Question

Q2

(2) Problem 12.6 from Griffiths, Introduction to Electrodynamics, 3rd ed Problem 12.6 from Griffiths, Introduction to Electrodynamics, 4th ed Every two years, more or less, the New York Times publishes an article in which someone claims to have found an object traveling faster than the speed of light. Many of these reports result from a failure to distinguish what is seen from what is observed that is, from a failure to account for light travel time. Here's an example: A star is traveling with speed v at an angle θ to the line of sight (see figure below). What is its apparent speed across the sky? (Suppose the light signal from b reaches the earth at a time Δι after the signal from a, and the star has meanwhile advanced a distance As across the celestial sphere; by apparent speed" I mean Δs/M.) What angle θ gives the maximum apparent speed? Show that the apparent speed can be much greater than c, even if v itself is less than To earth

Answer 1

Let d_a is distance from a to earth and d_b is distance from point b to earth. Let time signal leaves a at time t_a' and arrives at earth at time t_a then t_a = t_a' + d_a/c and Let time signal leaves b at time t'_b. then it arrives at earth t_b than t_b = t_b + d_b/c than delta t (time interval in which light signal from b reach after the signal reach from a) is delta t = t_b - t_a = t'_b - t'_a + (d_b - d_a/c) = delta t' + (-v cos theta middot delta t'/c) delta t = delta t' [1 - v/c cos theta] than From figure delta s = (v sin theta) delta t' = v sin theta delta t/[1 - v cos theta/c] from (1) than by delta s = u middot delta t = > u = v sin theta/1 - v cos theta/c u is apperent speed then du/d theta = v cos theta [1 -
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