## B2 Secundo: Fast Clocks Tick More Slowly

It gets even worse: not only does the concept of simultaneousness become meaningless when we observe events from two separate inertial frames which are moving with respect to each other, but the river of time itself flows at different speeds! For the derivation of this difference we need only the Pythagorean Theorem.

We assume the constancy and universality of c and consider “light clocks” with the following design: In a pipe a flash of light travels from the bottom up to a mirror at the top, where it is reflected (“tick”) back to the bottom. There it is sensed by a photoelectric cell (“tock”) releasing a new flash and incrementing a counter by 2. The counter can be read at any time. Take a moment and consider why the clock should be 30 cm in length and why the counter is increased each time by 2. Imagine that we have several such clocks and that some are set up and synchronized along the x axis of our coordinate system at known distances. A further identically constructed clock moves with speed v past these “resting” clocks (only this “fast” clock is shown below at three positions in the diagram!). How much time elapses in the resting system, during the time the “fast” clock makes a single “tick”? |

The distance light travels in the moving clock (call it the prime system with measured time t’) is 30 cm or, in general, c·∆t'. But what is the distance this light travels as seen from the resting system (call it the non-prime system with measured time t) and in relation to which the moving clock travels along the x-axis with velocity v? Given the constancy of the speed of light this will be, of course, c·∆t. These two distances are however not equal and thus the time intervals, ∆t and ∆t', must differ! The Pythagorean Theorem provides us with the relationship between these two values:

Obviously more time elapsed in the resting, non-prime-system, than in the moving, prime-system, since the light travelled a longer distance in that system. Thus:

(c·∆t)

^{2}= (v·∆t)

^{2}+ (c·∆t’)

^{2 }; c

^{2}·(∆t)

^{2}= v

^{2}·(∆t)

^{2}+ c

^{2}·(∆t’)

^{2 }; (∆t’)

^{2}= (∆t)

^{2}·(1- v

^{2}/c

^{2}) and we get

Thus moving clocks tick more slowly, compared to a set of clocks at rest. One calls this effect ‘time dilation’. Search the Internet with keywords ‘light clock’ or ‘time dilation’ and you will find innumerable nice animations concerning this consequence of holding to **M** and **R**.

In the prime system we measure the *proper time* of a “tick”. A proper time interval is measured by a single clock between two events that occur at the same place as the clock. In the non-prime system we need two synchronized clocks to measure the time interval corresponding to a single "tick" of the moving clock.

The proper time interval is always the *longest* time interval measured by a single clock between two events, that mark the beginning and the end of a given process. Section **H4** is entitled with "The Principle of Maximum Proper Time (Eigenzeit)". From the point of view of the prime system, the non-prime clocks are moving, and all of them tick more slowly than its own clock which is at rest. So each of those clocks would measure a shorter time interval for the same process. How do we explain however that we just measured a longer duration in the non-prime system? Does this not contradict our principle of maximum proper time? Perhaps you already see how to resolve this apparent contradiction: The important fact is that in order to make the measurement in the non-prime system, we need at least two distant clocks… We will clarify this point completely towards the end of section **B6**!

Thus little remains of Newton's absolute time! It makes sense to draw a diagram showing the time indicated by identical, perfect (!) clocks in relationship to the time indicated on my perfect clock:

black: my clock and all other perfect clocks synchronized with it in my inertial frame

green: a correct, but badly synchronized clock at rest in my inertial frame

red: “fast” clocks, which can be synchronized in their own inertial frame

Try adding to the diagram an imperfect, but more or less well synchronized clock at rest. Also add a second clock at rest, which is synchronized at time point 10, but then runs fast!

We have now described the relativity of objective time measurement. That subjective time experience is ‘malleable’ is well-known. Salvador Dali’s ‘The Persistence of Memory’ fits both perspectives quite well:

Einstein once illustrated the subjectivity of time experience in the following way:

An hour sitting with a pretty girl on a park bench passes like a minute, but a minute sitting on a hot stove seems like an hour. [17-247]