Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, but I'm having trouble understanding that.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
+1 Nice, I like. :] – Killercam Nov 9 '11 at 7:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.