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Announcements 12/3/10 Prayer Wednesday next week: Project Show & Tell a. a.5 extra credit points for volunteering, 10 points if I pick you b. b.Applications.

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Presentation on theme: "Announcements 12/3/10 Prayer Wednesday next week: Project Show & Tell a. a.5 extra credit points for volunteering, 10 points if I pick you b. b.Applications."— Presentation transcript:

1 Announcements 12/3/10 Prayer Wednesday next week: Project Show & Tell a. a.5 extra credit points for volunteering, 10 points if I pick you b. b.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) = … a. a.(0.5, 2) b. b.(0, 2) c. c.(-1, 2) d. d.(-2, 2) Dr. Colton’s rocket comes by going 0.5 c in the positive x direction. Where/when does he measure these events?  = 1.1547,  = 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: a. a.“Linear” transformation: Notice that lines transform into lines b. b.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) c. c.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. d. d.Causality! point “e” = (5.773, -1.155)

4 Worked problem #2 Lee is running past Cathy at  = +0.5 (  = 1.155). He passes her at t = 0. Cathy is holding the left end of a meterstick, length = 1 m. a. a.In Cathy’s frame: draw the world lines of Cathy, Lee, and the right end of the meterstick. b. b.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 a. a.Find slope of new line (which is inverse of  ) Result: v John-Cathy = 0.8 c General formula: Compare to “Galilean”: “1-3” = “of object 1 with respect to object 3” 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.)

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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