Unit Tangent Vector in a Scalar Field

[auditt241]

Hello,
I am attempting to calculate unit normal and tangent vectors for a scalar field I have, Φ(x,y). For my unit normal, I simply used:

\hat{n}=\frac{\nabla \phi}{|\nabla \phi|}​

However, I'm struggling with using this approach to calculate the unit tangent. I need to express it in terms of the gradient of the scalar field but I am having a hard time visualizing this.
Thanks!

 

[ShayanJ]

Take a linear combination of basis vectors with unknown coefficients. Equate its dot product with $\hat n$ to zero. You should be able to find two independent vectors as solutions. Those will span the tangent space to each point.

 

[auditt241]

Take a linear combination of basis vectors with unknown coefficients. Equate its dot product with $\hat n$ to zero. You should be able to find two independent vectors as solutions. Those will span the tangent space to each point.

Thanks for your response! I think I tried this: I used \hat{n} \cdot \hat{T} = 0, writing each in terms of an x- and y-component, and then solving for the x-component and y-component of the unit tangent \hat{T}. My T_x and T_y are written only in terms of unit normal components n_x and n_y (by defining \sqrt{T_x^2 + T_y^2} = 1. When I do this however, and plot my unit tensor field, I don't get clean, tangential vectors. Some are tangential, but they can be a bit of a mess. I am doing this in FEA software, could some of my issues be due to my mesh size? Or am I going about it the wrong way?

 

[ShayanJ]

One obvious problem I see, is that you're assuming the problem is two dimensional, Its not! You should work in three dimensions. Otherwise $\hat n\cdot \vec T=0$ and $|\vec T|=1$ will completely determine the components and you'll get only one vector as a solution which we know isn't right.
In fact you should've known this from the start because you are considering a function of two variables which can only be a surface in three dimensions.