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How to see this equality (Lagrangian mechanics)

  1. Jan 18, 2012 #1
    Hello,

    Can anybody tell me why [itex]\nabla_{\textbf q} \langle \textbf q, P \textbf q \rangle = P \textbf q[/itex]?

    Explanation of notations:
    q is an n-dimensional vector: [itex]\textbf q = (q_1, q_2, \cdots, q_n)^T[/itex]
    P is an nxn-dimensional, real, Hermitian matrix
    [itex]\nabla_{\textbf q} := \left(\frac{\partial}{\partial q_1}, \cdots, \frac{\partial}{\partial q_n} \right)^T[/itex]

    Background story: I know the (specific) Lagrangian is [itex]\mathcal L = \frac{m}{2} \langle \dot{\textbf q},\dot{\textbf q} \rangle - \frac{k}{2} \langle \textbf q, P \textbf q \rangle[/itex] and the book then tells me that [itex]\ddot{ \textbf q } = - \frac{k}{2m} P \textbf q[/itex] and I presume that they derived this from the Lagrangian equations [itex]\frac{\mathrm d}{\mathrm dt} \left( \nabla_{\dot{\textbf q}} \mathcal L \right) = \nabla_{\textbf q} \mathcal L[/itex] but to do that I need to have the above equality, right?
     
  2. jcsd
  3. Jan 18, 2012 #2

    Matterwave

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    Gold Member

    Are you sure that 1/2 is there? I'm not getting the 1/2 in my calculations. It's much simpler for me to do the calculation in terms of components:

    [tex]L=\frac{1}{2}m\sum_k \dot{q}_k^2-\frac{1}{2}k\sum_{n,m}q_n P_{nm} q_m[/tex]

    Giving:
    [tex]\frac{\partial L}{\partial \dot{q}_i}=m\dot{q}_i[/tex]
    [tex]\frac{\partial L}{\partial q_i}=-\frac{1}{2}k\sum_{n,m} (\delta_{in}P_{nm}q_m+\delta_{im}q_nP_{nm})[/tex]

    Thus by the symmetry of P (real+hermitian = symmetric):
    [tex]\frac{\partial L}{\partial q_i}=-k\sum_{n} P_{in}q_n[/tex]

    I get then:
    [tex]\ddot{q}_i=-\frac{k}{m}\sum_{n} P_{in}q_n[/tex]

    This is the component form of your last equation, except I'm missing the factor of 1/2. I could have made a mistake somewhere though.
     
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