Distance Dilation

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    Shortly after leaving Fresno, John's northbound train crosses the San Joaquin River and then, about 5000 feet further on, crosses Avenue 7 of Madera County. But for whom does that distance span 5000 feet? That particular distance would have to come from someone measuring it in the inertial frame in which the San Joaquin River and Avenue 7 remain stationary or anyone else, such as Jane, occupying that inertial frame. That would include, in particular, a surveyor who goes out to the railroad track and, laying down a measuring rod alongside the track, counts the number of feet that span the distance from the center of the bridge over the river to the center of Avenue 7.

    In accordance with Lorentz Rule 2B, then, we expect that John will measure that distance as 2500 feet. For him the measurement of the distance takes less effort than it does for the surveyor. Sitting next to a window, John starts a stopwatch when his seat crosses the middle of the bridge over the San Joaquin River and he stops the watch when his seat crosses the centerline of Avenue 7. Multiplying the elapsed time thus measured (19.685 seconds) by the 127 feet per second at which the scenery slides by his window gives John the distance he wants to know, 2500 feet.

    Do you suspect that something is about to go wrong here? If so, you need not worry about it: the error is an extremely subtle one. Over the span of our lives we become so thoroughly accustomed to measuring distances between places in much the way that the surveyor does that we encounter great difficulty in remembering that in Relativity we can only legitimately measure distances between events, even if we must contrive the events for that purpose.

    Imagine the following scenario: just north of the San Joaquin River bridge two naughty little boys have set up an ambush. They have spaced themselves 100 feet apart and they have agreed to throw mudballs at the train simultaneously. Of course they have their synchronized clocks with them to ensure that their throws will be truly simultaneous. Well, as you can imagine, the conductor does not like to have his train spattered, so he brings his complaint to the Madera County Small Claims Court. In court he presents this evidence:

1) the testimony of a witness who overheard two boys talking about throwing mudballs at a train, though he can't identify the boys he overheard;

2) a photograph that John took of two boys waving at the train; and

3) a photograph of the train, taken after the train had stopped in Modesto, showing the spatters.

    The boys' mother rises to their defense. She acknowledges that the boys in John's photograph are her sons and then she points out that they are standing 50 feet apart. The spatters in the second photograph are 200 feet apart, so she claims that her boys could not have thrown the mud at the train, not if the mudballs were thrown simultaneously.

    The conductor objects that the scene in the first photograph reflects the Lorentz-Fitzgerald contraction. The two boys in the picture are actually standing 100 feet apart in their frame.

    The mother acknowledges her error in interpreting the photograph. But even standing 100 feet apart, she claims, her boys could not have made spatters 200 feet apart. Is she right?

    If the conductor had known about the surveyor making measurements along the track, he could have called him as a witness who would have resolved the matter immediately. The surveyor would have testified that in his and the boys' frame the train was contracted and, thus, that points that were 200 feet apart on the train were actually 100 feet apart for the boys. However, that testimony is not available, so the conductor must prosecute his case on evidence obtained in the train's inertial frame.

    Because his job requires that he be familiar with the temporal distortions of Relativity, the conductor notices in the first photograph a fact that everyone else missed. The boys' clocks are out of synchrony with each other, the northern boy's clock showing a time 0.59 second ahead of the southern boy's clock. Accounting for time dilation, the conductor notes that 1.18 second would have elapsed on the train's clocks in the interval between the northern boy's throwing of his mudball and the southern boy's throwing of his mudball. With the scenery passing the train at 127 feet per second, that interval corresponds to a distance of 150 feet that the train moves between the throws. Adding to that distance the 50 feet between the boys, measured from John's photograph, yields a distance of 200 feet between the spatters in the train's frame.

    The judge accepts the conductor's reasoning and declares the boys guilty. The boys' mother, being a decent woman, agrees that her sons will spend one week of their summer vacation scrubbing down railroad carriages.

    But John is bewildered. He accepts the statement that the boys were 100 feet apart in their frame. But in his frame, he is being told, the boys are either 50 feet apart (according to his photograph) or 200 feet apart (according to the spatters). Apparently the boys' inertial frame either contracted or dilated due to its motion relative to the train. Which is correct?

