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In summary, the conversation discusses the concept of a covariant derivative and how it relates to other types of derivatives. The conversation also mentions the use of Christoffel symbols and how the covariant derivative acts on covectors. It is noted that the problem being discussed may be considered ill-posed from a textbook perspective.
  • #1
latentcorpse
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Edit: Solved
Don't know how to delete thread though!
 
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  • #2
A covariant derivative is an example of a derivation, that is a differential operator that satisfies the Leibnitz (product rule):

[tex] D ( v \otimes w ) = (Dv)\otimes w + v \otimes (Dw).[/tex]

Therefore the part that you proved above follows. Now, since [tex]l^2[/tex] is a scalar, it's covariant derivative reduces to the ordinary derivative. You can see this by explicit computation, since the signs in front of the Christoffel symbols are opposite when acting on covariant and contravariant vectors.
 
  • #3
Judging from a textbook's perspective, your problem is ill-posed, because the [itex] l^2 [/itex] for l a 4-vector is a genuine scalar, and the way the covariant derivative acts on the scalars is used as an axiom in deriving how the covariant derivative acts on covectors, knowing how it acts on scalars and 4-vectors.
 
  • #4
latentcorpse said:
Edit: Solved
Don't know how to delete thread though!

You don't delete your post after you receive help here. Check your PMs.
 
  • #5


Congratulations on solving the problem! As for deleting the thread, you can usually find an option to delete or close the thread on the website or forum you posted it on. If you are unable to find it, you can reach out to the website administrators for assistance. Keep up the good work in your scientific endeavors!
 

1. What is a covariant derivative?

A covariant derivative is a mathematical tool used in differential geometry to measure the rate of change of a vector field along a curve on a curved surface or manifold. It takes into account the curvature of the surface and ensures that the derivative is invariant under coordinate transformations.

2. How is a covariant derivative different from an ordinary derivative?

Unlike an ordinary derivative, which only considers the change of a vector in Euclidean space, a covariant derivative takes into account the curvature of the surface on which the vector is defined. It also accounts for the change in coordinate systems, making it a more general and powerful tool.

3. What is the importance of the covariant derivative in physics?

In physics, the covariant derivative is essential for understanding how quantities such as vectors and tensors behave in curved spacetime. It allows us to define the concept of parallel transport, which is crucial in general relativity for maintaining the consistency of physical laws in different frames of reference.

4. How is the covariant derivative calculated?

The covariant derivative is calculated using differential geometry techniques, such as the Christoffel symbols and metric tensor. It involves taking the partial derivative of a vector field with respect to each coordinate and adding terms that account for the curvature of the surface.

5. What are some real-life applications of the covariant derivative?

The covariant derivative has various applications in fields such as physics, engineering, and computer graphics. It is used to study the behavior of particles in general relativity, calculate trajectories of objects in electromagnetic fields, and model the movement of fluids in curved spaces. It is also used in computer graphics to simulate the deformation of surfaces and objects in 3D space.

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