# Geodesics on a twisted surface of revolution

How is Clairaut's Law modified to define geodesics in a twisted surface of revolution ( so not axi-symmetric)

$$u \cos v, u \sin v , f(u) + T \, v ,$$

where T is a twisting constant? It appears that Clairaut's Law $u \sin(..,T)$ = constant could be defined. How to use the fact that geodesic or Gauss curvature remains zero/same even after twisting towards twist generalization?

EDIT1:

Assumed that $T$ can be placed in the derived pde like for conserved $K$s:

$$\frac{u^3 f_1f_{11}- T^2}{[ u^2 (1+ f_1^2 )+ T^2 ]^2}= \frac{f_1f_{11}}{ u (1+ f_1^2)^2},$$

which simplifies to:

$$\frac{T^2}{(1+f_1^2)}+ u ( 2 u + \frac{(1+f_1^2)}{f_1f_{11} } )= 0$$

where partial differentiation w.r.t. $v$ does not occur.

When T=0, the following should reduce to Clairaut:

$$u + \frac{(1+f_1^2)}{ 2 f_1f_{11} } = 0$$

In a sense, "not at all". Clairaut's relation comes from intrinsic geometry, assuming a metric has the form $$g = E(u)\, du^{2} + G(u)\, dv^{2}. \tag{1}$$ The trick, from this perspective, is to find an isometric embedding of (1) into Euclidean $3$-space in a manner equivariant with respect to a helical group of isometries (i.e., so that intrinsic translation in $v$ is the same as extrinsic rotation-and-twisting about an axis).
The parametrization $$X(u, v) = (u\cos v, u\sin v, f(u) + Tv)$$ is inconvenient for studying geodesics, since the coordinate curves are non-orthogonal (i.e., $F = Tf'(u) \neq 0$). Instead of taking the "profile" curve to lie in a plane containing the "axis" of the surface, however, you can take the profile to be a "helix" along the surface that stays everywhere orthogonal to the $v$-coordinate curves.