Vector or not vector?


by BobbyFluffyPric
Tags: vector
BobbyFluffyPric
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Mar29-07, 04:24 PM
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If a vector is simply "any quantity having magnitude and direction", then how can a vector's components NOT transform according to the rule of transformation for the components of a vector when we apply a transformation of the coordinate system? Or alternatively, given three numbers (for the 3D case), how can we say that they are NOT the components of a vector in a given coordinate system, if we admit that they transform according to the rule of transformation for vector components when we change the coordinate system? I'm not sure anymore if a vector is simply "any quantity having magnitude and direction", I mean, in a given x-y coordinate system (2D case), I can define a vector, say (3,5) - but it's clear that (3,5) are NOT the components of my vector in another coordinate system. Does that mean that my vector is not a vector?
I know my question seems confusing. Please read my following analysis before answering:
Vector or not vector?
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HallsofIvy
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Mar30-07, 07:01 AM
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The pair of numbers, (3,5) does NOT have "magnitude and direction" except in some specific coordinate system so you haven't defined a vector! If you specify a coordinates system, then you can talk about (3,5) as a vector (more correctly, it is representing the vector). Of course, if you change to a different coordinate system, that same vector would be represented by a different pair of numbers.

When the author says "Given the vector field [itex]y\vec{i}- x\vec{j}+ z\vec{k}[/itex]" he is clearly referring to some given coordinate system. In a different coordinate system that vector field, in order to be referring to the same vectors, would have to have a different formula.

I cannot see why you say there cannot be a constant vector. Certainly I can talk about a vector having a specific length and direction. Yes, it have different components in different coordinate systems but that has nothing to do with it being "constant". We say that the speed of light is constant, even though it is represented by different numbers depending on whether we measure it in m/s or ft/s. It's the same thing.
robphy
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Mar30-07, 10:44 AM
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A vector is NOT "any quantity having magnitude and direction".
A vector is an element of a vector space, whose elements can be added to each other and multiplied by scalars, plus some additional rules.

Until one defines a norm (e.g. with a metric), there is no notion of "magnitude" to assign to a vector. The best one can do is to compare vectors that are proportional to each other.

christianjb
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Mar30-07, 11:53 AM
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Vector or not vector?


Quote Quote by robphy View Post
A vector is NOT "any quantity having magnitude and direction".
A vector is an element of a vector space, whose elements can be added to each other and multiplied by scalars, plus some additional rules.

Until one defines a norm (e.g. with a metric), there is no notion of "magnitude" to assign to a vector. The best one can do is to compare vectors that are proportional to each other.
That seems overly pedantic to me.
In N-dimensional Euclidean space, a vector pretty much is a magnitude and direction.

Yes, we can 'abstract' the notion of a vector to do clever things in Hilbert space etc., but let's not lose sight of the easy definition in x-y-z space that's good enough for most physicists.
robphy
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Mar30-07, 12:15 PM
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Quote Quote by christianjb View Post
That seems overly pedantic to me.
In N-dimensional Euclidean space, a vector pretty much is a magnitude and direction.

Yes, we can 'abstract' the notion of a vector to do clever things in Hilbert space etc., but let's not lose sight of the easy definition in x-y-z space that's good enough for most physicists.
Yep. If the context is Euclidean space, then you have the usual notion of magnitude... and you can get by the "magnitude and direction" description. In intro courses, that context is usually implied. But if one starts trying to question things, it's good to start with the real definitions.

Of course, when dealing with (say) relativity, the context is no longer Euclidean space... and some nonzero vectors have zero magnitude. While not every physicist needs to use that, I would hope that they would at least be aware of that.
radou
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Mar30-07, 12:21 PM
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Quote Quote by christianjb View Post
but let's not lose sight of the easy definition in x-y-z space that's good enough for most physicists.
In our language, we use the term "radius-vector" (i.e. position vector) for an "arrow whose tail is at the origin, and whose head is at the point with coordinates (x0, z0, z0)", so no confusion arises.
ObsessiveMathsFreak
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Mar30-07, 05:33 PM
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The problem here is that points are being confused with vectors. Though vectors seem to be equivilent to points, they are not. You can add and subtract vectors. You cannot add or subtract points!!! NB

But wait you say. I add add subtract points all the time. Not strictly true. What you are actually doing is adding and subtracting the vectors pointing from your origin to those points.

