Basis + covariant derivative question

In summary: Is that right?Using the covariant derivative makes the (partial) derivative of each coordinate unit vector zero …it would be really inconvenient not to have ∂x2/∂x1 = 0 etc …for example, div of ai + bj would not be ∂a/∂x + ∂b/∂y.
  • #1
7thSon
44
0
My apologies about lack of precision in nomenclature. So I wanted to know how to express a certain idea about choice of basis on a manifold...

Let's suppose I am solving a reaction-diffusion equation with finite elements. If I consider a surface that is constrained to lie in a flat plane or a volumetric solid in R3, the gradient operator and the diffusion tensor can be expressed in a Cartesian basis, and the gradient just reduces to partial derivatives.

It's come to my attention that if I am considering a non-planar surface embedded in R3, it is not valid to express the gradient operator in a Cartesian basis (i.e. d/dx, d/dy, d/dz); rather, you have to use covariant derivatives making use of surface coordinates.

My question is, what is the exact explanation as to why this is the case? Does it have something to do with the fact that the Cartesian basis will span all of R3, while the surface's tangent space is only a subset of R3? And, as such, the cartesian basis is an inappropriate choice of basis for the surface manifold?

Thanks for any help...
 
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  • #2
Hi 7thSon! :smile:
7thSon said:
It's come to my attention that if I am considering a non-planar surface embedded in R3, it is not valid to express the gradient operator in a Cartesian basis (i.e. d/dx, d/dy, d/dz); rather, you have to use covariant derivatives making use of surface coordinates.

My question is, what is the exact explanation as to why this is the case?

Using the covariant derivative makes the (partial) derivative of each coordinate unit vector zero …

it would be really inconvenient not to have ∂x2/∂x1 = 0 etc …

for example, div of ai + bj would not be ∂a/∂x + ∂b/∂y. :wink:
 
  • #3
Hey thanks for the reply. I get what you're saying, but I guess I wasn't being clear about the exact cause of my confusion.

So if I understand you correctly, I believe that yes, with a Cartesian basis the base vectors are not changing, so when taking the covariant derivative at that tangent space it reduces to partial derivatives because the derivative of the basis vectors is zero.

However, what I don't get is why for a general curved surface, if I write the gradient operator with the Cartesian base vectors and solve the correct linear system, it gives you the wrong answer for a diffusion on a surface, whereas the covariant derivative in surface coordinates gives you the right answer.

Why is the former approach wrong? The answer must have to do something with the choice of base vectors (E3) being inappropriate for solving a conservation law strictly on a surface.
 
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Related to Basis + covariant derivative question

What is a covariant derivative?

A covariant derivative is a mathematical operation that is used in differential geometry to calculate the rate of change of a vector or tensor field along a specific direction on a curved surface. It takes into account the curvature of the surface and how it affects the direction of the vector or tensor field.

How is a covariant derivative different from a regular derivative?

A regular derivative is used to calculate the rate of change of a function with respect to an independent variable. A covariant derivative, on the other hand, is used to calculate the rate of change of a vector or tensor field with respect to a specific direction on a curved surface. It takes into account the changing direction of the vector or tensor field due to the curvature of the surface.

What is the relationship between a basis and a covariant derivative?

A basis is a set of vectors that span a vector space, and a covariant derivative is a mathematical operation that calculates the rate of change of a vector field. The basis vectors are used in the calculation of the covariant derivative to determine the direction of the rate of change of the vector field.

How is a covariant derivative used in physics?

In physics, a covariant derivative is used to describe the motion of objects in curved space-time, such as in the theory of general relativity. It is used to calculate the acceleration of an object in a curved space and is an essential tool in understanding the behavior of particles in gravitational fields.

Are there any applications of covariant derivatives outside of mathematics and physics?

Yes, covariant derivatives are also used in other fields such as computer graphics, where they are used to describe the motion of objects in 3D space. They are also used in machine learning, specifically in the field of deep learning, to calculate the gradient of a neural network with respect to its parameters.

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