# Galilean transformations and Frenet Frame

How I can prove that the curvature and torsion of a curve are invariant under the Galilean transformations? In my physics book a hint is the isometries of Galilean transformations, but it's still difficult. I don't have any idea.

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## 2 Answers

A Galiliean transformation consists of a composite of any of the following transformations on spacetime:

1. Translation in time
2. Translation in space (time-independent)
3. Translation in space (time-dependent)
4. Rotation in space

These three transformations act on curves $\mathbf{x}(s)$ (where $s$ is the arc-length parameter) in the following ways:

1. $\mathbf{x}(s)\to \mathbf x'(s) =\mathbf{x}(s-\sigma)$
2. $\mathbf{x}(s)\to \mathbf x'(s)=\mathbf{x}(s) + \mathbf x_0$
3. $\mathbf{x}(s)\to \mathbf x'(s)=\mathbf{x}(s) + s \mathbf v$
4. $\mathbf{x}(s)\to \mathbf x'(s) =R\,\mathbf x(s)$

where $\sigma$ is a real number, $\mathbf x_0, \mathbf v$ are some constant vectors, and $R$ is a rotation (an orthogonal linear transformation that satisfies $R^t R = I$). The curvature of an arc-length parameterized curve $\mathbf x(s)$ is defined as

$\kappa = \Big|\frac{d^2 \mathbf x}{dt^2}\Big|$

Let's see what happens to this quantity under each of these types of transformations. Let $\kappa'$ denote the curvature of the transformed curve $\mathbf x'$, then we have

1. $\kappa'(s) = \Big|\frac{d^2\mathbf x'}{ds^2}(s)\Big| = \Big|\frac{d^2\mathbf x}{ds^2}(s-\sigma)\Big| = \kappa(s-\sigma)$

2. $\kappa'(s) = \Big|\frac{d^2\mathbf x'}{ds^2}(s)\Big| = \Big|\frac{d^2\mathbf x}{ds^2}(s)\Big| = \kappa(s)$

3. $\kappa'(s) = \Big|\frac{d^2\mathbf x'}{ds^2}(s)\Big| = \Big|\frac{d^2\mathbf x}{ds^2}(s)\Big| = \kappa(s)$

4. $\kappa'(s) = \Big|\frac{d^2\mathbf x'}{ds^2}(s)\Big| = \left(R\frac{d^2\mathbf x}{ds^2}(s)\right)\cdot\left(R\frac{d^2\mathbf x}{ds^2}(s)\right) = \left(\frac{d^2\mathbf x}{ds^2}(s)\right)\cdot\left(R^tR\frac{d^2\mathbf x}{ds^2}(s)\right) = \Big|\frac{d^2\mathbf x}{ds^2}(s)\Big| = \kappa(s)$

We see that under translation in time, the curvature has the same value along the new curve, but just at different times, and for the other three transformations, the curvature has the same value along the two curves for all times.

Composing all of these transformations, we find that the most general Galilean transformation has the form

$\mathbf x(s)\to\mathbf x'(s) =R\mathbf x(s-\sigma) +s\mathbf v + \mathbf x_0$

so in general, the curvature transforms as

$\kappa'(s) = \kappa(s-\sigma)$

The proofs for torsion $\tau$ are perhaps a little bit harder, but I'll leave them to you :)

Please let me know of any errors, questions or comments!

Cheers!

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For curves in space, it's very easy (too easy--- there's nothing to do) because a Galilean transformation is just a translation on any one time, it just shifts the space coordinates by a constant. For curves in space-time it is false. for example, if you are moving very very fast with acceleration g, your space-time path has nearly zero curvature, but if your coordinates go along with you, you become stationary with acceleration g, and then the radius of curvature is 1/g, not zero.

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I need a mathematical proof for curves in space. – stavros kassaras Oct 22 '12 at 8:22
@stavroskassaras:The mathematical proof is "the Galilean transformation is a translation at any fixed time. These quantities don't change under translation. QED" – Ron Maimon Oct 22 '12 at 13:07