spaceship

What Is the Bell Spaceship Paradox, and How Is It Resolved?

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Bell describes two spaceships that start out at rest relative to each other, with an elastic string between them, one end attached to each ship, which is initially just long enough to span the distance between the ships without being stretched.

bell-spaceship-description

The ships’ clocks are synchronized in their common initial rest frame K. At some time t by their clocks, both of them turn on their rockets and start accelerating in the positive x direction, with exactly the same constant proper acceleration (meaning that the crew of each ship feels the same constant acceleration). What happens to the string?

The correct answer is that, as the ships accelerate, the string is stretched more and more until it breaks. The “paradox” arises from the following (erroneous) line of argument: since both ships start accelerating at the same time, and they both have exactly the same acceleration, the distance between them should remain constant, so how can the string be stretched and break?

Of course, in relativity, whenever you see an argument using the terms “time” and “distance,” you know you need to be careful, since time and distance are frame-dependent. To resolve the “paradox,” we look at the distance between the ships, as seen from the instantaneous rest frame K’ of either ship. If you do this at different points along either ship’s worldline, you will see that the distance gradually increases, e.g., L2>L1 for the two lengths shown in the figure, as measured on the respective graph-paper grids. These lengths are measured parallel to the spacelike axes of frames K and K’, so that they are distances between events that are simultaneous as measured in these frames. Since the string remains attached to both ships, increasing distance between the ships in either ship’s instantaneous rest frame means the string has to stretch (and eventually it will stretch enough to break).

bell-spaceship-spacetime-diagram-4.

Some treatments of the paradox attribute the breaking of the string to “length contraction,” and you may be wondering how that comes into play, since the discussion so far has not mentioned it. In the article by John Bell in which he originally introduced the paradox, he gave an alternate line of reasoning to get the correct answer. The ships and the string are originally set up at rest relative to one another, with the string unstressed. As the ships accelerate, the string moves faster and faster relative to frame K, so its “natural” (i.e., unstressed) length, as measured in K, gets shorter and shorter due to length contraction. But its actual length as measured in K does stay the same, so the string will be increasingly stretched until it breaks. (It may be helpful to consider the case where an additional, identical string is attached only to the front spaceship. As the ships accelerate, it retains its original length, as measured in its own rest frame, and is therefore too short to span the increased distance between the ships. The shorter line marked L/γ in the diagram below can be thought of as representing this second string.)

bell-spaceship-natural-length

Bell’s argument is valid. However, note that what is “contracted” is not the actual length of the string in the original rest frame, but its “unstressed” length in that frame. The actual length of the string in the original rest frame stays the same, so the use of the term “length contraction” in this connection is a little unusual: normally that term is used to refer to a length that is actually measured in some frame. That’s why, if you ask whether the string in the Bell spaceship paradox breaks due to “length contraction,” the answer is not a straightforward “yes.”

References

J.S. Bell, “How to teach special relativity,” in Speakable and unspeakable in quantum mechanics, 1988, Cambridge University Press

http://en.wikipedia.org/wiki/Bell_spaceship_paradox

http://www.lightandmatter.com/sr/ (original source for diagrams, somewhat modified for this FAQ)

The following forum members contributed to this FAQ:

PeterDonis
WannabeNewton
PAllen
ZapperZ
jtbell
bcrowell
tiny-tim

PhD in physics. I teach physics at Fullerton College, a community college in Southern California. I enjoy writing, playing viola, brewing beer, climbing and mountaineering.

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  1. JD96
    JD96 says:

    Great article and since I am trying to get my head around this seeming paradox I found this insight to be very useful to get an overview. Nonetheless I would like this opportunity to ask a few additional questions:1. Since forces are involved, why does the rest length of the string stay the same in the K'? 2. Replacing the second spaceship with one whose proper acceleration has a different value, namely one corresponding to a hyperbolic worldline (where the center of the hyperbola is located in the origin), will the distance between the spaceships in K' now be constant? 3. In a relativity book I am reading a connection is drawn to the equivalence principle by showing that in K' the proper time of the second spaceship is ahead of the first one's proper time by the amount expected using the weak field formula for gravitational time dilation (when considering the first order approximation of the ratio of each spaceships proper time). This makes me curious whether using the equivalence principle the increasing distance in K' would be attributed to tidal force? I would be grateful for any help regarding one of the questions.Also thanks to all participants for the well written article.

  2. vanhees71
    vanhees71 says:

    Thanks for the very well written Insights.

    I put the beginning of my own SRT FAQ online now. It's not finished (only the kinematic part is nearly finished (I need to add a section on the socalled twin paradox) and the particle motion in an external field (also there I still have to write a section on Hamilton's principle and on the BMT equation discussing the Thomas precession). On the other hand, perhaps it's good when I get input from the forum before I put it (perhaps in several parts) also into the Insights section. You find a discussion of Bell's spaceship paradox (which of course is not a real paradox as all the socalled paradoxes of SRT) also there:

    http://fias.uni-frankfurt.de/~hees/pf-faq/srt.pdf

    PS: The idea to draw two grids for the two inertial frames is also great; I draw instead the hyperbolas of constant Minkowski "distances" 1, 2,… from the origin, but looking at your Minkowski diagrams, maybe the two linear grids are a better way, once one has explained the construction of the unit temporal and spatial distances with the hyperbolas first.

  3. Ken G
    Ken G says:

    What I think this "paradox" is best at bringing out is that there really does not exist a "correct" language to say "why" the string breaks.  The physics prediction is only that the string breaks– the why question simply does not have a unique answer.  One can say it is "because of length contraction", one can say it is "because of the equivalence principle", and one can even say it is "because the forward ship blasts off first," etc., it all depends on the reference frame chosen.  This is natural– we don't answer physics questions about "why" until after we choose a reference frame, indeed I would argue that the main reason we choose reference frames in the first place is to be able to create a language capable of answering "why" questions at a more experiential level.  Since relativity is the one place where we advance a certain agnosticism about reference frames, this gives "why" questions an especially interesting status there.

  4. PeterDonis
    PeterDonis says:
    JD96

    Since forces are involved, why does the rest length of the string stay the same in the K'?

    It doesn't; it increases. The length of the string stays the same in the original rest frame, K (no prime), but the string is not at rest in that frame once the ships start accelerating.

    JD96

    Replacing the second spaceship with one whose proper acceleration has a different value, namely one corresponding to a hyperbolic worldline (where the center of the hyperbola is located in the origin), will the distance between the spaceships in K' now be constant?

    In the scenario as given, both spaceships are following hyperbolic worldlines–but the hyperbolas are centered on different points. What I think you are trying to describe here is having both ships follow hyperbolic worldlines that are centered on the same point, which means the ship in front will have less proper acceleration than the ship in the rear. In this case, yes, the distance between the ships (and therefore the length of the string) in K', the momentary rest frame of the string, will be constant. And in the original rest frame, K (no prime), the distance between the ships will decrease (because the one in the rear is accelerating more and so catching up to the one in front).

    JD96

    In a relativity book I am reading a connection is drawn to the equivalence principle by showing that in K' the proper time of the second spaceship is ahead of the first one's proper time by the amount expected using the weak field formula for gravitational time dilation (when considering the first order approximation of the ratio of each spaceships proper time).

    I assume this is for the alternate scenario you proposed, correct? (In other words, the one in which the front ship has less proper acceleration than the rear ship, and the distance between them in K' remains constant.) If so, yes, this is a standard method for describing the equivalence principle, and the "rate of time flow" of the two ships differs accordingly. But this only works because the ships remain a constant distance apart in K'. It wouldn't work for the original Bell spaceship scenario, the one described in the Insights post.

    JD96

    This makes me curious whether using the equivalence principle the increasing distance in K' would be attributed to tidal force?

    I assume here you mean the original Bell spaceship scenario, correct? If so, no, the scenario is set in flat spacetime so there is no tidal force.

  5. PeterDonis
    PeterDonis says:
    Ken G

    one can even say it is "because the forward ship blasts off first,"

    You can't say this, because the two ships blast off simultaneously according to both ships; their clocks are synchronized in the original rest frame, and remain so until the instant they blast off, and that instant is stipulated to be the same instant, in the original rest frame, for both ships.

  6. Ken G
    Ken G says:

    Actually, you can say that, and be perfectly correct– it's all a question of reference frame. In the frame that the two spaceships end up in, assuming they accelerate rapidly and reach a final cruising speed, what you will see is two spaceships moving toward you, back end first, some distance L apart. Then the engine at the front fires, breaking the rope, and then the engine at the back fires. The two ships then slow to a stop in your frame, by which point they are not only farther apart than L, they are also farther apart than an un-length-contracted rope could span– which explains why the rope had to break. This is a perfectly good explanation of "why" the rope breaks: "because the rocket in front fired first, which caused the distance between the rockets to stretch farther than the decelerating rope could span without breaking." It's all a question of reference frame, no one's frame gets to talk about "absolute simultaneity."

    Simultaneity is never anything but a coordinate system, and we can choose our coordinates to say "what happened" in a completely arbitrary way. There isn't even a "reference frame of the rockets", because a reference frame is something local– language like "the reference frame of the rockets" is actually a coordinate system, not a reference frame, though this distinction is often blurred in special relativity. A reference frame should really mean "a perspective from which to make an observation", not a global coordinate system, though this distinction isn't really my point– my point is, "why" answers should avoid being interpreted as reifying some coordinate system.

  7. rede96
    rede96 says:
    PeterDonis

    You can't say this, because the two ships blast off simultaneously according to both ships

    A recent post I made looking at clock synchronisation and simultaneity (also looking at one way speed of light) seemed to suggest that it is impossible to prove simultaneity without assuming the speed of light is isotropic as there is no physical test that can be made to prove it. So if that logic applies, then it must equally apply to this paradox in so much as it must be impossible to say which ship moves first. So it must be impossible to say why the string breaks.

