In physics, the twin paradox is a thought experiment in special relativity, in which a twin who makes a journey into space in a high-speed rocket will return home to find he has aged less than his identical twin who stayed on Earth. This result appears puzzling on this basis: the laws of physics should exhibit symmetry. Each twin sees the other twin as traveling; so each should see the other aging more slowly. How can an absolute effect (one twin really does age less) result from a relative motion? Hence it is called a "paradox".

Starting with Paul Langevin in 1911, there have been numerous explanations of this paradox, all based upon there being no contradiction because there is no symmetry: only one twin has undergone acceleration and deceleration, thus differentiating the two cases. One version of the asymmetry argument made by Max von Laue in 1913 is that the traveling twin uses two inertial frames: one on the way up and the other on the way down. So switching frames is the cause of the difference, not acceleration per se.[1]

Other explanations account for the effects of acceleration. Einstein, Born and Møller invoked gravitational time dilation to explain the aging based upon the effects of acceleration.[2] Both gravitational time dilation and special relativity are needed to explain the Hafele-Keating experiment on time dilation using precise measurements of clocks flown in airplanes.

## History

In his famous work on special relativity in 1905, Albert Einstein predicted that when two clocks were brought together and synchronized, and then one was moved away and brought back, the clock which had undergone the traveling would be found to be lagging behind the clock which had stayed put. Einstein considered this to be a natural consequence of special relativity, not a paradox as some suggested, and in 1911, he restated and elaborated on this result in the following form:

If we placed a living organism in a box ... one could arrange that the organism, after any arbitrary lengthy flight, could be returned to its original spot in a scarcely altered condition, while corresponding organisms which had remained in their original positions had already long since given way to new generations. For the moving organism the lengthy time of the journey was a mere instant, provided the motion took place with approximately the speed of light. (in Resnick and Halliday, 1992)

In 1911, Paul Langevin made this concept more vivid and comprehensible by his now-iconic story / thought experiment of the twins, one of whom is an astronaut and the other a homebody. The astronaut brother undertakes a long space journey in a rocket moving at almost the speed of light, while the other remains on Earth. When the traveling brother finally returns to Earth, it is discovered that he is younger than his sibling, that is to say, if the brothers had been carrying the clocks mentioned above, the astronaut’s clock would be found to be lagging behind the clock which had stayed with the Earth-bound brother, meaning that less time had elapsed for the astronaut than for the other. Langevin explained the different aging rates as follows: “Only the traveler has undergone an acceleration that changed the direction of his velocity”. According to Langevin, acceleration is here "absolute", in the sense that it is the cause of the asymmetry (and not of the aging itself).[3]

As shown by Max von Laue in 1913, the process of acceleration is not as important as Langevin suggested, because the asymmetric aging is completely accounted by the fact that the astronaut twin travels in two separate frames, while the earth twin remains in one frame. Using Minkowski's spacetime formalism, Laue went on to demonstrate that the world lines of the inertially moving bodies maximize the proper time elapsed between two events.[4]

The significance of the “Twins Paradox” hinges on this one crucial detail of asymmetry between the twins.

• NOTE: Everyday English has "acceleration" as referring to "speeding up" only, but in scientific circles it equally refers to "slowing down" so that all the physical changes in speed and direction necessary to get the rocket to come back—slowing down, stopping, turning around, speeding up again—can be covered by the umbrella term "acceleration". This is the way the word is used in this article. In physics, the distinction between "acceleration" and "deceleration" is delineated, in the simplest scenarios, by a plus or minus. To expound: since acceleration in these simple scenarios is the rate of change in velocity (delta-V), acceleration is usually considered a positive number and deceleration a negative number with respect to velocity. So accelerating at a rate of 5 meters per second per second is considered a positive number, while decelerating at the same rate is considered a negative number (-5 meters per second per second).

It should be stressed that neither Einstein nor Langevin considered such results to be literally paradoxical: Einstein only called it "peculiar" while Langevin presented it as evidence for absolute motion. A paradox in logical and scientific usage refers to results which are inherently contradictory, that is, logically impossible, and both men argued that out of the time differential illustrated by the story of the twins no self-contradiction could be constructed. In other words, neither Einstein nor Langevin saw the story of the twins as constituting a challenge to the self-consistency of relativistic physics.

