Minkowski Diagram HW Question

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Minkowski Diagram HW Question

A friend of mine popped by and is desperately seeking an answer.

minkowski diagram
Consider the following thought experiment. Two spaceships are initially floating in a region of space far removed from other matter. They are at rest with respect to each other and with respect to sum inertial reference frame F there is a distance L between them. At some time t=0, as measured by this reference frame F they both turn on their engines and start accelerating very gently in the same direction. The two spaceships have identical engines and are both programmed by identical software to maintain this gentle constant acceleration for a long time T until they reach half the velocity of light, after which the engines are turned off simultaneously. Furthermore, the rockets are initially connected by a fragile thread, just long enough to cover the distance L between the two ships. The question is what will happen to this thread due to the relativistic length contraction.
a) sketch a minkowski diagram, clearly and completely labeled, of the world-lines of both spaceships
Here is what happens to the thread from the point of view of reference frame F: the ships started accelerating from rest with the same acceleration at time t=0 and thus, at all later times, gained the same speed. Hence the distance between them has not changed at time T, they are still at a distance L apart. However since they, and the thread that connects them, are now moving at very large velocity, relativity predicts that thread will suffer length contraction. It will become shorter than the distance L it needs to span, and build up tension and eventually break.

From what I can tell, I'm having trouble understanding the diagram itself. So what's a world line?

sardia

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Re: Minkowski Diagram HW Question

A path of an object in a Minkowski diagram

Bonus challenge: Describe the same situation from the view of an observer on one of the ships (NOT an inertial system!).
mfb

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Re: Minkowski Diagram HW Question

This is called Bell's spaceship paradox, and although Bell himself believed the string would break this seems to be somewhat controversial.
wikipedia wrote:Objections and counter-objections have been published to the above analysis. For example, Paul Nawrocki suggests that the string should not break,[3] while Edmond Dewan defends his original analysis from these objections in a reply.[4] Bell reported that he encountered much skepticism from "a distinguished experimentalist" when he presented the paradox. To attempt to resolve the dispute, an informal and non-systematic canvas was made of the CERN theory division. According to Bell, a "clear consensus" of the CERN theory division arrived at the answer that the string would not break. Bell goes on to add, "Of course, many people who get the wrong answer at first get the right answer on further reflection".

some_dude

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Re: Minkowski Diagram HW Question

At relativistic speeds, matter contracts? I'm a little out of my depth here.

Is this what the diagram looks like of the spaceships?

sardia

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Re: Minkowski Diagram HW Question

Why would the string break?

The spaceships (and presumably the string) start out at rest relative to eachother. They accelerate at the same speed. And they do this gently, so the string has time to match this speed. So all three will always be in the same frame of reference. Length contraction will thus never occur.

So why would it break?
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Re: Minkowski Diagram HW Question

In a reference frame that's undergoing uniform linear acceleration, the rate of acceleration felt at each point varies.

If the grease monkeys manning the engines down in the very stern of a one-light-year-long spaceship measure an acceleration of 1g, then the captain, up in her lofty cabin at the very nose one light year away, will feel an acceleration of only half a g. Crew members in between will feel accelerations between the extremes, according to their location along the length of the ship. The rate time passes also varies. If the engineers in the stern throw the switch to blast the engines for one year and then turn them off then afterwards it will be noted that the captain has aged two years to their one (and likewise varying age differences for the crew members in between).

As a consequence: if two tiny shuttles, one initially located alongside the large ship just outside the supreme grease monkey's cabin, while the other is initially located just outside the captain's captain each begin accelerating at exactly 1g at the same instant the large ship turns on its engines, then the shuttle alongside the captain will be accelerating at 1g while the captain herself is only accelerating at 0.5g. The lead shuttle will pull away.

To reverse the argument and prove this must be the case, consider the viewpoint of an observer in the (non accelerating) space dock. From this viewpoint, we know that as the ships accelerate up to some significant fraction of c, all three of them will undergo length contraction. But we also know that the two shuttles will, from the stay-at-home observer's viewpoint, remain always exactly one light year apart from each other - while the long ship shrinks to half a light year in length, then a third, then a fourth, and so on.

Or to put it more succinctly: the string breaks.
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Goemon

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Re: Minkowski Diagram HW Question

I assume you want the two spaceships aligned along the same direction as their acceleration, so they are “one behind the other” rather than “side by side”.

