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Euler-Langrange Dynamics

The dynamics of a mechanical system with coordinates \(q\), kinetic energy \(K\) and potential energy \(V\) which is subject to the forces \(f\) is governed by the Euler-Lagrange equations \[ \frac{d}{dt} \nabla_{\dot{q}} L(q, \dot{q}) - \nabla_{q} L (q, \dot{q}) = f \] where the Lagrangian \(L\) of the system is defined as \[ L(q, \dot{q}) = K(q, \dot{q}) - V(q). \]

Since the kinematic energy varies quadratically with the velocity, \[ K(q, \dot{q}) = \frac{1}{2} \dot{q}^t M(q) \dot{q} \] for some mass matrix \(M(q)\) which is symmetric and positive (semi-)definite. Using this expression in the Euler-Lagrange equations provides \[ M(q) \ddot{q} = f - \nabla_q V(q) + f^c \] where \(f^c\) are – by definition – the Coriolis and centrifugal forces: \[ f^c = - \dot{M}(q,\dot{q}) \dot{q} + \frac{1}{2} \nabla_q (\dot{q}^t M(q) \dot{q}) \]

Coriolis and Centrifugal Forces

Since \(f^c = -\dot{M}(q,\dot{q}) \dot{q} + \frac{1}{2} \nabla_q (\dot{q}^t M(q) \dot{q})\), every component \(f^c_i\) of the Coriolis and centrifugal forces varies quadratically with the velocity: we can find some matrix \(\Gamma_i\) such that \[ f^c_i = - \dot{q}^t \Gamma_i \dot{q} = - \sum_{j,k} \dot{q}_j \Gamma_{ijk} \dot{q}_k \]

Indeed, we have \[ f^c_i = -\sum_j \dot{M}_{ij} \dot{q}_j + \frac{1}{2} \frac{\partial}{\partial q_i} \sum_{j, k} \dot{q}_j M_{jk} \dot{q}_k \\ \] and thus \[ f^c_i = \sum_{j, k} \left( - \frac{\partial M_{ij}}{\partial q_k} \dot{q}_k \dot{q}_j + \frac{1}{2} \frac{\partial M_{jk}}{\partial q_i} \dot{q}_k \dot{q}_j \right) \]

Swapping the indices \(j\) and \(k\) yields \[ \sum_j \sum_k \frac{\partial M_{ij}}{\partial q_k} \dot{q}_k \dot{q}_j = \sum_k \sum_j \frac{\partial M_{ik}}{\partial q_j} \dot{q}_j \dot{q}_k = \sum_j \sum_k \frac{\partial M_{ik}}{\partial q_j} \dot{q}_k \dot{q}_j \] and thus we may also write \[ f^c_i = - \sum_{j, k} \frac{1}{2} \left( \frac{\partial M_{ij}}{\partial q_k} + \frac{\partial M_{ik}}{\partial q_j} - \frac{\partial M_{jk}}{\partial q_i} \right) \dot{q}_k \dot{q}_j \\ \] hence we may define \[ \Gamma_{ijk} = \frac{1}{2} \left( \frac{\partial M_{ij}}{\partial q_k} + \frac{\partial M_{ik}}{\partial q_j} - \frac{\partial M_{jk}}{\partial q_i} \right) \]

These \(\Gamma_{ijk}\) are called the Christoffel symbols of the first kind of \(M(q)\).

Mechanical Energy

Let \(C(q,\dot{q})\) be the matrix defined as \[ C_{ij} = \sum_{k} \Gamma_{ijk} \dot{q}_k. \] By construction \(f_c = - C(q, \dot{q}) \dot{q}\) thus the Euler-Lagrange equations become \[ M(q) \ddot{q} + C(q, \dot{q}) \dot{q} = f - \nabla_q V(q). \]

Consider the matrix \(S := \dot{M} - 2 C\). Since \[ \begin{split} S_{ij} = [\dot{M} - 2 C]_{ij} &= \sum_k \frac{\partial M_{ij}}{\partial q_k} \dot{q}_k - \sum_k \left( \frac{\partial M_{ij}}{\partial q_k} + \frac{\partial M_{ik}}{\partial q_j} - \frac{\partial M_{jk}}{\partial q_i} \right) \dot{q}_k \\ &= \sum_k \left( \frac{\partial M_{jk}}{\partial q_i} - \frac{\partial M_{ik}}{\partial q_j} \right) \dot{q}_k \end{split} \] it is skew-symmetric: \[ \forall \, i, \; \forall \, j, \; S_{ij} = - S_{ji} \] or equivalently¹, \[ \forall \, v \in \mathbb{R}^n, \; v^t S v = 0. \]

The mechanical energy of a Euler-Lagrange systems is defined as \[ E(q, \dot{q}) = K(q, \dot{q}) + V(q). \] It time derivative satisfies \[ \dot{E} = \dot{q}^t M(q) \ddot{q} + \frac{1}{2}\dot{q}^t \dot{M}(q) \dot{q} + \nabla V(q) \dot{q} \] and thus \[ \begin{split} \dot{E} &= \dot{q}^t \left(- C(q,\dot{q}) \dot{q}+ f - \nabla_q V(q)) \right) + \frac{1}{2}\dot{q}^t \dot{M}(q) \dot{q} + \nabla V(q) \dot{q} \\ &= f^t \dot{q} + \frac{1}{2} \dot{q}^t S \dot{q} \end{split} \] and since \(S\) is skew-symmetric \[ \dot{E} = f \cdot q. \] In plain words: the derivative of the mechanical energy is equal to the power transferred to the system by non-conservative external forces.