there is FacetNormal, but why there is no FacetTangent?


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1
3 months ago by
by using
n=FacetNormal(mesh)

on can obtain the normal vectors of a given mesh, but why there is no such a function 
FacetTangent(mesh)

?
Community: FEniCS Project
-1
Then, how about 2-dimensional cases?
written 3 months ago by Mingchao Cai  

2 Answers


5
3 months ago by
Because a facet tangent space of a three-dimensional cell is two-dimensional. A facet tangent of two-dimensional cell would have undefined direction.

More precisely, a facet tangent space is an Euclidian space of dimension tdim-1, where tdim is a topological dimension of underlying cell. So intended FacetTangent could return basis of the space. This basis could be orthogonalized but that are no natural directions which would be invariant of numbering of entities or similar conventions. By contrast FacetNormal has an invariant definition - exterior unit normal. Note that UFL is a language for describing variational forms in well-defined tensor algebra language.

You can define your own tangents as you need:

mesh = UnitCubeMesh(2048, 2048, 2048)
n = FacetNormal(mesh)
t0 = as_vector((0, n[2], -n[1]))
t1 = as_vector((n[2], 0, -n[0]))​
 

There's no reason why any such definition should be preferred and denoted as FacetTangents.
1
3 months ago by
pf4d  
As Jan so eloquently pointed out, you have infinite choices for FacetTangent, namely those vectors lying on the tangent plane to the facet.

If you want the tangent vector, you can make one yourself using a vector of your choice, the normal vector, and linear algebra.
1
Let me add that if you need a projector you can build it from the normals, i.e. \(\boldsymbol{P}=\boldsymbol{I}-\boldsymbol{n}\otimes \boldsymbol{n}\)
written 3 months ago by Marco Morandini  
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