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Orthonormal basis vectors for polar coordinate system

  1. Mar 23, 2013 #1
    Firstly; is there a difference between the "regular" polar coordinates that use [itex] \theta [/itex] and [itex] r [/itex] to describe a point (the one where the point [itex] (\sqrt{2}, \frac{\pi}{4}) [/itex] equals [itex] (1, 1) [/itex] in rectangular coordinates) and the ones that use the orthonormal basis vectors [itex] \hat{e}_r [/itex] and [itex] \hat{e}_\theta [/itex]? If there actually is a difference then that is probably the root of my confusion and I would love to get some input on that, but in the case that they are the same; I have a few questions..

    Okay; just like in the rectangular coordinate system, where the vector [itex] (x, y) [/itex] means the first basis vector, say [itex] \hat{e}_x [/itex], times [itex] x [/itex] plus the second basis vector, say [itex] \hat{e}_y [/itex], times y. Shouldn't [itex] \sqrt{2} [/itex] times [itex] \hat{e}_r [/itex] plus [itex] \frac{\pi}{4} [/itex] times [itex] \hat{e}_\theta [/itex] point to the same coordinate as [itex] (1, 1) [/itex] does in rectangular coordinates?

    But if the definition of [itex] \hat{e}_r [/itex] is that it is the unit vector in the direction of the point you're trying to specify, then [itex] r\hat{e}_r[/itex] already equals the point you're trying to specify. So what is the use of a second basis vector if the one basis vector you have times the first component of a vector represented in polar coordinates already gives you the point you want? What happens to [itex] \hat{e}_\theta [/itex]? It is always, by definition, orthogonal to the direction of the point you're trying to specify. Shouldn't it's component be always zero? Which means the [itex] \theta [/itex] component of a vector in polar coordinates is kinda useless?

    I am really confused, any help would be great!
     
  2. jcsd
  3. Mar 24, 2013 #2

    Simon Bridge

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    The notations are equivalent - but you have noticed the special nature of the polar coordinates. The extra "basis vectors" are "co-moving"... so they do not transform the same way as the rectangular components.

    So yep, ##\vec{r}=r\hat{e}_r## and we use the other entries ##\hat{e}_\theta## (with ##\hat{e}_\phi## for spherical) to keep track of just which direction ##\hat{e}_r## points in. Which is why they are not "useless".

    In a way it is the ##\hat{e}_r## that is a bit of an appendix - since it is a unit vector pointing in the direction specified by the angle components.
     
  4. Mar 24, 2013 #3
    right; so ## \vec{r}=r\hat{e}_r ##, but in the rectangular coordinates [itex] \vec{r}=x\hat{e}_x + y\hat{e}_y + z\hat{e}_z [/itex] so what I am still confused about is where does the [itex] \hat{e}_\theta [/itex] and [itex] \hat{e}_\phi [/itex] basis come in. And how they are related to the coordinate vector [itex] (r, \theta, \phi) [/itex]. because surely if I express the polar unit vectors in rectangular coordinates [itex] (\cos\theta, \sin\theta) [/itex] and [itex] (-\sin\theta, \cos\theta) [/itex] shouldn't those time [itex] r [/itex] and [itex] \theta [/itex], respectively, equal the point I would get in rectangular coordinates by treating [itex] r [/itex] as [itex] \sqrt{x^2+y^2} [/itex] and [itex] \theta [/itex] as [itex] arctan(\frac{y}{x})[/itex]?

    Except they don't..
     
  5. Mar 24, 2013 #4

    Simon Bridge

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    They are a convention in notation.
    Ah, your problem is that when you do
    ##r\hat{e}_r = r(\cos\theta\hat{e}_x + \sin\theta\hat{e}_y) = x\hat{e}_x+y\hat{e}_y## ... and ##\theta\hat{e}_\theta = \theta(\cos\theta\hat{e}_y - \sin\theta\hat{e}_x)##
    ... it suggests that ##r\hat{e}_r+\theta\hat{e}_\theta = \vec{r}+\theta(\cos\theta\hat{e}_y - \sin\theta\hat{e}_x)##
    ... see the following:
    http://www.engin.brown.edu/courses/en3/Notes/Vector_Web2/Vectors6a/Vectors6a.htm
    In particular II.1.4.
     
  6. Mar 24, 2013 #5
    So in short, the basis vectors in polar coordinates don't work the same way as the basis in rectangular coordinates do. I mean; the components of a vector are not the "weights" of which basis making the basis vectors sum up to a location in the coordinate system, right?

    It's more like phi and theta specify the direction of ## \hat{e}_r ## and then the component ## r ## scales the unit vector to specify a certain location.

