Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polar unit vectors form a basis?

  1. Aug 2, 2013 #1
    I keep reading about polar unit vectors, and I am a bit confused by what they mean.

    In the way I like to think about it, the n-tuple representation of a vector space is just a "list" of elements from the field that I have to combine (a.k.a. perform multiplication) with the n vectors in some subset of my vector space, the basis, to result in the particular vector from the vector space I am trying to represent.

    So if I have some subset B that is a basis of some vector space V, I can represent the vector λ in V as the list of scalars I have to multiply each basis vector to get λ in V.

    If the polar unit vectors ## \hat{e}_r ## and ## \hat{e}_{\theta} ## are a basis to ℝ2, how does that work. I mean IF it works..

    If I add the vector (1, π/4) to (1, π/4), I would expect to get (2, π/4). Since both vectors point in the same direction I should get a third vector pointing in the same direction as well. Or does the space get all distorted somehow. Like, does the equation r = θ that looks like a spiral in the cartesian coordinate system looks like a straight line with slope 1 through the origin in the polar coordinate system?

    I know that the polar basis vectors rotate with the angle θ, but if the point you're trying to specify is always in the direction of ## \hat{e}_r ## wouldn't everything always lie on the horizontal axis if my perspective was the polar coordinate system?

    I am confused :/
     
  2. jcsd
  3. Aug 2, 2013 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    You can't add coordinates in the way you described. Component addition is valid only in cartesian (x,y) coordinates.
     
  4. Aug 2, 2013 #3
    right.. so polar unit vector is a misnomer? because I should be able to add vectors.
     
  5. Aug 2, 2013 #4

    WannabeNewton

    User Avatar
    Science Advisor

    You can add any two vectors if they both originate at the same point. The basis vectors span a vector space at each point.
     
  6. Aug 3, 2013 #5

    mathman

    User Avatar
    Science Advisor
    Gold Member

    The point is, you can always add vectors. However, if you want to define unit vectors so you can add coefficients, they need to be orthonormal and have a fixed angle with the (x,y) axes.
     
  7. Aug 3, 2013 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    That is essentially correct. If you have a basis for a vectors space then it is a set of vectors [itex] {b_1, b_2,...} [/itex] such that each vector [itex] v [/itex] can be represented as a sum of scalar multiples of the basis vectors. If the basis is orthogonall, the scalars you use are unique. There is nothing in that concept that involves the concept of Cartesian coordinates. The scalars are the scalar components of the vector [itex] v [/itex]. They are not (necessarily) the Cartesian coordinates of [itex] v [/itex].

    Coordinates are a different concept that the representation a vector with respect to a basis of vectors. A coordinate system must define a map of an n-tuple of numbers to a "point" (or whatever kind thing is being coordinatized) and each point that is to be represented must have some n-tuple of numbers mapped to it. It is permissible for a coordinate system to map several different n-tuples of numbers to the same point.

    Among the many things that can be assigned coordinates are vectors. In physics books there are often some unstated assumptions about the relation between the coordinates of a vector and the basis that is being used to represent the vector. For example, a recent thread concerned a statement in a physics book that the scalar components of a vector don't change under a translation of the xyz coordinates system but they do change with respect to rotations of the xyz coordinate system. That claim is obviously false unless we make the assumption that a change in the coordinate system we use implies a change in the basis vectors we use. The axioms of a mathematical "vector space" don't address coordinate systems. They don't say that changing the coordinate system for a vector space must cause a change in the basis that is used for it.

    You are talking about polar coordinates not a set of vectors. If you have a coordinate system for vectors, there is usually some systematic formula for finding the coordinate of the sum of two vectors from the coordinates of the vectors that are added. However this formula need have no relation to the fact that you can find the scalar components of the sum of two vectors by adding the respective scalar components of the vectors being added.

    In summary, coordinate representation of a vector and representation of it with respect to a basis are two entirely different things. In applied math and physics there is often an assumption (sometime unstated) about a systematic relation between a coordinate system for vectors and the basis being used.
     
