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!
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!
