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I Polar coordinates and unit vectors

  1. Feb 1, 2019 #1
    Hello,

    I get that both polar unit vectors, ##\hat{r}## and ##\hat{\theta}##, are unit vectors whose directions varies from point to point in the plane. In polar coordinates, the location of an arbitrary point ##P## on the plane is solely given in terms of one of the unit vector, the vector ##\hat{r}##. Fo example, ##P=3 \hat{r}##. But how do we know where the point is? We only know it is 3 units away from the origin but don't know in which direction. Don't we need a component for the angular unit vector ##\hat{\theta}## as well? Shouldn't the position of point $$P=r \hat{r} + \theta \hat{\theta}$$ where the components are ##(r, \theta)##?
    I understand that the two unit vectors are orthogonal to each other and their direction depends on which point we are considering in the plane...

    thanks!
     
  2. jcsd
  3. Feb 1, 2019 #2

    Mark44

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    No, that's not true. The radial unit vector gives only the distance from the origin -- the point could be anywhere on a circle of that radius.
    We don't, unless we also know the angle.
     
  4. Feb 1, 2019 #3

    PeroK

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    You do need two coordinates to define the point. But, the radial unit vector at that point is a function of ##\theta##. In other words, ##\theta## implicitly determines the direction of ##\hat{r}##.
     
  5. Feb 1, 2019 #4
    Hi Perok,

    do you have a simple example to share? By just reading ##P= 3r \hat{r}##, I don't see how I can identify the specific point on the circumference of radius ##r=3##.

    I see how the two polar unit vectors go in pair and if we know where one points we automatically know the other. But how do we have enough information from just stating ##P=r \hat{r}## without the component for ## \hat{\theta}##?
     
  6. Feb 1, 2019 #5

    PeroK

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    Let's take acceleration in uniform circular motion about the origin. We have, at each point on the circle:

    ##\vec{a} = -a\hat{r}##

    However, ##\hat{r}## depends on the coordinates. So, if you want to refer to a specific point, you also need to specify ##\theta##.

    The acceleration vector in this case has zero component in the ##\hat{\theta}## direction.
     
  7. Feb 1, 2019 #6

    PeroK

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    Here's another way to look at it. First when we have:

    ##\vec{r} = x \hat{x} + y \hat{y}##

    Then ##x, y## are variables, depending on position, and the Cartesian unit vectors are constant. You must specify both variables to identify a specific vector.

    And, when we have:

    ##\vec{r} = r \hat{r}##

    Then both ##r## and ##\hat{r}## are variables. Again you must specify both variables to identify a specific vector. In this case however, it is the coordinate ##\theta ## that species ##\hat{r}##.
     
  8. Feb 1, 2019 #7
    I see, thanks. That makes sense.

    Writing the point as ## 0 \hat{x} +2 \hat{y}## involves the Cartesian coordinates ##(0,2)##. In polar, the two coordinates are ##(2, \pi/2)## but simply writing ##2 \hat{r}## would not be sufficient since the dependence of ##\hat{r}## on ##\theta## is implicit and not specified...
     
  9. Feb 1, 2019 #8
    So would the correct notation be $$r \hat{r}(\theta) = 2 \hat{r}(\theta = \pi/2) $$ to identify the point ##P## described above? I have never seen it written like this....
     
  10. Feb 13, 2019 at 4:09 PM #9
    Yes it would.

    The difficulties you are having simply go to show that in general "position vectors" can be difficult to define unless you are using suitable coordinate system. In polar coordinates, for example, you can define a vector that connects origin to some other point. But you cannot define a vector that connects the points ##\theta=0## and ##\theta=\pi/2## on a unit circle.

    In other coordinate systems, e.g. parabolic or bipolar, even the vector from origin to a point will no longer be definable. :-)
     
  11. Feb 13, 2019 at 6:10 PM #10
    I don't think "position" (a point in space) and "vectors" (a direction and magnitude) have any inherent relationship. You can use positions in space to describe a vector, like the 2-D vector from (0,0) to (1,1), but this vector isn't "at" (0,0). In this example it is the same vector as the vector described as connecting (0,1) to (1,2). The direction and length are the same, so the vectors are the same.
     
  12. Feb 13, 2019 at 7:13 PM #11
    That was the point I was trying to make :-).

    This logic works well on abstract level, but try to apply it to space addressed by, say, polar coordinates and spanned by polar basis vectors (which are position dependent). In fact, I would say that vectors are always "at" somewhere when it comes to space. More specifically, the only "safe" way I know how to define a vector is in the tangent space of a manifold (here Eucledian space). This definition specifically links vectors vectors to a point where the tangent space is defined
     
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