    The conductor explains that both are correct. The photograph depicts two events that are simultaneous in the train's frame, those events being the reflection from the boys of the light that goes into John's camera when the shutter opens. In that case Lorentz Rule 2B tells John what the distance between the boys must be in his frame. On the other hand, the spatters represent events that were simultaneous in the boys' frame. In that frame the Lorentz-Fitzgerald contraction has so shrunk the train that 200 feet of train fits into the 100 feet between the boys. Thus, in the train's frame the 100 feet between the boys' throwing of their mudballs has dilated to 200 feet. That analysis gives us now

LORENTZ RULE 4: If the distance between two events is measured in the frame in which they occur simultaneously and if that distance is measured parallel to the direction of motion of some other inertial frame, then the distance between the two events in the second frame will be equal to the distance measured in the first frame multiplied by the Lorentz factor between the two frames.

    Algebraically we need only consider John's observations and recall the imaginary experiment that led us to devise Equation 1 in the essay on inertial frames of reference. Boys throwing mudballs at a train and firecrackers smudging a white plank mark pairs of events in space and time, but now there's a significant difference: in the essay on inertial frames we feigned to believe that relative motion does not distort time. Now we know better and we know how to apply that knowledge.

    In their own frame the boys stand 100 feet apart (x = 100 ft) and in the train's frame the boys appear to stand 50 feet apart (X = x/L = 50 ft; L = 2). In the boys' frame the difference between the times displayed on the boys' clocks equals zero (that is, the clocks are synchronized with each other), but as seen from the train's frame the difference is 100 feet times 127 feet per second divided by the square of 146.7 feet per second, which is 0.59 second. But that's dilated time, which the train's clocks would register as 1.18 second, a time in which 150 feet of landscape would appear to pass the train. Now we have Equation 1 from our previous essay with the time interval taken to correspond to the temporal offset between the clocks;

(Eq'n 1)

in which we have made the appropriate substitutions for measurements made in the Fresno-Modesto frame. We factor out xL, substitute the definition of L, and obtain a rewritten equation,

(Eq'n 2)

Thus, simultaneous events that lie 50 feet apart in John's frame (reflection of a certain amount of light from the boys) lie 100 feet apart in the frame that moves relative to John's frame with Lorentz factor 2 (i.e. the boys' frame) and simultaneous events that lie 100 feet apart in the boys' frame (the throwing of the mudballs) lie 200 feet apart in the frame that moves relative to the boys' frame with Lorentz factor 2 (i.e. the train's frame). Equation 1 represents specifically the latter of those two calculations.

    Now we can solve our track inspector's problem from the essay on temporal offset. Someone tells the inspector that a train that spans 1000 feet will roll past him at a certain time. Measurements that he takes as the train passes his location enable him to infer that if he could see the clocks set on both ends of the train simultaneously (that is, from the same distance to cancel out the time lag due to the finite speed of light), he would see the rear clock showing a time 5.9 seconds ahead of the clock at the front of the train. Taking that temporal offset as if it represented a head start that the rear of the train was given over the front of the train, he then infers that the train will appear in his frame to be shrunk in the direction of relative motion, but he calculates the wrong length for the contracted train.

    But now we know that when we say that the train spans 1000 feet we mean that 1000 feet lie between two events that occur simultaneously at opposite ends of the train in the train's frame of reference. And we also know now that the length of the train in the track inspector's frame must come from the corresponding distance between those same two events as measured by rulers set alongside the track. By Lorentz Rule 4, then, the length of the train in the track inspector's frame must span 2000 feet.

    With that information the track inspector prepares to recalculate the apparent length of the train and he sees that he made another error in his first calculation. He neglected to take into account the fact that time displayed on the train's clocks is dilated relative to time elapsed on his clocks. In order to calculate correctly the shrinkage of the train, he must apply the fact that the 5.9 second temporal offset that he inferred between the train's clocks corresponds to 11.8 seconds on his clocks. With an 11.8 second head start at 127 feet per second, the clock at the rear of the train will overtake the clock at the front of the train by 1500 feet, so the inspector calculates that in his frame the train should appear to span 500 feet, which corresponds exactly to what Lorentz Rule 2B tells us.

    Now we have no more paradoxes to resolve in relativistic kinematics (the science of motion per se). After presenting you with a distillation of the material that I have laid out for you so far, I will extend the Lorentz Transformation to extra dimensions and then show you what relativistic dynamics looks like.


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