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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 $r=\sqrt{x^2+y^2}$? – robjohn Mar 11 '12 at 1:33
Dear @robjohn, the $\delta$ is just the radius of the open disc. – Mars Mar 11 '12 at 1:38
@rob: $r^2=x^2+y^2$, this is standard convention. – anon Mar 11 '12 at 1:41
@anon: then does the metric enter into this somewhere? – robjohn Mar 11 '12 at 1:43
Dear @anon, you are right :) – Mars Mar 11 '12 at 1:47
up vote 2 down vote accepted

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.$$ \frac{1}{g(r)^2}(dr^2 + r^2d\theta^2)$$

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Thanks, treble! – Mars Mar 11 '12 at 7:59
You're welcome :) – treble Mar 11 '12 at 8:09

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