If each spaceship accelerates such that the local acceleration felt onboard that ship is a known function of the local time on that ship (and importantly, the same function for both ships, and the ships themselves are of negligible mass such that neither they nor their thrust exhaust affects the other ship significantly) then the ships should behave as if they are part of the same object.

In other words, the outside inertial observer will see the length of the whole two-spaceship system contract. The outside observer will the see two ships get closer together, meaning the distance the string needs to cover will get shorter. The outside observer will also see the length of the string contract, by exactly the same amount.

In other words, you can imagine that there is one big gigantic (but still of negligible mass) spaceship with both of the little spaceships inside it. We can even nail the little spaceships to the side of the big spaceship. Then the big spaceship accelerates, and the little spaceships move right along with it. Inside the big spaceship, the two little spaceships are just nailed to a wall and connected by a string. No big deal.

The outside observer sees the big spaceship contract, and the little spaceships contract right along with it, and the distance between the little ships contracts just the same as the big ship, and so does the string. The string does not break.

The only thing that might cause tension in the string is the delay as the acceleration propagates through the string, since the string itself is only pulled along at the ends, and the middle part has to wait for the pressure wave to get there. If the acceleration is gentle enough then even that does not break the string.
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Qaanol

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Re: Minkowski Diagram HW Question

Qaanol wrote:I assume you want the two spaceships aligned along the same direction as their acceleration, so they are “one behind the other” rather than “side by side”.

If each spaceship accelerates such that the local acceleration felt onboard that ship is a known function of the local time on that ship (and importantly, the same function for both ships, and the ships themselves are of negligible mass such that neither they nor their thrust exhaust affects the other ship significantly) then the ships should behave as if they are part of the same object.

In other words, the outside inertial observer will see the length of the whole two-spaceship system contract. The outside observer will the see two ships get closer together, meaning the distance the string needs to cover will get shorter. The outside observer will also see the length of the string contract, by exactly the same amount.

In other words, you can imagine that there is one big gigantic (but still of negligible mass) spaceship with both of the little spaceships inside it. We can even nail the little spaceships to the side of the big spaceship. Then the big spaceship accelerates, and the little spaceships move right along with it. Inside the big spaceship, the two little spaceships are just nailed to a wall and connected by a string. No big deal.

The outside observer sees the big spaceship contract, and the little spaceships contract right along with it, and the distance between the little ships contracts just the same as the big ship, and so does the string. The string does not break.

The only thing that might cause tension in the string is the delay as the acceleration propagates through the string, since the string itself is only pulled along at the ends, and the middle part has to wait for the pressure wave to get there. If the acceleration is gentle enough then even that does not break the string.

That all seems very intuitive, unfortunately it is wrong. See above.

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Re: Minkowski Diagram HW Question

Goemon wrote:To reverse the argument and prove this must be the case, consider the viewpoint of an observer in the (non accelerating) space dock. From this viewpoint, we know that as the ships accelerate up to some significant fraction of c, all three of them will undergo length contraction. But we also know that the two shuttles will, from the stay-at-home observer's viewpoint, remain always exactly one light year apart from each other - while the long ship shrinks to half a light year in length, then a third, then a fourth, and so on.

Er, why? What makes you think the distance between the ship remains 1 lightyear in the frame of an inertial observer?

Your reasoning seems to break a number of symmetry laws. The ships start out in the same frame of reference. They apply the same force. They must experience the same acceleration. Anything else would break symmetry. It doesn't matter if they are connected or not. But if they have the same initial velocity and acceleration at time t they must have the same velocity at time t + dt. And so they must have the same velocity and acceleration at every moment in time. So in their frame of reference the distance between them does not change. The string is also in this frame of reference, so its length doesn't change, and it doesn't break.

Look at it from the viewpoint of the space dock again. It sees a ship accelerate and get smaller due to length contraction. Now partition this ship into smaller subsections. Either in your mind or literally with a bolt cutter. By your reasoning every individual section of the ship undergoes length contraction, but the distance between the subsections (and thus the total length of the ship) does not change. So the ship as a whole does not undergo length contraction. A contradiction.

