# The Lagrangian formulation of mechanics without going through variational principles.

In some texts on classical mechanics and not only, the Euler--Lagrange equations of motion are directly obtained as solution of variational problems.
On the other side, sometimes reading about hamiltonian mechanics, one find the expression that this latter formulation is preferred to the lagrangian one because of it does completely avoid the appeal to variational principles.

This observation suggested to myself the following question:

Is the variational approach to the Euler--Lagrange equations the only one viable?
If not, is there some reason that explain why the geometry of the Euler-Lagrange eqns is much more hidden than the geometry of the Hamilton eqns?

I was searching for suggestion of reading for best tackle this question.

As usual any feedback is welcome.

-
I'm not sure what you mean by "much more hidden" geometry. Both approaches are important and useful in geometry and in physics. They are also intimately connected to each other via the Legendre transform. – Spiro Karigiannis Nov 24 '11 at 14:30
@Spiro Karigiannis: With "much more hidden" geometry, I would have meant that while I know many exposition of the symplectic geometry as the proper basis of hamiltonian mechanics, I lack references for the geometry behind the Euler-Lagrange eqns. I would better comprehend the connection between the geometries behind these two approaches, so that I could take the best of the one or the other. – Giuseppe Tortorella Nov 24 '11 at 16:04

The Lagrangian (or variational) formulation of the Euler-Lagrange equations and the Hamiltonian formulation are equivalent. This equivalence can be made quite explicit and goes a bit deeper than the standard treatments show. The equivalence can be established in several steps, which I'll try to outline below with references.

### The Hamiltonian formalism is a special case of the Lagrangian formalism.

There is a generic operation that can be performed on variational problems: adjunction and elimination of auxiliary fields or variables. Given a Lagrangian $L(x,y)$, the variable $y$ (which could be vector valued) is called auxiliary if the Euler-Lagrange equations obtained from the variation of $y$ can be algebraically solved for solved for in terms of the remaining variables and their derivatives $y=y(x)$. The important point here is that we need not solve any differential equations to obtain $y(x)$. The Lagrangian $L'(x) = L(x,y(x))$ gives a new variational principle with the auxiliary field $y$ eliminated. The critical points of $L'(x)$ are in one-to-one correspondence with those of the original $L(x,y)$. The adjunction of an auxiliary field is the reverse operation. Given $L(x)$, we look for another Lagrangian $L'(x,y)$ where $y$ is auxiliary and its elimination gives $L'(x,y(x))=L(x)$.

It is straightforward to check that given a Lagrangian $L(x)$, with Hamiltonian $H(x,p)$, the new Lagrangian $L'(x,p) = p\dot{x} - H(x,p)$, associated to Hamilton's Least Action Principle, is a special case of adjoining some auxiliary fields, namely the momenta $p$. The elimination $p=p(x)$ is precisely the inverse Legendre transform.

The moral here is that the Legendre transform is not sacred. I learned this point of view from the following paper of Barnich, Henneaux and Schomblond (PRD, 1991). This point of view is particularly helpful in higher order field theories (multiple independent variables, second or higher order derivatives in the Lagrangian) where a unique notion of Legendre transform is lacking.

### The phase space is the space of solutions of the Euler-Lagrange equations.

When the Euler-Lagrange equations have a well-posed initial value problem. Then solutions can be put into one-to-one correspondence with initial data. The initial data are uniquely specified by the canonical position and momentum variables that are commonly used to define the canonical phase space in the Hamiltonian picture. This is a rather common identification nowadays and can be found in many places, so I won't give a specific reference.

If the Euler-Lagrange equations do not have a well-posed initial value problem, the definitions change somewhat on either side, but an equivalence can still be established. See the references in the next step for more details.

### Any Lagrangian defines a (pre)symplectic form (current).

This is, unfortunately, less well known than it should be. The (pre)symplectic structure of classical mechanics and field theory can be defined straightforwardly directly from the Lagrangian. In case the equations of motion are degenerate, the form is degenerate and hence only presymplectic, otherwise symplectic. There is more than one way to do this, but a particularly transparent one is referred to as the covariant phase space method. A very nice (though not original) reference is Lee & Wald (JMP, 1990). See also this nLab page, which also has a more extensive reference list.

Applying this method to the Lagrangian of Hamilton's Least Action Principle gives the standard symplectic form in terms of the canonical position and momentum coordinates.

