# 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

 $\displaystyle E$ $=$ $\displaystyle {\mathbf r}_\phi \cdot {\mathbf r}_\phi$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {- a \sin \phi, a \cos \phi, 0} \cdot \paren {- a \sin \phi, a \cos \phi, 0}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle a^2$ $\quad$ $\quad$
 $\displaystyle F$ $=$ $\displaystyle {\mathbf r}_\phi \cdot {\mathbf r}_z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {- a \sin \phi, a \cos \phi, 0} \cdot \paren {0, 0, 1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$
 $\displaystyle G$ $=$ $\displaystyle {\mathbf r}_z\cdot {\mathbf r}_z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {0, 0, 1} \cdot \paren {0, 0, 1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 1$ $\quad$ $\quad$

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

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

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

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 with respect to $\phi$ yields the desired result.

In other words, geodesics are helical lines.

$\blacksquare$