Talk:Vector field

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1 Coordinate transformations
2 Vector fields as operators
3 discussion at Wikipedia talk:WikiProject Mathematics/related articles
4 Question
- 4.1 Difference between scalar and vector field
5 expanded difference between scalar and vector fields
6 Vectors and coordinates
7 Vector and coordinates (more)
8 There are 3 types of lines that can be made from vector fields

[edit] Coordinate transformations

There seems to be a mistake in the relation given in Section 1.1 for translating the expression of a vector field from one coordinate system to another. I think it should read:

$V_y = \frac{\partial y}{\partial x} V$ .

This issue has been discussed in a Usenet thread. Could someone please verify this? Thanks. -- Pouya Tafti 20:56, 30 November 2006 (UTC)

The vector field is $V^i \partial_{x^i} = V^i \frac{\partial_{y^j}}{\partial_{x^i}} \partial_{y^j}$ so ${V_y}^j = V^i \frac{\partial_{y^j}}{\partial_{x^i}}$ or indeed $V_y = \frac{\partial y}{\partial x} V$ . The current transformation describes a covector field. I'll fix that. --MarSch 16:03, 4 December 2006 (UTC)

Thanks for the correction. I think example 2 should be modified as well, i.e. we should have $v_{\mathrm{Euclidean}}:x \mapsto 0.5$ . Also, if we have

ξ = 2 x

, then

x

changes by

0.5

(and not

2

, as in the article) units when

ξ

changes by

1

, right? —Pouya D. Tafti 17:20, 4 December 2006 (UTC)

Yes, you are completely correct. Let me fix that :) --MarSch 17:56, 4 December 2006 (UTC)

Ok fixed, rewritten to avoid factors of one half. --MarSch 18:06, 4 December 2006 (UTC)

[edit] Vector fields as operators

This page needs an explanation of vector fields acting as operators on functions since this concept is used elsewhere e.g. Lie algebra - Gauge 04:02, 3 Aug 2004 (UTC)

If you think a page should be expanded on then be bold and do it yourself. There is no need to discuss it on the talk page. On the other hand if you are completely restructering a page a short discussion on the talk page might be apropriate to prevent an edit war. MathMartin 10:08, 3 Aug 2004 (UTC)

This stuff is at Tangent_vector#Tangent_vectors_as_directional_derivatives. The problem is that we have too many pages with unclear relations to eachother. --MarSch 17:31, 11 Jun 2005 (UTC)

[edit] discussion at Wikipedia talk:WikiProject Mathematics/related articles

This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:05, 12 Jun 2005 (UTC)

[edit] Question

The following paragraph is now in the article:

[edit] Difference between scalar and vector field

The difference between a scalar and a vector field is in how their coordinates respond to coordinate transformations. Coordinates of scalars, by definition, don't transform at all. See example for Euclidean and cylindrical coordinates:

1 $(x, y) \mapsto x$

2 $(x, y) \mapsto x \boldsymbol{\partial}_x$

3 $(r, \theta) \mapsto r \cos \theta$

4 $(r, \theta) \mapsto r \cos \theta ( \frac{\partial r}{\partial x} \boldsymbol{\partial}_r + \frac{\partial \theta}{\partial x} \boldsymbol{\partial}_\theta )$

(numbering added). Now, let us ignore formulas 3 and 4. It seems to me that formulas 1 and 2 illustrate the difference between scalar fields and vector fields in Euclidean coordinates.

But then I defintely don't know what a vector field is. Formula 1 looks to me like a scalar field. But formula 2 does not look to me as a vector field, rather as a differential operator. Am I misunderstanding something? Oleg Alexandrov 00:24, 5 Jun 2005 (UTC)

Vectors are differential operators, hence the notation. $\boldsymbol{\partial}_x$ is thus the vector in the x direction, in this case of length 1. This is explained at Tangent_vector#Tangent_vectors_as_directional_derivatives. Basically when taking coordinates of a vector you throw away the operator. Thus the vector and scalar have the same Euclidean components. But those things you've thrown away transform, thus when you transform a vector and throw away the derivative operators, you get something different then when you transform a scalar and this happens in vector calculus already. --MarSch 00:47, 5 Jun 2005 (UTC)

I beg to contradict you. A vector is a quantity with n components, an element in R^n. Of course you can identify a vector with a directional derivative, but this is not what one means generally by vector.