## Specific example

Consider a space ship traveling from Earth to the nearest star system outside of our solar system: a distance d = 4.45 light years away, at a speed v = 0.866c (i.e., 86.6 percent of the speed of light). The Earth-based mission control reasons about the journey this way (for convenience in this thought experiment the ship is assumed to immediately attain its full speed upon departure): the round trip will take t = 2d / v = 10.28 years in Earth time (i.e. everybody on earth will be 10.28 years older when the ship returns). The amount of time as measured on the ship's clocks and the aging of the travelers during their trip will be reduced by the factor $\epsilon = \sqrt{1 - v^2/c^2}$, the reciprocal of the Lorentz factor. In this case $\epsilon = 0.500 \,$ and the travelers will have aged only 0.500×10.28 = 5.14 years when they return.

The ship's crew members also calculate the particulars of their trip from their perspective. They know that the distant star system and the Earth are moving relative to the ship at speed v during the trip. In their rest frame the distance between the Earth and the star system is εd = 0.5d = 2.23 light years (length contraction), for both the outward and return journeys. Each half of the journey takes 2.23 / v = 2.57 years, and the round trip takes 2×2.57 = 5.14 years. Their calculations show that they will arrive home having aged 5.14 years. The travelers' final calculation is in complete agreement with the calculations of those on Earth, though they experience the trip quite differently.

If a pair of twins are born on the day the ship leaves, and one goes on the journey while the other stays on Earth, they will meet again when the traveler is 5.14 years old and the stay-at-home twin is 10.28 years old. The calculation illustrates the usage of the phenomenon of length contraction and the experimentally verified phenomenon of time dilation to describe and calculate consequences and predictions of Einstein's special theory of relativity.

## Resolution of the paradox in special relativity

The standard textbook approach treats the twin paradox as a straightforward application of special relativity. Here the Earth and the ship are not in a symmetrical relationship: the ship has a "turnaround" in which it undergoes non-inertial motion, while the Earth has no such turnaround. Since there is no symmetry, it is not paradoxical if one twin is younger than the other. Nevertheless it is still useful to show that special relativity is self-consistent, and how the calculation is done from the standpoint of the traveling twin.

Special relativity does not claim that all observers are equivalent, only that all observers at rest in inertial reference frames are equivalent. But the space ship jumps frames (accelerates) when it performs a U-turn. In contrast, the twin who stays home remains in the same inertial frame for the whole duration of his brother's flight. No accelerating or decelerating forces apply to the homebound twin.

There are indeed not two but three relevant inertial frames: the one in which the stay-at-home twin remains at rest, the one in which the traveling twin is at rest on his outward trip, and the one in which he is at rest on his way home. It is during the acceleration at the U-turn that the traveling twin switches frames. That is when he must adjust his calculated age of the twin at rest.

In special relativity there is no concept of absolute present. A present is defined as a set of events that are simultaneous from the point of view of a given observer. The notion of simultaneity depends on the frame of reference (see relativity of simultaneity), so switching between frames requires an adjustment in the definition of the present. If one imagines a present as a (three-dimensional) simultaneity plane in Minkowski space, then switching frames results in changing the inclination of the plane.

In the spacetime diagram on the right, drawn for the reference frame of the stay-at-home twin, that twin's world line coincides with the vertical axis (his position is constant in space, moving only in time). On the first leg of the trip, the second twin moves to the right (black sloped line); and on the second leg, back to the left. Blue lines show the planes of simultaneity for the traveling twin during the first leg of the journey; red lines, during the second leg. Just before turnover, the traveling twin calculates the age of the resting twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnover, if he recalculates, he'll measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the resting twin. The traveling twin reckons that there has been a jump discontinuity in the age of the resting twin.

The twin paradox illustrates a feature of the special relativistic spacetime model, the Minkowski space. The world lines of the inertially moving bodies are the geodesics of Minkowskian spacetime. In Minkowski geometry the world lines of inertially moving bodies maximize the proper time elapsed between two events.

## What it looks like: the relativistic Doppler shift

Now, how would each twin observe the other during the trip? Or, if each twin always carried a clock indicating his age, what time would each see in the image of their distant twin and his clock? The solution to this observational problem can be found in the relativistic Doppler effect. The frequency of clock-ticks which one sees from a source with rest frequency frest is

$f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 - v/c}\right)/\left({1 + v/c}\right)}$

when the source is moving directly away (a reduction in frequency; "red-shifted"). When the source is coming directly back, the observed frequency is higher ("blue-shifted") and given by

$f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}$

This combines the effects of time dilation (reduction in source frequency due to motion by factor ε) and the Doppler shift in received frequency by factor (1 $\pm$ v/c)-1, which would apply even for velocity-independent clock rates. For the example case above where v / c = 0.866, the high and low frequencies received are 3.732 and 0.268 times the rest frequency. That is, both twins would see the images of their sibling aging at a rate only 0.268 times their own rate, or expressed the other way, they would both measure their own aging rate as being 3.732 that of their twin. In other words, each twin will see that for each hour that passes for them, their twin experiences just over 16 minutes.

