 We're now going to look at a general relativity twist on the twin paradox of special relativity, what I'll call the triplet paradox. Suppose we have a black hole with a van der Ryzen at r equals one. On the coordinate ring at r equals three, corresponding to the innermost stable circular orbit, suppose triplets are born, or maybe we just synchronize three clocks, one clock stays put on the coordinate ring, one goes off in a circular orbit, and one is thrown upwards so that it will eventually come to rest, fall back down, and reunite with the stationary clock. The bookkeeper, with his clock reading time t, observes and keeps track of all events. Let's call the readings of the clocks TLO for the stationary local observer, TOR for the orbiting triplet, and TUD for the up-down triplet. Suppose that the up-down triplet is launched with an initial velocity, such that he returns to the starting point at precisely the same time that the orbiting triplet does. Which of the triplets, if any, will be older or younger than the other two? We'll run the experiment numerically, showing the triplets as blue dots, with an adjacent box showing their individual clock readings. In a rectangular box, we'll show T equals the bookkeeper time. We have four different clocks synchronized to zero at the start of the experiment. As the experiment proceeds, we can see the clocks all running at different rates. Eventually the triplets reunite and we stop the clocks. We find that for the stationary local observer, the experiment lasted 37.7 seconds, for the orbiting triplet, 32.6 seconds, and for the up-down triplet, 40.3 seconds. The orbiting triplet aged the least, the up-down triplet the most, and the stationary triplet in between. The bookkeeper, who viewed the experiment from afar, measured a duration of 46.2 seconds. We can understand these results using the metric that tells us the elapsed time ds on a clock that moves between events separated in bookkeeper coordinates by dt, dr, and d phi. The local observer doesn't move, so her ds, which we call dt local, is square root of 1 minus 1 over r times bookkeeper time interval dt. The orbiting triplet moves in the angular direction phi, but not in the radial direction r. His ds, which we call dt orbit, has an additional phi velocity term under the square root. The up-down triplet moves in the radial direction r, but not in the angular direction phi. His ds, which we call dt up-down, therefore has an r velocity term instead of a phi velocity term. The bookkeeper measured the experiment lasting 46.2 seconds. The local observer should measure a factor of square root 1 minus 1 over 3, or about 82%, times the bookkeeper's 46.2 seconds, which is indeed the 37.7 seconds she measures. This is gravitational time dilation. Clocks further down in the gravitational potential run slower. The orbiting triplet has the same 1 minus 1 over 3 terms, and an additional negative velocity squared term, so he should measure even less time than the stationary triplet. R is 3, and d phi over dt is 2 pi radians in 46.2 seconds. Plugging those values in, we get a factor of about 71%, multiplying the bookkeeper's 46.2 seconds by this factor we obtain the orbiting triplet's 32.6-second clock reading. The up-down triplet has the 1 minus 1 over r terms, and a negative velocity squared term. However, unlike the other two triplets, he doesn't stay at a fixed r coordinate. Instead, as he travels away from the black hole, his r coordinate increases. 1 over r decreases, and 1 minus 1 over r increases. He moves to regions where there is less gravitational time dilation, and this more than offsets the negative velocity term. The result is that he measures a longer elapsed time than the stationary triplet, although still shorter than the bookkeeper's measurement. This brings up an interesting point. In relativity, the natural or freefall motion of particles is along spacetime geodesics. We've described this by saying, for all the waves a clock can move between two spacetime events, the natural motion is that which leads to the largest elapsed time. The local and up-down triplet results are consistent with this. The local observer was not in freefall. She felt her own weight, while the up-down triplet was in freefall. As expected, his clock reading was more than hers. But what about the orbiting triplet? He also fell freely between the two events, yet his clock reading was the smallest of all. Let's look at an analogy with spatial geodesics. Suppose your landscape has a hill. You lay a rope along an arbitrary path between two points. Assuming the rope is free to slide along the ground, if you pull on the rope's ends, its path will deform, continuously getting shorter until it finds the shortest path to be had, which is the geodesic between the two points. But suppose you had initially laid the rope along a path that went around the other side of the hill. Pulling other rope's ends will again cause its path to deform, continuously getting shorter until it finds the shortest path to be had. This is also a geodesic between the two points. It's the shortest path in a local sense, meaning there are no shorter paths to be found nearby. In a curved space, it's possible to have more than one geodesic between two points, and the geodesics can have different lengths. Pulling on the rope, path P1 will deform to geodesic