What is a vector?


by Mike2
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Mike2
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#1
Feb26-04, 07:41 AM
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Can a vector be described in the most general terms by two points on a manifold, as a line with a starting and ending point in some manifold?
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arivero
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Feb26-04, 08:24 AM
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Hmm depends on which vectors. The ones used in Differential Geometry do not live in the manifold but in the tangent bundle. So You specify them by given a supporting point in the manifold and a free vector in the linear space.

Then you have a bunch of theorems about how such vectors can be mapped to curves in the manifold across the supporting point.
HallsofIvy
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Mar12-04, 07:19 AM
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Originally posted by Mike2
Can a vector be described in the most general terms by two points on a manifold, as a line with a starting and ending point in some manifold?
NO! In Euclidean space you can get away with the "position vector" or the "vector between two points" but not in general differentiable manifolds or even surfaces. It is best to remember "a vector is a derivative"! The best way (indeed the only general way) to think of vectors on surfaces or manifolds is as tangent vectors to some curve on the surface or manifold: in other words the derivative.
A vector can only be defined at a single point. To have a notion of "moving" a vector from one point to another, you need some additional structure such as a metric tensor or a "Riemannian connection".

pmb_phy
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Mar14-04, 09:18 AM
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What is a vector?


Originally posted by Mike2
Can a vector be described in the most general terms by two points on a manifold, as a line with a starting and ending point in some manifold?
If the space is flat then yes. If the space is curved then no. For a curved manifold you can't arbitrarily take any two points and obtain a unique vector.

Hallsofivy wrote "a vector is a derivative"!. If by this you mean that a vector v = (v1, ... , vn) defines a directional derivative operator at a point P in the manifold and vice versa then I agree.

This is Cartan's notion of a vector if I recall correctly?

It is based on the notion that displacement vectors are in a one-to-one correspondence with directional derivative operators.

However a vector can also be defined in otherways too. E.g. a vector can be defined as any quantity v whosse components = (v1, ... , vn) transform as the coordinate displacements dxa which is almost the same thing. Or one can define a vector as a linear map of 1-forms to scalars which obeys the Leibnitz rule.
Hurkyl
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Mar14-04, 11:15 AM
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There's another way of thinking about it (that is 'natural', but logically strange):


An "arrow" is a one-dimensional manifold whose endpoints are different yet in the same place. (This is, of course, the odd part; some fancy logic tricks are needed to make this work)

Then, a vector is simply the image of an arrow under a smooth map.
HallsofIvy
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Mar15-04, 07:07 AM
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Originally posted by pmb_phy
If the space is flat then yes. If the space is curved then no. For a curved manifold you can't arbitrarily take any two points and obtain a unique vector.

Hallsofivy wrote "a vector is a derivative"!. If by this you mean that a vector v = (v1, ... , vn) defines a directional derivative operator at a point P in the manifold and vice versa then I agree.

This is Cartan's notion of a vector if I recall correctly?

It is based on the notion that displacement vectors are in a one-to-one correspondence with directional derivative operators.

However a vector can also be defined in otherways too. E.g. a vector can be defined as any quantity v whosse components = (v1, ... , vn) transform as the coordinate displacements dxa which is almost the same thing. Or one can define a vector as a linear map of 1-forms to scalars which obeys the Leibnitz rule.
Yes, that was exactly what I meant. I would also note that in order to "transform as coordinate displacements dxa" you need to use the chain rule so that you do need derivatives. I would just drop the word almost from "which is almost the same thing"! The crucial point, since the original question was about vectors in terms of "two points on a manifold" is that vectors exist in the tangent space at each point on a manifold, not on the manifold itself.
pmb_phy
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Mar15-04, 08:26 AM
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Originally posted by HallsofIvy
Yes, that was exactly what I meant. I would also note that in order to "transform as coordinate displacements dxa" you need to use the chain rule so that you do need derivatives. I would just drop the word almost from "which is almost the same thing"! The crucial point, since the original question was about vectors in terms of "two points on a manifold" is that vectors exist in the tangent space at each point on a manifold, not on the manifold itself.
You're discounting the special case of flat manifolds. There is important class of vectors and tensors which apply only to flat manifolds. One example, if we are dealing with flat spacetime in relativity, is that of Lorentz 4-vectors. They are things which apply to Lorentz coordinates and whose components transform in the same way as an event X = (ct, x, y, z). X is defined as a displacement from a point in the manifold chosen to be the origin. The transformation has little to do with derivtives. Instead of a quantity whose components transform as dxa it is a quantity whose components transform as xa (i.e. components of X), or if you like, as delta xa where the delta represents a finite difference rather than a differential displacement. In this case two seperate points are at two seperate and, prerhaps, remote places in the manifold. If one chooses to study and concern himself only with SR (by which I mean flat spacetime and only Lorentz coordinates) then Lorentz tensors are quite sufficient for all applications that I can readily think of

Many excellant text discuss these tensors. E.g.

Tensors, Differential Forms, and Variational Principles, Lovelock and Rund, page 53
Gravitation and Spacetime - Second Edition, Ohanian and Ruffini, page 308
Classical Electrodynanics - Second Edition, J.D. Jackson, page 535
Classical Field Theory, Francis E. Low, page 259


It's important to understand this when looking at something like Jackson since while he does use partial derivatives to define the transformation of components he never uses anything but the Lorentz coordinates and the Lorentz transformation. I.e. transformations from one set of Lorentz coordinates (ct, x, y, z) to another set of Lorentz coordinates (ct', x', y', z).

Lorentz coordinates are defined as (ct,x,y,z) in an inertial frame of reference.

Things which are Lorentz 4-tensors/vectors are not tensors/vectors in general. One has to be careful when reading Jackson. E.g. the components of 4-force in Lorentz coordinates appears in his text as dP^u/dT since the affine connection vanishes in Lorentz coordinates. In generalized coordinates the components of 4-force are DP^u/DT (T = proper time) which includes the affine connection

Lorentz coordinates and Lorentz 4-vectors are obviously a very widely used arena.

For those interested in the details of this please see the excelant online notes by Thorne and Blanchard at
http://www.pma.caltech.edu/Courses/p...p01/0201.2.pdf
mathwonk
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Aug10-04, 06:57 PM
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For some reason, probably the time of day, this reminds me of bill cosby's album "why is there air?"


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