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Arfken, parity operation on a point in polar coordinates

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the parity operation (reflection through the origin) on a point [itex] (\rho, \varphi, z) [/itex] relative to fixed (x, y, z) axes consists of the transformation:
    [tex] \rho \to \rho [/tex]
    [tex] \varphi \to \varphi \pm \pi [/tex]
    [tex] z \to -z [/tex]

    Also, show that the unit vectors of the cylindrical polar coordinate system [itex] \hat{e}_{\rho} [/itex] and [itex] \hat{e}_{\phi} [/itex] have odd parity while [itex] \hat{e}_{z} [/itex] has even parity
    2. Relevant equations
    Not sure


    3. The attempt at a solution

    All I know about Parity is it's definition. I know that for a vector (x,y,z) if you apply the parity operator you get (-x,-y,-z). So I thought maybe if you take the coordinates [itex] (\rho, \varphi, z) [/itex] and write it as:
    [tex] \rho = \sqrt{x^{2} + y^{2}} [/tex]
    [tex] \varphi = arctan(\frac{y}{x}) [/tex]
    [tex] z = z [/tex]

    And apply what I know about Parity I would get:

    [tex] \rho' = \sqrt{(-x)^{2} + (-y)^{2}} = \rho [/tex]
    [tex] \varphi' = arctan(\frac{-y}{-x}) = \arctan(\frac{y}{x}) = \varphi [/tex]
    [tex] z' = -z [/tex]

    Now I realize the definition of [itex] \varphi [/itex] depends on the x and y values, so I suppose I could reason that it should be [itex] \varphi \pm \pi [/itex] based on that, but I'm uncomfortable just outright stating it. I feel like there should be more to it.


    I haven't attempted the last part yet.
     
    Last edited: Sep 18, 2011
  2. jcsd
  3. Sep 18, 2011 #2

    vela

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    Show that those transformations cause x ➝ -x, y ➝ -y, and z ➝ -z.
     
  4. Sep 18, 2011 #3
    I'm not sure I follow...

    For [itex] z \to -z [/itex] it's obvious.

    For [itex] \rho \to \rho [/itex] there is no restriction on the sign of x or y so I supposed this could imply [itex] x \to -x [/itex] and [itex] y \to -y [/itex] but I'm not sure how I could definitively state that.

    and for [itex] \phi \to \phi \pm \pi [/itex] I'm completely lost as to how to approach it.

    edit - for [itex] \phi [/itex] would it be appropriate to look at the angle in the 4 quadrants and add/subtract pi to see how this changes the x and y value according to quadrant?
     
  5. Sep 18, 2011 #4

    vela

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    I meant use the inverse transformations, where x and y are written in terms of ρ and Φ. You avoid the complication that's introduced because tangent has period of only pi, as opposed to 2pi.
     
  6. Sep 18, 2011 #5
    Ok so if

    [itex] x = \rho cos(\phi) [/itex]
    and
    [itex] y = \rho sin(\phi) [/itex]

    and we take [itex] \phi \to \phi + \pi [/itex]
    we get
    [itex] x = \rho cos(\phi + \pi) = \rho (-cos(\phi)) = -x [/itex]
    and
    [itex] y = \rho sin(\phi + \pi) = \rho (-sin(\phi)) = -y [/itex]

    and the same follows for negative pi.

    This is a sufficient way to answer the problem? To 'work backwards' in a sense?
     
  7. Sep 18, 2011 #6

    vela

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    Yeah. You're showing that that change does in fact change r to -r, which is all the problem asked you to do.
     
  8. Sep 18, 2011 #7
    Oh alright, perfect, thank you. Now the last part should then fall out as follows:

    Since the unit vectors are defined as:

    [tex] \hat{e}_{\rho} = [cos(\phi), sin(\phi), 0] [/tex]
    [tex] \hat{e}_{\phi} = [-sin(\phi), cos(\phi), 0] [/tex]
    [tex] \hat{e}_{z} = [0,0,1] [/tex]

    Applying the operation [itex] \phi \to \phi \pm \pi [/itex] gives us:
    [tex] \hat{e}_{\rho} = [cos(\phi), sin(\phi), 0] \to [-cos(\phi), -sin(\phi), 0] = -\hat{e}_{\rho} [/tex]
    [tex] \hat{e}_{\phi} = [-sin(\phi), cos(\phi), 0] \to [sin(\phi), -cos(\phi), 0] = -\hat{e}_{\phi} [/tex]
    [tex] \hat{e}_{z} = [0,0,1] \to [0, 0, 1] = \hat{e}_{z} [/tex]

    Since the z unit vector didn't change it has even parity, and since the other 2 changed in sign they have odd parity.

    I just wanted to post the rest of the solution for anyone else who might come across this problem.
     
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