    But just one last question: in II.1.4 the article says "Let ## \vec{a}=a_r\hat{e}_r+a_\theta\hat{e}_\theta+a_\phi\hat{e}_\phi## be a vector." What does that mean? If the basis vectors are orthonormal doesn't that mean that if ## \hat{e}_\theta ## and ##\hat{e}_\phi## aren't zero, ## \hat{e}_r ## doesn't point in the direction of the point I am trying to specify anymore? Because that point now has components that are orthogonal to that direction?
     
  7. Mar 24, 2013 #6
    This type of thing confuse me too for a long time. Now, my understanding of basis vectors in the polar coordinate system is this: [itex]r[/itex] and [itex]\theta[/itex] define a point in space, while [itex]\hat{e}_r[/itex] and [itex]\hat{e}_{\theta}[/itex] describe a basis which happens to be associated with that point in space. Essentially, for every point with an angle [itex]\theta[/itex] we define a vector [itex]\hat{e}_r[/itex] which points from the origin to that point, and a vector [itex]\hat{e}_{\theta}[/itex] which is orthogonal to [itex]\hat{e}_r[/itex]. To help make it very clear that there are different basis vectors for each value [itex]\theta[/itex], let's write these explicitly as functions of [itex]\theta[/itex], so [itex]\hat{e}_r(\theta)[/itex] and [itex]\hat{e}_{\theta}(\theta)[/itex] are the basis vectors associated with points at an angle [itex]\theta[/itex].

    Now there's two things I'd like to mention which hopefully will clear things up for you. First of all, not all vectors are position vectors. You can have a vector associated with a point in space [itex](r,\theta)[/itex], but whose components are not determined by [itex]r[/itex] and [itex]\theta[/itex]. As an example, suppose a particle is located at the point [itex](r,\theta)[/itex]. Because of how [itex]\hat{e}_r(\theta)[/itex] and [itex]\hat{e}_{\theta}(\theta)[/itex] are defined, the position of that particle is always [itex]\vec{r} = r\hat{e}_r(\theta)[/itex]. But let's say there's some force [itex]\vec{F}[/itex] acting on that particle. [itex]\vec{F}[/itex] is not a position vector, but we can still write it in terms of the basis [itex]\left\{ \hat{e}_r(\theta), \hat{e}_{\theta}(\theta) \right\}[/itex] basis. i.e. [itex]\vec{F} = F_r \hat{e}_r(\theta) + F_{\theta} \hat{e}_{\theta}(\theta)[/itex]. There's no reason that [itex]\vec{F}[/itex] has to point towards or away from the origin, so unlike the position vector expression, we have both an [itex]F_r[/itex] term and an [itex]F_{\theta}[/itex] term. That's why it makes sense to write something like [itex]\vec{a} = a_r \hat{e}_r + a_{\theta} \hat{e}_{\theta}[/itex]. Basically what this says is "I'm looking at a point [itex](r,\theta)[/itex], and I've defined a basis [itex]\left\{ \hat{e}_r, \hat{e}_{\theta} \right\}[/itex] associated with that point. Now I have a vector [itex]\vec{a}[/itex] which is somehow associated with that point (maybe it's a force or the velocity of a particle at that point) and I'm expressing it in terms of that basis."

    The second thing I'd like to mention is this. Say we have two points [itex]P_1 = (r_1,\theta_1)[/itex] and [itex]P_2 = (r_2, \theta_2)[/itex]. The position vector for [itex]P_1[/itex] is [itex]\vec{r}_1 = r_1 \hat{e}_r(\theta_1)[/itex] and the position vector for [itex]P_2[/itex] is [itex]\vec{r}_2 = r_2 \hat{e}_r(\theta_2)[/itex]. Now you can hopefully see why I insisted on writing [itex]\hat{e}_r[/itex] as a function of [itex]\theta[/itex]. All position vectors can be written as something like [itex]r\hat{e}_r[/itex], but it's important to remember that this [itex]\hat{e}_r[/itex] is not the same for all position vectors! For example, if I wanted to find [itex]\vec{r}_{12} = \vec{r}_2 - \vec{r}_1[/itex], hopefully you now can see why it's very wrong to say that [itex]\vec{r}_{12} = r_2 \hat{e}_r - r_1 \hat{e}_r = (r_2 – r_1)\hat{e}_r[/itex]. To subtract [itex]\vec{r}_1[/itex] from [itex]\vec{r}_2[/itex], you need to write them in the same basis. For example, I could write [itex]\vec{r}_2[/itex] in terms of [itex]P_1[/itex]’s basis. i.e. [itex]\vec{r}_2 = \alpha \cdot \hat{e}_r(\theta_1) + \beta \cdot \hat{e}_{\theta}(\theta_1)[/itex], and then I could subtract them. The key point here is that "[itex]\vec{r} = r \hat{e}_r[/itex]" is not the only way to write a position vector. If you write it in terms of the basis associated with the point it's describing, it will look like that, but if you write it in terms of the basis for a different point (which you might do if you're subtracting vectors), it may have an [itex]\hat{e}_r[/itex] component and an [itex]\hat{e}_{\theta}[/itex] component.

