Length contraction in a boxcar problem

Question

I'm trying to understand length contraction with an eye to the relativity of simultaneity, but things didn't go well. Let's say Alice and Bob both try to find the length of a boxcar moving to the right with speed $v$. Alice is on the boxcar, while Bob is standing still on the ground. As is known to many of us, moving objects are shortened. So Bob would measure the length of the boxcar to be a number $L$ less than the reading $L_0$ taken by Alice. I'm trying to explain how the contracted length arose, by employing the relativity of simultaneity: in Alice's rest frame, Bob is moving to the left; thus, Alice would say that Bob measured incorrectly because he didn't take the readings of both ends simultaneously. In fact, according to my textbook, Alice would complain that Bob measured the front end first and then the back end. Why is that? I can understand the complaint immediately by using the transformation formula $$\Delta t'=\gamma(\Delta t-\frac{v}{c^2}\Delta x)$$ with $\Delta t=0$ and noting that the prime denotes the boxcar reference. But, is there any other approach that doesn't involve the Lorentz transformation? Thank you.

Answer 1

The key to understanding length contraction is to realise that to measure the length of a moving object you have to note the positions of its two ends at the same instant, and then work out the distance between them. If you note the position of one end and then the other, the second end will have moved in the time since you measured the first, so you will get the wrong answer.

The cause of length contraction is that reference frames that are moving relative to each other have planes of simultaneity that are tilted relative to each other. What that means is that events that are simultaneous to Alice happen at different times according to Bob, and vice versa.

When measuring the length of the moving box car, Bob notes the positions of the two ends of the car at a given instant in his frame- as far as he is concerned, he pinned down the two positions simultaneously, so it was a valid measurement. However, in Alice's frame the two events were not simultaneous- in her frame, Bob noted the position of the rear of the train a moment after he had noted the position of the front, which gave the rear of the train time to move forward slightly before its position was noted, thus giving a shortened result for the length of the car.

There is a nice thought experiment you can do to remember how the planes of simultaneity tilt for moving observers. Imagine Alice and Bob stand together, and Alice shines a flash of light to the left and to the right. Alice stands still but Bob walks after the light to the right.

After a nanosecond in Alice's frame, the light will have travelled (roughly) a foot away in each direction. In her plane of simultaneity there are two simultaneous events- namely the light being a foot to her left and the light being a foot to her right. In Bob's frame, however, as he is walking after the right hand flash, at the instant the two flashes are a foot from Alice, the right hand flash is slightly less than a foot from him, and the left hand flash is slightly more than a foot from him. Since the speed of light is the same in all directions, from Bob's perspective both flashes should be equidistant from him at any given instant, so the only way in which the right hand flash can be nearer to him than the left hand flash is if the positions of the two flashes are being noted at different times. From his perspective the right hand flash was a foot away from Alice (and just under a foot away from him) at one moment, then the left hand flash was a foot away from Alice (and just over a foot away from him) at a later moment. The two events that were simultaneous for Alice happened at different tines for Bob.

You should also be able to work out from the though experiment that from Bob's perspective Alice's clocks are getting progressively out of synch in each direction- appearing to be set ahead of time in the direction he is headed, and set behind time in the opposite direction.

Of course, the entire set-up is totally symmetrical. Alice thinks Bob's clocks are out by the same amount in the other direction.

Answer 2

"The professor has NOT introduced anything about the Lorentz transformation and spacetime diagrams up to his solution. So I believe he is using some elementary facts instead of invoking any transformation formula or any weird diagram."