Case in point. Take the points (1,0) and (0,1). The "addition" of these points is taken to be "the point" (1,1). But what we really did was add the vectors pointing from (0,0) to (1,0) and from (0,0) to (0,1) and obtained the vector pointing from (0,0) to (1,1). What's the distinction here? Well, let's take a change of coordinates!

Move the origin to the left by one unit. So the point that was at (1,0), is now at (2,0) and the point that was at (0,1) is now at (1,1). The point (1,1) is now at (2,1). But adding these two new points (2,0) and (1,1) gives the "point" (3,1). When we move our origin, the addition of points changes!! The addition of points is not invariant under a transformation.

But the addition of vectors is. If we take the vector from the "old" origin, now at (1,0), to each of the points, we will find that our results will be equal in both coordinate systems. Vector addition is invariant under translation.

A subtle point, and only one that becomes apparent under a change of coordinates, but one that should be noted nonetheless. Remembe, without an origin, you cannot define the "addition" of points at all, so it is natural that the placement of the origin should affect the result of your point "additions".

However, there is one operation on points that does remain invariant under translation. You would call it the "subtraction" of points p-q, the vector pointing from the point q to the point p. The operation of taking the vector pointing from one point to another is invariant under translation.

Hope that helps.
radou
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Mar30-07, 06:51 PM
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Quote Quote by ObsessiveMathsFreak View Post
Move the origin to the left by one unit. So the point that was at (1,0), is now at (2,0) and the point that was at (0,1) is now at (1,1). The point (1,1) is now at (2,1). But adding these two new points (2,0) and (1,1) gives the "point" (3,1). When we move our origin, the addition of points changes!! The addition of points is not invariant under a transformation.

But the addition of vectors is. If we take the vector from the "old" origin, now at (1,0), to each of the points, we will find that our results will be equal in both coordinate systems. Vector addition is invariant under translation.
What?!
Hurkyl
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Mar30-07, 07:27 PM
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Quote Quote by radou View Post
What?!
Suppose we apply the translation x ---> x + v.

If x + y = z, then it's not true that (x+v) + (y+v) = (z + v); addition is not preserved by translation.

on the other hand,

If (x - O) + (y - O) = (z - O), then after translating we see that
((x + v) - (O + v)) + ((y + v) - (O + v)) = ((z + v) - (O + v))
is true. This property is preserved by translation.
pmb_phy
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Mar30-07, 09:48 PM
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Quote Quote by robphy View Post
A vector is NOT "any quantity having magnitude and direction".
A vector is an element of a vector space, whose elements can be added to each other and multiplied by scalars, plus some additional rules.
I disagree. Consider the position of a particle in the Euclidean plane. This position can be defined by a vector. It is a vector because it satisfies certain transformation properties. Some types of vectors may belong to a vector space but they don't all need to be.

To determine the definition of a geometrical vector let's due the usual and turn to the literature on this. From A first course in general relativity, by Bernard F. Schutz, page 39 Eq. (2.7) is the criteria for something to be a vector and that is a transformation rule. The components of a vector are defined according to the position vector (or displacement vector) and how such a vector transforms from one coordinate system to another.


Pete
Hurkyl
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Mar30-07, 11:25 PM
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Quote Quote by pmb_phy View Post
To determine the definition of a geometrical vector let's due the usual and turn to the literature on this. From A first course in general relativity, by Bernard F. Schutz, page 39 Eq. (2.7) is the criteria for something to be a vector and that is a transformation rule. The components of a vector are defined according to the position vector (or displacement vector) and how such a vector transforms from one coordinate system to another.
You are describing one construction of the tangent space to the manifold at a point, or maybe of the tangent bundle -- both of which are vector spaces over R.
christianjb
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Mar30-07, 11:45 PM
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Here's a different view on things- which many here might disagree with.

