We see variational principles coming into play in different places such as Classical Mechanics (Hamilton's principle which gives rise to the Euler-Lagrange equations), Optics (in the form of Fermat's principle) and even General Relativity (we get Einstein's equation from the Einstein-Hilbert action). However, how do we explain this very principle, i.e., more mathematically, I want to ask the following:

If I am given a set of generalized positions and velocities, say, $\{q_{i}, \dot{q}_{i}\}$, which describes a classical system with known dynamics (equations of motion), then, how do I rigorously show that there always exists an action functional $A$, where $$A ~=~ \int L(q_{i}, \dot{q}_{i})dt,$$ such that $\delta A = 0$ gives the correct equations of motion and trajectory of the system?

I presume historically, the motivation came from Optics: i.e., light rays travel along a path where $S = \int_{A}^{B} n ds$ is minimized (or at least stationary). (Here, $ds$ is the differential element along the path). I don't mind some symplectic geometry talk if that is needed at all.

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/20188/2451 $\endgroup$ – Qmechanic Jan 31 '12 at 12:25
  • $\begingroup$ This is, in full generality, an open problem, by the name of "inverse problem of calculus", see Chapter 5 in "Olver - Applications of Lie Groups to Differential Equations". $\endgroup$ – ungerade Dec 6 '19 at 19:31

I) Not all equations of motion (eom) are variational. A famous example is the self-dual five-form in type IIB superstring theory. In classical point mechanics, frictional forces typically lead to non-variational problems.

II) Consider for instance $n$ variable $q^i$ and $n$ eoms,

$$\tag{1} E_i~\approx~ 0, \qquad i~\in~\{1, \ldots, n\}. $$

A simplified version of OP's problem (v3) is the following:

Does there exist an action $$\tag{2} S[q] ~=~\int{\rm d}t~L$$ such that the Euler-Lagrange derivatives $$\tag{3} \frac{\delta S}{\delta q^i}~=~E_i $$ precisely become the given $E_i$-functions?

The above restricted problem is relatively easy to answer once and for all, because one may differentiate the known $E_i$-functions to arrive at a set of consistency conditions. Let us for simplicity assume that the functions $E_i=E_i(q)$ do not involve generalized velocities $\dot{q}^i$, accelerations $\ddot{q}^i$, and so forth. Then we may assume that the Lagrangian $L$ does not depend on time derivatives of $q^i$ as well. So the question becomes if

$$\tag{4} \frac{\partial L}{\partial q^i}~=~E_i ? $$

We can collect the information of the eoms in a one-form

$$\tag{5} E~:=~E_i ~{\rm d}q^i.$$

The question rewrites as

$$\tag{6} {\rm d}L~=~E? $$

Hence the Lagrangian $L$ exists if $E$ is an exact one-form.

III) However, the above discussion is in many ways oversimplified. The eoms (1) do not have a unique form! E.g. one may multiply the given $E_i$-functions with an invertible $q$-dependent matrix $A^i{}_j$ such that the eoms (1) equivalently read

$$\tag{7} \sum_{i=1}^n E_i A^i{}_j~\approx~ 0. $$

Or perhaps the system variables $q^i$ should be viewed as a subsystem of a larger system with more dynamical or auxiliary variables?

Ultimately, the main question is whether the eoms have an action principle or not; the particular form of the eoms (that the Euler-Lagrange equations spit out) is not important in this context.

This opens up a lot of possibilities, and it can be very difficult to systematically find an action principle; or conversely, to prove a no-go theorem that a given set of eoms is not variational.

  • $\begingroup$ Is it true that frictional forces lead to non-variational problems? By allowing time-dependent Lagrangians they can be included, right? $\endgroup$ – dan-ros Apr 6 '16 at 10:47
  • $\begingroup$ Generic frictional forces lead to non-variational problems. In certain simple situations, an unconventional approach exists, cf. e.g. my Phys.SE answer here. $\endgroup$ – Qmechanic Apr 6 '16 at 11:56
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    $\begingroup$ @Qmechanic It looks like your comment regarding the issue with the self-dual five-form in Type IIB supergravity is now outdated. This paper (arxiv.org/abs/1511.08220) shows how to construct an action for Type IIB SUGRA without applying any additional constraints at the level of the equations of motion. I'm interested in the topic of your answer, so would you happen to have any other examples of non-variational equations of motion, aside from frictional forces, you could add? $\endgroup$ – Kenny H Mar 21 '18 at 16:37
  • $\begingroup$ @Kenny H: Thanks for the feedback. I plan an update. $\endgroup$ – Qmechanic Mar 21 '18 at 16:56
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    $\begingroup$ @glS: The answer should be viewed as a simplified appetizer for the full theory behind the variational bicomplex. $\endgroup$ – Qmechanic Apr 26 at 14:18

