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

From ProofWiki
Jump to: navigation, search

Theorem

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

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

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

$\mathbf r = \tuple {a \cos \phi, a \sin \phi, z}$

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 arbitrary constants.


Proof

From the given parametrization it follows that

\(\displaystyle E\) \(=\) \(\displaystyle {\mathbf r}_\phi \cdot {\mathbf r}_\phi\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {-a \sin \phi, a \cos \phi, 0} \cdot \tuple {-a \sin \phi, a \cos \phi, 0}\)
\(\displaystyle \) \(=\) \(\displaystyle a^2\)
\(\displaystyle F\) \(=\) \(\displaystyle {\mathbf r}_\phi \cdot {\mathbf r}_z\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {-a \sin \phi, a \cos \phi, 0} \cdot \tuple {0, 0, 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)
\(\displaystyle G\) \(=\) \(\displaystyle {\mathbf r}_z\cdot {\mathbf r}_z\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {0, 0, 1} \cdot \tuple {0, 0, 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

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.

Then geodesic equations read:

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

Integrate these differential equations once:

$\dfrac {a^2 \phi'} {\sqrt {a^2 \phi'^2 + z'^2} } = b_1$
$\dfrac {z'} {\sqrt {a^2 \phi'^2 + z'^2} } = b_2$

where $b_1, b_2$ are real arbitrary constants.

Divide the first equation by the second one:

$\dfrac {a^2 \phi'} {z'} = \dfrac {b_1} {b_2}$

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

$C_1 = \dfrac {a^2 b_2} {b_1}$

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.

$\blacksquare$


Sources