# 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.

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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:

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

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.

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A metric tensor1 does not correspond to a coordinate transformation, it corresponds to a coordinate system. The initial coordinate system, with coordinates $(t',x',y',z')$, corresponds to one metric tensor $g_{\mu\nu}'$, and the final coordinate system, with coordinates $(t,x,y,z)$, corresponds to a different metric tensor $g_{\mu\nu}$. When you transform from the primed coordinate system to the unprimed coordinate system, you also transform the metric tensor from the primed $g_{\mu\nu}'$ to the unprimed $g_{\mu\nu}$.

The rule by which you transform the metric tensor when changing from one coordinate system to another is

$$g_{\mu\nu} = \frac{\partial {x^\mu}'}{\partial x^\mu}\frac{\partial {x^\nu}'}{\partial x^\nu}g_{\mu\nu}'$$

If your initial (primed) coordinate system is the Cartesian system of Minkowski space, then it corresponds to a metric tensor of $\pm\operatorname{diag}(-1,1,1,1)$, and you get

$$g_{\mu\nu} = -\frac{\partial t'}{\partial x^\mu}\frac{\partial t'}{\partial x^\nu} + \frac{\partial x'}{\partial x^\mu}\frac{\partial x'}{\partial x^\nu} + \frac{\partial y'}{\partial x^\mu}\frac{\partial y'}{\partial x^\nu} + \frac{\partial z'}{\partial x^\mu}\frac{\partial z'}{\partial x^\nu}$$

In this case it may look like the metric tensor corresponds to the transformation, but that's only because you started from a particularly simple coordinate system.

1In the above post, when I say "metric tensor" I actually mean "matrix representation of the metric tensor". Technically, a tensor itself is an object which exists independent of any coordinate system, and in particular the metric tensor is a property of the underlying space. Once you choose a particular coordinate system, you can represent the tensor in that coordinate system by using a matrix. That matrix representation of the tensor is what changes as you transform from one coordinate system to another.

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+1 Nice, I like. :] –  Killercam Nov 9 '11 at 7:59