What is the relationship between a metric tensor and a metric?

Question

I have seen both terms many times, but never understood how the two are related (if they are). Could you explain the relationship? Concrete examples, and not just general explanations, would be appreciated. I have a basic understanding of metrics and tensors, but would enjoy a thorough explanation nonetheless.

Edit:

Wolfram alpha (http://mathworld.wolfram.com/MetricTensor.html) mentions that, for an orthogonal coordinate system with (I presume) coordinate variables

$$q_{1}, q_{2}, q_{3}$$

it is the case that the line element in three-space becomes

$$ds^2 = g_{11}dq_{1}^2 + g_{22}dq_{2}^2 + g_{33}dq_{3}^2 \tag{1}$$

where the $g$s follow a relationship labeled (31) on the linked site.

What I want to know is:

A) Do (1) and (31) imply that a metric tensor is a tensor with scale factors as its components, and that the tensor contains data on how the metric differs from the ordinary Euclidean metric?

B) What would (1) look like in a coordinate system without orthogonality?

Answer 1

Suppose I have a surface---the surface of the Earth, say. It's annoying to do calculations directly on the surface of the Earth, since it's a 2D object sitting in 3D space, so that not all points $(x,y,z)$ correspond to valid points on the surface. For calculations it's easier to work on a chart of part of the Earth: for example I can represent the northern hemisphere of the Earth by a unit disk in the plane, together with a map $$\phi_1: D_1 \subset \mathbb{R}^2 \to \mathbb{R}^3 \quad (x,y) \mapsto \left(Rx, Ry, R\sqrt{1-x^2-y^2}\right)$$ (where I've idealized the Earth as a sphere of radius $R$).

I could draw a second chart, of a region near the equator, using spherical coordinates: $$\phi_2: D_2 \subset \mathbb{R}^2 \to \mathbb{R}^3, (\theta,\phi) \mapsto \left(R\sin\theta \cos \phi, R\sin\theta\sin\phi, R\cos\theta\right).$$

The following concept is absolutely critical: the object I actually care about is the Earth. It exists independent of any charts I draw of parts of the Earth. But because it's more convenient to compute on a 2D parameterization of the Earth rather than on the Earth itself, I can create these maps $\phi_1$ and $\phi_2$. But there is nothing natural or canonical about these charts: just as in real-world charts of the Earth, I can draw many different charts in addition to these two, and they could all have different kinds of area distortions, angle distortions, etc.

If I draw a curve on $D_1$, I can't really tell just by looking at the chart $D_1$ how long the curve is on the sphere. In fact it's not even true that a straight line on $D_1$ corresponds to a "straight" curve (geodesic) on the Earth! If I want to compute anything geometric, like angles, or lengths, I need to (1) copy what I'm trying to measure from $D_1$ onto the actual Earth, using the $\phi_1$ map; (2) measure things on the Earth; (3) copy the result back to $D_1$.

For example, let's say I draw a vector at some point $(x,y)$ on $D_1$. I want to calculate its length: not the length of the arrow I've draw on the piece of paper with the chart of the Earth, but the length of the corresponding tangent vector on the actual surface of the Earth. There is some function $L(\mathbf{v})$ which takes in tangent vectors on the Earth and returns their lengths: and this is a function that's intrinsic to the Earth, independent of any chart.

The metric is a way of encoding lengths: the metric $g$ is a bilinear function $g(\mathbf{v},\mathbf{w})$ associated to the tangent plane at every point on the Earth, which when given two copies of the same vector, computes the squared length of the vector, $$g(\mathbf{v},\mathbf{v}) = L(\mathbf{v})^2.$$ It turns out this property uniquely determines $g(\mathbf{v},\mathbf{w})$ even when $\mathbf{v}\neq \mathbf{w}$, by the parallelogram law.

