Twin paradox

From Wikipedia, the free encyclopedia

Jump to: navigation, search

In physics, the twin paradox refers to a thought experiment in Special Relativity, in which a person who makes a journey into space in a high-speed rocket will return home to find they have aged less than an identical twin who stayed on Earth. This result appears puzzling, since the situation seems symmetrical, as the latter twin can be considered to have done the travelling with respect to the former. Hence it is called a "paradox". The seeming contradiction is explained within the framework of relativity theory and has been verified experimentally using precise measurements of clocks flown in airplanes.

Contents

[hide]

[edit] History

In his famous work on Special Relativity in 1905, Albert Einstein predicted that when two clocks were brought together and synchronised, 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 traveller 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).

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.)

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 the time differential illustrated by the story of the twins was an entirely natural and explainable phenomenon.

[edit] Specific example

Consider a space ship traveling from Earth to the nearest star 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, relative to the Earth). 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 flow of time on the ship and aging of the travelers during their trip will be slowed by the factor \epsilon = \sqrt{1 - v^2/c^2}, the reciprocal of the Lorentz factor. In this case \epsilon = 0.5 \, and the travelers will have aged only 0.5×10.28 = 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 is 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.

[edit] 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 feels inertial forces, 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 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.

Twins paradox Minkowski diagram
Twins paradox Minkowski diagram

In the spacetime diagram on the right, the first twin's lifeline 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 lifeline of the resting twin. The resting twin has suddenly "aged" very fast, in the reckoning of the traveling 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.

[edit] 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. Note that 3.732 = 1/0.268, that is, they are reciprocals of each other.

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
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 x − t (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). The first diagram shows the image carrying signals sent from Earth to Ship, while the second shows the signals sent from Ship to Earth. 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 twin receiving these signals, they are not being received at equal intervals, according to their own clock.

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.

[edit] 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 and more red-shifted (slow aging) images are received by Earth. 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, you 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.

[edit] 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 is 10.28 years old).

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).

[edit] 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.

[edit] 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.

[edit] 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[1]. 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, the clock rate of one clock versus another is given by ν' = ν(1 + Φ / c2) where Φ is the difference in gravitational potential for the location of the two clocks and ν is clock rate (i.e., the number of ticks of a clock per unit time as measured by some standard clock). 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.[2]

[edit] Accelerated rocket calculation

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 Δt' of clock K' will be given by the integral

\Delta t' = \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:

Phase 1 :\quad c / a \ \text{arcsinh}( a A/c )\,
Phase 2 :\quad T \ \sqrt{ 1 - v^2/c^2 }
Phase 3 :\quad c / a \ \text{arcsinh}( a A/c )\,
Phase 4 :\quad c / a \ \text{arcsinh}( a A/c )\,
Phase 5 :\quad T \ \sqrt{ 1 - v^2/c^2 }
Phase 6 :\quad c / a \ \text{arcsinh}( 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 t' = 2 T \sqrt{ 1 - v^2/c^2 } + 4 c / a \ \text{arcsinh}( a A/c )

which can be expressed as

\Delta t' = 2 T / \sqrt{ 1 + (aA/c)^2 } + 4 c / a \ \text{arcsinh}( 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 t'\,

[edit] See also

[edit] Notes

  1. ^ 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)
  2. ^ Jones, Preston; Wanex, L.F. (February 2006). "The clock paradox in a static homogeneous gravitational field". Foundations of Physics Letters 19 (1): 75–85. 
  3. ^ A Misunderstood Rebellion: The Twin-Paradox Controversy and Herbert Dingle's Vision Of Science by H. Chang, Studies In History and Philosophy of Science, Vol 24 (1993), pp 741-790.

[edit] References

Wikibooks
Wikibooks has a book on the topic of

[edit] External links

Retrieved from "http://en.wikipedia.org/wiki/Twin_paradox"
Personal tools