Temporal Offset
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.
habg