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2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...


1

There is already a good answer from John Rennie. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See ...


11

Let's start with what we mean by a horizon: The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists. A null geodesic is the path followed by a light ray, so the horizon marks the surface at ...


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You are standing on surface of a planet of mass $M$ and radius $R$. With which velocity $v$ you need to throw away from the planet an object of mass $m$ that it will not return? The gravitational force is $F=G\frac{mM}{r^2}$. The work to be done to move the object in the gravitational field of the planet from distance R to infinity is $A = ...


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There is an amusing "trick" (so is not meant as an answer to your question) in Newtonian gravity (plus a bit of special relativity) to derive that the Schwarzschild radius should be "special" Since self-gravitational energy of massive bodies is always negative (at least in Newtonian gravity) $$ - \frac{G M^2}{R} $$ And massive bodies have positive rest ...


5

The geometry of spacetime is described by an equation called the metric. This is analogous to Pythagoras' theorem but with some key differences. Start with a 2d plane, where we identify positions of points by their $(x, y)$ coordinates. Suppose you move a distance $dx$ then a distance $dy$, then the distance from your starting point, $ds$, is given by ...


0

For trying to give a conclusive answer it seems necessary to rigorously characterize the geometry (the "conicidence structure", incl. the "light-cone structure") of the region under consideration (arguably with exception of "the singularity itself"). Unfortunately, this seems complicated (as may be gathered from efforts to tackle related problems at least ...


1

What you learned is correct. More simply, it's a consequence of the "time reversal symmetry" of most of fundamental physics. This symmetry is still present in general relativity. But, it's obscured by the standard system of coordinates. When you transform these coordinates into the Kruskal coordinate system, you not only have a black hole, you also have ...


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I learned in school (very long ago) that the path followed by light is independent of its orientation on this path. There's 2 things going on there. One is that our knowledge has changed since you went to school (or more correctly when the textbook author went to school), and the other is that these statements tend to hold true only at human timescales ...


1

I"m pretty sure that this discussion does appear in Hawking and Ellis, though I admit that it's been a while since I looked. It's not done through a Penrose diagram, though. The argument really comes down to the fact that for sufficiently small $r$, $d\phi$ is timelike. But, by construction, the orbits of $\phi$ are closed curves. When $d\phi$ is ...


1

This is really one google search away, see e.g. page 26 (marked 64) here. As already noted by John Rennie, Penrose diagrams are not suited for the analysis of Kerr CTCs because they show a $\phi = const., \theta=\pi/2$ slice of the global structure. The $r<0$ region is however accessible only through $\theta \neq \pi/2$. The Boyer-Lindquist coordinates ...


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Timaeus : I can't answer your question, but would like to comment. However what I want to say is to big for a comment, so I'm using the answer facility. Apologies in advance, feel free to downvote. The region beyond the ring singularity in the maximal Kerr spacetime is described as having closed timelike curves. Let's imagine we're out in space, at a safe ...


1

I agree that you would not "notice" or "feel" anything as you "fall" into a black hole. As far as matter is concerned, the event horizon is not a line of demarcation were "strange" things happen on one side but not on the other. As far as you are concerned, all your molecules would be accelerated the same before and after crossing the event horizon, and ...


1

I think I found a coordinate transformation that shines more light on this. Instead of looking for a conical singularity in $g_{tt}$ alone I must look at conical singularities in any part of the metric (apart from $g_{rr}$) First of all the metric can be expressed in terms of $r_\pm$ and Wick rotated $t\to it_E$ such that $$ ds_{E}^2 = \frac{ ...


-2

According to the relativistic doppler effect light shifts into blue or red depending on the relative velocity of object and observer. Thus, looking at a redshifted light ray which source is close to the event horizon of a SMBH it must be possible to shift its light back into blue just by accelerating the speed of the observer by the relative amount. ...


2

For a non-rotating uncharged black hole (the Schwarzschild metric) once you have crossed the event horizon there is no timelike trajectory that will increase the distance between you and the singularity. By this I mean that to increase your distance from the singularity would require you to be moving faster than light, which is of course impossible. So all ...


4

This is largely the same answer as Rob's, though rather than use Eddington-Finkelstein coordinates I'm going to use Kruskal-Szekeres coordinates because I think this makes the argument easier to understand. This is what the situation looks like in Kruskal-Szekeres coordinates: For the non-nerd the Kruskal-Szekeres coordinates seem formidably complicated, ...


0

Therefore I'd like to ask a related question in which pings are plainly the main point... OK. I would refer you to Einstein talking about the speed of light varying with gravitational potential. And to Irwin Shapiro, who was involved in pinging radar signals to Venus and back, saying "the speed of a light wave depends on the strength of the gravitational ...


4

The answer must be closely related to my answer in Thought experiment - would you notice if you fell into a black hole? You can certainly use a similar Eddington-Finkelstein coordinate diagram to consider it (the E-F coordinates transform away the coordinate singularity at the event horizon). NB: This considers only GR and a non-rotating black hole (and ...


1

For a non rotating spherical black hole the event horizon is at a circle of circumference $4\pi MG/c^2$. And gravity gets weaker the farther away you get. At that location even moving straight outwards at light speed won't let you travel to someplace where gravity is weaker. But if you aren't moving straight away then you will get trapped even farther out. ...



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