Light paths for images exchanged during trip
Left: Earth to ship.              Right: Ship to Earth.
Red lines indicate low frequency images are received
Blue lines indicate high frequency images are received

The xt (space-time) diagrams at left 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 his age-clock to the other twin. The vertical black line is the Earth's path through space time and the other two sides of the triangle show the ship's path through space time (as in the Minkowski diagram above). As far as the sender is concerned, he transmits these at equal intervals (say, once an hour) according to his 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 0.866 c, each twin would see 1 second pass in the received image of the other twin for every 3.73 seconds of his own time. That is, each would see the image of the other's clock going slow, not just slow by the ε factor, but even slower because of the Doppler observational effect. 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 3.73 seconds pass in the image for every second of his 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 asymmetry in the Doppler shifted images

The asymmetry between the earth and the space ship is manifested in this diagram by the fact that more blue-shifted (fast aging) images are received by the Ship. Put another way, the space ship sees the image change from a red-shift (slower aging of the image) to a blue-shift (faster aging of the image) at the mid-point of its trip (at the turnaround, 2.57 years after departure); the Earth sees the image of the ship change from red-shift to blue shift after 9.59 years (almost at the end of the period that the ship is absent). In the next section, one will see another asymmetry in the images: the Earth twin sees the ship twin age by the same amount in the red and blue shifted images; the ship twin sees the Earth twin age by different amounts in the red and blue shifted images.

## Calculation of elapsed time from the Doppler diagram

The twin on the ship sees low frequency (red) images for 2.57 years. During that time, he would see the Earth twin in the image grow older by 2.57/3.73 = 0.69 years. He then sees high frequency (blue) images for the remaining 2.57 years of his trip. During that time, he would see the Earth twin in the image grow older by 2.57×3.73 = 9.59 years. When the journey is finished, the image of the Earth twin has aged by 0.69 + 9.59 = 10.28 years.

The Earth twin sees 9.59 years of slow (red) images of the ship twin, during which the ship twin ages (in the image) by 9.58/3.73 = 2.57 years. He then sees fast (blue) images for the remaining 0.69 years until the ship returns. In the fast images, the ship twin ages by 0.69×3.73 = 2.57 years. The total aging of the ship twin in the images received by Earth is 2.57+2.57 = 5.14 years, so the ship twin returns younger (5.14 years as opposed to 10.28 years on Earth).

### The distinction between what they see and what they calculate

To avoid confusion, note the distinction between what each twin sees, and what each would calculate. Each sees an image of his twin which he knows originated at a previous time and which he knows is Doppler shifted. He does not take the elapsed time in the image, as the age of his twin now. And he does not confuse the rate at which the image is aging with the rate at which his twin was aging when the image was transmitted.

• If he wants to calculate when his twin was the age shown in the image (i.e. how old he himself was then), he has to determine how far away his twin was, when the signal was emitted—in other words, he has to consider simultaneity for a distant event.
• If he wants to calculate how fast his twin was aging when the image was transmitted he adjusts for the Doppler shift. For example, when he receives high frequency images (showing his twin aging rapidly), with frequency $f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}$, he does not conclude that the twin was aging that rapidly when the image was generated, any more than he concludes that the siren of an ambulance is emitting the frequency he hears. He knows that the Doppler effect has increased the image frequency by the factor $1/\left(1 - v/c\right)$. He calculates therefore that his twin was aging at the rate of
$f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}\times \left(1 - v/c\right) = f_\mathrm{rest}\sqrt{1 - v^2/c^2}\equiv\epsilon f_\mathrm{rest}$

when the image was emitted. A similar calculation reveals that his twin was aging at the same reduced rate of $\epsilon f_\mathrm{rest}\,$ in all low frequency images.

