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How can we change variables from $(x,y)$ to $(r,\theta)$ for the metric on the open disc $r<\delta$ defined by $(dx^2+dy^2)\over g(\sqrt{x^2+y^2})^2$ where $g(\sqrt{x^2+y^2})>0$ $\forall r<\delta$?

I am tempted to say the transformed metric is $dr^2\over g(r)^2$, but There might be some monkey business with Jacobians or such?

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    You mention that $r<\delta$, but neither appears in $(dx^2+dy^2)\over g(\sqrt{x^2+y^2})^2$ or $g(\sqrt{x^2+y^2})>0$. Does that mean that $r^2=\frac{(dx^2+dy^2)}{g(\sqrt{x^2+y^2})^2}$ or $r=\sqrt{x^2+y^2}$?2012-03-11
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    Dear @robjohn, the $\delta$ is just the radius of the open disc.2012-03-11
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    @rob: $r^2=x^2+y^2$, this is standard convention.2012-03-11
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    @anon: then does the metric enter into this somewhere?2012-03-11
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    Dear @anon, you are right :)2012-03-11
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    Dear @robjohn, Sorry about the ambiguities. I am looking for an expression of the metric in polar coordinates instead of cartesians.2012-03-11
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    @rob: The open disc of the Cartesian plane is a coordinate chart for some piece of differentiable manifold with that given metric; Mars wants to change to a different coordinate chart which requires one to change the metric accordingly. At face value it looks like it should be $dr^2/g(r)^2$...2012-03-11
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    @anon: okay, between you and Mars, I understand the question now. :-)2012-03-11

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To change from cartesion to polar coordinates, write $$x = r\cos\theta$$ $$y = r\sin\theta$$ and then calculate $dx$ and $dy$ in terms of $dr$ and $d\theta$ the way we usually do in calculus, i.e. $$dx =dr\cos\theta - r\sin\theta d\theta$$ Then you may calculate $$dx^2 + dy^2 = dr^2 + r^2 d\theta^2$$ and you then have your expression of the metric tensor in polar coordinates, i.e.$$ \frac{1}{g(r)^2}(dr^2 + r^2d\theta^2)$$