One of the well-known paradoxes or really pseudo-paradoxes, of the special theory of relativity is the so-called spaceships on a rope paradox, so let's set this up. We have an Observer A here, could be Alice, say. She's right in the middle between two spaceships, and the spaceships are connected by a rope of some kind. Obviously, this is a thought experiment. And so we have Spaceship B and Spaceship C. The distance between them is D. And the idea is, on a signal from the Observer A here, so the observer will send out a simultaneous light signal to both spaceships. And we'll assume they're on automatic pilot, so that when they receive the signal, spaceships B and C will accelerate at the same time, such that the distance D between them remains constant to A. So in other words, they accelerate here, and A observing them will always see the same distance between the spaceships. And the question is, does the rope break? And there's actually an interesting story behind this paradox. It was developed in the 1950s independently by several people, and one of those people was, for those of you who have perhaps read some in quantum mechanics, you may have heard of Bell's inequality. And so a scientist named J.S. Bell who was working at the CERN, C-E-R-N, high energy physics laboratory, in France, Switzerland in Europe. And he was working there, and he came up with this idea ,and he asked some of his colleagues, presented to them and said, do you think the rope will break or not? And many of them, at first response say no, of course, it's not going to break because distance D here is always constant no matter. You can accelerate up the relativistic speeds, but if you hold the distance constant to A, then it shouldn't break. Well, he actually showed that a careful analysis is that the rope would break. So let's see if we can understand that. By the way, if you want to read more about his story here, he has a book called Speakable and Unspeakable in Quantum Mechanics and has several of his article and essays and things like that in it. So good book to read if you're interested in those sorts of things. So anyway, Bell and others came up with this, and since then around 1960, there's been a fair amount of debate on it, but most people agree that the rope will break. So let's see if we can understand why it would break because on first glance, it seems if everything remain constant here, why would the rope break? And we're assuming that the rope is just at a tension between the spaceship such that if it gets pulled on too hard, then it would break. Okay, well, let's look at it from the perspective of Observer A. Because anytime you see a paradox or question posed in relativity and you see the words at the same time in there or identically or anything like that, immediately there should be a red flag thrown up. You say we need to think about that a little more because, of course, relativity of simultaneity, the relativity of synchronization. Anytime you say that at the same time, you have to be very careful about what's going on. So let's think about this, Observer A sends a signal, the spaceships start accelerating. So they accelerate up to a certain velocity, V. And they're accelerating identically, that's the idea here, to give this distance D constant. So that's given in the problem, it's defined. This D between the spaceships will remain constant. Let's think about what has to happen for that distance D to remain constant as the spaceships start accelerating and gaining some velocity. Well, we know that as soon as they have some velocity, then they're in a different reference frame. Note that, B and C are in the same reference frame here because they're accelerating identically with the same velocity here, according to Observer A. Observer A then is in his or her own reference frame. So as soon as these spaceships get some velocity going, you can almost imagine them one big spaceship there if you want, attached. But what happens as soon as that occurs. Remember, leading clocks lag. If we have a clock on Spaceship C here, okay, so we just draw a clock up here. Spaceship C has a clock. Spaceship B has a clock. They're synchronized in their BC frame of reference. Because again, at any given instant, they will be at the same velocity. What does A see however, leading clocks lag. So as soon as they have some velocity, this clock, the lead clock as it moves this way. So we'll just draw, the velocity V is going this way now, will lag this clock. In other words, this clock is ahead. And just to make it a little more concrete, let's give some numbers here. Let's imagine the situation such that when this clock say, reads 10 seconds, then this clock reads 7 seconds, it's 3 seconds behind. Now we could, remember the leading clock lag figure is quantity is DV/C squared, so to get a lag of 3 seconds, both D and V have to be pretty big there. But we'll assume that that's what the lag is just so we can visualize a little bit better. Or you could have 7 nanoseconds there and 10 nanoseconds, if you wanted to do it that way, to make the numbers slightly more realistic, perhaps. So 7 seconds here, 10 seconds here. What that means is, if you think about accelerating identically, at the same time. So remember to B and C, they're accelerating right at 10 seconds, okay, according to their system of clocks. Everything has to maintain that distance D. So to them, their velocity is always the same as they go along. But notice what that means for Observer A. If Spaceship B is accelerating at 10 seconds, okay, so now, it's accelerating a little bit more, it's moving this way. Spaceship C though is only a 7 seconds according to Observer A, so it's not accelerating yet. Let's just assume that every 5 seconds, they do a little bit acceleration, so at 5 seconds, they did some acceleration. And then at 