1. Jun 24, 2007

duordi

I broke down and bought Edwin F. Taylor’s Space time Physics
I am having trouble understanding the detonator paradox.

The book states that there will be an explosion but it doesn’t explain why.
Am I suppose to say the story violates one of these principles.

Special relativity problems intended to be paradoxes usually involve some careful thinking. First of all, for special relativity to apply, the two postulates proposed by Einstein have to be true:
1) The bodies involved must have uniform velocity. No acceleration is permitted.
2) The speed of light is the same for all observers in all frames.

Or is this explanation correct I found it on the web.

Fom Edwin F. Taylor's Spacetime Physics, P.185, problem 6-5. A U-shaped structure contains a detonator switch connected by wire to one ton of the TNT. A T-shaped structure fits inside the U, with the arm of the T not quite long enough to reach the detonator switch when both structures are at rest. Now the T is removed to the left and accelerated to high speed. It is Lorentz- contracted along its direction of motion. As a result, its long arm is not long enough to reach the detonator switch when the two collide. However, look at the same situation in the rest frame of the T structure. In this frame the arm of T has its rest length, while the two arms of the U structure are Lorentz-contracted. Therefore the arm of the T will certainly strike the detonator switch. The question is whether there will be an explosion. Figure: | =======+ |====== S|~~~~TNT (S denotes the switch) | =======+ <--x-> <--y--> The answer is there will be an explosion. Someone said that due to there is no absolutely rigid body so the arm of T will continue its motion for some time even the cap of T strike the arms of U. But can we prove the two frames agree quantitatively? In the rest frame of T, the sufficient condition for the explosion is x>=y*Sqrt [1-v^2], but in the rest frame of U, the condition depends on when the arm stop and I can't make it agree with the condition in the rest frame of T exactly. How do you resolve it?. How do you resolve it?

2. Jun 24, 2007

JesseM

Yes, that's correct.
One simple proof is just to note that if you treat the bottom of the arm of the T and the cap of the T as separate objects and calculate when each one hits the U, there is a spacelike separation between the two hit-events (remember that the spacetime interval between events is the same in every frame, so if it's spacelike in one frame it's spacelike in every frame)--this should be sufficient to show that there's no possible way that the event of the cap hitting could have any causal influnce on the end of the arm of the T until after the arm hits the U.

Last edited: Jun 25, 2007
3. Jun 26, 2007

duordi

As viewed by the observer on the U, does T getting longer undo the Lorenz contraction of T as the T slows to the speed of U?

Duane

4. Jun 26, 2007

jambaugh

Of course the answer is, it doesn't matter! If the U and T collide at relativistic speeds you get an explosion anyway. All that kinetic energy has to go somewhere!

5. Jun 26, 2007

JesseM

Once the cap hits the U, the T is no longer a rigid object since different parts are moving at different speeds, I'm not sure it even makes sense to talk about "Lorenz contraction" for a non-rigid object. And in U's rest frame the T must actually end up stretched to longer than its rest length. Whether the different parts of the T oscillate a little and finally lose their relative velocity so the T looks rigid again, or whether the T breaks apart during the impact, or whether something else happens, all depends on the detailed physical properties of the T, but it's not important for solving the problem.

6. Jun 26, 2007

Ich

In the rest frame of U, x>=y*(1-v²). The bottom of the T can't stop before a light signal from the "earlier" collision reaches it. That means xmax>=(1-v²)/(1-v)>1.