Posted by on Jan 15, 2013 in Science | 0 comments

The name, metric tensor, is enough to strike fear into the hearts of any physics student, associated as it is with General Relativity. I’m here to tell you that, in its heart of hearts, the metric tensor is not much more complex than the standard vector dot product, of which it is a generalization. In what follows, I hope to remove some of the tension surrounding the metric tensor.

That raises a pretty basic question right off the bat: if we already have the machinery of the dot product, why do we need a metric tensor? Where does the dot product break down? Recall that one way to compute the dot product is to take the product of the magnitudes of the two vectors times the cosine of the angle between them. But in my last post, on parallel transport, it became apparent that the notion of the absolute angle of a vector with respect to a coordinate system in a curved space was highly location dependent. It is of the essence that the laws of physics, and hence any calculations that one does with them, should not depend on our subjective choice of a coordinate system.

So, if we’re considering questions of vector analysis in curved spaces or on curved surfaces, it is natural to challenge one’s assumptions about fundamentals like the angles between pairs of vectors. The results of a calculation should never depend upon the choice of a coordinate system, or if they do, as in the case of moving frames of reference in Special Relativity, we want to find some coordinate transformations that identify conserved or invariant quantities that do not depend upon the choice of coordinates.

What is this metric tensor thingy then? Well, our problem begins with the idea of the length of a path on some curved surface. The surface might be a sphere, a cone, a saddle or some much more arbitrary object. The sort of spaces in which we’ll be interested are often referred to as manifolds, a fairly impressive name for something right off the bat. When I was younger, a manifold was a bunch of pipes to get the exhaust out of an engine; but it’s not that kind of manifold I’m talking about now. Instead, as a mathematical entity, it refers to a set of points that have certain properties. One basic property is that points in a manifold have neighborhoods, which you can think of intuitively as the name suggests: a point has a set of other points that are close to it in some defined sense. Being oriented to axioms and theorems, mathematicians make this concept as abstract as possible in order to come up with the minimum possible set of definitions and axioms with which to construct the idea of a manifold. For my purposes here, it will be sufficient to associate the idea of a neighborhood with a a distance measure. A neighborhood of some point will be taken to be a set of points within a certain distance of the reference.

Another aspect of a manifold is that in the vicinity of any given point, it behaves like some n-dimensional Euclidean (flat) space. While globally, the manifold may be curved all to heck, one can always work closely enough to any given point and find that a Euclidean treatment of, particularly, differentials, is not problematic. This implies that the manifold is smooth and differentiable at least to second order, and that properties like convergence and continuity make sense.

Enough of that digression. We’re working on an n-dimensional manifold, which again, for now, is some curved 2-dimensional surface in a 3-dimensional space. We can certainly come up with a 3-parameter expression for our curved surface; for example, we could write for a sphere of radius $R$: $x^2 + y^2 + z^2 = R^2$

However, this is one too many parameters, since all we need are two parameters to specify any point on the surface. For a sphere of given radius, all we need are the two angles for latitude and longitude, which are basically equivalent to spherical coordinates with $r=R$ constant. Granted, not every surface admits such an easy parameterization; but assume that one exists for now, no matter how complex. Label the two parameters for our 2-D manifold as $u, v$. For the sphere, we could write the surface as $\overset{\rightharpoonup }{r}=(R cos\theta sin\phi, R sin\theta sin\phi, R cos\phi)$

with the identification $u = \theta$ and $v = \phi$.

For some path on the surface, we can introduce another parameter, $t$, which might be the time, and then consider the path to be a function of $t$ in the form $u = u(t), v=v(t)$. For the example of the sphere, we might set $\theta(t) = t$ and $\phi(t) = 0$ just to pick something specific. This would be a curve around the “Equator” of the sphere that increased directly as $t$.

