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This is a post inspired by Gauss Bonnet theorem validation with hyperbolic circles

I was thinking also about curvature and how to measure it

And was puzzling with the following :

Suppose some critters land on a hyperbolic plane (a surface with a constant negative curvature) and they want to measure the curvature of it.

(critters is a nod to an old question about critters on a hyperbolic plane :)

I was thinking on how they could measure the curvature and I found one the following way:

The critters have brought their own measure m (think meters/ miles/yards /feet) and construct an isosceles right triangle with the top angle as the right angle

The legs of this triangle are both length $a m$ and the hypotenuse has length $b m$ ,

(In Euclidean geometry the base angles are $45^o$ and the length of the hypotenuse is always $bm = \sqrt {2} a m$ )

In hyperbolic geometry the base angles are $< 45^o$ and length of the hypotenuse is between $\sqrt {2} a m $ and $ 2 a m $

Even better from this construction you can estimate what the hyperbolic absolute lengths of $a m$ and $b m $ are.

$a m$ and $b m$ are also related by $\cosh(b m) = \cosh(a m) \cosh(a m) =\cosh^2(a m)$

So for every $\frac{b m}{a m} = \frac{b}{a } $ there is only one pair of absolute lengths $l_h = am$ and $h_h = b m$

And from this $m$ and the curvature can be estimated .

To make it all a bit clear in a table: ( I am still myself puzzeling with this )

 b/a   -> l     -> h
 1.43  0.372    0.532
 1.44  0.479    0.690
 1.45  0.567    0.822
 1.5   0.919    1,378
 1.6   1.523    2.438
 1.7   2.224    3.816

So for example: if the by de critters measured lengths are $a m= 2m$ and $bm = 3m$ then the lengths in absolute hyperbolic lengths are 0.919 and 1,378 meaning $m=\frac{0.919}{2}=\frac{1,378}{3}= 0.459$

But then:

What is the curvature ? Is it $ -m$ ,$ -m^2$, $\frac{1}{-m}$ or $\frac{1}{-m^{2}} $?

Is there a more direct way to calculate $l_h$ and $h_h$ from a/b? (Instead of estimating it by looking it up in a table)

I like this way of calculating curvature (it is quite simple just a simple triangle an measuring of lenghts ) are there an even more simple ones ?

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    $\begingroup$ Not sure what you mean by an equilateral right triangle in hyperbolic geometry (equilateral triangles have equal internal angles, all less than $\pi/3$...?), but curvature is "angular defect per unit area". Precisely, if $T$ is a geodesic triangle in a space of constant curvature $K$, then $$K = \frac{\text{sum of internal angles} - \pi}{\operatorname{Area}(T)}.$$ $\endgroup$ – Andrew D. Hwang Aug 5 '17 at 11:18
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    $\begingroup$ Sorry mend isosceles triangle and the angle between them is right) $\endgroup$ – Willemien Aug 5 '17 at 12:06
  • $\begingroup$ You say in other words an isosceles right hyperbolic triangle has invariant side lengths and $k_g$ curvatures of its sides by non-euclidean slidings arbitrarily on a pseudospherical surface... Is that correct? $\endgroup$ – Narasimham Aug 24 '17 at 8:30
  • $\begingroup$ @narasimham not sure what you mean , my idea is that on a pseudo spherical surface you can calculate the curvature by constructing an isosceles right angled triangle and measuring the lengths of the legs and the hypotenuse $\endgroup$ – Willemien Aug 24 '17 at 12:24
  • $\begingroup$ If geodesic curvature $\kappa_g$ is constant in Poincare models,they should be zero in $\mathbb R^3$ to apply here into GB theorem, Right? $\endgroup$ – Narasimham Aug 24 '17 at 21:28
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Did not fully understood what geodetic information the critters get, however, am attempting a basic answer hoping it would serve as background for possible discussion / insight with 3D hyperbolic triangles constructed by pseudospherical trigonometry measurements.

Given: $\,C= \pi/2,a=b$

To Find:

A relation between $(a,A)$ for hyperbolic isosceles right triangles to find other angles and sides.

