How to define a vector field?

[Mike_bb]

Hello!

In one book I saw that function $V$ of 3 variables $V_x, V_y, V_z$ (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point $(0,0,0)$ such way:

I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true?
And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong: higher-order terms weren't equal to 0)

Thanks!

 

[martinbn]

Which book?

 

[Mike_bb]

Which book?

Russian book "Theory of electricity".

https://scask.ru/k_book_tel.php?id=19
page 38

 

[fresh_42]

I wouldn't use "higher terms equal 0". Instead, think of the standard example of a vector field, a tangent field. It consists of tangents, either to flows or to a manifold, consisting of flows. They are first-order derivatives by definition: the infinitesimal change of location. Those partial derivatives in your formulas are the local coordinates of the tangent space.

 

[Mike_bb]

I wouldn't use "higher terms equal 0". Instead, think of the standard example of a vector field, a tangent field. It consists of tangents, either to flows or to a manifold, consisting of flows. They are first-order derivatives by definition: the infinitesimal change of location. Those partial derivatives in your formulas are the local coordinates of the tangent space.

The book doesn't use such terms as "tangent space" and "manifold" quite. The book contains a chapter that is devoted to an introduction to vectors and that's all.

 

[fresh_42]

The book doesn't use such terms as "tangent space" and "manifold" quite. The book contains a chapter that is devoted to an introduction to vectors and that's all.

In its most abstract form, a vector field is a basis set $B$ of points and a vector $\vec{v}_p$ attached to each point $p$ of $B$:
$$
V=\coprod_{p\in B}\vec{v}_p=B\times \{v_p\,|\,p\in B\}.
$$
But this is a very abstract concept, and you might ask where this concept comes from. Answering this question leads you directly to tangent spaces. And if you try to avoid the term tangent, what do the partial derivatives in your formulas represent if not a tangent vector?

Draw a function graph, e.g., $x\longmapsto 2x^3-3x^2+1$ and attach a tangent line at each point. The set of all these lines will then be a vector field.

 

[Mike_bb]

In its most abstract form, a vector field is a basis set $B$ of points and a vector $\vec{v}_p$ attached to each point $p$ of $B$:
$$
V=\coprod_{p\in B}\vec{v}_p=B\times \{v_p\,|\,p\in B\}.
$$
But this is a very abstract concept, and you might ask where this concept comes from. Answering this question leads you directly to tangent spaces. And if you try to avoid the term tangent, what do the partial derivatives in your formulas represent if not a tangent vector?

Draw a function graph, e.g., $x\longmapsto 2x^3-3x^2+1$ and attach a tangent line at each point. The set of all these lines will then be a vector field.

You missed one thing that I wrote in post #1. I mentioned about Taylor series as it was written in the book.

 

[fresh_42]

You missed one thing that I wrote in post #1. I mentioned about Taylor series as it was written in the book.

I didn't miss it. Your question was all about the higher terms, and I answered that instead of looking at higher-order terms and artificially set them to zero, it is better to remain at first-order terms defining the tangents: the term of order zero represents the point $p=(x,y,z)$ and the term of order one represents the tangent vector $\vec{v}_p$. There is no need to consider the entire series and then try to find a reason to cut it off. But if you like, then the whole series represents the function, and the first two terms represent a linear approximation, the tangents. There is no "setting to zero". We simply only consider the linear approximation, which are the first two terms.

 

[Mike_bb]

I didn't miss it. Your question was all about the higher terms, and I answered that instead of looking at higher-order terms and artificially set them to zero, it is better to remain at first-order terms defining the tangents: the term of order zero represents the point $p=(x,y,z)$ and the term of order one represents the tangent vector $\vec{v}_p$. There is no need to consider the entire series and then try to find a reason to cut it off. But if you like, then the whole series represents the function, and the first two terms represent a linear approximation, the tangents. There is no "setting to zero". We simply only consider the linear approximation, which are the first two terms.

I'm not sure if it's so but the book uses such terms as "very small element f" = "infinitesimal element f" in context of vector $V$ circulation on element f. Perhaps it's the reason why we say about tangent vector as approximation of vector $V$ in infinitesimal scale of element f (or infinitesimal vector). Is this right?

 

[fresh_42]

I'm not sure if it's so but the book uses such terms as "very small element f" = "infinitesimal element f" in context of vector $V$ circulation on element f. Perhaps it's the reason why we say about tangent vector as approximation of vector $V$ in infinitesimal scale of element f (or infinitesimal vector). Is this right?

Yes. I like Weierstrass's notation of a derivative:

$f$ is differentiable at $x_0$ if there is a linear map $J$, such that $$f(x_0+v)=f(x_0)+J(v)+r(v)$$ where the error or remainder function $r$ has the property, that it converges faster to zero than linear, which means
$$\lim_{v \to 0}\dfrac{r(v)}{\|v\|}=0$$

Source: https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

$J(v)$ is the Jacobi-matrix, or the differential, or the derivative, or the tangent vector, or the infinitesimal change in direction $v,$ and $r(v)$ is what you want to set zero, or which I want to neglect in the first place. It is a matter of the point of view. But the formula summarizes all aspects: the function $f,$ the location $x_0,$ the direction $v,$ the derivative $J,$ which is a linear function of the direction, and finally the vanishing error $r.$

If we consider a vector field, then we restrict our perspective to location and direction:
$$
\left(x_0\, , \,f(x_0) + J(v)\right)
$$
Whether you consider this as neglecting the error $r$ or "setting it zero" is mute. I wouldn't write $r(v)=0$ because, as you noticed, it isn't zero. We just do not consider it if we talk about vector fields.

