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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!

Lou
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  • The hyperbolic angle subtended by the entire hyperbola is infinite. Physically, this is related to the rapidity of the speed of light being infinite. But how to answer this depends on the starting point of "hyperbolic angle"--can one start with the parametrization $(ct,x) = (\cosh\eta,\sinh\eta)$, which makes this trivial, or must one relate to the area of a hyperbolic sector? – Stan Liou Feb 05 '14 at 03:29
  • I've actually tried there, but I haven't had any luck. – Lou Feb 05 '14 at 03:29
  • Honestly, I think the parametrization would work.. I want the area between the asymptotes on the right hyperbola and the the hyperbola. – Lou Feb 05 '14 at 03:31
  • How, exactly, do I use the parametrization? – Lou Feb 05 '14 at 04:09

2 Answers2

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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.

Stan Liou
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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 $\alpha=\infty$.

Lou
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