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In: Physics

A firecracker goes off in Houston, Texas. A time of 0.03 seconds later (as measured on...

A firecracker goes off in Houston, Texas. A time of 0.03 seconds later (as measured on synchroniz... A firecracker goes off in Houston, Texas. A time of 0.03 seconds later (as measured on synchronized earth clocks), another firecracker goes off in Great Falls, Montana, 2400 kilometers away (as measured on Earth). (a) Draw a sketch of this context. (b) How fast must a rocket ship travel if it is to be present at both events? (c) What will the rocketship pilot note for the time interval between the events? (Be careful to express all speeds in the same units; here, meters/second is preferable to kilometers/second. (d) What is the numerical difference in the two times, that measured on earth versus that measured in the rocket ship? Is the difference consistent with the verbal, qualitative statement: "The context is a pair of events. An observer for whom the events happen at the same place measures the least elapsed time."

Solutions

Expert Solution

If both firecrackers went off at the same spatial point in the Earth frame then you could just use the time dilation formula to get the time interval in the rocket. However they go off at different spatial points and this means you have to do the full calculation. Let's see how this works.

We'll choose our origin so that in the Earth frame the first firecracker goes off at (0,0)(0,0), then the second firecracker goes off at (t,d)(t,d), where t=2μst=2μs and d=300md=300m, but let's keep it general for now. We'll choose the coordinate system of the rocket so its origin coincides with the Earth frame, then the point (0,0)(0,0) is the same in both frames. We just need to find where the point (t,d)(t,d) is in the rocket frame.

The Lorentz transformations tell us:

t′x′=γ(t−vxc2)=γ(x−vt)t′=γ(t−vxc2)x′=γ(x−vt)

Actually we are only asked for the time difference in the rocket frame so we just need t′t′, and this is:

t′=γ(t−vdc2)t′=γ(t−vdc2)

Now if the two firecrackers had been in the same place dd would be equal to zero and we would just have:

t′=γtt′=γt

So you could just use the usual time dilation equation. It's because d≠0d≠0 that we can't do this.


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