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Let's be specific. What time is it in Modesto when John's train arrives in that city's station? If the train leaves Fresno at noon, then Jane will figure that it will arrive in Modesto at one o'clock. But John, seeing that only half an hour has elapsed on his watch and being aware of time dilation, figures that the Modesto clock will show twelve-fifteen when he first sees it. Which of those calculations is wrong and how can we set it right?
Identifying tacit assumptions is a good way to start. In her calculation Jane has tacitly assumed that the station clock in Modesto shows noon when the station clock in Fresno does. Is that a good assumption? It is, but only if we make it good.
Imagine that halfway between Fresno and Modesto there is a high tower, one high enough to be visible from the equally high clock towers of the Fresno and Modesto train stations. At about eleven-thirty every night the station clocks are stopped, set to midnight, and connected to photocells that will start them when illuminated with a flash of light. At twenty-six minutes and six seconds before midnight a lamp in the midway tower emits an intense pulse of light toward a wedge-shaped mirror that splits the pulse into two pulses going in opposite directions, one toward Fresno and the other toward Modesto. Because those pulses originated at the same point at the same time and traveled equal distances at the same speed, the Fresno and Modesto clocks will start simultaneously. That synchronization procedure defines what simultaneity means in Relativity. Thus, if Jane orders a fresh cup of coffee at one o'clock, that act will be simultaneous with John's arrival in Modesto.
That statement won't be true in the inertial frame occupied by John's train. The second postulate of Relativity guarantees that clocks at rest in any inertial frame can be synchronized by the procedure described above. But that means that moving clocks won't be synchronized, even if they are synchronized in their own inertial frame.
Imagine that around midnight a pilot flies a small airplane north parallel to the Southern Pacific tracks at 87 miles per hour. Flying along a line some miles west of the tracks, the airplane occupies the same inertial frame that John's train will occupy. A thin tule fog fills the Valley, so scattering light that the pilot can follow the progress of the synchronization pulses from the midway tower. The fog is thin enough, though, that the pilot can see the lights of Modesto and Fresno clearly.
One way to visualize Einstein's second postulate is to imagine a flash of light being emitted in all directions as a kind of luminous aetherial balloon being inflated at tremendous speed. In any inertial frame you care to designate that balloon will be perfectly spherical and will be centered on a point that is stationary in that frame. Our synchronization pulses can be regarded as two bright spots on such a light balloon, the rest of whose area is dimmed to nonvisibility. In the Fresno-Modesto inertial frame those pulses expand away from the emission point, which remains motionless atop the midway tower, but in the pilot's inertial frame the pulses move away at 100 miles per hour from an emission point that remains motionless alongside the airplane.
But in the pilot's frame Fresno and Modesto are both moving south at a little less than 87 miles per hour. In the pilot's view, then, the Modesto train station is rushing to meet its pulse while the Fresno train station seems to be fleeing its pulse. The consequence is that the Modesto clock starts before the Fresno clock does. The two clocks, well synchronized in the Fresno-Modesto frame, are definitely not in synchrony in the pilot's frame. We now want to know by how much the Modesto clock is running "fast" relative to the Fresno clock. In order that the calculation come out right we must unround our numbers: 87 miles (per hour) must be replaced by 86.6 miles (per hour).
In the pilot's frame the distance between Modesto and Fresno has been Lorentz-Fitzgerald contracted to 43.3 miles, so the distance between the midway tower and either city is 21.65 miles. The time it takes the northbound pulse to reach Modesto is just that distance divided by the relative speed between the pulse and Modesto. However, we can't use the actual relative speed, because that would put us into the Fresno-Modesto frame, so we use the fiction that the relative speed is the sum of the individual speeds; that is, 186.6 miles per hour. Of course, this is purely an "as if" speed used for the purpose of our calculation: no two objects can ever have between them a relative speed that is greater than the speed of light. So we calculate the time the pulse needs to go from emission to absorption in Modesto as 21.65 miles divided by 186.6 miles per hour. Likewise the time needed by the southbound pulse to reach Fresno is equal to 21.65 miles divided by 13.4 miles per hour. The amount by which the Modesto clock leads the Fresno clock is just the difference between those two fractions, so we give them a common denominator (13.4 x 186.6 = 2500 (after I drop the 44/100 that comes from my not having unrounded my numbers to enough decimal places)) and carry out the subtraction. We thus have 1.499912 hours (which should be 1.5 hour precisely if I had unrounded my numbers sufficiently), so the pilot calculates that the Modesto clock started one hour and a half before the Fresno clock started.
