Twin paradox

 ( Relativistic Paradoxes ) 

Shortly after departure, the traveling twin sees the stay-at-home twin with no time delay. At arrival, the image in the ship screen shows the staying twin as 1 year after launch, because radio emitted from Earth 1 year after launch gets to the other star 4 years afterwards and meets the ship there. During this leg of the trip, the traveling twin sees their own clock advance 3 years and the clock in the screen advance 1 year, so it seems to advance at the normal rate, just 20 image seconds per ship minute. This combines the effects of time dilation due to motion (by factor , five years on Earth are 3 years on ship) and the effect of increasing light-time-delay (which grows from 0 to 4 years).

Of course, the observed frequency of the transmission is also the frequency of the transmitter (a reduction in frequency; "red-shifted"). This is called the relativistic Doppler effect. The frequency of clock-ticks (or of wavefronts) which one sees from a source with rest frequency f_rest is

$f\_\backslash mathrm\{obs\}\; =\; f\_\backslash mathrm\{rest\}\backslash sqrt\{\backslash left(\{1\; -\; v/c\}\backslash right)/\backslash left(\{1\; +\; v/c\}\backslash right)\}$

The stay-at-home twin gets a slowed signal from the ship for 9 years, at a frequency the transmitter frequency. During these 9 years, the clock of the traveling twin in the screen seems to advance 3 years, so both twins see the image of their sibling aging at a rate only their own rate. Expressed in other way, they would both see the other's clock run at their own clock speed. If they factor out of the calculation the fact that the light-time delay of the transmission is increasing at a rate of 0.8 seconds per second, both can work out that the other twin is aging slower, at 60% rate.

Then the ship turns back toward home. The clock of the staying twin shows "1 year after launch" in the screen of the ship, and during the 3 years of the trip back it increases up to "10 years after launch", so the clock in the screen seems to be advancing 3 times faster than usual.

When the source is moving towards the observer, the observed frequency is higher ("blue-shifted") and given by

$f\_\backslash mathrm\{obs\}\; =\; f\_\backslash mathrm\{rest\}\backslash sqrt\{\backslash left(\{1\; +\; v/c\}\backslash right)/\backslash left(\{1\; -\; v/c\}\backslash right)\}$

As for the screen on Earth, it shows that trip back beginning 9 years after launch, and the traveling clock in the screen shows that 3 years have passed on the ship. One year later, the ship is back home and the clock shows 6 years. So, during the trip back, both twins see their sibling's clock going 3 times faster than their own. Factoring out the fact that the light-time-delay is decreasing by 0.8 seconds every second, each twin calculates that the other twin is aging at 60% their own aging speed.

[[Image:rstd4.gif|thumb|Light paths for images exchanged during tripLeft: Earth to ship. Right: Ship to Earth. Red lines indicate low frequency images are received, blue lines indicate high frequency images are received]] The x– t (space–time) diagrams at right show the paths of light signals traveling between Earth and ship (1st diagram) and between ship and Earth (2nd diagram). These signals carry the images of each twin and their age-clock to the other twin. The vertical black line is the Earth's path through spacetime and the other two sides of the triangle show the ship's path through spacetime (as in the Minkowski diagram above). As far as the sender is concerned, they transmit these at equal intervals (say, once an hour) according to their own clock; but according to the clock of the twin receiving these signals, they are not being received at equal intervals.

After the ship has reached its cruising speed of 0.8 c, each twin would see 1 second pass in the received image of the other twin for every 3 seconds of their own time. That is, each would see the image of the other's clock going slow, not just slow by the factor 0.6, but even slower because light-time-delay is increasing 0.8 seconds per second. This is shown in the figures by red light paths. At some point, the images received by each twin change so that each would see 3 seconds pass in the image for every second of their own time. That is, the received signal has been increased in frequency by the Doppler shift. These high frequency images are shown in the figures by blue light paths.

