Exercise 1

For the motion of a particle of mass \(m\) in a one-dimensional space (position coordinate \(x\)) with potential \(V\) the following applies to the kinetic energy \(T\):

\[T = \frac{1}{2}m \dot{x}^2\] \[V = V(x)\]

Hence, the Lagrangian \(L=T-V\) is \[L(x, \dot{x}) = \frac{1}{2}m \dot{x}^2 - V(x)\] \[\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{x}} = \frac{\mathrm{d}}{\mathrm{d}t} m \dot{x} = m \ddot{x} = ma \]

\[\frac{\partial L}{\partial x} = \frac{\partial (T-V)}{\partial x} = - \frac{\partial V}{\partial x} = F\]

Using the Euler-Lagrange equation wie get Newton’s 2nd law: \[\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = ma-F = 0\]