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

d\textbf{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 d\textbf{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 d\textbf{R} \cdot d\textbf{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

    \[ \begin{pmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \end{pmatrix} = \begin{pmatrix} g_{11} & g_{12} & g_{13} \\ g_{21} & g_{22} & g_{23} \\ g_{31} & g_{32} & g_{33} \\ \end{pmatrix} \]

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

    \[ \begin{pmatrix} 1 & 0 & 0 \\ 0& \rho^2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \]