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Unformatted text preview: Physics 200 Problem Set 7 Solution Quick overview: Although relativity can be a little bewildering, this problem set uses just a few ideas over and over again, namely 1. Coordinates ( x,t ) in one frame are related to coordinates ( x ,t ) in another frame by the Lorentz transformation formulas. 2. Similarly, space and time intervals (Δ x, Δ t ) in one frame are related to inter vals (Δ x , Δ t ) in another frame by the same Lorentz transformation formu las. Note that time dilation and length contraction are just special cases: it is timedilation if Δ x = 0 and length contraction if Δ t = 0. 3. The spacetime interval (Δ s ) 2 = ( c Δ t ) 2 (Δ x ) 2 between two events is the same in every frame. 4. Energy and momentum are always conserved, and we can make efficient use of this fact by writing them together in an energymomentum vector P = ( E/c, p ) with the property P 2 = m 2 c 2 . In particular, if the mass is zero then P 2 = 0. 1. The earth and sun are 8.3 lightminutes apart. Ignore their relative motion for this problem and assume they live in a single inertial frame, the EarthSun frame. Events A and B occur at t = 0 on the earth and at 2 minutes on the sun respectively. Find the time difference between the events according to an observer moving at u = 0 . 8 c from Earth to Sun. Repeat if observer is moving in the opposite direction at u = 0 . 8 c . Answer: According to the formula for a Lorentz transformation, Δ t observer = γ ‡ Δ t EarthSun u c 2 Δ x EarthSun · , γ = 1 p 1 ( u/c ) 2 . Plugging in the numbers gives (notice that the c implicit in “lightminute” cancels the extra factor of c , which is why it’s nice to measure distances in terms of the speed of light) Δ t observer = 2 min . 8(8 . 3 min) √ 1 . 8 2 = 7 . 7 min , which means that according to the observer, event B happened before event A ! If we reverse the sign of u then Δ t observer 2 = 2 min + 0 . 8(8 . 3 min) √ 1 . 8 2 = 14 min . 2. Return to the EarthSun case above. (a) What is the speed of a spacecraft that makes the trip from the Sun to the Earth in 5 minutes according to the on board clocks? (b) What is the trip time in the EarthSun frame? (c) Find the square of the spacetime interval between them in lightminutes. (You may need to come back to part (c) after I do spacetime intervals in class. Do not just jump in and use some formula. Think in terms of events, assign as many possible spacetime coordinates as you can to each event in any frame and use the LT. Measure time in minutes, distance in lightminutes. Imagine a rod going from earth to the sun, if that helps. 1 Note that the spatial coordinate difference between events in the spacecraft frame are not the same as the distance between Earth and Sun in that frame. Even preEinstein, if I sit in my car going at 60 mph, I leave New Haven at t = 0 (Event 1) and arrive at Boston at t = 2 hrs (Event 2), the two events have the same coordinate in my frame (i.e., where I am in the car), Δtwo events have the same coordinate in my frame (i....
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 Fall '08
 RAMAMURTISHANKAR
 Physics, Special Relativity, Frame, XR

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