The "paradox": Special relativity says that physics is the same in reference frames that move at a uniform velocity relative to one another. Observers in any two frames moving relative to one another should not be able to make any observation that indicates which one is "actually" in motion. So, how is it that one of the twins (Prime) is younger than the other? Does this mean one of the twins was "actually" in motion while the other was not?
Like all such paradoxes the paradox is only apparent. In this case, Prime has actually taken a trip. That is, Prime has left the Frame of Reference of Unprime (feeling an initial acceleration at the beginning of the trip), then, after some time, Prime again feels an acceleration as Prime slows to a halt and turns around (boards the return train), again accelerating up to speed, then finally halting back at the original station. Prime has felt lots of accelerations; Unprime has felt no accelerations. Therefore, Prime did actually do something "out of the ordinary" --- it might not be unreasonable to find that Prime and Unprime have had different experiences. Indeed, it appears that Prime has experienced less time duration during the separation of the twins than Unprime has experienced.
However, it might be objected that although the two twins have had different experiences, it is not at all clear that one should be YOUNGER than the other! My objective here is to explain in detail exactly why Prime is not as old as Unprime when they meet up at the end of Prime's trip. The explanation is a bit involved, so don't be alarmed if you don't understand it all at first. Reread it a few times. You may need to consult an elementary text on Special Relativity for explanations of the Time Dilation, Lorentz Contraction, and Relativity of Simultaneity effects that I will use without explanation. In the explanation that follows, lengths and times as measured by Prime will have a "prime", or apostrophe attached to them to distinguish these measurements from those of Unprime. The bottom line of the explanation: after comparing notes about what they observe, both Unprime and Prime will agree that Prime has aged less than Unprime.
Let's ignore the duration of time during the accelerations undergone by Prime. Let's say these acceleration occur almost instantly, so we don't need to keep track of how long they take. (Of course, in reality such sudden accelerations would give Prime a very different experience than Unprime --- Prime would be killed! But let's ignore that problem.) As you will see, the accelerations therefore have NO BEARING on the quantitative difference in duration of the separation experienced by Prime and Unprime.
At the start, both Prime's and Unprime's wrist watches read the same time. Let's call that time t = t' = 0. In order to have some numbers to play with, let's say L = 1 light-year (a very, very distant train station!) and v = 0.80 c, in other words, v is 80% of the speed of light (c = the speed of light = 186,000 miles per second). The trains will be very fast.
When Prime momentarily stops and is about to change trains for the return trip, a photo is also taken of Prime's watch; this photo is also transmitted to Unprime (at the speed of light) so Unprime can take a look at it. The photo shows that Prime's watch reads t' = 0.75 years. Unprime understands this result since Unprime observed that Prime's wrist watch was running slowly while Prime was moving (at the start of the trip). So, of course less time would have elapsed on Prime's watch. Unprime even knows why: "I observed Prime's watch to be running slowly. Any clock I see moving, I would observe to be running slowly; it's an effect called Time Dilation. I can calculate exactly how much time should have elapsed on Prime's watch during Prime's trip to the distant station. The result is t'=(L/v)/g, where g is the Lorentz factor, and g = 1/sqrt(1-v^2/c^2) = 1/sqrt(1-0.8^2) = 1.6667. So, t' should be (1/0.8)/1.6667 = 0.75 years, exactly as I observe in this photo of Prime's watch." (I will use "sqrt" to mean "square root of" and I will use "^2" to mean "raised to the 2nd power".)
Unprime now looks at Prime's watch. It reads t' = 2 x 0.75 years = 1.5 years. Again, Unprime understands this result perfectly. Unprime saw the photo of Prime's watch taken at the distant station; it showed t' = 0.75 years. So, now, at the end of the total trip it should read t' = 1.5 years. Unprime understands exactly why t' is not the same as t: Unprime chalks it all up to Time Dilation.
Unprime summarizes it all: "My watch says the trip took t = 2L/v = 2.5 years, because that is how long it takes a train to travel a distance L = 1 light year at speed v = 0.8c, and return at the same speed. Prime's watch recorded only t' = 1.5 years for the trip, because Prime's watch was running slowly compared to mine."
Prime also observes the clock at the station reading t = 1.25 years. He is momentarily baffled by this clock's reading, but then remembers his Special Relativity. The Time Dilation and Lorentz Contraction results do not sum up all there is to Special Relativity. For example, there is also the "Relativity of Simultaneity" --- events that are simultaneous in one observer's frame of reference will not be simultaneous in another observer's frame of reference (if the two observers are moving relative to each other, and if the two events have some separation in space along the direction of the relative motion).
