Natural basis and dual basis of a circular paraboloid

In summary, the conversation discussed the natural and dual basis of a circular paraboloid parametrized by equations for x, y, and z, with an inverse relationship for V and U. The natural basis was found to be different from the dual basis, possibly due to the embedding of the 2-dimensional surface in a 3-dimensional space. To obtain the correct dual basis, a third coordinate W was added and the components of the gradient of U and V were taken orthogonal to the gradient of W. The full metric forms in 3-space were also provided.
  • #1
Adrian555
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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!
 
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  • #2
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.
 
  • #3
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:
[tex] \mathbf{g} \sim \left( \begin{array}{ccc} (1+\tfrac{1}{4U}) & 0 & -1\\ 0 & U & 0\\ -1 & 0 & 1\end{array}\right)[/tex]
and dual metric is:
[tex] \mathbf{g}^* = \left(\begin{array}{ccc}4U & 0 & 4U\\ 0 & 1/U & 0\\ 4U & 0 & 4U +1\end{array}\right)[/tex]

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]
 
Last edited:

Related to Natural basis and dual basis of a circular paraboloid

1. What is the natural basis of a circular paraboloid?

The natural basis of a circular paraboloid is a set of two vectors that are tangent to the surface of the paraboloid at a given point and perpendicular to each other. These vectors are known as the principal directions and they define the directions of maximum and minimum curvature at that point.

2. How is the natural basis determined for a circular paraboloid?

The natural basis for a circular paraboloid is determined by finding the partial derivatives of the paraboloid's equation with respect to the two independent variables. These partial derivatives represent the slopes of the surface in the x and y directions, which can then be used to calculate the principal directions.

3. What is the dual basis of a circular paraboloid?

The dual basis of a circular paraboloid is a set of two vectors that are perpendicular to the principal directions and lie on the tangent plane at a given point. These vectors are used to calculate the coefficients of the second fundamental form, which describes the curvature of the paraboloid's surface.

4. How is the dual basis related to the natural basis of a circular paraboloid?

The dual basis and natural basis are closely related, as they both describe the curvature of the circular paraboloid's surface. The principal directions of the natural basis are perpendicular to each other and lie on the surface, while the dual basis vectors are perpendicular to the principal directions and lie on the tangent plane.

5. Why are the natural basis and dual basis important for studying a circular paraboloid?

The natural basis and dual basis provide important information about the curvature of a circular paraboloid's surface, which is essential for understanding its geometric properties. These bases are also used in the calculation of important quantities such as surface area, volume, and curvature, making them crucial for further analysis and applications of the paraboloid.

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