In a translation of Einstein's article about special relativity, we can read,

If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighbourhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an " A time " and a " B time." We have not defined a common " time " for A and B, for the latter cannot be defined at all unless we establish by definition that the " time " required by light to travel from A to B equals the " time" it requires to travel from B to A.

I read this whole paragraph and I still don't know what "common time" is. What is a clear definition of common time in special relativity?

  • $\begingroup$ So what is your question? $\endgroup$ – Dale Mar 10 at 0:14
  • $\begingroup$ Like the ether? $\endgroup$ – Cinaed Simson Mar 11 at 2:35

Two (or more) spatially separated clocks show the “same time” or “common time” if these clocks have previously been synchronized. Einstein in his article gives the definition of common time (synchronicity of spatially separated clocks) in the special theory of relativity. This definition is known as Einstein synchrony convention (Einstein synchronization).

Einstein indicated that the question of whether or not two spatially separated events were simultaneous did not necessarily have a definite answer, but instead depended on the adoption of a convention for its resolution.

It is very convenient to synchronize clocks at points A and B by means of light, since nothing can move faster that light. For example, at certain moment (when clock A shows 0) someone sends light signal towards clock B. When this light pulse reaches clock B we can adjust clock B, but to do that we need to know how much time it took for the light pulse to cover the distance AB.

Hence, we need to know one – way speed of light from A to B. However, the "one-way" speed of light, from a source to a detector, cannot be measured independently of a convention as to how to synchronize the clocks at the source and the detector, so there is a circular reasoning. What can however be experimentally measured is the round-trip speed (or "two-way" speed of light ) from the source to the detector and back again. Measured round trip speed of light is always equal precisely to constant c.

Albert Einstein chose a synchronization convention (see Einstein synchronization) that made the one-way speed equal to the two-way speed. The constancy of the one-way speed in any given inertial frame is the basis of his special theory of relativity.

According to Albert Einstein's prescription from 1905, a light signal is sent at time $\tau_1$ from clock 1 to clock 2 and immediately back, e.g. by means of a mirror. Its arrival time back at clock 1 is $\tau_2$. This synchronisation convention sets clock 2 so that the time $\tau_3$ of signal reflection is defined to be

$$\tau_3 = \tau_1 + \tfrac{1}{2}(\tau_2 - \tau_1) = \tfrac{1}{2}(\tau_1 + \tau_2).$$

Einstein synchronization is only a special case of a more broader synchronization scheme (Non – standard or Reichenbach‘s) , which leaves the two-way speed of light invariant, but allows for different one-way speeds. For example, speed of light from point A to point B can be infinitely large and from point B to point A infinitely close to c/2.

One can imagine reference frame of an observer as a lattice of synchronized clocks (see here fig. 1-1) ; these clocks have previously been synchronized. Hence, in a reference frame of this observer all these clocks show „the same“ time, that is related to the certain (standard or non-standard) synchronization scheme.

Another article gives some more information .

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  • $\begingroup$ Thank you so much for this great answer and all these references. $\endgroup$ – Hilbert Mar 12 at 19:02

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