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Based on this page, suppose we have the following parametric curve:

$$v(t) = (cos(t), sin(t), t)$$

The arc-length parametrization would be:

$$\tilde{v}(t) = (cos(\frac{t}{\sqrt{2}}),sin(\frac{t}{\sqrt{2}}),\frac{t}{\sqrt{2}}) $$ Two questions

(1) Is it true that these two curves only differ in their speed? That is, they have the same image but they just travel along the curve at different speeds?

(2) What's the point of reparametrizing the curve like this? Why is having a speed of $1$ important?

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1.- You are considering $v(t)$ as the position vector of each point of the curve. Irrespective of whether or not the parametrization is arc length, we have:

$v(t(s))=\bar v(s)$, so is, for each $s$ there is some valor for $t$ being $v$ the same as $\bar v$ is. e.g.

$t(s=1)=1/\sqrt{2}$

$v(1/\sqrt{2}) = (\cos(1/\sqrt{2}), \sin(1/\sqrt{2}), 1/\sqrt{2})=\bar v(1)$

2.-It is a natural way to describe de curve and, as you say, the speed is always $1$ for arc-length parametrized curves. From here, some formulas describing properties for curves are very simplified, e.g. the Frenet-Serret formulas (compare expressions for each parametrization).

In General Relativity the usual way to describe the world line (the curve a particle follows in the space-time) of a particle is parametrizing this world line by the particle's proper time, that is, the arc length of the world line in the space-time geometry. In this parametrization, the magnitude for its four velocity (the "speed" of the particle through the space-time) is always c, the speed of light in vacuum, or $1$ measuring time in distance units.

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