Natural basis and dual basis of a circular paraboloid

[Adrian555]

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!

 

[jambaugh]

I believe that the problem you're having is due to embedding the 2 dim surface in 3 dimensional space. Your procedure would work if you added a 3rd coordinate W and was looking at a coordinate basis and dual basis of 3 space.

Keep in mind that the dual basis is really representing a basis of the dual space. These vectors in the dual basis correspond to dual vectors when appended to the dot product. The dual vectors you found have an extra component normal to the surface and you must project this out.

Noting that the paraboloid equation is $x^2 + y^2 -z = 0$ try adding the third coordinate $W=x^2+y^2-z$. Note that the third dual basis vector will be normal to the surface. Then take the components of $\nabla U$ and $\nabla V$ orthogonal to $\nabla W$.

I *** THINK *** this will give you the right dual basis for the surface. I'll think about it some more and update my reply if needed.

 

[jambaugh]

Now that my power's back on... (Irma's fault)... Yep, here are the full metric forms in 3-space if you add the W coordinate:
\mathbf{g} \sim \left( \begin{array}{ccc} (1+\tfrac{1}{4U}) & 0 & -1\\ 0 & U & 0\\ -1 & 0 & 1\end{array}\right)
and dual metric is:
\mathbf{g}^* = \left(\begin{array}{ccc}4U & 0 & 4U\\ 0 & 1/U & 0\\ 4U & 0 & 4U +1\end{array}\right)

The problem here, to get the 2-dimensional metric and dual metric is that the tangent space (spanned by $e_U,e_V$ [edited]) is of course different at every point and the dual vectors are not in that tangent space. One must define a tangent component of the 3-dim gradient.
[To Be Continued]