The “Method” of Small Displacements

October 14, 2022 Shreyas Comments 1 comment

1. Introduction

One of the most important things about calculus and its subfields is the ability to think about “very small” changes in certain variables and pull out information from them. The derivative of a variable $x$ with respect to another variable $t$ essentially tells us the relation between a very small change in $t$ (termed $dt$ ) and the very small change in $x$ (termed $dx$ ) that results. But how do we define “very small?” Limits are a powerful way of doing so: we can take $\Delta t$ , a finite and nonzero change in $t$ , and find its resulting $\Delta x$ . Most likely, $\Delta x$ will not be a simple scalar multiple of $\Delta t$ , but we can take the limit of the ratio $\frac{\Delta x}{\Delta t}$ as $\Delta t$ approaches 0: in this way, we effectively wash out any higher order complexities until $\Delta x$ basically starts behaving as a fixed scalar multiple of $\Delta t$ .

Why am I saying all of this? I found an interesting way of applying the concept of “small displacements” to something we don’t often think of as similar to calculus: Euclidean geometry. Essentially, the idea is that given a fixed point $X$ and a point $P$ , we can displace $P$ by a small amount in various directions, and see how quantities like the distance $PX$ is affected as a result.

2. The Basic Cases

Let $L(P)$ denote the distance $PX$ treated as a function of $P$ . The simplest case of this situation is when we displace from $P$ to $P'$ by an infinitesimal distance $dP$ along the line of $PX$ :
In this case, the distance $L$ clearly just increases or decreases by $dP$ depending on whether $P'$ is farther or closer to $X$ than $P$ .
Now, we handle a slightly trickier case: what if the displacement is perpendicular to $PX$ ?

We may express the new distance $P'X$ as $\sqrt{PX^2+dP^2}$ by the Pythagorean theorem. Thus, the difference $dL$ equals

$\begin{equation*} dL = P'X - PX = \sqrt{PX^2 + dP^2} - PX \end{equation*}$

We would like to find how much $dP$ influences $dL$ , so we need a way to work with $dP$ alone. We will use difference of squares in the following way:

$\begin{equation*} dL = \sqrt{PX^2 + dP^2} - PX \end{equation*}$

$\begin{equation*} =\frac{\sqrt{PX^2 + dP^2} - PX}{1} * \frac{\sqrt{PX^2 + dP^2} + PX}{\sqrt{PX^2 + dP^2} + PX} \end{equation*}$

$\begin{equation*} =\frac{PX^2 + dP^2 - PX^2}{\sqrt{PX^2 + dP^2} + PX} \end{equation*}$

$\begin{equation*} \frac{dP^2}{\sqrt{PX^2 + dP^2} - PX} \end{equation*}$

Thus, we have

$\begin{equation*} \frac{dL}{dP} = \frac{dP}{\sqrt{PX^2 + dP^2} + PX} \end{equation*}$

As we let $dP$ approach $0$ , we see that the denominator on the right hand side simply approaches $\sqrt{PX^2+0^2}+PX=2PX \neq 0$ . The numerator, on the other hand approaches $0$ , so we conclude $\frac{dL}{dP}=0$ . This will be a very important fact: displacing the point $P$ an infinitesimal amount perpendicular to $PX$ does not change the length $L$ , even in proportion to the displacement $dP$ .

3. The General Case

Now, suppose the point $P$ is displaced in a general direction, which we will encode by the angle $t$ between the extended ray $XP$ and $PP'$ as shown.

Splitting the displacement PP’ into components

We can thus split the displacement $dP$ into a component that is along $XP$ with magnitude $dP\cos(t)$ , and a component perpendicular to $XP$ with magnitude $dP\sin(t)$ . We consider their effects separately: the parallel displacement increases $L$ by the same amount of $dP\cos(t)$ , while the perpendicular displacement makes no contribution as we have seen. Thus, we have

$\begin{equation*} dL = dP\cos(t) \rightarrow \frac{dL}{dP} = \cos(t) \end{equation*}$

Although a fairly elementary result, it can prove quite powerful in the most extraordinary of circumstances!

4. Mirrors Everywhere

Let us now apply our result to a seemingly disparate problem: how do parabolic mirrors reflect light? Parabolic mirrors can be found everywhere, and for a very good reason: they are able to focus parallel incoming rays onto a single point (and vice versa!).
Specifically, we consider the geometric definition of a parabola: the set of points that are equidistant to a fixed point (the focus) and a fixed line (the directrix). The power of parabolic mirrors is that any incoming ray perpendicular to the directrix will be reflected onto the focus.

