Position Vectors in Cartesian XYZ Coordinates

[Ratzinger]

Given the typical cartesian xyz- coordinate system, is it correct to speak of a position vector? Isn't (x,y, z) just shothand for the coordinates? Distance vectors, force, velocity are real vectors with magnitude and direction in position space, but what is with a position vector in position space?

I'm confused, help needed

 

[whozum]

The position vector points from the origin to the position of the particle. For your point it is < x , y , z >

Positions that vary with time are expressed as functions of x y and z, and you get parametric functions:

\vec{r}_t = &lt; x(t) , y(t), z(t) &gt;

 

[Ratzinger]

yes, but isn't what you describe rather a parameterized curve, not these geometrical objects with head and tail that you can move parallel around?
I mean vectors in position coordinates without any time parameter mentioned.

 

[FredGarvin]

Your question is a bit confusing to me. A specified position vector without any time reference just means that the function that produces the curve for the thing's position in space has been evaluated at some specific time. Ultimately the position has to be a function of time or how can you come up with velocity and acceleration?

 

[pmb_phy]

Given the typical cartesian xyz- coordinate system, is it correct to speak of a position vector? Isn't (x,y, z) just shothand for the coordinates? Distance vectors, force, velocity are real vectors with magnitude and direction in position space, but what is with a position vector in position space?
I'm confused, help needed

A displacement vector is a vector which has its tail at one point and its tip at another point. If we refer to the point where the tail is as the "origin" then we refer to the displacement vector as the "position vector" of the point at the tip. We can label that point in any way we please, not only as R = (x, y, z). Quite often you'll see the position vector expressed as

R = (r, \theta, \phi)

This means that R has the value

R = A_r e_r + A_{\theta} e_{\theta} + A_{\phi} e_{\phi}

where e_k is a unit vector pointing in the direction of the increasing kth coordinate. The coeficients have the value A_k = R*e_k ("*" = dot product). The position vector is a bound vector, i.e. attached to the space in which it lies.
Pete

 

[vaishakh]

position vector is just a standard way of showing vectors and there is nothing like being attached. just the vector diagram has no physical meaning. the aim of physics is to solve the vector equations and while doing so we sometimes take the help of diagram.

 

[Ratzinger]

thanks for the replies, I found this

An important difference between a position vector R and a general vector such as delta R is that the components of R are x, y and z, whereas for delta R the components are delta x, delta y and delta z. It is important to distinguish between true vectors and position vectors. A true vector does not depend on coordinate system but only on the difference between one end of the vector and the other. A position vector, in contrast, does depend on the coordinate system, because it is used to locate a position relative to a specified reference point.

 

[pmb_phy]

It is important to distinguish between true vectors and position vectors. A true vector does not depend on coordinate system but only on the difference between one end of the vector and the other.

Where did you get this definition from?? A vector is a geometrical object which does not depend on the coordinate system. Any vector may be expressed in terms of other vectors which are related to a particular coordinate system. But that doesn't mean that it is defined by the coordinate system. I thought I made that clear above but I guess not. I did explain that a vector is an arrow. Some vectors are called "Bound vectors" while others are called "free vectors." This is an important distinction that you should learn. The position vector does not depend on any coordinate system whatsoever. It depends on a geometric object, i.e. a "point." This point I speak of is known as the "reference point."

A position vector, in contrast, does depend on the coordinate system, because it is used to locate a position relative to a specified reference point.

Wrong. You're confusing "reference point" with "coordinate system."
They are very different things.

For details on what I've been talking about see Thorne and Blanchard's online notes at

http://www.pma.caltech.edu/Courses/ph136/yr2004/0401.1.K.pdf

Pete

 

[Ratzinger]

thanks Pete for the great notes you linked

So we have bound and free vectors. That's a nice distinction.

Free vectors can be parallel moved around or can be represented in different coordinate systems, they keep their meaning. Bound vectors do not, because they can clearly not be moved around and keep meaning and when coordinates systems are changed we talk about distance vectors. Bound vectors are bound to a coordinate system. A position vector gives only for his coordinate system information.

 

[pmb_phy]

thanks Pete for the great notes you linked
So we have bound and free vectors. That's a nice distinction.
Free vectors can be parallel moved around or can be represented in different coordinate systems, they keep their meaning. Bound vectors do not, because they can clearly not be moved around and keep meaning and when coordinates systems are changed we talk about distance vectors. Bound vectors are bound to a coordinate system. A position vector gives only for his coordinate system information.

