# Dog leads, mirrors and Hermann Minkowski

Donovan Young looks at the shapes made when two cones collide

… it seemed to us a garden full of flowers. In it,
we enjoyed looking for hidden pathways and discovered many a new
perspective that appealed to our sense of beauty, and when one of us
showed it to the other and we marvelled over it together, our joy was
complete.

—David Hilbert’s obituary for Hermann Minkowski (1910)

You know those extendable dog leads—the ones where the dog can go really far? Imagine you’re standing still and the dog on the other end is running around. How fast is the lead unspooling? At this point I’ll issue a spoiler alert and encourage you to go away and think about this problem for a while, and when you’re satisfied, turn the page and read on.

## Vectors

Vectors are objects that have a size and a direction, and are widely used in geometry, physics, and elsewhere. In this article, we use two-dimensional vectors, for example
$\boldsymbol{r}=\begin{pmatrix}x\\y\end{pmatrix}=(x,y).$
This is the vector that represents a movement of $x$ units horizontally and $y$ units vertically.

Using Pythagoras’s theorem, we can write the magnitude of a vector as
$\left|\boldsymbol{r}\right|=\sqrt{x^2+y^2}.$
We write $\widehat{\boldsymbol{r}}$ for the unit vector pointing in the same direction as $\boldsymbol{r}$, ie
$\widehat{\boldsymbol{r}}=\frac{\boldsymbol{r}}{\left|\boldsymbol{r}\right|}.$
We can compute the dot product of two vectors by multiplying the pairs of components together then adding up the results, ie
$\begin{pmatrix}x_1\\y_1\end{pmatrix}\cdot \begin{pmatrix}x_2\\y_2\end{pmatrix}=x_1x_2+y_1y_2.$

Now that you’ve had time to think about the question yourself, we can take a look together.

If the dog is running around you in a circle, the lead isn’t unspooling at all. If the dog is running directly away from you, in any direction, we would expect the rate of unspooling to be the greatest (for a fixed speed). In fact, we only care about the component of the dog’s velocity in the direction radially away from the origin. That component is the unspooling rate.

The story I’m about to tell in this article has nothing to do with dogs, but hinges on the answer to this problem, which, expressed mathematically, says that in a small time $\mathrm{d}\hspace{0.7pt}t$, in which the $x$-coordinate of the dog moves from $x$ to $x+\mathrm{d}x$ (and similarly $y$ to $y + \mathrm{d}y$), exactly this much lead,
$\mathrm{d} \sqrt{x^2 + y^2} = \frac{x\,\mathrm{d}x + y\, \mathrm{d}y}{\sqrt{x^2 + y^2}},$
is unspooled. In vector language this may be expressed as
$\mathrm{d}|\boldsymbol{r}| = \widehat{\boldsymbol{r}}\cdot\mathrm{d}\hspace{0.7pt}\boldsymbol{r},$
where I have used the dot product to select the component of the vector $\mathrm{d}\hspace{0.7pt}\boldsymbol{r}$ which is in the direction of $\boldsymbol{r}$.

## Hocus focus! No smoke, but a curved mirror

To start with, we want to find a curved mirror with the following property: a laser pointer shone from the origin would be reflected through the point $(c,0)$, no matter the direction that the laser pointer is pointed. (If you already know the answer, bear with me. I hope to shed a slightly different perspective on a well-known curve.) Already we can see that this mirror must wrap around the person at the origin, so it will be a closed surface. It will also have rotational symmetry around the $x$-axis, which means we can reduce the problem to a closed curve in the $xy$ plane which is then rotated about the $x$-axis to produce the surface of the mirror.

All lasers shone from the origin should be reflected through point $(c,0)$.

Let $(x,y)$ be a point on the curve. The vector $(\mathrm{d}x,\mathrm{d}y)$ is tangent to the curve at that point. The law of reflection states that the angles the incoming and outgoing rays make with the tangent are the same.