The length of the path can be found be integrating over $t$ from some initial condition $t_1$ to a final condition $t_2$ as follows: $s = \int _{t_1}^{t_2}\left|\frac{d}{dt}\overset{\rightharpoonup }{r}(u(t),v(t))\right|dt$

So, expanding in terms of the Euclidean norm, $s = \int_{t_1}^{t_2} \sqrt{\left(\frac{du}{dt}\right)^2\frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial u}+2\frac{du}{dt}\frac{dv}{dt}\frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v}+\left(\frac{dv}{dt}\right)^2\frac{\partial \overset{\rightharpoonup }{r}}{\partial v}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v}} \, dt$

In taking these steps, we are taking advantage of the fact the we are working on manifold, in which differential elements behave as in a Euclidean space. As a result, we can write the differential element, $ds^2$ as $\text{ds}^2= \frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial u}(du)^2+\frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v}2dudv+\frac{\partial \overset{\rightharpoonup }{r}}{\partial v}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v}(dv)^2$

This can be expressed in matrix/vector form as $\text{ds}^2 = \left(
\begin{array}{cc}
\text{du} & \text{dv}
\end{array}
\right)\left(
\begin{array}{cc}
\frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial u} & \frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v} \\
\frac{\partial \overset{\rightharpoonup }{r}}{\partial u}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v} & \frac{\partial \overset{\rightharpoonup }{r}}{\partial v}\cdot \frac{\partial \overset{\rightharpoonup }{r}}{\partial v}
\end{array}
\right)\left(
\begin{array}{c}
\text{du} \\
\text{dv}
\end{array}
\right)$

The matrix of partial derivatives in this last equation is just the metric tensor, $g$. In this form, it is apparent that the introduction of the metric tensor is a generalization of what would otherwise just be a dot product of the differential vector, $(du, dv)$. The metric tensor itself is formed of the dot product of partial derivatives of the surface vector with respect to the two variables that parameterize the surface. Hence, each component is itself a scalar.

Let’s jump back for a moment and consider our running example of the sphere where we set $u = \theta$ and $v = \phi$. It is just a matter of the calculation of the dot products of the partial derivatives (what a great application for Mathematica!) for the equation of the surface of the sphere that we gave earlier in order to get the result: $g_{\text{uv}}= \left(
\begin{array}{cc}
R^2 \sin (v )^2 & 0 \\
0 & R^2
\end{array}
\right)$

remembering that we’ve set $u = \theta$ and $v = \phi$.

We’ve been calling this object a tensor, but the test of this assertion depends on how it transforms with a change of coordinates. Suppose we change from $u, v$ to $u', v'$. By the chain rule, $\frac{\partial \overset{\rightharpoonup }{r}}{\partial u'}=\frac{\partial u}{\partial u'}\frac{\partial \overset{\rightharpoonup }{r}}{\partial u}$

so, it is straightforward, if somewhat tedious to get the metric tensor in the primed coordinates as $g_{u'v' }= \left(
\begin{array}{cc}
\frac{\partial u}{\partial u'} & \frac{\partial u}{\partial v'} \\
\frac{\partial v}{\partial u'} & \frac{\partial v}{\partial v'}
\end{array}
\right)^Tg_{\text{uv}}\left(
\begin{array}{cc}
\frac{\partial u}{\partial u'} & \frac{\partial u}{\partial v'} \\
\frac{\partial v}{\partial u'} & \frac{\partial v}{\partial v'}
\end{array}
\right)$

where the $T$ indicates matrix transpose. This equation confirms that $g_{uv}$ transforms as a covariant tensor; in the partial derivatives for the transformation, the original coordinates appear in the numerator.

If you think now about these differential elements, $du$ and $dv$, since they are parameters of a generally curved surface, any significant extensions of these can be considered to fall into a tangent plane at a given point, $(u, v)$. So another interpretation of the metric tensor is that it provides a means to compute lengths and angles of tangent vectors at a point on the surface. For any point on the surface, unit vectors in the directions defined by $du$ and $dv$ will span the tangent plane. If we have the metric tensor, $g_{uv}$ at this point, then for any tangent vector, $\overset{\rightharpoonup }{a }= a_u\overset{\rightharpoonup }{r_u} + a_v\overset{\rightharpoonup }{r_v}$

its length will be $\left|\overset{\rightharpoonup }{a}\right|=\sqrt{\overset{\rightharpoonup }{a}\cdot \overset{\rightharpoonup }{a}}=\sqrt{\left(
\begin{array}{cc}
a_1 & a_2
\end{array}
\right)g\left(
\begin{array}{c}
a_1 \\
a_2
\end{array}
\right)}$

