Angular momentum in cartesian coordinates (Lagrangian)

In summary, the conversation discusses the problem of finding expressions for the cartesian components and magnitude of the angular momentum of a particle in cylindrical coordinates. The equations used are ##p_i = mx_i## and ##\vec{L} = \vec{r} \times \vec{p}##. The person attempting to solve the problem is confused by the use of ##r## instead of the usual ##\rho## in cylindrical coordinates. They set up ##\vec{r}## in the ##x,y##-plane with the rotation axis along the ##z## direction. They then provide the equations for ##x,y,z## and their derivatives, as well as the expressions for ##p_x,p_y,p_z## and ##
  • #1
JulienB
408
12

Homework Statement



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)##.

Homework Equations



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

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.
 
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  • #2
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 going to 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.
 

Related to Angular momentum in cartesian coordinates (Lagrangian)

1. What is angular momentum in cartesian coordinates?

Angular momentum in cartesian coordinates is a measure of the rotational momentum of a system. It is defined as the cross product of the position vector and the linear momentum of an object.

2. How is angular momentum related to Lagrangian mechanics?

In Lagrangian mechanics, the angular momentum of a system is conserved if there is no external torque acting on the system. This is known as the law of conservation of angular momentum.

3. What is the equation for angular momentum in cartesian coordinates?

The equation for angular momentum in cartesian coordinates is L = r x p, where L is the angular momentum, r is the position vector, and p is the linear momentum. This equation can also be expressed in terms of mass and velocity as L = mvr, where m is the mass and v is the velocity.

4. How is angular momentum in cartesian coordinates different from angular momentum in polar coordinates?

In cartesian coordinates, angular momentum is expressed as a vector quantity, while in polar coordinates it is expressed as a scalar quantity. Additionally, the equations for angular momentum differ in these two coordinate systems.

5. What are some real-life examples of angular momentum in cartesian coordinates?

Some real-life examples of angular momentum in cartesian coordinates include the rotation of a spinning top, the motion of a gyroscope, and the orbit of planets around the sun.

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