# Bell's Spaceship Paradox & Length Contraction

## Main Question or Discussion Point

You and two identical spaceships are all at rest with respect to each other. You note that the two engines start up at the same time, and the thrust curve and acceleration profile of both spaceships are identical. As the ships pick up speed, would you measure the ships to be shorter than their rest length?

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Nelli

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PeterDonis
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For him lengths of the rockets are shortened. The distance between the ships are kept constant. The thread which is moving and accelerating is shortened so torn apart.

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Ibix
As the ships pick up speed, would you measure the ships to be shorter than their rest length?
Of course (with reasonable assumptions about the ship design). But the string's length does not change until it snaps.

Ibix
The thread which is moving and accelerating is shortened so torn apart.
A key point is that the thread does not shorten. In the original rest frame of the ships you said it yourself: The distance between the ships are kept constant. So the thread does not shorten in this frame - it stays the same length. In each successive momentarily comoving inertial frame of the thread (to the extent such a thing can be defined) the spacing between the ships is larger than the moment before, and the thread actually lengthens.

The point you are trying to make, I think, is that the original rest frame of the ships regards the thread as moving. Hence the thread's unstressed length would decrease over time. However, stresses build up and stretch it, countering this effect. The stresses cause the thread to break.

sweet springs
Thanks. In other words, say the thread is tightened to the front ship but loose and not tightened to the back ship, the thread is shortened as the rockets accelerate so the distance between the loose end of the thread and the back rocket increases from zero.

Ibix
PeroK
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You and two identical spaceships are all at rest with respect to each other. You note that the two engines start up at the same time, and the thrust curve and acceleration profile of both spaceships are identical. As the ships pick up speed, would you measure the ships to be shorter than their rest length?
You have the same issue with each spaceship. If, in your frame, the acceleration profiles of the front and rear of a spaceship are identical, then they stay the same distance apart in your frame, hence the ship must be stretching in its rest frame, such as can be defined.

If, however, the ship moves rigidly in its rest frame, then it must contract in your frame, hence the acceleration profile of the ship varies (slightly) from front to rear in your frame.

David Lewis and Ibix
...the acceleration profile of the ship varies (slightly) from front to rear in your frame.
This is exactly the point I was having trouble with. Many thanks.
The distance between the ships are kept constant.
Then I take it you'd measure distance a as remaining the same. Would you measure the length of the thread (b) getting shorter due to length contraction of the ships?

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Re:＃８ your figure shows that $$b-a$$ = length of a ship = $$l_0\sqrt{1-\frac{v^2}{c^2}}$$.
It does not matter whether there is a thread between the ships or not as far as elastic property of thread does not harm designed acceleration of rockets.

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pervect
Staff Emeritus
You and two identical spaceships are all at rest with respect to each other. You note that the two engines start up at the same time, and the thrust curve and acceleration profile of both spaceships are identical. As the ships pick up speed, would you measure the ships to be shorter than their rest length?
It's not clear to me what frame of reference "you" are in. I'm guessing that "you" are in an inertial frame of reference, and that when 'you' (David Lewis) say that both space-ships start up "at the same time", you mean that they start up the same time in said inertial frame of reference.

If I'm interpreting it correctly, and if we also assume that the spaceships are Born rigid, the answer is yes, from the inertial frame of reference one observes both spaceships to length contract.

The main point I have to make is that it matters - "you" are not being to clear about what frame of reference the other "you" is in, and this is important information to give an accurate answer.

A secondary but also important point is that one needs to know if the spaceships are rigid. The applicable notion of rigidity in special relativity is Born rigidity. If the spaceships are not rigid, then their length can change so the question cannot be answered without more information. As others have mentioned, if the spaceships are rigid (Born rigid), their bow and stern have different acceleration profiles.

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PeterDonis
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Would you measure the length of the thread (b) getting shorter due to length contraction of the ships?
No, because the thread is being stretched, which in the frame you are using means its measured length stays the same, instead of contracting as it would if it were not connected to both ships.

Again, have you read the PF FAQ entry on the Bell Spaceship Paradox, which I linked to above? It discusses all of these points. (In particular, it compares the length of the thread connected to both ships, to the length of a thread connected only to the front ship, as the ships accelerate.) So do many other sources that discuss the paradox, including Bell's original paper.

PeterDonis
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I'm guessing that "you" are in an inertial frame of reference, and that when 'you' (David Lewis) say that both space-ships start up "at the same time", you mean that they start up the same time in said inertial frame of reference.
This is my intepretation of the OP's intent. But you are right that in any relativity problem, it's better not to leave this up to interpretation, but to be explicit about what frame is being used.

PeterDonis
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The thread which is moving and accelerating is shortened
No, it isn't, it stays the same length. What shortens is the unstressed length of the thread. See the FAQ entry I linked to.

PeterDonis
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say the thread is tightened to the front ship but loose and not tightened to the back ship, the thread is shortened as the rockets accelerate so the distance between the loose end of the thread and the back rocket increases from zero.
Again, see the FAQ entry I linked to, it discusses precisely this point.