10 seconds, they'll do some acceleration, 15 seconds, they do some more acceleration and build up the velocity that way to break it up. So at 10 seconds now, Spaceship B does its acceleration. But Spaceship C according to Observer A isn't accelerating yet. It still has 3 seconds to go before it gets to the time where it's supposed to accelerate. And what that means then, is Spaceship B here would get closer, according to Observer A. In other words, Spaceship B would creep in a little bit, compared to Spaceship C. But from the statement of the problem, we said the distance between them has to remain constant. That's how we set up the situation. So the only way that Observer A can see that distance remain constant for them is if when Spaceship B starts accelerating at 10 seconds, Spaceship C starts accelerating at 7 seconds, okay. So then as far as Observer A is concerned, they are accelerating identically at that point and building up their velocity. And for some of you who might be thinking well, wait a minute, I thought we didn't deal with accelerations in the special theory of relativity. It's only inertial frames of reference, constant velocity of motion. Yes, you're absolutely right. But we can break up a situation like this in terms of acceleration to just little impulses of acceleration, just a little push, and then constant velocity of motion then another little push, constant velocity of motion and analyze it that way. And again, here we're not going to go too precisely into all the details, we want to get the idea of what's going on in this so-called spaceships on a rope paradox and why the rope breaks. So from this analysis, we say okay, if D is going to remain constant to Observer A, if the distance there, once you have the leading clocks lag situation going on, then if Spaceship B is doing a little acceleration at 10 seconds, Spaceship C to Observer A has to be accelerating at 7 seconds. Otherwise, that distance will decrease as Spaceship B gets a little closer to Spaceship C. It's going up to a little higher velocity there. So what does that mean? Well, okay so, if Observer A is seeing this accelerate at 7 seconds and this one at 10 seconds, we can certainly take a photo snapshot right there and see the rockets turn on and see the clock at 7 seconds. And we take a photo snapshot here, and see the rockets turn on at 10 seconds, okay. Well, think about that, that means Spaceship C, in terms of the BC reference frame, Spaceship C is accelerating at 7 seconds. Spaceship B is not accelerating at 7 seconds. It's still waiting to get to 10 seconds. So what happens? Spaceship C moves ahead a little bit, and the rope snaps at that point. So from the BC frame of reference, we can see why you get the rope breaking. Because as Observer A, again if we're going to mandate that Observer A always sees constant distance D between them, the only way that can occur, as the velocity gets up there a little bit or really any velocity in principle. But certainly, you get more of the effects as you get their relativistic velocities. But for that distance D to remain constant to Observer A, Spaceship C has to start accelerating before Spaceship B does, at any given instant, if we're doing those sort of space accelerations. And therefore, if C starts to accelerate at 7 seconds, and back here Spaceship B, according to Observer A, is not accelerating yet. Then, this one will pull and break the rope. I should say actually, let me back up be very precise about this. In the BC frame of reference, okay, again, we take photographs of this. Everybody has to bring the photographs. According to Observer A, to maintain that distance, we concluded Spaceship C has to accelerate before Spaceship B because of the leading clocks lag factor. What that means in the BC frame of reference, once again, is that if Spaceship C is accelerating at 7 seconds, Spaceship B is not accelerating yet because it's going to accelerate again at 10 seconds. And therefore, Spaceship C moves ahead a little bit, Spaceship B is still waiting for the next interval of acceleration, and the rope breaks. Okay, so that's the analysis from the Spaceship BC frame of reference. What about Observer A? How do they explain that the rope breaks? Well, actually for Observer A, it's fairly straightforward according to special theory of relativity. Because for Observer A what you have, you have length contraction. And if this distance D, if they start at rest and the rest length, the proper length of the rope is D, then as soon as you get some velocity going, we know that to A, L sub A, is 1 over gamma, L sub BC frame of reference, the rest frame of reference for that rope. And therefore, you'll have a length contraction effect. In a previous video, we actually talked about how metal rods, in that case, can do a Lorentz contraction. Really, we're using a very similar type of analysis here, based on the fact that leading clocks lag, and you get this unequalness in time, of when accelerations occur, and that leads to the rope breaking. But for Observer A, they simply say, hey if we maintain this distance D, if I mandate that, then the rope however, is going to be undergoing length contraction. And so if this distance maintains at D, so we could say LBC here, so this becomes 1 over gamma times the distance D, then the length of the rope gets smaller, and it breaks. It exceeds its tensile strength, we would say. Okay, so that's the so-called spaceship on a rope paradox. Here again, pseudo-paradox because using the special theory of relativity and the concepts, key concepts we've had such as length contraction and leading clocks lag, we can see arguing from both perspectives, both the Observer A frame of reference and the Spaceship BC frame of reference, they both will conclude the rope breaks although different analyses for that.