Do the same equations of motion imply the same Lagrangians? [duplicate]

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

marked as duplicate by ACuriousMind♦, Kyle Kanos, Kyle Oman, Ryan Unger, MartinMay 4 '15 at 13:47

• What exactly do you mean by 'equivalent'? – AV23 May 1 '15 at 13:51
• Same functional form up to a constant or total derivative – SuperCiocia May 1 '15 at 14:28
• This question (v2) is essentially a duplicate of physics.stackexchange.com/q/131925/2451 – Qmechanic May 1 '15 at 15:39

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.

• In the field-theoretic context, one allows for a total divergence $\partial_\mu \Phi$ rather than just a time derivative (one also integrates the density over space). – Danu May 1 '15 at 13:55
• I feel like this doesn't really answer the question. It gives a sufficient condition for generating the same equations of motion, but not (as AV23's answer shows) a necessary one. – user12029 May 2 '15 at 17:24

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).

• What the physical connection between the two Lagrangians? Which one should we choose? – SuperCiocia May 1 '15 at 21:52
• The 'physical connection' in the first case is through the energies, and in the second case, well, one is the square of the other (effectively square distance vs distance). We normally choose $L_1$ and $L$ as opposed to $L_2$ and $L'$ respectively because of the ease of combining systems: for example, if you want to include electromagnetism, you simply add terms to $L_1$ and $L$. You can't do it so easily for $L_2$. – AV23 May 2 '15 at 8:58
• Minor comment to the last example in the answer (v3): Technically speaking, the square-root Lagrangian and the non-square-root Lagrangian lead to slightly different EL eqs. As is well-known, the solutions are for both eqs. (parametrized) geodesics, but where the former eq. produces all parametrized geodesics, the latter eq. only produces affinely parametrized geodesics, cf. e.g. my Phys.SE answer here. – Qmechanic May 2 '15 at 13:08
• @Qmechanic Hmmm...yes, the correction for the general parametrization in terms of $\lambda$ would then be an extra factor of $(\frac{ds}{d\lambda})^{-1}$ (constant for affine parametrization) in the Lagrangian $L'$, practically making it equivalent to $L$. This is perhaps not the best example then. – AV23 May 2 '15 at 17:08