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I came across this exercise in Elementary General Relativity by Alan MacDonald:

A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. Let $ \Delta t_e $ be the time between the emission of pulses, and $ \Delta t_o $ be the time between their reception at the observer. Show that $ \Delta t_o = \Delta t_e + v\Delta t_e $.

Based on my understanding of special relativity, the space-time interval between two events as measured from two inertial frames of reference should be the same. Therefore, $$ \Delta t_e^2 = \Delta t_o^2 - \Delta x^2 $$ $$ \implies \Delta t_e^2 = \Delta t_o^2 - v^2\Delta t_o^2 $$ $$ \implies \Delta t_o = (1 - v^2)^{-1/2}\Delta t_e $$

which is not the same relation. What is wrong with my reasoning?

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If nothing specified shouldn't the pulses move with speed $c$? – Jorge Jun 15 '13 at 14:03
@Nivalth c=1 here – user997712 Jun 15 '13 at 15:33
up vote 0 down vote accepted

Your answer is right assuming $\Delta t_e$ is the interval between emission as measured by the emitting source itself. The given answer is right assuming $\Delta t_e$ is the time between emission as measured by the observer. It seems as though this problem is aiming at a lower level than your current understanding of relativity; you put too much thought into it.

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But it worked it out in another way: let $ E_1 $ and $ E_2 $ be the pulse emission events, $ O_1 $ and $ O_2 $ be the observation events. Since c = 1, by definition, the interval between $ O_1 $ and $ E_1 $ is zero (they are light-like separated). Same for $ O_2 $ and $ E_2 $. This gives $ (t_{O_1} - t_{E_1})^2 = x_1 ^2 $ and $ (t_{O_2} - t_{E_2})^2 = x_2 ^2 $. Taking square roots on both sides and subtracting the equations, the expected result appears. – user997712 Jun 15 '13 at 15:29
Yes, because in this case all your times are measured in the observer's frame. The $\Delta t_e$ you get in the above comment is not the same as the $\Delta t_e$ you used in the derivation in your question. Remember, the number used for a time depends not only on the event, but on whose watch is being used to measure the event. – Chris White Jun 15 '13 at 15:35
i have thought about this and i am still confused about the basics: How is $ t_{E_1} $, for instance, different as observed by the emitter and the observer? If the setting in the question were to be a real experiment, how would we measure $ t_{E_1} $ as measured by the two different reference frames? I am a complete newbie to relativity so any explanations would be helpful. – user997712 Jul 13 '13 at 19:13
@user997712 That might be good to ask as a separate question, with a link back to this one. – Chris White Jul 13 '13 at 20:52

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