# General Relativity Paradox: Holding a string across a gravitational gradient

The paradox I envision involves two objects that exist in very different gravitational potentials -- one very high (eg: close to a black hole) and one very low (eg: far away from a black hole).

Now, imagine these objects are actually little spaceships that are blasting their engines so that they remain a constant distance away from the black hole. Also, imagine they are lined up with each other (the black hole, the near spaceship, and the far spaceship form a line).

Would it be possible to lower a string from the further spaceship to the spaceship which is closer? That is, will the distance between the two spaceships be measured to be a constant?

If it's not.. please explain this! It seems odd that the distance between the two ships wouldn't be constant.

If it is, then it seems like there would be a paradox. First, let's label the spaceship nearer to the blackhole "A" and the one further "B". Next, let's say the string's length is exactly 1 light second.

Now, let's start a light clock between "A" and "B" -- "A" will send a light pulse to "B", and upon receiving this, "B" will send a light pulse to "A", and so forth. Since both "A" and "B" know that their distance is 1 light second apart, each time they receive the pulse, they will add 2 seconds to their clock. While we're at it, they might as well send along their recorded time with the light pulse.

The paradox is that from the perspective of B viewing A, A should appear to be moving in "slow motion" -- everything will be redshifted, and everything will transpire slower. From the perspective of A viewing B, B will be blueshifted and moving in "fast motion". Yet, this cannot be the case, because both will be receiving ticks at 2 second intervals, and each tick will represent 2 seconds of elapsed time on the other party.

So, what gives? I'm assuming you cannot have a constant distance between two objects if their gravitational potentials are different. But that just seems very strange.

Thanks.

Edit: It would be very helpful if the answers could give an example of what it would be like to be onboard A and onboard B. If you're on A, and you send the pulse, how much time will you measure before you hear back from B? If you're on B, and you send the pulse, how much time will you measure before you hear back from A?

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## 2 Answers

It certainly is possible for two observers to remain at rest near a black hole, as long as they are both outside the horizon. They can stretch a string from one to the other, and they can bounce light signals back and forth to each other as you describe.

It is not true, however, that the elapsed time as measured by either observer between light signals will be 2 seconds (for a one-light-second string length). The reason is that, in curved spacetime, the very definitions of things like length, time, and speed only make sense (or to be precise only correspond to our special-relativity notions) locally.

In particular, even a rule like "light travels at speed $c$" only makes sense locally in general relativity. Here "locally," to be precise, means "in an infinitesimal region," although in practice you can stretch it to mean "in a small region," where "small" is defined relative to the length scales associated with curvature of spacetime.

Imagine stationing a bunch of observers all along the length of the string. If each observer uses his own rulers and clocks to measure the speed of a light pulse as it goes by, each will find that its speed is $c$ as expected. (These are "local" measurements.) But each observer's clocks are ticking at a different rate (as measured by a distant observer), so it's not the case that the elapsed time, as measured by A (or B) equals the sum of the elapsed times as measured by all of the other observers.

Over a long path (long enough to be "not local"), the speed of light is not necessarily $c$. In fact, the speed (of light or anything else) is not even well-defined in this situation! The reason is that in curved spacetime measurements of distances and times depend on the choice of coordinates one adopts. In flat spacetime, there are a bunch of natural coordinate systems to use, which we call "inertial reference frames." With these coordinates, the special relativity rules apply. But when spacetime is curved, there are no inertial frames anymore (except locally). In fact, the nonexistence of inertial frames is one way do define what we mean by "curved spacetime."

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Something seems odd about this. From either spaceship, one could measure the length of the string. Why, then, would it take longer or shorter for the light to travel to the other ship? Also, it seems like this situation would be symmetric, it should take the same amount of local time for the light to reach the other ship: 1 second. –  user3243 Apr 22 '11 at 21:03
I don't know what to add except to emphasize that lengths and times are funny in curved spacetime. For one thing, you actually can't measure the length "from either spaceship." To measure the length, you'd have to crawl along the string from one end to the other! You can only trust your ruler over short enough distances that it lies in a single inertial frame. –  Ted Bunn Apr 22 '11 at 21:13
Interesting. The reason I find this difficult to believe is that I was told by a physicist that it wouldn't be possible to hold the string between the two, as it would be stretched apart since the spaceships are receding apart from one another, since the blackhole acts as a gravitational "sink." Hopefully another answerer will take that stance. I would agree that this is probably above my level of understanding. To clarify: you're telling me that A could measure 10 seconds between pulses, and B measure 1 second, for example? –  user3243 Apr 22 '11 at 21:27
@Sam: it sounds like that physicist was talking about the tidal effect as it applies to spaceships in freefall (i.e. not firing their engines, and therefore not maintaining constant distance from the black hole). –  David Z Apr 22 '11 at 22:23
@David: He definitely was not. –  user3243 Apr 22 '11 at 23:20

Dear Sam, it's just not true that they will both receive ticks in 2-second intervals. I think that the assumption you used to derive this incorrect conclusion was "flatness of the whole spacetime", which is nothing else than the denial of the basic point of the whole general theory of relativity.

I think that you just assumed that the distance is $x=ct$ but this is only valid for a uniform motion in a flat spacetime or, equivalently, for infinitesimal pieces of the trajectory (because infinitesimal volumes of spacetime are flat at an arbitrarily good accuracy). But there's surely no law that would say that in a general curved spacetime, some integrated proper distance $s$ is equal to $ct$ where $t$ is some proper time measured along a different path in spacetime. Different paths in a general curved spacetime have different proper lengths or proper times - and the only case in which the simple linear relationship may hold is the case when the spacetime is linear.

The proper times measured at places "A" and "B" in the static configuration that you defined are related exactly by the multiplicative factor - the ratio of the red shifts - and you seem to know very well what it is. So it's surprising that you also want to convince yourself that it is not the case even though you don't have any rational reason to think it is not the case.

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So, then, what would it be like to be onboard spaceship A, for example? When you send the pulse, how long will it take for you to hear back from B? What's it like to be on spaceship B, how long does it take to hear back? –  user3243 Apr 22 '11 at 21:10
@Sam: Assuming that the spaceships are fixed above a Schwarzschild blackhole (firing rockets "straight down"). Here we use the usual Boyer-Lindquist coordinates $(t,r)$, with event horizon at $r = 2M$. Assume spaceship A is at radius $r = R_A \gt 2M$ and spaceship B at $r = R_B \gt R_A$. Then the $t$ coordinate difference between sending a signal and receiving a response is the same for the two spaceships, and is equal to $\delta t = 2R_B - 2R_A + 4M \ln (R_A/R_B)$. However, in terms of proper time (time as observed on the space-ships), the observed times are ... –  Willie Wong Apr 22 '11 at 22:38
(contd) $\sqrt{1 - 2M / R_A}\cdot\delta t$ at spaceship A, and $\sqrt{1-2M/R_B}\cdot\delta t$ at spaceship B. So at spaceship A the time between sending and receiving will be perceived as shorter than at spaceship B, being consistent with gravitational red shift. –  Willie Wong Apr 22 '11 at 22:42
Thanks, @Willie. @Sam, to summarize, it's a nontrivial task in general. You must choose specific coordinates in your spacetime, remember the point A1 where spacecraft A has sent a signal, find the trajectory that the light rays take on their round trip, find the event A2 in spacetime where spaceship A hears the reply, and calculate the proper time on the trajectory between A1 and A2 that was actually taken by spacecraft A. There is no universal trivial shortcut that would replace the calculation by a naive proportionality law. Proper lengths of different paths in spacetime are different. –  Luboš Motl Apr 23 '11 at 4:51