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Ellipsoidal motion (Lagrange multipliers)

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose you have an object of mass m that is constrained to move on an ellipsoid with a constraint function [itex]f(x,y,z) = x^2+4y^2+4z^2 -1=0[/itex]. Aside from the force of constraint, the only force acting on the mass is an elastic force [itex]\vec{F}=-kx\hat{x}[/itex]. Find the Lagrangian, the Hamiltonian and the integrals of motion.

    2. Relevant equations
    Euler-Lagrange (EL) equation, Lagrange multipliers, Legendre transform, any equations related to classical mechanics..

    3. The attempt at a solution
    The kinetic energy is [itex]K = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\dot{y}^2 + \frac{1}{2}m\dot{z}^2[/itex]. The potential energy is just the elastic potential energy [itex]U=\frac{1}{2}kx^2[/itex]. The Lagrangian L can be found in the usual way as [itex]L=K-U[/itex]. By applying a Legendre transform I get a Hamiltonian [itex]H = p_x\dot{x}+p_y\dot{y}+p_z\dot{z} - L[/itex] where I can find the generalised momenta by solving [itex]p_i = \frac{\partial L}{\partial q_i}[/itex] for p. Plugging p back into H I can summarize what I have found so far:
    [tex]
    L = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\dot{y}^2 + \frac{1}{2}m\dot{z}^2 - \frac{1}{2}kx^2 \\
    H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} + \frac{1}{2}kx^2
    [/tex]

    Ok, so in order to find the integrals of motion, I basically have to find the equations of motion and integrate them if I'm not mistaken. I plug in the Lagrangian into the EL equation, taking into account the constraint with the help of a Lagrange multiplier
    [tex]
    \frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} = \lambda\frac{\partial f}{\partial q_i}
    [/tex]
    which gives the equations of motion
    [tex]
    m\ddot{x}+kx+2\lambda x = 0 \\
    m\ddot{y} + 8\lambda y = 0 \\
    m\ddot{z} + 8\lambda z = 0.
    [/tex]
    I can find the integrals of motion if I multiply each EOM with [itex]Q_i(q_i) = \dot{q}_i[/itex], rearranging, integrating etc..

    The problem for me lies in finding [itex]\lambda[/itex]; it's just a mess of fractions and radicals. Is there perhaps a way to describe the system in a nicer set of coordinates or maybe some other method altogether?
     
  2. jcsd
  3. Dec 1, 2015 #2

    Orodruin

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    You are making things more complicated than they are. You do not need to find ##\lambda## because you are not asked to find the constraint force. Your constraint is holonomic and you can use it to rewrite your system with three degrees of freedom as a system with two degrees of freedom. I suggest you look at the symmetries of your problem to figure out what coordinates are suitable.
     
    Last edited: Dec 1, 2015
  4. Dec 1, 2015 #3
    Oh right, I can just eliminate a variable straight away. Well, there is obviously some symmetry since the constraint can be thought of as a deformed sphere so some variant of spherical coordinates could perhaps work? I was thinking something like (2)-(5) here http://mathworld.wolfram.com/Ellipsoid.html.
     
  5. Dec 1, 2015 #4

    Orodruin

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    I think taking the equivalent of spherical coordinates will mess up your kinetic terms a bit. However, you might want to note that the ##y## and ##z## directions are completely equivalent. This gives you full rotational symmetry around the ##x##-axis which seems to suggest the use of ...
     
  6. Dec 1, 2015 #5
    I'm gonna go with what is cylindrical coordinates?
     
  7. Dec 1, 2015 #6

    Orodruin

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    Why don't you try it and see how it works out?
     
