Ladies and Gentlemen: Pythagoras!

  Picking up exactly where we left off, we’ve now solved the problem of how to ensure that you see the light pulse even though, when the pulse makes it back to Shirley’s floor, the old girl is two kilometers down the line to your right.  At risk of belaboring the point, the solution is illustrated (again) in the figure above.   The pulse travels up the cylinder to the top while Shirley travels one kilometer away to the right; it travels back down while she moves another kilometer to the right.  Meanwhile, the glass in Shirley’s floor has reconfigured itself to reflect the light right back to your eyes.   Your stopwatch automatically subtracts the extra time it takes the pulse to cover the distance from Shirley back to you. 

The white arrows inside Shirley, showing the path of the light pulse from George’s perspective, illustrate the other important fact that I briefly touched on last time.  When the pulse of light reaches the floor, in George’s reference frame 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 if George shone 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.  Viewed another way, from George’s perspective, Shirley is perfectly stationary. 

From your perspective, though, Shirley is displaced two kilometers when the flash returns to the floor.  That simple geometric fact dictates one of the linchpins behind General Relativity and the whole phenomenon of time dilation: From where you’re sitting (actually, from where you’re lying, since you’re flat on the ground), the light pulse cannot have traveled straight up and down!  To see the path it’s had to take, look at the really narrow triangle on the left side of the figure to the right.  When the light pulse reaches her ceiling, Shirley is physically one kilometer down to your right.  To get to the ceiling, therefore, the pulse HAS to have traveled up the slanted side (the hypotenuse) of a very, very, narrow (very acute) right-angle triangle.  Then, after the pulse is reflected, it has to travel down the hypotenuse of an identical (but mirror-image) triangle to get back to the floor.  As the figure illustrates, the hypotenuse of the two right-angle triangles, which I’m going to label Line A, is ever so slightly longer than the straight line that defines the path that the pulse follows for George.  I’ve labeled the line in the triangle that is the equivalent length to George’s path as Line C.  That leaves the side of the triangle created by Shirley’s motion down the line to your right as Line B.

Now here’s the crux of the issue.  The speed of light is the same everywhere, in all frames of reference.  Since the light pulse has to cover more distance for you than it does for George, the pulse has to take more time to complete its journey.  How much more time is determined by how much longer Line A is than Line C.  Employing some simple grade-school geometry, let’s now compute that distance.

To work it out, we need to use the handy old Pythagorean theorem that you hopefully remember from boring high school math courses.  In plain English, the Pythagorean theorem provides a way to calculate the length of Line A based on the lengths of Lines B and C.   As shown in the formula below, you multiply the length of Line B by itself, and then multiply the length of Line C by itself, and then add (sum) those two values together.   Line A’s length is just the square root of that sum:

 For the triangle we’re concerned with, the length of Line B is just the distance Shirley travels in one second, or one kilometer.  Line C is the same distance that the light pulse covers from George’s perspective, or 300,000 kilometers.  300,000 multiplied by itself 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 extremely 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.  Thus, the extra distance along Line A compared to Line C is miniscule. 

Even thought these calculations may seem quite obvious, working through them enables me to draw your attention to a couple of things that will become important later on.  First, note how the math jives completely with the physical shape of the triangle in the figure.  My illustration doesn’t begin to do justice to the small size of the slope of line A; in reality, the slope is so small that if they were superimposed, A would overlay C almost perfectly.  Clearly, when line C is super long and B is super short, virtually all of the length of the hypotenuse (Line A) is determined (though I actually prefer the connotations of the word “donated”) by the much longer line C.

Turning this around, and comparing the figure to the math, the visual impression that Line A is almost equal in length to Line C jives perfectly with the Pythagoran theorem.  When you square a number that’s huge to begin with, you end up with a humungous number; when you square a very small number, the result stays pretty small.  When you add them together (and then compute the square root), the big number (the long line) contributes virtually everything to the result, while the small number (the short line) contributes virtually nothing.

Forgive me, patient readers because, at this point, I’d like to take a short detour from our main thread to more thoroughly illustrate how the relative lengths of B and C determine what proportion of the length of Line A is controlled by the one line compared to the other.  In the triangle we’ve been discussing up to now, Line B is very short compared to Line C.  But now, please direct your attention to the triangles depicted in the figure below, where Lines B and C are equal in length (both 300,000 kilometers long).  Since now, Line B equals the length of Line C, B squared and C squared are both 90 billion, and when you sum the two, for a total of 180 billion, and compute the square root, you find that they contribute exactly equal amounts to the length of side A (which turns out to be 424,264 kilometers).  In other words, of Line A’s total length, B and C both “donate” exactly half, 212,132 kilometers, or just over 71 percent of their own lengths. The crucial take-home message:  Once their lengths are equalized, Lines B and C contribute equal amounts to the length of Line A. 

Don’t believe me? Pick some examples of your own, say a triangle where B and C are just one kilometer long, run the numbers through the theorem, and confirm it for yourself!   And suppose you went back to our target triangle, where Line B is a tiny fraction of Line C, and used the Pythagorean theorem to calculate the length of Line A while gradually increasing the length of Line B.  What you’d find, of course, is that the longer Line B got relative to Line C, the more Line B would “donate” to the length of Line A.  

The importance of what happens when Line B approaches (and then exceeds) the length of Line C won’t become obvious until later blogs.  For now, let’s return to our main thread, where Shirley is traveling only one kilometer a second, and Line B is miniscule compared to Line C.  As I mentioned earlier, the tiny extra distance “donated” to Line A by the presence of Line B takes the light pulse such a tiny amount of extra time to traverse that it is way below the sensitivity of you and your stopwatch to measure.  For all extents and purposes, both you and Astronaut George are still in the same “time zone”, and Einstein has still not really shown up to the party. 

But that’s about to change.  In the next blog, you and George are going to crank up Shirley’s speed to something far more substantial: fully half the speed of light (150,000 kilometers per second).  That’s just absurdly fast, almost 2,000 times faster than as the fastest spacecraft ever launched.  It is so fast, in fact, that in just the single second it takes for the light pulse to travel up to Shirley’s ceiling, the old gal travels 150,000 kilometers along the line to your right (and creates a Line B that’s 150,000 kilometers long).  Add the additional second it takes the pulse to return to the spacecraft floor, and Shirley will travel a total of 300,000 kilometers away to your right! 

What will such a blinding speed do to the shape of our all-important triangles?  What will it do to the relative lengths of Lines B and C?   If you get really ambitious, you could easily run the numbers through the Pythagorean theorem yourself, and then compute the distance the light pulse will now have to travel.  What value will you, stationary on the ground, register on your stopwatch for how long it takes the pulse to complete its journey?  And what value will George register?

Stay tuned!