Confusing a vector with a triplet of numbers is like confusing a statue with its shadow. No triplet of numbers IS the vector, though some triplets may be equal to the projections of a vector onto a particular set of x-y-z axes. However, if we are told that a=(2,1,3) is a vector projection and b=(3,1,4) is a vector projection- then we can use rules, like the dot product which are in correspondence to real manipulation of the actual vectors. In particular- all manipulations of real vectors are completely independent of the directions of any coordinate set- and so the rules vector operations on the components only allow operations which would give the same end result in any x-y-z basis.

In some ways it's a philosophical (Platonic, and even Taoist) concept. It's easy to confuse the names and properties of things with the things themselves.
robphy
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Mar30-07, 11:50 PM
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From Schutz's "Geometrical Methods of Mathematical Physics" (p.13),
A set V is a vector space (over the real numbers) if it has a binary operation called + with which it is an Abelian group and if multiplication (.) by real numbers is defined to satisfy the following axioms (in which [itex]\tilde x[/itex] and [itex]\tilde y[/itex] are vectors and [itex]a[/itex] and [itex]b[/itex] are real numbers):
[tex]a \cdot (\tilde x+ \tilde y) = (a \cdot \tilde x) + (a \cdot \tilde y)[/tex]
[tex](a+b) \cdot \tilde x = (a \cdot \tilde x)+ (b \cdot \tilde x)[/tex]
[tex](ab) \cdot \tilde x = a \cdot ( b \cdot \tilde x)[/tex]
[tex]1 \cdot \tilde x = \tilde x[/tex]
(p. 14)
So far nothing has been said about inner products or magnitudes of vectors. These are additional concepts which may or may not be useful in particular applications involving vectors: there is no necessity to impose them on a vector space.
BobbyFluffyPric
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#14
Mar31-07, 03:41 AM
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I want to thank all of you for taking time to help out with your views. Ill try and draw some conclusions.

HallsofIvy, thank you so much for your reply, it is much appreciated. Yes, I think I understood everything you said, but perhaps I didn't express myself too well concerning the "constant" vector. About the constant vector (3,5) - I agree, it doesn't define a vector unless a specific coordinate system is involved, as christianjb has also pointed out:
Confusing a vector with a triplet of numbers is like confusing a statue with its shadow. No triplet of numbers IS the vector, though some triplets may be equal to the projections of a vector onto a particular set of x-y-z axes.
However, the same author who provided me with the example of the vector field a = y i x j + z k in a chapter dedicated to Vector Calculus, in a later chapter dedicated to Tensors defines a (Cartesian) tensor in the following way (which I think is the definition that pmb_phy has pointed out from A first course in general relativity, by Bernard F. Schutz):

we may consider any set of three quantities vi, which are directly or indirectly functions of the coordinates xi and possibly involve some constants, and ask how their values are changed by any rotation of the Cartesian axes. The specific question to be answered is whether the specific forms vi in the new variables can be obtained from the old ones vi using vi = Lijvj . If so, the vi are said to form the components of a vector or first-order Cartesian tensor v.
Then he asks,
Which of the following pairs (v1, v2) form the components of a first-order Cartesian tensor in two dimensions?:
(i) (x2,−x1), (ii) (x2, x1)
Answer:
(i) Here v1 = x2 and v2 = −x1, referred to the old axes. In terms of the new coordinates they will be v1 = x2 and v2 = −x1, i.e.
v1 = x2 = −sin x1 + cos x2
v2 = −x1 = −cos x1 − sin x2.
Now if we start again and evaluate v1 and v2 as given by vi = Lijvj we find that
v1 = L11v1 + L12v2 = cos x2 + sin (−x1)
v2 = L21v1 + L22v2 = − sin (x2) + cos (−x1).
which coincides with
v1 = x2
v2 = −x1
and thus the pair (x2,−x1) is a first-order Cartesian tensor.
(ii) Here v1 = x2 and v2 = x1. Following the same procedure,
v1 = x2 = − sin x1 + cos x2
v2 = x1 = cos x1 + sin x2.
But, by vi = Lijvj for a Cartesian tensor we must have
v1 = cos v1 + sin v2 = cos x2 + sin x1
v2 = (−sin)v1 + cos v2 = − sin x2 + cos x1.
These two sets of expressions do not agree and thus the pair (x2, x1) is not a first-order
Cartesian tensor.
A similar analysis of (y,-x,z) will show that these are NOT the components of a vector, and yet, we seem to agree that a = y i x j + z k is a vector. I think that this is because of what you pointed out, a = y i x j + z k defines vector a but with reference to a particular coordinate system. In another coordinate system with basis vectors i,j,k , the components of a will be given by different formulae, according to vi = Lijvj , and they wont necessarily be y, -x, z, ie. we shall not have a = y i x j + z k . Nevertheless, as a geometrical entity that exists regardless of the coordinate system, a IS a vector, and its components DO transform according to ai = Lijaj. The reason then that (y, -x, z) are not the components of a vector, though they do coincide with the components of a in a certain coordinate system, is that (y, -x, z) is not referred to any particular frame of reference, and thus if they did define a vector then the vector would have to have these components in any coordinate system, which, as we have seen, is not true, since the (a1, a2, a3)=(y. x, z) is not obtained from Lijaj.

