Given a Newtonian system (with forces that may depend position, velocity, and time) with solutions $\vec{q}(t)$, can we always derive a Lagrangian system such that the following are satisfied:

  1. It yields the same trajectories $\vec{q}(t)$.
  2. It is possibly time-dependent.
  3. It has no generalized forces, so the solutions are generated by the action principle and they satisfy the original Euler-Lagrange equations $\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{i}}) - \frac{\partial L}{\partial q_{i}} = 0$.

If the answer is yes, is there a proof that demonstrates this to be the case? If the answer is no, is there an explicit counterexample that can be given?

For background info, I've looked over this post and this post, which explain that some Newtonian systems can be converted to Lagrangian systems with generalized forces, but this isn't what I'm looking for since it doesn't satisfy (3). I've also seen this interesting webpage, but the post operates under the restriction that the Lagrangian cannot be time-dependent, so the webpage doesn't satisfy (2).

Whether the notions of momenta, energy, and generalized forces in the Lagrangian system match the corresponding notions in Newtonian mechanics is not important here (they may fail to equal if the Lagrangian is time-dependent I believe). I am only interested in satisfying (1) (along with (2) and (3)), which is that the two systems generate the same trajectories $\vec{q}(t)$.

I thought maybe a damped oscillator would have served as a counterexample, but apparently I found the following:

This is sort of why I want a concrete counterexample with a proof that it has no Lagrangian.

  • 1
    $\begingroup$ Yes, e.g. (possibly time-dependent) dissipative or non-holonomic systems are generically counterexamples. $\endgroup$
    – Qmechanic
    Commented Apr 27, 2021 at 7:25
  • $\begingroup$ @Qmechanic I appreciate the suggestions, but if you see my edit note at the end of the post, it seems dissipative systems can have a Lagrangian. This is sort of why I want an explicit counterexample with proof that no Lagrangian can produce the same trajectories. (Also, I'm not familiar with the holonomic vs non-holonomic distinction.) $\endgroup$ Commented May 19, 2021 at 22:42
  • $\begingroup$ I didn't address this back when I wrote my answer, but I cannot put the statement 'it has no generalized forces' into context. What I mean is: the concept 'generalized force' subsumes the more specific 'force expressed in cartesian coordinates*. Its not that there is some dichotomy between 'cartesian coordinates' on one side and 'generalized coordinates' on the other side. To my understanding: generalized coordinates is a higher level set of which cartesian coordinates is a subset. It would appear you are using the expression 'generalized force' in a sense that I am unaware of. $\endgroup$
    – Cleonis
    Commented Dec 23, 2021 at 22:22
  • $\begingroup$ About the case of a damped oscillator. As we know: at the level of atoms no such thing as damping of an oscillator exists. So it seems to me that whether or not the case of a damped oscillator has a Lagrangian counterpart has no bearing on our understanding of physics. (I'm assuming here that all physics at macrosopic level arises from physical properties at microscopic level.) $\endgroup$
    – Cleonis
    Commented Dec 23, 2021 at 22:49

2 Answers 2


About narrowing down search to possible counterexamples:

In order to define a potential (as a function of spatial coordinates) the integral of force over distance must be uniquely defined.

Example of satisfying that requirement: the inverse square law of gravity. The difference in potential between two altitudes is independent of how an object moves from one altitude to the other. the change of velocity caused by the gravitational field is independent of the already existing velocity.

A way to thwart the Lagrangian formulation is to mess with the definability of the integral. Example: rolling while disallowing slipping. For instance, a marble on an incline, rolling without slipping. Then the specific path that the marble takes from point A to point B determines what happens to the orientation of the marble, so there is no uniquely defined integral.

Then again, 'rolling without slipping' is rather contrived, and it is exclusive to the macroscopic world; I don't think rolling without slipping is possible at all at atomic scale.

The Foucault pendulum setup is - to my knowledge - always treated in terms of the newtonian formulation. I've seen the Foucault pendulum setup described as a 'non-holonomic system'. When the platform has gone through a complete rotation the pendulum plane of swing is not back to the initial orientation.

One can readily visualize a molecular counterpart of a Foucault pendulum setup. You take a large molecule, with one group sticking out, almost like an antenna. That antenna will have vibration modes, and the molecule as a whole will have rotation modes. There will be coupling between the rotational modes and the vibrational modes.

