16
$\begingroup$

I stumbled upon the 2d rotation matrix $$R(\theta)=\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$

which has determinant 1 because $$ \cos^2(\theta) + \sin^2(\theta) =1$$ So I thought what would happen if I replace the trig functions with hyperbolic ones, and when you do that you end up with determinant $$ \cosh^2(t) + \sinh^2(t) $$ but that tends to infinity so instead of having $$ -\sinh(t)$$ in the top right corner I replaced it with the positive version which gives us for determinant $$\cosh^2(t) - \sinh^2(t)$$ which is nicely equal to 1, but what's the name and purpose of this matrix?

$\endgroup$
19
$\begingroup$

First of all, these matrices

$$\underbrace{\begin{pmatrix} \cosh(\alpha) & \sinh(\alpha)\\ \sinh(\alpha) & \cosh(\alpha) \end{pmatrix}}_{M_{\alpha}}$$

are called "hyperbolic rotations" ; they have several applications as well in mathematics and physics.

They share a common property (which is a "defining property" : see remark 4 below) : they "preserve" the value of quadratic form $x^2-y^2$ (signature $(+,-)$). In an explicit way :

$$\underbrace{\begin{pmatrix} \cosh(\alpha) & \sinh(\alpha)\\ \sinh(\alpha) & \cosh(\alpha) \end{pmatrix}}_{M_{\alpha}}\begin{pmatrix} x_1\\y_1 \end{pmatrix}=\begin{pmatrix} x_2\\y_2 \end{pmatrix} \implies \ \ x_1^2-y_1^2=x_2^2-y_2^2\tag{1}$$

(in physics, it is the quantity $x^2-c^2t^2$ which will be preserved ; take the speed light $c$ as the unit if you want to stick to the above invariant).

But what is the "natural" (physical ?) meaning of $\alpha$ ? Is it an angle ? And why "hyperbolic" ?

(see figure 1) Let us consider the right branch of equilateral hyperbola $(H)$ with equation $x^2-y^2=1$, playing the same rôle for this trigonometry as the unit circle for ordinary (circular...) trigonometry. In fact, the intrinsic meaning of $a$ is the area of the "triangle" $OA_1A_2$, with side $A_1A_2$ taken not as a straight line but as arc $A_1A_2$ on $(H)$.

The formula is plainly

$$\text{doubled area} \ \ 2a \ \ = \ \ a_2-a_1\tag{2}$$

The proof of (2) is easy ; indeed, (1) can be written :

$$\begin{pmatrix} \cosh(2a) & \sinh(2a)\\ \sinh(2a) & \cosh(2a) \end{pmatrix}\underbrace{\begin{pmatrix} \cosh(a_1)\\ \sinh(a_1)\end{pmatrix}}_{A_1}=\underbrace{\begin{pmatrix} \cosh(a_2)\\ \sinh(a_2) \end{pmatrix}}_{A_2}$$

which can be written, using the hyperbolic addition formulas :

$$\begin{pmatrix} \cosh(2a+a_1)\\ \sinh(2a+a_1)\end{pmatrix}=\begin{pmatrix} \cosh(a_2)\\ \sinh(a_2) \end{pmatrix}\tag{3}$$

Using the bijectivity of $\sinh$, one can deduce from (3) that $2a+a_1=a_2$ : we have thus proved (2).

enter image description here

Fig. 1 : Hyperbolic trigonometry with equilateral hyperbola (H).

A comparison with usual trigonometry is enlightening. Take a look at figure 2. We could consider the association of the rotation sending $A_1$ to $A_2$ with the area of the angular sector, instead of the length of arc $A_1A_2$ (recall that it's the definition of the measure in radians of the "angle" $A_1OA_2$). This would be a good alternative to the measure in radian (with a measure $\pi$ instead of $2 \pi$ for the complete turn (think to formulas $\pi R^2$ and $2 \pi R$). It is a consequence of the formula $a=\tfrac12 \alpha R^2$ ; see here for the area of a circular sector and the measure $\alpha$ in radians of its angle. As a conclusion, we have the same formula (2) as for the hyperbolic case !

enter image description here

Fig. 2. : Circular trigonometry... with unit circle. The area of the circular sector is half the measure $(a_2-a_1)$ in radians of angle $A_1OA_2$.

Now, let us consider another curve, the parabola. It is not well known that one can build a very sound geometry called 'equiaffine" or 'centroaffine' geometry (a reference) where the distance between the two points $A_1$ and $A_2$ is ... the cubic root of the area of triangle $A_1A_2B$ (fig. 3) obtained with point $B$ defined as the intersection of tangents to the parabola at $A_1$ and $A_2$. The fact that we need a cubic root should not be so surprising because we deal here in fact with "contact elements" (see legend of Fig. 3) which are characterized by 3 real numbers (two for the position and one for angle direction).

enter image description here

Fig. 3 : Equiaffine geometry : the equiaffine distance between two "contact elements" (contact element = a point and a direction) can be defined as $\sqrt[3]{a}$ where the featured parabola is the unique parabola with these "contact elements".

Remarks :

1) Here is a simple presentation of the mathematical apparatus of hyperbolic rotations and their physical interpretation (http://www.physicsinsights.org/hyperbolic_rotations.html)

2) As can be seen in the article of remark 1, the connection between circular and hyperbolic trigonometry can be understood using "complexification" (= introducing complex numbers in the play), due to formulas :

$$\cos(ia)=\cosh(a) \ \ \ \text{and} \ \ \ \sin(ia)=i\sinh(ia)$$

(slightly misnamed "taking a pure imaginary angle") with a nice 3D interpretation : see for example this article.

3) Don't miss this informative and well written article.

4) Relationship (1) is a defining property with the following meaning. Matrices $M_{\alpha}$ are the only matrices $M$ with unit determinant such that :

$$MQM^T=Q \ \ \ \text{where} \ \ \ Q=\begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}$$

5) Basic properties of matrices $M_{\alpha}$, and in particular the fact that they form a so called "one-parameter group" $O(1,1)$ comes from the fact that :

$$M_{\alpha}=\exp(\alpha J) \ \ \ \ \text{where} \ \ \ \ J=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$$

(in perfect parallelism with the fact that a rotation matrix with angle $\alpha$ is $R_{\alpha}=\exp(\alpha K)$ where $K$ is matrix $J$ in which the upper right entry is changed into $-1$).

6) An interesting historical perspective on the discovery of centro-affine geometry by G. Tzitzeica can be found there.

| cite | improve this answer | |
$\endgroup$
10
$\begingroup$

What you've found is a Lorentz boost. Informally they are just rotations in Minkowski spacetime as well as the matrix you gave is the one of a rotation in euclidian space.

The first one is an orthogonal transformation and it leaves unchanged the euclidian metric. The latter leaves unchanged the Minkowski metric.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ In other words, the 2D rotation matrix applied to a vector $(x, y)$ leaves the quantity $x^2 + y^2$ unchanged; and the "hyperbolic rotation" matrix applied to a vector $(t, x)$ leaves the quantity $t^2 - x^2$ unchanged. In other, other words, the rotation matrix leaves distances in Euclidean space unchanged and the hyperbolic rotation matrix leaves distances in Minkowski "spacetime" unchanged. In other, other, other words, it's the proper time to watch fun YouTube videos. $\endgroup$ – Mateen Ulhaq Mar 1 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.