I would like to point out though, that though a is a vector, since one cannot write its components via formulae that are valid for any coordinate system (as could be done with the vector defined in example (i) ), a cannot in general represent a physical property of space. For example, the position vector of a particle cannot be defined without first specifying an origin, and thus, though a vector, is does not represent a physical field that describes some property of space (this same thing occurs with any constant vector). On the contrary, the velocity field is simply defined as (dx1/dt, dx2/dt, dx3/dt), and this definition is independent of the coordinate system. Thus, this is a true field (eg. the velocity field of a fluid).

To rhobphy - thank you for your insight, though I should have said that when I was talking about vectors I meant in the particular case of being elements of a metric space (if that's what you call it). I think this is what is referred to as "spatial vectors" on Wikipedia.

To christianjb thank you for helping out. Your radius vector (position vector) is another thing I was wondering about in the context of my question: you see, if defined as (x,y,z), this generic coordinate definition actually requires an origin to be previously fixed, and only holds for rotations of Cartesian Systems. If you define r = x i + y j + z k given a particular Cartesian Coordinate System, then in a coordinate system that is not Cartesian, for example, the formulae will be different (eg, r = rer in Sphericals), whereas in another Cartesian System with the same origin the components will simply be (x,y,z). Thus, (x,y,z) are the components of a Cartesian Vector, but they are not, in general, the components of a vector, even if they are the components of vector r in Cartesian Systems.

To ObsessiveMathsFreak: I think you hit the key of my problem. In fact, the definition of a vector as an entity whose components transform according to vi = Lijvj (for the particular case of rotations of Cartesian Systems) is saying that the components must transform the same way as the points transform, so that the vector itself remains unchanged. Again though, this brings me back to my original dilemma: take, for example, your vector (1,1) which points form the origin O of a given coordinate system to the point whose coordinates are (1,1) in the same system. This vector, which in general points to a point with coordinates (x,y), has components (x,y) in this coordinate system this is the position vector of a point. Now, if you move the origin to the left by one unit, the coordinates of the points change as you described, but if my vector which points from the OLD origin to each of the points is to remain unchanged, it is no longer the position vector in the new coordinate system. Thus, we cannot define the position vector without first defining an origin. However, if we restrict ourselves to rotations of Cartesian Systems, then the components (x,y) are always the components of the position vector of a point, where x and y are the respective coordinates of the point in whichever coordinate system we choose. But for general transformations, you cannot define the position vector without first defining a coordinate system. Hence, for each coordinate system we can define vectors that act as position vectors of points, but these are not in general THE position vector of a point in space. Thus, as a physical object THE position vector is not a vector! On the other hand, the velocity vector of a point (say, of a fluid particle), is completely independent of any coordinate system of reference (it exists on its own right), and thus is a vector.

These are more or less my conclusions Id like to know if what I say makes sense to anyone or not.

