What can be inferred about this particle from a Lagrangian?

Question

If Lagrangian, $\mathscr L = \dot{q}^2 - q \dot{q}$. Then what can be inferred about the particle?

Simply that it is a free particle or something else?

Answer 1

It is well known that adding a total time derivative to the Lagrangian does not change equations of motion.

The Lagrangian above adds a term $$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$ (a total time derivative) to the free particle Lagrangian $\dot q^2$. It is thus fully equivalent to to the standard free particle Lagrangian (up to an irrelevant overall factor).

Answer 2

\begin{equation} \mathcal{L}\left( q,\dot{q},t\right)= \dot{q}^{2} - q\dot{q} \tag{01} \end{equation}

\begin{equation} \dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{q}}\right)-\dfrac{\partial \mathcal{L}}{\partial q}=0 \tag{02} \end{equation}

\begin{equation} \dfrac{d}{dt}\left[\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial \dot{q}}\right]-\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial q}=0 \tag{03} \end{equation}

\begin{equation} \dfrac{d}{dt}\left(2\dot{q}-q \right)- \left(-\dot{q}\right)=0 \tag{04} \end{equation}

\begin{equation} 2\ddot{q}-\dot{q}+\dot{q}=0 \tag{05} \end{equation}

\begin{equation} \ddot{q}=0 \Longrightarrow \dot{q}= \text{constant} \tag{06} \end{equation}

If generalized $\:q\:$ is Cartesian coordinate then : free particle . But if not, for example is an angle coordinate : something else but not free particle.