THE LENGTH PARADOX

Back to Contents

    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? Why, for Jane, of course, and anyone else occupying her inertial frame, the one that's stationary relative to the Fresno train station and relative to the ground between the San Joaquin River and Avenue 7. 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 is even easier than it is for the surveyor. Sitting next to a window, John starts a stopwatch when his seat crosses the middle of the bridge 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. We are 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 is not pleased to have his train spattered, so he brings his complaint to the Madera County Small Claims Court. The evidence that he presents is this:

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.

    Thus, simultaneous events that are 50 feet apart in John's frame (reflection of a certain amount of light from the boys) are 100 feet apart in the frame that's moving relative to John's frame with Lorentz factor 2 (i.e. the boys' frame) and simultaneous events that are 100 feet apart in the boys' frame (the throwing of the mudballs) are 200 feet apart in the frame that's moving relative to the boys' frame with Lorentz factor 2 (i.e. the train's frame).

    Now we can solve our track inspector's problem from the previous essay. A train that he is told is 1000 feet long rolls past him and measurements that he takes as the train passes 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.

    Imagine that an airplane is pacing the train, flying north parallel to the railroad track at 86.6 miles per hour, and that a passenger in the airplane takes a picture of the train. At the instant that the passenger presses the shutter release a straight line drawn perpendicular to the railroad track from the camera strikes the train exactly halfway between the two clocks; thus, the distances between the clocks and the camera are equal to each other and the light that goes into making the picture was reflected from the clocks simultaneously. Simple geometry then enables the passenger to calculate the length of the train as 1000 feet, that length actually representing the distance between two simultaneous events in the passenger's inertial frame. By Lorentz Rule 4, then, the length of the train in the track inspector's frame must be 2000 feet.

    With that information the track inspector prepares to recalculate the apparent length of the train and sees that he made an 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 be 500 feet long, which is just 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 show you what relativistic dynamics looks like.

efefefaaabbbefefef

Back to Contents