  8. Aug 4, 2013 #7
    I think the others who answered are a bit confused because usually polar coordinate refer to specifying a point by its distance from the origin and (if you make a line from the origin to the point), the angle the line makes with the positive [itex] x [/itex] axis. But there is another convention where you use the basis of unit vectors you are talking about and it is more useful in physics. The unusual feature of this polar coordinate basis is that the actual vectors are different for different values of the angle [itex] \theta [/itex] (as you pointed out). You can see the dependence on [itex] \theta [/itex] as follows:
    [tex]
    \hat{e}_r = \begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix}, \,\,\,\,
    \hat{e}_\theta = \begin{pmatrix} -\sin(\theta) \\ \cos(\theta) \end{pmatrix}
    [/tex] So the basis vectors are more of a function of [itex] \theta [/itex]. But your main confusion is that when you work with this basis, then a vector [itex] (a, b)^t [/itex] means that [itex] a [/itex] is the distance from the origin and [itex] b [/itex] is the angle. This is NOT the case. [itex] a [/itex] is still the horizontal component and [itex] b [/itex] is the vertical component. This basis just allows you to change the notation in terms of [itex] r [/itex] and [itex] \theta [/itex] as you can see in the definition of the vectors above (it is polar in that sense only). Working with this basis is very useful for in physics especially for circular motion. You can specify the position of a particle in general by [itex] \vec{r} = r\hat{e}_r [/itex] (remember the basis vector depends on the angle the particle makes with the positive [itex] x [/itex] axis so [itex] \hat{e}_r [/itex] is the unit vector pointing towards the particle and you just multiply [itex] r [/itex] to get the position vector for the particle). To do stuff with velocities and acceleration we need their derivatives:
    [tex]
    \frac{d}{d\theta} \hat{e}_r = \begin{pmatrix} \cos'(\theta) \\ \sin'(\theta) \end{pmatrix} = \begin{pmatrix} -\sin(\theta) \\ \cos(\theta) \end{pmatrix} = \hat{e}_\theta, \,\,\,\,
    \frac{d}{d\theta} \hat{e}_\theta = \begin{pmatrix} -\sin'(\theta) \\ \cos'(\theta) \end{pmatrix} = \begin{pmatrix} -\cos(\theta) \\ -\sin(\theta) \end{pmatrix} = -\hat{e}_r
    [/tex] So, to summarise [itex] \frac{d}{d\theta} \hat{e}_r = \hat{e}_\theta [/itex] and [itex] \frac{d}{d\theta} \hat{e}_\theta = -\hat{e}_r [/itex]. Now to get the velocity just differentiate [itex] \vec{r} [/itex] with respect to [itex] t [/itex] and use product rule and chain rule to get [itex] \vec{v} = \dot{\vec{r}} = \dot{r}\hat{e}_r + r\left(\frac{d}{dt}\hat{e}_r\right) = \dot{r}\hat{e}_r + r\dot{\theta}\left(\frac{d}{d\theta}\hat{e}_r\right) = \dot{r}\hat{e}_r + r\left(\frac{d}{dt}\hat{e}_r\right) = \dot{r}\hat{e}_r + r\dot{\theta}\hat{e}_\theta [/itex]. For uniform circular motion, [itex] r [/itex] and [itex] \dot{\theta} [/itex] is constant so [itex] \dot{r} = 0 [/itex] and letting [itex] \omega = \dot{\theta} [/itex], we have [itex] \vec{v} = r\omega\hat{e}_\theta [/itex] which makes sense because it has constant speed [itex] r\omega [/itex] in [itex] \hat{e}_\theta [/itex] direction. You could have of course just worked with the actual vector (with the sines and cosines) but advantages of these unit vectors are the cleaner notation (you have to write less) and the easy to remember derivatives etc. Hope this cleared up any confusion and showed you the advantages of using these unit vectors. You can read more about it here (they include acceleration which I didn't go into): http://mathworld.wolfram.com/PolarCoordinates.html (in the bottom half) and http://en.wikipedia.org/wiki/Mechanics_of_planar_particle_motion (in the section called Polar coordinates in an inertial frame of reference)
     
  9. Aug 4, 2013 #8

    WannabeNewton

    User Avatar
    Science Advisor

    There is no confusion here. There is a simple reason for why you cannot simply add, component wise, two vectors in polar coordinates that originate at different points: you cannot parallel transport vectors in polar coordinates while keeping the components of the vectors the same like you can in Cartesian coordinates.
     
