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Suppose $\vec{\phi}(x^1,x^2) = (x^1,x^2,(x^1)^2+(x^2)^2.$ The metric tensor induced by $\vec{\phi}$ is given by:

$$g = \begin{pmatrix}1+4(x^2)^2 & 4x^1x^2 \\\ 4x^1x^2 & 1+4(x^2)^2\end{pmatrix}$$

Let $x^1 = \bar{x}^1\cos\bar{x}^2$

$x^2 = \bar{x}^1\sin\bar{x}^2$

Find the components of the metric tensor $\bar{g}_{12}=\bar{g}_{21},\bar{g}_{22}$ in the $\bar{x}^1,\bar{x}^2$ coordinate system.

This is what I am thinking so far, but I am confused on how to solve the problem.

$$\bar{g}_{kl} = \sum_{j=1}^2 \sum_{i=1}^2 g_{ij} \frac{\partial x^i}{\partial \bar{x}^k} \frac{\partial x^j}{\partial \bar{x}^l}$$

So in the case of $\bar{g}_{12} = \bar{g}_{21}$:

$$\bar{g}_{12} = \sum_{j=1}^2 \sum_{i=1}^2 g_{ij} \frac{\partial x^i}{\partial \bar{x}^1} \frac{\partial x^j}{\partial \bar{x}^2}$$

$$ = g_{11} \frac{\partial x^1}{\partial\bar{x}^1}\frac{\partial x^1}{\partial\bar{x}^1}+ g_{12} \frac{\partial x^1}{\partial\bar{x}^1}\frac{\partial x^2}{\partial\bar{x}^1}+g_{21} \frac{\partial x^2}{\partial\bar{x}^1}\frac{\partial x^1}{\partial\bar{x}^1}+g_{22} \frac{\partial x^2}{\partial\bar{x}^1}\frac{\partial x^2}{\partial\bar{x}^1}$$

$\frac{\partial x^1}{\partial\bar{x}^1} = \cos \bar{x}^2$

$\frac{\partial x^2}{\partial\bar{x}^1} = \sin \bar{x}^2$

$\frac{\partial x^1}{\partial\bar{x}^2} = -\bar{x}^1 \sin \bar{x}^2$

$\frac{\partial x^2}{\partial\bar{x}^2} = \bar{x}^1\cos \bar{x}^2$

So $\bar{g}_{12} = (1+4(x^1)^2)(\cos\bar{x}^2)^2+2(4x^1x^2)(\cos \bar{x}^2)(\sin \bar{x}^2)+(1+4(x^2)^2)(\sin\bar{x}^2)^2$

user8290579
  • 886

1 Answers1

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$\newcommand\d[1]{{\rm d} #1}$You can use the transformation law that you correctly stated, and be patient. You messed up the indexes in red, by the way: $$\overline{g}_{\color{blue}{1}\color{red}{2}}= g_{\color{green}{11}} \frac{\partial x^\color{green}{1}}{\partial\bar{x}^\color{blue}{1}}\frac{\partial x^\color{green}{1}}{\partial\bar{x}^\color{red}{2}}+ g_{\color{green}{12}} \frac{\partial x^\color{green}{1}}{\partial\bar{x}^\color{blue}{1}}\frac{\partial x^\color{green}{2}}{\partial\bar{x}^\color{red}{2}}+g_{\color{green}{21}} \frac{\partial x^\color{green}{2}}{\partial\bar{x}^\color{blue}{1}}\frac{\partial x^\color{green}{1}}{\partial\bar{x}^\color{red}{2}}+g_{\color{green}{22}} \frac{\partial x^\color{green}{2}}{\partial\bar{x}^\color{blue}{1}}\frac{\partial x^\color{green}{2}}{\partial\bar{x}^\color{red}{2}}.$$Another way is to write $$g = (1 + 4(x^2)^2)((\d{x^1})^2+(\d{x^2})^2) + 8x^1x^2 \d{x^1}\d{x^2},$$compute $$\begin{align}\d{x}^1 &= \cos \overline{x}^2\d{\overline{x}^1} - \overline{x}^1 \sin \overline{x}^2 \d{\overline{x}^2}, \\ \d{x^2}&= \sin \overline{x}^2 \d{\overline{x}^2} + \overline{x}^1 \cos \overline{x}^2 \d{\overline{x}^2}, \end{align}$$and plug in everything in the expression for $g$. You compute squares and products formally, as in $$(\d{x}^1)^2 = \cos^2 \overline{x}^2 (\d{\overline{x}^1})^2 - 2\overline{x}^1 \sin \overline{x}^2 \cos \overline{x}^2 \d{\overline{x}^1}\d{\overline{x}^2} + (\overline{x}^1)^2\sin^2 \overline{x}^2 (\d{\overline{x}^2})^2,$$etc.. This is a pain, and you must be patient.

Ivo Terek
  • 77,665