# Non-uniqueness of the Lagrange Function

### From bio-physics-wiki

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.

The Lagrangian is determined only up to multiplication by a constant and addition of a function containing the total derivative with respect to time.

Further Reading:

- L. D. Landau & E. M. Lifschitz - Lehrbuch der Theoretischen Physik I: Mechanik