Conservative vector field; classification of derivative

In summary, the speaker is asking for help with some fundamental questions related to a dynamical, autonomous and conservative system. They are unsure if the system is a proper manifold and also question the accuracy of a Taylor-expansion. They are specifically interested in knowing if a specific space is a proper manifold.
  • #1
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Dear forum-members,

Pestered by many (in my opinion, fundamental) questions and no literature at hand to answer them, I resort to posing my questions here. Let me start with the following. (Hopefully I have the correct subsection.)

I am inspecting a dynamical, autonomous and conservative system driven by a 'conservative' vector-field:
[tex]\left\|\dot{\mathbf{x}}\right\|_1 = 0[/tex]
(No need to specify this any further).
If we start off with
[tex]\left\|\mathbf{x}\right\|_1 = n[/tex]
Then the system we inspect is a vector-field on an n-simplex.

To be honest I am in doubt about this being a proper manifold, since the tangent has to abide the conservation constraints, while the neighborhood of any point in the state space is not perfectly Euclidean.
In line with this: how accurate is a Taylor-expansion of a subspace cut off at the m-th term?

Any response would be very much appreciated!
 
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  • #2
Let me address one topic at a time, and let me put the first issue differently.

Is a space
[tex]\Delta^r = \{ \mathbf{y} \; | \mathbf{y} \in R^r, \sum_i y_i = n \; \}[/tex]
a proper manifold?
 

1. What is a conservative vector field?

A conservative vector field is a type of vector field where the line integral of the vector field along any closed path is equal to zero. This means that the work done by the vector field on a particle moving along a closed path is independent of the path taken.

2. How is a conservative vector field classified?

A conservative vector field is classified based on its derivative. If the derivative of a vector field is equal to zero, then it is considered conservative. If the derivative is not equal to zero, then the vector field is considered non-conservative.

3. What is the significance of a conservative vector field?

A conservative vector field is significant because it represents a type of vector field that obeys the law of conservation of energy. This means that energy is conserved along any path in the field, making it a fundamental concept in physics and engineering.

4. How is the classification of derivative used in real-world applications?

The classification of derivative in conservative vector fields is used in many real-world applications, such as in fluid dynamics, electromagnetism, and mechanics. It helps to predict the behavior of systems and understand the underlying principles of energy conservation.

5. Can a vector field be both conservative and non-conservative?

No, a vector field cannot be both conservative and non-conservative at the same time. If the derivative of a vector field is equal to zero, it is considered conservative, and if the derivative is not equal to zero, it is considered non-conservative.

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