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Springs, Normal Modes, and Center of Mass coordinates

  1. Apr 14, 2008 #1
    [SOLVED] Springs, Normal Modes, and Center of Mass coordinates

    1.) Problem

    The problem of the linear triatomic molecule can be reduced to one of two degrees of freedom by introducing coordinates [tex]y_{1} = x_{2} - x_{1}, y_{2} = x_{3} - x_{2}[/tex], and eliminating [tex]x_{2}[/tex] by requiring that the center of mass remain at rest. Obtain the frequencies of the normal modes in these coordinates and show that they agree with the results of Section 6.4.

    -Classical Mechanics, Goldstein, 3rd Edition, pg 272

    2.) Useful Formulae and Context

    I've attached a picture I drew of the set-up.

    In section 6.4, the normal modes of the triatomic molecule are found by

    1. writing out the potential and kinetic energy,
    2. converting to coordinates relative to the equilibrium positions,
    3. expressing them (V and T) as tensors, and
    4. solving the characteristic equation [tex]|V - \omega^{2} T| = 0[/tex]

    Explicitly,

    [tex]V = \frac{k}{2} (x_{2} - x_{1} - b)^{2} + \frac{k}{2} (x_{3} - x_{2} - b)^{2} [/tex]

    The coordinates relative to the equilibrium positions are introduced:

    [tex]\eta_{i} = x_{i} - x_{0i}[/tex]

    where

    [tex]x_{02} - x_{01} = b = x_{03} - x_{02}[/tex]

    So the potential energy becomes,

    [tex]V = \frac{k}{2} (\eta_{2} - \eta_{1})^{2} + \frac{k}{2} (\eta_{3} - \eta_{2})^{2}[/tex]
    [tex]V = \frac{k}{2} (\eta_{1}^{2} + 2\eta_{2}^{2} + \eta_{3}^{2} - 2\eta_{1}\eta_{2} - 2\eta_{2}\eta_{3})[/tex]

    which can pretty easily be written in tensor form. A similar thing is done with kinetic energy.

    3.) Attempt at the Solution

    Goldstein writes that we should impose the constraint that "the center of mass remain stationary at the origin:"

    [tex]m(x_{1} + x_{3}) + M x_{2} = 0[/tex]

    and that this equation should be used to eliminate one of the coordinates from V and T.

    Clearly, this coordinate should be [tex]x_{2}[/tex], since it appears in both [tex]y_{1}, y_{2}[/tex], right? I've scribbled and rearranged these equations over and over, and can't figure out how express V and T only in terms of [tex]y_{1}, y_{2}[/tex].

    The "Center of Mass" is described by:

    [tex]R = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m(x_{1} + x_{3}) + M x_{2}}{2m + M}[/tex]

    How does knowing this help me? If someone could just point the way, or give me the smallest hint, I'm sure I could push this through - I'm just having a block on this. Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2008 #2
    Let me just add specifically some steps I've taken:

    [tex]R = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m(x_{1} + x_{3}) + M x_{2}}{2m + M} = 0[/tex]
    [tex]\rightarrow m(x_{1} + x_{3}) + M x_{2} = 0[/tex]
    [tex]\rightarrow x_{2} = - \frac{m}{M} (x_{1} + x_{3}) [/tex]

    Now using,

    [tex]y_{1} = x_{2} - x_{1}[/tex]
    [tex]y_{2} = x_{3} - x_{2}[/tex]

    I should be able to find some combination of those that would give me:

    [tex]x_{1} + x_{3}[/tex]

    so that I could insert it into the equation up above for [itex]x_{2}[/itex], thereby eliminating it. But I can't find such a combination.
     
    Last edited: Apr 14, 2008
  4. Apr 21, 2008 #3
    Hi all - it's been a week, and I still haven't made any progress on this problem.

    There must be some simple algebra trick I'm missing here. Or maybe my method is completely off? Any thoughts at all?
     
  5. Apr 25, 2008 #4
    Progress Report

    Alright, I was clearly looking at this incorrectly. (Perhaps that's why I've had 120+ views, and zero replies in almost 2 weeks.)

    Again, the given equations:

    Energies for Triatomic Model in Cartesian Coordinates:
    [tex]V=\frac{k}{2}(x_{2}-x_{1}-b)^{2}+\frac{k}{2}(x_{3}-x_{2}-b)^{2}[/tex]
    [tex]T=\frac{m}{2}(\dot{x_{1}}^{2}+\dot{x_{3}}^{2})+\frac{M}{2}\dot{x_{2}}^{2}[/tex]

    Transformation to Internal Coordinates:
    [tex]y_{1}=x_{2}-x_{1}[/tex]
    [tex]y_{2}=x_{3}-x_{2}[/tex]

    Stipulation of Stationary Center of Mass:
    [tex]m(\dot{x_{1}}+\dot{x_{3}})+M(\dot{x_{2}})=0[/tex]

    Now, transforming V to internal coordinates is easy if you apply a new transformation:
    [tex]y_{1}-b=\gamma_{1}[/tex]
    [tex]y_{2}-b=\gamma_{2}[/tex]

    So we have,
    [tex]V=\frac{k}{2}(\gamma_{1}^{2}+\gamma_{2}^{2})[/tex]

    or, written as a tensor:
    [tex]\bar{V}=\frac{k}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)[/tex]

    The Kinetic Energy is where I'm having trouble. Can anyone help me transform [itex]\dot{x_{1}},\dot{x_{2}},\dot{x_{3}}[/itex] to [itex]\dot{\gamma_{1}},\dot{\gamma_{2}}[/itex]? I have pages of work, and I've seem to be getting no where. If no one is replying because I'm breaking a rule, could someone let me know, too? Thanks much!
     
  6. Apr 25, 2008 #5
    Alright, it took 5 more pages of algebra, but I got it.
     
  7. Mar 8, 2010 #6
    Re: [SOLVED] Springs, Normal Modes, and Center of Mass coordinates

    Can u help me how u solved it? its due tomorrow, so please respond asap. how did u express T in terms of gammas?

    thanks in advance
     
  8. Nov 29, 2011 #7
    Re: [SOLVED] Springs, Normal Modes, and Center of Mass coordinates

    Solve for x3 and x1 in terms of y1, y2, and x2. Then plug into center of mass equation and solve for x2. You then can get x1, x2, and x3 in terms of y1 and y2. Then plug away...
     
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