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# Nanairo Batake
## Solutions to TRTR Ch. 2 (An ancient theorem and a modern question) 2.7-2.8
- *NOTE**: these are my personal solutions. I have made them as logically sound as I can but they may still contain mistakes. Also there might be alternative ways of solving a problem. I will try to point those out whenever I am aware.
### 2.7 (hard)
In Fig. 2.19(a), Penrose illustrated a lattice of squares in hyperbolic space. Note that at every vertex there are 5 squares coming together, rather than 4 in the Euclidean case. It is not hard to imagine that, if we make the size of these squares a little smaller while tilting them to keep their edges touching, these aforementioned vertices would expand into little pentagons. If we continue doing this, the squares will shrink and the pentagons will expand more, until the squares shrink to a point and the hyperbolic plane becomes tessellated entirely by pentagons, four of which come together at a vertex. During this process, the inner angle of a square increases from $72^\circ$ to $90^\circ$ whereas the inner angle of a pentagon decreases from $108^\circ$ to $90^\circ$. The sum of these to angles always equals $180^\circ$.
- *Fig. IV** Tesellation of hyperpolic plane with squares and pentagons, inner angle of square is $74^\circ$. Conformal representation.
- *Fig. V** Tesellation of hyperpolic plane with squares and pentagons, inner angle of square is $88^\circ$. Conformal representation.
Fig. IV and Fig. V depict two different stages in this process, one with small pentagons and one with larger pentagons. Detailed knowledge about hyperbolic triangles and boosts (which are analogous to rotations of a sphere, but the entity being "rotated" now is the hyperbolic plane) is necessary to produce these plots. Here I will list some useful results without any further explanation.
For a hyperbolic triangle $\mathrm{ABC}$ with inner angles $A$, $B$, $C$ and their respective opposite sides $a$, $b$ and $c$ (sides are measured with something known as "rapidity" in special relativity), the following analogues to the cosine theorem and the sine theorem hold:
$\cosh a = \cosh b \cosh c - \sinh b \sinh c \cos A\ ,$
$\frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}\ .$
In particular, if an isosceles hyperbolic triangle has inner angles $\beta$, $\alpha$ and $\alpha$, then the length of one side opposite to $\alpha$ is
$\cosh^{-1}\left({\frac{1}{\tan\alpha\tan{\frac{\beta}{2}}}}\right)\ .$
On the hyperbolic plane one can use the "hyperbolic polar coordinates" $(\chi, \phi)$ which is the hyperbolic analogue to "spherical polar coordiates". $\chi$ is the distance from origin measured in rapidity, and $\phi$ is the good old azimuth angle. A boost that transforms the origin to point $(\kappa, \psi)$ will transform $(\chi, \phi)$ to $(\chi', \phi')$ with
$\chi' = \cosh^{-1}(\cosh\chi \cosh\kappa + \sinh\chi \sinh\kappa \cos(\phi-\psi))\ ;$
$\phi' = \mathrm{atan2}(\sinh\chi\sin(\phi-\psi),~ \sinh\chi\cosh\kappa\cos(\phi-\psi)+\cosh\chi\sinh\kappa) + \psi\ ,$
where "atan2" is the `atan2` function in e.g. Matlab etc.
The point $(\chi, \phi)$ corresponds to, in plane polar coordinates,
$\left(\cosh\chi\left/\sqrt{\cosh^2\chi + 1}\right., \phi \right)$
in the projective (Klein) representation.
## 2.8 (hard)
Although this is labelled as hard, it's actually quite straightforward. Consider the segment of sphere bounded by the those two great circle arcs that are the two sides of $\alpha$. It is shaped like two watermelon peels, and obviously has area $4\pi R^2\cdot(\alpha/\pi) = 4\alpha R^2$, given that the sphere has total area $4\pi$. Consider the same for $\beta$ and $\gamma$, and we now have three "double-watermelon-peel" shapes whose areas are $4\alpha R^2$, $4\beta R^2$ and $4\gamma R^2$ respectively. When we add these together, the sphere is entirely covered, but the triangle and its polar opposite are covered thrice. I.e., the sum of these "double-watermelon-peel" areas equal the total area of the sphere plus 4 times the area of the triangle. Therefore
$\Delta = \frac{1}{4}(4\alpha R^2+4\beta R^2 +4\gamma R^2 -4\pi R^2) = R^2(\alpha+\beta+\gamma-\pi)\ .$