Grinkle
Gold Member
Is the problem equivalent to a hammock hanging at right angles to the direction of acceleration in a perfectly rigid accelerating spaceship? The hammock would eventually break if its made of breakable material?

PeterDonis
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Is the problem equivalent to a hammock hanging at right angles to the direction of acceleration in a perfectly rigid accelerating spaceship?
No, because, as I'm understanding your description, both suspension points of the hammock would be at the same height in the ship, so they would have the same proper acceleration. The hammock would hang down in the ship just as it would in a room at rest on the surface of a planet, and would not stretch (except for a brief period when the acceleration first began, while the whole system was coming to a new equilibrium).

More generally, if the spaceship is Born rigid (which is the closest thing to "perfectly rigid" that you can have in relativity), any object inside it will not stretch, because the ship itself will not stretch. (So, for example, you could "hang" the hammock vertically, with one suspension point on the ceiling of the ship and the other on the floor, and it would not stretch, just as it wouldn't if hung vertically in a room at rest on a planet.) Actually, the usual way this is defined in relativity is to turn that around: the definition of the ship "not stretching" (and by extension everything inside the ship not stretching) is that its motion is Born rigid. There are plenty of ways to flesh this out with technical details, but that goes beyond the scope of a "B" level thread.

Grinkle
Ibix
This is exactly the point I was having trouble with. Many thanks.
Then I take it you'd measure distance a as remaining the same. Would you measure the length of the thread (b) getting shorter due to length contraction of the ships?
Assuming that the ships are of identical design, then in the original rest frame equivalent points will remain at constant separation. The obvious equivalent point is the rocket exhausts on the two ships, since this is the point where the thrust is applied. Since identical ships will react identically to identical thrust, though, the noses will remain at constant separation, as will the attachment pylons for the thread. The nose-to-tail separation of the ships will vary slightly because the length of the ships varies as measured in this frame.

So, on your diagram, and assuming identical ships with identical proper acceleration profiles, and that we're working in the original rest frame of the rocket, (b) remains constant. (a) does not.

David Lewis and sweet springs
It's not clear to me what frame of reference "you" are in. I'm guessing that "you" are in an inertial frame of reference, and that when 'you' (David Lewis) say that both space-ships start up "at the same time", you mean that they start up the same time in said inertial frame of reference.
Correct. "You" is the spaceman depicted in the diagram attached to Post #1. I labeled him "OBSERVER".

No, because the thread is being stretched, which in the frame you are using means its measured length stays the same, instead of contracting as it would if it were not connected to both ships.

Again, have you read the PF FAQ entry on the Bell Spaceship Paradox, which I linked to above? It discusses all of these points. (In particular, it compares the length of the thread connected to both ships, to the length of a thread connected only to the front ship, as the ships accelerate.) So do many other sources that discuss the paradox, including Bell's original paper.
Re: the bold (my emphasis). Am I to understand this to mean that the stretching of the thread exactly compensates for length contraction until the moment it breaks?

jbriggs444
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Am I to understand this to mean that the stretching of the thread exactly compensates for length contraction until the moment it breaks?
Well, yes. It is held in place between two space craft whose positions are at a fixed separation given by the problem setup. It must remain the same length because it stretches between those two end points. Until it breaks, of course.

FactChecker
Gold Member
I think I am missing something. It's not clear to me how this is different from the forces within a long rocket of the same total length (from the nose of the lead rocket to the tail of the trailing rocket). Would such a rocket experience different acceleration forces at the front and back? If not, wouldn't it also be torn apart just like the thread is broken?

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Ibix
I think I am missing something. It's not clear to me how this is different from the forces within a long rocket of the same total length (from the nose of the lead rocket to the tail of the trailing rocket). Would such a rocket experience different acceleration forces at the front and back? If not, wouldn't it also be torn apart just like the thread is broken?
A long rocket only has an engine at the back, so it isn't stretched. Bell's ships scenario has a rocket at front and back which can move under their own power. If you had a long rocket with engines at front and back (don't think too hard about where the exhaust from the front engine goes...) it could tear itself apart too.

From the perspective of either ship this is actually what's happening. The other ship's engine is mis-set, so it moves away and stretches the thread until it breaks.

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FactChecker
stevendaryl
Staff Emeritus
A long rocket only has an engine at the back, so it isn't stretched. Bell's ships have rockets at front and back which can move under their own power.
I didn't think of Bell's scenario as implying that the ships themselves had rockets on both ends. He wasn't making a point about the stresses inside the ships, but about the distance between the ships. Of course, the case you're talking about, with the front and rear having independent rockets, is basically equivalent to having multiple connected ships.

FactChecker
Nugatory
Mentor
I didn't think of Bell's scenario as implying that the ships themselves had rockets on both ends. He wasn't making a point about the stresses inside the ships, but about the distance between the ships. Of course, the case you're talking about, with the front and rear having independent rockets, is basically equivalent to having multiple connected ships.
I was understanding @Ibix as saying, in response to @FactChecker, that Bell's setup (two independently powered ships connected) is equivalent to a single long ship with independent motors at each end.

FactChecker and Ibix
stevendaryl
Staff Emeritus