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If an observer approaches a clock at a significant fraction of the speed of light, would they see the clock's hands moving at a faster or slower than usual rate?

I figure there are two competing effects at play - time dilation and diminishing distance.

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3 Answers 3

up vote 5 down vote accepted

You can see this as an example of the relativistic Doppler shift (for equations, see eg: http://en.wikipedia.org/wiki/Relativistic_Doppler_effect ).

The hands of the clock are moving with some angular frequency and you are moving with a velocity v towards the clock. It follows that the frequency you are seeing will be higher, thus the clocks hands will move faster.

Conceptually this makes sense. Suppose a picture of the clock is emitted each second. Since you're moving towards the clock, you will pick up one of those pictures more often than once per second, thus making the clock seem to run faster.

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Here's a simple animation that shows both time dilation and the doppler effect: tinyurl.com/lo34vat –  John O Oct 28 '14 at 0:45

Time dilation says moving clocks run slow, therefore, in the reference frame of the observer, the clock (that is moving toward him/her) appears to run more slowly than his/her own.

I don't think the doppler effect is important here, because it is a traditionally longitudinal effect (and the so-called relativistic transverse doppler effect is simply time dilation).

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The rate the clock ticks in the observer's reference frame is indeed slowed, but that's not the same as the rate the observer sees visually if they are getting light-signals from the clock, because the clock's distance is changing and thus the time delay between the signals being sent and received is different for each tick. Only after compensating in some way for signal delays can the observer determine the rate the clock was ticking in their own frame. –  Hypnosifl Oct 28 '14 at 0:59
Yes this is true, the observer also has to account for the time delay between signals sent and received due to the changing distance between the observer and sender. –  kotozna Oct 28 '14 at 7:54

Yes, you're right, there are two competing effects at play. The relativistic doppler formula describes how frequencies seen at a detector change with the speed and angle-of-approach of the emitter of the signal. In this case, the frequency is the ticking of the clock.

According to the motion of the emitter relative to the detector, that frequency can increase, decrease, or, stay the same.

Reference: link

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