Uh oh. Another Whabby Wednesday? Yep… afraid so! When we left off last week, we’d solved the problem of how to let you see the light pulse even though, when it gets back to Shirley’s floor, the old gal is two kilometers down to your right. This is illustrated in the figure above. The pulse travels up the cylinder to the top while your spacecraft is moving one kilometer to the right; it travels back down while the craft moves another kilometer to the right. By then, the glass in Shirley’s floor has angled itself to reflect the light right back into your eyes. Moreover, your stopwatch is smart enough to automatically subtract the extra time it takes the pulse to cover the distance from Shirley’s floor back to you.
The figure illustrates another important fact. When the pulse of light reaches the floor, from George’s perspective, it has traveled along a perfectly straight path up to the ceiling and back down, just as it did when Shirley wasn’t moving. This is because George is moving right along with Shirley. It would be just like shining George a flashlight on the ceiling of a moving train from within the dining car. Both the flashlight and the dining car are moving together, at the same speed, so the beam from the flashlight would go directly up and down.
From your perspective, though, Shirley is displaced two kilometers when the flash returns to the floor. From your frame of reference, the pulse cannot have traveled straight up and straight down! To see the path it’s had to travel, look at the really narrow triangle on the left side of the figure below. Shirley is one kilometer to your right after one second. To reach the ceiling, therefore, the pulse has had to travel up the slanted side (the hypotenuse) of a very, very, very narrow right-angle triangle. Then, after the pulse is reflected, it has to travel down the hypotenuse of an identical (but mirror-image) right-angle triangle to get back to the floor. As you can see from the figure, the slanted side of a right-angle triangle, which I’ve labeled Line A, is ever so slightly longer than the straight up-and-down line which is the path the pulse follows for George. I’ve labeled that line “Line C” (bonacci will know why I’ve done that). The side of the triangle created by Shirley’s motion to your right is Line B.
Now, the speed of light is the same everywhere, in all frames of reference. Since the pulse of light has to cover more distance for you than it does for naked George, the pulse has to take more time to complete its journey for you than for George. How much more time is determined by exactly how much extra distance there is along Line A compared to Line C. With the use of some simple grade-school geometry, we will compute that distance, and work out precisely how much additional time the light pulse took to traverse it.
To work out that distance, we must turn to the handy old Pythagorean theorem.
In plain English, the Pythagorean theorem is a way of calculating the length of Line A when you know the lengths of Lines B and C. As shown in the formula below, you multiply the length of Line B by itself, then multiply the length of Line C by itself, then add (sum) those two values together. Line A is just the square root of that sum.
B2 + C2
A =
For the triangle we’re considering, Line C, the distance that the light beam covers from the viewpoint of naked George, is 300,000 kilometers. Line B is just the distance Shirley has traveled in one second, or one kilometer. 300,000 multiplied by 300,000 is a ridiculously large number, 90 billion in fact. On the other hand, as far as Line B is concerned, multiplying the number one by itself equals… LOL… one! The sum of these two numbers is 90 billion and one. Obviously, being so, so very close to 90 billion, the square root of 90 billion and one is virtually identical to the square root of 90 billion itself, or 300,000.
Even thought this all may seem quite obvious, I want to draw your attention to a couple of things. First, note how this simple math jives completely with the physical shape of the triangle in the figure. The illustration doesn’t begin to do justice to the small size of the slope of line A; in reality, that slope is so small that A almost overlays C directly; they are almost exactly the same length (or almost exactly the same line). Clearly, when Line C is super long and Line B is super short, virtually all of the length of the hypotenuse (Line A) is determined by the length of the much longer line C.
Turning this around, the visual impression from the figure, that Line A is almost equal to Line C, jives perfectly with the Pythagoran theorem. When you square a number that’s quite large to begin with, you end up with a humungous number; on the other hand, when you square a very small number, the result is still pretty small. When you add them together and take the square root of the resultant, the long line contributes virtually everything, and the short line contributes virtually nothing.
Compare that situation to the right-angle triangle depicted in the figure below, where Lines B and C are both 300,000 kilometers long. Since Line B = Line C, B squared and C squared are the identical number (90 billion), and since you add the one to the other, for a total of 180 billion, before taking the square root, they contribute exactly equal amounts to the length of side A (which turns out to be 424,264 kilometers). Of that amount, each line contributes exactly 212,132 kilometers, or just over 71 percent of its own length. As long as they are of equal length, each side of a right-angle triangle always contributes exactly 71% of its length to Side A. Don’t believe me? Pick some examples of your own, say B and C are both one kilometer long, do the Pythagorean math, and confirm it for yourself!
The purpose of this little geometry lesson won’t become obvious until the later blogs. For now, let’s return to our main thread. When Shirley was traveling one kilometer a second, mathematically, the tiny extra distance “donated” to Line A by the presence of Line B took the light pulse such a tiny amount of extra time to traverse that it was way below the sensitivity of you and your stopwatch to measure. For all extents and purposes, both you and your confederate were still in the same “time zone”. But what if you increased Shirley’s speed to something much more substantial, say half the speed of light, or 150,000 kilometers per second? That’s just absurdly quick; 2000 times as fast as the fastest spacecraft we’ve ever launched, in fact. In just the single second that it takes for the light pulse to stream away from George’s searchlight and travel up to the ceiling, the spacecraft will cover 150,000 kilometers along the line to your right add
the additional second it takes the pulse to return to the spacecraft floor, and the spacecraft reaches a point 300,000 kilometers away!
My question to you for today’s comment section is: What does that do to the shape of the all-important right angle triangle? What does it do to the lengths of Lines A and B? What is YOUR estimate of the time you will measure now for the light pulse to complete its journey? And why?
Or, if you wish, you can just wait until next Wednesday’s blog for the answer!