The historic Foucault setup is a pendulum, but it doesn't have to be a pendulum, it can also be a vibrating rod. In fact, the idea for the Foucault pendulum occurred to Foucault as follows. He had placed a long stiff rod in the jaws of a lathe, he had twanged the rod, and when he turned the spindle of the lathe the the plane of vibration remained in the same plane. A tabletop demonstration of the Foucault effect will use either a rod or a high tension coil spring.

For example, within months after the opening of the demonstration in the Pantheon Charles Wheatstone published a note titled: Note relating to M. Foucault's new mechanical proof of the Rotation of the Earth

Wheatstone describes his tabletop device, and the rotational-vibrational coupling effect that he observed.

  • $\begingroup$ You may be interested in looking at the answer I gave to myself. I was looking for something more direct. However, I appreciate this post. $\endgroup$ Commented Dec 23, 2021 at 19:18
  • $\begingroup$ @MaximalIdeal The treatment by Jesse Douglas is entirely in the abstract. That is valid of course, but it's an approach that for my purpose is out of scope. (Also, I don't have the mathematical skill to understand the article by Douglas.) My focus: physical interpretation. (Example of physical interpretation focus: the youtube video about Heisenberg uncertainty by Grant Sanderson.) As to the inverse problem: which differential equations can be restated in variational form says something about the physics. My interest is the physics ramifications. $\endgroup$
    – Cleonis
    Commented Dec 23, 2021 at 22:03

I've found some material that essentially answers my question. The problem of finding a Lagrangian for a system of second-order ordinary differential equations is called the the inverse problem of Lagrangian mechanics. According to wikipedia, the question of whether a Lagrangian exists was solved by Jesse Douglas (although I've seen some conflicting information on this for some reason).

I'll provide some relevant links and references here.

I want to warn readers that the inverse problem for Lagrangian mechanics can come in two flavors: one specific and another more broad. I present these in terms of two questions.

Question 1. Given a system $\ddot{x}_{i} = F_{i}(x, \dot{x}, t)$, does there exist a Lagrangian $L = L(x, \dot{x}, t)$ such that $$\frac{d}{dt}\frac{\partial L}{\partial\dot{x}_{i}} - \frac{\partial L}{\partial x_{i}} = \ddot{x}_{i} - F_{i}(x, \dot{x}, t)$$ for all $i$?

Question 2. Given a system $\ddot{x}_{i} = F_{i}(x, \dot{x}, t)$, does there exist a Lagrangian $L = L(x, \dot{x}, t)$ such that when setting the Euler-Lagrange expressions to zero we obtain differential equations that yield the same solutions as the original differential equations?

The two questions are not the same. See the answers to my post here for an example. Be careful in determining which question is being answered in any paper you read!!

Example. Anyways, Jesse Douglas himself provided two examples of systems with no Lagrangian on page 81 (page 11 in the PDF). I combined the two examples and changed the notation. The example is \begin{align*} \ddot{x}_{A} &= x_{A}^{2} + x_{B}^{2}, \\ \ddot{x}_{B} &= \alpha x_{A} \end{align*} where $\alpha$ is any real number.

According to the wikipedia page, a Lagrangian exists if and only if there is a non-singular symmetric matrix $$ g = \begin{pmatrix} g_{11} & g_{D} \\ g_{D} & g_{22} \end{pmatrix} $$ such that the Helmholtz conditions $(\text{H}1)$-$(\text{H}3)$ are satisfied. In our example, we have $$ \Phi = \begin{pmatrix} -2x_{A} & -2x_{B} \\ -\alpha & 0 \end{pmatrix}. $$ Because our system has no explicit time dependence, $(\text{H}2)$ significantly simplifies things by telling us that $dg/dt = 0$ in our example. This implies $g$ has no dependence on time, positions, or velocities, and so if it exists, it consists of constants.

By looking at condition $(\text{H}1)$ and doing some work, I find that I need to satisfy $$ x_{A}g_{D} + x_{B}g_{22} = \frac{\alpha g_{11}}{2}. $$ Since the RHS has no $x_{A}$- or $x_{B}$-dependence, we must have $g_{D} = 0$ and $g_{22} = 0$ (this further implies that either $\alpha = 0$ or $g_{11} = 0$ but that actually doesn't matter to us). But then $g$ is a singular matrix.

Therefore, the desired matrix $g$ satisfying the Helmholtz conditions does not exist and our system has no Lagrangian counterpart.


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