Likewise, for any two tangent vectors, the cosine of the angle between them will be $cos \theta = \frac{\left(
\begin{array}{cc}
a_1 & a_2
\end{array}
\right)g\left(
\begin{array}{c}
b_1 \\
b_2
\end{array}
\right)}{\sqrt{\left[\left(
\begin{array}{cc}
a_1 & a_2
\end{array}
\right)g\left(
\begin{array}{c}
a_1 \\
a_2
\end{array}
\right)\right] \left[\left(
\begin{array}{cc}
b_1 & b_2
\end{array}
\right)g\left(
\begin{array}{c}
b_1 \\
b_2
\end{array}
\right)\right]}}$

If each of the vector multiplications through $g$ is seen as a generalized dot product, then this equation reduces precisely to the simple equation of vector algebra for the cosine of the angle between to vectors in a Euclidean space with Cartesian coordinates.

Thinking about that raises and answers the question of the metric tensor for just such a situation; namely, a Euclidean space with Cartesian coordinates, $x_m$. The answer is that $g_{mn} = \delta_{mn}$

where $\delta_{mn}$ is the Kroenecker delta which is $1$ if $m=n$ and $0$ otherwise. As a matrix, the Kroenecker delta is just the identity matrix. Substituting back into the equations just given simply yields the usual equations for vector norm and the cosine of angles between vectors.

This provides a hint at the physical meaning of the metric tensor; namely, it introduces exactly the correct conversion factors, in terms of the appropriate partial derivatives, to compensate locally for the non-linearity of coordinates on a curved surface. There are really two factors in play here: non-Euclidean spaces and non-Cartesian coordinates. One can, of course, introduce non-Cartesian coordinates within a Euclidean space; for example, polar coordinates in the Euclidean plane. In such a case, there will always be some coordinate transformation that can bring the metric tensor back to $\delta_{mn}$. Therefore, the metric tensor given in these non-Cartesian coordinates is just some covariant transformation of the Kroenecker delta. But in a truly non-Euclidean space, no such transformation is possible; and the metric tensor on such a space will represent an irreducible curvature of the space itself, not just a non-linearity of the coordinates.

For example, we could imagine the Euclidean plane embedded in 3-space given by the constraint that $z=0$. Then, we could parameterize this surface with polar coordinates, $r, \theta$. The equation for the surface is just $\overset{\rightharpoonup }{r}= (x, y, 0) = (R cos \theta, R sin \theta)$

With the identification that $u = R$ and $v = \theta$, the metric tensor is $g = \left(
\begin{array}{cc}
1 & 0 \\
0 & R^2
\end{array}
\right)$

The Jacobian matrix $\left(
\begin{array}{cc}
1 & 0 \\
0 & R^{-1}
\end{array}
\right)$

transforms this metric tensor into the identity matrix, which is equivalent to the Kroenecker delta. That is, $\left(
\begin{array}{cc}
1 & 0 \\
0 & R^{-1}
\end{array}
\right)\left(
\begin{array}{cc}
1 & 0 \\
0 & R^2
\end{array}
\right)\left(
\begin{array}{cc}
1 & 0 \\
0 & R^{-1}
\end{array}
\right)=\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)$

## Summary

So, we’ve seen that we can create a mathematical object, commonly called the metric tensor, which reduces to the identity matrix for Cartesian coordinates on Euclidean spaces, but which is generally more complex for curved surfaces with non-Cartesian coordinates. It is composed of the partial derivatives of the equations of the surface in terms of a parameterization of it aka a coordinate system for the surface. This object transforms into other coordinate systems for the surface like a covariant tensor. Under such transformations, the square of the length of a differential line element, $ds^2 = d{s'}^2$ is preserved. As a consequence of this property, the length of line segments and therefore vectors is also invariant when the metric tensor is included in the generalized dot product. The same is true of the angles between line segments and vectors. The utility of the metric tensor is that it can be used to make distance, angles, areas, and other commonly employed geometric measures invariant to coordinate transformations on curved spaces. 