- #1

Adrian555

- 7

- 0

Member warned that the homework template must be used

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!

$$x = \sqrt U cos(V)$$

$$y = \sqrt U sen(V)$$

$$z = U$$

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!

Last edited by a moderator: