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Position vector

  1. Nov 10, 2005 #1
    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
     
  2. jcsd
  3. Nov 10, 2005 #2
    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:

    [itex] \vec{r}_t = < x(t) , y(t), z(t) > [/itex]
     
  4. Nov 10, 2005 #3
    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.
     
  5. Nov 10, 2005 #4

    FredGarvin

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    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?
     
  6. Nov 10, 2005 #5
    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

    [tex]R = (r, \theta, \phi)[/tex]

    This means that R has the value

    [tex]R = A_r e_r + A_{\theta} e_{\theta} + A_{\phi} e_{\phi}[/tex]

    where ek is a unit vector pointing in the direction of the increasing kth coordinate. The coeficients have the value Ak = R*ek ("*" = dot product). The position vector is a bound vector, i.e. attached to the space in which it lies.
    Pete
     
    Last edited: Nov 10, 2005
  7. Nov 10, 2005 #6
    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.
     
  8. Nov 11, 2005 #7
    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.
     
  9. Nov 11, 2005 #8
    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."
    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
     
  10. Nov 11, 2005 #9
    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.
     
  11. Nov 11, 2005 #10
    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
     
  12. Nov 11, 2005 #11
    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.
     
  13. Nov 11, 2005 #12

    HallsofIvy

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    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".
     
  14. Nov 12, 2005 #13
    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
     
  15. Nov 12, 2005 #14

    robphy

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    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.)
     
  16. Nov 12, 2005 #15

    lightgrav

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    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.
     
  17. Nov 13, 2005 #16
    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
     
  18. Nov 15, 2005 #17
    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
     
  19. Nov 15, 2005 #18

    HallsofIvy

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    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.
     
  20. Nov 15, 2005 #19

    HallsofIvy

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    You're right. I referred to "coordinate systems" when I really meant the space itself.
     
  21. Nov 15, 2005 #20

    robphy

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    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.)
     
  22. Nov 16, 2005 #21
    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
     
    Last edited: Nov 16, 2005
  23. Nov 16, 2005 #22
    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.
    [tex]\[
    (x,y,z) = x\left( \begin{array}{l}
    1 \\
    0 \\
    0 \\
    \end{array} \right) + y\left( \begin{array}{l}
    0 \\
    1 \\
    0 \\
    \end{array} \right) + z\left( \begin{array}{l}
    0 \\
    0 \\
    1 \\
    \end{array} \right)
    \][/tex]
    Here (x,y,z) are components and three orthogonal unit vectors the basis.
    But in my quantum mechanis text I find the following.
    [tex]
    \[
    \left| {x_i \rangle } \right. \leftrightarrow \left[ \begin{array}{l}
    0 \\
    0 \\
    . \\
    . \\
    1 \\
    . \\
    . \\
    0 \\
    \end{array} \right] \leftrightarrow i{\rm{ th place}}
    \]
    [/tex]
    Which is the (infinite dimensional) basic vector. And the three dimensional position basis is defined like that.
    [tex]
    \[
    \left| {\rm{x}} \right\rangle \otimes \left| y \right\rangle \otimes \left| y \right\rangle = \left| {xyz} \right\rangle = \left| r \right\rangle
    \]
    [/tex]
    And it says that this one basis vector.
    What shall I make of all that? What now are coordintes, base vectors, coordinate bases???
     
    Last edited: Nov 16, 2005
  24. Nov 16, 2005 #23
    That is sufficient but it is not necessary. I can introduce an origin without introducing any coordinate system whatsoever.
    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 er = r(r, [itex]\theta[/itex], [itex]\phi[/itex]). er is a unit vector which points in the direction of the position vector and is a function of [itex]\theta[/itex] and [itex]\phi[/itex].

    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
     
    Last edited: Nov 16, 2005
  25. Nov 16, 2005 #24

    robphy

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    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 https://www.amazon.com/exec/obidos/tg/detail/-/0521406498/

    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.
     
    Last edited by a moderator: Apr 21, 2017
  26. Nov 16, 2005 #25

    robphy

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    As a postscript, look at this paper by Wald on teaching General Relativity: http://arxiv.org/PS_cache/gr-qc/pdf/0511/0511073.pdf [Broken]
    In particular, the first paragraph says...
    ..and on page 4:
    (Spin_network posted links to papers in the SR/GR forum https://www.physicsforums.com/showthread.php?t=100021 .)
     
    Last edited by a moderator: May 2, 2017
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