Obviously one can mathematically cook up equations of motion that would not arise from an action principle.

The original motivation for believing that Nature obeys a Law of Least Action was metaphysical, and then it turned out that in reality, one could only guarantee that the action was stationary, not necessarily minimal, which ruined the metaphysics... besides, one must be cautious about postulating that Nature has to do something which we have deduced based on philosophical motivations.

But ever since Hertz and Einstein, there has been another motivation. (Whether it will stand the test of time better than string theory, remains to be seen...) Gauss, Hertz, and after them, Klein (see Whittaker, Analytical Dynamics, p. 254ff. and Hertz, The Principles of Mechanics, http://www.archive.org/details/principlesofmech00hertuoft ) reformulated Newtonian Mechanics in terms of an abstract curved space on which all particles followed geodesics. The metric on the space was cooked up from the forces acting on the system, and all the laws of mechanics reduced to Hertz's principle of least curvature instead of least action. Now after Einstein we know that if we interpret gravity as the metric of space-time, then particles under the influence of gravity follow a geodesic. This is a generalisation of the very old principle of inertia: with Newton it was stated as, a particle not acted on by a force travels in a straight line, i.e., a geodesic in flat Newtonian space. Einstein re-formulated this as above stated. The quest for a (non-quantum) unified field theory was always motivated by this: define a geometry on space-time based on the forces of Nature so that all trajectories will be geodesics. The physical insight here is the same as that underlying the original law of inertia: natural, unconstrained motion is straight, i.e., geodesic. But geodesics always do obey some variational principle.

If we take Einstein's point of view seriously, and think it will survive when treated quantum-mechanically, then the answer to your question would be: if the set of trajectories arise as the set of geodesics from some metric on the relevant space, then there is a physically significant action principle which governs the dynamics.

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    $\begingroup$ Can you give a mathematically cooked up example of an equation of motion that does not arise from an action principle? $\endgroup$ – dan-ros Apr 6 '16 at 11:06

This question is rather old but as I see there is no complete answer, or at least an answer that would be reasonably complete. This is a topic I have been interested in for a long time and I'd like to present a reasonably complete answer here.

The problem OP was inquiring about is called the inverse problem to the calculus of variations. This problem does not currently have a general solution even locally, but at least locally it is possible to give an explicit solution to a restricted problem.

Notation and basics:

Let's set up some notation. Since in this answer I will only attack this problem locally, we can work on (open sets of) euclidean spaces. Let $\phi(x)$ be an $m$-component function of $n$ independent variables. In index notation, the independent variables are written as $x^\mu$ and greek indices are used to label them. The dependent variables are written as $u^a=\phi^a(x)$ and latin indices are used to label them. Functional arguments shall be denoted using square brackets: $F[\phi]$.

Suppose that $E_a(x)[\phi]$ are a set of $m$ functional expressions of the field $\phi$ and - when evaluated on a given field $\phi$ - they are themselves functions of the independent variables.

The (restricted) inverse problem asks if there is a functional $S[\phi]$ of the field such that $$ E_a(x)[\phi]=\frac{\delta S[\phi]}{\delta\phi^a(x)}, $$ where the RHS is the functional derivative (to be defined properly later), and if there is such a functional, how to find it.


This has been touched on by Qmechanics' answer, but it is nonetheless useful to repeat here. I have called the above the restricted inverse problem. For the most part, we are not actually interested in the restricted inverse problem.