The Earth twin sees 9 years of slow (red) images of the ship twin, during which the ship twin ages (in the image) by Then, fast (blue) images are seen from Earth for the remaining 1 year until the ship returns. In the fast images, the ship twin ages by The total aging of the ship twin in the images received by Earth is , so the ship twin returns younger (6 years as opposed to 10 years on Earth).

• If they want to calculate when their twin was the age shown in the image ( i.e. how old their twin themself was then), they have to determine how far away their twin was when the signal was emitted—in other words, one has to consider simultaneity for a distant event.
• If they want to calculate how fast their twin was aging when the image was transmitted, they adjusts for the Doppler shift. For example, when receiving high frequency images (showing their twin aging rapidly) with frequency $\backslash scriptstyle\{f\_\backslash mathrm\{rest\}\backslash sqrt\{\backslash left(\{1\; +\; v/c\}\backslash right)/\backslash left(\{1\; -\; v/c\}\backslash right)\}\}$, they do not conclude that the twin was aging that rapidly when the image was generated, any more than they conclude that the siren of an ambulance is emitting the frequency they hear. They know that the Doppler effect has increased the image frequency by the factor 1 / (1 − v/ c). Therefore, one calculates that the twin was aging at the rate of

$f\_\backslash mathrm\{rest\}\backslash sqrt\{\backslash left(\{1\; +\; v/c\}\backslash right)/\backslash left(\{1\; -\; v/c\}\backslash right)\}\backslash times\; \backslash left(1\; -\; v/c\backslash right)\; =\; f\_\backslash mathrm\{rest\}\backslash sqrt\{1\; -\; v^2/c^2\}\backslash equiv\backslash epsilon\; f\_\backslash mathrm\{rest\}$

The mechanism for the advancing of the stay-at-home twin's clock is gravitational time dilation. When an observer finds that inertially moving objects are being accelerated with respect to themselves, those objects are in a gravitational field insofar as relativity is concerned. For the traveling twin at turnaround, this gravitational field fills the universe. In a weak field approximation, clocks tick at a rate of where Φ is the difference in gravitational potential. In this case, where g is the acceleration of the traveling observer during turnaround and h is the distance to the stay-at-home twin. The rocket is firing towards the stay-at-home twin, thereby placing that twin at a higher gravitational potential. Due to the large distance between the twins, the stay-at-home twin's clocks will appear to be sped up enough to account for the difference in proper times experienced by the twins. It is no accident that this speed-up is enough to account for the simultaneity shift described above. The general relativity solution for a static homogeneous gravitational field and the special relativity solution for finite acceleration produce identical results.

Other calculations have been done for the traveling twin (or for any observer who sometimes accelerates), which do not involve the equivalence principle, and which do not involve any gravitational fields. Such calculations are based only on the special theory, not the general theory, of relativity. One approach calculates surfaces of simultaneity by considering light pulses, in accordance with Hermann Bondi's idea of the k-calculus. A second approach calculates a straightforward but technically complicated integral to determine how the traveling twin measures the elapsed time on the stay-at-home clock. An outline of this second approach is given in a .

• 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 K' be associated with the rocket that makes the trip. At the departure event both clocks are set to 0.

Phase 1: Rocket (with clock K') embarks with constant proper acceleration a during a time T_a as measured by clock K until it reaches some velocity V.

Phase 2: Rocket keeps coasting at velocity V during some time T_c according to clock K.

Phase 3: Rocket fires its engines in the opposite direction of K during a time T_a 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_a according to clock K, until K' 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_c 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_a, 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 K' will be given by the integral function of coordinate time Δ t

$\backslash Delta\; \backslash tau\; =\; \backslash int\; \backslash sqrt\{\; 1\; -\; (v(t)/c)^2\; \}\; \backslash \; dt\; \backslash$

$v(t)\; =\; \backslash frac\{a\; t\}\{\; \backslash sqrt\{1+\; \backslash left(\; \backslash frac\{a\; t\}\{c\}\; \backslash right)^2\}\}.$

This integral can be calculated for the 6 phases:C. Lagoute and E. Davoust (1995) The interstellar traveler, Am. J. Phys. 63:221-227