Again, this object exists on the Earth itself. It doesn't depend on which chart you use for navigating on the Earth. Of course, it would be nice to actually calculate lengths, given a vector $(v_x,v_y)$ at $(x,y)$ on the chart $D_1$. Fortunately, we can do this easily: the tangent vector on the surface of the Earth corresponding to the vector on the chart is given by $$\left[J\phi_1(x,y)\right]\begin{bmatrix}v_x\\v_y\end{bmatrix} = \begin{bmatrix}R & 0 \\ 0 & R\\ -\frac{R x}{\sqrt{1-x^2-y^2}} & -\frac{R y}{\sqrt{1-x^2-y^2}}\end{bmatrix}\begin{bmatrix}v_x\\v_y\end{bmatrix},$$ where $J\phi_1$ is the Jacobian of $\phi_1$. It follows that the metric evaluated on two vector $\mathbf{v}, \mathbf{w}$ at a point $(x,y)$ on $D_1$ can be written in terms of matrix multiplication (as can all bilinear functions, of course):

$$g(\mathbf{v},\mathbf{w}) = \begin{bmatrix}v_x & v_y\end{bmatrix}^T\begin{bmatrix} \frac{R^2(y^2+1)}{1-x^2-y^2} & \frac{R^2xy}{1-x^2-y^2}\\ \frac{R^2xy}{1-x^2-y^2} & \frac{R^2(x^2-1)}{1-x^2-y^2}\end{bmatrix}\begin{bmatrix}w_x\\w_y\end{bmatrix} = \begin{bmatrix}v_x & v_y\end{bmatrix}^Tg_1 \begin{bmatrix}w_x & w_y\end{bmatrix}.$$

Observe that the left-hand side involves the intrinsic metric, and does not involve any charts. The right-hand side expresses the same thing in coordinates of the chart $D_1$, with the matrix $g_1$, which is a coordinate representation of $g$, the metric tensor.

Notice, incidentally, that $g_1$ depends on the position $(x,y)$. This makes perfect sense: the farther from the north pole, the more length distortion on your chart $D_1$.

Different charts will give you different expressions for the metric tensor. For $\phi_2$ you will get

$$g(\mathbf{v},\mathbf{w}) = \begin{bmatrix}v_{\theta} & v_{\phi}\end{bmatrix}^T\begin{bmatrix} R^2 & 0\\ 0 & R^2\sin(\theta)^2\end{bmatrix}\begin{bmatrix}w_{\theta}\\w_{\phi}\end{bmatrix} = \begin{bmatrix}v_{\theta} & v_{\phi}\end{bmatrix}^Tg_2 \begin{bmatrix}w_{\theta} & w_{\phi}\end{bmatrix}.$$

As you note, some authors like to write $$g_2 = R^2 d\theta^2 + 0 d\theta d\phi + R^2\sin(\theta)^2 d\phi^2$$ but don't be fooled: it's just different notation for the exact same concept: a representation of the metric in coordinates as a $2\times 2$ bilinear operator. Some people just prefer quadratic form notation over matrix notation.

I hope that clears things up. In your question you also ask about orthogonal coordinate systems, but that's a red herring: all of the above applies regardless of whether the metric tensor turns out to be diagonal or not.

Answer 2

In short:
A metric is "macroscopic" in that it gives a distance between points however far away they are, while a metric tensor is "microscopic" in that it only gives a distance between (infinitesimally) close points.

The metric tensor $g_{ab}$ defines a metric in a connected space, $d(p_1, p_2) = \inf_\gamma \int_\gamma ds,$ where $ds = \sqrt{\sum_{a,b} g_{ab} \, dx^a \, dx^b} = \sqrt{\sum_{a,b} g_{ab} \dot x^a(t) \, \dot x^b(t)} \, dt$ for some parametrization $x(t) = (x^a(t))_{a=1}^{n}$ ($n$ being the dimension of the space), and the infimum is taken over all smooth enough paths between $p_1$ and $p_2$.

Like many tensors, the metric tensor is often given in some coordinate system. For example, $ds^2 = dx^2 + dy^2$ and $ds^2 = dr^2 + r^2 \, d\theta^2$ are both representations of the same metric in different coordinate systems (Cartesian and polar, respectively).

In a non-orthogonal coordinate system there are non-diagonal terms, e.g. $$ds^2 = \underbrace{dr^2 + r^2 \, d\theta^2}_{\text{diagonal}} - \underbrace{2 r \, dr \, d\theta}_{\text{non-diagonal}}$$