# Non-uniqueness of the Lagrange Function

From the Lagrange Function $L$ can be multiplied by an arbitrary constant $c$, without changing the Lagrange Equation since the constant can be factored out and canceled. \begin{align} L'=c \cdot L \end{align} \begin{align} \frac{\partial L'}{\partial q_i}-\frac{d}{dt}\frac{\partial L'}{\partial \dot{q_i}} =0 \end{align} \begin{align} c \cdot \left( \frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}} \right) =0 \end{align} Hence the Lagrange function is not unique. Two Lagrange functions that differ only by a multiplied constant lead to the same equations of motion. Moreover, two Lagrange functions that differ by $\frac{d}{dt} f(q,t)$ do lead to the same equations of motion. The Lagrangian \begin{align} L'(q,\dot{q},t)=L(q,\dot{q},t) + \frac{d}{dt} f(q,t) \end{align} has the action \begin{align} S'=\int_{t_1}^{t_2} L'(q,\dot{q},t) \, dt = \int_{t_1}^{t_2} L(q,\dot{q},t) + \frac{d}{dt} f(q,t) \, dt=S + \left[ f(q,t) \right]_{t_1}^{t_2} \end{align} so the action $S'$ derived from $L'$ differs from $S$ derived from $L$ only by the term $f(q(t_2),t_2)-f(q(t_1),t_1)$. Since the end and starting point $q(t_2)$ and $q(t_1)$ are given as boundary conditions, they do not change during variation of $q(t)$. This means $\delta S= \delta S'$ and both Lagrangians lead to the equations of motion like we claimed above.