# Angular momentum in cartesian coordinates (Lagrangian)

1. May 28, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I would like to discuss with you a problem that I am wondering if I understand it correctly:

Find expressions for the cartesian components and for the magnitude of the angular momentum of a particle in cylindrical coordinates $(r,\varphi,z)$.

2. Relevant equations

$p_i = mx_i$, $\vec{L} = \vec{r} \times \vec{p}$

3. The attempt at a solution

The $r$ confuses me a lot, since we normally use $\rho$ in cylindrical coordinates. I've set up $\vec{r}$ laying in the $x,y$-plane (i.e. $z=0$) and $z$ as the rotation axis, but am I allowed to do that? If so, the rest seemed pretty simple:

$x = r \cos \varphi$
$y = r \sin \varphi$
$z = 0$

$\dot{x} = \dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi$
$\dot{y} = \dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi$
$\dot{z} = 0$

$p_x = m\dot{x} = m(\dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi)$
$p_y = m\dot{y} = m(\dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi)$
$p_z = m\dot{z} = 0$

$\vec{L} = \vec{r} \times \vec{p}$

$L_x = y p_z - z p_y = 0$
$L_y = z p_x - x p_z = 0$
$L_z = x p_y - y p_x = mx (\dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi) - my(\dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi)$

As expected from my initial conditions, the angular momentum vector points in the z-direction, and after simplification I get the well-known expression:

$L_z = mr^2\dot{\varphi}$

But I find it strange to set up the $z$ component to $0$, and if it was not the case the angular momentum vector could be pointing basically any direction depending on what angle $\theta$ it is leaning (in spherical coordinates). I've tried to solve the problem that way, but I can't get my hands on a proper expression for $x$,$y$ and $z$ that does not involve $\theta$ or a very unpractical arctan or I don't know what. Any suggestion?

Julien.

2. May 28, 2016

### JulienB

Actually I just found in other pages that apparently for many people the $r$ refers to the $\rho$ I was mentioning. Then my $x_i$ and $\dot{x}_i$ stay the same as before (except with $\rho$ instead of $r$), and I get those expression for $\vec{L}$:

$L_x = m[\dot{z}y - z(\dot{\rho} \sin \varphi + \rho \dot{\varphi} \cos \varphi)]$
$L_y = m[z(\dot{\rho} \cos \varphi - \rho \dot{\varphi} \sin \varphi) - \dot{z}x]$
$L_z = mr^2 \dot{\varphi}$

And the magnitude is very long:

$|\vec{L}| = m \sqrt{\dot{z}^2\rho^2 - 2 z\dot{z} \bigg[x\big(\dot{\rho} \cos \varphi - \rho \dot{\varphi} \sin \varphi\big) + y\big(\dot{\rho} \sin \varphi + \rho \dot{\varphi} \cos \varphi\big) \bigg] + z^2 (\dot{\rho}^2 + \rho^2 \dot{\varphi}^2) + \rho^4 \dot{\varphi}^2}$

Could that be? I really thought it was gonna be something of the form $I \omega$, maybe I cannot see it because of the cylindrical coordinates.

EDIT: I just saw I posted in "Advanced Physics" by mistake. I hope it's not too bad, sorry for that.

Julien.