Suppose $\Sigma^2\subset \mathbb R^3$ is a surface of revolution, i.e. we consider \begin{align}I &\xrightarrow{(r,h)} \mathbb R^+\times\mathbb R\\ t&\mapsto (r(t),h(t))=:(r,h)\end{align}
And the corresponding parametrization of the surface $\Sigma^2=:\Sigma$ given by \begin{align} (r,\varphi)\stackrel{\Sigma}{\longmapsto} (r\cos\varphi,r\sin\varphi,h)\end{align} for $\varphi\in [-\pi,\pi]$.
Let $H= \frac{\partial}{\partial \varphi}$ and $V= \frac{\partial}{\partial t}$ be the coordinate vector fields on $\Sigma$. That's all the data I have.
I would like to compute some quantities to study the geometry of $\Sigma$, in particular determining
- the unit normal vector field on $\Sigma$
- the shape tensor and its
- principal curvature.
However, I do not know how to determine the unit normal vector field. What would be the "obvious" way in this scenario?
The following two approaches seemed reasonable to me, but I didn't manage to succeed.
Idea 1:
We know that the unit normal field $N$ is orthogonal to both $H$ and $V$ with respect to the $g$-metric in $\mathbb R^3$ (I suppose we are required to choose cylindral coordinates for $\mathbb R^3$).
Thus solving the system of linear equations \begin{align} \langle H,N\rangle = 0 \\ \langle H,V\rangle = 0\end{align} where $g = dr^2+r^2d\varphi+dz$.
Idea 2:
We know from elementary linear algebra, that (up to sign) the normal vector to $\Sigma$ at a point $p \in \Sigma$ is determined by the cross product of the two tangent (basis) vectors of the tangent plane to $p$. But I don't really see how to apply this to the given scenario.
Could someone help me understanding how we can find the unit normal vector field $N$? Because if I can't manage to determine $N$, I basically can not really continue determining the shape tensor and other quantities, can I?
I appreciate any help!
Let $\phi: U \longrightarrow \mathbb{R}^3$, $\phi(x_1,x_2) = (\phi_1(x_1,x_2),\phi_2(x_1,x_2),\phi_3(x_1,x_2))$ be a local parametrisation of some surface. Let $x\in U$ and $p = \phi(x)$. The basis $\{e_1, e_2\}$ for $T_x\mathbb{R}^2\cong \mathbb{R}^2$ pushes forward to a basis $\{E_1=d\phi_x(e_1),E_2=d\phi_x(e_2)\}$ for $T_{p}\phi(U)$. These vectors are just the partial derivatives $\frac{\partial \phi}{\partial x^1}$ and $\frac{\partial \phi}{\partial x^2}$. You want a unit normal vector to $T_{\phi(x)}\phi(U)$, so the easiest choice is to take the cross product and normalise it. $$ N(p) = \frac{E_1\times E_2}{|E_1\times E_2|}\,. $$
Edit: It seems that you are having difficulty understanding the derivative and how it acts on vectors. Just to give an example from multivariable calculus, if $\phi(x,y) = (x,y,f(x,y))$ is the graph of a function $f\in C^\infty$, then $$ Df = \begin{pmatrix} 1 & 0 & \frac{\partial f}{\partial x}\\ 0 & 1 & \frac{\partial f}{\partial y} \end{pmatrix} $$
Then, given the basis vectors $e_1 = (1,0)$ and $e_2= (0,1)$ what do you get when you compute $Df(e_1)$ and $Df(e_2)$? Notice that they are linearly independent and tangent to the surface! It might help to use Mathematica or something to try plotting some surfaces and the corresponding vector fields $Df(e_1)$ and $Df(e_2)$.
Also this definition of derivative is often more enlightening: Let $\phi: U \longrightarrow \mathbb{R}^3$ be a local parametrisation with $\phi(x) = p$. Then given a vector $v\in T_xU$, let $\gamma: (-1,1) \longrightarrow U$ be a curve with $\gamma(0) = x$ and $\gamma'(0) = v$. Then $$ d\phi_x(v) = (\phi\circ\gamma)'(0)\,. $$ For the basis vectors $e_1$ and $e_2$ you can choose the curves $\gamma_i(t) = x+ te_i$. So in the graph example I gave above, you can imagine the derivative as being the tangent vector of the image of the curve.