# Contents

## 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.

# Notes

1. If $$v^t S v = 0$$ for every $$v$$, then $$v = e_i$$ yields $$S_{ii}=0$$ and $$v=e_i + e_j$$ yields $$S_{ii} + S_{jj} + S_{ij} + S_{ji} = 0$$ and thus $$S_{ij} = -S_{ji}$$. On the other hand, if $$S_{ij} = -S_{ji}$$ for every $$i$$ and $$j$$, then since every vectory $$v$$ may be represented as a combination $$v = \sum_i \lambda_i e_i$$, we have $\begin{split} v^t S v = \sum_{i,j} S_{ij} \lambda_i \lambda_j &= \sum_{i < j} S_{ij} \lambda_i \lambda_j + \sum_{i} S_{ii} (\lambda_i)^2 + \sum_{i > j} S_{ij} \lambda_i \lambda_j \\ &= \sum_{i < j} S_{ij} \lambda_i \lambda_j + \sum_{i} 0 \times (\lambda_i)^2 - \sum_{j < i} S_{ji} \lambda_j \lambda_i \\ &= 0. \end{split}$