Exercise 2

This exercise is very similar to Exercise 1, but now we deal with more than one particle and more than one degree of freedom.

The index \(i\) is used to numerate particles as well as degrees of freedom.

\[T = \frac{1}{2} \sum_{i=1}^n m_i \dot{x}_i^2\] \[V = V(x_1, x_2, \ldots, x_n)\]

Hence, we get for the Lagrangian \(L=T-V\):

\[L(x_1, x_2, \ldots, x_n, \dot{x}_1, \dot{x}_2, \ldots, \dot{x}_n) = \frac{1}{2} \sum_{i=1}^n m_i \dot{x}_i^2 - V(x_1, x_2, \ldots, x_n)\] \[\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{x}_i} = \frac{\mathrm{d}}{\mathrm{d}t} m_i \dot{x}_i = m_i \ddot{x}_i = m_i a_i \] \[\frac{\partial L}{\partial x_i} = \frac{\partial (T-V)}{\partial x_i} = - \frac{\partial V}{\partial x_i} = F_i\]

Again, Newton’s 2nd law can be derived from the Euler-Lagrange equation: \[\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{x}_i} - \frac{\partial L}{\partial x_i} = m_i a_i - F_i = 0\]