 

[martinbn]

The book defines vectora and vector fields in the begining. This part is to define the curl of a vector field or the rot (вихрь). The higher oder terms in the Taylor series are not zero, but you can ignor them because in the limit of the integral they will contribute zero. If you find it not rigorus enough just skip it and take the definitio of rot as given.

 

[Mike_bb]

fresh_42
martinbn

Big thanks!!

 

[haushofer]

Ah, the one book, from the one author. My favourite.

 

[wrobel]

You can define a vector field as follows. Assume that in any coordinates $x=(x^1,\ldots,x^m)$ in a domain $D\subset\mathbb{R}^m$ we have an ordered set of functions $v=(v^1(x),\ldots, v^m(x))$ and if an ordered set $(v'^1(x'),\ldots,v'^m(x'))$ corresponds to other coordinates $x'=(x'^1,\ldots,x'^m)$ then
$$v'^i=\frac{\partial x'^i}{\partial x^j}v^j.$$
In this case we say that the vector field $v$ is defined in the domain $D$
Actually any tensor field can be defined in such a way and this is enough for most needs of engineers.

 

[Mike_bb]

$$v'^i=\frac{\partial x'^i}{\partial x^j}v^j.$$

Could you explain how to derive this formula?

 

[wrobel]

It depends on definitions. You can take this formula as a definition.

On the other hand we can start with an ordinary differential equation $\dot x^i=v^i(x),\quad i=1,\ldots,m.$ Then we have a family of the integral curves of this equation and the field of velocities along these curves is exactly the vector field $v$. Then make a change of variables: $x=x(y)$. Substitute this formula into the differential equation:
$$\frac{\partial x^i}{\partial y^k}\dot y^k=v^i(x(y)).$$
This is it.

 

[fresh_42]

This only works if the vector field is a tangent field and the coordinate transformation is differentiable:

assumption 1: $\dot x^i =v^i(x)$
assumption 2: $\dfrac{\partial x^i}{\partial y^k}$ exists

Neither has to be given if we define an abstract vector field: A vector field is a section of a vector bundle.

 

[wrobel]

In its most abstract form, a vector field is a basis set $B$ of points and a vector $\vec{v}_p$ attached to each point $p$ of $B$:
$$
V=\coprod_{p\in B}\vec{v}_p=B\times \{v_p\,|\,p\in B\}.
$$

The set $B\times \{v_p\,|\,p\in B\}$ contains an element $(p',v_p)$ with $p'\in B,\quad p'\ne p$. Such a definition of a vector field looks strange.

 

[fresh_42]

The set $B\times \{v_p\,|\,p\in B\}$ contains an element $(p',v_p)$ with $p'\in B,\quad p'\ne p$. Such a definition of a vector field looks strange.

It was the most basic version I could imagine: a set of points with vectors attached to each. It was meant by me as a visualization of the concept. From here, we get to tangent vector fields, vector fields of differential equations, or finally, sections of bundles. Once we get the picture, we can not only attach vectors, but also vector spaces or even Lie algebras. My point was that a vector field is not automatically a tangent field, even though these are the most important examples, and, admittedly, given the photo from the book, the ones the author had in mind.

 

[wrobel]

My point was that a vector field is not automatically a tangent field,

I do not understand what this means. By definition a vector field on a manifold $M$ is a function $v:M\to TM$ such that $v(x)\in T_xM$

 

[martinbn]

It was the most basic version I could imagine: a set of points with vectors attached to each.

His point is that the way you wrote down the set it isn't points and vectors attached to them, but points and vectors attached to some pounts possibly different.

 

[Jaime Rudas]

Ah, the one book, from the one author. My favourite.

Wikipedia lists ten books written by that author.

 

[WWGD]

I didn't miss it. Your question was all about the higher terms, and I answered that instead of looking at higher-order terms and artificially set them to zero, it is better to remain at first-order terms defining the tangents: the term of order zero represents the point $p=(x,y,z)$ and the term of order one represents the tangent vector $\vec{v}_p$. There is no need to consider the entire series and then try to find a reason to cut it off. But if you like, then the whole series represents the function, and the first two terms represent a linear approximation, the tangents. There is no "setting to zero". We simply only consider the linear approximation, which are the first two terms.

Hope not to derail the post, but, how would you define higher/partial derivatives in manifolds through bundles? Is that what Jet Bundles are about?

 

[weirdoguy]

Is that what Jet Bundles are about?

Yes. That's why they are used in classical filed theories. There are also k-tangent bundles that kind of deal with derivatives of order k.