Suppose you place a number of rotating black holes in linear sequence (rotating around the same axis) between two stars at distance $d$ (assume as tightly packed as practical for purposes of calculation). If a ship enters each ergosphere and goes out into the next one, can it make the roundtrip back to home in less time than $2d/c$ (home local time) ?

What would be the simplest calculation to see that it cannot?

Edit to make the ergosphere overlapping region more symmetric, imagine there is a symmetric sequence of black holes in front of the treadmill arranged like this:

a kerr blackhole treadmill with an overlapping region of ergospheres

the symmetry in this arrangement should cancel the angular components (at least in a small region in the middle of overlapping region). Obviously the geometry for this thing is highly unstable.

  • $\begingroup$ So all the black holes have the same angular momentum vector, but I imagine that the angular momentum vector is perpendicular to the line on which they all lie. Is this correct? $\endgroup$ – Alan Rominger Nov 14 '11 at 20:12
  • $\begingroup$ yes, perpendicular so the interesting frame dragging equatorial plane happens in the longitudinal direction $\endgroup$ – lurscher Nov 14 '11 at 20:13
  • $\begingroup$ I don't see a clear reason that you cannot go faster than light here--- the black holes pick out a frame. You wouldn't clearly be able to use this as a time machine. Gravity can certainly focus light, and past the focusing point, a massive object can catch up to the light, so in some sense it is travelling faster than the light. $\endgroup$ – Ron Maimon Nov 15 '11 at 0:05
  • $\begingroup$ Dear @Ron, the reason why you can't exceed the speed of light in the vacuum is called the special theory of relativity and it was found by Albert Einstein in 1905. Black holes and gravity may locally disturb the behavior of objects in the spacetime but as long as the long-distance structure of the spacetime approaches the Minkowski spacetime, special relativity still holds for all the particles interacting with the objects in the middle, whether or not they may also be interpreted as agents making the spacetime curved. $\endgroup$ – Luboš Motl Nov 15 '11 at 6:55
  • $\begingroup$ @Lubos: The special theory of relativity does not imply that you cannot catch up to a light beam which starts at your position, if that light beam is bent by gravity. For a trivial example, shine a light so that it is deflected by 180 degrees by a black hole, and just move a little to catch it on the way back. $\endgroup$ – Ron Maimon Nov 15 '11 at 7:33

This is impossible when the treadmill is not infinite in extent in the z direction and matter obeys the null energy condition (so it is impossible using Kerr black holes), but the argument I see is subtler than Lubos's answer suggests, although the physical intuition in his answer is impeccable.

Gott time machine

First, an example that actually works--- this is the cosmic string. If you slowly bring in an infinite deficit-angle type relativistic cosmic string in the z direction so that it is close to inbetween two far-away points in the x-y plane, the angular wedge cut out by the cosmic string will allow a particle going around the cosmic string to outrun light going straight across from A to B. This is pure geometry, but it is not particularly enlightening, because the reason it works is by the gross scattering of particles by the string. The actual real causal distance between A and B is decreased by the cosmic string the moment A starts to see a lensed image of B, and this lensed image is closer.

To verify these statements, just draw a straight line from A to B, cut it at some point, place a cosmic string vertex somewhere below the line, and a cut-out wedge pointing straight down of some angle less than 180 degrees. The properties are obvious from the geodesics in the picture you get, when you remember that the spacetime is locally flat, with a global time, with gluing conditions at the boundaries of the wedge. This is an exact solution of 2+1 gravity, or of 3+1 cosmic strings, and so it is possible to outrun light in this bogus way, which has nothing to do with your suggestion.

By boosting such wedges, Gott tried to turn this outrunning-light property into a time machine. This doesn't work for subtle reasons, as established by t'Hooft, but this is a somewhat different question.

Decaying sources and infinite light-planes

Given these cosmic string shenanigans with the actual distance between two points, to give a good answer to this question which can be turned into a theorem, you have to choose points A and B which are asymptotic, so that they are in the Minkowski region of the localized gravitational solution you are examining. This is impossible for cosmic strings, because the deficit angle makes the spacetime only ALE, not asymptotically Minkowski, it still cuts out an angle no matter how far out you go.

If you have a solution which decays as m/r at long distances, then the question becomes the following: looking at an infinitely distant point A, and link it to an infinitely distant point B, can you find paths in the gravitational solution which allows certain paths from A to B which pass through the gravitational region to outrun a light ray which is nearly straight, but bends an infinitesimal amount along the way so as to completely avoid the gravitating region, staying in the asymptotic region where the field is small.

The light cone from A entering the gravitational region is a light front sheet by the time it gets close, and similarly, the incoming light cone to B is a light front sheet. The question is about the evolution of the light-front sheet in an arbitrary gravitational metric. If there were a crazy path that outran the light front, then the boundary of the light-front along this path would spreading out from this path, to cover regions which it can reach which had not yet been reached by the rest of the front, and this would lead to a momentary increase in area of the front, as the blob spread out.

