A. John Mallinckrodt Professor of Physics, Cal Poly Pomona  
The socalled "Twin Paradox"

An explanation using spacetime diagrams of how the socalled "twin paradox" of special relativity is resolved.


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Time dilation: The basis for the socalled "twin paradox" Early in the study of special relativity students learn about the phenomenon of time dilation, i.e., that "moving clocks run slow." Those who have properly appreciated the fact that all motion is relative and who have properly appreciated that clocks measure the passage of physical (including biological) time, ought necessarily to experience some mental discomfort at this result. After all when two observers pass each other, the prediction is that both will find the other's clocks to run slower than their own. This might well seem to be impossible on its face and, therefore, to invalidate the entire theory. Nevertheless, a detailed analysis that properly takes into account two other equally peculiar relativistic effects—the Lorentz contraction and the relativity of simultaneity—shows that one can build a perfectly sensible relativistic world in which all observers agree on the only things that they must agree on including the details of local events (e.g., what everybody's watches read in a group picture) and the temporal order of causal sequences (e.g., which came first, the lightning or the thunder?) The statement of the socalled "twin paradox" The confusion caused by the phenomenon of time dilation has long been encapsulated in the socalled "twin paradox" stated as follows*:
* Because much confusion persists to this day about the implications of the socalled "twin paradox," I hasten to emphasize unequivocally that the socalled "twin paradox" is not a paradox, that there is no controversy about its resolution (as we will see), and that it does not in any way cast suspicion on—let alone invalidate—the theory of special relativity. Why there is no paradox The socalled "twin paradox" is easily resolved by noting that there is a physically meaningful disinction between the experiences of the two twins during the trip. The Earthbound twin remains in a single constant velocity reference frame the entire time while the traveling twin must accelerate to turn around and come home. The acceleration causes the traveling twin to change from one constant velocity reference frame to another and produces effects that can be measured locally by the traveling twin in the form of inertial forces that can knock things over, compress springs, and generally endow objects with weight. As a result of the fact that their experiences are different, there is no a priori reason for them to come to the same conclusion. NB: The "same conclusion" I refer to here—the one they do not come to!—is the one referred to in the penultimate sentence of the statement of the socalled "twin paradox" above, that the other twin will be found to be younger. Indeed, it would be intolerable for special relativity or any other physical theory to predict that utterly incomprehensible finding. On the other hand, because both twins, in fact, must agree on what they find, we demand that any successful theory should be able to account for that finding from either point of view. As it turns out, the result is that the traveling twin is younger upon return than the Earthbound twin. This is readily understood from the point of view of the Earthbound twin who remains at all time in a single, constant velocity reference frame with respect to which the traveling twin's clocks are always (except for one instant during the turnaround) running slow. But how do we understand that conclusion from the point of view of the traveling twin? Spacetime diagrams One of the most illuminating ways of understanding the resolution of the socalled "twin paradox" is by analyzing carefully drawn, detailed spacetime diagrams for specific choices of trip distance and velocity. I have done so below for a trip of three lightyears undertaken at a speed of 3/5 c (giving a relativistic factor γ = 5/4) in both directions and with a "turnaround time" of negligible duration. (The approximation of negligible turnaround time may very well lead to anatomically unrealistic "g forces" (!) and can be relaxed at the expense of additional computational complexity, but it makes no qualititative difference in the result.) In this case the Earthbound twin (EBT) finds that it takes the traveling twin (TT) five years to reach the destination and five years to return for a total of ten years. During this time the TT's clocks run slow by a factor of 1/γ = 4/5 so that the TT ages by eight years, four years on each leg of the journey, and is, therefore, two years younger at the reunion. The view from the reference frame of the Earth The left panel of the figure below (which you can click on to open a larger version in a new window), shows the worldlines of the EBT and the TT in the reference frame of Earth. Note that the TT reaches the destination at a distance of three lightyears after an elapsed time of five years in this frame and that the TT has aged only four years at that point. Note also that the scale of the x and t axes are such that light travels along lines at a 45 