Line e Surface Infinitesimal

Jhenrique

I think you know definition of line infinitesimal:
$$[ds]^2 = \begin{bmatrix} dx & dy & dz \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}^2 \begin{bmatrix} dx\\ dy\\ dz\\ \end{bmatrix} = \begin{bmatrix} dr & d\theta & dz \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & r & 0\\ 0 & 0 & 1\\ \end{bmatrix}^2 \begin{bmatrix} dr\\ d\theta\\ dz\\ \end{bmatrix} = \begin{bmatrix} d\rho & d\phi & d\theta \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & \rho & 0\\ 0 & 0 & \rho\;sin(\phi)\\ \end{bmatrix}^2 \begin{bmatrix} d\rho\\ d\phi\\ d\theta\\ \end{bmatrix}$$

From this, is correct if I deduce the formula to surface infinitesimal like this?
$$[d^2S]^2 = \begin{bmatrix} dydz & dxdz & dxdy \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}^2 \begin{bmatrix} dydz\\ dxdz\\ dxdy\\ \end{bmatrix} = \begin{bmatrix} d\theta dz & drdz & drd\theta \end{bmatrix} \begin{bmatrix} r & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & r\\ \end{bmatrix}^2 \begin{bmatrix} d\theta dz\\ drdz\\ drd\theta\\ \end{bmatrix} = \begin{bmatrix} d\phi d\theta & d\rho d\theta & d\rho d\phi \end{bmatrix} \begin{bmatrix} \rho^2\;sin(\phi) & 0 & 0\\ 0 & \rho\;sin(\phi) & 0\\ 0 & 0 & \rho\\ \end{bmatrix}^2 \begin{bmatrix} d\phi d\theta\\ d\rho d\theta\\ d\rho d\phi\\ \end{bmatrix}$$

And more one second question: dxdy is equal d²xy ?

Related Differential Geometry News on Phys.org

ChrisVer

Gold Member
you can always try the "old way" of doing things...
Find the infinitesimal tangent vectors on your surface, and take the cross product