# Vector/Talk

The main page needs quite a lot of attention.

However...

It would be appropriate to retain (somewhere) the information on the current page because it would be very useful to beginners in physics and mechanics. While it is true that a vector is simply an element of a linear space, it is also true that vectors are very useful, and such uses can be remarkable in their own right.

I changed some things, notably:

• Position is not an example of a vector in our sense, but distance is: position doesn't have direction and magnitude, but distance does. Adding positions doesn't make sense but adding distances does.
• The graphical method of vector operations has to be introduced before one can talk about coordinates, because coordinate systems use the vector operations.
• Mention that scalar product works in all dimensions but vector product only in dimension three.
• Scalar product is called "symmetric", not "commutative".
• Use bold lower case letters for vectors throughout.
• It needs to be mentioned that vectors form a vector space.

--AxelBoldt.

Position, at least in 3-space, is represented as: (x, y, z), right? What is the difference between using that and [x y z]? Position, you might say, is implicitly a vector. Distance is not. If a position vector is r then distance = |r| or |Δr| (Δr, by the way, is displacement).
I agree, distance is the wrong word and is indeed a non-negative scalar. Displacement is the word I was looking for and it has since been changed in the main article. While it is true that position is typically described with three coordinates, one cannot really call it a vector in the sense of our article: what is the direction of a position? what is the length or magnitude of a position? If I'm in Saint Paul, MN and you are in Houston, TX, where is the sum of our two positions? --AxelBoldt
Position, as I admitted below, is really a bound vector. You might think of position as displacement with respect to the origin. Since position is bound to the origin, you cannot add two positions, but you can add a position and a displacement. The reason is, as another author's name, bound vector, for it suggests, is that these vectors are bound to the origin, and thus cannot be added to any other bound vector that does not end on its tip, since addition of vectors requires the ability to place them tip to tail.
Lemme think for a minute on the adding part, but displacement is defined as the difference between 2 position vectors. All of this stuff is pure horse puckey unless you know where all the manifolds are (See below). Then it all makes sense. Maybe one day, when I don't have to write boring sockets code to pay the rent, I will write a nice monograph on the topic. In the meantime, to paraphrase Janich, one may spend a pleasant afternoon pondering the origin with respect to what?

Whether or not position is a vector depends on how you treat it. For a generic manifold, velocity is a vector belonging to the tangent space of that manifold at whatever point, and position is a point which need not be a vector. When no origin is specified, the same applies to flat space, but once one is chosen the space becomes a vector space and position becomes a vector in a natural way. Note that a position vector may be defined as the displacement between the origin and a given point, so the above argument doesn't quite work.

I fail to see how one can speak of a position without assuming the existence of a coordinate system. On manifolds we have charts, and you can't have a tangent space to a point unless that point is defined in terms of at least one chart.

Ordered n-tuples of coordinates and vectors are not the same thing, though. Vectors are defined in a coordinate-free way all the time, and n-tuples don't define vectors unless they form a linear space in some natural way, which isn't how we usually think of charts on spheres. And, btw, the charts are only there to establish a concept of differentiability - for manifolds where this concept arises naturally, like Lie groups, they can be ignored entirely.

Which is a very good point, should probably be mentioned on the main page, and is interesting to think about.

Position may certainly be expressed as a vector. Distance, however, is not. Distance is a scalar induced by the action of a distance function working on elements of a metric space. Distance is by definition equal to or greater than 0. Since there is no negative distance, distance cannot satisfy the definition of elements of vector spaces.

Isn't the commutative property of multiplication: ab = ba? I recall my math prof referring to it as commutative, plus, the properties of the vector product are named in terms of commutative and assosiative, so I felt it fitting.

Yeah, for binary operations it's called commutative, for scalar products and such it's called symmetric. Probably because it is closely related to symmetric matrices. --AxelBoldt

Do we really need to make the definition of the cross product dependent on the handedness of the coordinate system? I find that really ugly since coordinate systems are just crutches that shouldn't enter the theory. Why not simply declare that a, b, a×b always form a right-handed system? --AxelBoldt

Because in a left-handed system the cross product, the way it is usually worked, does take on the opposite value. Remember a cross product isn't a vector, it is a pseudovector. The two can be mapped to each other easy enough, but there are two possibilities related by a reflection - in short, though the coordinate system is incidental, the handedness isn't. And remeber that mathematically there's no a priori reason to prefer one handedness over the other.

