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In (mathematical) physics the equations of motion of a system of particles are often interpreted as Euler-Lagrange equations for appropriate Lagrangian $L=L(x,\dot x,t)$ where $x$ is a collection of variables.

As far as I understand in physics usually the equations of motion are experimentally determined first, and $L$ is chosen a posteriori to satisfy these equations and may be some other natural symmetries. In classical (not quantum) physics $L$ has no physical meaning other than the Euler-Lagrange equations.

Hence the problem of uniqueness of the Lagrangian arises: Can one classify all Lagrangians $L$ such that its Euler-Lagrange equations are equivalent to a given system of equations of motion? A reference would be very helpful.

Remark. (1) My question is somewhat vague since I am not sure what does it mean exactly "equivalent". To make it precise is a part of the question. One obvious option to define the equivalence is to say "the two systems of differential equations have exactly the same solutions".

(2) It is well known that the following two kinds of transformations of $L$ lead to equivalent (in any sense) systems of equations:

(a) $L\to aL+b$ where $a,b$ are constants.

(b)$L\to L+\frac{\partial F}{\partial x}\dot x+\frac{\partial F}{\partial t}$ where $F=F(x,t)$ is a function.

(3) I am far from this field and may not aware even of basic results in this direction.

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    $\begingroup$ Just a comment on your second paragraph: Arriving at the correct theory is usually a much more messy, iterative process. Initially, one might just experimentally observe a couple of symmetries. Then one might write down an ansatz for a Lagrangian that has those symmetries, derive equations of motion and make predictions. Test them with more experiments to further narrow down the ansatz for $L$. Iterate. $\endgroup$
    – Michael Engelhardt
    Commented Sep 24, 2020 at 13:33
  • $\begingroup$ @MichaelEngelhardt: Thank you. Do you have a reference where this procedure is explained for well known theories? At the moment I am particularly interested in Lagrangians of classical mechanics and classical electrodynamics. In all the literature I know the Lagrangians are chosen without commenting at all on their non-uniqueness and why they are chosen in that particular way. For me this would be very important to understand, but I thought this is not appropriate for the mathematical site. $\endgroup$
    – asv
    Commented Sep 24, 2020 at 13:40
  • $\begingroup$ Since you never know what experiment is going to tell you, there isn't really anything more specific to say than that one follows the scientific method. Now, if you go back far enough in history, such as the interesting sequence of developments that finally led to Maxwell's equations, you might indeed get the impression that one first guesses equations of motion and then formulates a Lagrangian - but this is only because the idea of basing theories on variational principles took a while to take hold. Nowadays, I think you'd try to guess a Lagrangian first ... (cont) $\endgroup$
    – Michael Engelhardt
    Commented Sep 24, 2020 at 13:59
  • $\begingroup$ ... because it's much more transparent and simpler to incorporate symmetries and other shreds of information you may have. From the physics point of view, it's secondary that the choice of $L$ is not unique, since in the end what matters are the predictions you get. You'd probably just stick with the simplest one using Occam's razor. $\endgroup$
    – Michael Engelhardt
    Commented Sep 24, 2020 at 14:02
  • $\begingroup$ @MichaelEngelhardt: "From the physics point of view, it's secondary that the choice of L is not unique, since in the end what matters are the predictions you get." I completely agree with this point. But this is true for classical equations only. The interesting thing happens when one tries to quantize the theory. Then the choice of $L$ is important, as far as I understand. A particular choice has to be made, and this step is never explained in the literature I know. $\endgroup$
    – asv
    Commented Sep 24, 2020 at 14:10

2 Answers 2

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In a sense, all the Lagrangians giving the same Euler-Lagrange equations are exhausted by transformations of your type (b), which adds a total derivative/total divergence/boundary term/... Transformations of your type (a) can alter the Euler-Lagrange equations, for instance if $a\ne 1$, then the EL equations get rescaled by the same constant $a$. Perhaps you don't care about such a rescaling. In that case, why care about any kind of violence that can be done to the EL equations, as long as they keep the same solutions. This point of view changes the question substantially, to the point where the general answer is not known.

The name given to your question is the "inverse problem of the calculus of variations" (also just mentioned in a comment by Robert Bryant!). The literature is vast, but there is one core result, which corresponds to the first part of the previous paragraph: (a) A Lagrangian $L(x,t)$ has vanishing equations of motion iff it is locally (in both independent and dependent variables) a total divergence. (b) $F(x,t)=0$ is locally an EL equation for some Lagrangian iff the linearization of $F(x,t)$ is a formally self-adjoint linear differential operator. Condition (b) is known as the Helmholtz condition.

To find references, a good start is typing in "inverse problem of the calculus of variations" into Google. Previously, it has come up on MO here, where you can also find some references, but they are not very up to date, since this field is still evolving:

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This problem is discussed in Bryant, Griffiths, Hsu, Exterior Differential Systems and Euler-Lagrange Partial Differential Equations, for Lagrangians for scalar fields.

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    $\begingroup$ If you want to look at the literature on this (which is vast), you should look for "the inverse problem in the calculus of variations". $\endgroup$
    – Robert Bryant
    Commented Sep 24, 2020 at 15:56

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