A unit sphere can be parametrized as:
$$\vec r(u,v)=\langle\cos(u)\sin(v), \sin(u)\sin(v), \cos(v)\rangle$$
Now if one wishes to study the arclength accross the meridians and the parallels we get that, for the meridians the rate of change of the length is:
$$\left|\vec{r'_{m}}\right| = 1$$
The rate of change with respect to $v$ of the arclength of a meridian is 1, so there is no "deformation" along a meridian, or meridians form arclength parametrized curves. (I didn't go into the details, you can verify this if you want)
However let's study the parallels now, a parallel can be thought as a curve of one variable if we hold $v$ constant, and so we get:
$$\begin{align}x'&=-\sin(u)\sin(v) \\ y' & =\cos(u)\sin(v)\\ z' & =0\end{align}$$
So we get that the arclength becomes:
$$\int^u_0 \sqrt{\sin^2(u)\sin^2(v)+\cos^2(u)\sin^2(v)}\,du$$ $$=\int^u_0\sin(v)du$$ $$=u\sin(v)$$
So we get: $S=u\sin(v)\iff u=\frac{S}{\sin(v)}$
The re-parametrization of one of these curves should thus be:
$\vec r_v(u)=\left\langle\cos\left(\frac{u}{\sin(v)}\right)\sin(v), \sin\left(\frac{u}{\sin(v)}\right)\sin(v), \cos(v)\right\rangle$
Now we obviously get a huge instability near the poles where $\sin(v)=0$ but we will ignore them for now and focus exclusively on the regions of the Sphere were this new parametrization is well defined.
Consider the original parametrization of a sphere. In this case if one projects a straight line onto the sphere there is a deformation appearing from the fact that the parallels are "squished":
By straigth line we mean, take 2 points $P_1, P_2$ such that their coordinates are within the correct ranges for $(u,v)$ and linearly interpolate between the 2 (i.e $(1-t)P_1 + tP_2$).
The new parametrization however, should have corrected this deformation, and so we should see something close to a geodesic. However this is what I am observing:
Could anybody explain to me where on my analysis I went wrong, and ideally, how to correct this deformation?
If you could reparametrize any open subset of a sphere so that the coordinate vector fields have constant length, then that open subset would be flat (zero curvature). But, as we can see in various ways, the sphere has constant positive curvature.