  8. Dec 1, 2015 #7
    Cylindrical coords. would give me [itex]x=x,\ y = \rho cos\theta, z = \rho sin\theta[/itex] where [itex]y^2+z^2 = \rho^2,\ -1<x<1,\ 0<\theta < 2\pi[/itex]. The constraint would then become
    [tex]
    x^2 + 4\rho^2 = 1.
    [/tex]
    The velocity squared for my mass is then
    [tex]
    v^2 = \rho^2 + \rho^2\dot{\theta}^2 + \dot{x}^2
    [/tex]
    which gives me L
    [tex]
    L = \frac{1}{2}m(\dot{\rho}^2 + \rho^2\dot{\theta}^2 + \dot{x}^2) - \frac{1}{2}kx^2.
    [/tex]
    Now I can use the constraint to eliminate [itex]x[/itex].for example. If I plug this into L and then calculate EL eq. for [itex]\rho[/itex] I find, after a whole bunch of differentiation using chain rule etc, a messier expression than I found with cartesian coords.
     
  9. Dec 2, 2015 #8
    Hm.. I can't seem to get anywhere with cylindrical coords. Whenever I try to calculate the total time derivative in EL eq. I tend to get terms with mixed time derivatives and of varying orders that I don't really know how to deal with. It doesn't seem to matter what substitution I make into L. : [
     
  10. Dec 2, 2015 #9

    Orodruin

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    Did you write down the Lagrangian with the constraints taken into account? What did you get?

    You do not need to write down the EL equations to find the integrals of motion.
     
  11. Dec 2, 2015 #10
    Oh okay, I was under the impression that an integral of motion would be an integration of the EOM's. In any case, the Lagrangian is as stated in #7. If I express [itex]x=x(\rho(t))\rightarrow dx/dt = \partial x / \partial \rho \cdot d\rho/dt[/itex] I get
    [tex]
    L = \frac{1}{2}m(\dot{\rho}^2 + \rho^2\dot{\theta}^2 + \frac{16\rho^2\dot{\rho}^2}{1-4\rho^2}) - \frac{1}{2}k(1-4\rho^2).
    [/tex]
    How then would I go about finding the integrals of motion without EL eqs?
     
  12. Dec 2, 2015 #11

    Orodruin

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    It is, but you do not need to write out the full EL equations in order to do that.

    So, this Lagrangian is not explicitly dependent on several things. Are you familiar with how to use this fact in order to find first integrals?
     
  13. Dec 2, 2015 #12
    Right, so it is explicitly independent of [itex]\theta[/itex] and time. If I recall, a Lagrangian that doesn't change explicitly with time would mean that neither the kinetic term or the potential term changes explicitly with time, so the total energy is conserved? If we consider the system to be closed, then the Hamiltonian remains constant (our constant of motion?).

    As for the [itex]\theta[/itex] independence.. I assume it has something to do with angular momentum conservation. Is it as simple as to show that [itex]\vec{L} = \vec{r}\times \vec{p}[/itex] is constant?
     
  14. Dec 2, 2015 #13

    Orodruin

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    They are not explicitly dependent on time, but they do change with time due to the implicit dependence. What is constant is given by the Beltrami identity. (It will be the total energy in this case.)

    The angular momentum here is not conserved (only one of its components is!). I suggest you write down the EL equations for the ##\theta## direction without inserting the explicit form of the Lagrangian and only use that ##\partial L/\partial\theta = 0##. What do you get?
     
  15. Dec 2, 2015 #14
    Ah yes, I assume you mean [itex]\frac{d}{dt}(m\rho^2\dot{\theta}) = 0 \rightarrow \rho\cdot m\rho\dot{\theta} = \rho \cdot p_\theta = C[/itex], i.e. the angular momentum in the x-direction (or around the x-axis, whatever you call it..) is conserved.

    edit: So the conserved quantities/integrals of motion in this case would be the total energy H=E and what I wrote above?
     
  16. Dec 2, 2015 #15

    Orodruin

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    Correct.
     
  17. Dec 2, 2015 #16
    Ok, great! I'll try to solve some more problems like these on my own. Thanks for pointing me in the right direction, I really appreciate your help and patience!
     
  18. Dec 2, 2015 #17

    Ray Vickson

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    If I were doing the problem I would use "almost-spherical" coordinates:
    [tex] \begin{array}{rcl}
    x &=& \cos(\theta) \\
    y &=& \frac{1}{2} \sin(\theta) \cos(\phi) \\
    z &=& \frac{1}{2} \sin(\theta) \sin(\phi)
    \end{array}
    [/tex]
     
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