Do the same equations of motion imply the same Lagrangians?

Question

If two Lagrangian (densities) $\mathcal{L}$ give the same equations of motion, are they equivalent?

Answer 1

As it happens, it is not necessary that two Lagrangians that have the same equation of motion have the same functional form. Consider the Lagrangians $L_1 = T-V$ and $$L_2 = \frac{1}{3}T^2 + 2TV - V^2$$ where $T = \frac{1}{2}m\dot{x}^2$ and $V(x)$ is the potential energy. They both lead to the same equation of motion: $$m\ddot{x} = -\frac{dV}{dx}$$

This is worked out in detail in this answer.

To compare the two further, let us obtain the Hamiltonian of $L_2$ (The Hamiltonian of $L_1$ is obviously $H_1 = T+V$). We observe that (without bothering to express the Hamiltonian as a function of the momentum) $$\frac{\partial L_2}{\partial \dot{x}} = \frac{1}{3}m^2\dot{x}^3 + 2m\dot{x}V$$ $$H_2 = \frac{1}{3}m^2\dot{x}^4 + 2m\dot{x}^2V - \frac{1}{12}m^2\dot{x}^4 - m\dot{x}^2V+V^2$$ $$\implies H_2 = \frac{1}{4}m^2\dot{x}^4 + m\dot{x}^2V + V^2 = \left(T+V\right)^2$$ i.e. $$H_2 = (H_1)^2$$ Thus, while the relation between the two in the Lagrangian framework is not very obvious, we can easily see how they are related by comparing the Hamiltonians. Both Lagrangians do not explicitly depend on time, and therefore conserve $T+V$ or $(T+V)^2$ respectively, which is essentially the same thing.

Another famous example (though of a less serious nature in the context of this question) is the Lagrangian of a free relativistic particle, which is given by $$L = \sqrt{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}$$ The equations of motion are then the geodesic equations. It is well known that the square of this Lagrangian, $$L' = g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$$ also leads to the same equations of motion (for affine parametrizations of the path - see Qmechanic's comment below).

Answer 2

If by 'equivalent' you mean equal, then no. They can clearly differ by a constant, but they moreover can differ by a total time derivative. So if two lagrangians $L_{1}$ and $L_{2}$ are such that $L_{1} - L_{2} = \frac{\mathrm{d} \Phi}{\mathrm{d}t}$ for some function $\Phi$, then they lead to the same equations of motion. You can find a proof of this in Jose and Saletan's Classical Dynamics: A Contemporary Approach.

EDIT: Although if I recall correctly, they only deal with Lagrangians of two freedoms (a generalized position and velocity), leaving the more general case as an exercise.