Bound vectors are not bound to a coordinate system. They're bound to the space that the vector is in and that is independant of the coordinate system. Give me any position vector and I can easily represent it in three different coordinate systems. I clarified this point here as I recall

http://www.geocities.com/physics_world/ma/coord_system.htm

Pete

 

[Ratzinger]

Distance, velocity, force -all coordinate-independent quantities. But to ask what the position is of a point clearly needs a coordinate system. Position is only meaningful with a reference system.

Position vector in position space is a misnomer (in any position coordinates). As much as a momentum vector in momentum space.

 

[HallsofIvy]

Your original question is a very good one. In a Cartesian coordinate system, we can think of the "position vector" as the vector from the origin to the point, so that, at one instant, the tip of that vector is the point and the curve the point moves on is the curve sketched by the tip of the vector.

However, in non-Cartesian coordinates, especially in situations, such as are common in General Relativity, where we have curved surfaces that admit no Cartesian coordinate system, there is no such thing as a "position vector". (I used to worry about what vectors on the surface of a sphere 'looked like"!)
It is much better to think in terms of tangent vectors at every point. I like to think "a vector is a derivative".

 

[pmb_phy]

Distance, velocity, force -all coordinate-independent quantities. But to ask what the position is of a point clearly needs a coordinate system. Position is only meaningful with a reference system.
Position vector in position space is a misnomer (in any position coordinates). As much as a momentum vector in momentum space.

Given a reference point the position of an arbitrary point requires only a magntitude and a direction - i.e. a vector.

HallsofIvy - I believe that you're confusing coordinate spaces with the space itself. A position vector can be defined in all spaces which have no curvature. Curvature exists independant of a coordinate system. There is no position vector in a curved space.

Pete

 

[robphy]

I think the issue underlying the OP's questions is the distinction between a vector space and an affine space.
http://mathworld.wolfram.com/AffineSpace.html
http://en.wikipedia.org/wiki/Affine_space
A vector space has an "origin", whereas an affine space does not.
Positions are elements of an affine space.
Displacements (the difference of two positions) are elements of a vector space.
(Only after one assigns a norm can one talk of a "magnitude" of a vector.)

 

[lightgrav]

WHY is the distinction between free vectors and bound vectors IMPORTANT?

I can add and subtract position vectors, even tho they're "bound" to origin.
Yet, the Torque due to a Force does NOT treat the Force as a "free" vector.

 

[pmb_phy]

WHY is the distinction between free vectors and bound vectors IMPORTANT?

I can add and subtract position vectors, even tho they're "bound" to origin.
Yet, the Torque due to a Force does NOT treat the Force as a "free" vector.

The physical meaning of adding position vectors is to start with one vector whose tail is at the origin. The next position vector added to this one has its tail at the tip of the first one. So this vector is bound as well and is bound at a different place. Tourque is the cross product of a bound vector and a free vector making it a free vector.

Other physical vectors are things like the center of mass vector.

Pete

 

[pmb_phy]

I think the issue underlying the OP's questions is the distinction between a vector space and an affine space.
http://mathworld.wolfram.com/AffineSpace.html
http://en.wikipedia.org/wiki/Affine_space
A vector space has an "origin", whereas an affine space does not.
Positions are elements of an affine space.
Displacements (the difference of two positions) are elements of a vector space.
(Only after one assigns a norm can one talk of a "magnitude" of a vector.)

Rob - There is no requirement for a vector space to have an origin. The term "Origin" refers to a particular point that an observer uses as a reference point. The user may also use other points in order to clearly define directions. There is no unique point in any space which demands to be an origin. From your links I don't see what you mean by "affine space." Can you elaborate for me? Thanks.

Pete

 

[HallsofIvy]

Rob - There is no requirement for a vector space to have an origin. The term "Origin" refers to a particular point that an observer uses as a reference point. The user may also use other points in order to clearly define directions. There is no unique point in any space which demands to be an origin. From your links I don't see what you mean by "affine space." Can you elaborate for me? Thanks.
Pete

Yes, there is a requirement for a vector space to have an "origin". One of the requirements for a vector space is that there be a 0 vector. That is what you are referring to as an "origin". An "affine space" is a set of points such that any line through one of the points is contained in the space. You can think of it as a plane or 3d space or any Rⁿ without a coordinate system. Once you add a coordinate system (so that you have an origin), you can make it a vector space. An affine space is what you seem to be thinking of as a "vector space". You can add vectors, you can't add points in an affine space.

 

[HallsofIvy]

Given a reference point the position of an arbitrary point requires only a magntitude and a direction - i.e. a vector.
HallsofIvy - I believe that you're confusing coordinate spaces with the space itself. A position vector can be defined in all spaces which have no curvature. Curvature exists independant of a coordinate system. There is no position vector in a curved space.
Pete

You're right. I referred to "coordinate systems" when I really meant the space itself.

