My question is to do with parametrisation of arclengths.
As part of a course on mechanics (with mathematical focus), I have covered intrinsic coordinates in a plane curve. In class and homeworks etc, we covered ideas such as determining the shape of a chain hanging under its own weight.
We also learned arclength parametrisation. in three dimensions (although it loses the use of $\psi$ and instead parametrises in terms of arclength only using the i,j,k basis vectors), and its application to something like a helix for example.
I have a question that sprung to mind however: Could we extend this idea of parametrising a curve in terms of its arclength be extended to a 4 dimensional analogue or higher, so we let:
$(\frac{dw}{ds})^2+(\frac{dx}{ds})^2+(\frac{dy}{ds})^2+(\frac{dz}{ds})^2$=1
And attempt to parametrise from there?
Does this still "make sense"? Or is there an issue with it?
If so, what would be a good example of something in 4D space which isn't just a straight line for example that could be parametrised in terms of its arclength? Preferably something written in parametric form (i.e w($\phi$)=... x($\phi$)=...y($\phi$)=... z($\phi$)=...) AND a parametrisation in terms of arclength so I can try to observe the behaviour a little better!
Yes, both the measurement of arclenght and the reparameterization by arclength are well-defined in $n$ dimensions, where $n$ coul be $4$, as in your question, or much larger.
As an example, the set of points $(x, y, z, w)$ satisfying the relations $$ x^2 + y^2 = 1 \\ z^2 + w^2 = 1 $$ constitute something called "the flat torus". A typical path on that surface is something like $$ \gamma(t) = (t, \sqrt{1-t^2},t, \sqrt{1-t^2}) $$ whose tangent vector is ... um ... $$ \gamma'(t) = (1, \frac{t}{\sqrt{1-t^2}},1, \frac{t}{\sqrt{1-t^2}}) $$ whose squared length is $$ 1^2 + \frac{t^2}{1-t^2} + 1^2 + \frac{t^2}{1-t^2} $$ which is certainly not $1$, so this curve is not parameterized by arclength.
(It's also only defined for $-1 \le t \le 1$.)
We can find a different path that traverses the same points, but does have constant speed, namely $$ \mu(s)= (\sin(s/2), \cos(s/2), \sin(s/2), \cos(s/2)) $$ where in this case $s$ ranges from $-\pi$ to $0$. I'll let you check that the length of $\mu'(s)$ is in fact $1$ everywhere. To see that $\mu$ is a reparameterization of $\gamma$, consider that $$ \mu(s) = \gamma(\sin(s/2)) $$
In general, reparameterizing exactly is very difficult -- the integral you have to compute is often messy, and inverting the resulting function is messier. I built this example so that it happened to be obvious, because I've played with the flat torus in the past. The proof that a reparameterization exists is exactly the same in $n$ dimensions as it is in 2D: for a non-stopping curve (i.e., one whose tangent always has nonzero length), you show that integrating the length of the tangent from some starting point, i.e., computing $$ L(t) = \int_{t_0}^t \| \gamma'(s) \| ds $$ gives you an increasing function $L$; that means that $L$ must have an inverse ... call it $H$. And then defining $$ \mu(s) = \gamma(H(s)) $$ gives you a function which, by the chain rule and fundamental theorem of calculus, has unit-length tangent everywhere, and traverses the same point as does $\gamma$. The only problem is that you can seldom write down the formula for $H$, or for $\mu$, so all you know is that in the abstract, some such thing exists.