: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). |
: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). |
:::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:
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]):
Maybe this will clear things up.--BlackGriffen |
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.
--AxelBoldt.
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.
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.
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.
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.
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.
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.
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.
The page had a logical contradiction previously:
" 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)?
I also re-added position as a bound vector (see above for reasons why).
When transforming from one reference frame to another that is related to the first by only a translation, the following occurse:
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]):
"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