In comment at https://math.stackexchange.com/a/1381829/88985 at
Distance of two hyperbolic lines
is says (as i interpreted it) that the distance between two points $(a,r)$ and $(a, R)$ in the Poincaré half-plane model ( https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model ) is $ \ln(R) – \ln(r) $.
I tried to deduct this formula from the formulas:
$$ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) = \operatorname{arcosh} \left( 1 + \frac{ {(R – r)}^2 }{ 2 r R } \right) $$
and
$$\operatorname{arcosh} {x} =\ln \left(x + \sqrt{x^2 – 1} \right) $$
but failed miserably, (especially the bit under the square root didn't want to simpify)
can somebody show me the deduction?
ADDED LATER:
following https://math.stackexchange.com/users/208255/user24142
's suggestion (below) and other comments I come to
$$ \operatorname{dist} (\langle a, 1 \rangle, \langle a, R \rangle) = \operatorname{arcosh} \left(\frac{1}{2} \left( R + \frac{1}{R} \right) \right)$$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\sqrt{\frac{1}{4}\left( R + \frac{1}{R} \right)^2 – 1 } \right) = $$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\frac{1}{2}\sqrt{ R^2 – 2 +\frac{1}{R^2} \ } \right) = $$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\frac{1}{2}\sqrt{ (R – \frac{1}{R} )^2 } \right) = $$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\frac{1}{2} \left( R – \frac{1}{R} \right) \right) = \ln (R)$$
THANKS
Best Answer
$$ {\cosh ^{-1}} \frac{(R+ 1/R)}{2} $$
$$= {\cosh ^{-1}} \frac{ e^ {\log (R)}+ e^{-\log(R)}}{2} $$
$$ = {\cosh ^{-1}} [ \cosh (\log R)] = \log R . $$
EDIT1:
BTW, why do we assume unit (abs value) Gauss curvature? Should it not appear symbolically at least in a formula ?
$$ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) = \operatorname{ a \cdot \,arcosh} \left( 1 + \frac{ {(R - r)}^2 }{ 2 r R } \right) $$