∆x = x2– x1= –720 m. If we set ∆x'= 0 in Eq. 37-25, we find 0720500 106=−=−−×−γγ()(.∆∆xvtvms)chwhich yields v= –1.44 ×108m/s, or /0.480vcβ==. (b) The negative sign in part (a) implies that frame S'must be moving in the –xdirection. (c) Eq. 37-28 leads to ∆ttvxc'.(.FHGIKJ=×−−×−×FHGIKJ−2688144 102 998 10sm / s)( 720m)m/s)2which turns out to be positive (regardless of the specific value of γ). Thus, the order of the flashes is the same in the S'frame as it is in the Sframe (where ∆tis also positive).
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.