Physics 012 1st Edition Lecture 37 Outline of Last Lecture I. Special Relativitya. Newtonian mechanics fails to describe the motion of very fast moving objects (close to the speed of light).b. Event: a physical happening that occurs at a certain place at a certain timei. An observer need three spatial and one temporal dimension (x, y, z, t)c. For special relativity, we need to observe from inertial frames of reference (observational reference frames in which Newton’s Laws of motion are valid).i. Accelerated frame of reference is not inertial.d. Postulates of Special Relativityi. Relativity postulate: the laws of physics are the same in every inertial reference frameii. Speed of light postulate: the speed of light in a vacuum “c” as measured in any inertial frame of reference always has the same value regardless of how fast the source of light is movinge. Interpretationi. Any inertial reference frame is as good as any other one. There is no absolute velocity.ii. In relativistic mechanics, there is no such thing as absolute length or an absolute time interval.iii. Two events that are observed as simultaneous in one frame of reference are, in general, not simultaneous in another frame moving relative to the first.II. Time Dilationa. Ex) person in a moving train (at velocity v) holding a flashlight jumping up and down to a height of hi. Time for light to go up and down for person on train = proper time = Δtp = 2d/cii. Time for light to go up and down for stationary person watching train = Δt= (2d)/(c√(1-[v2/c2])) = Δtp/√(1-[v2/c2])Outline of Current Lecture III. Time DilationThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. 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
View Full Document
Unlocking...