis of the light cones in these two different coordinate systems. On the one hand, when done in Schwarzschild coordinates, light cones close up and align with the horizon, while in E-F coordinates it is clear that there are geodesics into the horizon, only none that come out.

To be clear, I’m not asking whether light can or cannot enter an horizon formed by a black hole. I’m looking for some clarity on why the analysis in Schwarzschild coordinates cannot be trusted (fully), but apparently the one done in the Eddington-Finkelstein coordinates can.

  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or were otherwise inappropriate, as well as responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Jun 28 '20 at 22:04

The whole point of GR is that physics is independent of the specific coordinate system you are using. So if a phenomenon occurs for a certain coordinate system (i.e. the $r = 2GM$ singularity in Schwarzschild coordinates) but does not occur for a different one (i.e. Eddington-Finkelstein coordinates) then chances are it is not a physical effect but rather an artifact of the maths.

So basically if you see a singularity, try to see whether it is still a singularity when using different coordinate systems. If it does not, then it was a coordinate singularity (not physical, mathematical artifact). if it does, chances are it is an intrinsic (physical) singularity. So you should "trust" that $r=0$ is weird because multiple coordinate systems find it weird, but you should not "trust" $r = 2GM$ as weird because some systems are ok with it.

This reasoning is similar to removable or essential singularities/poles in complex analysis.

Also, you have to be careful about the "physical interpretation" of the light cones. In that what you are plotting versus what.

The typical Schwarzschild coordinates plot is the following (from here), where the $y$ axis is time $t$ (geometrised constants, $c=1$ etc.) and the $x$ axis is $r$. In this basis, the metric entries $g_{00}$ and $g_{11}$ that control the dot product contributions $\mathrm{d}t^2$ and $\mathrm{d}r^2$ are plotted (below) and you can see that their signs change. This is why some people may say that "time and position" change places past the event horizon. But it's because of the interpretation in this specific coordinate system.

enter image description here enter image description here

On the other hand, the typical plot in Eddington-Finkelstein coordinates is the following. Note how the axes are now $v-r$ and $r$. $v$ is related to $t$ but it's not exactly the same.

enter image description here

Finally, another set of coordinates usually employed for spherical non-rotating black holes are Kruskal–Szekeres coordinates, $u$ and $v$ (different $v$ from before), shown below. The Kruskal-Szekeres parameterisation is useful as it is the unique maximal extension of Schwarzschild spacetime. Also, they are known as they lend themselves to introducing Einstein-Rosen bridges (wormholes).

enter image description here

  • $\begingroup$ @Rolk Please do read the comments under your answer 'cause it seems that my understanding is incomplete $\endgroup$
    – SuperCiocia
    Jun 28 '20 at 20:11
  • $\begingroup$ The problem is that I fail to understand that the physics described in the Schwarzschild coordinates appear to describe a situation that light cannot reach the horizon, while the Eddington-Finkelstein diagram seems to imply that they can, but cannot ever leave. Two different coordinate sets imply different physical results. The question is then, can we trust any of these lightcone analyses, and if so, how do we know which are physically meaningful? I’ll check the referenced papers from the comments too. $\endgroup$
    – Rolk
    Jun 28 '20 at 20:14
  • 1
    $\begingroup$ Why do you say that the Scharzschild coordinates suggest the light cone can never reach the horizon? It can. $\endgroup$
    – SuperCiocia
    Jun 28 '20 at 21:45
  • $\begingroup$ Because in those coordinates the light cones close up, i.e. the geodesic of inward light and outward light get nearer ad nearer as you approach the horizon. In the limit, they align with the horizon, and there is only time motion, and no more spatial motion. $\endgroup$
    – Rolk
    Jun 29 '20 at 6:04
  • $\begingroup$ Ah...it’s a bit like Zeno’s paradox. Even if it seems like it gets harder to get nearer the horizon, light can still reach it in finite time. $\endgroup$
    – Rolk
    Jun 29 '20 at 6:38

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.