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Natural basis and dual basis of a circular paraboloid
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[QUOTE="Adrian555, post: 5838203, member: 570058"] Hi everyone!I'm trying to obtain the natural and dual basis of a circular paraboloid parametrized by: $$x = \sqrt U cos(V)$$ $$y = \sqrt U sen(V)$$ $$z = U$$ with the inverse relationship: $$V = \arctan \frac{y}{x}$$ $$U = z$$ The natural basis is: $$e_U = \frac{\partial \overrightarrow{r}} {\partial U }=\frac{\partial x} {\partial U }\hat {i} + \frac{\partial y} {\partial U }\hat {j} + \frac{\partial z} {\partial U }\hat {k}=$$ $$ = \frac {cos(V)}{2\sqrt{U}} \hat {i} + \frac {sen(V)}{2\sqrt{U}} \hat {j} + \hat {k}$$ $$e_V = \frac{\partial \overrightarrow{r}} {\partial V }=\frac{\partial x} {\partial V }\hat {i} + \frac{\partial y} {\partial V }\hat {j} + \frac{\partial z} {\partial V }\hat {k}=$$ $$ = -\sqrt{U} sen(V) \hat {i} + \sqrt{U} cos(V) \hat {j} + 0\hat {k}$$ Which gives the following metric tensor: $$e_U\cdot e_U=g_{UU}=1+\frac{1}{4U}$$ $$e_U\cdot e_V=g_{UV}=g_{VU}=0$$ $$e_V\cdot e_V=g_{VV}=U$$ What respects to the dual basis: $$e^U = \nabla U=\frac{\partial U} {\partial x }\hat {i} + \frac{\partial U} {\partial y }\hat {j} + \frac{\partial U} {\partial z }\hat {k}=$$ $$ = 0\hat {i} + 0\hat {j} + \hat {k}$$ $$e^V = \nabla V=\frac{\partial V} {\partial x }\hat {i} + \frac{\partial V} {\partial y }\hat {j} + \frac{\partial V} {\partial z }\hat {k}=$$ $$ = \frac{-y}{x^2+y^2}\hat {i} + \frac{x}{x^2+y^2}\hat {j} + 0\hat {k}=-\frac{\sqrt{U}sen(V)}{U}\hat {i} + \frac{\sqrt{U}cos(V)}{U}\hat {j} + 0\hat {k}$$ However, in this case, the metric tensor is: $$e^U\cdot e^U=g^{UU}=1$$ $$e^U\cdot e^V=g^{UV}=g^{VU}=0$$ $$e^V\cdot e^V=g^{VV}=\frac{1}{U}$$ which is definitely not the inverse of the one I obtained from the natural basis, because the component $$g^{UU}$$ should be $$\frac{1}{1+\frac{1}{4U}}$$ and not 1 (The metric tensor is diagonal) Could anyone tell me where am I wrong?Thanks for your help! [/QUOTE]
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Natural basis and dual basis of a circular paraboloid
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