TITLE: What is the relationship between a metric tensor and a metric?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.
$$ds^2 = \underbrace{dr^2 + r^2 \, d\theta^2}_{\text{diagonal}} - \underbrace{2 r \, dr \, d\theta}_{\text{non-diagonal}}$$