Cross products are usually' defined as positive in right-handed frames. Really, a cross product is just a specific exterior form, tied to a certain orientation of E^3. The exterior forms are much more powerful, and do not depend on any particular choice of coordinate system. The conundrum here is really that the handedness doesn't really matter, what matters is the sign of the result. Thus, cross-products (exterior forms really) make really spiffy integration functions for any domain well-approximated by peicewise linear boundaries.

I will confess that I don't know what a pseudovector is. I agree that there's no reason to prefer right handedness over left handedness, but for example Encyclopedia Britannica defines the cross product so that a, b and a×b always form a right handed system; some convention apparently is needed. I don't really understand what you mean when you say "in a left handed system, the cross product takes on the opposite value". How about if I have two vectors, thought off as arrows in space, and I form their cross product. This should be possible without talking about coordinate systems at all, no? After all, what if I happen to choose a left-handed coordinate system and you choose a right handed one, will we compute different cross products? And what about the third guy who doesn't like coordinate systems and prefers to work geometrically? --AxelBoldt

Ok, the usual differential geometry approach is this, and you can decide how relevant it is. The pseudovectors are simply the antisymmetric 3x3 tensors, which expressed in coordinates are simply the arrays of this form:

```[  0  a  b ]
[ -a  0  c ]
[ -b -c  0 ]
```

So these form a 3-D space separate from those of ordinary vectors [ x y z ]. Under a linear transformations of positive determinant (or the corresponding transform for the space of tensors) the components a,b,c vary exactly like the components x,y,z, but under a transformation of negative determinant they end up with opposite values. Of course these are just those given components - the underlying objects are coordinate-independent.

Now the space of pseudovectors and the space of vectors have the same dimension, so it is tempting to associate the two, which would for instance allow one to represent pseudovectors by an arrow. Thanks to the above difference, though, this can't be done without disallowing reflections, and in that case one ends up with two entirely equivalent representations. Once coordinates are introduced these can be seen to related to the use of right-handed and of left-handed systems, so we don't talk about either being the vector associated with our pseudovector in a coordinate independent way. I hope that makes sense. I suppose we might want to overlook it, but it's more important than the is-position-a-vector stuff above.

I don't think we should overlook it, but I tend towards putting this stuff into the Tensor article (which needs to be rewritten), in order to keep this vector article a little self contained introduction to vectors for engineers and starting physics students. Mathematicians look at all this in horror anyway. --AxelBoldt

Another way to describe "vectors" such as moment (i.e. cross-product) is by the term bound vector. These vectors can't be slid around arbitrarily like a regular vector. They have to be fixed with respect to some point.

The difference between a left handed coordinate system and a right handed one is literally whether you use your left or right hand to figure out the cross product. Also note that the cyclic permutation of unit vectors in the cross product I put in the initial article holds true in either a left or right handed system. As for the guy who likes to work in geometry, he'll probably have to face indecision as to where the vector points, or worse, arbitrarily pick the direction of the cross product. The deal of handedness is simply all about consistency. From what I understand of a vector, one of it's properties is that it transforms a certain way under a change of coordinate systems. Specifically, the vectors must still be equal. Because the cross product relies on the handedness of the coordinate system, the direction of the vector may change under such a transform. I must admit that position is another such psuedo vector, though displacement is not.
This isn't what pseudovector means - see above.

In a sense, position is a bound vector, because it depends on the choice of coordinate systems. That is to say, position with respect to what? The displacement may also be a bound vector in the sense that displacement must be taken with respect to a local origin, it becomes a position vector with respect to its initial position. The key point is that bound vectors do not all reside in the same vector space.