### Simultaneity in the Doppler shift calculation

It may be difficult to see where simultaneity came into the Doppler shift calculation, and indeed the calculation is often preferred because one does not have to worry about simultaneity. As seen above, the ship twin can convert his received Doppler-shifted rate to a slower rate of the clock of the distant clock for both red and blue images. If he ignores simultaneity, he might say his twin was aging at the reduced rate throughout the journey and therefore should be younger than him. He is now back to square one, and has to take into account the change in his notion of simultaneity at the turn around. The rate he can calculate for the image (corrected for Doppler effect) is the rate of the Earth twin's clock at the moment it was sent, not at the moment it was received. Since he receives an unequal number of red and blue shifted images, he should realize that the red and blue shifted emissions were not emitted over equal time periods for the Earth twin, and therefore he must account for simultaneity at a distance.

## Resolution of the paradox in general relativity

The issue in the general relativity solution is how the traveling twin perceives the situation during the acceleration for the turn-around. This issue is well described in Einstein's twin paradox solution of 1918.[5] In this solution it was noted that from the viewpoint of the traveler, the calculation for each separate leg equals that of special relativity, in which the Earth clocks age less than the traveler. For example, if the Earth clocks age 1 day less on each leg, the amount that the Earth clocks will lag behind due to speed alone amounts to 2 days. Now the accelerated frame is regarded as truly stationary, and the physical description of what happens at turn-around has to produce a contrary effect of double that amount: 4 days' advancing of the Earth clocks. Then the traveler's clock will end up with a 2-day delay on the Earth clocks, just as special relativity stipulates.

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 turn-around, this gravitational field fills the universe. (It should be emphasized that according to Einstein's explanation, this gravitational field is just as "real" as any other field, but in modern interpretation it is only perceptual because it is caused by the traveling twin's acceleration). In a gravitational field, clocks tick at a rate of t' = t(1 + Φ / c2) where Φ is the difference in gravitational potential. In this case, Φ = gh where g is the acceleration of the traveling observer during turnaround and h is the distance to the stay-at-home twin. h is a positive value in this case since 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.

Although this is called a "general relativity" solution, in fact it is done using findings related to special relativity for accelerated observers that Einstein described as early as 1907 (namely the equivalence principle and gravitational time dilation). So it could be called the "accelerated observer viewpoint" instead. It can be shown that the general relativity solution for a static homogeneous gravitational field and the special relativity solution for finite acceleration produce identical results.[6]

## Difference in elapsed time as a result of differences in twins' spacetime paths

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 two 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 A as measured by clock K until it reaches some velocity v.
Phase 2: Rocket keeps coasting at velocity v during some time T according to clock K.
Phase 3: Rocket fires its engines in the opposite direction of K during a time 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 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 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 A, still according to clock K, until both clocks reunite.

Knowing that the clock K remains inertial (stationary), the total accumulated proper time Δτ of clock K' will be given by the integral function of coordinate time Δt

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

where v(t) is the velocity of clock K' as a function of t according to clock K.

This integral can be calculated for the 6 phases:[7]

Phase 1 $:\quad c / a \ \text{arsinh}( a A/c )\,$
Phase 2 $:\quad T \ \sqrt{ 1 - v^2/c^2 }$
Phase 3 $:\quad c / a \ \text{arsinh}( a A/c )\,$
Phase 4 $:\quad c / a \ \text{arsinh}( a A/c )\,$
Phase 5 $:\quad T \ \sqrt{ 1 - v^2/c^2 }$
Phase 6 $:\quad c / a \ \text{arsinh}( a A/c )\,$

where a is the proper acceleration, felt by clock K' during the acceleration phase(s) and where the following relations hold between v, a and A:

$v = a A / \sqrt{ 1 + (a A/c)^2 }$
$a A = v / \sqrt{ 1 - v^2/c^2 }$

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

$\Delta \tau = 2 T \sqrt{ 1 - v^2/c^2 } + 4 c / a \ \text{arsinh}( a A/c )$

which can be expressed as

$\Delta \tau = 2 T / \sqrt{ 1 + (aA/c)^2 } + 4 c / a \ \text{arsinh}( a A/c )$

whereas the stationary clock K shows an elapsed time of

$\Delta t = 2 T + 4 A\,$

which is, for every possible value of a, A, T and v, larger than the reading of clock K':