It's not just objects that contract along the direction of motion. Everything does. The entire universe contracts, both objects and the space between objects. Travel fast enough and the entire universe becomes a pancake.
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Re: Minkowski Diagram HW Question

Qaanol wrote:the two little spaceships are just nailed to a wall

This wall violates the "same acceleration at the same time for the non-accelerating observer"-thing.

If v1(t)=v2(t) is the same for our outside observer and we have acceleration, then v'1(t) != v'2(t) from the view of the spaceships, as "at the same time" means something different for them.
Therefore, when viewed from the spaceships, their distance is increasing (the sign is not clear from that argument alone, but can be derived).
mfb

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Re: Minkowski Diagram HW Question

Physics is simple in an inertial reference frame. The simple, well known equations of Special Relativity are "Special" because they come with a restriction: they can be applied only by an observer in uniform motion. The equations applicable to an accelerated reference frame must be deduced from the conclusions reached by first analyzing a problem from the viewpoint of an inertial observer. Bearing this in mind, I'll take the liberty of twisting the words above to suit my own purpose, if I may:

Diadem wrote:The [two small] ships start out in the same frame of reference. They apply the same force. They must experience the same acceleration. Anything else would break symmetry. It doesn't matter if they are connected or not. But if they have the same initial velocity and acceleration at time t they must have the same velocity at time t + dt. And so they must have the same velocity and acceleration at every moment in time. So in their the space dock's frame of reference the distance between them does not change.

All of the reasoning you're applying is valid - from the viewpoint of those who remain on the space dock. On the other hand, we do not know what sort of reasoning may or may not be valid from the viewpoint of any of the ships, who are undergoing continuous acceleration and therefore cannot apply "simple" physics. We know straight off the bat that the ordinary equations of Special Relativity will not apply correctly to their observations; they have to deal with "fictitious" forces that cause objects which should float to "mysteriously" plunge to the floor, and so on. We can't know what happens from their point of view unless we start with the space dock's point of view and work backwards.

And from the fact that the space dock observers will see the two shuttles remain 1ly apart according to their measurements, and will also see the single very long spaceship shrink to a fraction of a light year in length, we must conclude that the nose of the long spaceship can't be accelerating as rapidly as the shuttles.
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Goemon

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Re: Minkowski Diagram HW Question

From the viewpoint of an inertial observer at the space dock, the ships stay the same distance apart, but because of length contraction, the "natural" length of the string is getting smaller, so it would have to be stretched to keep the same length, so it would break.

On board the ships, it's harder to solve directly, as they're accelerating reference frames, but in broad strokes relativity still applies... in particlar, relativity of simultaneity means that they won't see the ships as moving at the same speed at the same time, as the observer on the dock does. Instead, they will both see the front ship as moving faster than the rear ship, and thus the two ships moving apart, breaking the string.
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phlip
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Re: Minkowski Diagram HW Question

Goemon wrote:Physics is simple in an inertial reference frame. The simple, well known equations of Special Relativity are "Special" because they come with a restriction: they can be applied only by an observer in uniform motion. The equations applicable to an accelerated reference frame must be deduced from the conclusions reached by first analyzing a problem from the viewpoint of an inertial observer.

True. But the equations of special relativity still apply at any given point in time.

Diadem wrote:The [two small] ships start out in the same frame of reference. They apply the same force. They must experience the same acceleration. Anything else would break symmetry. It doesn't matter if they are connected or not. But if they have the same initial velocity and acceleration at time t they must have the same velocity at time t + dt. And so they must have the same velocity and acceleration at every moment in time. So in their the space dock's frame of reference the distance between them does not change.

All of the reasoning you're applying is valid - from the viewpoint of those who remain on the space dock.

When I said 'their frame of reference' I admit I was sloppy. Their frame of reference is not an inertial frame. But we can at any point of time look at the inertial frame in which the ships are at rest. Let's look at one such inertial frame. Both ships are at rest (You may argue that they are not. But you have to agree that they at least start out in this situation) in this frame. Both have equal engines that apply equal force. By symmetry both must have equal acceleration. Thus a time dt later (in this frame) they must both have velocity dv (in this frame). Now let's look at the reference frame that moves at dv relative to the previous frame. Now both ships are again at rest. And since they still have equal engines, they must again have speed dv after a time dt in this new reference frame. We can repeat this process and conclude that at any point in time the ships have the same speed and acceleration, as viewed from any inertial reference frame along the path of the ships.

I must admit it's a long time since I last did SR. But I can see no holes in my logic.

