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Assume usual polar coordinates equations:

$$ x=r\cos(\theta) $$ $$ y=r\sin(\theta) $$

In what I think is called covariant we have a Jacobian of:

$$ J=\begin{pmatrix} \frac{\partial{x}}{\partial{r}} & \frac{\partial{x}}{\partial{\theta}} \\ \frac{\partial{y}}{\partial{r}} & \frac{\partial{y}}{\partial{\theta}} \\ \end{pmatrix}=\begin{pmatrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \\ \end{pmatrix} $$

and a metric tensor of: $$ g_{ij}=J^T J=\begin{pmatrix} 1 & 0 \\ 0 & r^2 \\ \end{pmatrix} $$

My problems appears when I try the contravariant part:

$$ J=\begin{pmatrix} \frac{\partial{r}}{\partial{x}} & \frac{\partial{r}}{\partial{y}} \\ \frac{\partial{\theta}}{\partial{x}} & \frac{\partial{\theta}}{\partial{y}} \\ \end{pmatrix}=\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\frac{\sin(\theta)}{r} & \frac{\cos(\theta)}{r} \\ \end{pmatrix} $$

because the metric tensor: $$ g^{ij}=J^T J=\begin{pmatrix} \cos^2(\theta)+\frac{\sin^2(\theta)}{r} & ... \\ ... & ... \\ \end{pmatrix} $$

results different of the expected one: $$\begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{r^2} \\ \end{pmatrix} $$

why the expected result is not the one I obtain?

( I obtain the expected only if I change the product from $J^T J$ to $J J^T$ ).

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  • $\begingroup$ Sorry so what is the question exactly? $\endgroup$ – IAmNoOne Jan 3 at 13:13
  • $\begingroup$ @IAmNoOne: edited and clarified. Why the expected result is not obtained? $\endgroup$ – pasaba por aqui Jan 3 at 13:15
  • $\begingroup$ You are right, my bad. $\endgroup$ – IAmNoOne Jan 4 at 1:59
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Maybe a bit of a preamble will be useful here. If $\xi$ are some coordinates defining the local metric $\eta$ then, under the transformation $x^\alpha = x^{\alpha}(\xi)$ the metric becomes

$$ g_{\mu\nu} = \frac{\partial \xi^{\alpha}}{\partial x^\mu}\frac{\partial \xi^{\beta}}{\partial x^\nu} \eta_{\mu\nu} \tag{1} $$

So for example, if you take $\xi^1 = x$ and $\xi^2 = y$ the cartesian coordinates, then the local matrix is the identity $\eta_{\mu\nu} = \delta_{\mu\nu}$ and Eq. (1) becomes

$$ g_{\mu\nu} = J^{\alpha}_{\;\mu}J^{\beta}_{\;\nu}\delta_{\alpha\beta} = (J^T J)_{\mu\nu} \tag{2} $$

where the entries of the matrix $J$ are defined as

$$ J^{\alpha}_{\;\beta} = \frac{\partial \xi^\alpha}{\partial x^\beta} \tag{3} $$

As an example take $x^1 = r$ and $x^2 = \theta$ (your case), so that $\xi^1 = r\cos\theta = x^1 \cos x^2$ and $\xi^2 = x^1\sin x^2$, it is easy to calculate

$$ \frac{\partial \xi^1}{\partial x^1} = \frac{\partial }{\partial x^1}(x^1 \cos x^2) = \cos x^2 ~~~(\cdots) $$

so when you evaluate that in (1) you get

$$ g_{11} = 1, ~ g_{12} = g_{21} = 0 ~\mbox{and}~ g_{22} = (x^1)^2 $$

It is important to keep track of the order of things here. In this representation we are labeling the rows with the super-index $\alpha$ and the columns with the sub-index $\beta$.

Now, the contravariant version of Eq. (2) is

$$ g^{\mu\nu} = \eta^{\alpha \beta}\frac{\partial x^\mu}{\partial \xi^\alpha} \frac{\partial x^\nu}{\partial \xi^\beta} \tag{4} $$

and again, under the same assumptions as before, this transforms to

$$ g^{\mu\nu} = \delta^{\alpha\beta}\mathcal{J}^\mu_{\;\alpha}\mathcal{J}^\nu_{\;\beta} = (\mathcal{J}\mathcal{J}^T)^{\mu\nu} \tag{5} $$

where

$$ \mathcal{J}^{\alpha}_{\;\beta} = \frac{\partial x^\alpha}{\partial \xi^\beta} \tag{6} $$

Taking the same changes of coordinates you will get $x^1 = [(\xi^1)^2 + (\xi^2)^2]^{1/2}$ and $x^2 = \arctan(\xi^2/\xi^1)$ so that

$$ \frac{\partial x^1}{\partial \xi^1} = \frac{\xi^1}{[(\xi^1)^2 + (\xi^2)^2]^{1/2}} ~(\cdots) $$

which results in

$$ g^{11} = 1, ~ g^{12} = g^{21} = 0, ~\mbox{and}~ g^{22} = \frac{1}{(\xi^1)^2 + (\xi^2)^2} = \frac{1}{(x^1)^2} $$

as expected

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  • $\begingroup$ Thanks for your answer. Question: in (1) and (3), no need for $\xi$, like in (4) and (6) ? Moreover, if you add some references to the case of polar coordinates (by example, the values of $x_1$, $\xi_1$, $J_1^1$, $g_{11}$ and $g^{11}$ in this case), then this one will be an answer to accept. $\endgroup$ – pasaba por aqui Jan 3 at 15:03
  • $\begingroup$ @pasabaporaqui Sorry, I had a bunch of typos in there. I hope it is clearer now $\endgroup$ – caverac Jan 3 at 15:20
  • $\begingroup$ Nice answer, thanks a lot. About tensors, I do not find examples that helps me to understand. $\endgroup$ – pasaba por aqui Jan 3 at 15:25
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    $\begingroup$ @pasabaporaqui If I may give you a piece of (unsolicited) advice: keep track of your indices, and that's pretty much it. I found the notation is very convenient if you are consistent with the labels. Also, feel free to ask, there's plenty of people willing to help you when you get stuck! $\endgroup$ – caverac Jan 3 at 15:31

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