The vector which joins the origin to $(x,y)$ is $(x,y)$, while that which joins $(x,y)$ to $(c,0)$ is $(c-x,-y)$. The equal angles can be expressed this way, using the dot product:
$\widehat{(x,y)} \cdot (\mathrm{d}x,\mathrm{d}y) = \widehat{(c-x,-y)} \cdot (\mathrm{d}x,\mathrm{d}y),$
or
$\frac{x\,\mathrm{d}x + y\, \mathrm{d}y}{\sqrt{x^2 + y^2}} = -\frac{(x-c)\,\mathrm{d}x + y\, \mathrm{d}y}{\sqrt{(x-c)^2 + y^2}}.$

But wait! We know this expression from thinking about dog leads! This is
$\mathrm{d} \sqrt{x^2 + y^2} = -\mathrm{d} \sqrt{(x-c)^2 + y^2},$
or
$\mathrm{d} \left(\sqrt{x^2 + y^2} + \sqrt{(x-c)^2 + y^2}\right) = 0.$

## Conic sections

The conic sections are a family of curves that can be obtained by taking slices through a cone.

• A circle can be obtained by taking a horizontal slice through the cone.
• An ellipse can be obtained by taking a non-horizontal slice that is less steep than the side of the cone.
• A parabola can be obtained by taking a slice with the same steepness as the cone.
• A hyperbola can be obtained by taking a slice steeper than the steepness of the cone.

This is one of the easiest differential equations you will ever see: it says that
$\sqrt{x^2 + y^2} + \sqrt{(x-c)^2 + y^2} = k,$
where $k$ is a constant. We asked for the curve with the reflective property explained earlier, and we’ve found a curve with another property: the sum of the distances from the points $(0,0)$ and $(c,0)$ to any point $(x,y)$ on the curve is a constant $k$.

If you are familiar with the conic sections, I hope you don’t feel too cheated right now, because the curve we’ve just found is called an ellipse, and it is usually thought of as coming from the intersection between a cone and a tilted plane. It is famous for having these properties.

## Spacetime diagrams and intersecting light cones

I want to reconsider our solution
$\sqrt{x^2 + y^2} = k-\sqrt{(x-c)^2 + y^2},$
and think of it as the intersection of the two surfaces
$z=\sqrt{x^2 + y^2},$
and
$z= k-\sqrt{(x-c)^2 + y^2}.$

When viewed from above, the intersection of these two cones is an ellipse.

What are these surfaces? They are cones, both with an opening angle of 45 degrees. The first opens upwards from the origin. The second opens downwards from $(c,0,k)$.

This is interesting: we have found a famous conic section, the ellipse, from the intersection of two cones, neither of which is the cone used in the usual conic section construction! What do these cones represent?

Another way to think about light, rather than as rays, is as wavefronts. These are circles which begin as points at their source, and then expand out homogeneously at a fixed speed (the speed of light)—like ripples on a pond when a rain drop falls in. In this language, our mirror problem is asking that wavefronts emanating from the origin are reflected and ‘zero back in’ on the point $(c,0)$. Hermann Minkowski invented a lovely way to visualise expanding wavefronts, a device now called a spacetime diagram. He needed this to think about Einstein’s theory of relativity, but our use will be slightly more mundane (but no less beautiful!).

In a spacetime diagram, we place the $xy$ plane in its usual position (ie horizontal). The vertical axis is for time. What would an expanding circular wavefront look like in such a diagram? In the $xy$ plane, ie when the time $t=0$, we would have a point at the origin—the nascent wavefront waiting to expand outwards. At some time $T$ later, ie on the plane $t=T$, we would have a circle of radius the speed of light times $T$. So an expanding wavefront traces out a cone. We’ll assume natural units—this means we take the speed of light to be 1 and the opening angle of the light cone is 45 degrees.

