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Angular momentum in cartesian coordinates (Lagrangian)

  1. May 28, 2016 #1
    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. jcsd
  3. May 28, 2016 #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 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.
     
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