# geodesic is parametrised by a constant multiple of arc length

• Dec 22nd 2010, 10:53 AM
slevvio
geodesic is parametrised by a constant multiple of arc length
Let $\displaystyle S\in \mathbb{R}^3$ be a regular surface, and $\displaystyle \gamma: I \rightarrow S$ a geodesic, where $\displaystyle I$ is an open interval in $\displaystyle \mathbb{R}$. We have that $\displaystyle \frac{d}{dt}|\gamma ' (t) | = 0 \implies |\gamma '(t)| = c$, a constant , i.e.

$\displaystyle \frac{ds}{dt} = | \gamma' (t) | = c$ ..................(1)

where s is the arc length parameter $\displaystyle s(x) = \int_a^x |\gamma'(p)| dp$, $\displaystyle a\in I$. I understand what is happening up to here but then my notes claim that $\displaystyle t=\frac{s}{c}$. However do we not get that upon rearranging (1)

$\displaystyle ds=cdt \implies s = ct + A$ where A is some constant? Why is A zero? Any help would be appreciated
• Dec 23rd 2010, 01:58 PM
Rebesques
The constant is omitted in your notes, though it shouldn't be. Affine transformations of the parameter do not alter geodesy.
• Dec 27th 2010, 09:47 AM
HallsofIvy
The choice of the added constant, A, is just a choice of where to start measuring the arclength. It is always possible to choose the starting point so that A= 0. I expect that it was left out here because the starting point is not relevant.
• Dec 27th 2010, 03:04 PM
Rebesques
Quote:

it was left out here because the starting point is not relevant