In a thought experiment, you might try and use the fact that the speed of light is the same for both observers to come to this conclusion. Suppose Bob has already measured the length of a moving boxcar to be $l_\text{train}^\text{Bob}$, and wanted to confirm this measurement and discuss it with Alice. Bob creates the following setup: two light sources are placed alongside the tracks the distance $l_\text{device}^\text{Bob} = l_\text{train}^\text{Bob}$ apart, and a detector is placed exactly in the middle. When the boxcar aligns with the two sources (when the front and the back pass their corresponding post), a mechanism triggers, and they each emit short light pulse - these signals travel at the speed of light, and hit the detector on each side (they have traveled the same distance). Alice will record her own observations of events, and also determine these lengths in her own frame: $l_\text{device}^\text{Alice}, l_\text{train}^\text{Alice}$

Why this setup? Because Bob wants to know what Alice observes in her frame, and this setup has a number of distinct single-place events that both Alice and Bob will agree upon. An event (which is what we call a point in space and time) is as such agreed upon by all observers. They might assign it different local coordinates, but a thing that happened at a particular place, happened - there's no way around it. It's just that observers might disagree if two distinct events (things that are, in any frame, separated in space and/or time) are simultaneous or not.

So the unique events of interest here are:

1. the front end of the boxcar is at the same place¹ & time as the front light (and the signal is triggered at this exact moment),

2. the back end of the boxcar is at the same place¹ & time as the rear light, (and the signal is triggered at this exact moment),

3. the two signals arrive to the location of the detector (so at a particular place) at the same time (both observers must agree on what the detector shows, and the detector indicates simultaneous arrival). This is considered a single event (single point in spacetime - one set of coordinates), not two distinct, but simultaneous events.

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¹ They occupy different points in 3D space, but since the effects of special relativity only occur in the direction of motion, we can suppress the coordinate perpendicular to it, and treat the problem as two-dimensional. So, in the 2D picture, the front end of the boxcar, the front light & the emission triggering event all have the exact same coordinates (t, x, y).

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From Alice's perspective, the whole setup is moving past her in the opposite direction. The event (1) happens when the light source on the right passes the front of the boxcar, the event (2) happens when the light source on the left passes the back of the boxcar. So from her perspective, the signals are emitted from the locations of the two ends of the boxcar (both emission points are stationary; it's just a very quick pulse, and the origin does not move with the source). The signals themselves are traveling at the speed of light. Note that I haven't said anything yet about when each emission happened.

Now, since the detector is in motion, if the signals were emitted simultaneously (from Alice's point of view), the signal coming from the back of the boxcar would hit the detector first (because the detector is moving towards it), and the signal from the front would catch up some time later.

But the detector was triggered, and it shows that the signals arrived at the same time, in accordance with (3) - both Alice and Bob must agree on this.

This can only happen if, in Alice's frame, the device was shorter than the boxcar, so that (1) triggers before (2) in such a way as to allow the signal emitted from the front end to catch up to the detector just as the detector slams into the signal coming from the back.

(Note that in the image below, the origin of the arrow is an imaginary point of reference; it tells you from where the signal was emitted so that you can see the travel distance. To Bob, the signal appears to originate from the tilted black line. The the tip of the arrow, though, corresponds to the propagation of the wavefront, which is physical.)

This then tells you (a) that Alice and Bob disagree on simultaneity of distinct events, and (b) that both Alice and Bob see the moving object as shortened (Alice sees the measuring contraption as shortened, and Bob sees the boxcar as shorter than it is in Alice's frame):

$$ l_\text{device}^\text{Alice} < l_\text{device}^\text{Bob} \\ \text & \\ l_\text{train}^\text{Alice} > l_\text{train}^\text{Bob} $$

The back end of the boxcar that Bob experiences as existing at the same time as the front end is in Alice's future. Suppose the back end explodes at the moment the endpoints of Bob's device light up; in Bob's view, the state of the boxcar is that it is a wreck with a hole at the back end. For Alice, who is leaning against the wall at the front end, and who happens to be giving a quick yawn at the moment when the first light source emits the signal, the explosion hasn't happened yet. Even so, Bob sees this same Alice standing and yawning next to a developing explosion. They are literally not experiencing the same time-slice of the world; their simultaneity hyperplanes (what they experience as "now") are tilted in time with respect to one another.