# Why do such different Lagrangians give the same Euler-Lagrange equations?

The Lagrangians $$L_\pm:=\frac{1}{2}\left( \dot{q}_1^2\pm\dot{q}_2^2\right)-\frac{1}{2}m^2\left( q_1^2\pm q_2^2\right)$$ each have Euler-Lagrange equations $$\ddot{q}_i=-m^2q_i,\,i\in\left\{ 1,\,2\right\}$$. The obvious reason is that these equations are uncoupled, so any linear combination of "one-equation" Lagrangians, including these $$\pm$$ choices, will give the same pair.

The Lagrangian $$L_0:=\dot{q}_1\dot{q}_2-m^2q_1q_2$$ has the same ELEs as well, so it doesn't couple the $$q_i$$ even though it looks like it might. What's the physical interpretation of $$L_0$$ obtaining the same ELEs as $$L_+$$? It doesn't seem to be of the form $$a_+L_++a_-L_-+b+\dot{f}$$.

Thinking in terms of Hamiltonians doesn't make it any clearer. With $$L_0$$ as our choice of Lagrangian, the Hamiltonian is $$H_0:=p_1p_2+m^2q_1q_2$$, which bears no obvious equivalence to $$H_\pm:=\frac{1}{2}\left( p_1^2\pm p_2^2\right)+\frac{1}{2}m^2\left( q_1^2\pm q_2^2\right)$$ (The momenta have different definitions in terms of the $$\dot{q}_i$$ in the two cases, but of course we still get the same equations of motion in all cases.)

So is there a more general principle that explains why $$L_\pm$$ is equivalent to $$L_0$$, or $$H_\pm$$ to $$H_0$$?

• One may rightly ask about the physics of a negative kinetic energy $-\frac{1}{2}\dot{q}_2^2$. Mar 27 '17 at 16:24
• @ZeroTheHero I think the other sign change on the potential makes up for it. "What's this, a negative KE & a PE bounded above instead of below?" "You have the graph upside down."
– J.G.
Mar 27 '17 at 16:29
• Take your second Lagrangian, do the transformation $q_1 = Q_1+Q_2$, $q_2 = Q_1-Q_2$. Mar 27 '17 at 16:53
• I think the idea is that your original system is like two particles on separate springs with coincidentally the exact same angular velocity. The second system is then basically equivalent to putting an extra (massive) spring connecting the two particles, as can be seen by adding and subtracting $q_1 + q_2$ terms, but this has no physical effect. If the angular velocities were different, it would. Mar 27 '17 at 16:55
• @knzhou Perhaps I can visualise that, if two springs are joined by a third while they move in lockstep, nothing changes. Suppose we instead ask about $\partial_\mu\phi_1\partial^\mu\phi_2-m^2\phi_1\phi_2$. What would be the equivalent of an unimportant "coupling" between the fields?
– J.G.
Mar 27 '17 at 16:58

Having thought about this, I think I've found the key point. A Lagrangian of the form $$\frac{1}{2}A\dot{q}^2-\frac{1}{2}Bq^2$$ with $$1$$-dimensional $$q$$ and numbers $$A,\,B$$ with $$A\ne 0$$ has equations of motion that depend only on $$A^{-1}B$$, as scaling the Lagrangian is physically irrelevant. In particular, scaling the Lagrangian can be thought of as a scaling of the coefficients. For multi-dimensional $$\mathbf{q}$$, we can generalise this result. In a Lagrangian $$L_{AB}:=\frac{1}{2}\dot{\mathbf{q}}^TA\dot{\mathbf{q}}-\frac{1}{2}\mathbf{q}^TB\mathbf{q}$$ we can assume without loss of generality that $$A,\,B$$ are symmetric, in which case $$\mathbf{p}=A\dot{\mathbf{q}}$$ so the equation of motion is $$A\ddot{\mathbf{q}}=-B\mathbf{q}$$. If $$A$$ is invertible this simplifies to $$\ddot{\mathbf{q}}=-A^{-1}B\mathbf{q}$$, which is invariant under a "scaling" of the matrix coefficients of the form $$A,\,B\to KA,\,KB$$, with $$K$$ an arbitrary invertible square matrix conformable with $$A,\,B$$ for which $$KA,\,KB$$ are symmetric. We can find the same result in the Hamiltonian formalism, viz $$H_{AB}:=\frac{1}{2}\mathbf{p}^TA^{-1}\mathbf{p}+\frac{1}{2}\mathbf{q}^TB\mathbf{q}$$.