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

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• Abhishek11235
In summary, the unit normal to a surface is a vector that is perpendicular to the tangent plane at a chosen point on the surface. It is found by computing the gradient vector and dividing it by its own length. This can be represented using the metric tensor, which is the dual to the differential. In coordinates, the gradient is given by ##\nabla \phi = g^{r\alpha}\frac{\partial\phi}{\partial x^\alpha}\frac{\partial}{\partial x^r}## and the norm squared is ##g^{uv}\frac{\partial\phi}{\partial x^u}\frac{\partial\phi}{\partial x^v}##. Therefore, the two expressions are equivalent.
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?

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

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
martinbn said:
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.

## 1. What is a normal to coordinate curve?

A normal to coordinate curve is a line that is perpendicular to a curve at a specific point on the curve. It represents the instantaneous direction of the curve at that point.

## 2. How is a normal to coordinate curve calculated?

A normal to coordinate curve can be calculated by finding the derivative of the curve at the specific point and then finding the negative reciprocal of that derivative.

## 3. What is the significance of a normal to coordinate curve?

A normal to coordinate curve helps us understand the behavior of a curve at a specific point. It can also be used to find the equation of a tangent line at that point.

## 4. Can a normal to coordinate curve have multiple points of intersection with a curve?

Yes, a normal to coordinate curve can intersect a curve at multiple points. This can occur if the original curve has multiple inflection points or if the normal curve is shifted along the x or y axis.

## 5. How is a normal to coordinate curve used in real life?

A normal to coordinate curve is commonly used in physics and engineering to analyze the motion of objects in space. It can also be used in computer graphics to create smooth and realistic curves on a screen. Additionally, it has applications in fields such as biology, economics, and statistics.

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