I am reading Mark Tuckerman's Statistical Mechanics and I am going through his derivation of the Lagrangian in generalized coordinates, using the mass metric tensor $G$. He defines kinetic energy as \begin{align} K(q,\dot{q}) &= \frac{1}{2}\sum_{\alpha=1}^{3N}\sum_{\beta =1}^{3N}\left[\sum_{i=1}^N m_i \frac{\partial r_i}{\partial q_{\alpha}} \frac{\partial r_i}{\partial q_{\alpha}} \right]\dot{q_{\alpha}}\dot{q_{\beta}} \\ &= \frac{1}{2}\sum_{\alpha=1}^{3N}\sum_{\beta =1}^{3N} G_{\alpha \beta}(q_1,q_2,...)\dot{q_{\alpha}}\dot{q_{\beta}} \end{align} And potential energy is simply $$U\equiv U(q_1, q_2, ...)$$

So the Lagrangian is \begin{align} L = K-U \end{align} Now plugging in the above definitions into the Euler-Lagrange equation gives: \begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_{\gamma}}} \right) - \frac{\partial L}{\partial q_{\gamma}} = 0 \end{align} The potential energy part is trivial, $\frac{\partial U}{\partial q_{\gamma}}$. Unpacking, the KE part of the above equation, I get, \begin{align} \frac{d}{dt}\left(\frac{\partial K}{\partial \dot{q_{\gamma}}}\right)&= \frac{d}{dt}\left(\frac{\partial }{\partial \dot{q_{\gamma}}} \frac{1}{2} G_{\alpha \beta}(q_1,q_2,...)\dot{q_{\alpha}}\dot{q_{\beta}}\right)\\ &=\frac{d}{dt}\left( \frac{1}{2}(G_{\alpha \beta} \delta_{\alpha \gamma}\dot{q}_{\beta} + G_{\alpha \beta} \dot{q}_{\alpha}\delta_{\gamma \beta}) \right) \\ &= \frac{1}{2}\frac{d}{dt}\left( G_{\gamma \beta} \dot{q}_{\beta} + G_{\alpha \gamma} \dot{q}_{\alpha}\right) \end{align} This is where I actually don't know how to simplify further to get,

\begin{align} \sum _{\beta=1}^{3N}G_{\gamma \beta}\ddot{q}_{\beta}+\sum_{\alpha=1}^{3N}\sum_{\beta=1}^{3N} \left[ \frac{\partial G_{\gamma \beta}}{\partial q_{\alpha}} - \frac{1}{2}\frac{\partial G_{\alpha \beta}}{\partial q_{\gamma} } \right]\dot{q}_{\alpha} \dot{q}_{\beta} \end{align} which is what Tuckermann has in his book. How do I get to this result?

  • $\begingroup$ Note that the kinetic energy seems to depend on the $q$'s through the mass metric tensor. EDIT: Therefore, you have to take into account also the kinetic part in order to compute $\partial_{q_{\gamma}} L$ $\endgroup$
    – Javi
    Commented Aug 21, 2021 at 20:50
  • $\begingroup$ Thanks @Javi for the reply. Could you possibly expand on that? $\endgroup$
    – megamence
    Commented Aug 21, 2021 at 20:52

1 Answer 1


Notice that the mass metric tensor $G_{\alpha\beta}$ is symmetric, that is $G_{\alpha\beta}=G_{\beta\alpha}.$ So your kinetic part $$\frac{1}{2}\frac{d}{dt}(G_{\gamma\beta}\dot{q}_{\beta}+G_{\alpha\gamma}\dot{q}_{\alpha})=\frac{1}{2}\frac{d}{dt}(G_{\gamma\beta}\dot{q}_{\beta}+G_{\gamma\alpha}\dot{q}_{\alpha}).$$ Since $\alpha$ and $\beta$ are dummy indices, they can be summed up to give $\frac{dG_{\gamma\beta}\dot{q}_{\beta}}{dt}.$ Then the second term in Eq.(1.4.19) is simply the partial derivative of $G_{\alpha\beta}$ with respect to $q_{\gamma}$.


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