Announcements 12/3/10
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Prayer
Wednesday next week: Project Show & Tell
a. 5 extra credit points for volunteering, 10 points if I pick you
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:
1
v


1  2
c
Lorentz transformations:
x2   x1   (ct1 )
ct2   x1   (ct1 )
or also
x  
  
 ct  2  
  x 
 
  ct 
1
 ct   
  
 x  2  
  ct 
 
  x 
1
Worked problem #1
Four “simultaneous” events: viewed by Earth, (x, ct) = …
a. (0.5, 2)
b. (0, 2)
c. (-1, 2)
d. (-2, 2)
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Dr. Colton’s rocket comes by going 0.5 c in the positive x
direction. Where/when does he measure these events?
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 = 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
Worked problem, cont
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Some things to notice:
a. “Linear” transformation: Notice that lines transform into lines
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)
point “e” = (5.773, -1.155)
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. Causality!
Worked problem #2
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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. In Cathy’s frame: draw the world lines of
Cathy, Lee, and the right end of the
meterstick.
b. In Lee’s frame: draw the same worldlines.
Lee’s program
Velocity transformations
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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
Usetransform
this instead
book eqns
39.16 and 39.18.
Then:
to of
Cathy’s
frame
Far simpler; works every time!
a. Find Caution:
slope of terms
new line
is inverse
of )
are (which
sometimes
negative.
Result:
vJohn-Cathy
0.8 c velocity formula, eqn 39.17.)
(Don’t
need
to know =
transverse
General formula:
12   23
13 
1  12  23
“1-3” = “of object 1 with respect to object 3”
Compare to “Galilean”:
v13  v12  v23
Worked Problem #3
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HW 39-2 (the one that got canceled)
0.99687
Answers: (a) 53.0; (b) 0.083; (c) 53.0
Worked Problem #4
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Optional problem from HW 40
Worked Problem #5
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Optional problem from HW 40