degree angle, i.e., 1 year per light year, and that the TT's worldlines have a larger slope (representing a lower speed) of 5/3 years per light year. Because the figure is drawn from the frame of reference of Earth, horizontal lines represent collections of events that occur at the same time, i.e., "lines of simultaneity" for the EBT. The figure also, however, includes a few lines of simultaneity in the reference frame of the TT as shown in gray. Because of the relativity of simultaneity, these lines are tilted and run from lower left to upper right during the outbound leg and from lower right to upper left during the inbound leg. In both cases the slope is the inverse of that of the TT's worldline, 3/5 years per light year. For instance, note that one of these lines indicates that at the moment the TT sends the third yearly signal, the TT would say that the EBT's clock reads 2.4 years as should be expected since the TT says the EBT's clock is "moving" and, therefore, running slow. Note finally that there are two lines of simultaneity linking the turnaround point, one for the outbound leg and one for the inbound leg. They indicate that the EBT's instant of simultaneity in the TT's frame jumps quickly from 3.2 years to 6.8 years during the turnaround as the TT moves from one reference frame to another. Both the EBT and the TT send light signals to each other at yearly intervals as shown by the red and blue lines respectively and each transmission is marked with the number of the year at which it was sent. Note that the TT receives only the first two signals from the EBT on the outbound leg and recieves all eight remaining signals on the inbound leg with the final, tenth, signal received at the moment of the reunion. Note also that the EBT receives the last of the four outbound signals (including the fourth at the turnaround point) at year eight and receives the four inbound signals during only the last two years. Thus, we see that the EBT receives signals at the rate of one every two years for the first eight years and then at a rate of two per year for the final two years. This amounts to a total of (1/2)*8 + (2)*2 = 8 signals received from the TT. On the other hand the TT receives signals also at the rate of one every two years for only the first four years and then at a rate of two per year for the final four years. This amounts to a total of (1/2)*4 + (2)*4 = 10 signals received from the EBT. It is well worth noting the fact that both twins agree that they receive signals at a rate of 1/2 per year (low frequency) when those signals reflect relative motion away from each other and both twins also agree that they receive signals at a rate of 2 per year (high frequency) when those signals reflect relative motion toward each from each other. The difference is that, for the TT, the low and high frequency signals occupy equal portions of the trip while, for the EBT, the low frequency signals are observed during 80% of the trip. The view from the reference frame of the outbound traveler The right panel of the figure, shows the worldlines of the EBT and the TT in the outbound reference frame. Note that, in this reference frame, the reunion takes place 12.5 years after the departure in keeping with the fact that the EBT's clocks run slow the entire time. The TT, however, leaves this reference frame at the turnaraound point four years into the trip when the EBT's clock reads 3.2 years as previously noted. Note also that the EBT is 2.4 lightyears away at this time, the largest separation of the trip in this frame, properly reflecting the Lorentz contraction of the 3 lightyear distance observed in the reference frame of Earth. On the inbound leg, the TT moves at a speed of 15/17 c as obtained from the relativistic addition of the speeds 3/5 c and 3/5 c. During this time the TT's clocks run slow by a factor of 1/γ = 17/8, which is reflected in the fact that it takes just over two years for the TT's clock to advance each additional year. The worldlines for the light signals are again shown along with the lines of simultaneity for the TT and they confirm every feature previously noted in the spacetime diagram for the reference frame of Earth. Relationship to the relativistic Doppler effect and another way to predict the amount of differential aging It is most interesting to note that, no matter what speed is used for the outbound and inbound legs, the TT will receive signals at a rate of f_{out} < f_{o} (the transmission frequency) for half the trip and at a rate of f_{in} = 1/f_{out} > f_{o} for the other half. Thus, the average frequency is (f_{out}+f_{in})/2 which is readily shown to be greater than f_{o}. This insures that the TT will receive more signals during the trip then he sends and will, therefore, expect to find the EBT to be older at the reunion, in perfect harmony with what the EBT expects to find. Indeed, one can use the relativistic Doppler formula f_{out} = f_{o} [(1v/c)/(1+v/c)]^{1/2} directly to predict the relative age differential. For instance, at a speed of 12/13 c, f_{out} = f_{o} [(1v/c)/(1+v/c)]^{1/2} = (1/5)f_{o}. Accordingly, the average frequency received by the TT is (1/5+5)f_{o}/2 = (13/5)f_{o} implying (correctly) that the EBT will have aged 13/5 as much as the TT.


