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Announcements 12/3/10

Announcements 12/3/10. Prayer Wednesday next week: Project Show & Tell 5 extra credit points for volunteering, 10 points if I pick you Applications due tomorrow night; 5 volunteers so far Lee’s program: see email TA ratings: see email Survey about Optics HW problems: extended until tomorrow

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Announcements 12/3/10

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  1. Announcements 12/3/10 • Prayer • Wednesday next week: Project Show & Tell • 5 extra credit points for volunteering, 10 points if I pick you • Applications due tomorrow night; 5 volunteers so far • Lee’s program: see email • TA ratings: see email • Survey about Optics HW problems: extended until tomorrow • Definitions: Lorentz transformations: or also

  2. Worked problem #1 • Four “simultaneous” events: viewed by Earth, (x, ct) = … • (0.5, 2) • (0, 2) • (-1, 2) • (-2, 2) • Dr. Colton’s rocket comes by going 0.5 c in the positive x direction. Where/when does he measure these events? g = 1.1547, bg = 0.5774 a = (-0.5774, 2.0207); b = (-1.1547, 2.3094); c = (-2.3094, 2.8868); d = (-3.4642, 3.4642) Lee’s program

  3. Worked problem, cont • Some things to notice: • “Linear” transformation: Notice that lines transform into lines • This case: downward sloping line. There will be some point having ct=2, that (in B’s frame) is at negative time! • Let’s try transforming point “e” = (6, 2) • Turns out… • If a point is outside the light cone (“spacelike”), you can always find some observer that sees it happen at a negative time. • If a point is inside the light cone (“timelike”), then no observer can see it happen at negative time. • Causality! point “e” = (5.773, -1.155)

  4. Worked problem #2 • Lee is running past Cathy at b = +0.5 (g = 1.155). He passes her at t = 0. Cathy is holding the left end of a meterstick, length = 1 m. • In Cathy’s frame: draw the world lines of Cathy, Lee, and the right end of the meterstick. • In Lee’s frame: draw the same worldlines. Lee’s program

  5. Velocity transformations • Lee is standing on a train going past Cathy at +0.5 c. John is also on the train, running past Lee at +0.5 c. What is John’s speed relative to Cathy? (NOT 1.0 c!) • First: draw diagram from Lee’s frame • Then: transform to Cathy’s frame • Find slope of new line (which is inverse of b) • Result: vJohn-Cathy = 0.8 c • General formula: Use this instead of book eqns 39.16 and 39.18. Far simpler; works every time! Caution: terms are sometimes negative. (Don’t need to know transverse velocity formula, eqn 39.17.) Compare to “Galilean”: “1-3” = “of object 1 with respect to object 3”

  6. Worked Problem #3 • HW 39-2 (the one that got canceled) 0.99687 Answers: (a) 53.0; (b) 0.083; (c) 53.0

  7. Worked Problem #4 • Optional problem from HW 40

  8. Worked Problem #5 • Optional problem from HW 40

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