Forces in relativistic rolling motion

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jartsa said:
OK then, I'll do some math.

So we have a spacetime diagram with one mostly vertical world line, describing how an object at rest is given two opposite impulses. It looks like this:
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My task is to draw a spacetime diagram where that same world line is tilted. Convert the diagram to other frame, or whatever the correct idiom is.

How do I do that? It will be a tilted line with two angles, is there a nice simple formula to convert the angles?

Ok, let's give the events some labels in this inertial reference frame (S). The start of the worldline is e1, the first kink to the left is e2, the second kink is e3 and the top of the worldline is e4. During the impulse the object moves a distance ##\Delta x## in a time interval ##\Delta t##. x must be less that t as this is a physical object. Now use the Lorentz transforms to find ##\Delta x'## and ##\Delta t'## of events e2' and e3' in another reference frame (S') moving to the right with velocity v relative to frame S. Once you have the deltas it is easy enough to find the angles using simple trigonometry.

I think you will better off demonstrate to yourself that if the equal impulses occur simultaneously in frame S' where the rod is moving, that they do not occur simultaneously in the original rest frame S of the rod and so the centre of the rod will be accelerated in inertial reference frames.
 
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jartsa said:
So we have a spacetime diagram with one mostly vertical world line, describing how an object at rest is given two opposite impulses. It looks like this:

An object at rest does not have a world line.

Each individual point in the object has a world line, and all of these worldlines together form the world sheet of the object; the individual world lines may converge, diverge, and jog independently.
 
Nugatory said:
An object at rest does not have a world line.

Each individual point in the object has a world line, and all of these worldlines together form the world sheet of the object; the individual world lines may converge, diverge, and jog independently.
I think jartsa is talking about a single point marked at the centre of the object to simplify things.
 
yuiop said:
I think jartsa is talking about a single point marked at the centre of the object to simplify things.

I'm sure that he is, but that simplification only works if all parts of the body move in unison - and that's not applicable in all of this discussion about impulses being applied to opposite ends of the object at the same or different times.
 
Nugatory said:
I'm sure that he is, but that simplification only works if all parts of the body move in unison - and that's not applicable in all of this discussion about impulses being applied to opposite ends of the object at the same or different times.
True, but we can determine if the centre of the object moves in its original rest frame by considering the compression waves in isolation. If they arrive simultaneously at the centre in the rest frame, the centre does not move (for equal impulses) and if they do not arrive simultaneously, then it does move. Don't want to overcomplicate things at this stage.
 
Let's consider a T-shaped object, which we will call T, moving to the right very fast, with two unstable particles on both ends of the horizontal bar. Both particles decay to two photons simulteneously in T's frame. Then two photons travel from the end points of the horizontal bar to the vertical bar, into which the photons are absorbed.

Now we observe these events from that frame where the T was moving to the right very fast.

We see the left side particle decaying first, then one of the decay products travels to the right as a high energy photon.

Then we see the right side particle decaying, then one of the decay products travels to the left as a low energy photon.Some of T's stuff traveled from the left to the right, smaller amount of T's stuff traveled from the right to the left.

Conclusion: In this case where opposite impulses pushed the T, a shift to the right happened, as we'd expect when the right pushing impulse preceeds the left pushing impulse.Clarification:

We consider the photons that leave the T to not be part of the the T, while the other two photons we consider to be part of the T, also the two unstable particles we consider to be part of the T.

So when the left side particle decays, T loses a small part of itself, when the right side particle decays, T loses a larger part of itself, later the balance between left and right is restored by a net mass-energy flow from left to right.
 
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This is pointless. Until you actually go through the effort to mathematically analyze a simple situation using the correct mathematical tools which have been identified there is no point in analyzing progressively more complicated scenarios.

Please try to do some actual work on your own. If you get stuck then post it to a new thread, this one is closed.