PHYSICS 131 hw-2-soln

PHYSICS 131 hw-2-soln - HOMEWORK SET 2, PHYS. 131 Solutions...

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HOMEWORK SET 2, PHYS. 131 Solutions originally by Matt Pillsbury Instructor: Anthony Zee Email: zee@kitp.ucsb.edu TA: Kevin Moore Email: kmoore@physics.ucsb.edu 5.4. Work out the components of the four-vector a d u /dτ in terms of the three-velocity ± V and the three-acceleration ±a = d ± V /dt to obtain expressions analogous to (5.28). Using this expression and (5.28), verify explicitly that a · u = 0. First, note that dt = 1 (1 - ± V 2 ) 3 ± V · d ± V dt = γ 3 ± V · With (5.28) on p. 84 this gives a = dt d dt ( γ, γ ± V ) = γ 2 ± γ 2 ( · ± V ) , γ 2 ( · ± V ) ± V + ² and a · u = - γ 5 ( ± V · )+ γ 3 ( ± A · ± V γ 5 ( ± A · ± V ) ± V 2 = γ 5 ( · ± V ) ³ - 1+ ± V 2 ´ + γ 3 ( · ± V ) = - γ 3 ( ± A · ± V γ 3 ( ± A · ± V )=0 1
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5.7. A particle is moving along the x -axis. It is uniformly accelerated in the sense that the acceleration measured in its instantaneous rest frame is always g , a constant. Find x and t as functions of proper time τ assuming that the particle passes through x 0 at time t = 0 with zero velocity. Draw the world line of the particle on a spacetime diagram. In the particle’s instantaneous rest frame, the last problem gives a t ± =0 ,a x ± = g Boost to the observer frame (where the particle has velocity v ), using (5.9) from p. 80 and (5.28) to get a t = gγv = gu x x = = t which is equivalent to d ± u t u x ² = ± 0 g g 0 ²± u t u x ² This system of ode s is solved by u t ( τ )= A cosh( )+ B sinh( ) ,u x ( A sinh( B sinh( ) The particle is at rest at τ = 0, so A = 1 and B = 0. Then we have t ( τ ³ dτ u t = g - 1 sinh( ) x ( τ ³
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PHYSICS 131 hw-2-soln - HOMEWORK SET 2, PHYS. 131 Solutions...

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