Now I see your point. In your view, a vector field X is basically the differential operator:

$\nabla_X$ .

But you see, again, this identification of a vector field with the differential operator acting on the space of scalar functions and taking values in the space scalar functions, is not what people mean by a vector field (unless you are deep in differential geometry). And you did not explain this identification. It is no surprise that I did not understand the section you just added. Oleg Alexandrov 00:58, 5 Jun 2005 (UTC)

As you said, this is not the place to discuss that vectors are differential operators. I just used the notation coming from this, which is handy in remembering how vectors transform, since it is just the product rule. Vector are not members of Rⁿ, they are members of TRⁿ, its tangent space. These are usually identified because Rⁿ has a preferred basis. But switching from these Euclidean coordinates to cylindrical coordinates changes that identification. This makes vectors different from scalars. I have tried explaining this also at scalar, perhaps you can have a look there also.

This difference occurs already in flat space, like in special relativity, which I think is the realm of vector calculus and not differential geometry. $\nabla_X$ is the extension of the differential operator to all tensors. You need a connection to do this. Specifically, if X and Y are two vectors, it is not clear what $\nabla_X Y$ is, You have some freedom to define it. This is just more complication that neither of us wants here. Perhaps you could try explaining why vectors transform as they do when regarded as a bunch of numbers. This might help me understand better your point of view. Of course you can simply define how they are supposed to transform, but that is rather unilluminating.--MarSch 11:19, 5 Jun 2005 (UTC)

Hey, you are right. I remember the notation $\nabla_X Y$ from a long while ago, but I forgot what it meant.

Now, about the section "Difference between scalar and vector field" which you inserted. Again, nobody will understand it, because the necessary context is missing. I believe it should not be in that article, or otherwise significantly expanded. Even better, a new article as I suggested, which is differential geometry from the very beginning. Oleg Alexandrov 15:35, 5 Jun 2005 (UTC)

Okay, let me be blunt. If anything should be in this article it should be an explanation of what a scalar field is. My explanation concerns vector calculus already as I've said many times now, not only differential geometry. Why do you insist on ignoring this? That said, we can use different notation or even various notations and perhaps some picture about decomposing a vector into two different bases, which is simple enough I think. --MarSch 14:48, 6 Jun 2005 (UTC)

My main point is that you wrote things without providing the necessary background for the reader to understand what you are up to. Except for that, and merging the two articles, please feel free to do whatever you wish. Oleg Alexandrov 15:25, 6 Jun 2005 (UTC)

Hmmm, rereading your next to last post I see that one of the option you favour is significantly expanding the section. I'm sorry for the confusion of my last post, although it does explain why moving to a pure diff geo article is not an option. I will try to expand the section so it is comprehensible. Please continue to give me feedback. --MarSch 17:23, 11 Jun 2005 (UTC)

A scalar field is a tensor of rank (0,0). However, the function (x,y)-> x is not a tensor and thus not a scalar field. (It is not invariant under rotations, but a scalar field must be invariant, because a tensor of rank (0,0) is by definition invariant under rotations.) — MFH: Talk 16:32, 20 Jun 2005 (UTC)

All tensors have an intrinsic geometric meaning which is independent of coordinates. Thus you _might_ say that these geometric objects are invariant under arbitrary coordinate transformations. What is not true is that any coordinate representation of a tensor is invariant. (x,y) -> x is a function and can thus be used to represent a scalar field. If you transform to new coordinates u:=x-y, v:=x+y, then the new representation would be (u,v) -> (u+v)/2, which is a different function than (x,y) -> x, but it represents the same scalar field in different coordinates. There is no problem. --MarSch 12:21, 30 August 2005 (UTC)

[edit] expanded difference between scalar and vector fields

Okay, a full explanation is now in place. Perhaps you really like it. I feel it is rather unencyclopedic at the moment, but maybe that is only because of the popular way I phrased it and you disagree. --MarSch 18:43, 11 Jun 2005 (UTC)

[edit] Vectors and coordinates

I moved some stuff contributed by MarSch in scalar field to here, as it has more to do with vectors than with scalars.

From what I know, a vector does not change if you change coordinates, only its components change. This unlike what the section ===Example 2=== in this article seems to imply.

