# A Lagrangian that is a vector... Can I apply the Euler Lagrangian equations to get equations of motion?

In physics, one only sees Lagrangian that are scalar-valued:

$$L:t,q,\dot{q}\to \mathbb{R}$$

whose integral over time is the action

$$S=\int_0^tL(t,q,\dot{q})dt$$

In my own amusement research I have ended up with a Lagrangian that is vector valued:

$$L:t,q,\dot{q}\to \mathbb{R}^n$$

For instance

$$\mathbf{L}=e_0L_t(t,q,\dot{q})+ e_1L_x(t,q,\dot{q})+e_2L_y(t,q,\dot{q})+e_3L_z(t,q,\dot{q})$$

A-priori, there doesn't seem to be any reason why this shouldn't be workable in some sense. But, what would be the interpretation of such a thing: each orthogonal axis are 'independent' sub-systems and the master system cannot 'rotate' between them?

Is there any literature which investigate such Lagrangian with perhaps physical applications?

Does the Euler-Lagrangian equations apply to it - can I get $$n$$ independent equations of motions via:

$$e_1\frac{\partial L_1}{\partial q}(t,q,\dot{q})-e_1\frac{d}{dt}\frac{\partial L_2}{\partial \dot{q}}(t,q,\dot{q}) =0\\ \vdots\\ e_2\frac{\partial L_2}{\partial q}(t,q,\dot{q})-e_2\frac{d}{dt}\frac{\partial L_2}{\partial \dot{q}}(t,q,\dot{q})=0$$

• What would be the analogue of the action in that case? If it's a vector, it's not at all clear what it would mean to minimize such a vector function. If the action is scalar, then the Lagrangian is just scalar again. Commented Feb 24, 2021 at 13:18
• A vector Lagrangian is going to make a Hamiltonian a bit funny, too. Commented Feb 24, 2021 at 13:18
• @Semiclassical The action would remain a vector. $S=\int (e_1 L_1+e_2L_2) dt=e_1\int L_1 dt+e_2\int L_2dt$ Commented Feb 24, 2021 at 13:36
• @Semiclassical Wouldn't you just minimized each part of the vector independently? Commented Feb 24, 2021 at 13:54
• That's not possible in general. Take the functions $f(x)=x^2$, $g(x)=(x-1)^2$. You can minimize either of them independently ($x=0$ for the first, $x=1$ for the second). You can also minimize any linear combination of them. But there's no $x$ which minimizes both of them simultaneously. Commented Feb 24, 2021 at 13:57