Compute the hyperbolic angle subtended to the origin by the unit hyperbola through (ct, x) = (0, 1)

Question

I'm trying to find the angle subtended by the unit hyperbola through the point $(ct,x)=(1,0)$. I think that I should be integrating something, but I'm not sure how to set it up. I've been trying to think of this as it would be related to a unit circle, where we would have $R=1$ and then the following $$ \int_0^{\theta_{0}} R^2(\sin^2\theta+\cos^2\theta)d\theta=R^2\int_0^{2\pi}d\theta=(1)(2\pi)=2\pi $$ So the angle subtended would just be $2\pi$. I know that $1=\cosh^2 x-\sinh^2 x,$ but as I'm only interested in the right hyperbola, I'm not sure I can use the same trick. I do believe this is related to the total proper time along the hyperbola. Beyond this, I'm stuck, and I feel I'm going about this the wrong way. Any ideas? Thank you!

Answer 1

How, exactly, do I use the parametrization?

If you let $(x,y) = (r\cosh\eta,r\sinh\eta)$ and consider the Jacobian determinant, you can show that the area element transforms as $$\mathrm{d}A = \mathrm{d}x\,\mathrm{d}y = r\,\mathrm{d}r\,\mathrm{d}\eta\text{.}$$ Alternatively, you can transform the metric $\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$ directly and get its determinant for $\mathrm{d}A^2$, but that's slightly more work.

Integrating $\mathrm{d}A$ between the asymptotes involves $\eta\in(-\infty,+\infty)$, and so obviously gives infinite area.

Answer 2

This is what I've come up with:

We begin by first parametrizing our coordinate system with the following transformations. Let $$ x=\cosh\alpha,\\ ct=\sinh\alpha. $$ Then, from the unit hyperbola, we want the right-hand hyperbola. The area swept out between the hyperbola and its asymptote is given by $$ \int_0^{\alpha_0}\left(\cosh^2\alpha-\sinh^2\alpha\right)\text{d}\alpha=\left.\alpha\right|_{\alpha=0}^{\alpha_0}=\alpha_0. $$ Hence, as $\alpha_0\in(-\infty,\infty)$, the angle subtended to the origin is $\alpha=\infty$.