From the perspective of the space station it also fits. It will slowly see the entire ship-thread-ship system contract, just as it would if it was one giant spaceship. This means the two ships do not have the same acceleration. That's not wrong. They start out with the same acceleration, but as they accelerate what is simultanious for them is no longer simultanious for the space station (since they are a distance apart). So as seen from the space station the ships at the same only at t=0.

You may worry that I silently reformulated the question. The question says they start accelerating at the same time as measured from the space station, not as measured from the ships. But I can simply initially take the space station to be at rest relative to the ships, and an equal distance from both ships too, if you want, so that t=0 on the space station corresponds to t=0 on both ships. This does not lose generality, since if the thread does not break in one frame of reference, it does not break in any of them.
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Re: Minkowski Diagram HW Question

Diadem wrote:But we can at any point of time look at the inertial frame in which the ships are at rest.

There is no inertial frame in which that is the case. If you have a system where one ship is at rest, you know that for the space dock, both ships have the same velocity. However, in your frame of your ship, "simultanous" refers to a different time for the other ship, so it has a different velocity (that means !=0). If you are in the leading ship, "at the same time" refers to an earlier time (relative to space dock) for the following ship, so that appears slower (moving away from you). If you are in the following ship, "at the same time" refers to a later time (relative to space dock) for the leading ship, so that appears faster (moving away from you).

If you are in one of the ships, the other ship is moving away from you, as long as the acceleration is there.
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Re: Minkowski Diagram HW Question

mfb wrote:
Diadem wrote:But we can at any point of time look at the inertial frame in which the ships are at rest.

There is no inertial frame in which that is the case. If you have a system where one ship is at rest, you know that for the space dock, both ships have the same velocity. However, in your frame of your ship, "simultanous" refers to a different time for the other ship, so it has a different velocity (that means !=0).

At least initially the ships are at rest with respect to eachother. This is given in the formulation of the question. So that we can have a frame where both are at rest.

My argument is by induction. If it's true at some t, it's true at time t+dt. And it's true at t=0. So it's true for all t. That logic is mathematically correct. So if the conclusion is false, the problem must be in the premises. The second one is given in the exercise, so it it most certainly correct. And I can see no fault in the first one.
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Re: Minkowski Diagram HW Question

Tass wrote:
Qaanol wrote:If each spaceship accelerates such that the local acceleration felt onboard that ship is a known function of the local time on that ship (and importantly, the same function for both ships, and the ships themselves are of negligible mass such that neither they nor their thrust exhaust affects the other ship significantly) then the ships should behave as if they are part of the same object.

That all seems very intuitive, unfortunately it is wrong. See above.

Are you sure? I deliberately used a different situation from the original problem. In the original, the spaceships accelerate at equal rates as viewed from a given fixed inertial frame. In mine, spaceship A views its own acceleration as f(tA), and spaceship B views its own acceleration as f(tB), where the subscript denotes time as measured aboard the corresponding ship, synchronized from a common reference time when they were in the same inertial frame.
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Qaanol

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Re: Minkowski Diagram HW Question

Diadem wrote:If it's true at some t, it's true at time t+dt.

This is your faulty premise. Between time t and time t+dt, the ships have accelerated by dt*a, so they're travelling at a larger velocity, so relativity of simultaneity has a stronger effect. In particular, the effect is zero at t=0, but non-zero at t=0+dt.
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phlip
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Re: Minkowski Diagram HW Question

phlip wrote:
Diadem wrote:If it's true at some t, it's true at time t+dt.

This is your faulty premise. Between time t and time t+dt, the ships have accelerated by dt*a, so they're travelling at a larger velocity, so relativity of simultaneity has a stronger effect. In particular, the effect is zero at t=0, but non-zero at t=0+dt.

But simultaneous events in one frame of reference are the simultaneous in *that same* frame of reference. Both ships accelerate by the same amount at the same time in both of their frames of reference (since it's the same frame).

Ok let me take a different angle. What happens if the ships decelerate? Does the thread go slack? It must, right? Imagine they accelerate for a while, then cruise along for a bit, and then decelerate back to their original velocity. Your argument is that the distance between them as seen from the space station never changes. Logically then the thread (if it is elastic so it doesn't break) is back to its original length in the final situation. So if it went taut while accelerating, it must have gone slack while decelerating.

