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Unformatted text preview: PHYSZ70 Homework #5 Due in recitation week of April 25, 2011 1. Some time, in the not too distant future, relativistic Purple Line trolleys will be
zipping across our campus. Consider two such trolleys as they pass by the Stamp Union.
Trolley 1 is traveling at 0.60 to the right (increasing 1:) and Trolley 2 is traveling at 0.8c
to the left as measured by someone on the steps of the Stamp Union. Each trolley is 10m
long in its rest frame (back at the Trolley round house). Define event #1 to be the front of
the trolleys pass each other, and define event #2 to be the back of the trolleys pass each
other. How long is each trolley in the Stamp Union frame? Suppose Event #1 occurs at (X=0, t=0) in the Stamp frame. When and where does
event #2 occur in the Stamp frame. Now calculate the coordinates of these events in each of the trolley frames. What is the velocity of Trolley 2 in the frame of Trolley 1? and vice versa? FD?“ .50 Em“ +‘Ct2_l_.!m..w:l\ be .éflPr‘7L9“ h 1%....
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Far" obs8W9 (3n. ecu». Mug? ﬂ+mﬁw,§ Symmfvrc , {9+ (1.1: Veloci'h; 0‘" be../.L§ #1. ..:?L.¢??'¢p. /eJ— V I: n “ rmibwﬁﬂl n" . “ 2.0.6.9... Oxa; {felon(+7 aLF 19’0le {11 ha TVDIM> EL! 70%,” E arv .ﬂ *0‘.gc_.0lbc..___._ __ —*—’ —~ :2. “4 7.20.4!95; Véfod‘h 0+ walla aw rut 11/73”? "#2. ﬂame : +0c‘i‘tée~~~% 2. (Bonus) You are going to derive the Lorentz transformation from a few simple
assumptions. Let us suppose we consider all possible linear transformations of X and t to
find x’ and t’. x’=Ax+Br
t':Ct+Dx where A, B, C, and D are constants. We restrict ourselves to linear transformations
because space and time are homogeneous. For example the transformation of the
distance between two events that are separated by 1 m should be the same whether the
two events are at x21 and x=0, or at x=51 and X=50. Thus there should be no X2 or 12 or
even higher power terms in the transformation. What are the conditions imposed on the constants A, B, C, and D by the following
requirements? An object moving with coordinate X=vt in S should be stationary in S’.
An object stationary in S should be moving with X’=—vt’in 8’. By now you should know that B=VA, and B=—vC, so A=C. Now we need two more conditions to fix the values of the constants. These relate to the
speed of light being the same in both frames. That is: when x=ct, X’:ct’, and when x=~ct, then x’=—ct’. Show that these give A=C='\{, B=—vy, and D=—a{v/cz. x , 4d. A" [Himkéikkﬁu 9% —— 4449M] . .. ,, . mm war] AT «tiWm W29???) 3 , 9 ﬂ”; —'"‘ ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.
 Fall '08
 drake

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