This is just something that I've made up to see if I understand the method.

If I have the line element: $$ds^2 = dr^2 + r^2\,d\phi^2$$ and I want to carry out a transformation with $r = \dfrac{a^2}{r^\prime}$.

Note $$g_{xx} = \dfrac{\partial x^\alpha}{\partial x}\dfrac{\partial x^\alpha}{\partial x}g_{\alpha\alpha} = \left(\dfrac{\partial x^\alpha}{\partial x}\right)^2$$

Would the following be correct?

$$x^\alpha = (r,\phi).\quad x = r\cos\phi = \dfrac{a^2}{r^\prime}\cos\phi,\quad y = r\sin\phi = \dfrac{a^2}{r^\prime}\sin\phi$$ Calculating the metric coefficients $$g_{11} = \left(\dfrac{\partial x^\alpha}{\partial r^\prime}\right)^2 = \dfrac{a^4}{r^{\prime4}}\cos^2\phi + \dfrac{a^4}{r^{\prime4}}\sin^2\phi = \dfrac{a^4}{r^{\prime4}}$$ $$g_{22} = \left(\dfrac{\partial x^\alpha}{\partial \phi}\right)^2 = \dfrac{a^4}{r^{\prime2}}\sin^2\phi + \dfrac{a^4}{r^{\prime2}}\cos^2\phi = \dfrac{a^4}{r^{\prime2}}$$ Giving the final result $$ds^2 = \dfrac{a^4}{r^{\prime4}}\left(dr^{\prime2} + r^{\prime2}\,d\phi^2\right)$$

  • $\begingroup$ Your $g_{xx}$ equation doesn’t make sense. You should not be introducing $x$ and $y$. $g$ is not the “connection” but the “metric”. Your result is incorrect. $\endgroup$
    – G. Smith
    May 19, 2019 at 18:29
  • $\begingroup$ You can calculate the line element like this. $\begin{aligned}x=f\left( r',\varphi \right) \\ dx=\dfrac {\partial f}{\partial r'}dr'+\dfrac {\partial f}{\partial \varphi }d\varphi \\ y=g\left( r',\varphi \right) \\ dy=\ldots \\ ds^{2}=dx^{2}+dy^{2}\end{aligned} $ $\endgroup$
    – Eli
    May 20, 2019 at 7:06
  • $\begingroup$ @G.Smith $g_{xx}$ is the covariant tensor transformation law for when $i=j$. As I said in my question I am trying to understand it better so if you do not wish to contribute to helping me understand then please do not interact with the question $\endgroup$
    – MRT
    May 20, 2019 at 18:25
  • $\begingroup$ You can’t have four contracted indices. $\endgroup$
    – G. Smith
    May 20, 2019 at 18:28
  • $\begingroup$ @G.Smith then why does $$\Large \left(\dfrac{\partial x^\alpha}{\partial r}\right)^2 = \cos^2\phi + \sin^2\phi = 1$$ $$\Large \left(\dfrac{\partial x^\alpha}{\partial \phi}\right)^2 = r^2\sin^2\phi + r^2\cos^2\phi = r^2$$ Leading to the correct standard result $$\Large ds^2 = g_{ij}\dot{x}^i\dot{x}^j = dr^2 + r^2\,d\phi^2$$ $\endgroup$
    – MRT
    May 20, 2019 at 18:36

1 Answer 1


To express a line element in different coordinates, you don't need any transformation formulas because it doesn't transform. It is an invariant, the same in all coordinate systems. So all you do is substitute for the old coordinates and their differentials in terms of the new coordinates and their differentials.

In your example, you want to express the line element


in terms of a new radial coordinate, $r'$, where


Simply compute


and find


There is no need to think in terms of the components of the metric tensor and how they transform. You can do it that way; it's just needlessly harder.

Your introduction of the flat coordinates $x$ and $y$ is not going to possible when you are dealing with a metric for a space with curvature.

An expression like

$$\dfrac{\partial x^\alpha}{\partial x}\dfrac{\partial x^\alpha}{\partial x}g_{\alpha\alpha}$$

where four indices are contracted is never correct. Only two indices ever get contracted together. Of course, you can contract multiple pairs of indices, but you must write them as different symbols so that you can tell what is being contracted with what.

If you really want to transform the metric tensor, then don't introduce an intermediate set of coordinates like $(x, y)$. Go directly from the original coordinates


to the new coordinates


using the transformation rule for a 2-index covariant tensor:

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

For example,

$$\begin{align} g_{r'r'}&=\left(\frac{\partial r}{\partial r'}\right)^2g_{rr}+2\frac{\partial r}{\partial r'}\frac{\partial\phi}{\partial r'}g_{r\phi}+\left(\frac{\partial \phi}{\partial r'}\right)^2g_{\phi\phi}\\ &=\left(\frac{-a^2}{r'^2}\right)^2g_{rr}\\ &=\left(\frac{a}{r'}\right)^4 \end{align}.$$

You may be used to seeing primes on the vectors or tensors, not on their indices. That notation can end up being more confusing, because it leads to ambiguous-looking things like ${g′}_{rr}$ rather than clear things like $g_{r'r'}$.

  • $\begingroup$ Thank you, this is very clear. I was taking a lazy short hand because of how only the $g_{r^\prime r^\prime}$ term doesn't go to zero. But I see now how going from $$\large r = \dfrac{a^2}{r^\prime}\quad\rightarrow\quad dr = -\dfrac{a^2}{r^{\prime2}}\,dr^\prime$$ is an easier approach. Much more convenient $\endgroup$
    – MRT
    May 22, 2019 at 9:30

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