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Polar to cartesian using e(theta) and e(r)

  1. Jan 9, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-1-9_10-28-40.png

    2. Relevant equations
    NONE

    3. The attempt at a solution
    I'm trying to understand why the unit vector in the y direction is that formula. I get that e(theta) and e(r) are unit vectors used with polar coordinates that define direction and are perpendicular to each other always. What I don't get is their relationship with theta to get cartesian unit vectors.
     
  2. jcsd
  3. Jan 9, 2017 #2

    BvU

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    A simple rotation over the angle ##\theta##. Your second picture shows it clearly for ##\bf\hat\jmath##.
    For ##\bf \hat\imath## you see ##e_\theta## points in the negative ##\hat\imath##-direction so you get ##\hat\imath = \cos\theta \; e_r - \sin\theta \; e_\theta##
     
  4. Jan 9, 2017 #3

    PeroK

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    What do you think it should be?
     
  5. Jan 9, 2017 #4
    I don't follow.

    I don't know what it should be. I originally thought the j vector was just the projection of e(theta) but that clearly cannot be because both j and e(theta) have length 1. Is there a general relationship that might explain how two perpendicular unit vectors in a plane can somehow form this relationship?
     
  6. Jan 9, 2017 #5

    PeroK

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    Your diagram in post #1 only needs you to read off the sine and cosine components to get this relationship. However, another way is to note that (in Cartesian coordinates):

    ##j = (0, 1), \ \hat{r} = (\cos \theta, \sin \theta), \ \hat \theta = (-\sin \theta, \cos \theta)##

    Then, you simply have to use a bit of vector algebra to express ##j## as a ,linear combination of ##\hat{r}## and ##\hat \theta##.
     
  7. Jan 9, 2017 #6

    BvU

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    You can check it from the expressions I gave. A rotation preserves vector lengths
     
  8. Jan 9, 2017 #7
    Suppose that ##\theta = 0##. What would j be equal to, ##i_r## or ##i_{\theta}##?

    Suppose that ##\theta = 90\ degrees##. What would j be equal to, ##i_r## or ##i_{\theta}##?
     
  9. Jan 9, 2017 #8
    I follow this but I wanted the derivation for it.
    Ok that first part makes sense. How can you at a glance determine that those two vectors r and theta would lineraly combine to form a vector of length 1?
     
  10. Jan 10, 2017 #9

    PeroK

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    ##\hat{r}## and ##\hat \theta## are linearly independent so form a basis (in fact, they form an orthonormal basis). Any vector, therefore, can be expressed as a linear combination of them.

    You can use the basic techniques of solving simultaneous equations to get the coefficients. In this case, we look for coefficients ##a, b## such that:

    ##a \hat{r} + b \hat \theta = j##

    Hence

    ##a(\cos \theta, \sin \theta) + b(-\sin \theta, \cos \theta) = (0, 1)##

    Which implies:

    ##a \cos \theta - b \sin \theta = 0##
    ##a \sin \theta + b \cos \theta = 1##

    Which leads to the solution ##a = \sin \theta, \ b = \cos \theta##.
     
  11. Jan 10, 2017 #10
    Just draw a right triangle anywhere in the domain with j as the hypotenuse, and the other two sides parallel to the local ir and itheta. Then determine the lengths of those other two sides.
     
  12. Jan 10, 2017 #11
    Now it makes sense. Thanks!

    I see it now. Thanks!
     
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