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This is very simple. Consider a cylinder in $\mathbb{R}^{3}$. Let the axis of the cylinder coincide with the $z$-axis. Allow the cylinder to be paramterized as follows: \begin{align*} x(\varphi,h) &= a \cos \varphi \\ y(\varphi,h) &= a \sin \varphi \\ z(\varphi,h)&= h \\ \end{align*} The metric $g$ for this surface is \begin{equation*} g = a^{2} \text{d}\varphi^{2} + \text{d}h^{2} \end{equation*} How do we know this is the metric? Like, derive it for me please from the given parameterization. Thanks.

Stan Shunpike
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2 Answers2

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Hint: Let $(a,\phi,h) = (\theta^1,\theta^2,\theta^3)$. Write

$$dx = \frac{\partial x}{\partial \theta^i}d\theta^i\ \ dy = \frac{\partial x}{\partial \theta^i}d\theta^i\ \ dz = \frac{\partial z}{\partial \theta^i}d\theta^i \tag{1}$$

we are using Einstein Notation in (1).

Compute $dx^2$, $dy^2$ and $dz^2$.

Finally the metric is given by $$g_{\mathbb{R}^3} = dx^2+dy^2+dz^2$$

Aaron Maroja
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  • Feel free to ask if you have any questions. – Aaron Maroja Dec 10 '14 at 20:55
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    That was very helpful. I'm new to tensors and didn't organized the problem neatly. Once I saw how nicely you laid it out with the Einstein summation convention it was straightforward to solve. Must remember: when working with tensors, don't lose track of my indices! – Stan Shunpike Dec 10 '14 at 22:22
  • I'm glad I could help. You may use the same approach to find other metrics as well. That's right! – Aaron Maroja Dec 10 '14 at 22:25
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If you choose the usual Euclidean (Pythagorean) metric, then the metric tensor (and its pull-back to cylindrical coordinates) is $g=dx^2+dy^2+dz^2=\\(\cos\phi da-a\sin\phi d\phi)^2+(\sin\phi da+a\cos\phi d\phi)^2+(dh)^2=\\da^2+a^2d\phi^2+dh^2.$

rych
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    This should be the chosen answer. The question is asking how we find the pullback of the Euclidean metric to the cylindrical coordinate system – Liam Clink Apr 25 '21 at 13:39