Circular angles vs. hyperbolic angles

Question

I have been interested recently about hyperbolic trigonometric functions and their angles.

With regular trigonometric functions, we conceptualize the angle $\theta$ in radians as the arc length $L$ in the unit circle, or the ratio of the arc length $L$ divided by the radius $r$. So we have

$$\theta = \frac{L}{r}$$

It is also possible to visualize the angle as twice the area of the bounded sector by the same angle, we have

So the area of the bounded sector is directly proportional the angle in radians.

We can extend this definition for the hyperbola where an hyperbolic angle is the half of the area defined by a bounded sector

Is it also possible the conceptualize the hyperbolic angle (in hyperbolic radians) as a ratio of the length of an hyperbolic arc over the radius of the hyperbola as we do for the circle, thus obtaining a proportionality between the hyperbolic sector area and the hyperbolic angle in hyperbolic radians?

Answer 1

You can, but it requires a little ingenuity.

It should be obvious that if you try to define the length of the arc via the usual Euclidean measure

$L=\int\sqrt{dy^2+dx^2}=\int{\sqrt{(dy/dx)^2+1}}~dx$

you will get nowhere near where you want to go.

Instead define an alternative, non-Euclidean metric:

$L=\int\sqrt{dy^2-dx^2}=\int{\sqrt{(dy/dx)^2-1}}~dx$

Let us see what this metric gives for the hyperbola defined by $x=\cosh t, y=\sinh t$:

$dy^2-dx^2=(\cosh^2t-\sinh^2t)dt^2=dt^2$

So we may render the differential length as $dt$ and then integrating from $t=0$ to $t=\alpha$ gives, indeed, $\alpha$ "hyperbolic radians".

You may want to look up how proper time and distance intervals are defined in Einstein's Theory of Special Relativity:

In special relativity, however, the interweaving of spatial and temporal coordinates generates the concept of an invariant interval, denoted as ${\displaystyle \Delta s^{2}}$:

${\displaystyle \Delta s^{2}\;{\overset {def}{=}}\;c^{2}\Delta t^{2}-(\Delta x^{2}+\Delta y^{2}+\Delta z^{2})}$[note 6]

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Given the comments, the question is posed: Can we derive the hyperbolic sine and cosine a priori from the arc length definition used above?

Our problem may be stated as follows: How do $x$ and $y$ vary with arc length, defined by the metric

$ds^2=dy^2-dx^2$

from an initial point $(1,0)$ along the curve

$x^2-y^2=1$?

Begin by differentiating the equation of the hyperbola:

$2xdx-2ydy =0$

$\dfrac{dx}{dy}=\dfrac{y}{x}$

Substituting this into the metric definition and isolating the derivative leads to

$(\dfrac{ds}{dy})^2=\dfrac{1}{y^2+1}$

$(\dfrac{dy}{ds})^2={y^2+1}$

To solve this last equation we can differentiate it. Using the Chain Rule:

$2\dfrac{dy}{ds}\dfrac{d^2y}{ds^2}=2y\dfrac{dy}{ds}$

We cannot have $dy/ds=\sqrt{y^2+1}=0$, so:

$\dfrac{d^2y}{ds^2}=y$

and by the usual methodology for linear differential equations with constant coefficients

$y=Ae^s+Be^{-s}$

We need two initial conditions. First from the problem statement we must have $y=0$ at $s=0$. Second, $(\dfrac{dy}{ds})^2={y^2+1}$ implies $dy/ds=1$ at $y=0$ which in turn was just matched with $s=0$ (positive $s$ is taken to be positive $y$, which is basically just a sign convention). From these conditions is obtained

$\color{blue}{y=\dfrac{e^s-e^{-s}}{2}\overset{\rm def}{=}\sinh s}$

And then it's all algebra, using the fact that $(e^s+e^{-s})^2-(e^s-e^{-s})^2=4$ and the curve is confined to positive $x$ by construction:

$\color{blue}{x=\sqrt{1+y^2}=\dfrac{e^s+e^{-s}}{2}\overset{\rm def}{=}\cosh s}$

So, the result of a displacement of length $L$ from $(1,0)$ along $x^2-y^2=1$ may indeed be rendered as $(\cosh L, \sinh L)$.

Answer 2

The red region in your diagram can be parameterized as $x=\rho\cosh\phi,\,y=\rho\sinh\phi$ for $\rho\in[0,\,1],\,\phi\in[0,\,a]$. We've thus related Cartesian coordinates to another coordinate system, with Jacobian matrix$$J=\left(\begin{array}{cc} x_{\rho} & x_{\phi}\\ y_{\rho} & y_{\phi} \end{array}\right)=\left(\begin{array}{cc} \cosh\phi & \rho\sinh\phi\\ \sinh\phi & \rho\cosh\phi \end{array}\right),$$of determinant $\rho$, so $dxdy=\rho d\rho d\phi$. So the red area is$$\int_0^1\rho d\rho\int_0^a d\phi=\frac12a,$$but you already knew that. Meanwhile, the conditions$$dx=\cosh\phi d\rho+\rho\sinh\phi d\phi,\,dy=\sinh\phi d\rho+\rho\cosh\phi d\phi$$simplify on the $\rho=1$ arc to$$dx=\sinh\phi d\phi,\,dy=\cosh\phi d\phi\implies ds=\sqrt{\cosh 2\phi}d\phi.$$The Euclidean arc length is therefore$$\int_0^a\sqrt{\cosh 2\phi}d\phi=\int_0^a\sqrt{1+2\sinh^2\phi}d\phi=-2iE\bigg(\frac{ia}{2}\bigg|2\bigg)$$in terms of elliptic integrals. By contrast, $ds=\sqrt{dy^2-dx^2}$ gives the desired result with a Lorentzian pseudometric, as @OscarLanzi explained.