How to see this equality (Lagrangian mechanics)

1. Jan 18, 2012

nonequilibrium

Hello,

Can anybody tell me why $\nabla_{\textbf q} \langle \textbf q, P \textbf q \rangle = P \textbf q$?

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

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

2. Jan 18, 2012

Matterwave

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:

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

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

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

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

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.