Various incoming rays, all perpendicular to the directrix (blue line), all being reflected to the focus (F).

But how can we mathematically prove that this happens? First, we must understand a basic fact about reflection: when an incoming ray strikes a planar surface, the reflected ray will remain on the same side of that surface, and will form an equal angle with the surface as it leaves. If a light ray strikes a curved surface like our parabola, then it will reflect as if it struck a line tangent to the parabola at the point of contact. So, in order to understand the properties of the reflected ray, we will need to find some way to geometrically express the tangent line to the parabola at any point.

Pick an arbitrary point $P$ on the parabola, at which an incoming ray will reflect. Let $L_D$ be the distance from our point $P$ to the directrix, and $L_F$ be the distance to the focus. Because $P$ is assumed to be on the parabola to start with, we have $L_F=L_D$ . Now, roughly speaking, we will find out which direction to “move” a small amount in so that we always stay on the parabola, hence finding the tangent line. If we displace $P$ to $P'$ by an infinitesimal amount $dP$ : then, we know that the resulting infinitesimal change in the distance $L_F$ must satisfy $dL_F=dP\cos(t_1)$ where $t_1$ is the marked angle from the ray $FP$ .

Now, convince yourself that $dL_D=dP\cos(t_2)$ , where $t_2$ is the marked angle from the displacement to the vertical. This should be easier than finding $dL_F$ : we effectively just need to know how much the displacement $dP$ adds to the vertical height (or y-coordinate) of $P$ .

Here is the crucial part: in order for $P'$ to still be on the parabola after the displacement from $P$ , the new distances $L'_F$ and $L'_D$ must be the same. Since we originally had $L_F=L_D$ , we conclude that $dL_F$ and $dL_D$ must also have been the same. So

$\begin{equation*} dL_F=dL_D \rightarrow dP\cos(t_1)=dP\cos(t_2) \rightarrow \cos(t_1)=\cos(t_2) \rightarrow t_1=t_2 \end{equation*}$

We have shown that the marked angles are actually equal; in other words, the tangent to the parabola at the point $P$ is really the angle bisector of $FP$ and the vertical from $P$ . But now the reflection result is trivial! If a vertical ray strikes $P$ , it will deflect with equal angles off the tangent to the parabola at $P$ , which happens to bisect the angle to the focus. Thus, the resultant ray will always intersect the focus.

`Incoming light ray reflecting with equal angles onto the focus.`

5. Ellipses

Other curves, especially conic sections, also have interesting properties when it comes to reflections. Consider an ellipse: the set of points with a constant distance sum to two foci, $F_1$ and $F_2$ . We will apply the same technique to find a geometric interpretation for the tangent to an ellipse at a point $P$ . If we displace $P$ by $dP$ , then the distance $PF_1$ will change by $dP\cos(t_1)$ and the distance $PF_2$ will change by $dP\cos(t_2)$ .

Since we want both the original point $P$ and the displaced point $P'$ to be on the ellipse, we want the sum of distances to the foci not to change; in other words

$\begin{equation*} dL_1+dL_2=dP\cos(t_1)+dP\cos(t_2)=0 \end{equation*}$

$\begin{equation*} \rightarrow \cos(t_1)+\cos(t_2)=0 \end{equation*}$

$\begin{equation*} \rightarrow t_1+t_2=180^\circ \rightarrow \frac{t_1+t_2}{2}=90^\circ. \end{equation*}$

How can we interpret this geometrically? $\frac{t_2+t_2}{2}$ is simply the angle formed by the bisector of $PF_1$ and $PF_2$ . Hence, we see that the tangent line to the ellipse at a point $P$ is perpendicular to the angle bisector of the rays from $P$ to each focus.

But what does this mean for reflections? If you stare at the above diagram long enough, you may notice a very interesting result: any ray of light that originates at one focus will always reflect onto the other focus! This is a direct consequence of the fact that the tangent (and the normal) of the ellipse from the point $P$ always makes equal angles with the lines $PF_1$ and $PF_2$ .

One can also derive interesting properties regarding reflections on other conics like hyperbolas, which the reader is encouraged to investigate. However, this “method” is far more versatile than I once thought, as you will see in the next post…see you soon!

Shreyas' blog – Ignited Minds

I love math and science. This is a blog about things I think of, or work on, in my spare time.