The Twin Paradox
The “twin paradox” is not a paradox in the sense of a logical contradiction that falsifies relativity but rather a very curious puzzle. Traditionally, the twin paradox is concerned with the strange result that if one of two twin brothers leaves the other and embarks on a high-speed journey to a remote point and back again, the twins will no longer be the same age. Let’s call these hypothetical twin brothers A and B. For this discussion, we’ll stipulate that A stays home while B travels away from his brother at a speed of 60 percent of the speed of light (0.6c, where c is the speed of light, nearly 300 million meters per second). B travels for fifteen years by A’s reckoning then quickly decelerates to a stop, turns around, and quickly accelerates back to 0.6c in the direction toward his brother, A. After another fifteen years (again, by A’s reckoning), B arrives home, decelerates, and rejoins his brother, who has aged thirty years since he last saw B. The “paradox” is that, even though A’s velocity relative to B is the same as B’s velocity relative to A, B will have experienced only twenty-four years of travel and find himself six years younger than his twin brother, A.
While this is indeed puzzling, it is not a logical flaw in relativity. The twins do not have similar experiences during B’s long journey, and that resolves the “paradox.” (While the fiction of very short deceleration/acceleration periods is useful to keep this discussion from getting into general relativity theory, it should be noted that such accelerations would almost certainly reduce twin B to a thin red puddle. It would take weeks to make the velocity changes at tolerable accelerations, say 5 to 10 g. See my accompanying sidebar “On Problems with Near-light-speed Travel” for more on this type of difficulty.) The journey of B, as viewed by twin A, is depicted in figure 1.
The workings of the “Twin Paradox” can be explained with the aid of space-time diagrams. A space-time diagram for the stay-at-home twin, A, appears in the left half of figure 2. The grid marks show years on the vertical axis and distance in light-years on the horizontal axis. The thick lines represent A’s and B’s positions over time, while the thin lines with arrows represent the paths of light beams sent between the twins. During the fifteen years (in A’s frame of reference) of outbound travel by twin B, B gets out to a distance of nine light-years (0.6c315 years) from twin A. However, signals or light rays sent from B’s turnaround point won’t even reach A for another nine years, or until twenty-four years (15+9) after B’s departure. That is, A will see his brother B recede for twenty-four years, and then approach for just six years, arriving thirty years after his initial departure.
This is in marked contrast to B’s observations: B will see his stay-home brother recede for twelve years. After B turns around, he will see A approaching for twelve years and will return a total of twenty-four years after his departure. However, the same interval is thirty years by A’s calendar. The difference is that, during the short but intense accelerations experienced by B, B’s velocity relative to the universe (and to A) is changing. Twin B effectively “loses synch” with the rest of the universe, including his twin brother, A. Twin B is not in an inertial reference frame over the entire trip—and his bouts with intense accelerations will certainly remind him of that fact. Of course, A won’t be aware of B’s velocity changes until many years later.
The space-time diagrams for B’s journey appear on the right of figure 2. These can’t be represented as a single diagram, because they are views of two different inertial frames (B outbound versus B inbound). The twin that undergoes acceleration will be the one who returns home younger than his stay-at-home brother. The loss of synchronization due to acceleration is the key and the reason it’s not a logical “paradox.”
Figure 2: Twin paradox space-time diagrams for stay-at-home twin A (left) and traveler B (right).
This point is crucial: the time discrepancies between the twins are absolutely real. Here is a quick example, presented with the “radar method”: since any radar beams sent from A meet the target (B) at only one point in space-time, those beams must spend equal times outbound and inbound with respect to the sender. Figure 2 shows that a radar beam emitted by twin A at his time of two years will be reflected from B at some unknown time, and received again by A when his (A’s) calendar reads eight years. Likewise, a beam emitted by twin A at four years will be reflected from B and received by A when his calendar reads sixteen years.
Twin A can calculate the time and distance (in A’s frame of reference) of reflections from B, knowing only his own sending and receiving times and that the signals propagate at the speed of light. Since A’s two-year pulse returns at eight years, the reflection occurred (by A’s calendar) at the midpoint of the send/receive times, (2+8)/2=5 years. Since A’s four-year pulse returns at sixteen years, the reflection occurred at (4+16)/2=10 years by A’s calendar. Therefore, A measures the interval between these reflections (at five years and ten years) as being five years long.
Because the twins are separating rapidly, there will be a delay in B’s receipt of A’s transmissions. In particular, while A’s transmissions were sent two years apart by his clock, they were received by B over an interval longer than two years, say, K*2 years, where K is a factor greater than 1. However, the same must hold true for B’s “transmissions” back to A: whatever period separates the reflections from B’s craft, A’s measurement of receiving times will be longer—in fact, precisely K times longer—since B is moving away from A exactly as fast as A recedes from B (“relativity”). So, A’s original pulses were sent two years apart; these were received by B at K*2 years apart and received again by A at K*K*2 years apart, or eight years. Clearly, K must equal 2, and B’s interval between receipt of A’s two signals must be 2*2=4 years, while A’s measurement of the time for the pulses to return from B is K*4=8 years, as required. This is how “Time Dilation” comes to be measured by twin A: the five-year interval that A experiences in his own frame of reference takes only four years in B’s frame of reference.