General Relativity Paradox: Holding a string across a gravitational gradient

Question

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?

Answer 1

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."

Answer 2

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.