Editing Twin paradox (section)

==Difference in elapsed time as a result of differences in twins' spacetime paths==
{{Further|Hyperbolic motion (relativity)}}

[[Image:TwinParadoxProperAcceleration.svg|thumb|right|250px|Twin paradox employing a rocket following an acceleration profile in terms of coordinate time T and by setting c=1: Phase 1 (a=0.6, T=2); Phase 2 (a=0, T=2); Phase 3-4 (a=-0.6, 2T=4); Phase 5 (a=0, T=2); Phase 6 (a=0.6, T=2). The twins meet at T=12 and τ=9.33. The blue numbers indicate the coordinate time T in the inertial frame of the stay-at-home-twin, the red numbers the proper time τ of the rocket-twin, and "a" is the proper acceleration. The thin red lines represent lines of simultaneity in terms of the different momentary inertial frames of the rocket-twin. The points marked by blue numbers 2, 4, 8 and 10 indicate the times when the acceleration changes direction.]]

The following paragraph shows several things:
*how to employ a precise mathematical approach in calculating the differences in the elapsed time
*how to prove exactly the dependency of the elapsed time on the different paths taken through spacetime by the twins
*how to quantify the differences in elapsed time
*how to calculate proper time as a function (integral) of [[coordinate time]]

Let clock ''K'' be associated with the "stay at home twin".
Let clock <var>K'</var> be associated with the rocket that makes the trip.
At the departure event both clocks are set to 0.

:Phase 1: Rocket (with clock <var>K'</var>) embarks with constant [[proper acceleration]] ''a'' during a time ''T''<sub>a</sub> as measured by clock ''K'' until it reaches some velocity ''V''.
:Phase 2: Rocket keeps coasting at velocity ''V'' during some time ''T''<sub>c</sub> according to clock ''K''.
:Phase 3: Rocket fires its engines in the opposite direction of ''K'' during a time ''T''<sub>a</sub> according to clock ''K'' until it is at rest with respect to clock ''K''. The constant proper acceleration has the value −''a'', in other words the rocket is ''decelerating''.
:Phase 4: Rocket keeps firing its engines in the opposite direction of ''K'', during the same time ''T''<sub>a</sub> according to clock ''K'', until <var>K'</var> regains the same speed ''V'' with respect to ''K'', but now towards ''K'' (with velocity −''V'').
:Phase 5: Rocket keeps coasting towards ''K'' at speed ''V'' during the same time ''T''<sub>c</sub> according to clock ''K''.
:Phase 6: Rocket again fires its engines in the direction of ''K'', so it decelerates with a constant proper acceleration ''a'' during a time ''T''<sub>a</sub>, still according to clock ''K'', until both clocks reunite.

Knowing that the clock ''K'' remains inertial (stationary), the total accumulated [[proper time]] Δ''τ'' of clock <var>K'</var> will be given by the integral function of [[coordinate time]] Δ''t''
:<math>\Delta \tau = \int \sqrt{ 1 - (v(t)/c)^2 } \ dt \ </math>
where ''v''(''t'') is the ''coordinate velocity'' of clock <var>K'</var> as a function of ''t'' according to clock ''K'', and, e.g. during phase 1, given by
:<math>v(t) = \frac{a t}{ \sqrt{1+  \left( \frac{a t}{c} \right)^2}}.</math>

This integral can be calculated for the 6 phases:<ref>C. Lagoute and E. Davoust (1995) The interstellar traveler, ''Am. J. Phys.'' '''63''':221-227</ref>
:Phase 1 <math>:\quad c / a \ \text{arsinh}( a \ T_a/c )\,</math>
:Phase 2 <math>:\quad T_c \ \sqrt{ 1 - V^2/c^2 }</math>
:Phase 3 <math>:\quad c / a \ \text{arsinh}( a \ T_a/c )\,</math>
:Phase 4 <math>:\quad c / a \ \text{arsinh}( a \ T_a/c )\,</math>
:Phase 5 <math>:\quad T_c \ \sqrt{ 1 - V^2/c^2 }</math>
:Phase 6 <math>:\quad c / a \ \text{arsinh}( a \ T_a/c )\,</math>

where ''a'' is the proper acceleration, felt by clock <var>K'</var> during the acceleration phase(s) and where the following relations hold between ''V'', ''a'' and ''T''<sub>a</sub>:
:<math>V = a \ T_a / \sqrt{ 1 + (a \ T_a/c)^2 }</math>
:<math>a \ T_a = V / \sqrt{ 1 - V^2/c^2 }</math>

So the traveling clock <var>K'</var> will show an elapsed time of
:<math>\Delta \tau = 2 T_c \sqrt{ 1 - V^2/c^2 } + 4 c / a \ \text{arsinh}( a \ T_a/c )</math>
which can be expressed as
:<math>\Delta \tau = 2 T_c / \sqrt{ 1 + (a \ T_a/c)^2 } + 4 c / a \ \text{arsinh}( a \ T_a/c )</math>

whereas the stationary clock ''K'' shows an elapsed time of
:<math>\Delta t = 2 T_c + 4 T_a\,</math>
which is, for every possible value of ''a'', ''T''<sub>a</sub>, ''T''<sub>c</sub> and ''V'', larger than the reading of clock <var>K'</var>:
:<math>\Delta t > \Delta \tau\,</math>