# Lagrangian of a Free Particle

In the article on Galilean Transformation we found that the Lagrangian is a function of $v^2$ only. To find the exact form of the Lagrange function for a free particle, we use the invariance of the Lagrange function under Galilean Transformation. We transform from inertial system $K$ to the inertial system $K'$ where the particle has the velocity $v'=v + \mathbf{\varepsilon}$ for an infinitesimal velocity $\mathbf{\varepsilon}$. We compute the transformation \begin{align} L \rightarrow L' \end{align} \begin{align} L'(v'^2)=L'(v + \mathbf{\varepsilon})^2=L(v^2 + 2v\mathbf{\varepsilon} + \mathbf{\varepsilon}^2) \end{align} We develop the series for $\mathbf{\varepsilon}$ around $v^2$ we have \begin{align} L'(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}(2v\mathbf{\varepsilon}+\mathbf{\varepsilon}^2) \tag{1} \end{align} where the rule for very small epsilon is $\mathbf{\varepsilon}^2=0$. We showed that the Galilean Transformation allows only constant velocities between different inertial systems, thus $\frac{\partial L}{\partial v^2}$ must be constant since (1) already contains the velocity $v$. Integrating \begin{align} \frac{\partial L}{\partial v^2}=const. \end{align} gives \begin{align} L(v^2)=a\cdot v^2 \end{align} This Lagrangian is indeed invariant under the infinitesimal transformation $\mathfrak{v}$ \begin{align} L'=a\cdot v'^2=a(v+\mathfrak{v})^2=av^2+2av\mathfrak{v}+a\mathfrak{v}^2 \end{align} or \begin{align} L'=L+\frac{d}{dt}(2ar\mathfrak{v}+a\mathfrak{v}^2t) \end{align} By the Non-uniqueness of the Lagrange Function up to the addition of total derivatives $L$ and $L'$ are equal.