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I would like to apply the coordinate transformation $x^{\bar{1}} = 2x^1$, $x^{\bar{2}} = x^2$ in the 2D Cartesian plane. The metric in the barred frame is $g_{\overline{ij}} = \Lambda^i_{\bar{i}} \Lambda^j_{\bar{j}} g_{ij}$, where $\Lambda^i_{\bar{i}}$ is a 2 x 2 inverse transformation matrix, and $g_{ij}$ appears to be the identity matrix:

$$g_{\overline{ij}} = \left( \begin{matrix} \frac{1}{2} & 0 \\ 0 & 1 \end{matrix} \right) \Lambda^j_{\bar{j}} \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)$$

My problem is that I have no idea what $\Lambda^j_{\bar{j}}$ represents. The text I am learning from is The Standard Model in a Nutshell. Any guidance you can offer in obtaining $g_{\overline{ij}}$ will be appreciated.

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For a rank-2 tensor (like the metric) and a general coordinate transformation $x\to \bar x(x)$ you have the transformation law: $$g_{ij}\to g_{\overline{ij}}=\frac{\partial x^i}{\partial x^{\bar i}}\frac{\partial x^j}{\partial x^{\bar j}}g_{ij}$$ In your case you have $g_{ij}=g^{ij}=\mathrm{diag}(1, 1)$, as you indicated. You also know exactly how each coordinate transforms so you should have no problem in calculating $g_{\overline{ij}}$ now.

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  • $\begingroup$ Thanks for the input. I've edited my question to be more clear with respect to the missing transformation matrix. And I apologize for being somewhat dense ... I have not played around with inertial reference frames and coordinate transformations since Lorentz boosts. I'm just not used to thinking about them in this way... $\endgroup$ – Joel DeWitt Sep 5 '18 at 1:36

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