Once you have u τ you can get x by integrating dxdτ u dtdτ Think about the

# Once you have u τ you can get x by integrating dxdτ

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. Once you have u ( τ ), you can get x by integrating dx/dτ = u ( dt/dτ ). (Think about the chain rule.) (d) Since the car moves, it doesn’t experience as much proper time as coordinate time t . Find the total time lag Δ over all time, which is defined as Δ = lim t →∞ [ t - τ ( t )] - lim t →−∞ [ t - τ ( t )] . 2. Velocity Addition In the lecture notes, we argued that a boost by v 1 along x followed by a boost of v 2 also along x is the same as a boost along x by v 3 = v 1 + v 2 1 + v 1 v 2 /c 2 , (1) but we didn’t finish the proof. (a) We needed to show that γ ( v 3 ) = γ ( v 1 ) γ ( v 2 )(1 + v 1 v 2 /c 2 ). To do this, prove the following: 1 - v 2 3 c 2 = (1 - v 2 1 /c 2 )(1 - v 2 2 /c 2 ) (1 + v 1 v 2 /c 2 ) 2 , (2) using v 3 as given in (1). This is 1 ( v 3 ) 2 , so it proves what we want. (b) If v 1 c and v 2 c , can v 3 ever exceed c ? (c) Give a quick argument relating this velocity addition to the transformation of particle velocities to prove (1) indirectly. 3. Relative Velocities I’m sitting in a lab in a particle accelerator, and I see two protons approaching each other on the x axis. One has velocity u 1 = 3 c/ 4 and the other has velocity u 2 = - 3 c/ 4 in my frame. (a) At what speed do I see them approach each other? To think about this, you might want to find their positions x 1 ( t ) and x 2 ( t ) and then figure out d ( x 2 - x 1 ) /dt just to get things straight in your head.

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(b) You might be surprised at your answer, since it should be bigger than c . Is anything real moving faster than c ? (c) Find the velocity u of one particle relative to the other (that is, the velocity of one particle in the rest frame of the other). Page 2

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