Instead the problem we are interested in is as follows. Suppose we have a functional expression $E_a(x)[\phi]$ like above, which we interpret as equations of motion (EoMs) as $E_a(x)[\phi]=0$. Another set of EoMs $E^\prime_a(x)[\phi]$ are said to be equivalent if $$ E_a(x)[\phi]=0\Longleftrightarrow E^\prime_a(x)[\phi]=0, $$ i.e. the functional zeros of the two EoM expressions agree. The actual problem we are interested is whether there is a functional $S[\phi]$ such that $$ E^\prime_a(x)[\phi]=\frac{\delta S[\phi]}{\delta\phi^a(x)}, $$ i.e. is there an equivalent set of EoMs that come from a variational principle.

This is very difficult to answer. Probably the simplest case is when so-called variational integrating factors exist. Suppose that $A^a_{\ b}(x)[\phi]$ is a matrix that is a functional of the field $\phi$ and when evaluated on a field, is a function of the independent variables, and in addition for each field and each independent variable, $A^a_{\ b}$ is invertible.

The EoM $E_a(x)[\phi]$ has a variational integrating factor if there is such an $A^a_{\ b}$ such that $$ E_a(x)[\phi]=A^b_{\ a}(x)[\phi]E^\prime_b(x)[\phi], $$ and $E^\prime_b(x)[\phi]$ is variational (i.e. it is the functional derivative of some functional).

We do not even have a method to determine if for a given EoM there is a variational integrating factor or not, although if the restricted inverse problem can we solved, it can also be used to try to look for simple integrating factors.

And the "equivalence of EoMs" can be much more general than this, including the case where an equivalent EoM depends on more dependent variables than the original EoM.

To give several examples of variational integrating factors, consider first the functional $$S_1[\phi^1,\phi^2]=\int\mathrm dx\ \frac{1}{2}((\dot\phi^1)^2+(\dot\phi^2)^2),$$ which is just the action functional for a free particle in two dimensions with unit mass, and the functional $$ S_2[\phi^1,\phi^2]=\int\mathrm dx\ \dot\phi^1\dot\phi^2. $$

Both of these functional give as their EoMs the vanishing of the accelerations of $\phi^1$ and $\phi^2$, but the two Lagrangians do not differ by a total derivative, and in fact the two EoMs are not the same, because the EoMs of the first functional is $$ (E_1)_1(x)[\phi]=-\ddot\phi^1(x),\ (E_1)_2(x)[\phi]=-\ddot\phi^2(x), $$ while the EoMs of the second functional are $$ (E_2)_1(x)[\phi]=-\ddot\phi^2(x),\ (E_2)_2(x)[\phi]=-\ddot\phi^1(x), $$ i.e. the indices are exchanged, and we have $$ (E_1)_a(x)[\phi]=\pi^b_{\ a}(E_2)_b(x)[\phi], $$ where $$ (\pi^b_{\ a})=\left(\begin{matrix} 0 && 1 \\ 1 && 0 \end{matrix}\right). $$

As a second example, if $$S[g]=\int\mathrm d^4x\sqrt{-\mathfrak g}R$$ is the Einstein-Hilbert action, we have $$ \frac{\delta S[g]}{\delta g_{\mu\nu}}=-\sqrt{-\mathfrak g}G^{\mu\nu}, $$ however if one wanted to do apply the inverse problem to this and forgot that the $\sqrt{-\mathfrak g}$ density is also part of the EoMs, then it would turn out that the Einstein tensor $G^{\mu\nu}$ is not variational, however $\sqrt{-\mathfrak g}G^{\mu\nu}$ is, so $\sqrt{-\mathfrak g}$ plays the role of a variational integrating factor.

"Low tech" functional method:

Now that the limitations to the inverse problem has been discussed, let us try to find ways to solve the restricted inverse problem, namely, if an EoM $E_a(x)[\phi]$ is given, determine whether this expression (exactly as it is) comes from the functional differentiation of a functional and if so, how to construct that functional.

If $S[\phi]$ is a functional of the field, let us define the functional derivative at a field value as follows. For a given field value $\phi$, the functional derivative of $S$ is the function(s) $\delta S[\phi]/\delta\phi^a(x)$ such that $$ \delta S[\phi]=\int\mathrm d^4x\ \frac{\delta S[\phi]}{\delta\phi^a(x)}\delta\phi^a(x), $$ where the integration is over the entirety of the coordinate space we are working on and this formula is valid for all field variations $\delta\phi^a$ with compact support. So by this definition the Einstein-Hilbert action has a functional derivative even without the Gibbons-Hawking-York boundary term.

