Physics 012 1st Edition Lecture 38 Outline of Last Lecture I. Time Dilationa. Problem: A pendulum has a period of 3 seconds in the reference frame of the pendulum. What is the period of the pendulum within the reference frame of an observer moving at a constant velocity of 0.95 c?i. γ = 1/√(1-[(0.95c)2/c2]) = 1/√(1-[0.952]) = 3.2ii. Δt = Δt0γ = (3)(3.2) = 9.6 secondsb. Problem: In the frame of reference of a stationary person on Earth, the distance between Earth and Alpha Centauri is 4.3 light years. A rocket leaves Alpha Centauri at 0.95 c heading toward Earth. How much will passengers age by the time they reach Earth?i. Passengers on ship measure proper time, Earth observers measure dilated time.ii. Δt = d/v = 4.3 lyr/ 0.95c = 4.3(1yr)(c)/0.95c = 4.53 yrsiii. Δt0 = Δt/γ = Δt √(1-[v2/c2]) = 4.53 √(1-[0.952]) = 1.4 yrsOutline of Current Lecture II. Length Contractiona. Proper length (L0): distance between two points measured by an observer at rest relative to these two pointsi. L = L0/γii. Ex) Pole in a barn paradox: A runner is moving at 0.75c carrying a pole 15m long toward a barn that is 10 m wide. A ground observer (stationary relative to the barn) can open and close both doors simultaneously via remote control. Will the pole fit in the barn?1. For ground observer: Lpole = L0/γ = 15 m √(1-[(0.75c)2/c2]) = 9.9 ma. Pole will fit in barn (9.9 m < 15 m).2. For runner: Lbarn = L0/γ = 10 m √(1-[(0.75c)2/c2]) = 6.6 ma. Pole will not fit in barn (10m > 6.6m).3. Simultaneity is not constant, but is relative. Even though the barn doors close simultaneously for the stationary observer, they do not close simultaneously for the runner. For the runner, there is always one door open where the pole can stick out of.b. Relativistic Addition of Velocitiesi. vAC = (vAB + vBC)/(1+[vABvBC/c2]These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a
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