Also, a vector is not a differential operator. In some situations it is convenient to identify a vector with the partial derivative in the direction of the vector, but this identification is by no means universal in mathematics. And without this identification (which was not made explicit), the section ===Example 2=== from this article does not make sence. Any comments? Oleg Alexandrov 04:45, 25 Jun 2005 (UTC)

First of all. I think that the section makes a valid point, at least in the proper context: the components of a scalar field do not change under coordinate transformations, while the components of a vector field do change. I do not see where in the text a vector is identified with a partial derivative. To be honest, I do not understand where Oleg's confusion comes from. Oleg, do you agree with ===Example 1===, and if so, what is the different with the ===Example 2===?

The differential operator identification is what MarSch mentions way above on this talk page (which I moved from the talk page of scalar field).

I have problems with ===Example 1=== too, but ===Example 2=== is simpler and easier to make a point about.

My point still stays the same, that being that it was not explicit what is meant by a vector, and then these examples don't make sence. Any help rewriting this part and making it fit in the article? Oleg Alexandrov 16:53, 25 Jun 2005 (UTC)

However, while I agree that the section fits here better than in scalar field, I do question the placement of the section. In the previous parts of the article, a vector field is a function from (a subset of) Rⁿ to Rⁿ, that is, basically as a bunch of scalar fields. Then, the point of view suddenly shifts to the differential geometry, where vector fields are sections of the tangent bundle. MarSch's section seems to be based on the definition vector fields by how they behave under coordinate transformations (what I would call the physicist's definition).

Summarizing, I have no problems with the section, save that it could probably be explained better, but I think we should first explain the magic incantation "vector fields are sections of the tangent bundle". -- Jitse Niesen (talk) 12:52, 25 Jun 2005 (UTC)

the view doesn't change to differential geometry, because all of this is valid and very relevant for Euclidean space itself. It isn't 100% rigorous, because a vector is an arrow on Euclidean space and this is never defined. Intuitively it is very clear however what a vector is. If you want rigor you define a vector as a bunch of coordinates that transform in the way derived in my examples (the physicists' def as jitse calls it). There is no need for the tangent bundle approach here nor for the interpretation of vectors as differential operators. Although that is a natural isomorphism so it is pretty universal if you ask me. Oleg, I'm troubled by the fact that you still have (big) problems with these sections, because I thought they were very clear. Perhaps you have never been taught this stuff and have trouble believing it, but otherwise please start from the beginning and criticize every single letter I wrote. Perhaps I should say what I mean by a vector, I see I've glossed over this, so perhaps you assumed I was using the differential operator def. Yes I think this is it. I will try to implement this. --MarSch 19:44, 25 Jun 2005 (UTC)

This sentence was supposed to explain one of the vector fields I define :"Thus we have a scalar field which has the value 1 everywhere and a vector field which attaches a vector in the r-direction with length 1 to each point." Is this unclear? --MarSch 19:48, 25 Jun 2005 (UTC)

You are very right that I don't know this material (this makes me a good guinea pig in whether this text is understandable, don't you think so? :)

No, a vector is not just an arrow. A vector is very precisely defined as an element of the Euclidean space Rⁿ.

So, you, Jitse and me agree that what is missing is what you mean by a vector, because what you mean by that is definitely not what is meant usually, or what is meant everywhere else in this article. So, good luck working on that. :)

And one more request. The material you put feels a bit out of place in this page. Some integration would be good, so that the article reads as one. Thanks a lot, Oleg Alexandrov 22:31, 25 Jun 2005 (UTC)

Yes, if you say this text is not understandable, then few of the intended audience will understand it. What I hoped my examples would explain is _why_ vector fields are not just a bunch of functions. Therefore I have to use an intuitive notion of vector field and not a rigorous definition. I am trying to explain what some properties of vector fields are, which a rigorous definition of a vector field ought to satisfy. I am trying to _justify_ a rigorous definition. Therefore I cannot start by giving a rigorous definition of a vector field. Thus in my explanation I am talking about the intuitive concept of a vector or arrow field. What you picture in your mind or would draw in a picture.

Let's for a moment imagine you are standing on a big flat plane. If I give you a vector ( an arrow pointing from your feet into some direction ) and then ask you to walk in the direction of the vector, that would be something you could do. Now instead of giving you a vector I give you an element of R^2 ( a vector in the vector space R^2), could you walk in the direction of the vector? Not if I don't first tell you how the directions in R^2 correspond to the directions in the plane you are standing on.

