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If you set up a clock that sends out a light pulse every second, and move towards it at a speed of .866c, will the clock appear to run faster?

Krel
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2 Answers2

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In special relativity the rule of thumb is 

Moving Clocks run slowly.

So in your inertial reference frame. The clock is moving towards you at 0.866c and thus is running slower than a clock you keep in your frame of reference. This means the clock will pulse out the light less often.

With the light coming towards you at 1.00c and you moving towards it at 0.866c, some calculations need to be done to determine your relative velocity with respect to the light. Check out this hyperphysics tutorial on it

jkeuhlen
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  • Take note this rule applies to the timing computed after lightspeed delays have been figured in. For a fast approaching source the pulses may arrive more frequently then every second (and pulses from receding sources will come in even less frequently). Most intro texts have a section of "apparent velocity" where this is discussed. – dmckee --- ex-moderator kitten Jul 04 '14 at 16:19
  • The link to other question says the opposite. The disparity here is quite frustrating as I have read velocity is irrelevant and that only speed matters, meaning approaching a clock at relativistic speeds would make it appear slower, though I have also read that due to the relativistic doppler shift the clock would appear to tick faster. So is time running slower and the clock appears to tick faster or is time actually running faster? – Krel Jul 04 '14 at 16:25
  • @Krel Your problem here is that you are failing to distinguish between the arrival frequency of the light fronts and the frequency with which the light fronts were generated. At significant relative speeds these can differ by a lot. – dmckee --- ex-moderator kitten Jul 04 '14 at 16:35
  • @dmckee is dead on here. The clock will pulse more slowly. However, because of dilation you will encounter them at a different speed. NOTHING (not even light) can exceed the speed of light. So when you travel towards the clock at 0.866c and the light from the clock comes toward you at 1.00c there is a relativistic velocity addition problem that needs to be solve to determine how fast you are moving relative to the light. – jkeuhlen Jul 04 '14 at 16:48
  • @Krel I edited my answer hope that helps clear up your confusion. – jkeuhlen Jul 04 '14 at 16:50
  • @dmckee still confused. From the frame of reference of the observer moving towards the clock there would be no distinction between the rate at which pulses arrive and the rate at which they are sent out. C is constant even if I'm moving relative to the source, so the only way I as a moving observer could explain an increase in pulse frequency would not be that I am moving towards the wave front but rather the pulses have actually increased in frequency. Where am I going wrong here? – Krel Jul 06 '14 at 18:23
  • Assume that you have a way to know the range from which each signal comes (parallax, interferometry, working backwards from a constant blue-shift and when they pass you, whatever). Then you can work out when&where each signal was sent, and the difference in the whens will be different than the time period between the pulse arrivals (because each successive pulse is sent from closer and therefore has less travel time). So there are two difference periods: that of generation and that of reception. Jkeuhlen's answer refers to the period of generation; the answers to 19370 to that of reception. – dmckee --- ex-moderator kitten Jul 06 '14 at 18:33
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First, one must be careful to distinguish observation in Special Relativity from what I think it is you have in mind.

It's true that the clock will be observed (in the SR sense of the word) to run slow.

However, if what you're interested in is your proper time between reception of the light pulses, keep in mind that this is a different notion than observe in SR."

Draw the spacetime diagram and the answer will be plain.

The world line of the observer moving towards the clock will intersect the world lines of the light pulses from the clock. The proper time between the intersection events is straightforward to calculate.

Alfred Centauri
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