Everything I was taught about vectors in statics, dynamics and physics is a load of horse puckey. In those classes it was just memorize a bunch of rules. No rhyme or reason. In point of fact, all of these objects may reside in different vector spaces and the operations governing their interaction are precise. Moments are in the product space between force and position. Work is a linear functional between displacement (position vector wrt to a point on a manifold) and its dual, the space of "force" vectors. Unfortunately, a sensible theory of geometry is not taught to undergrads in the US.

The main part of the page should probably stand pretty close as is, because most casual readers and students will need go no further. There should probably be a link to differential geometry somewhere though.

" Examples are displacement?, velocity, momentum and acceleration. One also consideres bound or fixed vectors which are characterized by magnitude, direction and base point. Examples of these are force, torque and [angular momentum]?."

If acceleration is not a bound vector, how can force be if force is the product of a scalar and an acceleration (well, you might argue that definition is precisely backwards, acceleration is the division of force by a scalar)?

Ok, that makes sense, but what I'm concerned about is: if you have two forces acting on a body and if the forces were unbound vectors, you could just add them to get the resultant force. But in fact you cannot do that; it's only allowed if the forces attack at the same point. Doesn't that make forces bound vectors? --AxelBoldt
That conflict comes from oversimplifying a solid body. Forces proper are only applied to point masses, and the point masses then interact to produce the "problem" above. In a sense you might say that the forces are then bound to the point masses, but then acceleration would also have to be bound because there is no such thing as a acceleration without force ( and vice-versa). Also, they don't have to have the same point of attack, they just have to have no net torque (a bound vector that may be what's causing the confusion) about the center of mass (or some other significant rotational axis).--BlackGriffen

I also re-added position as a bound vector (see above for reasons why).

But what does it mean to "add" two positions? What is the length and direction of a position? Position doesn't seem to fit the definition of "quantity characterized by magnitude and direction". --AxelBoldt
Position is a displacement who's tail is bound to the origin. Think about the geometric interpretation of adding vectors: it requires you to move the vectors tip to tail. Doing so changes one position vector so that it doesn't have the special title "position" any more. Positions can, however, be subtracted from each other and added to displacements. Also, defining position as a bound vector explains why torque and angular momentum are bound (the cross product is no sufficient for this. A bound vector will transform differently from a vector [read: vectors proper don't transform at all] if there is any change of coordinates, cross products only change if the handedness of the system changes).--BlackGriffen
It almost seems to me as if position is a third kind of vector, in addition to bound and free. Because two torques bound to the same point can easily be added or subtracted and the result is a meaningful torque, still bound to the same point. But two positions cannot really be added meaningfully, and their difference yields a free vector (displacement).
Maybe the confusion stems from the fact that positions, if represented in Cartesian coordinates, are of course elements of a vector space, and as such "vectors". That however is not the concept of geometrical vector the article talks about.
Is there any other way to defin position but as desplacement from the origin?
I suppose that position is unique in that respect (not being additive), because I tried to come up with a good mathematical definition of bound vector from the example, and this is the best I could come up with:

When transforming from one reference frame to another that is related to the first by only a translation, the following occurse:

1. True vectors do not transform.
2. Pure pseudovectors do not transform.
3. Pure bound vectors transform thus:
anew = aold - btransform

When the transformation only involves a change in handedness (i.e. the transformation matrix is something of the form [±1 0 0] [0 ±1 0] [0 0 ±1]):

1. True vectors transform via multipliction with the matrix
2. Pure bound vectors do the same
3. pseudo-vectors do not transform (even though they point in the opposite direction).

Maybe this will clear things up.--BlackGriffen

"Everything I was taught about vectors in statics, dynamics and physics is a load of horse puckey. In those classes it was just memorize a bunch of rules. No rhyme or reason."

Not really. The rules were memorized because they accurately describe what the physical world does. Emperical rules are in one sense more fundamental than mathematically derived ones because, as non-Euclidian geometries proved, mathematics requires that one knows basic postulates in order to derive everything within mathematics correctly. Some would say that one must accept the postulates a priori, but they are, in fact, just imperical rules. For vectors it just so happens that these rules have a mathematical proof from other emperical rules.--BlackGriffen