At a given location L in K, at (coordinate) time t, the corresponding time t' in K' is read off from the clock in K' that is momentarily located at L. Since coordinate time is defined separately in each single inertial frame, it's not guaranteed that t and t' are equal, not even if K and K' were at rest relative to each other.

If the observation at L spans a duration Δt = t_{end} – t_{begin} in K, then the corresponding duration in K' is Δt' = t'_{end} – t'_{begin}. If K and K' were at rest relative to each other, it would be guaranteed that Δt = Δt' holds. Because of causality, it is true in any case that:

**(Theorem)** Δt > 0 implies Δt' > 0.

**(Proof)** Let there be a particle at L, at rest relative to K. Then, the beginning and the end (events) of the observation happen to the same particle.

We'll show now that Δt' depends solely on Δt:

**(Theorem)** If two observations, at L_{1} and L_{2} in K, beginning at t_{begin1 }and t_{begin2}, respectively, span the same duration Δt, the corresponding durations Δt_{1}' and Δt_{2}' in K' are equal too.

**(Proof)** There must exist an inertial frame K'', also moving at velocity **v** relative to K, such that for the observation at L_{2} in K, the corresponding duration Δt_{2}'' in K'' is equal to that of Δt_{1}' in K'. The rationale is that if such a K' can exist for L_{1} and t_{begin1} of K, there is no reason why a similar K'' would not exist for L_{2} and t_{begin2}. And as K' and K'' are at rest relative to each other, the durations Δt_{2}'' and Δt_{2}', and thus Δt_{1}' and Δt_{2}' too, must be equal.

This means that for any positive integer n, if an observation takes n · Δt time at L in K, it will have a corresponding duration of n · Δt' in K'. Due to continuity it follows that:

**(Theorem)** Δt' = λ · Δt, where λ > 0 is a constant.

**Notes:** (a) the value of λ can only depend on the choice of K and K', or rather only on v due to symmetry reasons, (b) in classical mechanics, λ = 1; in special relativity, we don't yet know the exact value.

Since v = v', the situation of K and K' is symmetrical, so the theorem will be valid with the very same λ if the roles of K and K' are switched.

Let A and B be two points on a straight line l in K parallel to **v**, and let Δx = x_{B} – x_{A} denote the signed distance between them. Then, at any time t in K, for the corresponding time values in K':

**(Theorem)** t'_{B} – t'_{A} = (Δx / v) · (1 / λ – λ).

**(Proof)** Let A' denote the point in K' that corresponds to A at time t_{1} in K. A' needs Δt = t_{2} – t_{1} = Δx / v time to get from A to B in K. An observer at A in K will see that a corresponding λ · Δt time has elapsed in K'. To spell it out, at time t_{2} in K, at location A, the corresponding time in K' is t'_{A} = t_{1}' + λ · Δt. On the other hand, an observer at A' in K' will see that a corresponding Δt time has elapsed in K, and thus that Δt / λ time has elapsed in K'. To spell it out, at time t_{2} in K, at location B, the corresponding time in K' is t'_{B} = t_{1}' + Δt / λ.

Therefore, unless λ = 1, the corresponding time t' in K' along l changes linearly with the x-coordinate, i.e. Δt' ~ Δx. This has a curious implication:

**(Theorem)** If λ ≠ 1, the speed of particles has a finite upper bound.

**(Proof)** Otherwise, if e.g. t'_{A} > t'_{B} holds, a fast enough particle could get from A to B within such a short time that on its arrival at B, the corresponding time in K' would still be less than t'_{A}, and that would violate causality.

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