The way in which light fronts evolve does not allow spreading out, only collapse. The reason is the null energy condition--- the null geodesics that define the light front start out parallel, so they can only focus with their neighbors, and when they do, they are inside the light front and behind it, never ahead. The proof is essentially due to Hawking, it is contained in the proof that the area of a black hole horizon does not increase. This theorem is proved here: Second Law of Black Hole Thermodynamics . You can interpret the light front as the horizon of an enormous infinite black hole at infinity, and the light front passing over the solution is just the solution falling into the black hole. If there were rays that outran the front, this would be a violation of the null energy condition--- it would force a region of defocusing of geodesics.

Although this argument seems more involved than Lubos's, I must point out that the end conclusion is exactly the same--- when you can outrun light, there is a violation of a condition very closely related to the sub-extremality condition for black holes (the weak energy condition). So it is clear that for Kerr black holes which are not nakedly singular, there cannot be outrunning of light from infinity.

Gubser's preprint on superluminal neutrinos

About a month ago, Gubser argued that superluminal neutrino travel in an extra dimension scenario requires a violation of the weak energy condition (null energy condition--- positive-stress-energy/positive-focusing along null geodesics). I only skimmed his paper for the purpose of evaluating the OPERA results, but his argument is certainly parallel to the one above.

  • $\begingroup$ @actually i think Lubos derivation of a top speed is incorrect, see my comment to his answer $\endgroup$ – lurscher May 30 '12 at 19:44

Well, the simplest way to show that this method to overcome the speed-of-light barrier cannot work is to remove the portion of the spacetime occupied by the "black holes employed as accelerators" and look at the whole hypothetical superluminal trip from a different reference frame that is sufficiently boosted (we need to use different frames for the two halves of the trip). Just like in all superluminal motion, such a trip will look like a trip backwards in time which would lead to logical contradictions.

More microscopically, one may also easily mention where your intuition is going wrong if you think that the black holes accelerate you in average. Aside from the angular momentum, black holes also have a positive mass (which is greater than a certain function of the angular momentum, the extremality bound). So you want to know how the massive black holes influence the geometry and whether they may "speed up" others. Look at the ordinary Schwarzschild geometry: $$ ds^2 = -(1-2M/R) dt^2 + \frac{dr^2}{1-2M/R} + d\theta^2+ \sin^2\theta \,d\phi^2 $$ Just think about an astronaut moving in this geometry and ask what is the maximum allowed "coordinate speed" $dr/dt$. Well, the condition comes from causality i.e. the timelikeness of the interval, $$-(1-2M/R) dt^2 + \frac{dr^2}{1-2M/R} < 0.$$ That's equivalent to $$\frac{dr}{dt} < c (1-2GM/R) $$ I restored $c$ and $G$ to get to the usual units. So in the presence of a black hole, the maximum coordinate speed is actually smaller than the speed of light in the vacuum, as long as the mass is positive, and it has to be.

You're right that rotating black holes may help you to speed up but the rotation is never "too high" because you would breach the extremality bound. The maximum you may probably do is to fly "near the horizon" of an extremal rotating black hole, and then the black hole doesn't slow you down at least in some way of counting the coordinate distances. All other black holes (and all other trajectories near the extremal one) slow you down.

  • $\begingroup$ Your answer is that the local speed of light is always bigger than the local speed of matter--- big deal. This just means that light traveling along your path will outrun you. It doesn't mean that light travelling along your path doesn't outrun parallel light far away. A closed path which comes back into the future in one frame will come back into the future in all frames. The thing you want to do is boost the whole system in one direction, boost the holes very fast to go from B to A as you go from A to B, and then boost them very fast the other way on returning. But your conclusion is ok. $\endgroup$ – Ron Maimon Nov 15 '11 at 7:57
  • $\begingroup$ I am impressed with the physical intuition in your answer: where does it come from? Is it just from knowing the asymptotic scattering of geodesics in the Schwartschild metric, and the delay of light along these? Or is there more to it? It seems that you can smell a null energy violation without thinking about light fronts--- can you explain the reasoning? Perhaps its as simple as sensing causality violation, but I don't think it's so straightforward, because null energy violation is not automatically causality violation, or is it? Thanks for the interesting answer. +1. $\endgroup$ – Ron Maimon Nov 15 '11 at 9:41
  • $\begingroup$ That Schwarzschild line element is missing $r$ factors in the angular components so your derivation for the top speed does not seem to depend on the proximity to the equatorial bulge in the ergosphere, a result that does not make physical sense $\endgroup$ – lurscher May 30 '12 at 19:42
  • 1
    $\begingroup$ @RonMaimon, i'm sorry but i don't understand what defocusing means exactly in this context, "superluminal speed" for me only has meaning as local four-velocity. The fact that the ergosphere observers seem to move like that to outside observers is not that interesting. Locally, observers in the treadmill will be regular inertial frames (possibly with ample tidal forces far from the symmetry axis, but thats another story). $\endgroup$ – lurscher May 31 '12 at 6:18
  • 1
    $\begingroup$ @RonMaimon, also, i don't get how you can sustain arguments about global measurements without knowing the details of the local geometry. As an example of global measurement that is relevant to my questions: "did the test particle went and returned and forth through two anti-parallel treadmills a distance $2d$ in less than $\frac{2d}{c}$ time from the point of view of the origin?". The metric is curved, the space doesn't look minkowski except far away. $\endgroup$ – lurscher May 31 '12 at 17:19

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.