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Unformatted text preview: PHY 251 Fall 2009: homework problem set 1, due PHY 251 drop box in room A129 by noon on Friday, Sep. 11. Some general comments: you’ll see that I always work out an algebraic solution first and then plug in the numbers at the end. Also, if you put mks (meter · kilogram · second) units in, you get an answer in mks units out. Therefore I tend to be lazy about writing the units in the numerical calculations. 1. Show that if one uses the Galilean relativity transformation of x 2 = x 1 − vt 1 y 1 = y 2 z 1 = z 2 t 1 = t 2 (1) and the relationships for expansion of light spheres of x 2 1 + y 2 1 + z 2 1 − c 2 t 2 1 = 0 (2) x 2 2 + y 2 2 + z 2 2 − c 2 t 2 2 = 0 . (3) that you get a nongeneral and inconsistentwithclassicalphysics result. Answer: Let’s do the substitutions in the second equation: ( x 1 − vt 1 ) 2 + y 2 1 + z 2 1 − c 2 t 2 1 = 0 x 2 1 + y 2 1 + z 2 1 − c 2 t 2 1 + v 2 t 2 1 − 2 vx 1 t 1 = 0 But x 2 1 + y 2 1 + z 2 1 − c 2 t 2 1 = 0 so v 2 t 2 1 − 2 vx 1 t 1 = 0 giving v = 2 x 1 t 1 which implies that Galilean relativity works only at a particular relationship between distance and time which differs by a factor of 2 from the usual v = Δ x/ Δ t . 2. Show how classical relativity affects speed in an airplane flight with either headwinds and tailwinds, or crosswinds. Let’s say that the wind is blowing east west at a velocity of v = 10 m/s, and that you’re in an airplane that travels at a speed of c = 250 m/s. Consider a round trip journey where you fly 300 km out and back as measured on the ground, going east then west. Then consider the case where you go straight south then north as measured from the ground (what compass heading should you fly the plane at to accomplish this?). What’s the time difference between the two trips? Work out an algebraic answer approximated to lowest order in β = v/c before you plug in numbers. Answer: Here’s a diagram: Wind speed v cv c+v L L c c v v 1 When the airplane is flying against and with the wind, its net travel time t = distance/velocity is t 1 = L c − v + L c + v = L c parenleftBigg 1 1 − β + 1 1 + β parenrightBigg with β ≡ v/c . Now for small x the binomial expansion says (1 + x ) r ≃ 1 + rx + 1 2 r ( r − 1) x 2 + . . . so we have t 1 = L c parenleftBig (1 − β ) − 1 + (1 + β ) − 1 parenrightBig ≃ L c parenleftbigg 1 + ( − 1)( − β ) + 1 2 ( − 1)( − 1 − 1)( − β ) 2 + . . . + 1 + ( − 1)( β ) + 1 2 ( − 1)( − 1 − 1)( β ) 2 + . . . parenrightbigg ≃ L c parenleftBig 1 + β + β 2 + . . . + 1 − β + β 2 + . . . parenrightBig = 2 L c parenleftBig 1 + β 2 parenrightBig . When the plane is traveling across the wind, it has to fly along a compass bearing of tan θ = v/c (or θ = tan − 1 (10 / 250) = 2 . 3 ◦ in this case) to travel due north or south as seen on ground. Its speed c ′ along the straight line distance from start to destination is found from c 2 = c ′ 2 + v 2 which gives c ′ = √ c 2 − v 2 = c radicalBig...
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This note was uploaded on 05/28/2011 for the course PHY 251 taught by Professor Rijssenbeek during the Fall '01 term at SUNY Stony Brook.
 Fall '01
 Rijssenbeek
 Physics, Work

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