A light clock consists of two parallel mirrors a distance apart
PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Special relativity: spacetime and kinematics · Special relativity: spacetime and kinematics
Relativity and Gravitation, Ch The postulates and the Lorentz transformation
Problem
A light clock consists of two parallel mirrors a distance d=1.20m apart, with a short light pulse bouncing between them; one tick is one round trip. The mirrors are mounted so that the line joining them is perpendicular to the direction of the clock's motion, and the separation perpendicular to the motion is the same in both frames. The clock is at rest in frame S′, which moves at speed v along the +x axis of frame S. Both frames are inertial and the light travels in vacuum at c=2.998×108ms−1.
((a)) Using Pythagoras' theorem on the zigzag light path, derive an expression for the duration T of one tick measured in S in terms of d, c and v, and show that T=γ(2d/c) with γ=(1−v2/c2)−1/2.
((b)) For v=0.800c, evaluate the tick measured in the clock's own frame S′, the tick measured in S, and the total length of the light path in one tick as measured in S.
((c)) The same clock is carried by a satellite whose speed in S is 7.60kms−1; over one tick, treat its motion as uniform and ignore gravity. Using the binomial expansion to first order in v2/c2, obtain an expression for T−T0, where T0 is the tick in the clock's frame, evaluate it, and say what this tells you about the Galilean assumption t′=t at everyday speeds.
((d)) A body moves at velocity u′ along the x′ axis of S′, and its velocity in S is u=(u′+v)/(1+u′v/c2). Show that 1−c2u2=(1+u′v/c2)2(1−u′2/c2)(1−v2/c2), and deduce that ∣u∣<c whenever ∣u′∣<c and ∣v∣<c. What does the identity give when u′=c?