# Arfken, parity operation on a point in polar coordinates

1. Sep 18, 2011

### Clever-Name

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

Also, show that the unit vectors of the cylindrical polar coordinate system $\hat{e}_{\rho}$ and $\hat{e}_{\phi}$ have odd parity while $\hat{e}_{z}$ 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 $(\rho, \varphi, z)$ and write it as:
$$\rho = \sqrt{x^{2} + y^{2}}$$
$$\varphi = arctan(\frac{y}{x})$$
$$z = z$$

And apply what I know about Parity I would get:

$$\rho' = \sqrt{(-x)^{2} + (-y)^{2}} = \rho$$
$$\varphi' = arctan(\frac{-y}{-x}) = \arctan(\frac{y}{x}) = \varphi$$
$$z' = -z$$

Now I realize the definition of $\varphi$ depends on the x and y values, so I suppose I could reason that it should be $\varphi \pm \pi$ 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. Sep 18, 2011

### vela

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

3. Sep 18, 2011

### Clever-Name

I'm not sure I follow...

For $z \to -z$ it's obvious.

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

and for $\phi \to \phi \pm \pi$ I'm completely lost as to how to approach it.

edit - for $\phi$ 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?

4. Sep 18, 2011

### vela

Staff Emeritus
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.

5. Sep 18, 2011

### Clever-Name

Ok so if

$x = \rho cos(\phi)$
and
$y = \rho sin(\phi)$

and we take $\phi \to \phi + \pi$
we get
$x = \rho cos(\phi + \pi) = \rho (-cos(\phi)) = -x$
and
$y = \rho sin(\phi + \pi) = \rho (-sin(\phi)) = -y$

and the same follows for negative pi.

This is a sufficient way to answer the problem? To 'work backwards' in a sense?

6. Sep 18, 2011

### vela

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

7. Sep 18, 2011

### Clever-Name

Oh alright, perfect, thank you. Now the last part should then fall out as follows:

Since the unit vectors are defined as:

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

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

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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