Phase 1 $:\backslash quad\; c\; /\; a\; \backslash \; \backslash text\{arsinh\}(\; a\; \backslash \; T\_a/c\; )\backslash ,$

Phase 2 $:\backslash quad\; T\_c\; \backslash \; \backslash sqrt\{\; 1\; -\; V^2/c^2\; \}$

Phase 3 $:\backslash quad\; c\; /\; a\; \backslash \; \backslash text\{arsinh\}(\; a\; \backslash \; T\_a/c\; )\backslash ,$

Phase 4 $:\backslash quad\; c\; /\; a\; \backslash \; \backslash text\{arsinh\}(\; a\; \backslash \; T\_a/c\; )\backslash ,$

Phase 5 $:\backslash quad\; T\_c\; \backslash \; \backslash sqrt\{\; 1\; -\; V^2/c^2\; \}$

Phase 6 $:\backslash quad\; c\; /\; a\; \backslash \; \backslash text\{arsinh\}(\; a\; \backslash \; T\_a/c\; )\backslash ,$

$V\; =\; a\; \backslash \; T\_a\; /\; \backslash sqrt\{\; 1\; +\; (a\; \backslash \; T\_a/c)^2\; \}$

$a\; \backslash \; T\_a\; =\; V\; /\; \backslash sqrt\{\; 1\; -\; V^2/c^2\; \}$

So the traveling clock K' will show an elapsed time of

$\backslash Delta\; \backslash tau\; =\; 2\; T\_c\; \backslash sqrt\{\; 1\; -\; V^2/c^2\; \}\; +\; 4\; c\; /\; a\; \backslash \; \backslash text\{arsinh\}(\; a\; \backslash \; T\_a/c\; )$

$\backslash Delta\; \backslash tau\; =\; 2\; T\_c\; /\; \backslash sqrt\{\; 1\; +\; (a\; \backslash \; T\_a/c)^2\; \}\; +\; 4\; c\; /\; a\; \backslash \; \backslash text\{arsinh\}(\; a\; \backslash \; T\_a/c\; )$

$\backslash Delta\; t\; =\; 2\; T\_c\; +\; 4\; T\_a\backslash ,$

$\backslash Delta\; t\; >\; \backslash Delta\; \backslash tau\backslash ,$

$\backslash Delta\; \backslash tau\; =\; \backslash int\_0^\{\backslash Delta\; t\}\; \backslash sqrt\{\; 1\; -\; \backslash left(\backslash frac\{v(t)\}\{c\}\backslash right)^2\; \}\; \backslash \; dt,\; \backslash$

Δ τ represents the time of the non-inertial (travelling) observer K' as a function of the elapsed time Δ t of the inertial (stay-at-home) observer K for whom observer K' 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 K', where only quantities measured by K' are accessible, the following formula can be used:E. Minguzzi (2005) - Differential aging from acceleration: An explicit formula - Am. J. Phys. 73: 876-880 arXiv:physics/0411233 (Notation of source variables was adapted to match this article's.)

$\backslash Delta\; t^2\; =\; \backslash left\; \backslash ,\backslash left\backslash int^\{\backslash Delta,\; \backslash$

$\backslash begin\{align\}$

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:

$\backslash Delta\; t\; =\; \backslash frac\{1\}\{\backslash sqrt\{1-\backslash tfrac\{v^2\}\{c^2\}\}\}\; \backslash Delta\backslash tau\; .\backslash$

In the case where the accelerated observer K' departs from K with zero initial velocity, the general equation reduces to the simpler form:

$\backslash Delta\; t\; =\; \backslash int^\{\backslash Delta\backslash tau\}\_0\; e^\{\backslash pm\backslash int^\{\backslash bar\{\backslash tau\}\}\_0\; a(\backslash tau\text{'})d\; \backslash tau\text{'}\}\; \backslash ,\; d\; \backslash bar\backslash tau\; ,\; \backslash$