B6 Quantitative Aspects of the Relativity of Simultaneousness
In B1 we concluded that the synchronization of a set of clocks in different inertial frames must fail. Even if the synchronization were possible for a given time point it would make little sense considering B2. It is however possible to exactly say by how much 2 clocks which are synchronized in the red system B are desynchronized from the point of view of the black system A. We will now derive this formula. It appears only in a few books concerning STR. It is however indispensable, if one really wants to put all the pieces together. This will be clear from the set of example problems presented at the end of this section.
We introduce three clocks U1, Um and U2 moving in relationship to each other like Epstein’s small fleet, that is, at constant distance from each other with velocity v in the x-direction of the black, at rest, non-prime system A. x' is the distance of neighboring clocks in the red, fast-moving, prime system B; x is the corresponding value measured in the black system. x' is larger than x.
At the exact point when Um flies past the zero point A of the black system, a flash of light is released. We call this time point 0. In the red system the two clocks U1and U2 are thereby synchronized (Um remains in the middle of U1 and U2 whether or not Lorentz contraction takes place!) However when are U1 and U2 for the black system triggered by this flash of light?
U1 is flying toward the flash; U1 will encounter the flash at time point t1 yielding
t1 • c = x – t1 • v ; and thus t1 = x / (c + v)
U2 is flying away from the flash; U2 will encounter the flash at time point t2 yielding
t2 • c = x + t2 •v ; and thus t2 = x / (c – v)
The clock in front, U2, will be started for the black system with the following delay:
t1 – t2 = ... (do the math!) ... = - 2 • v • x / (c2 – v2)
That is the time difference for the black system, which knows however that the red clocks run more slowly than its own. We discover the time difference of the red clocks only if we multiply this value by our radical:
∆t’ = (t1 – t2) • √ = ... (do the math!) ... = - 2 • x • (v / c2) / √
2 • x / √ is exactly the distance ∆x' of the clocks U1 and U2! Thus we have the quite simple result
The red clocks are synchronized in their own system and have distance ∆x' in the direction of the relative motion. However, these clocks are desynchronized in the black system by the amount ∆t'. One can write the result differently, seeing more clearly that the formula is as simple as possible:
The factor c on the left of the equals sign serves only to convert times into lengths. The de-synchronization is thus proportional to the distance between the red clocks in the direction of motion and also to the ratio v / c.
That one cannot do without this formula, if one wants to present the whole situation without contradiction, will be clarified through a careful study of the following
Sample Problem and Solution
A particle moves with v = 0.8 • c through a 12 m long pipe, which is equipped with detectors at both ends, which in turn contain clocks, allowing one to measure the transit flight time precisely. The pipe is at rest in the black system. Let the particle's rest system be the red system. We pose and then answer the following questions:
- How long does the transit flight of the particle through the pipe last for black?
- How much time elapses thereby in the red system (of the particle) from the point of view of the black?
- How long is the pipe for red?
- How long does it last for red, until the pipe has raced over the particle?
- How much time elapses from the point of view of red for this flyby on each clock of black?
- How does red explain the measured value of black??
The questions 5 and 6 are omitted in most text books, whereby they form the capstone of understanding of the STR.
- Time is distance divided by velocity: ∆t = ∆x/v = 12 m / (0.8•3•108 m/s) = 50 ns
- Because of the time dilation red will measure a shorter duration: ∆t’ = ∆t•√ = 50 ns • 0.6 = 30 ns
- Red sees the pipe as Lorentz-contracted: ∆x’ = ∆x•√ = 12 m • 0.6 = 7.2 m
- Until the 7.2 m long pipe has completed flying over red: ∆t’ = ∆x’/v = 7.2 m / (0.8•3•108 m/s) = 30 ns
(in complete agreement with black!)
- The fast-moving clocks of black tick naturally more slowly for red than they do in their own frame.
Thus the flyby in the black system as seen by red lasts only: ∆t = ∆t’•√ = 30 ns • 0.6 = 18 ns ( !! )
- Also red knows that black measured 50 ns, however red attributes it to the fact that the two clocks of black are de-synchronized by
∆t = ∆x • v / c2, which numerically constitutes exactly 12m • 0.8 / (3•108 m/s) = 32 ns. The 18 ns duration plus 32 ns desynchronization yield together the 50 ns that black measured with his two clocks!! Check that the sign is also correct.