Using Equation 8, we can also see why it takes an enormously fast train, or whatever object, to yield a detectable time dilation. Our first train, travelling at an excruciatingly slow 60 feet per second, moved 16 million times slower than the speed of light. If we plug v / c = 1/16 million into Equation 8, we get a time dilation of only 1 part in 500 trillion. No wonder no one ever noticed this effect before the 20th century!
One more thing: there is a fundamental difference between light and ordinary objects such as tennis balls. The ball in Figure 1 and the light burst in Figure 2 take exactly the same path, but you don’t get time dilation on the slower train in Figure 1, because it’s not a basic law of nature that tennis balls always travel at 45 feet per second. Once the train gets moving, you accept that the tennis ball travels at 75 feet per second (from the point of view of the observer on the ground). It has to, because it travels a greater distance in the same amount of time.
In Figure 2, the light also travels a greater distance, but in this case, it is a basic law of nature that light always travels at the same speed. Something has to give, and Einstein—equipped with the results of experiment and theory—decided it was time that had to blink first.
Now, let’s once again return to our train, moving at 0.8c. This time, let’s put the mirror at the front of the boxcar, and the laser and the detector at the back, so that from your perspective, riding the train, the burst of light now travels the length of the car twice. If the car is 45 feet long, it travels twice 45 feet, or 90 feet. That takes 90 nanoseconds—this should be getting easy by now!
So—how long does it take from the outside observer’s perspective?
From the moment the burst of light leaves the source at the back of the boxcar, it travels, of course, at the speed of light, c. But since the boxcar itself is travelling at 0.8c, the light is only “gaining” on the front of the boxcar at c—0.8c = 0.2c, one-fifth the speed of light. Ordinarily, at the speed of light, it would take 45 nanoseconds to catch up to the front of the boxcar, but at only one-fifth that speed, it takes 5 times longer, or 225 nanoseconds.
After bouncing off the mirror, the light now heads toward the back of the boxcar. But instead of the light having to chase the boxcar, this time the boxcar is rushing headlong to intercept the light, and the outside observer sees detector and light meet at c + 0.8c = 1.8c, or nine-fifths the speed of light. Again, ordinarily, it would take 45 nanoseconds, but now, at nine-fifths that speed, it takes five-ninths as long, or 25 nanoseconds.
The round trip time is therefore 225 nanoseconds plus 25 nanoseconds, or 250 nanoseconds. Now, we know that 250 nanoseconds as measured by the outside observer doesn’t take 250 nanoseconds on board the train. No, on the train, clocks are slowed down according to Equation 8; inside the boxcar, you should measure the interval as
(250 nanoseconds) sqrt(1—0.82) = 150 nanoseconds
But wait—that’s not the time that you actually measured. As we said, you measure it as 90 nanoseconds. So, despite taking into account the time dilation effect, we still have a discrepancy. Where did we go wrong?
Einstein again decided that the analysis was just fine, it was another Newtonian assumption at fault. Which one was it this time? Again, put yourself in Einstein’s shoes and see if you can figure out which assumption to abolish.
He decided that it was the notion of absolute length that was the problem. Aboard the train, you measure the length of the boxcar as the old 45 feet. But in order for the times to match, it must somehow be the case that from the outside observer’s perspective, the boxcar is compressed in the direction of motion! Compressed by how much? In this case, to correct the 150 nanoseconds down to 90 nanoseconds, the length must also be compressed to 90/150 times its former value. Since the original length was 45 feet, the compressed length must be (90/150)(45 feet) = 27 feet. That must be the length of the boxcar from the outside observer’s perspective.
How long would it be in general? In our example, 90 nanoseconds was the time it took for light to travel twice the length of the boxcar as measured on the train itself. Call the length of the boxcar at rest, Lo. Then, the time it takes light to travel from the back to the front, and then back to the back is
(9) to = 2Lo / c
If the train is in motion, however, then we’ve shown that it must be shortened to some length L, and we have to figure out what L is in terms of Lo. If it’s moving at velocity v, then the light catches up with the front of the boxcar at speed c—v. The time it takes to do that is just L / (c—v). (This is just Equation 1 again!) Then, after bouncing off the mirror, the outside observer sees the light meet the back of the boxcar at speed c + v, and the time it takes to do that is L / (c + v). The total round trip time is therefore
(10) t = L / (c—v) + L / (c + v) = 2L / c (1—(v / c)2)
From Equation 8, we know that this time is reduced on board the train to
(11) to = t / (t / to) = t sqrt(1—(v / c)2)
Combining Equations 10 and 11, we get
(12) to = 2L / c sqrt(1—(v / c)2)
Finally, since the to in Equation 9 has to be equal to the to in Equation 12, we can write
(13) 2Lo / c = 2L / c sqrt(1—(v / c)2)
or, by simplifying and rearranging terms,
(14) L / Lo = sqrt(1—(v / c)2) = 1 / y
And this is another part of the Lorentz transform equations, so Einstein had another hint here for length compression. As a matter of fact, Einstein wasn’t the first to suggest length compression. Lorentz and the Irish physicist George Fitzgerald (1851-1901) tried to sustain the aether theory by supposing that as objects such as the earth plowed through the aether, they were dragged and thereby compressed. Everything was compressed, including the Michelson-Morley apparatus, so—as Lorentz and Fitzgerald claimed—the speed of light really did slow down in the face of the “aether headwind,” but since the apparatus was shortened by aether drag by exactly the same amount, no change in the speed of light was recorded!
Aether drag was therefore very similar to what Einstein proposed, but it differed in one significant respect. Aether drag still required that the compression take place relative to an all-pervasive aether, whereas in Einstein’s formulation, compression took place between any two different frames of reference. Experiment eventually proved Einstein right, and aether drag went the way of the dodo.
Again, this isn’t an optical illusion caused by the train moving so fast past the observer on the ground that he underestimates the length. For the laws of nature to make sense, and for light to always have the same speed everywhere, the train must actually shrink in the direction of motion! Does it make any sense? No, but logically that’s what has to happen, and experiment has demonstrated this effect time and again.
Something that we’ve left unstated here, but which you may have guessed at, might be making you a bit uncomfortable. We’ve been assuming all along that it’s the observer on the ground who is “at rest,” and you on the train moving, relative to the ground. But isn’t the whole deal with special relativity that, well, it’s all relative?
Consider this old conundrum. Suppose you’re running at a high speed—again, say, four-fifths the speed of light—carrying a 10-foot pole parallel to the ground. You’re running toward a 10-foot shack with front and back doors, both of which are controlled by a remote actuator, which I control. At a push of a button, I can make both doors close at precisely the same time.
My goal is to trap you and your pole in the shack. Since your pole is 10 feet long, this seems like a tricky task, requiring absolutely precise timing. But I take advantage of the fact that you’re running so fast. At a rate of four-fifths the speed of light, Equation 14 tells me that your 10-foot pole will be compressed to a length of 6 feet, so I have plenty of time to capture you and your pole in the shack. (I have quick reflexes.) No problem!
But you see things differently. From your point of view, it’s the shack that’s moving relative to you, and is therefore compressed to a depth of 6 feet. Your pole is safely longer than that, so there’s no way I can possibly succeed. Again, no problem! Of course there’s no problem. But who’s the one with no problem?