Distance formula for points in the Poincare half plane model on a "vertical geodesic".

Question

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

Answer 1

$$ {\cosh ^{-1}} \frac{(R+ 1/R)}{2} $$

$$= {\cosh ^{-1}} \frac{ e^ {\log (R)}+ e^{-\log(R)}}{2} $$

$$ = {\cosh ^{-1}} [ \cosh (\log R)] = \log R . $$

EDIT1:

BTW, why do we assume unit (abs value) Gauss curvature? Should it not appear symbolically at least in a formula ?

$$ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) = \operatorname{ a \cdot \,arcosh} \left( 1 + \frac{ {(R - r)}^2 }{ 2 r R } \right) $$

Answer 2

Well, to being with, assume that $r= 1$ and $R>1$. Now,

$$\begin{align*} \operatorname{dist} (\langle a, 1 \rangle, \langle a, R \rangle) &= \operatorname{arcosh} \left( 1 + \frac{ {(R - 1)}^2 }{ 2R } \right)\\ &= \operatorname{arcosh} \left(\frac{1}{2} \left( R + \frac{1}{R} \right) \right)\\ &= \ln(R) \end{align*}$$

where the last line is accomplished by observation and knowledge of the $\cosh$ function. I claim that the case $R<1$ yields $-\ln(R)$, and that together these give the general result.

Just to round this out, we know that the distance function is continuous, and that it must be invariant under $z\mapsto az$ and $z\mapsto z+b$. From this, its possible to derive that the distance function along any vertical must be the same as the distance function along any other vertical, and that it must be the logarithm wrt some base. Our choice of base here guarantees that we have a curvature of $-1$, which guarantees several other nice things, like perfect triangles having area $\pi$, but just the isometries give us most of the important geometric facts. The specific distance function is, to some extent, a distraction imo.

Answer 3

Compute first the inverse of

$$y=\operatorname {arcosh}(x)=\frac{e^x+e^{-x}}2$$

for positive $x$'s.

Whit the substitution $u=e^x$ we have $2y=u+\frac1u$. Multiplying both sides by $u$ we get the following quadratic equation:

$$u^2-2yu+1=0,$$

the solutions of which are

$$u^+=y+\sqrt{y^2-1}\ \text{ and } \ u^-=y-\sqrt{y^2-1}.$$

We need only the positive part

$$e^x=y+\sqrt{y^2-1}$$

and if $y\ge 1$ then we'll have real results.

Taking the natural logarithm of both sides we get

$$x=\ln\left(y+\sqrt{y^2-1}\right), \ y\ge 1.$$

In our case $y= 1 + \frac{ {(R - 1)}^2 }{ 2R } \ge 1$, so we may substitute it in $y+\sqrt{y^2-1}.$ The result will be $R$.

This is why we may say that for $R\ge0$

$$\operatorname{arcosh}\left(1 + \frac{ {(R - 1)}^2 }{ 2R }\right)=\ln(R).$$