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Center of mass in Lagrangian mechanics

  1. Jun 9, 2012 #1
    We all know the proof, from Newtonian mechanics, that the motion of the center of mass of a system of particles can be found by treating the center of mass as a particle with all the external forces acting on it. I want to prove the same think, but within the framework of Lagrangian mechanics, and I',m having some trouble.

    So to start, the Lagrangian of a system of N particles is
    [tex]L=\sum_{i=1}^N \left(\frac{1}{2}m_i v_i^2 - U_{\mathrm{ext.}, i}(\vec{r}_i)\right) - U_{\mathrm{int.}}(\vec{r}_1, ..., \vec{r}_N).[/tex]
    [itex]U_{\mathrm{ext.},i}[/itex] is the potential energy due to any external forces on the ith particle. With a little algebra, we can rewrite this as

    [tex]L=\frac{1}{2}MV^2 - U_{\mathrm{ext.}} + \sum_{i=1}^N \frac{1}{2}m_i \tilde{v}_i^2 - U_{\mathrm{int.}}[/tex]

    where [itex]M=\sum m_i[/itex] is the total mass, V is the speed of the center of mass, [itex]U_{\mathrm{ext.}} = \sum U_{\mathrm{ext.},i}[/itex], and [itex]\tilde{v}_i[/itex] is the speed of the ith particle relative to the center of mass.

    I don't know where to go from here. Help?
  2. jcsd
  3. Jun 9, 2012 #2
    You have the Lagrangian; what do you usually do with it? Invoke the Euler-Lagrange equations. With a set of generalized coordinates [itex]Q, \bar q_i[/itex], you should be ready to go.
  4. Jun 10, 2012 #3
    I thought of that, but I had a problem: the number of degrees of freedom seems to go up, is that a problem?
  5. Jun 10, 2012 #4


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    I'd use Noether's theorem in extended form. You have a Lagrangian where explicit Galilean symmetry is broken by considering a part of the system described by an external potential. Otherwise the Lagrangian is invariant under Galileo boosts.

    From the partial symmetry of the Lagrangian under Galileo boosts you should get the equation of motion for the center of mass

    [tex]M \dot{V}=-\partial_X U_{\text{ext}}(X).[/tex]
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