Shadows of Objects

Surely you have also played with shadows. What shapes can be created with one hand in this way? Shadow theatre is about exactly that. This question is also very interesting from a mathematical point of view: Given is a geometric three-dimensional body. What shadow can this cast with different rotations in relation to the light source?

A cube, for example, can cast a square, but also the shadow of a regular hexagon. A sphere, on the other hand, always casts the shadow of a circle. What shadows can a given triangle cast? This is the subject of the exhibit “All triangles are equal“. After some trial and error, you realise that any triangle can cast the shadow of an equilateral triangle (with suitable rotation relative to the light source).

And now … the mathematics:

Mathematically speaking, behind a shadow is a so-called projection. Given a point $p$ in three-dimensional space $\mathbb R^3$ and a projection plane $E$ that does not contain $p$ . The point $p$ corresponds to the (ideal) light source and $E$ is the screen onto which the projection is made. Any point $x\in\mathbb R^3$ is now projected onto the canvas $E$ by the ray $s$ starting from $p$ and passing through $x$ (it must still be required that $x\neq p$ , otherwise the ray just mentioned is not unique). We thus obtain a mapping

$\pi\colon\mathbb R^3\setminus\{p\}\to E,$

where $\pi(x)$ is defined as the just-mentioned intersection $E\cap s$ of the ray emanating from $p$ and passing through $x$ . This mapping is called a central projection at point $p$ onto the plane $E$ . This is exactly what happens, for example, when the sun casts a shadow on a house wall. Here, however, the sun is very far away. Idealising this, we would say that the point $p$ is no longer itself in the surrounding three-dimensional space, but at an infinite distance (in the so-called projective closure, of the three-dimensional space $\mathbb R^3$ ). Then all rays emanating from $p$ run parallel, which is why $\pi$ , is then called a parallel projection. The mapping $\pi$ is also idempotent, i.e. $\pi(\pi(x))=\pi(x)$ . This is simply because every point $x\in E$ that lies in the plane of projection is mapped onto itself. But now for the shadows. If $B\subseteq\mathbb R^3$ is a body in space (e.g. a sphere or a cube), the shadow cast on the screen $E$ by the light source located at point $p$ is exactly given as the image

$\pi[B]=\{\pi(x)\mid x\in B\}$

of the set $B$ under the image $\pi$ .We want to examine in the following the images that can appear as shadows for interesting bodies $B$ :

The sphere

Let $B$ be a sphere with radius $r$ centred at point $x_0$ . What shadow can this cast on the canvas? Now, the rays emanating from point $p$ and passing through a point $x\in B$ of the sphere obviously form a cone $K$ with apex at point $p$ and axis of rotation $\overline {px_0}$ . If we now choose a ray $s$ that starts from $p$ and is only tangent to the surface of the sphere $B$ — let’s say it touches it at point $y$ –, then $x_0yp$ forms a right triangle (with a right angle at $y$ ), so that the angle $\alpha=\angle ypx_0$ is half the opening angle of the cone. So if we set $d\coloneqq|\overline {px_0} |$ , then according to the definition of the sine, we get that $\alpha=\arcsin(d/r)$ . But what shadow does the sphere $B$ cast on the canvas $E$ ? Now, the boundary of this shadow is obviously the intersection curve $K$ of the lateral surface of the cone $K$ with the plane $E$ , i.e. an ellipse (see also the exhibits “Circle and Ellipse“, “Cone Sections“, “Ellipse Mountains“). We can even calculate the two semi-axes of this ellipse $K$ .

