# Is there a universal algorithm that allows to find the Lagrangian/Hamiltonian, given **any** physical system that admits them?

Every time I have seen someone ask what the universally valid definition of Lagrangian/Hamiltonian is, the answer has often been one of the following two kinds:

• Partially or fully incorrect, i.e. not rigorous nor universal, but only valid for a subset of cases (e.g. the infamous "definition" $$L=T-V$$);
• Rooted in circular reasoning, e.g.: using Lagrangian, Hamiltonian, Action, Euler-Lagrange equations etc to define each other, in a "self-contained way" that does not connect to any other part of physics (which is something that -for example- $$L=T-V$$ would instead successfully achieve, if only it would be universally true).

At this point, in my attempts to understand the definition (or at least the universal characteristics) of these functions, my idea is just that these functions are (i) derived from empirical observation and (ii) case by case. Meaning that when we want to find these functions, there is no universally valid definition, but we procede in the following way, specific for each different physical system:

• Taking the principle of stationary action as for granted (fundamental & universal), elevating it as an axiom;

• Trying to tune our Lagrangian/Hamiltonian in such a way that the equations of motions that we have empirically observed come out of it.

This experimental approach has been the only universal characteristic I have found yet of these functions.

My question is: is it the only one, and if not, is there a universal algorithm that, given any physical system that admits them, allows to find its Lagrangian/Hamiltonian?

Edit: clarified that my question is about all the physical systems that admit Lagrangian/Hamiltonian formalism.

• Of course there is no universal algorithm… if there was, we could just declare victory and never do an experiment again. There are principles that can be more or less useful in each individual situation, but those principles are motivated by experiment as well. Commented Feb 17, 2022 at 19:21
• Given that there are plenty of physical systems that do not have a Lagrangian or Hamiltonian formulation (most notably dissipative physical systems), I would say no. Commented Feb 17, 2022 at 19:22
• @MichaelSeifert edited my question after your feedback Commented Feb 17, 2022 at 19:29
• Related/Possible duplicates: physics.stackexchange.com/q/20298/50583, physics.stackexchange.com/q/357775/50583, physics.stackexchange.com/q/429276/50583 Commented Feb 17, 2022 at 19:33
• Forget about Lagrangians and Hamiltonians. Do you know of a universal algorithm that would tell you what formula to plug in for the force in $F=ma$? We know of examples like the gravitational inverse-square law, the Lorentz force, and Hooke's law, but wouldn't you agree that ultimately these are justified empirically? The algorithm you're looking for would need to be able to deal with this question as a special case. Commented Feb 17, 2022 at 19:44

Your question is closely related to the inverse problem of Lagrangian mechanics. That article gives the Helmholtz conditions, which hold if and only if a given system of ODEs can arise as the Euler-Lagrange equations from some Lagrangian.

However, the Helmholtz conditions boil down to "there exists a matrix depending on $$q$$ and $$\dot{q}$$ satisfying certain properties", without necessarily giving you any clue how to find such a matrix. Since this proof is not constructive, it may not be the algorithmic sort of method you're looking for.

• Related on Mathematics (which is where I learned about the Helmholtz conditions): Is there a Lagrangian that produces these equations? How can I find one if it exists? Commented Feb 17, 2022 at 19:47
• I want to remark that there appears to be an ambiguity in what authors call the Helmholtz conditions. There are also Helmholtz conditions which are algorithmically calculatable, but only tell you if the given diff. eq. is precisely an Euler-Lagrange equation, and in this case a (local) Lagrangian can also be constructed by quadrature via the so-called Vainberg-Tonti formula. The "unknown" matrix mainly appears when you want to show if the given diff. eq. is equivalent to an EL equation (rather than "precisely the same"). There are some recent methods that can be helpful there as well: Commented Feb 17, 2022 at 23:04
• arxiv.org/abs/1210.0802 Commented Feb 17, 2022 at 23:04

No, L = T+V together with the Legendre transformation comes close (only for classical systems). But, for example, the standard model Lagrangian or the QED Lagrangian cannot be summarized in one class. Otherwise, it would be easy :).

Since the nature of the question seems to blend computing and physics, one way to look into it is from the perspective of the former. Not sure if the following quite makes sense, but I think it is worthwhile to see where this way of thinking leads to.

It is certainly possible to have a recursively enumerable procedure for the purpose, something that would propose a Lagrangian given a description of a physical situation (which would then be tested against an experiment, and accepted or rejected accordingly). Here, we only assume that a given physical situation (problem) can be encoded with a computable function (to define the input for the above). Note that it does not matter whether this function itself is correct or not.

