I have a metric $g_{i,j}$, in two dimensions for example. How do I find the surface that it represents, in paramatrized form, as a function $z=f(x,y)$, etc. ?
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It is not possible to obtain a unique function if given only the metric. I will work the problem backwards to illustrate why. I am assuming this a perfectly orientable surface embedded in $\mathbb{R}^3$. $${(ds)^2}={(dx)^2}+{(dy)^2}+{(dz)^2}$$ Write out the constraint $${dz}={(∂f/∂x)}{dx}+{(∂f/∂y)}{dy}$$ Plug it into the metric and you get $${(ds)^2}={(1+(∂f/∂x)^2)}{(dx)^2}+{(1+(∂f/∂y)^2)}{(dy)^2}+{2(∂f/∂x)(∂f/∂y)}{dx}{dy}$$ You now have: $${g_{12}}={(∂f/∂x)(∂f/∂y)}=g_{21}$$ $${g_{11}}={(1+(∂f/∂x)^2)}$$ $${g_{22}}={(1+(∂f/∂y)^2)}$$
Given an arbitrary functions as metric components, there may not be any solution possible. For example, if ${g_{11}}=cos(x^2+y^2)/2$ it be impossible to find a surface in this case. Supposing we do find a solution, $f(x,y)$, against all odds, it is not unique. For example $z=f(x,y)+c$ where $c$ is a constant is also a valid surface in such a case.
orange_soda
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1(+1) @dj_mummy this example looks much better. – achille hui Sep 25 '13 at 09:25
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In general: you don't. There are metrics without a corresponding surface.
Michael
Michael Hoppe
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