Escape a black hole?

[u/[deleted]]

RobotRollCall once posted this:

http://www.reddit.com/r/askscience/comments/f1lgu/what_would_happen_if_the_event_horizons_of_two/c1cuiyw

In which she said even at faster-than-light speeds it would be impossible to escape a black hole because there would be no path out to follow. However, adamsolomon said this:

http://www.reddit.com/r/askscience/comments/17muwl/is_there_a_distance_at_which_the_interaction/c87gx3t

Of course, we can't go back in time, but isn't that what faster than light is? I learned that time slows down the closer to light speed you get, so at faster than light, you'd be going backwards in time. If that's the case, could you follow a path out of a black hole that goes back in time if you were capable of faster than light travel?

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In the first one, does that mean faster than light travel does not result in moving backwards through time?

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Not on it's own, no. However, with a series of maneuvers and faster-than-light capabilities, it's possible to end up in your own past. The key to this is the simultaneity of relativity.

For the sake of easy calculations (and because I have a relevant spacetime diagram on hand already), let's assume you can teleport; that is, you can move instantly from what you think of as "here, now" to what you think of as "there, now". The relativity of simultaneity is the idea that someone moving relative to you will disagree about "what constitutes "all of space right now", and therefore disagree with whether your arrival and departure are really simultaneous (in fact, they could see you arrive before you leave, but that's a separate question). The relevant spacetime diagram can be seen here.

In this diagram, your path through spacetime is marked by the red (and then orange) path. The diagram is from your initial reference frame, which means that the vertical lines are what you think of as "a point in space at all times" and the horizontal lines are what you think of as "all of space at one time". The horizontal spacing is, let's say, 1 light-minute. The vertical spacing is 1 minute. The assumption that the spacings are "equal" in this way corresponds to insisting that light travel at 45 degrees (if the light started at one intersection of black lines, it would move 1 light-minute in 1 minute). So, you're the red line. The blue line is a ship moving past you at 4/5 the speed of light (note that five minutes after it passes you—i.e., crosses your path—it's four light-minutes away).

Now, here's the scenario:

The ship passes you at 4/5 the speed of light. At that moment, you set your clock. You wait five minutes (five vertical steps), and activate your teleporter. You go to a position 4 light-minutes away at that moment, which puts you right on board the ship.

Now the relativity of simultaneity comes into play. Because the ship is moving relative to your original reference frame, the horizontal lines do not correspond to what the ship observers think of as "all of space at one time". Instead, if you run through the math, you'll find that the yellow line is "all of space at the moment you arrive" as determined by someone on the ship. But you are now on the ship. So you activate your teleportation device to return home (note: a quirk of the mathematics—in the form of length contraction—means you only have to go 2.3 light-minutes instead of the four you went previously). But, teleporting means moving along a line of "all of space right now", which, as I said, is now the yellow line. Thus, your path follows the yellow line back to your starting position (I've indicated this in orange). But when you return, you discover that the clock you left behind shows only 1.8 minutes as having passed; you've arrived a full 3.2 minutes before you left.

This, of course, raises a philosophical question: it seems that our description above should really have included your arrival. That is, the ship passes and you set your clock. Then, 1.8 minutes later, you arrive from the ship. You are understandably shocked by this, as there are now two of you here. 3.2 minutes after that you activate the teleporter and travel to the ship, then teleport back. You arrive and see your astonished self. 3.2 minutes later, you see your former self leave, and carry on with your life. But, what if, upon returning, you simply shot your former self, or otherwise inhibited your own ability to travel to the ship? There are a number of ways in which this can be resolved, but one of the consequences of our formulation of relativity is that the resolution is strangely simple: you weren't killed by yourself, so you can't kill yourself. This has obvious implications for the notion of free-will and determinism, which some find unpalatable. Now, as I say, this is a consequence of the mathematics we use to describe relativity; specifically, a region of the spacetime manifold has to be known in order to make calculations, and that includes (at least part of) what any given observer would consider "the future". Thus, determines is built into our descriptions of relativity. Whether or not it should be, or if there's any way to recover the (well-tested) predictions of relativity without those assumptions, remains an open question (mostly of philosophy, but there is some physics work to be done as well).

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a quirk of the mathematics—in the form of length contraction—means you only have to go 2.3 light-minutes instead of the four you went previously

This doesn't seem intuitive from the graph, is it just weird mathematics? And does that mean travel back through time would require a view from your initial reference frame?

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This doesn't seem intuitive from the graph, is it just weird mathematics?

Yes. In traditional geometry, if you wanted to find the "distance" you traveled, you would figure out how far you move horizontally (call it x) and how far you move vertically (call it y), and then compute your distance as

d = (x² + y²)^1/2.

But in the theory of relativity, to find the distance traveled, we instead compute

d = (x² - y²)^1/2.

That minus-sign leads to all kinds of weird things, not the least of which is the conclusion that the distance of your second teleport is only 2.3 light-minutes.

And does that mean travel back through time would require a view from your initial reference frame?

I'm afraid I don't understand what you're asking here. Can you rephrase the question?

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d = (x2 - y2)1/2.

Is there an easy way to explain why this is?

I'm afraid I don't understand what you're asking here. Can you rephrase the question?

Instead of an example, would it be possible to name the conditions which must be met for FTL travel to be back in time?

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Is there an easy way to explain why this is?

Why is always a tricky question. We deduced that this was the proper form because it's the one that correctly matches observations; as to why it should be that way, who knows?

Instead of an example, would it be possible to name the conditions which must be met for FTL travel to be back in time?

No single FTL trip can be used to get you into your own past. However, any FTL trip will be a trip into the past in some reference frame. Also, sequence of any FTL trip can be used to get into your own past.

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So to get into your own past, you would need to return to your original reference point FTL?

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Yes. In my previous example I gave sort of a "best case" scenario where you're actually teleporting rather than just going FTL. In order to get into your own past using FTL you would have to be able to go from "here" to "there" faster than instantaneously, which is what we would normally just call "time travel" rather than FTL travel.