Allright, now you are standing in the topological space R^2 itself. I give you an element of R^2. This is where I am going to try to teach you something. You still don't know how to walk in the direction of that "vector". "Why the hell not? Just watch me", you say ;) . The problem is that you have to make an extra (obvious?) assumption which you didn't even realize you made. This assumption is how you get from coordinates to a (tangent) vector. Let me (try to) explain. Suppose you are standing at the point (3, 4) and I give you the coordinates (1, 0). Then "obviously" I want you to walk towards (4, 4). But why? Basically what you do is turn the point (1, 0) into a vector from (0, 0) to (1, 0) and then transport that vector from (0, 0) to (3, 4), so you know which way to walk. But perhaps I meant for you to turn (1, 0) into the vector (of length one) pointing from the origin in the direction of (3, 4) where you are standing. Then transport this vector to your feet and follow it. Perhaps I meant for you to walk towards (1, 0). How would you know? What you have to understand is that R^2 is just a set of points. We can make an assumption of how to turn it into a vector space, by identifying points with vectors from (0, 0) to that point. But if we want to have a vector at every point of R^2 ( a vector field) then we need a vectorspace stucture for each point with that point as the origin. When you transport a vector from (0, 0) to another point, you make an assumption. You choose an isomorphism between these vectorspaces. Or equivalently you have chosen isomorphisms from R^2 to each of those vectorspace structures. The difference between Euclidean coordinates and spherical coordinates is in which isomorphisms I choose. The difficulty is that the Euclidean choice seems so obvious or indeed the only possible choice. Just try to imagine that you know only spherical coordinates and nothing about Euclidean coordinates. Wouldn't you still describe vectors by a bunch of numbers? What would (1, 0) mean as a vector field?

My example of standing in R^2 itself may be debatable, but maybe it does provoke you to some ponderings. Basically if we want to describe a the plane and vectorfields on it, we use R^2 to describe the plane and all of its tangent vector space structures. I think this is the most useful way to think about it. When you think these things are all the same, you are using Euclidean coordinates. --MarSch 12:40, 26 Jun 2005 (UTC)

Please correct me if I misunderstand something. So, the most elementary definition of a 2d vector is the arrow pointing from (0,0) to a point P. All of us know that, and there is no disagreement of this matter altogether.Oleg Alexandrov 21:44, 26 Jun 2005 (UTC)

I look at R^2 as an affine space. If you give me two points then I can draw an arrow from one to the other and that is a vector. Vectors starting at (0, 0) are a special case of this. --MarSch 28 June 2005 11:47 (UTC)

So, what you are saying is that by default R^2 is just a manifold. You can't talk about vectors till you are given an origin, or otherwise a coordinate system. Did I understand you correctly? Oleg Alexandrov 21:44, 26 Jun 2005 (UTC)

No, this is not what I am saying. You can talk about vectors as going from one point to another. Only if you want to use points of R^2 to represent vectors, than you have to say what isomorphism you are using, implicitly or explicitly. --MarSch 28 June 2005 11:47 (UTC)

[edit] Vector and coordinates (more)

OF COURSE a vector field is not the same as a bunch of scalar fields! If a manifold M has dimension n, then the tangent space to M is not isomorphic to the manifold

$M\times \mathbb{R}^n$

even if both of them have dimension 2n.

But OF COURSE that is not clear from the section ==Difference between vector fields and scalar fields== because MarSch never bothered to say that a vector field no longer goes from R^n to R^n, but rather, from R^n to its tangent space. Oleg Alexandrov 28 June 2005 15:56 (UTC)

No, this is not what I am talking about. That is the domain of differential geometry. In vector calculus the tangent bundle _is_ a product space. It is beside the point--MarSch 28 June 2005 17:21 (UTC)

Good point. Thank you for your patience so far. I belive I am gradually getting it. Oleg Alexandrov 29 June 2005 05:03 (UTC)

No problem. How about those suggestions for improvement? ;) --MarSch 29 June 2005 12:50 (UTC)

[edit] There are 3 types of lines that can be made from vector fields

streaklines, fieldlines and pathlines. They seem all the same to me. Can anyone explain their difference? --MarSch 12:48, 30 August 2005 (UTC)

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