So what is Prime's explanation of the t = 1.25 years versus the t' = 0.75 years? According to Unprime, his watch is synchronized with the clock at the distant station. But according to Prime, not only are Unprime's watch and the clock on the distant station running slowly, but they are NOT synchronized. More specifically, according to Special Relativity, the clock at the distant station is ahead of Unprime's watch, according to Prime! Prime could verify this by enlisting an elaborate system of cohorts, each with a clock, spread out along the very long train in which Prime sits. But Prime also knows how to calculate this effect and the result is that the clock at the station is ahead of Unprime's watch by an amount vL/c^2. That clock STARTED ahead of Unprime's watch by this amount (when Prime noticed that Unprime was racing off at v = 0.8c), so of course it reads a later time than 0.75 years when the distant station arrives at Prime's location. Prime calculates the time that will be displayed on the station clock as t = (((L/g)/v)/g) + (vL/c^2). The first term in this result is ((L/g)/v)/g and is equal to the distance the stations travel as observed by Prime (L/g) divided by the speed of their travel (v), all as observed by Prime, but then reduced by the Lorentz factor (g) to take account of the slow rate of the station clock (the Time Dilation effect). The second term is the initial amount by which the station clock was ahead at the beginning of the motion. So, Prime says, "The time reading on the station clock is 1.25 years, and I understand why. It's a combination of Time Dilation and the Relativity of Simultaneity. I can calculate that the reading on the station clock should be t = (((L/g)/v)/g) + (vL/c^2) = (L/v) x ( (1/g^2) + (v^2/c^2) ) = (L/v) x ( 1 - (v^2/c^2) + (v^2/c^2) ) = L/v = 1 light-year / 0.8c = 1.25 years. That's exactly what the station clock reads. Yet, of course, my clock only reads 0.75 years, and I understand why that is also."
Prime now looks at Unprime's watch and sees that it reads t = 2.5 years, and Prime is not surprised by that since he observed that the station clock read t = 1.25 years at the halfway point of the experiment, and since Prime understands why (as we just explained above).
Prime summarizes things: "On my watch it took 0.75 years for the distant station to reach me, and another 0.75 years for the original station to get back to me. I understand exactly why since the stations, I observed, were separated by 0.6 light-years, and were moving at 0.8c. Unprime observed a longer time for the trip. His result (2.5 years) can be understood as a result of his clocks running slowly and not being synchronized (as I observed them)."
Note, on the return trip, because of the Relativity of Simultaneity effect, Prime observes Unprime's watch to be ahead of the distant station's clock!! That, with Time Dilation, explains the t = 1.25 years reading for the return trip on Unprime's watch. If you are baffled by how, as observed by Prime, the distant station clock was at one point ahead of Unprime's watch, and then later behind Unprime's watch, the reason lies not in a some mistake in the synchronization of the two clocks by Unprime, but instead is the fact that Prime observes those clocks from DIFFERENT frames of reference on the way out and on the way back. Remember, way back at the beginning, BEFORE any acceleration by the train, Prime and Unprime's watches were synchronized (at t = t' = 0), and the distant station's clock was synchronized with Unprime's watch and Prime's watch. But then Prime changed his frame of reference: he accelerated (momentarily) into a new uniform velocity frame. Back in Unprime's frame (Prime's original frame), Unprime still noted that all clocks in his frame were synchronized. It is Prime that saw the clocks in Unprime's frame to be out of synchronization. Since Prime was then moving toward the distant clock, he observed it to be ahead. Once Prime stopped at the distant station, he rejoined Unprime's frame of reference: Prime saw all the clocks in that frame to be synchronized again (but reading 1.25 years, while his own watch said 0.75 years). Then Prime started up again, now traveling in the reverse direction: Prime then had to observe Unprime's watch to be ahead of the distant station's clock, and by just the right amount so that when Prime arrives at the original station, Unprime's watch will display 2.5 years.
If you think this whole affair is difficult to understand, then you're not alone. But it has been observed (in particle experiments), and can be explained (as, for example, in the above story), through Special Relativity.
Instead of having Prime start at rest, next to Unprime, then accelerate to speed v, then stop at the distant station, then accelerate once again until moving at speed v back toward the initial station, etc., imagine the following demonstration. Unprime sits still at the original station. According to Unprime, Prime is moving along (and always has been) at the speed v, and passes Unprime at the moment that Unprime's and Prime's watches read t = t' = 0. Prime continues on, ultimately passing the distant station. At the moment he passes the distant station, Prime's watch reads t'=0.75 years (for reasons he understands, and were explained above). At that exact moment, Doubleprime (another person), passes the same distant station in another train heading back toward the original station at speed v. Both Prime and Doubleprime notice that the station clock reads t = 1.25 years. The also notice that both their watches display the time t' = t'' = 0.75 years (Doubleprime's watch is just coincidently equal to 0.75 years). According to Unprime (or his cohorts spread all over the place) Doubleprime has always been moving this way. Eventually Doubleprime passes the original station. As he passes, Doubleprime notices that his own watch reads 1.5 years, and that the watch on Unprime's wrist reads 2.5 years. Unprime notices the two watches also. Unprime's explanation is Time Dilation (just as in the above explanation). Doubleprime's explanation is that Unprime's clocks were running slowly and out of synchronization (again, just as in the above explanation).
The resulting difference between the reading on Unprime's watch and that on Doubleprime's watch is not the result of any accelerations experienced by anyone (nobody experienced any accelerations). But notice that the full duration measured by Unprime was, of course, measured in one reference frame; the duration for the full experiment recorded by Prime and Doubleprime required the combined results acquired in two different reference frames. That is the source of the asymmetry in the results. This explanation of the twin paradox (without accelerations) shows that it takes TWO different reference frames to keep track of the time duration experienced by the twin who actually takes the trip, while it take only one frame to keep track of the duration for the twin who stays at home. Their situations are fundamentally different, and the different time durations they experience are the result.
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