It's clear to me that in acquiring the equations of motion using the Euler-Lagrange equations and the Lagrangian, defined as$^\dagger$ $$L \equiv T - V,$$ the potential energy may have an explicit time dependence if the generalized coordinates depend on the regular coordinates and time. It's not clear to me whether or not the potential energy is allowed to depend explicitly on time even if we're using regular coordinates, i.e., it's not clear whether or not a potential energy of the form $$V = V\left(\mathbf{r}_1, \dots, \mathbf{r}_N, t\right)$$ can be used to arrive at the equations of motion the "normal way".$^\ddagger$ Can such a potential energy be used? More generally, what kind of potential energies are allowed?

$^\dagger$ For the purposes of my question, this is the definition, though I am aware that there are other Lagrangians.

$^\ddagger$ By the "normal way" I mean via application of the Euler-Lagrange equations, $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} = \frac{\partial L}{\partial q_i}.$$

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    $\begingroup$ With a time-dependent $V$, Lagrangian formalism still works fine, and it still gives the equations of motion in the usual way, but there is no conserved energy (if "energy" is defined as usual via Noether's theorem). Intuitively, if $V$ depends on time (other than via the coordinates $\mathbf{r}_n$), then the laws of physics (aka equations of motion) don't have time-translation symmetry. That ruins the conservation of energy. (Posting this as a comment because it doesn't answer the question "what kind of potential energies are allowed?") $\endgroup$ Apr 8, 2019 at 0:46
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    $\begingroup$ @ChiralAnomaly, think a massive charged particle in a time-varying EM field. No conservation of mechanical energy, of course, but still a physical setup. $\endgroup$
    – kkm
    Apr 8, 2019 at 2:23
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    $\begingroup$ @kkm Good example. The important distinction is between time-dependence that enters only via the dynamic variables (the things governed by the equations of motion [EOMs], such as the $\mathbf{r}_n$ in the OP's example) versus time-dependence that enters via non-dynamic structures (the things that define the EOMs, such as the background EM field in your example). One model's background field can be another model's dynamic field, like the metric field in special relativity (background) versus general relativity (dynamic, unless considering test-particle motion in a given curved metric). $\endgroup$ Apr 8, 2019 at 13:05
  • $\begingroup$ This has the same issue I pointed out in the other bountied question -- you're asking answerers to use an artificially restrictive definition, then asking what would happen if that restrictive definition was applied to situations where it doesn't work well. As a class of question, it doesn't make sense. The right questions to ask are: what is the more general definition, and how does it reduce to the more restrictive definition in simple cases, and when does that reduction fail? $\endgroup$
    – knzhou
    Feb 10, 2020 at 8:09
  • $\begingroup$ @knzhou I see where you're coming from, but at this stage I'm not concerned with understanding more general stuff, I just want to understand the limits of applicability of the method I know. $\endgroup$
    – user113773
    Feb 10, 2020 at 8:34

1 Answer 1

  1. Within Lagrangian point mechanics, a local Lagrangian $L(q,\dot{q}, \ddot{q}, \ldots;t)$ could in principle be higher-order, but often it is assumed to only depend on generalized positions $q$, generalized velocities $\dot{q}$, and time $t$, cf. e.g. this & this Phys.SE posts.

  2. The Lagrangian $L$ does not need to be of the form of kinetic energy minus potential energy $T-U$, cf. e.g. this Phys.SE post.

  3. A generalized potential $U(q,\dot{q},t)$ could in principle depend on all variables. Think e.g. of a charged point particle in a time-dependent E&M background, cf. my Phys.SE answer here and above comment by user kkm.

  4. A system with an external time-dependent background source is a prime example of a potential $U$ with explicit time-dependence, cf. e.g. this Phys.SE post.

  • $\begingroup$ For the purposes of my question, a Lagrangian is how I've defined it. What I'm asking is under what assumptions is $$\frac{d}{dt}\frac{\partial L}{\partial \dot q} = \frac{\partial L}{\partial q}$$ valid? I'll edit my question to be clearer. $\endgroup$
    – user113773
    Feb 10, 2020 at 7:54

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