    In summary, there is a basis [itex]\left\{ \hat{e}_r(\theta), \hat{e}_{\theta}(\theta) \right\}[/itex] associated with every point [itex](r,\theta)[/itex], and you can write any vector (position or otherwise) in terms of the basis at any point. By convention, we usually write vectors in terms of the basis for the point with which they're associated, but you always have to remember what the notation is really hiding.
     
  8. Mar 24, 2013 #7
    Thank you!! That was very helpful! I think what did it for me was to realize I could write other vectors in terms of the basis defined by ## \theta ##.

    Anyway, in response to my first question: they are different things. The whole idea of "polar basis vectors" is actually kind of wrong.. All you're doing is transforming the orthonormal rectangular coordinate system by rotating it around the origin. The only difference is that the basis you get after the transformation occurs is a parameter of some variable ## \theta##. Of course; the "standard" polar coordinates are kind of related as in they're useful for working in this forever transforming basis, but these basis are not the actual basis of THE polar coordinate system.

    Just as an aside: would the basis of the "actual" polar coordinate system be some kind of ## \hat{r} ## and ## \hat{\theta} ## that when properly "weighted" by, for instance, ## (\sqrt{2}, \frac{\pi}{4}) ## equal the point ## (1, 1) ## in rectangular coordinates? If that is right, are these basis orthonormal. I mean; I could think of an orthonormal basis ## \hat{r} ## and ## \hat{\theta} ## that when I write the equation ## r = \theta ## it looks like a regular linear equation with slope 1. But when I represent it in rectangular coordinates it looks like the actual "polar spiral". Is that the right way to go?
     
  9. Mar 24, 2013 #8
    I think you're maybe confused about the difference between basis vectors and coordinates. The trouble is that in rectangular coordinates, there's no real need for distinction. In rectangular, the pair of numbers, or coordinates, (##x## and ##y##) that we use to represent a point are exactly the same as the components of a position vector. That is, we have basis vectors ##\hat{e}_x## and ##\hat{e}_y## so that the point ##(x,y)## corresponds to the position vector ##x\hat{e}_x + y\hat{e}_y##. For that reason, we tend to use the pair of coordinates ##(x,y)## interchangeably with the vector components ##x## and ##y## because they have the same numerical value. What they represent, though, is actually different.

    That difference is important in polar coordinates because the coordinates of a point and components of a position vector corresponding to that point do NOT have the same value anymore. The pair of coordinates ##(r,\theta)## represents the point which is a distance ##r## from the origin and forms an angle ##\theta## with the ##x##-axis (or whatever your reference line is). Now suppose we try to define some basis vectors ##\hat{e}_r## and ##\hat{e}_{\theta}## in such a way that the position vector for the point ##(r,\theta)## is given by ##r\hat{e}_r + \theta\hat{e}_{\theta}##. It can't be done! The values ##r## and ##\theta## do not represent components of a vector, and so there's no "actual" basis for the polar coordinate system in that sense. If you want to transform the pair of values ##(r,\theta)## into a position vector, you have to do something like ##r\cos(\theta)\hat{e}_x + r\sin(\theta)\hat{e}_y##. (I used ##\hat{e}_x## and ##\hat{e}_y## to illustrate my point, but with appropriate modification, you could use any basis. There's no "correct" or "actual" basis.) What I just describes is a vector with components ##r\cos(\theta)## and ##r\sin(\theta)##, and it's associated with basis vectors. ##r## and ##\theta## by themselves are not vector components, and so there are no basis vectors associated with them.
     
  10. Mar 24, 2013 #9
    Wow, thanks!! That actually makes a lot of sense! So just because an ordered array of values specifies a position in a coordinate system it doesn't necessarily mean that they have anything to do with basis vectors or vectors at all.. right?
     
  11. Mar 24, 2013 #10
    Right. Vectors can be written as an ordered list of values, but to be a vector, ordered lists have to satisfy certain conditions/rules. For example, you can add two vectors by adding their components. If you try that with points in polar coordinates, it doesn't work. ##(r_1,\theta_1) + (r_2,\theta_2) = (r_1 + r_2,\theta_1 + \theta_2)## doesn't make much sense.
     
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