I hope you now see that our intersecting cones are just the light cones of the outgoing and reflected wavefronts. Presto chango, our ‘$z$’ coordinate is time. We’ve ‘lifted’ the ellipse from a plane curve to a three-dimensional one. This new curve represents the history of when the various parts of the outgoing wavefront encountered the elliptical mirror and were reflected. The time it takes for any part of the wavefront to go from the original point (the origin) to the final point at $(c,0)$ is constant (and equal to $k$). The ancient Greeks, and most famously Apollonius of Perga, knew a lot about the conic sections, but now I think they had the wrong cones. Let’s see how much more we can do!

## Two more mirrors…

In the case of the ellipse, the reflected ray travelled from the point of reflection $(x,y)$ to the point $(c,0)$. A close cousin of the ellipse is obtained by asking that the reflected ray, continued backwards from the point of reflection, passes through $(c,0)$, where now we must impose $c < 0$.

The law of reflection now says
$\widehat{(x-c,y)} \cdot (\mathrm{d}x,\mathrm{d}y)=\widehat{(x,y)} \cdot (\mathrm{d}x,\mathrm{d}y) ,$ or $\frac{(x-c)\,\mathrm{d}x + y\, \mathrm{d}y}{\sqrt{(x-c)^2 + y^2}}= \frac{x\,\mathrm{d}x + y\, \mathrm{d}y}{\sqrt{x^2 + y^2}} ,$ or $\mathrm{d}\left( \sqrt{(x-c)^2 + y^2}-\sqrt{x^2 + y^2} \right) = 0.$
The solution of this is

$\sqrt{(x-c)^2 + y^2}-\sqrt{x^2 + y^2} = k.$
This curve is called a hyperbola and is also a conic section, traditionally gotten by cutting the cone with a steeply tilted plane. But now, with our new eyes, we can see it as the intersection of the two light cones,
$t=\sqrt{x^2 + y^2}$
and
$t= -k+\sqrt{(x-c)^2 + y^2}.$

The backwards continuation of the reflected ray (dashed red line) goes through the point $(c,0)$.

There is another curve which we can think of as a sort of midway point between the ellipse and the hyperbola. For this mirror we ask that the reflected ray is always horizontal.

The law of reflection says

$\widehat{(x,y)} \cdot (\mathrm{d}x,\mathrm{d}y) = (1,0)\cdot (\mathrm{d}x,\mathrm{d}y),$
or
$\frac{x\,\mathrm{d}x + y\, \mathrm{d}y}{\sqrt{x^2 + y^2}} = \mathrm{d}x,$

When viewed from above, the intersection of these two cones is an hyperbola

or
$\mathrm{d}\left(\sqrt{x^2 + y^2}-x\right) = 0.$
The solution of this is
$\sqrt{x^2 + y^2}-x = k.$

All of the reflected rays are horizontal

This curve is a parabola (which is also a conic section): this time, the cone is cut with a plane tilted at the opening angle of the cone. Here, the traditional plane-cuts-cone construction, and our intersecting light cone nearly coincide. We can consider our solution above as the intersection of the light cone $t = \sqrt{x^2 + y^2}$ a

When viewed from above, the intersection of these two light cones is a parabola

nd the plane $t = k + x$.

How is this plane a light cone? Well, when we asked for the rays to be reflected horizontally, we were asking for the wavefronts to be vertical lines rather than concentric circles. This can be thought of as taking the limit of the ellipse, where the second focus $(c,0)$ is moved infinitely far down the $x$-axis, so that the reflected rays reach that point at some infinite time in the future. This means that the light cone representing the light leaving the origin intersects only a small portion of the reflected light cone, small enough to be approximated by a plane. The difference with the usual construction is that here the parabola we’re interested in isn’t the one in the plane cutting the cone (that’s a parabola too), but rather the one projected down onto the $xy$ plane.

