special relativity - Metric tensor of coordinate transformation

How do you find a metric tensor given a coordinate transformation, $(t', x', y', z') \rightarrow (t, x, y, z)$? Our textbook gives a somewhat vague example as it skips some steps making it difficult to understand. What's the general definition for a metric tensor of a given transformation? The closest I could find was http://en.wikipedia.org/wiki/Metric_tensor#Coordinate_transformations, but I'm having trouble understanding that.

Answer

You look at the distance between two infinitesimally different points. Let the two coordinate systems be x and y, where x is four numbers and y is four numbers. Consider an infinitesimal displacement from y to y+dy. You know this distance in the x coordinates, so you find the two endpoints of the displacement

$$x(y)$$ $$x^i(y + dy) = x^i(x') + {\partial x^i \over \partial y^j} dy^j$$

This is using the Einstein summation convention--- repeated upper/lower indices are summed automatically, and an upper index in the denominator of a differential expression becomes a lower index, and vice-versa. The distance between these two infinitesimally separated points is:

$$g_{ij}(x) {\partial x^i \over \partial y^k} {\partial x^j \over \partial y^l} dy^k dy^l$$

And from this, you read off the metric tensor coefficients--- since this is the quadratic expression for the distance between y and y+dy.

$$g'_{kl}(y) = g_{ij}(x(y)) {\partial x^i \over \partial y^k} {\partial x^j \over \partial y^l}$$

This is a special case of the tensor transformation law--- every lower index transforms by getting contracted with a Jacobian inverse, and every upper index by getting contracted with a Jacobian.