Finding the unit Normal to a surface using the metric tensor.

[Abhishek11235]

Let $$\phi(x^1,x^2...,x^n) =c$$ be a surface. What is unit Normal to the surface?

I know how to find equation of normal to a surface. It is given by:
$$\hat{e_{n}}=\frac{\nabla\phi}{|\nabla\phi|}$$However the answer is given using metric tensor which I am not able to derive. Here is the answer

$$({g^{\alpha\beta}}{\frac{\partial\phi}{\partial x^{u}}}{\frac{\partial\phi}{\partial x^{v}}})^{-1/2} {g^{r\alpha}}{\frac{\partial\phi}{\partial x^{\alpha}}}$$

How can I Derive this?

 

[jedishrfu]

The unit normal to the surface has a length of one. It is a vector that is perpendicular to vectors in the tangent plane at that chosen point. The gradient vector at that point is also perpendicular to the tangent plane and so by computing the gradient and dividing the gradient vector by its own length we get the unit normal vector to the surface.

It’s explained in more detail in this video lecture

 

[martinbn]

The gradient $\nabla \phi$ is the vector dual to the differential. In coordinates you have $d\phi=\frac{\partial\phi}{\partial x^\alpha}dx^\alpha$, so for the gradient you get $\nabla \phi = g^{r\alpha}\frac{\partial\phi}{\partial x^\alpha}\frac{\partial}{\partial x^r}$. The norm squared will be $g^{uv}\frac{\partial\phi}{\partial x^u}\frac{\partial\phi}{\partial x^v}$. So the two expressions are the same.

 

[Abhishek11235]

The gradient $\nabla \phi$ is the vector dual to the differential. In coordinates you have $d\phi=\frac{\partial\phi}{\partial x^\alpha}dx^\alpha$, so for the gradient you get $\nabla \phi = g^{r\alpha}\frac{\partial\phi}{\partial x^\alpha}\frac{\partial}{\partial x^r}$. The norm squared will be $g^{uv}\frac{\partial\phi}{\partial x^u}\frac{\partial\phi}{\partial x^v}$. So the two expressions are the same.

Thankyou. But can you elaborate it in more detail. Please. I started Tensors few days ago.