## … and two lenses

As I mentioned earlier, the Greeks knew about these optical properties of conic section mirrors. In the late 10th century, in Persia, Ibn Sahl furthered that knowledge using refraction, rather than reflection, to change the course of the light rays. If you don’t know what refraction is, it is the bending of light rays when they travel between two different materials. In his book On the burning instruments, Sahl gives the first account of the law of refraction.

Sahl’s law

In modern language, the law considers the unit vectors $\widehat{\boldsymbol{r}}_1$ and $\widehat{\boldsymbol{r}}_2$ along the incoming and bent rays. It says that the ratio of their components parallel to the interface (between the two materials) is a constant. That constant depends on the two materials in question, and we now know that it is equal to the ratio of the speeds of light, $c_1$ and $c_2$, in each one. It reads as follows,
$\frac{{\boldsymbol{f}} \cdot \widehat{\boldsymbol{r}}_1 }{{\boldsymbol{f}}\cdot \widehat{\boldsymbol{r}}_2 }= \frac{\sin\theta_1}{\sin\theta_2} = \frac{c_1}{c_2},$
where ${\boldsymbol{f}}$ is a vector parallel to the interface.

If you know it, you probably know it as Snell’s Law—but that’s the historic forces of Eurocentrism at work: it was known centuries before its rediscovery in the 1600s. We should start calling it Sahl’s Law.

Parallel lenses are all refracted to the origin

In his book, Sahl finds what we might like to call a ‘refractive parabola’. It’s a lens that takes horizontal rays, and bends them so that they all go through a given point (or, if we run time backwards, takes rays emanating from a point and bends them to be horizontal). If you wanted to make such a lens, what shape should the glass be ground to? Let’s find out!

Now we’ll be a bit more cavalier, to emphasise the power of the method, and skip over the detail we used when finding the mirrors. We’ll just take the two light cones representing our parallel and focused rays, and bosh them together. Where they intersect (projected down onto the $xy$ plane) is our desired curve. The parallel rays have a light ‘plane’ (as we saw for the parabola), given by
$t = n x.$

When viewed from above, the intersection of these two light cones is an hyperbola

The $n$ is $1$ over the speed of light in glass, in units where the speed of light in air is $1$. It is usually called the index of refraction. Since light moves slower in the glass, $n > 1$, and so this light plane is not tilted at 45 degrees as was the case for the parabolic mirror, but more steeply, according to the fact that these wavefronts move more slowly. The light cone for the rays focusing on the origin is $t = k-\sqrt{x^2+y^2}$. Voilà—the shape of our lens is
$n x = k-\sqrt{x^2+y^2},$
and happens to be an hyperbola (for $k<0$).

Wait, we forgot to use Sahl’s Law! I told you we were going to be cavalier. You can go ahead and use it to derive a differential equation like we did earlier for the mirrors, but you’ll find the same answer. The law of refraction is all taken care of by the light cones!

When viewed from above, the intersection of these two light cones is…

Rays from $(c,0)$ appear to be coming from the origin

What would a ‘refractive hyperbola’ look like? This would be a sort of lens which would take rays emanating from one point within the glass, and bend them as they exited so that they appeared as if they were emanating from a different point.

… the refractive hyperbola (for $k=4$, $n = 1.15$ and $c = -6.5$)

We’ll take the first point to be $(c,0)$, and the apparent point the origin. Our light cones are $t = n\sqrt{(x-c)^2+y^2}$ and $t=k+\sqrt{x^2+y^2}$. Where they intersect is our magical curve. This one wasn’t something known to the Greeks, or to Sahl, as far as I know.

When I first began thinking about the conic sections this way, I just wanted to see whether I could construct them, starting with their focusing properties. When I realised I could re-express them as conic intersections I was surprised, and delighted. When I then realised these were the light cones of the original and reflected wavefronts, to paraphrase Hilbert, “my joy was complete”.

Donovan is a maths teacher who is transfixed by the science and mathematics of light and vision.

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