 

[robphy]

Here's some physical motivations on this issue concerning an affine space vs. a vector space.

A space of positions in a plane (or the space of times on a line) is an affine space. With no point being physically distinguished from any other, an affine space is more natural than a vector space. In an affine space, there is no sense of addition of elements of this space. (If you attempt to add two elements, the sum depends on the choice of an "origin" [which does not exist in an affine space]. As was mentioned, one can introduce an origin by introducing a coordinate system. Then, the sum now depends on a choice of coordinate system.) There is, however, a sense of subtraction... the "difference of two positions [in an affine space]" is a vector... the displacement vector. (The difference does not depend on a choice of [or existence of] an origin.)

 

[pmb_phy]

Yes, there is a requirement for a vector space to have an "origin". One of the requirements for a vector space is that there be a 0 vector.

I believe you're confusing the usage of the term vector space here. Rob used the term "vector space" in this thread, a thread whose topic is the position vector. I believe this is a misuse of the term "vector space." There is no requirement that the vector being discussed here to be an element of a vector space in the formal sense of that term. In fact this could prove to be trouble in certain case. Take a square as an example of a space. Then the sum of two position vectors may not lie in the area of the space and thus the square fails to be a vector space. We are not defining "vector" here as an element of a vector space. We are defining it in the geometric sense of the term. The space in R³ does not have a unique point which is called origin. The origin is an arbitrarily chosen point in the space and is not inherent to the space.

The definition of vector, both displacement and tangent, is found on a web page I'm building. I put it online so that the definition can be seen

http://www.geocities.com/physics_world/glossary_of_terms.htm

I need to correct that because there is another way to define "vector" as in "element of a vector space."

Pete

 

[Ratzinger]

A space of positions in a plane (or the space of times on a line) is an affine space. With no point being physically distinguished from any other, an affine space is more natural than a vector space. In an affine space, there is no sense of addition of elements of this space. (If you attempt to add two elements, the sum depends on the choice of an "origin" [which does not exist in an affine space]. As was mentioned, one can introduce an origin by introducing a coordinate system. Then, the sum now depends on a choice of coordinate system.) There is, however, a sense of subtraction... the "difference of two positions [in an affine space]" is a vector... the displacement vector. (The difference does not depend on a choice of [or existence of] an origin.)

But what does that imply for my initial question. What is the nature of position vectors? Is the term position vector inappropriate since position is not a vector quantity but a coordinate –dependent concept?
And then this.
\[<br /> (x,y,z) = x\left( \begin{array}{l}<br /> 1 \\ <br /> 0 \\ <br /> 0 \\ <br /> \end{array} \right) + y\left( \begin{array}{l}<br /> 0 \\ <br /> 1 \\ <br /> 0 \\ <br /> \end{array} \right) + z\left( \begin{array}{l}<br /> 0 \\ <br /> 0 \\ <br /> 1 \\ <br /> \end{array} \right)<br /> \]
Here (x,y,z) are components and three orthogonal unit vectors the basis.
But in my quantum mechanis text I find the following.
<br /> \[<br /> \left| {x_i \rangle } \right. \leftrightarrow \left[ \begin{array}{l}<br /> 0 \\ <br /> 0 \\ <br /> . \\ <br /> . \\ <br /> 1 \\ <br /> . \\ <br /> . \\ <br /> 0 \\ <br /> \end{array} \right] \leftrightarrow i{\rm{ th place}}<br /> \]<br />
Which is the (infinite dimensional) basic vector. And the three dimensional position basis is defined like that.
<br /> \[<br /> \left| {\rm{x}} \right\rangle \otimes \left| y \right\rangle \otimes \left| y \right\rangle = \left| {xyz} \right\rangle = \left| r \right\rangle <br /> \]<br />
And it says that this one basis vector.
What shall I make of all that? What now are coordintes, base vectors, coordinate bases?

 

[pmb_phy]

As was mentioned, one can introduce an origin by introducing a coordinate system.

That is sufficient but it is not necessary. I can introduce an origin without introducing any coordinate system whatsoever.

But what does that imply for my initial question. What is the nature of position vectors? Is the term position vector inappropriate since position is not a vector quantity but a coordinate–dependent concept?

