Temporal Offset

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Leaving his twin sister Jane in the Fresno train station, John has taken a train moving at Lorentz factor two (about 87 miles per hour in our little fantasy world) and gone 87 miles north to Modesto. The train left Fresno at noon, so what time do the clocks in Modesto tell when John's train arrives in that city's station? 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 knowing about time dilation, figures that the Modesto clock will show twelve-fifteen when he first sees it. Which of those calculations has gone wrong and how can we set it right?

Identifying tacit assumptions makes a good start. In her calculation Jane has tacitly assumed that the station clock in Modesto shows noon when the station clock in Fresno does. Can we accept that as a good assumption? We can, but only if we make it good.

Imagine that halfway between Fresno and Modesto someone has built a high tower, one high enough that an observer at its top with a good pair of binoculars can see the equally high clock towers of the Fresno and Modesto train stations. At about eleven-thirty every night the time-keepers stop the station clocks, set them to show midnight, and then connect them to photocells that will start them when a flash of light illuminates them. 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 occur simultaneously with John's arrival in Modesto.

We cannot take that fact as true to Reality in the inertial frame that John's train occupies. Yes, the second postulate of Relativity guarantees that anyone can synchronize clocks at rest in any inertial frame by using the procedure described above. But it also means that moving clocks won't appear synchronized if someone synchronized them in their own inertial frame. Here's a simple image that makes that statement clear.

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. But the fog fills the Valley thinly enough, though, that the pilot can see the lights of Modesto and Fresno clearly.

You can visualize Einstein's second postulate by imagining a flash of light emitted in all directions as a kind of luminous aetherial balloon inflating at tremendous speed. In any inertial frame you care to designate the light in that flash will comprise the skin of a perfectly spherical balloon whose center remains motionless in that frame, regardless of the motion of the body that emitted the flash. We can thus imagine our synchronization pulses as two bright spots on such a light balloon, the rest of whose area remains dimmed to non-visibility. 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 begins at the tower and then moves north, remaining motionless alongside the airplane.

Now in the pilot's frame Fresno and Modesto both move south at a little less than 87 miles per hour. In the pilot's view, then, the Modesto train station rushes to meet its pulse while the Fresno train station flees its pulse. We have the consequence, then, that the Modesto clock starts before the Fresno clock does. The two clocks, well synchronized in the Fresno-Modesto frame, definitely did not start in synchrony in the pilot's frame. We now want to know by how much the Modesto clock runs "fast" relative to the Fresno clock. In order that the calculation come out right we must unround our numbers: we must replace 87 miles (per hour) by 86.6 miles (per hour).

In the pilot's frame the Lorentz-Fitzgerald contraction has shrunk the distance between Modesto and Fresno to 43.3 miles, so we have a distance of 21.65 miles between the midway tower and either city's train-station clock tower. To reach Modesto the northbound pulse takes an amount of time equal to 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 between the pulse and Modesto equals the sum of the individual speeds; that is, 186.6 miles per hour. Of course, this is purely an "as if" speed that we use for the purpose of our calculation: no two objects can ever have between them a relative speed that exceeds 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 equals 21.65 miles divided by 13.4 miles per hour. We calculate the amount of time by which the Modesto clock leads the Fresno clock by taking 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 elapsed on the pilot's clock, in the pilot's frame. The Modesto and Fresno clocks tick off dilated time in that 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 goes into an inertial frame, moving south at 86.6 miles per hour, in which that temporal offset between distant clocks appears reversed; that is, he goes into a frame in which the Fresno clock already shows one fifty-five. 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.

Our pilot calculated the temporal offset between the Modesto and Fresno clocks straightforwardly enough that we can see how easily we could 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 someone synchronizes two clocks in the inertial frame in which they both stay at rest, then in any inertial frame in which an observer sees them moving, the observer will see the following clock 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 move divided by the square of the speed of light.

We now want to describe the pilot's calculation algebraically so that we can see that process of condensation. We will take the distance between the midway tower and the Modesto clock or the Fresno clock as x/2 in the Fresno-Modesto frame and X/2 = x/2L in the pilot's frame. Thus we have x = 86.6 miles and X = 43.3 miles if L = 2 (as we have made it in our example). The Modesto clock and the trigger pulse move toward the point in the pilot's frame where they will meet each other at the speeds v and c respectively, so we can calculate the time interval between emission of the pulse and its reception by the Modesto clock as if the pulse were flying toward a stationary Modesto clock at the speed c + v; that is, we calculate the time

(Eq'n 1)

Likewise, the Fresno clock seems to flee from its pulse at a speed v, so the time the pulse requires to overtake it can be calculated as if the clock were stationary and the pulse were flying toward it at the speed c - v; that is, we calculate the time

(Eq'n 2)

Now we can calculate the amount by which the Modesto clock will appear to lead the Fresno clock by subtracting TM from TF. That difference is

(Eq'n 3)

But X = x/L, so we can express that offset in terms of distances measured in the Modesto-Fresno frame. We obtain, at last,

(Eq'n 4)

which we must now so modify that it will properly reflect the fact that in the pilot's frame the Modesto clock will tick off dilated time as the southward flying pulse continues its pursuit of the Fresno clock. We use the time dilation equation in the form T = tL and obtain

(Eq'n 5)

which is just the algebraic expression of Lorentz Rule 3.

That rule gives us now the means to understand how the Lorentz-Fitzgerald contraction comes about. Imagine that someone has mounted two clocks 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 in the frame in which the train is at rest a distance of one thousand feet lies between those clocks. The train leaves the Fresno station at 86.6 miles per (which also corresponds to 127 feet per second in our little fantasy world in which light flies, for comparison, at 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 runs fast relative to the forward clock by 5.9 seconds. That inference means that if the inspector had 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, as if the train's motion has somehow shoved the rear clock 5.9 seconds into the future relative to the forward clock. But in the track inspector's frame both clocks move, which means that the rear clock will partly overtake the forward clock, as if the train had given it a 5.9 second head start coming out of the Fresno station. We must pro-rate that effect over the full length of the train, so the train appears 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 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 equals 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 deed.

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