Needless to say, this is utterly a theoretical construct and not useful in any way in practice.

But, let us try to translate the entire idea into the computing domain and suppose that there is an algorithm (Turing machine) that can spit out a correct Lagrangian for the given physical situation. Call this machine A. (That this is unlikely to be possible is indicated by the fact that even a seemingly simpler problem of whether a Hamiltonian is gapped or gapless is in general undecidable. (There are some technical assumptions that go with this, of course.)) In the absence of a physical experiment, we need to define correct. By that, we mean that when a given physical situation (experiment) is encoded into a Turing machine (which always can be done), call it X, we can certify that the output Lagrangian from A is equivalent to the one encoded in the Turing machine X. But, then, we would be able to test a non-trivial property of an arbitrary Turing machine (because X can be arbitrary), which is impossible. (That's a standard argument from computer science.)

Obviously, this is plausible (or, something along those lines if there is an error in the argument above) to the extent that one believes that every physical function (experiment, thought of as a function from preparation/input to outcomes/output) is computable, in the spirit of Church-Turing thesis. Note that any assumption of quantumness or (ontic) randomness in the description of X plays no role in the validity (or falsehood, for that matter) of the argument, because, as is well known, we can still simulate those processes using conventional (classical) algorithms.

• I was going to complain that this was unhelpful, but I reread the question and realized that technically the original post asks for a method find a Lagrangian when it exists and did not ask to be able to determine when one does not exist. But, I suspect that this is a distinction that only a computer scientist would love. Commented Feb 18, 2022 at 11:24
• Well, the last statement in the OP is: "... is there a universal algorithm that, given any physical system that admits them, allows to find its Lagrangian/Hamiltonian?" So, I gave an outline of an idea of how one might go about proving that it is impossible to have one. The OP used the notion of the algorithm in a very broad sense -- I believe exactly what a computable function is normally understood in computer science. There are special sub-categories of physical systems when a more precise answer can be had, such as the case of the inverse problem in the Langrangian mechanics. Commented Feb 18, 2022 at 11:51

My expectation is that inherently no such universal algorithm exists.

That said: there are patterns.

I will discuss statics, and classical mechanics. (For discussion of classical dynamics I refer to an answer I submitted in october 2021, on the subject of Hamilton's stationary action )

I will first discuss the catenary problem. I start with setting up the differential equation for the catenary, then I will discuss how to construct the corresponding Lagrangian. Then I will discuss how the variational approach finds the catenary

The catenary

The image shows a hanging chain in a state of force equilibrium.

The image is stylized, the idea is that at the cusps the chain can move freely. Imagine there are frictionless rollers at the cusps.
The length of the vertical sections is set up to provide the amount of tension force that is required for force equilibrium with the length of chain hanging between the cusps. (I use the line integral to compute the length of the chain, and I used the derivative to compute the angle at the cusps.)

(The image is a screenshot of an interactive diagram that is on my own website.)

As we know: the graph of the hyperbolic cosine gives the shape of the hanging chain.

First:
To set up the differential equation for the shape of the catenary.

Since the shape is symmetric it is sufficient to evaluate from the midpoint to the cusp.

The catenary is in force equilibrium at every point along its length. Hence at every point along its length the local tension force is tangent to the local slope.

With:
$$T_H$$ The horizontal component of the tension
$$λ$$ The weight per unit of length
$$L$$ the length of the chain from the midpoint to the x-coordinate.

The weight that has to be supported at coordinate x is given by multiplying the length L with the weight per unit of length: $$λL$$

The differential equation setup here uses the following property: gravity is acting in the vertical direction therefore there is no gradient in the horizontal component of the tension. That is: the horizontal component of the tension is not a function of the x-coordinate, but constant

In preparation: from midpoint to cusp the slope of the curve increases; the length of chain per unit of x-coordinate increases accordingly. (1.1) gives an expression for dL/dx, which will later be used.

$$(dL)^2=(dx)^2+(dy)^2 \quad \Leftrightarrow \quad \frac{dL}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \tag{1.1}$$

At every point along the catenary the slope of the curve is equal to the ratio of horizontal tension component and vertical tension component:

$$\frac{dy}{dx} = \frac{\lambda L}{T_H} \tag{1.2}$$

We need an expression with the derivative of $$L$$ with respect to $$x$$, so that (1.1) can be put to use. $$T_H$$ and $$λ$$ are constants, so we can treat $$λ$$/$$T_H$$ as a constant multiplication factor.

