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Unformatted text preview: 1.Solution: Take the spacecraft as the rest frame and the earth as the moving frame Assume the length of the antenna is L in the rest frame (spacecraft frame), Because the velocity of the spacecraft has no length contraction effect on the direction which is perpendicular to the direction of the velocity, In the velocity direction, we can have, 2 2 1 0.70 cos10 1 cos10 1 m v c L L L c c = = o o In the direction which is perpendicular to the direction of the velocity, we can have, 2 sin10 m L L = o So in the earth frame (moving frame) 2 2 1 sin10 tan10 tan 0.2469 0.51 0.70 cos10 1 m m L L L c L c θ = = = ≈  o o o namely, the angle is ( 29 1 tan 0.2469 14 θ = ≈ o 2.Solution: now we measure the length at 2 vL t c γ ′ = , So we must let the event A at the time 2 A vL t c γ ′ = , not at A t ′ = Let see what happen for the event A in the rest frame, Set A x = , using the Lorenzt Transformation, we have 2 2 2 A A A A vL vL vx t t t c c c γ γ ′ = = ⇒ =  then, ( 29 ( 29 2 2 A A A v L x x vt vt c γ γ γ ′ = = = So the length in the s ′ frame is 2 2 2 2 2 2 2 2 2 2 (1 ) (1 ) 1 1 B A v L v L v v c L x x L L L c c c v c γ γ γ ′ ′ ′ = = = = = Length contraction proved! 3.Solution: The Lorentz Transformation is ( 29 2 x x vt y y z z vx t t c γ γ ′ = ′ = ′ = ′ = and ( 29 2 x x vt y y z z vx t t c γ γ ′ ′ = +...
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 Spring '11
 HoBunCHAN
 Physics, Special Relativity, Frame, Lorentz Transformation, c

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