$\Delta t > \Delta \tau\,$

## Difference in elapsed times: how to calculate it from the ship

Note that the above formula

$\Delta \tau = \int_t^{t+\Delta t} \sqrt{ 1 - (v(t)/c)^2 } \ dt \$

(here τ is another customary notation for the time of the non-inertial observer K') gives us the dependence of the proper time τ on the time elapsed in the inertial frame K, assumed the velocity v(t) is known. Actually, this equation answers the problem of the dependence between proper time and inertial time for the inertial observer at rest in K but not for the moving observer. Indeed, the data v(t) is not always easily accessible from the ship. The observer in the non-inertial frame K' may find it difficult to measure its velocity with respect to K, not to mention that this velocity has to be parametrized with respect to the inertial time of K, a step which needs the dependence between proper and inertial time to be known in advance. Can the non-inertial observer predict what will be the time elapsed for the twin which remained at home using only easily accessible data? The answer is affirmative and at least for a unidirectional motion has a simple answer given by[8]

$\Delta t^2 = \left[ \int^{\tau+\Delta\tau}_{\tau} e^{\int^{\bar{\tau}}_{\tau} a(\tau')d \tau'} \, d \bar\tau\right] \,\left[\int^{\tau+\Delta \tau}_{\tau} e^{-\int^{\bar\tau}_{\tau} a(\tau')d \tau'} \, d \bar\tau \right] \$

where a(τ) is the acceleration of the non-inertial observer as measured by himself (for instance with an accelerometer) in the whole round-trip and should not be confused with the non-relativistic acceleration d2x / dt2. It can also be shown that the inequality Δt > Δτ follows from the previous expression by using the Cauchy-Schwarz inequality:

$\Delta t^2=\left[ \int^{\tau+\Delta\tau}_{\tau} e^{\int^{\bar{\tau}}_{\tau} a(\tau')d \tau'} \, d \bar\tau\right] \,\left[\int^{\tau+\Delta \tau}_{\tau} e^{-\int^{\bar\tau}_{\tau} a(\tau')d \tau'} \, d \bar\tau \right] > \left[ \int^{\tau+\Delta\tau}_{\tau} e^{\int^{\bar{\tau}}_{\tau} a(\tau')d \tau'} \, e^{-\int^{\bar\tau}_{\tau} a(\tau')d \tau'} \, d \bar\tau \right]^2=\Delta \tau^2 \$

## A rotational version

Twins Bob and Alice inhabit a space station in circular orbit around a massive body in space. Bob departs the station and uses a rocket to hover in the fixed position where he left Alice, while she stays in the station. When the station returns to Bob, he rejoins Alice. Alice is now younger than Bob.[9] In addition to rotational acceleration, Bob must decelerate to become stationary and then accelerate again to match the orbital speed of the space station.

## Notes

1. ^ Miller, Arthur I. (1981). Albert Einstein’s special theory of relativity. Emergence (1905) and early interpretation (1905–1911). Reading: Addison–Wesley. pp. 257-264. ISBN 0-201-04679-2.
2. ^ Max Jammer (2006). Concepts of Simultaneity: From Antiquity to Einstein and Beyond. The Johns Hopkins University Press. p. 165. ISBN 0801884225.
3. ^ *Langevin, Paul (1911). "L’évolution de l’espace et du temps". Scientia 10: 31-54.
4. ^ Laue, Max von (1913). Das Relativitätsprinzip (2 ed.). Braunschweig: Vieweg.
5. ^ Einstein, A. (1918) "Dialog über Einwände gegen die Relativitätstheorie", Die Naturwissenschaften 48, pp697-702, 29 November 1918 (English translation: dialog about objections against the theory of relativity)
6. ^ Jones, Preston; Wanex, L.F. (February 2006). "The clock paradox in a static homogeneous gravitational field". Foundations of Physics Letters 19 (1): 75–85. doi:10.1007/s10702-006-1850-3.
7. ^ C. Lagoute and E. Davoust (1995) The interstellar traveler, Am. J. Phys. 63:221-227
8. ^ E. Minguzzi (2005) - Differential aging from acceleration: An explicit formula - Am. J. Phys. 73: 876-880 arXiv:physics/0411233
9. ^ Michael Paul Hobson, George Efstathiou, Anthony N. Lasenby (2006). General Relativity: An Introduction for Physicists. Cambridge University Press. p. 227. ISBN 0521829518.  See exercise 9.25 on page 227.

## References

The ideal clock

The ideal clock is a clock whose action depends only on its instantaneous velocity, and is independent of any acceleration of the clock. Wolfgang Rindler (2006). "Time dilation". Relativity: Special, General, and Cosmological. Oxford University Press. p. 43. ISBN 0198567316.

Gravitational time dilation; time dilation in circular motion