Let $y_0$ be the intersection of the rotation axis $\overline {px_0}$ with the canvas $E$ .We set $D\coloneqq|\overline {py_0} |$ . The small semi-axis $b$ now encloses a right angle with the line $\overline {py_0}$ . In the corresponding right triangle, the interior angle at point $p$ is equal to half the opening angle $\alpha$ . So $b=D\tan(\alpha)$ results. To determine the major semi-axis $a$ , we need to know the angle $\beta$ at which the axis of rotation $\overline {py_0}$ intersects the plane $E$ . If this angle is $\beta=90^\circ$ , then $K$ is a circle and thus $a=b$ . So let us assume that $\beta\lt 90^\circ$ . Then there is a unique plane $F$ that is perpendicular to the canvas $E$ and contains the axis of rotation $\overline {py_0}$ . The great semi-axis $a$ and the point $p$ now form a triangle whose interior angle at $p$ equals $\alpha$ , and at $y_0$ equals $\beta$ . Thus we can use the sine theorem to determine $a$ :

$\frac {a} {\sin(\alpha)}=\frac {D} {\sin(180^\circ-\alpha-\beta)}.$

So we get

$a=\frac{D\sin(\alpha)}{\sin(180^\circ-\alpha-\beta)}.$

That $p$ is an infinitely distant point would be expressed in the equation by $\alpha=0^\circ$ . Then for $b=D\tan(\alpha)$ one obtains the expression $0\cdot\infty$ , since the distance $D$ is then equal to infinity. But it is easy to see that $b=r$ must then apply. For $a$ we get: $s=r/\sin(\beta)$ .

The cube

What shadow a cube can cast is a much more interesting question than for a sphere. To do this, we assume that $p$ is infinitely far away, so we are dealing with a parallel projection. We want to make use of the vector calculus. For this purpose, we further assume that the screen $E$ is exactly the $xy$ -plane and the light is incident vertically from above, i.e. in the direction of the negative $z$ -axis. The projection image $\pi$ from above, is then simply given by setting the $z$ coordinate to zero:

$\pi\begin {pmatrix} x\\ y\\ z\end {pmatrix} =\begin {pmatrix} x\\ y\\ 0\end {pmatrix} .$

It’s certainly OK if we just forget the last coordinate, so $\pi(x,y,z)=(x,y)$ . We now want to find out what shadow a cube can produce under this figure. What we observe is that the mapping $\pi$ is linear, so the vector addition and scaling of vectors obtains:

$\pi(x_1+x_2,y_1+y_2,z_1+z_2)=(x_1+x_2,y_1+y_2)=(x_1,y_1)+(x_2,y_2)=\pi(x_1,y_1,z_1)+\pi(x_2,y_2,z_2)$

and

$\pi(\lambda x,\lambda y,\lambda z)=(\lambda x,\lambda y)=\lambda(x,y)=\lambda\pi(x,y,z).$

A cube $W$ is now twisted in space and projected vertically from above onto our screen $E$ , the $xy$ -plane. Let’s say for simplicity that the cube $W$ has edge length one. We now fix a corner point $p$ of $W$ . Three perpendicular edges emanate from this, forming a coordinate system that is twisted in space, a so-called tripod. Let us call the corresponding three unit vectors attached in $p$ $v_1,v_2,v_3$ . Then the (filled) cube $W$ is exactly the set

$W=\{(av_1+bv_2+cv_3)+p\,|\,a,b,c\in [0,1] \}.$

The cube $W$ is thus completely described by the knowledge of the point $p$ and the three unit vectors $v_1,v_2,v_3$ . To better understand the shadow of this given cube, we can now apply the projection $\pi$ to the representation just mentioned and exploit its linearity:

$\begin{align*}\pi[W]&=\{\pi((av_1+bv_2+cv_3)+p)\,|\,a,b,c\in [0,1]\}\\ &=\{a\pi(v_1)+b\pi(v_2)+c\pi(v_3)+\pi(p)\,|\,a,b,c\in [0,1] \}.\end{align*}$

So, to study the shadow of $W$ , it is enough to know the projected vectors $\pi(v_1),\pi(v_2),\pi(v_3)$ and the projected footpoint $\pi(p)$ . If we shift $p$ so that it lies on the $z$ axis, then $\pi(p)=(0,0)$ . Surely this is not a disadvantage for our understanding of the shadow $\pi[W]$ . This is now exactly the Minkowski sum of the distances corresponding to the vectors $\pi(v_1)$ , $\pi(v_2)$ and $\pi(v_3)$ (which are attached at the origin. It is thus easy to consider that the shadow $\pi[W]$ is thus either a rectangle (if two of these vectors $\pi(v_1),\pi(v_2),\pi(v_3)$ are parallel, i.e. one side face of the cube is perpendicular to the canvas $E$ ), or a point-symmetric hexagon. It is relatively easy to clarify exactly which rectangles occur: One edge length $a$ is exactly $1$ , and the other edge length $b$ satisfies the condition $1\leq b\leq\sqrt {2}$ . Exactly which point-symmetrical hexagons occur is much more complicated and we will not go into it here.

