Talk:Vector field

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n-dimensional real vector space

Hey, up to isomorphism there is only one real n-dimensional vector space. Up to isomorphism there is only one real-dimensional Slawomir Bialy. Of course, Slawomir Bialy is always changing, but there is a reasonable equivalence class, and we call the equivalence class Slawomir Bialy. There are many different elements of the equivalence class called the `the n-dimensional real vector space' where equivalence is isomorphism. This is common usage in mathematics. In any case, it's rude to delete undo all changes while objecting only to one. — Preceding unsigned comment added by Lost-n-translation (talkcontribs) 16:49, 1 February 2011 (UTC)

This is true, but of course a vector field is not just a function into another vector space up to isomorphism, it is a mapping into a particular vector space (in mathematics, the tangent space at a point; in physics, it is characterized by a contravariant transformation law). Either way, it's misleading to say that a vector field is a mapping into the n-dimensional real vector space; there is strictly more structure than that. Sławomir Biały (talk) 17:17, 1 February 2011 (UTC)

Example 1

If we write the vectors of the vector field as (1, 0) in polar coordinates, that means we're using different coordinate systems for these vectors at different positions in the original R2. The theta=zero direction of each tangent space would be identified with the r direction of the original space. Is that really a good choice? In the second part using cartesian coordinates, the coordinate systems of all tangent spaces are parallel to that of the original space, which means that in most of the tangent spaces we don't have the usual transformation between polar and cartesian coordinates. Isn't that a bit confusing? — HHHIPPO 20:33, 26 September 2012 (UTC)

Careful, the polar coordinates refer to polar coordinates on . The tangent spaces carry the linear coordinates induced by the differentials . Sławomir Biały (talk) 23:53, 26 September 2012 (UTC)
Ahh, I see. That is actually a very obvious choice of coordinates. I hadn't fully digested that the codomain of a vector field is a tangent bundle, which only for Cartesian coordinates looks like a single vector space. Thanks! — HHHIPPO 18:57, 27 September 2012 (UTC)

Picture of vector field on sphere

Normally the indicated map in the picture of the sphere with the red arrows would not be regarded as vector field, since on the equator some arrows seem to be orthogonal to the surface rather than tangent to it. They must be tangent otherwise its not a vector field. — Preceding unsigned comment added by (talk) 21:36, 16 November 2012 (UTC)

The picture is very bad. Someone should upload a better one. Sławomir Biały (talk) 22:00, 16 November 2012 (UTC)

Definition needs work.

The definition claims (fV)(p):= f(p)V(p) is scalar multiplication (for real valued function f:S → R). This may be an example of scalar multiplication, it is not scalar multiplication: scalar multiplication is multiplication by a scalar, not by a function.

I have tried to understand why the definition is so awkward, and I can't figure it out. You can NOT define a vector field in terms of two vector fields! Come on!

A vector field is the mathematical structure in which each element in the field has associated with it a direction and magnitude. A vector field is a ring (a commutative (Abelian) group having (at least) two operations, addition and multiplication, defined over it. Addition is abelian, while multiplication is monoid (that is, every element may not have a multiplicative inverse).

I don't know the exact definition of a vector field (if I did, I wouldn't be looking it up). All of the vector fields I am familiar with have the following additional characteristics which may or may not be required: Scalar multiplication is defined. The direction of the field is in the same space as the field.

The definition here fails to explain these things, as well as what the common properties are: (commutation, distribution of multiplication over addition, associativity, inverse, division, subtraction, (in)finiteness, closedness, magnitude, dot product, inner product, outer product, cross product...)

I KNOW that the vector fields are NOT limited to Euclidean Spaces, since I am currently working with them in Special Relativity. The lede is misleading or simply wrong.

All in all, this needs a bit of work...whether it goes from the narrow case (Euclidean) to the general case or vice versa is a matter or pedagogy, (I personally prefer general to specific then back to general, but I am not competent to write it). (talk) 20:19, 18 February 2014 (UTC)

It's not clear to me what perspective you're coming from, but it's very likely the source of your confusion is not this article. Vector fields do not form a ring. The scalar fields are just functions. While it's true that Euclidean space is not the most general setting for vector fields, it is one that we can expect all readers of this article at least to be familiar with. Sławomir Biały (talk) 17:55, 19 February 2014 (UTC)