Editing Twin paradox (section)

==Difference in elapsed times: how to calculate it from the ship==
[[Image:TwinParadoxProperAcceleration2.svg|thumb|right|250px|Twin paradox employing a rocket following an acceleration profile in terms of proper time τ and by setting c=1: Phase 1 (a=0.6, τ=2); Phase 2 (a=0, τ=2); Phase 3-4 (a=-0.6, 2τ=4); Phase 5 (a=0, τ=2); Phase 6 (a=0.6, τ=2). The twins meet at T=17.3 and τ=12. {{br}}This is a different voyage than the one shown above, as both schemes take the same assumed total ''point-of-view time'': T=12 (stay-at-home), resp τ=12 (ship), so the results of the calculated ''other-one's times'' must be different: τ=9.33 (ship), resp T=17.3 (stay at home).]]

In the standard proper time formula
:<math>\Delta \tau = \int_0^{\Delta t} \sqrt{ 1 - \left(\frac{v(t)}{c}\right)^2 } \ dt, \ </math>

Δ''τ'' represents the time of the non-inertial (travelling) observer <var>K'</var> as a function of the elapsed time Δ''t'' of the inertial (stay-at-home) observer ''K'' for whom observer <var>K'</var> has velocity ''v''(''t'') at time ''t''.

To calculate the elapsed time Δ''t'' of the inertial observer ''K'' as a function of the elapsed time Δ''τ'' of the non-inertial observer <var>K'</var>, where only quantities measured by <var>K'</var> are accessible, the following formula can be used:<ref name='Minguzzi'>E. Minguzzi (2005) - Differential aging from acceleration: An explicit formula - ''Am. J. Phys.'' '''73''': 876-880 [https://arxiv.org/abs/physics/0411233 arXiv:physics/0411233] (Notation of source variables was adapted to match this article's.)</ref>
:<math>\Delta t^2 = \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d \tau'} \, d \bar\tau\right] \,\left[\int^{\Delta \tau}_0 e^{-\int^{\bar\tau}_0 a(\tau')d \tau'} \, d \bar\tau \right], \ </math>
where ''a(τ)'' is the [[proper acceleration]] of the non-inertial observer <var>K'</var> as measured by himself (for instance with an accelerometer) during the whole round-trip. The [[Cauchy–Schwarz inequality]] can be used to show that the inequality {{nowrap|Δ''t'' &gt; Δ''τ''}} follows from the previous expression:
:<math>\begin{align}
\Delta t^2 & = \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d\tau'} \, d \bar\tau\right] \,\left[\int^{\Delta \tau}_0 e^{-\int^{\bar\tau}_0 a(\tau')d \tau'} \, d \bar\tau \right] \\
& > \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d\tau'} \, e^{-\int^{\bar\tau}_0 a(\tau') \, d \tau'} \, d \bar\tau \right]^2 = \left[ \int^{\Delta\tau}_0 d \bar\tau \right]^2 = \Delta \tau^2.
\end{align}</math>

Using the [[Dirac delta function]] to model the infinite acceleration phase in the standard case of the traveller having constant speed ''v'' during the outbound and the inbound trip, the formula produces the known result:
:<math>\Delta t = \frac{1}{\sqrt{1-\tfrac{v^2}{c^2}}} \Delta\tau .\ </math>

In the case where the accelerated observer <var>K'</var> departs from ''K'' with zero initial velocity, the general equation reduces to the simpler form:
:<math>\Delta t = \int^{\Delta\tau}_0 e^{\pm\int^{\bar{\tau}}_0 a(\tau')d \tau'} \, d \bar\tau , \ </math>
which, in the ''smooth'' version of the twin paradox where the traveller has constant proper acceleration phases, successively given by ''a'', −''a'', −''a'', ''a'', results in<ref name='Minguzzi'/>
:<math>\Delta t = \tfrac{4}{a} \sinh( \tfrac{a}{4} \Delta\tau) \ </math>
where the convention ''c'' = 1 is used, in accordance with the above expression with acceleration phases {{nowrap|''T''<sub>a</sub> {{=}} Δ''t''/4}} and inertial (coasting) phases {{nowrap|''T''<sub>c</sub> {{=}} 0.}}