# Contravariant Metric Tensor

Okay, so I have exactly ZERO experience with tensors and this project I am working on involves tensors. I have looked through a bunch of online resources, and attempted to look for textbooks (not available to me) and I am getting really confused. The extent to which I have picked up is that I understand how to find the metric tensor for the spherical coordinate transformation. So I was hoping someone could run me through the following example or give some advice on how to approach this.

Curvilinear coordinate system: $v=x$ $u = y-a(x)$ $a(x)$ arbitrary $w=z$

I am looking for the contravariant metric tensor for this (and possibly what the interpretation is). I have the result; \begin{pmatrix}1 & -a'(x) & 0 \\ -a'(x) & 1+a'(x) & 0 \\ 0 & 0 & 1\end{pmatrix}

Thanks

This is a process that can feel very arbitrary, but using geometric principles, you should be able to develop an intuition about these problems.

Imagine the coordinate functions $v, u, w$ as scalar fields on the 3d space, assigning their respective coordinates to a given position. For all these coordinates, there are associated gradients: $\nabla v$ for $v$, and so on. These tell us the direction of greatest increase for each coordinate.

What we do then is use these gradient vectors as a basis for our space: a set of vectors $g^v, g^u, g^w$ such that $g^v = \nabla v$ and so on. The contravariant metric tensor just measures the dot products of these vectors, so we can have an idea of how to measure lengths with them.

For instance, take $u = y - a(x)$ as you gave us. Taking the gradient of $u$, we get

$$g^u = \nabla u = (g^x \partial_x + g^y \partial_y + g^z \partial_z) u(x,y,z) = g^y -a'(x) g^x$$

where $g^x, g^y, g^z$ are a Cartesian basis (thus, they are orthonormal), so it's easy to take the dot product:

$$g^{uu} = g^u \cdot g^u = [g^y - a'(x) g^x] \cdot [g^y - a'(x) g^x] = 1 + [a'(x)]^2$$

as you found. So if we have two vectors expressed using this basis of gradients, we can find the overall dot product using the contravariant metric, rather than having to go back and figure out the relationships between those gradients all over again.

• This makes more sense then anything I have seen so far. What is the covariant metric tensor? – yankeefan11 Jun 10 '13 at 14:46
• If $\mathbf r$ is the position vector, then there is a basis of "tangent vectors" $g_v, g_u, g_w$ such that $g_v = \partial \mathbf r/\partial v$ and so on. Geometrically, the logic is that you can put a curve of increasing $v$ and constant $u,w$ through any point, and the vector tangent to this curve is $g_v$. Practically, this means you have to solve for $x(v,u,w)$ and so on (inverting the equations you have) so you can take the derivative. Then the dot products of these tangent basis vectors form the covariant metric. – Muphrid Jun 10 '13 at 14:53