There is a theorem so powerful in geometry, that it lies at the heart of not only geometry, but also coordinate geometry, trigonometry, algebra, vector physics and more! I am talking about an ancient theorem known ages ago, called in modern day by a name none other than the Pythagorean theorem.
I am going to try to explain the beauty of this theorem and the breadth of its application in this post. This brings me to a key difference between mathematics and physics. In mathematics, one may never ask why, only if. You can ask if 11 is prime and how to determine if it is prime. But you may never meaningfully pose the question .. why is 11 prime. Yet I deeply believe, that all of human endeavor into the sciences relies on understanding the why. This is why I like physics so much, it allows us to ask why. My effort here is not really to prove such a well known theorem, but rather ask why the Pythagorean theorem is true. In the process, of course, it will be a non-rigorous proof.
So you may ask, am I claiming that prime numbers or other areas of relatively abstract math are unnatural? Not at all! In other words, what I am hinting at is the fact that we cannot answer why 11 is prime is just a sign of incomplete understanding. Even abstract math is often rooted in our observations of reality, if not specifically in nature. In fact, this is often the litmus test for something being the right way to define things or pick axioms.
So, let us start with a right triangle of side lengths a, b,c. Assume c is the hypotenuse, without loss of generality. Let us consider a light source at each vertex of the triangle, casting a vertical shadow. This is called a projection in physics. Specifically here, we are using an orthogonal projection. If it helps, one may imagine the hypotenuse as a ladder and the other two sides as a wall and the ground. So, with light sources present at each vertex, both sides cast shadows on the hypotenuse and the hypotenuse casts shadows onto each side from the light source at the opposite vertex. (In our example of the ladder, the ladder casts a shadow on the wall and on the ground. It is important to note that the shadows from the ladder fall vertically as the light sources are at the foot of the ladder and the top of the ladder)
Now we switch our attention to the shadows that fall from the sides onto the hyptoenuse. (Imagine a light source just behind the point where the ground meets the wall, the ground prevents part of the light from reaching the ladder and the other part of light is blocked by the wall, so no light reaches the ladder). Then we notice that the shadow of a and b fall onto c without any overlap!!! This is the key observation. So to get the side length of c , we need to simply add up the length of the shadows of a and b. This leads us to the question of determining the projection effect.
Note that when the shadow of c falls onto side a, it exactly covers the side a, so the shadow has length a. This leads to a length contraction. Now when the shadow of a falls back onto c, it is a second projection of the same contraction factor. A slight nuance is implicit: The projection factor depends only on the angle between two lines and not on which line is projected onto another. This is provable by symmetry, as if we extend the two lines to infinity, then both lines are indistinguishable from each other… we can reverse the construction to use the other line to cast the projection or shadow and the factor has to remain the same.
If c—> a in the first projection, then by the unitary method, 1–> a/c and consequentially, a —> a*a/c. But this is what happens in the second projection, when a casts a shadow on c! This gives us the length of the shadow cast by a. Repeating for side b, we get, c—> b ==> b—> b*b/c. Now adding up both shadows, we get c = a*a/c + b*b/c. Multiplying through by c, we get the usual form of the theorem!
So what have we learnt? We learnt that a good name for this theorem might be … the double projection theorem. The same factor by which the hypotenuse projects its shadow onto the sides is again the physical factor when the sides are projected back onto the hypotenuse. And the sides project back with no overlap, so they add up to the hypotenuse. This double projection or factor is the whole reason the exponent in the Pythagoras theorem is 2, and not some other power.
Now let’s study the generation of triples, which are basically sets of whole numbers which can be sides of a right triangle as they fit the equation, like (3,4,5). In general any set {(u-v)2, 4uv, (u+v)2} will suffice, if 4uv is a perfect square. This condition is definitely possible to meet by starting with non-prime equal u,v and factorizing it. Obviously, with equal u,v the product is a square. Then, exchange some of the factors, to make them unequal and still keep the product the same .. i.e, a perfect square. For example, if we start with u=v=6, then exchange factors to set u=9 and v=4, then the triple (as per the above formula) would be 13, 5 and the square root of 144, which is 12. There is a proof that this is the only way to construct triples! To be precise, the claim is that this construction covers every possible set, upto a factor, so all the co-prime ones would be covered at the very least.
In terms of application, we can show that any directional quantity (known as vectors) for instance, velocity, adds up in this way. If an object is moving to the left at 3 units velocity and forward at 4 units velocity, how far will the object travel in total?
Let’s break it down in time: In unit time, the object moves 3 units to the left and 4 units forward. There is no impact due to the order of translation, as the distance covered is the same (regardless of whether the translation in both left and forward direction is simultaneous or one direction is covered first followed by the other). But following the sequential step of motion, essentially the sides of a right triangle are traced out, if we mark the object’s original position and trace the path to a new position (And if the motion is simulatenous, we know that the starting position and final position must be the same as the sequential motion). The final translation will be the shortest distance between the initial and final position, which is the straight line betwen the points, which in turn is the hypotenuse of the right triangle with aforemntioned sides. THe hypotenuse has length 5, covered in unit time. So, the effective velocity is 5 units, equivalent to adding the impact of the two velocities by the Pythagoras theorem.
Let me also take the opportunity to explain why orthogonal directions are considered independent in physics. (Independent things/directions are called dimensions). It is because we can meaningfully specify any vector value (like velocity) for one direction separately from the other. On the other hand, if we have two directions at 60 degrees with each other, then 5 units velocity along one line is not consistent with, for example, zero velocity in the other direction. An object moving along one of the lines is by default also moving along the other direction. (An object having zero velocity in a direction is defined as: if we draw a line parallel to the direction, the closest point on the line to the object does not change). This is however possible, if the lines are orthogonal. The closest point on the orthogonal line does not change and there is no motion in the direction that the line is pointing toward.
This is how the Pythagorean theorem derives it’s great importance and utility. It is a formula to combine actions along multiple dimensions. It is more than just a triangle! (… in some sense, but in another sense, it is all just a triangle)
Cars and Physics 