The question was; Three events, A, B, C, are seen by observer O to occur in the order A B C. Another observer, O', sees the events to occur in the order C B A. Is it possible that a third observer O'' sees the events in the order A C B? Support your conclusion by drawing a spacetime diagram.

I couldn't see how to do this without a space time diagram, moreover to make life easier I created excel charts for them so that I could easily compare temporal orders.

I started with a diagram like this

It shows the

Next we have a boost of 0.16.

I couldn't see how to do this without a space time diagram, moreover to make life easier I created excel charts for them so that I could easily compare temporal orders.

I started with a diagram like this

It shows the

*x, t*coordinates that B must have to get the sequence ABC in O and CBA in O'. B must be inside the yellow parallelogram. The boost is 0.6.Next we have a boost of 0.16.

For the temporal order ABC in O and ACB in O'', B must be in the green triangle. (Actually it is a an open sided parallelogram with the lower side at t = t

^{a}.)
To fulfil both the conditions, B must be in the overlap of the yellow parallelogram and the green triangle, which is a small triangle, shown here:

By this time I had moved A to (0,0) to make life easier.

It then became possible to draw a fateful triangle for various boosts. Here are a few samples.

To see all the workings read the pdf here.

I believe you're right here, Axolotl example does not meet the requirements of the question. Applying Lorentz transformation, the condition to flip the sequence of 2 events is deltaT> V * deltaX, where V is the boose speed (T, X measured in original frame). This means we have to put point B above line AC, I believe.

ReplyDeleteSorry the inequality is flipped. Should be

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ReplyDelete(sorry for the spam, on a phone here, and did not logged in) I believe you're right here, Axolotl example does not meet the requirements of the question. Applying Lorentz transformation, the condition to flip the sequence of 2 events is deltaT< V * deltaX, where V is the boost speed (T, X measured in original frame). This means we have to put point B above line AC, so that the line BC has the least gradient and we can find a boost speed V that is larger than its gradient but less than gradients of AC and AB.

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