Vector or Not Vector: Analysis of Magnitude & Direction

[Hurkyl]

Do you or do you not agree with this. If not then please elaborate. Thanks.

It depends on the context.

In a setting that adopts the convention that the word "vector" is used exclusively to refer to tangent vectors of manifolds, and furthermore that tangent vectors are only to be thought of as coordinate-chart-dependent-tuples-of-'components', I would completely agree with what you have written in that post.

But the tangent vectors at a point P are elements of a vector space: the tangent space at P. So this isn't an example of vectors that are not elements of a vector space.

 

[pmb_phy]

The transformstion properties are derived from vector space properties, so vector space is more fundamental.

How would you derive the transformation properties for vector spaces? There are vector spaces in which no such requirement need be defined or cannot be defined.

Pete

 

[George Jones]

How would you derive the transformation properties for vector spaces? There are vector spaces in which no such requirement need be defined or cannot be defined.

Pete

Let V be an n-dimensional real vector space, and let \left\{ e_{1}, \dots, e_{n} \right\} be a basis for V. If L: V \rightarrow V is an invertible linear transformation on V, then \left\{ e'_{1}, \dots, e'_{n} \right\} with e'_{i} = Le_{i} is also a basis for V. Each member of the primed basis can be written as linear combintion of elements of the unprimed basis:

e&#039;_{i} = L^{j} {}_{i} e_{j}.[/tex]<br /> <br /> Any v in V can be expanded with respect to both bases:<br /> <br /> v = v^j e_j = v&amp;#039;^i e&amp;#039;_i = v&amp;#039;^i L^{j} {}_{i} e_{j}.<br /> <br /> Thus, for example,<br /> <br /> v^j = v&amp;#039;^i L^{j} {}_{i}.<br /> <br /> Some books blur the distinction between (tangent) vectors (at a point) and vector fields. The set of vector fields is not a vector space, but is a generalization (a module over the ring of scalar fierlds) of a vector space. Again, transformation properties are derived properties.<br /> <br /> Consider an n-dimensional real differentiable manifold, and suppose further that the bases above are coordinate tangent vector fields that arise from 2 overlapping charts:<br /> <br /> e_{i} =\frac{\partial}{\partial x^{i}}<br /> <br /> and<br /> <br /> e&amp;#039;_{i} =\frac{\partial}{\partial x&amp;#039;^{i}}.<br /> <br /> Then, by the chain rule, the change of basis relation is<br /> <br /> e&amp;#039;_{i} = L^{j} {}_{i} e_{j} = \frac{\partial x^{j}}{\partial x&amp;#039;^{i}} e_{j},<br /> <br /> and<br /> <br /> v^j = \frac{\partial x^{j}}{\partial x&amp;#039;^{i}} v&amp;#039;^i.

 

[pmb_phy]

Let V be an n-dimensional real vector space, and let \left\{ e_{1}, \dots, e_{n} \right\} be a basis for V. If L: V \rightarrow V is an invertible linear transformation on V, then \left\{ e&#039;_{1}, \dots, e&#039;_{n} \right\} with e&#039;_{i} = Le_{i} is also a basis for V. Each member of the primed basis can be written as linear combintion of elements of the unprimed basis:

e&#039;_{i} = L^{j} {}_{i} e_{j}.[/tex]

What gives you the idea that there are more than one coordinate systems for a particular vector space or that a vector is something that can be represented as an object with a single index? Rember that a vector in a vector space can be a lot of things like a second rank tensor or a 10x10 matrix. And why do you think vector transformations exists for all kinds of vectors?<br /> <br /> I&#039;ll get back after I do some more searching on this topic. The idea is to get a good example.

 

[George Jones]

What gives you the idea that there are more than one coordinate systems for a particular vector space

It's a basic fact of elementary linear algebra that there are infinitely many distinct bases for any vector space.

or that a vector is something that can be represented as an object with a single index? Rember that a vector in a vector space can be a lot of things like a second rank tensor or a 10x10 matrix.

All vector spaces be they spaces of 10x10 matrices or anything else, can be transformed in the way I outlined.

