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

up vote 4 down vote accepted

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

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

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

EDIT1:

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

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

  • The Poincaré half-plane model is only a model of an hyperbolic plane (for something that aproaches the real thing use a Pseudosphere en.wikipedia.org/wiki/Pseudosphere ) I presume that in this model the Gauss curvature is fixed at -1 i think you cannot even make an Poincaré half-plane model with a curvature of for example -2. There have been some questions about this, but never with a decent answer (so you could start a new one i guess :)) – Willemien Aug 8 '15 at 6:58
  • I am afraid your presumption could stand to scrutiny. If the maximum radius of cuspidal pseudosphere is $r_c=a$, then its Gauss curvature $K=−1/a^2$. If $r_c=a/√2$ then its K is doubled as also mentioned in the wiki. – Narasimham Aug 8 '15 at 10:37
  • maybe i misundersood, different tracioids can have different gaussian curvatures , but the curvature for the poincare half plane is always -1, (as i understand it) the wikipedia article is not very good in this , it more or less implies there is just one tracioid with a curvature of -1 – Willemien Aug 8 '15 at 11:20
  • The clarification I seek is, the "a" on the left hand side cannot simply vanish in any mathematical formula , that too in a model. Does it express that the RHS distance is independent of all radii "a" of the geodesics !! ? – Narasimham Aug 8 '15 at 17:21
  • not sure what you mean the $a$ in $ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) $ is just a location on the x axis, boundary line in the halfplane model , and yes it vanish. But I guess you refer to another formula (adopted from the wikipage) $ \operatorname{dist} ( \langle a, r \rangle, \langle a+ r\sin\phi , r\cos\phi \rangle ) = \operatorname{arcosh} \left( \operatorname{sec} \phi \right) $ where $a$ and $r$ vanish. but better to ask a new question about this. – Willemien Aug 8 '15 at 17:40

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.

  • thanks, but i still don't understand the bit $ \operatorname{arcosh} \left(\frac{1}{2} \left( R - \frac{1}{R} \right) \right) = \ln(R) $ (also added bit more to the question) – Willemien Aug 6 '15 at 20:51
  • obvious typo middle line – Narasimham Aug 6 '15 at 21:34
  • @Narasimham right you are. – user24142 Aug 7 '15 at 5:07

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 $R\ge0$

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

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