p38_016

# p38_016 - 6 The two events in frame S occur in the order:...

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16. The “coincidence” of x = x 0 =0at t = t 0 = 0 is important for Eq. 38-20 to apply without additional terms. We label the event coordinates with subscripts: ( x 1 ,t 1 )=(0 , 0) and ( x 2 ,t 2 ) = (3000 , 4 . 0 × 10 6 ) with SI units understood. Of course, we expect ( x 0 1 ,t 0 1 )=(0 , 0), and this may be veriFed using Eq. 38-20. We now compute ( x 0 2 ,t 0 2 ), assuming v =+0 . 60 c =+1 . 799 × 10 8 m / s (the sign of v is not made clear in the problem statement, but the ±igure referred to, ±ig. 38-9, shows the motion in the positive x direction). x 0 2 = x vt p 1 β 2 = 3000 ( 1 . 799 × 10 8 )( 4 . 0 × 10 6 ) p 1 (0 . 60) 2 =2 . 85 × 10 3 t 0 2 = t βx/c p 1 β 2 = 4 . 0 × 10 6 (0 . 60)(3000) / (2 . 998 × 10 8 ) p 1 (0 . 60) 2 = 2 . 5 × 10
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Unformatted text preview: 6 The two events in frame S occur in the order: Frst 1, then 2. However, in frame S where t 2 < 0, they occur in the reverse order: Frst 2, then 1. We note that the distances x 2 x 1 and x 2 x 1 are larger than how far light can travel during the respective times ( c ( t 2 t 1 ) = 1 . 2 km and c | t 2 t 1 | 750 m), so that no inconsistencies arise as a result of the order reversal (that is, no signal from event 1 could arrive at event 2 or vice versa)....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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