Seeing how I wrote a new post yesterday in the area of tensor analysis, I was reminded of the beautiful result below. It likely would have made it onto this blog at some point, so I thought I would quickly write it all out.

Here we go.


Imagine we are presented with some arbitrary curve in Euclidean space. This curve has no defining coordinate system, and so we can simply picture it like this:

Now, as should be emphasised more than once, without enforcing a coordinate system we’re going to want to parameterise this curve. Similar, perhaps, to how we might build a picture of a relativistic string from the ground up, we’re going to want to issue some generalised coordinates and piece together what we can glean about this string. So let’s use arc length, s, for parameter. Let’s also define some general coordinates, say, \gamma : (Z^1(s), Z^2(s)). But remember, these general coordinates can be Cartesian, Polar or whatever.

This is good. We are making progress.

With the stage now set, here’s the intriguing question I want you to think about (you have already seen this in a textbook): Can you, given the unit tangent to this arbitrary curve (image below), find the components of this unit tangent with respect to the covarient basis?

First, one might be inclined to say, “without defining a coordinate system I can’t even think about or imagine deriving the coordinates of the tangent basis!” Putting aside what we know of the power of tensor analysis, one can certainly sympathise with such a response. But what we have is, indeed, the power of tensor analysis and so we can proceed in a very elegant way.

The main objective here is that, given this unit tangent, we want to find some algebraic expression for it of the general form

\vec{T} = T^1 \vec{e}_1 + T^2 \vec{e}_2

Let me also say this: the desire here is to express T^1 and T^2 in terms of our general coordinates (Z^1(s), Z^2(s)).

So, how do we go about this? Recall, firstly, that as a standard the definition of the unit normal is dr ds.

Think of R as a function of s. As such, it follows that we can write: R(s) = \vec{R}(Z^1(s),Z^2(s)).

But our unit tangent can also be written as \vec{T}(s), noting that \vec{T}(s) = \frac{d \vec{R}(s)}{ds}.

This leads us directly to ask, “what is  \frac{d \vec{R}(s)}{ds}. Well, we can compute it as follows

    \[ \vec{T}(s) = \frac{d \vec{R}(s)}{ds} \implies T(s) = \frac{\partial R}{\partial Z^1}\frac{d Z^1}{ds} + \frac{\partial R}{\partial Z^2}\frac{d Z^2}{ds} \]

Ask yourself, what is \frac{\partial R}{\partial Z^1}? It is the covariant basis!

And this deserves celebration, as it is appeared in our investigation of this arbitrary curve without force!

Thus, and I will write it out again in full as the conclusion,

    \[ \vec{T}(s) = \frac{d\vec{R}(s)}{ds} \implies T(s) = \frac{\partial R}{\partial Z^1}\frac{d Z^1}{ds} + \frac{\partial R}{\partial Z^2}\frac{d Z^2}{ds} = \frac{d Z^1}{ds} e_1 + \frac{d Z^2}{ds} e_2 \]

Where, T^1 = \frac{d Z^1}{ds} and T^2 = \frac{d Z^2}{ds}.

There we have it, a beautiful result.