Post by Admin on Oct 8, 2016 12:46:51 GMT
Gary
This exercise is on page 300 and amounts to proving (33).
The length of the arc of a sector of a circle of radius $r$ subtending an angle of $\theta$ at the centre of the circle is $s=r\theta$.
In figure 21 on page 300 of the book we have an infinitesimal disc with diameter $ds$ centred at $(x+iy)$. Because the disc is infinitesimal we can treat the triangle in figure 21 as the sector of a circle of radius $y$ with an arc length equal to the diameter $ds$ of the infinitesimal disk, and write $ds=y\theta$ where $\theta$ is the angle subtended by the infinitesimal disk at the point $x$ on the real axis.
If we apply (31) on page 298 above we can calculate the hyperbolic diameter $d\widehat{s}$ of the disk in figure 21 as $ds/y=\theta$. Done.
So if your hyperbolic diameter $d\widehat{s}=\theta$ is to remain constant as you walk towards $q$, then since $y$ decreases, so must $ds$, and so your strides look shorter and shorter.
Vasco
This exercise is on page 300 and amounts to proving (33).
The length of the arc of a sector of a circle of radius $r$ subtending an angle of $\theta$ at the centre of the circle is $s=r\theta$.
In figure 21 on page 300 of the book we have an infinitesimal disc with diameter $ds$ centred at $(x+iy)$. Because the disc is infinitesimal we can treat the triangle in figure 21 as the sector of a circle of radius $y$ with an arc length equal to the diameter $ds$ of the infinitesimal disk, and write $ds=y\theta$ where $\theta$ is the angle subtended by the infinitesimal disk at the point $x$ on the real axis.
If we apply (31) on page 298 above we can calculate the hyperbolic diameter $d\widehat{s}$ of the disk in figure 21 as $ds/y=\theta$. Done.
So if your hyperbolic diameter $d\widehat{s}=\theta$ is to remain constant as you walk towards $q$, then since $y$ decreases, so must $ds$, and so your strides look shorter and shorter.
Vasco