view-notes-pcs110-011 - PART 6 MODERN PHYSICS CHAPTER 38...

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PART 6 MODERN PHYSICS CHAPTER 38 THE THEORY OF RELATIVITY Section 38-2: Matter, Motion, and the Ether Problem 1. Consider an airplane flying at 800 km/h airspeed between two points 1800 km apart. What is the round-trip travel time for the plane (a) if there is no wind? (b) if there is a wind blowing at 130 km/h perpendicular to a line joining the two points? (c) if there is a wind blowing at 130 km/h along a line joining the two points? Ignore relativistic effects. (Why are you justified in doing so?) Solution Since the velocities are small compared to c , we can use the non-relativistic Galilean transformation of velocities in Equation 3-10, u u v = + , where u is the velocity relative to the ground ( S ), u is that relative to the air ( ), S and v is that of S relative to S (in this case, the wind velocity). We used a notation consistent with that in Equations 38-11 and 12. (a) If v = 0 (no wind), then u u = (ground speed equals air speed), and the round-trip travel time is t d u a = = ÷ 2 2 1800 = ( ) km ( / ) . 800 4 5 km h h. = (b) If v is perpendicular to u , then = + u u 2 2 2 v , or u = - = 800 130 789 2 2 km h km h / / , and the round-trip travel time is t d u b = = 2 4 56 = . h. (c) If v is parallel or antiparallel to u on alternate legs of the round-trip, then u u = ′ ± v and the travel time is t d u d u c = ′ + + ′ - = + + - = v v 1800 800 130 1800 800 130 4 62 h h h. . Note that t t t a b c < < , as mentioned following Equation 38-2. Problem 2. What would be the difference in light travel times on the two legs of the Michelson-Morley experiment if the ether existed and if Earth moved relative to it at (a) its orbital speed relative to the Sun (Appendix E)? (b) 10 2 - c ? (c) 0 5 . ? c (d) 0.99 c ? Assume each light path is exactly 11 m in length, and that the paths are oriented parallel and perpendicular to the ether wind. Solution The difference between Equations 38-2 and 1 is t t t cL c L c L c c c = - = - - - = - - - F H G G I K J J || , 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 v v v = v = and 2 2 11 3 10 733 8 L c = = = × = ( ) ( / ) . m m s ns for the given interferometer. If v = 2 2 1 c ¿ , we can expand the denominators (Appendix A) to obtain t L c c c L c c ¼ ( )[ ( )] ( )( ). 2 1 1 2 2 2 2 2 2 2 2 2 = v = v = = v = + - + = (a) For the Earth’s orbital speed (30 km/s), v = 2 2 8 10 c = - , so that t = × = × - - 73 3 10 3 67 10 1 2 8 16 . . ns( ) s (a fraction of the period of visible light). (b) If v = 2 2 4 12 10 367 10 c t = = × - - , . s (a few thousand periods of visible light). (c) At relativistic speeds, we use the exact expression for t . If v = = v = = 2 2 1 4 2 2 1 1 2 3 c c = = - = , , g and t = - = 73 3 2 3 131 4 3 . . . ns( ) ns = (d) If v= c = 0 99 . , g = - = 1 1 0 98 7 09 = . . , and t L c = - = ( ) ( ) . . 2 1 317 = g g m s Problem 3. Figure 38-30 shows a plot of James Bradley’s data on the aberration of light from the star g Draconis, taken in
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896 CHAPTER 38 1727–1728. (a) From the data, determine the magnitude of Earth’s orbital velocity. (b) The data very nearly fit a perfect sine curve. What does this say about the shape of Earth’s orbit?
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This note was uploaded on 10/19/2010 for the course PHY 78647 taught by Professor Miller during the Spring '09 term at Albany College of Pharmacy and Health Sciences.

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view-notes-pcs110-011 - PART 6 MODERN PHYSICS CHAPTER 38...

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