  8. JD96
    JD96 says:
    PeterDonis

    It doesn't; it increases. The length of the string stays the same in the original rest frame, K (no prime), but the string is not at rest in that frame once the ships start accelerating.

    I am realizing I forgot to distinguish between unstressed and "actual" length as done in the insight post and probably that is where my confusion comes from. So when you write the length of the string stays constant in unprimed K, it's because each end of the string is always attached to one of the spaceships and since the problem is set up, so that the distance between the spaceship remains the same in K, the "actual" length of the string will also stay constant until it breaks. What I didn't understand is, when an observer in unprimed K wants to account for the breaking of the string using length contraction he should observe the unstressed length of the string be shortened by the lorentz factor, which I would argue means primed K should measure the unstressed rest length.

    Now if that is true and I haven't done an error in my thinking so far, how do we know the unstressed rest length stays constant (I hope this is a better way to put the question)?

    PeterDonis

    What I think you are trying to describe here is having both ships follow hyperbolic worldlines that are centered on the same point, which means the ship in front will have less proper acceleration than the ship in the rear

    Yes

    PeterDonis

    I assume this is for the alternate scenario you proposed, correct?

    Actually no, with the ratio of the proper time I mean the following (now I am referring to the minkowski diagram in the insight post): Calling the intersection between the simultaneity line of primed K with the left worldline event A and the intersection with the right worldline event B, the proper time from event (coordinates as given in unprimed K) (0|c[SUP]2[/SUP]/α) to event A along the left worldline is computed and similiary the proper time from event (0|c[SUP]2[/SUP]/α+L[SUB]0[/SUB]) (L[SUB]0[/SUB] means distance of the spaceships in unprimed K) to event B along the right worldline is computed and the ratio between the latter and the former proper time yields when taking the first order approximation τ[SUB]B[/SUB]/τ[SUB]A[/SUB]=1+αL[SUB]0[/SUB].
    Writing these lines I guess for small velocities of the left spaceships both scenarios don't significantly differ when it comes to the distance between both spaceships in primed K and the proper time along the paths, which is why talking about Bell's scenario also justifies using the equivalence principle in this way (?).

    PeterDonis

    I assume here you mean the original Bell spaceship scenario, correct? If so, no, the scenario is set in flat spacetime so there is no tidal force.

    Yes and I was asking the question because my poor understanding of the equivalence principle made me think an accelerating observer will justify him being at rest, despite feeling proper acceleration, via a gravitational field, so that from an accelerating observer's perspective there's automatically spacetime curvature and other general relativistic effects involved, but this is only what I expected from reading popular science articles (I should probably wait until I learn the math needed to get into general relativity before randomly asking questions based on speculation).

    Thanks already for your great help!

  9. PeterDonis
    PeterDonis says:
    Ken G

    In the frame that the two spaceships end up in, assuming they accelerate rapidly and reach a final cruising speed, what you will see is two spaceships moving toward you, back end first, some distance L apart. Then the engine at the front fires, breaking the rope, and then the engine at the back fires.

    Ah, I see. Yes, you're right, you can always make either one of a pair of spacelike separated events occur first by an appropriate choice of frame.

    The only point I see here that still might require clarification is that you say "the engine at the front fires, breaking the rope, and then the engine at the back fires". I don't think you can say definitely that, in the frame in which the two spaceships are finally at rest, the rope breaks before the engine of the rear ship fires. In fact, it seems unlikely to me that that would be the case. It will, however, always be the case that the front ship's engine fires first in this frame.

    (Btw, the Bell spaceship scenario does not require assuming that there is such a frame; the ships could keep accelerating indefinitely, or they could stop in such a way that they are not at rest in the same inertial frame.)

  10. PeterDonis
    PeterDonis says:
    rede96

    A recent post I made looking at clock synchronisation and simultaneity (also looking at one way speed of light) seemed to suggest that it is impossible to prove simultaneity without assuming the speed of light is isotropic as there is no physical test that can be made to prove it.

    The fact that the two ships fire their engines simultaneously, in the frame in which they are originally at rest, is not something that has to be proven for this scenario; it is assumed as a condition of the scenario. The definition of simultaneity used is Einstein clock synchronization.

    rede96

    if that logic applies, then it must equally apply to this paradox in so much as it must be impossible to say which ship moves first.

    Of course it's possible; you just need to specify what frame you are using. Specifying a frame means assuming a particular simultaneity convention; it doesn't require proving one (which is, of course, impossible since simultaneity is not absolute). Of course that also involves assuming that the speed of light is isotropic and that the one-way speed of light equals the two-way speed. All of these are assumptions, which are justified by the fact that the theory that makes them, special relativity, correctly predicts the results of experiments.

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