# Is time dilation due to the travel time of light?

I'm trying to think about special relativity without "spoiling" it by looking up the answer; I hope someone can offer some insight - or at least tell me I'm wrong.

Suppose I have an ordinary clock in front of me and I push it back with my hands. The force applied to the clock causes it to retreat away from me and after the push, it will travel away with uniform velocity. Suppose further, I can always see the clock clearly no matter how far away it is. Since the speed of light is constant, the light coming from the clock must travel a longer distance to reach my eye as it moves away. This would make time appear to slow down? If, on the other hand, the clock is moving towards me, the distance the light must travel to reach my eye becomes shorter and shorter, thus time would appear to speed up?

• The speed of light delay before you see things that happen far away is real and part of the description of how things look, but it is separate from relativistic time dilation. Sep 9, 2014 at 13:51
• Your intuition is about the doppler shifting of the clock ticks is incorrect. I believe that you think the perceived interval between ticks will increase as the clock moves further away, but it won't. The interval will increase as the velocity increases, but once the final velocity is reached, the interval will remain constant. Sep 9, 2014 at 14:24
• @mbeckish: Not true. If the clock and the observer were stationary wrt. each other than the intervals would remain constant (yet the time the observer would see would be shifted). However, if the clock is moving away, each consecutive light signal will have to travel longer than the previous one, which will produce apparent dilatation. Sep 9, 2014 at 14:30
• @brightmagus - Assuming 1 tick per second (in clock's reference frame), between each click the clock moves v * 1 second further away than where it was at the last click. So, the time difference between the two clicks is (1 second + v/c), which is constant. Sep 9, 2014 at 14:36
• @brightmagus - According to your own example, the clock ticks are observed at 2s, 4s, 6s, etc. So the time interval between each observed tick is a constant 2 seconds, no matter how far away the clock moves. Everyone agrees that the clock appears to be ticking slower than 1 tick per second. What I was saying is that it doesn't slow down more and more as the distance increases - the time between ticks is constant when the velocity is constant. Sep 9, 2014 at 15:12

Analyzing one moving clock from the perspective of one stationary person will be inadequate to derive special relativity from. With just that set-up, you aren't actually using the key fact that the speed of light is the same for all observers – all you're actually using is just the fact that the speed of light is finite. With just taking into account that the speed of light is finite, all you'll arrive at is the non-relativistic Doppler effect, which is different from time dilation.

• As an additional point: in a special relativistic treatment of the setup Michael Lee describes, both the Doppler effect and time dilation will influence the rate at which you see the clock tick. See perhaps this question: physics.stackexchange.com/q/99175. Sep 9, 2014 at 13:31
• @RedAct: "Analyzing one moving clock from the perspective of one stationary person will be inadequate to derive special relativity from." Perhaps I misunderstand you, but what about Einstein's train and the observer on the embankment? Sep 10, 2014 at 7:12
• @brightmagus you only learn from that when you consider what happens from the perspective of both the moving train and the stationary observer. The key input here is that light moves at speed 1 according to both observers. Sep 10, 2014 at 9:29
• @Holographer: Yes and no. From the perspectives of two frames of reference, always the other one is dilated. And you are right, it is really telling ... Sep 10, 2014 at 10:06
• I believe Tamm and Frank's explanation for Cherenkov radiation requires special relativity. Sep 11, 2014 at 14:13

Relativity is not needed: If you replace the clock with a strong laser which fires a ray of light every second with an atomic clock, you know that the ticks will not slow down because every tick will be followed by another after a second from the laser's perspecive . But how can you arrange that with the fact that the light move further and further away ?

Simple: As the clock is moving away from you with constant speed, every interval the distance will increase by an amount. This additional amount needs to be traversed by the light ray. Lets say the clock moves with 1 m/s. Then your measured time interval between the ticks increases from 1s to 1+(1/300 000 000) s, which is imperceptible. But this time accumulates and after 300 000 000 ticks, you have finally reached a 1s delay which is identical with the distance the clock is now away from you: 300 000 000 m.

So your measured time between the ticks does not slow down with distance, but increases from the start of releasing the clock and remains constant then.

• The retardation of the signal (the technical word for what you describe here) is separate from time dilation and is not enough to explain what would be observed at high separation velocities. Oct 25, 2014 at 5:40
• @dmckee The OT has a very specific problem with a thought experiment which could be effortlessly explained with classical physics and needs no special relativity. He did not ask what would be observed at high separation velocities to keep the problem simple. But for high separation velocities, the effect should be in fact identical: Because the observer sees the clock (light speed !), time dilation (atomic clock on laser slows down) and lorentz contraction (distance laser to observer) cancel each other out. This is to be expected by the principle of relativity. Oct 28, 2014 at 23:06

In special relativity there is a distinction between 'experiencing events' and the concept of an observer:

Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from

In particular the observed clock ticks at the same rate, whether the clock moves away or towards you, since you don't measure the ticks from your point of view, but rather as they appear relative to imaginary stationary clocks that you placed in advance in the moving clock's path.

Einstein explains this beautifully in his popular book Relativity; the special and general theory

And Feynman also has a great exposition in http://www.feynmanlectures.caltech.edu/I_15.html

No matter in which direction the clock moves (away or towards you) , the time will slow down in the clock. Time will always dilate as the clock moves faster and faster, but will be apparent to a human eye only once it reaches speeds close to the light speed. This is not taking into effect the doppler shift which is merely the increase/decrease in the frequency of light.

• The part about the Doppler effect is wrong. The Doppler effect influences observation of any time intervals (hence frequency too) - in this case it will influence both the frequency of the light and the frequency (accordingly time intervals too) of the clock ticks. --- It is nicely illustrated by this animated picture: en.wikipedia.org/wiki/Doppler_effect#mediaviewer/… As far as the speed of propagation of (information about) some events is finite the Doppler effect will show up. Sep 10, 2014 at 7:24

A photon is fired across the width of a space ship a distance of one unit “ud”. The space ships velocity v relative to a fixed observer is 0.8*c The observer sees the photon travel a distance of cT and knowing the width to be one unit of distance calculates that 1ud = T*(c^2-v^2)^0.5 1ud = Tc*(1-(v/c)^2)^0.5 1ud = 0.6 Tc A photon is now fired forward in the space ship and in time T the observer sees it travel a distance Tc. The space ship has moved forward a distance of Tv so the photon is seen to be in a position fwd of its origin by Tc - Tv 1ud/0.6 – 0.8ud/0.6 = 1/3 ud As the laterally moving photon travels a unit of distance, it appears that the longitudinal dimensions of the space craft has been reduced to one third of its length. Now a photon is fired aft and the observer sees it travel a distance Tc and relative to the space ship a distance of Tc + Tv 1ud/0.6 + 0.8ud/0.6 = 3ud As the laterally moving photon travels a unit of distance, it appears that the longitudinal dimensions of the space craft has been increased to three times its length. There is another way of looking at this issue. The difference in speed between the space ship and a photon travelling in the same direction (but not inside the ship) is 0.2*c. Our observer sees a photon inside the ship moving forward and says its speed relative to his position is c. The ship has a set of longitudinal markers unit distance apart. Our observer knows that the photons apparent speed relative to the ship is 0.2*c but a photon within the ship is travelling at c relative to the ship . In order to rationalize the observations t he observer concludes that the markers are 0.2 units apart from his perspective. Now there is photon travelling in the opposite direction to the ship and the relative speed as seen by the observer is 1.8*c and therefore the markers are 1.8 units apart