The term "position vector" is very appropriate because it is not a coordinate dependant concept. I fail to understand why you keep insisting that it is!? If you keep thinking that r = (x,y,z) then you're in big trouble and you've failed to understand this vector. (x,y,z) is only a representation of the position vector as expressed in Cartesian coordinate systems.
Many physicists actually use the position vector as the prototype of a vector when they choose to define "vector" in terms of transformation properties. So in that case the position vector is a vector by definition. In fact if you have Classical Electrodynamics - Second Edition by J.D. Jackson then turn to section 6.11 which starts on page 245. It will explain this. Basically a "vector" is an object whose Cartesian components transform in the same way as the Cartesian components of r = <x, y, z>.
So that you understand why I'd like to see you stop using <x, y, z> then express the position vector in spherical coordinates. You'll see that its simly r = r e_r = r(r, \theta, \phi). e_r is a unit vector which points in the direction of the position vector and is a function of \theta and \phi.

Have you studied vector analysis in other coordinate systems?

When you study special relativity then you'll see the 4-position which can be expressed in Lorentz coordinates as X = (ct, x, y, z). This is the prototype for Lorentz 4-vectors.

Pete

 

[robphy]

Read the discussion here http://www.acs.ucalgary.ca/~alphy/MAP/workinprogr/Pos/ItemDescr.htm
It raises an additional point that "positions" can't be scaled (i.e. scalar multiplied), as well as can't be added. Let P denote the position ("location") of an object. It makes no sense (that is, there is no physical, coordinate-independent interpretation) to multiply P by 2, or to add another position Q to form P+Q. Thus, the space of such P's is not a vector space. P is not a vector. [Note, I am not referring to assigning coordinates to P. If you wish, you can assign a triple of numbers to P, effectively introducing a coordinate system. Then, then triple (0,0,0), the origin of that imposed coordinate system, is a position, call it "oh" O. One can draw an arrow from O to P... this is a displacement vector, OP, from the origin of the imposed coordinate system. Of course, another assignment of coordinates locates the new origin at another point O', which yields a different displacement vector O'P. Certainly, 2(OP) is generally different from 2(O'P), and (OP+OQ) is generally different from (O'P+O'Q). However, (OQ-OP)=(O'Q-O'P)=PQ... the choice of coordinate origin and coordinate system is irrelevant when forming displacements between two positions P and Q.]

http://mathworld.wolfram.com/AffineCoordinates.html is probably relevant to this discussion.

If I recall correctly, there is a section in Bamberg and Sternberg's

Let me make a comment, which may or may not help:
not all "configuration spaces" are vector spaces... however, the "velocity space" (the tangent space) is a vector space.

 

[robphy]

As a postscript, look at this paper by Wald on teaching General Relativity: http://arxiv.org/PS_cache/gr-qc/pdf/0511/0511073.pdf
In particular, the first paragraph says...

goes against what students have been taught since high school (or earlier): namely, that
“space” has the natural structure of a vector space.

..and on page 4:

However, very few students have any inkling that, in nature, the points of space and/or
the events in spacetime fail to have any natural vector space structure. Indeed, the con-
cept of a “vector” is normally introduced to students early in their physics education
through the concept of “position vectors” representing the points of space! Students are
taught that, given the choice of a point to serve as an “origin”, it makes sense to add
and scalar multiply points of space. The only significant change introduced by special
relativity is the generalization of this vector space structure from space to spacetime: In
special relativity, the position vector ~x representing a point of space is replaced by the
“4-vector” xμ representing an event in spacetime. One can add and/or scalar multiply
4-vectors in special relativity in exactly the same way as one adds and/or scalar multiplies
ordinary position vectors in pre-relativity physics.
This situation changes dramatically in general relativity, since the vector space char-
acter of space and/or spacetime depends crucially on having a flat geometry. In general
relativity, it does not make any more sense to “add” two events in spacetime than it would
make sense to try to define a notion of addition of points on the surface of a potato.

(Spin_network posted links to papers in the SR/GR forum https://www.physicsforums.com/showthread.php?t=100021 .)

 

[Ratzinger]

thanks robphy for these very helpful posts and the nice links
(Hope Katrina left your house alone)

