I am trying to learn GR. In two of the books on tensors, there is an example of evaluating the inertia tensor in a primed coordinate system (for example, a rotated one) from that in an unprimed coordinate system using the eqn. ##I’ = R I R^{-1}## where R is the transformation matrix and ##R^{-1}## is its inverse, and ##I## is the inertia matrix.(adsbygoogle = window.adsbygoogle || []).push({});

Is this method of transformation of (inertia) tensor from one coordinate system to another applicable for all tensors? In particular, can I use this method to evaluate the metric tensor in the primed coordinate system, given the metric tensor in the unprimed system and the transformation matrix?

Assuming it is applicable, I attempted to evaluate the metric tensor in Cartesian coordinate system, from the 2D polar system (using ##g’(Cart) = T g(polar) T^{-1}##, where ##T## is the transformation matrix from polar to Cartesian), and was expecting an identity matrix with diagonal elements 1 and all others 0.

I am evaluating the metric tensor as follows:

\begin{equation}

g'(Cart) =

\begin{bmatrix}

cos\theta & -r\ sin\theta\\

sin\theta & r\ cos\theta \end{bmatrix} \times

\begin{bmatrix}

1 & 0\\

0 & r^2 \end{bmatrix} \times

\begin{bmatrix}

cos\theta & sin\theta\\

-sin\theta \over r & cos\theta \over r\end{bmatrix}

\end{equation}

Instead of the identity matrix for the metric tensor in Cartesian coordinates, I ended up with:

\begin{bmatrix}cos^2\theta + r^2 sin^2\theta&cos\theta\ sin\theta - r^2 cos\theta\ sin\theta\\

cos\theta\ sin\theta - r^2 cos\theta\ sin\theta&sin^2\theta+r^2 cos^2\theta \end{bmatrix}

Almost there, but not quite; the ##r^2## is messing it up.

What am I missing? Since the Euclidean space is flat, the Cartesian coordinate system has an identity matrix as a metric tensor, right?

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# I Evaluating metric tensor in a primed coordinate system

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