Thanks,
~Bobby
HallsofIvy
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Mar31-07, 06:44 AM
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Bobby, your quote refers specifically to a "Cartesian" vector, which is a special case of the general idea of "vector". A Cartesian vector must transform from one coordinate system to another by a rotation: the only coordinate systems allowed are those with straight line, orthogonal axes, ordered by the "right hand rule"- Cartesian coordinates.

The second "vector" in the example is not a Cartesian vector because it is derived from (x1, x2) by a reflection rather than a rotation. Conversely, you can think of it as "flipping" the coordinate system around the line x2= x1 making it a "lefthand" coordinate system.

The book you quote is not saying it is not a vector, just that it is not a Cartesian vector.

More general vectors (and tensors) allow other kinds of coordinate systems including curved axes.
BobbyFluffyPric
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Mar31-07, 07:30 AM
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HallsofIvy, as far as I had gathered, if the quantities ai transform, under a general transformation of the coordinate system xi=f(x1,x2,x3) (where f can be nonlinear), as ai=(δxi/δxj)aj, then they are the components of a vector. In the case of rotations (whether improper or not), the partial derivatives (δxi/δxi) are constant and equal to Lij (the cosine of the angle between the new axis Oi and the old axis Oj). Thus, I understand that if the quantities ai transform under any transformation as ai=(δxi/δxj)aj, then they certainly transform as ai=Lij aj as well in the case of rotations of Cartesian Systems. Thus, the vector whose components are the ai is a general vector, and a Cartesian one as well. However, the converse is not true: the components bi may transform as bi=Lij bj for rotations of Cartesian systems, but not as bi=(δxi/δxj)bj under more general transformations. Thus, the components bi define a vector only for Cartesian Systems, which is what we call a Cartesian Vector, but they dont really define a single vector because for a more general transformation the bi will represent some other vector. Thus, a Cartesian Vector is one that can be described via a set of components as long as they are referred to Cartesian Systems, but cannot be generally described by components. It just so happens that the components bi transform as they should, in order to represent the same vector, within rotations of the Cartesian System. Im not comfortable with (maybe because I dont understand) the term Cartesian vector (but I allow for it with the interpretation I have given), and I am neither am I comfortable with the term general vector I can only assume that it refers to vectors that can be described via their components through formulae that are valid regardless of which coordinate system we choose, Cartesian or not. But a vector is a vector regardless of coordinate systems (no?), and so a Cartesian vector is a vector, and a general vector is also a vector, right? The only reason I restricted myself earlier to Cartesian transformations is because it served me to make my point. Your last post confuses me. If what I have said is wrong, I'll be grateful if you could explain.
Thank you,
Bobby
pmb_phy
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Mar31-07, 11:05 AM
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Quote Quote by Hurkyl View Post
You are describing one construction of the tangent space to the manifold at a point, or maybe of the tangent bundle -- both of which are vector spaces over R.
Consider a flat space which consists only of a square. I.e. it is a finite manifold. Let the metric be the Euclidean one. I can still define position vectors on this space, however now there is no closure. I can't arbitrarily multiply any vector by an arbitrary number and expect the result to be within the space.

Pete
pmb_phy
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Mar31-07, 11:12 AM
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Quote Quote by robphy View Post
From Schutz's "Geometrical Methods of Mathematical Physics" (p.13),


(p. 14)
You haven't really said anything here my friend towards an example/proof in your favor. All you've done is shown that Schutz defined the term "vector space". He did not say that all types of geometric vectors are in any vector space. He even defined a geometric tensor (as a tangent vector) without mentioning the space that its in.

I'm 99.99% sure that I'm correct here. Just to make myself get more towards 100% sure I spoke with a friend of mine last night. He was a physics prof since 1960 until recently (1995?). You may have heard of him: Dr. Robert W. Brehme. His book is recommended reading by Schutz in his gr text. He's written tons of stuff on relativity, since that's his area. He tried to explain it to me in terms I could relay to you but I was certain such a relay wouldn't work. I.e. I was 100% sure that someone would come back with the response that hurkyl provided. I'll try to get a better example.

Best wishes

Pete


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