Has someone the patience to give me some hints on my second problem which went like this:


\[<br /> (x,y,z) = x\left( \begin{array}{l}<br /> 1 \\ <br /> 0 \\ <br /> 0 \\ <br /> \end{array} \right) + y\left( \begin{array}{l}<br /> 0 \\ <br /> 1 \\ <br /> 0 \\ <br /> \end{array} \right) + z\left( \begin{array}{l}<br /> 0 \\ <br /> 0 \\ <br /> 1 \\ <br /> \end{array} \right)<br /> \]
Here (x,y,z) are components and three orthogonal unit vectors the basis.
But in my quantum mechanis text I find the following.
<br /> \[<br /> \left| {x_i \rangle } \right. \leftrightarrow \left[ \begin{array}{l}<br /> 0 \\ <br /> 0 \\ <br /> . \\ <br /> . \\ <br /> 1 \\ <br /> . \\ <br /> . \\ <br /> 0 \\ <br /> \end{array} \right] \leftrightarrow i{\rm{ th place}}<br /> \]<br />
Which is the (infinite dimensional) basic vector. And the three dimensional position basis is defined like that.
<br /> \[<br /> \left| {\rm{x}} \right\rangle \otimes \left| y \right\rangle \otimes \left| y \right\rangle = \left| {xyz} \right\rangle = \left| r \right\rangle <br /> \]<br />
And it says that this one is the basis vector.
What shall I make of that? What now are coordinates, base vectors, coordinate bases?

 

[pmb_phy]

Read the discussion here http://www.acs.ucalgary.ca/~alphy/MAP/workinprogr/Pos/ItemDescr.htm
It raises an additional point that "positions" can't be scaled (i.e. scalar multiplied), as well as can't be added. Let P denote the position ("location") of an object. It makes no sense (that is, there is no physical, coordinate-independent interpretation) to multiply P by 2, or to add another position Q to form P+Q. Thus, the space of such P's is not a vector space. P is not a vector.

Rob - Why is it that you keep on insisting that geometric vectors must form a vector space? These are two very different kinds of animals. One is defined through geometry while the other kind is defined according to a mathematical definition according to whether they fit a certain mathematical criteria.

I guess I'll have to bow out of this thread since I can't seem to get an answer to my question. I'll therefore not read this thread again. If there is a response to my questions (which people don't seem to want to answer) then please PM me so that I can know I did get an answer.

Thanks folks

Pete

Note: Let me be more precise on this. There are two uses of the term "vector" in mathematics. They are as follows

(1) Vector - A directed line segment
(2) Vector - An element of a vector space

These two definitions are not identical. According to definition #1 a vector is defined if one specifies both a magnitude and a direction. There is nothing which says these kinds of vectors must be able to be added to each other and the result must be a vector. That is confusing the first definition with the second. However one can give the sum a meaning. The tip of one position vector is taken as a new origin for the position vector to be added. E.g. "A = West 1 mile" is a position vector. "B = East 1 mile" is a position vector. "A + B = Northeast \sqrt{2} miles" is a position vector. The physical meaning of this suming procedure we define as "Walk West 1 mile and then walk north one mile. You'll end up \sqrt{2} miles Northwest of where you started. Thus robs link above was written by someone who really doesn't grasp what a geometrical vector is since under definition #1 there is nothing in the definition of the vector which demands that addition must have a phyical/geometrical meaning, unlike definition #2. However we can give it meaning of which I just gave an example above.

According to definition #2 a quantity is a vector if and only if it is an element of a vector space.

rob - I've already said that I will not read anything else in this thread and will read only a PM sent to me. It's nothing personal. I had to set some rules for myself in oder to cut stress in my life. Stress has proved to be extremely dangerous for me at this time in my life. Once I make a rule for myself I refuse to break it. Sorry if that is inconvenient for anyone.

 

[robphy]

Pete,
The "geometric vector", apparently defined on your glossary as an ordered pair of points [highlighted by drawing a directed line segment] plus the condition that it "has both a magnitude and a direction", is a specialized type of vector. To define the "magnitude of a vector" one has to introduce a norm or inner-product to a vector space. The "magnitude and direction" definition of a vector is okay for intro physics... but not good enough for more mathematical treatments.

The key property of a vector is that it satisfies the parallelogram rule of addition. This property does not require any notion of magnitude.

 

[robphy]

Ratzinger, this is a notational issue.

The right-hand side of your first expression can be written symbolically as
x \hat x + y \hat y + z \hat z
or, to possibly reduce confusion,
x \hat \imath + y \hat \jmath + z \hat k.
To be fancy, one can write
x \hat e_x + y \hat e_y + z \hat e_z
or
r_1 \hat e_1 + r_2 \hat e_2 + r_3 \hat e_3, which lends itself to generalization to higher dimensions. In the spirit of the first form, one could write this as
x_1 \hat x_1 + x_2 \hat x_2 + x_3 \hat x_3
In each of the above, I'm sure you can pick out the scalar components from the basis vectors.

In QM, rather than the hat, one often uses the ket. So, the object \hat x_1 is the same as |x_1&gt;. Written in ket notation, the above is
x_1 | x_1&gt; + x_2 | x_2&gt; + x_3 | x_3 &gt;.

I'm not sure if this addresses your question.