But now imagine another space station, travelling at at a velocity relative to the first one equal to the velocity the spaceships have after accelerating. From its perspective, they first decelerate, come to rest for a while, then accelerate. So from its perspective, by your reasoning, the thread must first go slack, then taut again. But the laws of physics are the same in any inertial frame. So this is a contradiction.

I'm really open to the possibility of being wrong. SR was a long time ago for me. But the more I think of it the more convinced I get that I am right. Every new approach I take of looking at the problem brings me to the same conclusion.
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Re: Minkowski Diagram HW Question

Diadem wrote:But now imagine another space station, travelling at at a velocity relative to the first one equal to the velocity the spaceships have after accelerating. From its perspective, they first decelerate, come to rest for a while, then accelerate. So from its perspective, by your reasoning, the thread must first go slack, then taut again. But the laws of physics are the same in any inertial frame. So this is a contradiction.

No, because relativity of simultaneity.

For the following I'm going to name the ships, because calling them the "front" and "rear" ships is going to cause confusion. In our original space dock's point of view, we'll call the front ship "ship A" and the rear ship "ship B". So initially ship B is accelerating towards ship A, and ship A is accelerating away from ship B, and then in the second phase ship B accelerates away from ship A and ship B towards ship A.
From the point of view of our new space station, the ships are already moving in the opposite direction - ship A is already moving toward ship B. Now, the accelerations are the same - in the first phase ship B is accelerating towards ship A, etc. But from this point of view, ship A could be called the "rear" ship and ship B could be called the "front" ship.

Anyways, this second space station would see the ship A decelerate first, and fall back from ship B, and then ship B would decelerate after. When they're at "rest" (which they wouldn't be at the same time, but in that time range), they'd have increased their distance substantially - by such an extent that even after the removal of length contraction is taken into account, the string is still being stretched. and then, when they accelerate, again ship A moves first, and catches up towards ship B, and when they've both accelerated to their final speed, they're back to their original distances.
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phlip
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Re: Minkowski Diagram HW Question

Ok good point. So if they simultaneously accelerate the thread goes taut,and if they simultaneously decelerate the thread goes slack, without contradiction because of the relativity of simultaneity. Ok. Agreed.

However that just means this particular argument doesn't support my point, but it doesn't contradict it either. So throwing it out, I'm now back to my original reasoning.
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Re: Minkowski Diagram HW Question

Yes, it does. Because, as you said, switching observers doesn't change the result, and I just did an observer-switch to one where the ships definitely get much further apart than the proper length of the string, so it would snap regardless of the presence or absence of length contraction.

Unless your point isn't "the string wouldn't snap"... I must admit I'm a little confused as to exactly what you're trying to claim.
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phlip
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Re: Minkowski Diagram HW Question

Ok, if you want to do the analysis from the accelerated reference frame, and then work backwards to the inertial frame, we'll expand on phlip's comment:

Consider the viewpoint of the rearmost shuttle. In time dt, their own velocity changes by an amount dv = adt. They calculate that the velocity of the leading shuttle also changes by an amount dv = adt, all well and good.

BUT, now consider the effect of the change in reference frames. The rear shuttle has changed from the initial reference frame to a new one, with a velocity difference of dv. The axes of their reference frame rotate by an angle dtheta, proportional to the change in velocity. The rear shuttle's reference frame is of course centered on the rear shuttle. As the axis which represents "now" to the rear shuttle rotates about the rear shuttle, it sweeps "ahead" of the lead shuttle. The linear distance swept out, at the distance L between the shuttles, equals Ldtheta.

In other words, in the time dt as measured by the rear shuttle, the lead shuttle experiences a time dt, PLUS some additional time Ldtheta. The crew of the lead shuttle is aging faster than the trailing shuttle. And since dtheta is proportional to dv, the time experienced by the lead crew as measured by the trailing crew is dtau = dt + Ldv = dt + Ladt = (1+aL)dt. And therefore the velocity of the lead shuttle, as measured by the trailing shuttle, increases by a(1+aL)dt while their own velocity increases by only adt.

Spoiler:

From the viewpoint of the space dock - where physics is simple due to their remaining in a single reference frame - the velocity of both shuttles always changes by dv = adt. From this viewpoint, both shuttles always have the same speed and they remain the same distance apart.
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Goemon

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