$$\frac{dy}{dx} = L \frac{\lambda}{T_H} \tag{1.3}$$

Taking the derivative:

$$\frac{d^2y}{dx^2} = \frac{\lambda}{T_H} \frac{dL}{dx} \tag{1.4}$$

Combining (1.4) and (1.1) allows us to eliminate the quantity dL, arriving at a differential equation that is is purely in terms of the cartesian coordinates x and y:

$$\frac{T_H}{\lambda}\frac{d^2y}{dx^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \tag{1.5}$$

Given the form of the right hand side of (1.5) it is clear that if the differential equation has a solution at all it must involve the hyperbolic cosine; the sinh and cosh are related according to the equation: $$sinh^2 + 1 = cosh^2$$

(1.6) incorporates the ratio $$\tfrac{T_H}{λ}$$ such that it satisfies (1.5)

$$y = \tfrac{T_H}{\lambda} \cosh \left(\tfrac{\lambda}{T_H} x \right) \tag{1.6}$$

Variational approach

Take, as usual, the horizontal axis as the x-coordinate, and the vertical axis as the y-coordinate.

The objective is to find the y-coordinate as a function of the x-coordinate.

Differential approach is that you probe the function by introducing an infinitisimally small increment of the x-coordinate, and you examine how the y-coordinate responds to that.

Variational approach operates at right angles to that.
You probe the function by introducing an infinitisimally small increment, but that increment is increment of the y-coordinate.

The variation of variational calculus is always variation of the other coordinate.

In the case of the catenary: The variation is variation of height; the y-coordinate

In the case of classical dynamics:
The equation of motion gives position as a function of time. If you plot time coordinate and position coordinate in a cartesian coordinate system: the variation of height that is applied is variation of the coordinate that is at right angles to the time coordinate.

The relation between force equilibrium and potential energy level

The potential energy as a function of position is obtained by evaluating the integral of force with respect to the position coordinate.

That is the definition of potential energy: it is the integral of exerted force with respect to the position coordinate.

In the case of the catenary setup as represented in the image:
When the hanging part sags lower the length of the hanging part increases, and the vertical sections move up by that amount.

We have that the vertical section(s), and the catenary section, are each a repository of potential energy.

When the catenary section sags down the catenary section is doing work, raising the potential energy of the vertical section(s).

When the catenary section is lifted up the vertical section is doing work, raising the potential energy of the catenary section.

To find the equilibrium position you find the shape such that there is no longer any opportunity to do work.

Any shape of the catenary that still has opportunity for conversion of potential energy is not (yet) the shape of lowest possible potential energy. The catenary shape is there when there is no longer opportunity for conversion of potential energy; lowest possible potential energy.

If you probe with an infinitisimally small variation and you find that over that infinitisimally small variation an equal amount of work is done either way: that is the equilibrium point.

Variational approach

The variational approach consists of the following:
You take the derivative with respect to the position coordinate of each of the repositories , and then you identify the point where those derivatives have the same value.

In the case of the catenary: the variational approach takes the derivative of the potential energy with respect to the position coordinate.

As we know: since potential energy is the integral of exerted force with respect to the position coordinate: the derivative of the potential energy with respect to the position coordinate is the force.

Hence the variational approach does the same thing as the differential approach does: the procedure finds the force equilibrium.

As mentioned at the start:
For discussion of classical dynamics I refer to an answer I submitted in october 2021, on the subject of Hamilton's stationary action )

General discussion:
The variational approach takes the derivative with respect to the position coordinate. So: in order to do the same thing as the differential equation the inputs of variational approach must first be converted to the corresponding integral with respect to position coordinate.

That is how variational approach is applied in Statics. You integrate the force with respect to the spatial coordinate, and then - to find the point of lowest possible potential energy - you take the derivative of the potential energy with respect to the position coordinate.

How variational approach is applied in Classical mechanics:
The integral of $$F$$ (Force) with respect to the position coordinate is the potential energy.
The integral of $$ma$$ with respect to the position coordinate is $$\tfrac{1}{2}mv^2$$, kinetic energy. Your objective is to find the trajectory such that the rate of change potential energy is equal to the rate of change of kinetic energy. To identify that point of equal rate of change you take the derivative of the energy with respect to the spatial coordinate.