  10. Aug 4, 2013 #9
    The components DO remain the same at any point. Consider the vector [itex] (1, 0) [/itex]. When this vector is anywhere on the positive horizontal axis, it can be written as [itex] (1, 0) = \hat{e}_r [/itex] and when this vector is on the negative vertical axis, then it can be written as [itex] (1, 0) = \hat{e}_\theta [/itex]. But wherever it is, it is always the vector [itex] (1, 0) [/itex]! It may have different coefficients in terms of the polar unit vectors but the vectors are still the same. And yes you can add component wise because the the components in this notation are still horizontal and vertical components of the vector. They are NOT distance and angle.
     
  11. Aug 4, 2013 #10

    WannabeNewton

    User Avatar
    Science Advisor

    No you cannot. I don't want to go too far from the OP but the equation of parallel transport relies on the Christoffel symbols associated with a given coordinate basis. In the Cartesian basis the Christoffel symbols vanish identically and the equation of parallel transport is trivial: the components of vectors as represented in the Cartesian basis remain constant no matter how and where you move them. In the polar basis the Christoffel symbols do not vanish identically and the components of vectors as represented in the polar basis do not remain constant when you try to parallel transport them hence you cannot trivially add component-wise vectors at different points in the polar basis by parallel transporting them like you would in the Cartesian basis. This is because the basis vectors in polar coordinates change direction from point to point and hence are not constant whereas the basis vectors in Cartesian coordinates are everywhere the same. If you want to trivially add vectors at different points component wise then you must first convert back to Cartesian coordinates, you cannot do so in polar coordinates.
     
  12. Aug 4, 2013 #11
    Yes, I know the basis is different at different points, I wrote than in my previous post. To clarify what I was trying to say, denote [itex] \beta(\theta) [/itex] to be the basis at angle [itex] \theta [/itex] (radius doesn't matter). Now, continuing my previous example, if [itex] v = (1, 0) [/itex], then [itex] v = \hat{e}_r [/itex] for [itex] \theta = 0 [/itex] and [itex] v = \hat{e}_\theta [/itex] for [itex] \theta = -\frac{\pi}{2} [/itex]. So writing [itex] v [/itex] in terms of the basis, we have [itex] [v]_{\beta(0)} = (1, 0) [/itex] and [itex] [v]_{\beta(-\frac{\pi}{2})} = (0, 1) [/itex] (the vectors consisting of coefficients when writing [itex] v [/itex] in terms of [itex] \beta(\theta) [/itex]). Hence we have [itex] [v]_{\beta(0)} = (1, 0) \neq (0, 1) = [v]_{\beta(-\frac{\pi}{2})} [/itex]. In my notation, I think you mean [itex] [v]_{\beta(\theta)} [/itex] when you say "vector in polar coordinates" (in which case you are correct that you can't add two vectors component wise) whereas I was thinking [itex] v [/itex]. The thing is a vector is the same in any basis it is in because vectors are independent of any basis. Only their representations in terms of a basis is different ... what I wrote as [itex] [v]_{\beta(\theta)} [/itex] (which is the vector consisting of the coefficients when writing [itex] v [/itex] in terms of [itex] \beta(\theta) [/itex]). So I think we were in agreement but misunderstanding each other.
     
  13. Aug 4, 2013 #12

    WannabeNewton

    User Avatar
    Science Advisor

    It would seem so, upon reading your latest post. We were just saying the same thing in two different ways :wink:. Cheers!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Polar unit vectors form a basis?
  1. Forming basis of R^3 (Replies: 2)

Loading...