G1, and path P2 to geodesic G2. G2 is a geodesic, even though it's longer than path P1, which is not a geodesic. This doesn't disqualify G2 as a geodesic, because P1 is not in G2's neighborhood. There's no way to deform P1 into G2 such that the length continuously decreases. Spacetime geodesics are paths of longest time rather than shortest distance. We conclude that all spacetime paths close to the up-down triplets path will take less than 40.3 seconds. And all spacetime paths close to the orbiting triplets path will take less than 32.6 seconds. These spacetime geodesics are on opposite sides of the hill created by the black hole, so to speak. Because the local observer time is less than the up-down time, but more than the orbiting time, it seems likely that she is on the up-down triplets side of the hill. We can show this by considering how the up-down path can be continuously deformed into the local observer path and vice versa. We'll show bookkeeper time T horizontally and coordinate R vertically. According to the bookkeeper, the experiment lasts capital T equals 46.2 seconds. For the up-down path, the clock has an initial upward velocity, reaches a peak, and returns. This unforced motion results in a clock reading of 40.3 seconds. Imagine a variation of this where the clock stays still for time tau, and is then launched upward with a reduced velocity so that it still returns at the end of the experiment. For tau equals zero, we have the up-down path. For tau equals T, the clock would never be launched, and we have the stationary local observer path. For tau greater than zero, we find that S is less than 40.3. In fact, plotting S as a function of tau shows that as tau increases, S decreases. Starting with tau equals 46.2, and decreasing it to zero, we have a continuous transition from the local observer path to the up-down path in which the clock time continuously increases. In a similar fashion, we can show that the orbiting triplet's time of 32.6 seconds is also a local maximum. Suppose we have the clock stay still for tau seconds of bookkeeper time. Then it travels around the r equals 3 circle for the remaining T minus tau seconds. For non-zero tau, it will be going faster than it would in a natural orbit, so force will need to be applied to keep it on this path. An expression for the elapsed clock time can be written down, and plotting this versus tau, we get the curve shown here. The maximum is at tau equals zero, which is the natural orbital motion. Deviation from this results in a shorter elapsed clock time. So the orbiting and up-down paths are indeed spacetime geodesics. Each locally maximizes the elapsed time on a clock moving between the same two events. We now have the tools we need to calculate relativity's predictions for the global positioning system. Our black hole will have the Earth's mass, about 1.5 times 10 to the minus 11 seconds. Earth's equator is represented by a coordinate ring with radius r1 equal to Earth's radius of 0.0213 seconds. Earth rotates, so our clock on this ring goes around once every 86,200 seconds. For comparison, the orbital period at Earth's surface is 5,060 seconds. Our GPS clock will be in an equatorial orbit at r2 equals 0.0866 seconds. The resulting orbital period is half Earth's rotation period. For both clocks, we use the form of the metric from motion at fixed r value. That metric is shown here. Angle phi changes by 2 pi radians over period t, so we can replace d phi over dt by 2 pi over t. Under the square root, we have 1 minus very small quantities. An excellent approximation for square root 1 minus x is 1 minus 1 half x when x is small. So we can write our metric as ds equals 1 minus m over r minus 1 half the square of 2 pi r over t, all times dt. To start, we'll consider only gravitational effects by neglecting the motion term. Imagine a non-rotating Earth with the GPS satellite suspended above it, maybe mounted on a huge tower. For Earth, m over r is 6.96 times 10 to the minus 10. A clock on a non-rotating Earth runs slower than a faraway clock by about seven parts in 10 billion. At the GPS altitude, m over r is 1.67 times 10 to the minus 10th. The ratio of these tells us how much faster GPS clocks run than clocks on Earth, again neglecting motion. The answer is 5.29 parts in 10 billion. That factor over the course of one day adds up to 45.7 microseconds. That might sound small, but for electronics it's a huge number, equal to roughly 100,000 clock cycles of a typical CPU. Now we repeat the calculations including the motion term. For Earth, the change is tiny, 6.97 as opposed to 6.96. But for the satellite, the change is significant, 2.5 compared to 1.67. The ratio of clock rates is now 4.47 parts in 10 billion. Over one day, this totals 38.6 microseconds, which we can think of as due to 45.7 microseconds of gravitational effects, less 7.1 microseconds of motion effects. On the one hand, GPS clocks run faster because they are at a higher gravitational potential. On the other hand, they run slower because they are moving relatively faster. But the gravitational effect more than offsets the motion effect. This difference in clock rates is indeed observed and must be taken into account for the GPS system to properly function. Thus the GPS system provides continuous, precise, experimental verification of the general theory of relativity.