Instead, we want to point out an interesting but little-known theorem that can be called the shadow theorem. Designate $\pi'\colon\mathbb R^3\to\mathbb R$ the orthogonal projection onto the $z$ -axis. The shadow theorem now states that the shadow $\pi[W]$ of the cube under the orthogonal projection $\pi$ from above, and the shadow of $W$ under the orthogonal projection $\pi'$ onto the $z$ -axis from the side are equal. Here it is important that we work with a unit cube! The area of a square of dimension $1\times 1$ is set here equal to $1$ , as is the length of the interval $[0,1]$ . We want to give a short proof of the theorem. However, if you want to try it yourself first, then stop reading here now.

The proof goes as follows: Let us again assume the tripod $v_1,v_2,v_3$ attached to the vertex $p$ of the cube $W$ . Let us now glue a square side surface $F$ between $v_1$ and $v_2$ . We now claim that the area of the projected square $\pi[F]\subseteq E$ is exactly as large as the length of the projected distance $s = { av_3 \mid a \in [0,1]}$ belonging to $v_3$ , $\pi'[s]$ . Why is that?

To do this, we must first consider how large the area of $\pi[F]$ is. But this is not difficult, because if $\beta$ is the angle of intersection between the plane $E$ (i.e. the $xy$ -plane) and the plane containing the square lateral surface $F$ , it simply follows that the surface of $\pi[F]$ is exactly $|\cos(\beta)|$ . Similarly, consider that the length of the projected line $\pi'[s]$ is also equal to $|\cos(\beta)|$ , because $\beta$ is also the angle at which the $z$ axis and the line $s$ intersect.

But now the proof is almost finished: The area of the cube’s shadow in the $xy$ -plane is (as a geometrical consideration shows) exactly the sum of the areas of the projections of the three squares, each of which spans two of the vectors $v_1,v_2,v_3$ , and the length of its projection onto the $z$ -axis is exactly equal to the sum of the projection lengths of the three unit vectors $v_1,v_2,v_3$ onto the $z$ -axis. According to the above consideration, both are therefore of equal size.

The triangle

Finally, we look at the shadow that a triangle can cast. To do this, we again assume an orthogonal projection onto the $xy$ -plane as before. Let’s say our triangle to be projected $\Delta=abc$ is in the plane $F$ . We want to determine the possible shadows of $\Delta$ down to similarity. To do this, we first need to better understand the projection mapping $\pi$ . Let $s$ be the intersection of the planes $E$ and $F$ . It is certainly possible to move the plane $F$ so that the vertex $a$ of $\Delta$ lies at the origin and thus $s$ also passes through it. Let again $\beta$ be the intersection angle between $E$ and $F$ . If we now rotate the plane $F$ by $\beta$ it is transformed into the plane $E$ . If we identify $F$ with $E$ in this way, we can describe the mapping $\pi$ as follows: Along the straight line $s$ $\pi$ fixes every point. Along the perpendicular to $s$ in the plane $E$ (which has now been identified with $F$ ) $\pi$ compresses each vector by the factor $\cos(\beta)$ . So if we shift our coordinate system so that $a$ is at the origin and $s$ becomes the $x$ -axis, then $\pi$ (taking into account the identification of $E$ and $F$ ) can now be written as follows:

$\pi(x,y)=(x,\cos(\beta)y).$

So to find out what shadow a given triangle $\Delta$ produces, we simply have to apply $\pi$ (in the form just executed) to it. Here $\Delta$ is to be understood as a triangle in the $xy$ -plane, which has a corner point $a$ in the origin. However, the shadow $\pi[\Delta]$ does not change if we rotate it around the origin or stretch it centrically (i.e. the result is again similar to the original shadow). Similarly, we can also rotate and centrically stretch the triangle $\Delta$ itself. We can therefore — according to the reasoning just given — go from $\pi$ to $\lambda Q\pi R$ , where $R,Q\in\mathrm {SO} _2(\mathbb R)$ are orientation-preserving rotation matrices and $\lambda\in\mathbb R_+$ is a positive parameter. We now claim:

Assertion: Any $2\times 2$ matrix $M$ over the real numbers $\mathbb R$ other than the zero matrix can be written in the form $\lambda Q\pi R$ , for suitable $\lambda\in\mathbb R_+$ , $Q,R\in\mathrm {SO} _2(\mathbb R)$ and $\beta$ fitting.

Proof of the assertion: For this we need a little linear algebra: According to the existence of the polar decomposition there is a positive semidefinite symmetric matrix $S$ and an orthogonal matrix $O$ with $M=OS$ . Since every real symmetric matrix is orthogonally diagonalisable, there is now a rotation matrix $T\in\mathrm {SO} _2(\mathbb R)$ and a diagonal matrix $D$ such that

$S=T^ {-1} DT=T^\top DT.$

Since $S$ was positive semidefinite, $\lambda_1,\lambda_2\geq 0$ consequently holds, where $\lambda_1,\lambda_2\in\mathbb R$ are the eigenvalues of $D$ (i.e. the diagonal entries). Since $M$ was not the zero matrix, the larger of the eigenvalues $\lambda_1,\lambda_2$ must be greater than zero; without qualification, let this be $\lambda\coloneqq\lambda_1$ . Thus $D'\coloneqq\frac {1} {\lambda}D$ has the eigenvalues (diagonal entries) $1$ and $\lambda'\coloneqq\lambda_2/\lambda_1\leq 1$ . Thus there is exactly one angle $\beta\in [0^\circ,90^\circ]$ with $\cos(\beta)=\lambda'$ . If we now retrace all the steps, we get

$M=\lambda OT^\top D' T.$

Here $R\coloneqq T$ is orientation-preserving. If $OT^\top$ also has this property, we are done (with $\pi=D'$ and $Q=OT^\top$ ). If this is not the case, we replace $D'$ with $D''=\mathrm {diag} (1,-1)D'$ and set $Q\coloneqq OT^\top\mathrm {diag} (1,-1)$ . Then we have to choose $\beta\in [90^\circ,180^\circ]$ such that $\cos(\beta)=-\lambda'$ . This proves our assertion.

Proof that all triangles are “equal”: We wanted to transform our original triangle $\Delta\subseteq E$ with vertices $a,b,c$ , $a=(0,0)$ , into any other triangle $\Delta'\subseteq E$ with vertices $a',b',c'$ , $a'=(0,0)$ . For this we assume that $\Delta$ has not degenerated to a distance and $\Delta'$ does not fall on a point. Then, according to linear algebra, there is exactly one real $2\times 2$ matrix that maps $b$ to $b'$ and $c$ to $c'$ (and $a=a'=(0,0)$ fixed). However, this matrix $M$ — according to the assertion just shown — comes from a projection $\pi$ . Thus we can map $\Delta$ to $\Delta'$ (up to similarity) by means of a suitable projection. Thus, in fact — as the exhibit “All triangles are equal” claims — all triangles are equal in this sense!

Literature

[1] https://de.wikipedia.org/wiki/Projektion_(Lineare_Algebra)

[2] https://de.wikipedia.org/wiki/Funktion_(Mathematik)

[3] https://de.wikipedia.org/wiki/Projektiver_Raum

[4] https://de.wikipedia.org/wiki/Parallelprojektion

[5] https://de.wikipedia.org/wiki/Kegel_(Geometrie)

[6] https://www.youtube.com/watch?v=rAHcZGjKVvg

[7] https://www.youtube.com/watch?v=cEhLNS5AHss