And why do you think vector transformations exists for all kinds of vectors?

Because of the linear algebra fact I stated above.

Now, for the space of 10x10 matrices there are (at least) two types of transformations that can induced by a change of basis, depending on whether the change of basis is for the space of matrices (this treats the matrices as single-index objects), or for the space on which the matrices act (this treats the matrices as two-index objects).

This discussion is getting somewhat surreal, so I'll think I'll bow out now.

 

[pmb_phy]

I can't, for the life of me, understand why you think that any vector can be transformed from one coordinate system to another when such a transformation may be entirely meaningless for a particuar kind of vector. E.g. I meant for the matrices to be ten dimensional, not 10x10
(In case you were thinking about 4-D spacetime). As I said, I'm looking for such an example right now and will post it when I find it. But such a criteria never exist for a vector space where such a critera always exists for geometrical vectors.

A good example which comes to mind now is the vector space whose elements belong to Fourier series, i.e. a sequance of sines and cosines in increasing frequency which can approximate any function of a given interval. How would you transform these elements to another coordinate system when, even when you can write down something that lokks like a transformation (God only knows what that transformation looks like or means) such a transformation is meaningless.

Pete

 

[HallsofIvy]

How would you derive the transformation properties for vector spaces? There are vector spaces in which no such requirement need be defined or cannot be defined.

Pete

In any vector space there exist an infinite number of possible bases. "Transformation" properties are transformation that take a vector written in one basis to the same vector written in a different basis.

 

[pmb_phy]

You folks really don't get the part about the transformation properties that I speak of do you?

In any case I believe I understand the problem now. First off - I believe I was wrong [ know. Hard to believe, right? :) ]. A geometric vector is an element of a vector space. Adding on the requirement that it satisfies certain transformation properties makes it a geometric vector.

Good think I stuck it out with this of shoot of this thread or I never would have known the solution to my delema. However this doesn't at all mean that I'm right now. I'm checking double checking my assumptions off the web and I'll get back to you.

Pete

 

[robphy]

However you never know. To be a good physicist one must keep an open mind so I e-mailed Ohanian again and asked him about this. Due to our eight years of exchanging e-mail with the man I've grown a very deep repspect for him. So we'll see what happens. I have no need to be right, just a need to know what is right.

Pete

Make sure you ask him this specific question:

Must a vector [whether from linear algebra or geometry] be an element of a vector space?

 

[pmb_phy]

Make sure you ask him this specific question:

Must a vector [whether from linear algebra or geometry] be an element of a vector space?

It doesn't matter Rob. I've sent it already and I'm really not disagreeing with you now. But I'll PM the response (if I get one) to you if you'd like?

Pete

 

[Hurkyl]

Adding on the requirement that it satisfies certain transformation properties makes it a geometric vector.

It's not the tangent vector (or cotangent vector, or tensor, or whichever thing we're looking at) that transforms -- it's the coordinate representation of that vector that transforms when you change coordinates. The whole point of the "geometric" adjective is that the vector itself is entirely independent of your choice of coordinates.

 

[pmb_phy]

It's not the tangent vector (or cotangent vector, or tensor, or whichever thing we're looking at) that transforms -- it's the coordinate representation of that vector that transforms when you change coordinates. The whole point of the "geometric" adjective is that the vector itself is entirely independent of your choice of coordinates.

First off I never referred to a tangent vectors. I kept it simple by keeping it in a flat space and using the position vector as a prototype of a generalized vector. And when I said that Adding on the requirement that it satisfies certain transformation properties makes it a geometric vector. I meant that the components must obey the given transformation rule. And I know what I meant since I've been over with this with folks for 8 years now. I think you simply read me wrong. Now let's say we let this thread lay to rest. You win. Rejoice! Peace out my friend. Please don't destroy your victory by attempting to re-educate me about things I know all too well. Okay? :)

Oh. By the way. If I were you then I'd read page 40 of MTW.

Pete

 

[mathwonk]

the following link seems relevant:

https://www.amazon.com/dp/B0000062TI/?tag=pfamazon01-20