Simply beautiful: Finding the covariant basis from seemingly nothing


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{dvec{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.

Understanding metric elements for cylindrical coordinates


The metric tensor is ubiquitous when arriving at a certain level in one’s physics career. When it comes to cyndrilical coordinates, there is a useful way to remember its deeper meaning through a rather simple derivation – or at least through the use and construction of a series of definitions. (I say ‘simple’ in that this is something one can do ‘on the fly’, should they be required to remind themselves of the properties of the metric tensor). For a more general treatment of what follows, see also this document on special tensors.


To start, if we had some general curvilinear coordinate system, we could begin by writing

$dtextbf{R} = g_1 du^1 + g_2 du^2 + … = g_i du^i $

From this we can also immediately invoke the principle that $ds^2 = d textbf{R} cdot dtextbf{R} $. I won’t explain this for sake space, but any textbook will cover why $ds^2$ is equal to the dot product of our displacement vector with itself.

But from above we also can see that $dtextbf{R} cdot dtextbf{R} $ is the same as, in this example using two curvilinear coordinates, $g_i du^i cdot g_j du^j$. It follows,

$ds^2 = g_i du^i cdot g_j du^j = g_i cdot g_j du^i du^j$

As taking the dot product of our two tangent vectors, $g_1 du^1 + g_2 du^2$, is equal by standard relation to the metric tensor, we arrive at the following

$ds^2 = g_i cdot g_j du^i du^j = g_{ij} du^i du^j$

This identity, if I may describe it as such, is important to remember. But how can we issue further meaning to this result?


Let’s use cylindrical coordinates.


Now, for pedagogical purposes, I’ve labelled the unit vectors $e_p$,$e_phi$, and $e_z$. But, to generalise things, moving forward note that I have set $e_p = e_1$,$e_phi = e_2$, and $e_z = e_3$.

The first thing that one should notice is that our displacement vector, $textbf{R}$, can be written as

$textbf{R} = rho cos phi hat{i} + rho sin phi hat{j} + z hat{k}$

If this is not immediately clear, I suggest taking a few minutes to revise cylindrical coordinates (a generalisation of 2D polar coords.).

Now, here is a key step. Having written our equation for $textbf{R}$, we want to focus in on  $e_1$, $e_2$, and $e_3$ and how we might represent them with respect to $textbf{R}$. To do this, we’re going to take partials.

$e_1 = frac{partial textbf{R}}{partial rho} = cos phi hat{i} + sin phi hat{j}$

$e_2 = frac{partial textbf{R}}{partial phi} = – rho sin phi hat{i} + rho cos phi hat{j}$

$e_3 = frac{partial textbf{R}}{partial z} = hat{k}$

With expressions for $e_1$, $e_2$, and $e_3$ established, we can return to our previous definition of the metric tensor and look to derive a definition of our metric tensor for cylindrical coordinates.

Notice, if $e_i cdot e_j = g_{ij}$, in matrix form we have

[ begin{pmatrix}
e_{11} & e_{12} & e_{13} \
e_{21} & e_{22} & e_{23} \

e_{31} & e_{32} & e_{33}


g_{11} & g_{12} & g_{13} \

g_{21} & g_{22} & g_{23} \

g_{31} & g_{32} & g_{33} \

end{pmatrix} ]

What is this saying? Well, let’s look. We know, again, that $g_{ij} = e_i cdot e_j$. So if we take the dot product, as defined, we should then arrive at our metric tensor for our cylindrical system. And, indeed, this is exactly the result. For $g_{11}$, for instance, we’re just taking the dot product of $e_1$ with itself.

[ g_{11} =  (cos phi hat{i} + sin phi hat{j}) cdot (cos phi hat{i} + sin phi hat{j})

= cos^2 phi + sin^2 phi = 1 ]

We keep doing this for the entire matrix. But to save time, I will only work out the remaining diagonals. The non-diagonal components cancel to zero. You can do the dot product for each to see that this is indeed true. Thus,

[g_{22} = ( – rho sin phi hat{i} + rho cos phi hat{j}) cdot (- rho sin phi hat{i} + rho cos phi hat{j})

=rho^2 sin^2 phi + rho^2 cos^2 phi = rho^2 ]

[g_{33} = hat{k} cdot hat{k} = 1]

Therefore, we arrive at our desired metric

[ begin{pmatrix}
1 & 0 & 0 \

0& rho^2 & 0 \

0 & 0 & 1 \

end{pmatrix} ]