# What is the metric tensor for?

I am wondering how to use the metric tensor, in practice? I read the book and done the exercises in A student's guide to vectors and tensors by Dan Fleisch. The concept of a tensor and their applications are well defined.

In that book, is explained how to get the metric tensor for coordinate system transformation, such as from spherical coordinates to ordinary Cartesian coordinates or even from cylindrical coordinates to Cartesian coordinates; which are easy to obtain, given enough practice. But what do such metric tensors mean (in practice), how does one use such a tensor in an actual math/physics problem?

• A metric is usually associated with a particular coordinate system. Can you elaborate on how you are using one in a transformation? Usually that would be done by the Jacobian. – user10851 Jan 13 '16 at 0:09
• For example the trasformation from polar coordinates (r,θ)= (x^1,x^2) to Cartesian coordinates (x,y) goes as follows: x=rcos(θ) and y = rsin(θ) and the components of the metric are g ij = [(dx/x^i)(dx/d^j)+(dy/dx^i)(dy/d^j)] where i and j take on values [1 or 2]. This yeilds a 2 x 2 matrix in this case. – Investor Jan 13 '16 at 0:17
• @ChrisWhite - The metric is most certainly not associated with any coordinate system (no tensor is!). – Prahar Jan 13 '16 at 0:25
• @Prahar The components (you know what I mean). Certainly a metric is never associated with a coordinate transformation. – user10851 Jan 13 '16 at 0:26
• @ChrisWhite - Agreed. – Prahar Jan 13 '16 at 0:26

The metric measures lengths in various directions, and also angles between various directions. For example if $\vec{e}_{(1)}$ is the basis vector in the $x^1$-direction, it will have length (squared) given by $$\lVert \vec{e}_{(1)} \rVert^2 = g(\vec{e}_{(1)}, \vec{e}_{(1)}) = g_{11}.$$ If we also have the basis vector $\vec{e}_{(2)}$ in the $x^2$-direction, then the angle $\theta$ between these vectors obeys $$\lVert \vec{e}_{(1)} \rVert \cdot \lVert \vec{e}_{(2)} \rVert \cos\theta = \vec{e}_{(1)} \cdot \vec{e}_{(2)} = g(\vec{e}_{(1)}, \vec{e}_{(2)}) = g_{12}.$$

So far we haven't made any mention of transforming coordinates. Now coordinate transformations are something we'd like to be able to do, and the rule for the metric (or indeed any rank-(0,2) tensor) is $$g_{ij} = \sum_{\hat{\imath},\hat{\jmath}} \frac{\partial x^\hat{\imath}}{\partial x^i} \frac{\partial x^\hat{\jmath}}{\partial x^j} g_{\hat{\imath}\hat{\jmath}}. \qquad \text{(all coordinate transformations)}$$ If the hatted coordinate system is normal Euclidean space with normal Cartesian coordinates, $g_{\hat{\imath}\hat{\jmath}} = \delta_{\hat{\imath}\hat{\jmath}}$ and we are left with $$g_{ij} = \sum_{\hat{\imath}} \frac{\partial x^\hat{\imath}}{\partial x^i} \frac{\partial x^\hat{\imath}}{\partial x^j}. \qquad \text{(Cartesian hatted coordinates only)}$$ But this is just a rule for transforming the metric from one coordinate system to another. The real use of the metric is to calculate lengths and angles in a particular coordinate system (as above), or to describe the local geometry of space(time) in a concise, abstract way (in which case you don't even find its components in any particular coordinate system).

The metric is an important concept in general relativity.

In GR, vectors correspond to weighted directions in spacetime (by "weighted", I mean any scalar multiple of a vector corresponds to the same direction, but weighted differently). The metric tensor can then tell us about the angle between two directions or the magnitude of a given vector, which gives us a notion of length in spacetime.

The metric also appears in Einstein's equtaions, relating the distribution of energy and momentum through spacetime to the curvature--which involves the metric and its derivatives. That is, curvature--and therefore the metric--of spacetime are determined by the distribution of energy and momentum.

• But how do you use me metric in transformation from spherical coordinates to Cartesian coordinates, as taught in the book I read? – Investor Jan 13 '16 at 0:06
• Is that your question--whether the metric plays a role in the transformation from one set of coordinates to another? If so, then I would ask what you're trying to compare against. Do you want to look at the expressions for the spherical basis vectors in terms of the Cartesian ones? Do you want to look at the new metric compared to the old one? – Muphrid Jan 13 '16 at 0:16
• I added a comment above, if it can help. – Investor Jan 13 '16 at 0:19
• Sure, that computes the new metric (in polar coordinates) when the old metric (in Cartesian coordinates) is the identity matrix. That formula would not be correct if the metric in Cartesian coordinates were some other symmetric matrix (positive definite as well, for Euclidean spaces). – Muphrid Jan 13 '16 at 0:22
• It may mean curved space; it is not simple to tell if space is curved or not just by looking at the metric. The metric may be flat and still not identity (for instance, a multiple of the identity is still flat); conversely, even in polar coordinates the space is still flat. But in general, the space isn't necessarily flat. Consider, for example, the metric for a Schwarzschild black hole. Even if you considered the coordinates there to be related to Cartesian ones by the usual transformation, the resulting metric for those Cartesian coordinates would not be the identity. – Muphrid Jan 13 '16 at 0:30

The metric is a rank two tensor that defines several features of a differential manifold. It defines how to relate changes in distance to changes in coordinates, how to take take the inner product of two vectors. Given an inner product we have a way of measuring angles.

More indirectly, the metric describes geodesics in a manifold, the shortest distance between two points. Just as the shortest distance between two points on the surface of a sphere is not a straight line, this is also the case in a general non-flat space. Speaking of, the metric contains information about the curvature of a manifold as well as how coordinate basis vectors change within the manifold.

Given its relation with the inner product, the metric provides a means to generalize Gauss' and Stoke's Laws to higher dimensions.

This is an excellent introduction to these topics:A First Course in General Relativity by Schultz