The analogous finite dimensional problem is that if one is given $n$ functions $A_\mu(x)$ of $n$ variables, what is the condition for the existence of a function $\chi(x)$ such that $$ A_\mu(x)=\frac{\partial\chi}{\partial x^\mu}(x). $$

As it is well-known, the necessary condition is that $$ \frac{\partial A_\nu}{\partial x^\mu}-\frac{\partial A_\mu}{\partial x^\nu}=0, $$ because partial derivatives commute. This is also a sufficient condition, because if this is satisfied, then we can take $$ \chi(x)=\int_0^1\mathrm dt\ A_\mu(tx)x^\mu, $$ which will satisfy $A_\mu=\partial\chi/\partial x^\mu $ provided the integrability conditions above hold.

We can formally try to implement the same method "functionally" as follows. Since functional derivatives commute, a necessary condition for $$ E_a(x)[\phi]=\frac{\delta S[\phi]}{\delta\phi^a(x)} $$ is that $$ \frac{\delta E_a(x)[\phi]}{\delta\phi^b(x^\prime)}-\frac{\delta E_b(x^\prime)[\phi]}{\delta\phi^a(x)}=0. $$ Then for given EoMs, define $$ S[\phi]=\int_0^1\mathrm dt\int\mathrm d^nx\ E_a(x)[t\phi]\phi^a(x). $$

An explicit calculation will reveal that the functional derivative of this expression is $E_a(x)[\phi]$, once again, provided the above integrability conditions hold.

Now I present an explicit calculation of the Einstein-Hilbert action as an example. The Einstein-tensor $G^{\mu\nu}=G^{\mu\nu}(x)[g]$ is considered as a functional of the metric tensor. We must evaluate the Einstein-tensor at $tg_{\mu\nu}$ to use the above formula. It is useful now to look up how the curvature tensors change under a conformal transformation (for example in Wald), but take the simpler case when the conformal factor is a constant. We have $$ \bar R_{\mu\nu}=R_{\mu\nu},\quad \bar R=\frac{1}{t}R, $$ whenever $\bar g_{\mu\nu}=tg_{\mu\nu}$. From this $$ \bar G_{\mu\nu}=\bar R_{\mu\nu}-\frac{1}{2}\bar g_{\mu\nu}\bar R=R_{\mu\nu}-\frac{1}{2}tg_{\mu\nu}\frac{1}{t}R=G_{\mu\nu}, $$ and $$ \bar{\mathfrak g}=t^4\mathfrak g, $$ thus $$ \sqrt{-\bar{\mathfrak g}}\bar G^{\mu\nu}=t^2\sqrt{-\mathfrak g}t^{-2}G^{\mu\nu}=\sqrt{-\mathfrak g}G^{\mu\nu}, $$ thus if $E^{\mu\nu}[g]=\sqrt{-\mathfrak g}G^{\mu\nu}[g]$, we have $$ E^{\mu\nu}[tg]=E^{\mu\nu}[g]. $$ Now we use the previous formula: $$ S[g]=\int_0^1\mathrm dt\int\mathrm d^4x\ E^{\mu\nu}[tg]g_{\mu\nu}=\int\mathrm d^4x\int_0^1\mathrm dt E^{\mu\nu}[g]g_{\mu\nu}=\int\mathrm d^4x\sqrt{-\mathfrak g}G^{\mu\nu}g_{\mu\nu} \\ =\int\mathrm d^4x\sqrt{-\mathfrak g}(-R), $$ which is indeed the Einstein-Hilbert action, up to a constant multiple, but that's because the "real" EoMs are $-(16\pi G)^{-1}\sqrt{-\mathfrak g}G^{\mu\nu}$, and we calculated without the factor of $-(16\pi G)^{-1}$.

Mathematical rigour, or lack of thereof:

[Work in progress]

"High tech" differential geometric method:

[Work in progress]

(I intend to return to these last two sections and complete it, however what is present so far also answers OP's question)


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