Sinh and Cosh Laws in pseudospherical Trigonometry

(The term used from Roberto Bonola's Non-Euclidean Geometry text-book pp 137)

Let Gauss curvature be $ -1/T^2.$

By hyperbolic Law of (Sinhs)

$$ \frac{\sin A}{\sinh (a/T)}=\left(\frac{\sin B}{\sinh (b/T)}\right)=\frac{\sin C}{\sinh (c/T)}=\frac{1}{\sinh (a/T)}$$

$$ \frac{\sin A}{\sinh (a/T)}=\frac{1}{\sinh (c/T) } \tag1$$

From hyperbolic Law of Cosines (Coshs) triangle sides with a right angle can be generalized to "hyperbolic Pythagoras " theorem:

$$\cosh(c/T) = \cosh(a/T) \cosh(b/T) = \cosh^{2}(a/T) \tag2$$

Combine (1) and (2) and simplify to get a relation for isosceles hyperbolic right triangles as

$$\sin A \sqrt{ 1+ \cosh^{2} a/T }=1 \tag3$$

The graph of above trigonometric relation.. $ (a,A) $ plotted on $(x,y)$ axes respectively.. and representative 3D projection of triangles made on bisected sides $a=(1,2) $ is shown red and blue triangles. Also

$$\sin A \sqrt{ 1+ \cosh \, c/T }=1 \tag4$$

3D isosceles Hyper Trias

Eliminate $A$ from (3) and (4) side/hypotenuse ratio is found variable for right hyperbolic isosceles requirement but you are trying to fix it up .. so cannot understand how that could be done.

If $ a=1,c=\sqrt2 $ are given, dividing (3) and (4) numerical solution $ T \approx 7300,K=-1/T^2 \approx \frac{-1}{7300^2}$ the Gauss curvature is seen to be very low, near to a flat plane.

c=Sqrt[2.]; a=1. ;Plot[{Cosh[c/tt]/Cosh[a/tt]^2}-1., {tt,6500., 8000.}]

(a,c) relation

Numerical data

Clairaut's constant for all geodesics shown

$$r_i \sin \psi_i= 1/4 $$

$$ K=-1/T^2,\, (a,b,c)= (2,2,3.38021); \quad (a_{small},b_{small},c_{small})=(1,1,1.51337);\quad (A,B,C)= (0.249795,0.249795,\pi/2);\quad(for \ both) $$

When three arcs are plotted with these lengths and surface angles, a closed isosceles triangle forms checking accuracy of calculation.

An interesting take on this is about Integral curvature $\int\int K dA$ is same for all right isosceles triangles same $A$ when $a$ is partitioned in anyway.. as the pseudospherical deficit is same for the red and blue side triangles given above or for any other triangle constructed in a similar way.It evaluates to $ I.C.= (\pi/2-2A)$.

The set of hypotenuses are not parallel in any sense.

In response to your question in comments that I still cannot understand.. for all prescribed ratios of leg length $a$ to hypotenuse $c..$ by virtue of Equation (2) we can have only one trivial triangle solution $ a=c=0.$

In conclusion, ratio $c/a$ is a fixed number. In the last $\cosh$ graph $c/T,a/T$ relation is not a straight line through the origin. To me it appears the "ratio" of side/hypotenuse restriction is perhaps our hang-up from Euclidean trigonometry concept of angle definition. One has to abandon it .. large nonlinearities of non-Euclidean geometries demand it. Even in spherical trigonometry a construction in with different $c/a$ ratios produces different integral curvatures.

EDIT1:

The conclusion was not comprehensive enough.

We should be able to plot $K$ as a function of hypotenuse to side ratio $c/a$. The plot should include flat Euclidean special case as well as spherical geometry if possible. The following does precisely that. Dividing (3) by (4)

$$ \cosh (c \sqrt{|K|})/ \cosh ^2 (a \sqrt{|K|}) = 1 ,\quad \cos (c \sqrt{|K|})/ \cos ^2 (a \sqrt{|K|}) = 1 \tag5 $$

Non-linear Hyp/Sph Geom IsoscelesTrias

The hyperbolic triangle hypotenuse/side ratio$c/a$ should be $\sqrt 2 $ when $K \approx 0$ for flatter triangles and $c/a=2$ when the sides are deep and curved...

You cannot choose a ratio like 1.2 or 3.0 outside this interval.

That solves the puzzle perhaps more than earlier ..

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  • $\begingroup$ The critters know nothing more than 1) the ratio between between the legs and the hypotenuse, 2) they have some unit of lenght / distance and how long the sides are in these units) $\endgroup$ – Willemien Aug 28 '17 at 8:43
  • $\begingroup$ Maybe clearest can make it the ratio $c:a $ differs from the ratio $ c_{small} : a_{small} $ and from this difference the question follows. $\endgroup$ – Willemien Aug 30 '17 at 3:16
  • $\begingroup$ Each has a different integral curvature. $\endgroup$ – Narasimham Aug 30 '17 at 11:43
  • $\begingroup$ My question is not about the integral curvature but about the total (Gaussian) curvature ( the product of the principal curvatures at a point) $\endgroup$ – Willemien Aug 30 '17 at 12:47
  • $\begingroup$ Omg! By assumption it is the same for all the hyperbolic triangular patches $ K= -1/T^2, T=1 $ $\endgroup$ – Narasimham Aug 30 '17 at 13:45

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