Dis-ambiguate: derivative of a vector field Y on a curve is the covariant of Y

In summary, the conversation discusses the concept of covariant derivative and its applications on vector fields defined on a curve. The first part proves that the covariant of a vector field with respect to a tangent vector is equal to the derivative of the vector field restricted to the curve. The second part shows that this result can be extended to any curve with initial velocity, not just a straight line. The use of the word "almost" in the second part is due to the fact that a vector field on a curve is not defined on the entire R^3, but only on a specific region. Therefore, a slight modification is needed to define the covariant derivative in this case.
  • #1
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Homework Statement


This two-part problem is from O'Neill's Elementary Differential Geometry, section 2.5.

Let W be a vector field defined on a region containing a regular curve a(t). Then W(a(t)) is a vector field on a(t) called the restriction of W to a(t).

1. Prove that Cov W w.r.t. a'(t) equals (W(a))'(t) where "Cov W w.r.t a'(t)" reads "the covariant of W with respect to a'(t)."

2. Deduce that the straight line in Definition 5.1 (below) may be replaced by any curve with initial velocity v. Thus the derivative Y' of a vector field Y on a curve a(t) is (almost) Cov Y w.r.t. a'(t).

Definition 5.1. Let W be a vector field on R^3, and let v be a tangent vector to R^3 at the point p. The the covariant derivative of W with respect to v is the tangent vector (W(p+tv))'(0) at the point p.

The following definition is useful for part (2) since it distinguishes between a vector field and a vector field on a curve.

Definition 2.2. A vector field on a curve 'a' from I to R^3 is a function that assigns to each number t in I a tangent vector Y(t) to R^3 at the point a(t).


Homework Equations


Part (1) was fairly straight-forward, using the definitions of covariant derivative and what it means to differentiate a composition.

Part (2) has two parts. My approach to the first part is the following, and I believe it to be correct. The idea is to define a function on curves a(t) and show that it agrees with the covariant derivative with respect to a vector v at a point p for all curves a(t) such that a(0)=p and a'(0)=v. Part (1) can be used to show the function is well-defined and that it indeed equals the covariant. The second part, starting at "Thus" is where I'm having trouble. It's with the use of the word "almost." To me, if Y is a vector field on a curve a(t), then using the definition of covariant, straight-line or otherwise, makes no sense because Y(a(t)) is not defined (i.e., Y is not defined on R^3, only R). So I thought this might be the almost part. However, what's 'almost' about it? I was thinking that perhaps given Y on a(t) that a vector field Ybar could be defined such that Ybar equals Y(a(t)), and that then Y' would equal the covariant of Ybar instead of Y. However, I don't think I can 'always' define such a Ybar.


Thank you in advance for your guided help. This is my first posting.
 
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  • #2
In the "Relevant equations" section I meant to say "Ybar composed with 'a' equals Y," not the other way around.
 
  • #3
Attached is the PDF of an easier-to-read TeX version.
 

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Related to Dis-ambiguate: derivative of a vector field Y on a curve is the covariant of Y

1. What does it mean for a vector field Y to have a derivative on a curve?

A vector field Y having a derivative on a curve means that for every point on the curve, there exists a vector that represents the rate and direction of change of Y at that point.

2. What is the difference between a covariant and a derivative?

A covariant is a vector that represents the change in a given direction, while a derivative is a vector that represents the rate of change in a specific direction.

3. How is the covariant of Y related to its derivative on a curve?

The covariant of Y on a curve is the vector that best approximates the derivative of Y at each point on the curve. It represents the direction and magnitude of the change in Y along the curve.

4. Can the covariant of Y on a curve be calculated using the chain rule?

Yes, the covariant of Y on a curve can be calculated using the chain rule, which allows us to break down the curve into smaller segments and calculate the covariant for each segment, then combine them to find the overall covariant.

5. Why is it important to dis-ambiguate the derivative of a vector field on a curve?

Dis-ambiguating the derivative of a vector field on a curve helps us to clearly define and understand the direction and magnitude of change of the vector field at different points on the curve. It allows us to make accurate predictions and calculations in various scientific and mathematical applications.

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