Geodesic Equation/2d Surface Embedded in 3d Euclidean Space/Cylinder

Theorem
Let $\sigma$ be the surface of a cylinder.

Let $\sigma$ be embedded in 3-dimensional Euclidean space.

Let $\sigma$ be parametrised by $\tuple {\phi,z}$ as


 * $\displaystyle \mathbf r = \paren {a \cos \phi, a \sin \phi, z}$

where


 * $a > 0$

and


 * $z,\phi \in \R$

Then geodesics on $\sigma$ are of the following form:


 * $z = C_1 \phi + C_2$

where $C_1, C_2$ are real constants.

Proof
From the given parametrization it follows that

where $E, F, G$ are the functions of the first fundamental form.

Furthermore, all derivatives of $E, F, G$ $\phi$ and $z$ vanish.

Then geodesic equations read:


 * $\displaystyle \dfrac \d {\d t} \frac {a^2 \phi'} {\sqrt{a^2 \phi'^2 + z'^2} } = 0$


 * $\displaystyle \dfrac \d {\d t} \frac {z'} {\sqrt{a^2 \phi'^2 + z'^2} } = 0$

Integrate these differential equations once:


 * $\displaystyle \frac {a^2 \phi'} {\sqrt{a^2 \phi'^2 + z'^2} } = b_1$


 * $\displaystyle \frac {z'} {\sqrt{a^2 \phi'^2 + z'^2} } = b_2$

where $b_1, b_2$ are real constants.

Divide the first equation by the second one:


 * $\displaystyle \frac {a^2 \phi'} {z'} = \frac {b_1} {b_2}$

To solve this in terms of $z$ as a function of $\phi$, define


 * $\displaystyle C_1 = \frac {a^2 b_2} {b_1}$

and use the chain rule:


 * $\displaystyle \dfrac {\d z} {\d \phi} = C_1$

Integration $\phi$ yields the desired result.

In other words, geodesics are helical lines.