Is it at all possible to, by a change of variables, transform the metric $dx^2+dy^2\over g(r)^2$ where $g$ is a function and $r=\sqrt{x^2+y^2}$ to something of the form $du^2+f(u,v) dv^2$? Thank you.
Is it possible to achieve the following form?
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ordinary-differential-equations
differential-geometry
1 Answers
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Yes, it is always possible to do this, at least in a sufficiently small neighborhood about $0$. Such coordinates are called "normal coordinates" or "exponential coordinates." In such coordinates the function $u$ will always measure the distance (w.r.t. the metric $\frac{1}{g(r)^2}(dx^2 + dy^2)$) from the origin and $\frac{\partial}{\partial v}$ will be a vector field tangent to the level sets of $u$. To construct this as explicitly as your example allows, calculate the distance to the origin as a function $u$. Then $\frac{\partial}{\partial u}$ will be the gradient of $u$, and then you may choose any $\frac{\partial}{\partial v}$ that is orthogonal to $\frac{\partial}{\partial u}$.