Arfken, parity operation on a point in polar coordinates

[Clever-Name]

Homework Statement

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

Homework Equations

Not sure

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.

 

[vela]

Show that those transformations cause x ➝ -x, y ➝ -y, and z ➝ -z.

 

[Clever-Name]

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?

 

[vela]

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.

 

[Clever-Name]

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?

 

[vela]

Yeah. You're showing that that change does in fact change r to -r, which is all the problem asked you to do.

 

[Clever-Name]

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.