But that hour and a half is time elapsed on the pilot's clock. The Modesto and Fresno clocks are ticking off dilated time in the pilot's frame, each minute dilated to fill two minutes of the pilot's time. So one and a half hours on the pilot's clock corresponds to forty-five minutes elapsed on the Modesto and Fresno clocks, which means that in the pilot's frame the Modesto clock appears to be running forty-five minutes "fast" relative to the Fresno clock. That frame is just the frame that John's train occupies on its trip from Fresno to Modesto; thus, when the train leaves the Fresno station at 86.6 miles per hour, when the Fresno clock shows twelve noon, the Modesto clock, in the train's frame, shows twelve forty-five. Half an hour later, by John's watch, the train arrives in Modesto, where the station clock, having ticked off fifteen dilated minutes, shows one o'clock, in perfect agreement with Jane's calculation.
When John catches his southbound train at one ten, he is carried into an inertial frame, moving south at 86.6 miles per hour, in which that temporal offset between distant clocks is reversed; that is, he goes into a frame in which it's already one fifty-five in Fresno. When the train arrives in Fresno the station clock shows two ten, again in agreement with Jane's expectation. And when he sits down across the table from Jane, John expresses a fervent wish that he could live in a world in which the speed of light were vastly faster than one hundred miles per hour, so that he would be able to take a simple train ride without having to keep track of relativistic time zones.
The pilot's calculation of the temporal offset between the Modesto and Fresno clocks is straightforward enough that we can see how easy it would be to carry out the same calculation for another inertial frame or for a different pair of clocks. Indeed, if we were to lay out a description of the calculation as a set of rules, absent the numbers that we manipulate with them, (that is, if we use the algebraic representation of the calculation), we might be able to see a way in which we could condense those rules into a more compact and simpler rule. That condensation gives us
LORENTZ RULE 3: If two clocks are synchronized in the inertial frame in which they are at rest, then in any inertial frame in which they are moving the following clock will be found to be running "fast" relative to the leading clock by an interval equal to the product of the at-rest distance between the clocks (measured only in the direction of motion)and the speed at which the clocks are moving divided by the square of the speed of light.
That rule gives us now the means to understand how the Lorentz-Fitzgerald contraction comes about. Imagine that two clocks are mounted on top of the train that's going from Fresno to Modesto, one on top of the locomotive and the other atop the observation car at the end of the train. Let's say that the distance between those clocks, in the frame in which the train is at rest, is one thousand feet. The train leaves the Fresno station at 86.6 miles per (which is also 127 feet per second, the speed of light being, for comparison, 146.67 feet per second) and the train's crew synchronizes the clocks. Moments later the train passes a track inspector who has parked his speeder on a siding.
The inspector compares the readings on his watch with the reading on each of the train's clocks as those clocks pass his position. When he subtracts the respective readings to calculate the elapsed interval and accounts for time dilation, he discovers that the train's rear clock is running fast relative to the forward clock by 5.9 seconds. What that inference means is that if the inspector had been able to set up two cameras alongside the track in such a way that they could photograph the train's clocks simultaneously, the resulting pictures would show two clocks whose readings differ by 5.9 seconds. It's as if the rear clock has somehow been shoved 5.9 seconds into the future relative to the forward clock. But both clocks are moving, which means that the rear clock will partly overtake the forward clock, as if it had been given a 5.9 second head start coming out of the Fresno station. That effect is pro-rated the full length of the train, so the train appears to be uniformly shrunk in the direction of its motion. In 5.9 seconds an object moving 127 feet per second will travel 750 feet, so our train, nominally one thousand feet long, has been shrunk to 250 feet for the track inspector.
Well, that's just
plain wrong. The Lorentz factor between the train's frame and the track
inspector's frame is equal to two, so Lorentz Rule 2B tells us that the train
should shrink to 500 feet. The shrinkage due to the temporal offset is too big
by a factor equal (perhaps not coincidentally?) to our Lorentz factor. By now
you know what that means: we have one more problem to solve and one more Lorentz
rule to deduce and, when you feel up to reading the next essay, we shall do the
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