Briefly, and without going into the details of differential forms on jet spaces where this is most easily formalized, the construction is as follows. Denote by $d$ the space-time (which is just 1-dimensional if you only have time) exterior differential and by $\delta$ the field variation exterior differential. Without dropping boundary or total divergence terms, the total variation of the Lagrangian can be expressed as $\delta L(x) = \mathrm{EL}\delta x + d\theta$. Here the Lagrangian is a space-time volume form, $\mathrm{EL}$ denotes the Euler-Lagrange equations and $d\theta$ consists of all the terms that are usually dropped during partial integration. Clearly $\theta$ is of 1-form in terms of field variations and as a spacetime form of one degree lower than a volume form (aka a current). Taking another exterior field variation, we get $\omega=\delta\theta$, which is the desired (pre)symplectic current. If $\Sigma$ is a Cauchy surface (in particular it is codimension-1), the (pre)symplectic form is defined as $\Omega=\int_\Sigma \omega$, now a 2-form in terms of field variations, as expected. It can be shown that the (pre)symplectic form is closed, $d\omega=0$, when evaluated on solutions of the Euler-Lagrange equations. Hence, by Stokes' theorem, $\Omega$ is independent of the choice of $\Sigma$. One space-time is 1-dimensional, $\Omega$ is just $\omega$ evaluated at a particular time.

### A (pre)symplectic form (current) defines a Lagrangian.

As alluded to in the question, Hamilton's equations of motion are often expressed in a special form that highlights a certain geometrical structure that is not obvious in the original Lagrangian form. It is well known from the symplectic formulation of classical mechanics that this structure can be seen as a consequence of the fact that they correspond time evolution generated by a Hamiltonian via the Poisson bracket. The symplectic form is then preserved by the evolution. There are analogous statements for field theory. In fact, no variational formulation is necessary to discuss this geometrical structure of the equations.

What is quite remarkable is that a kind of converse to this statement holds as well. Namely, given a system of (partial) differential equations, if there is a conserved (pre)symplectic current $\omega$ on the space of solutions ($\omega$ is field dependent and $d\omega=0$ when evaluated on solutions, see previous step), then a subsystem of the equations is derived from a variational principle. There is a subtlety here. Even if there exists a conserved $\omega$, if it is degenerate (not symplectic) other independent equations need to be added to the Euler-Lagrange equations of the corresponding variational principle to obtain a system of equations equivalent to the original one. Onece the Lagrangian of this variational principle is known, the Hamiltonian and symplectic forms could be defined in the usual way and the original system of equations recast in the canonical Hamilton form.

To my knowledge, the above observation first appeared in Henneaux (AnnPhys, 1982) for ODEs and in Bridges, Hydon & Lawson (MathProcCPS, 2010) for PDEs. The calculation demonstrating this observation is given in a bit more detail on this nLab page.

Another way to look at this result is to consider a conserved (pre)symplectic form as a certificate for the solution of the inverse problem of the calculus of variations.

A final note about the usefulness of the Hamiltonian formulation. Despite the fact that as a consequence of the above discussion it's not strictly necessary. Any symplectic manifold has local coordinates in which the symplectic form is canonical (Darboux's theorem). The Legendre transform identifies this choice of coordinates explicitly.

-
"Any symplectic manifold has local coordinates in which the symplectic form is canonical (Darboux's theorem). The Legendre transform identifies this choice of coordinates explicitly." Sorry, can you clarify what you mean by this? I'm not seeing the connection between the Legendre transform and canonical coordinates. – user17945 Nov 25 '11 at 8:59
Consider the phase space coordinatized by initial data at time $t=0$, say for 1-dimensional particle motion. Then natural coordinates are $(x,\dot{x})$. In these coordinates the symplectic form will look like $\omega=\omega(x,\dot{x}) dx\wedge d\dot{x}$. The Legendre transform defines $p=p(x,\dot{x})$. The new coordinate system $(x,p)$ is special because $\omega=dp\wedge dq$ is now in canonical form, while in the $(x,\dot{x})$ coordinates it is not. Darboux's theorem guarantees that such special coordinates always exist (locally), but they need not always be easy to find. Here they are. – Igor Khavkine Nov 25 '11 at 10:28
Oh, ok, I see now what you're saying. I didn't realise you were starting with the symplectic form defined by the Lagrangian (although I should have). Thanks. – user17945 Nov 25 '11 at 14:14
This is a great answer! Thanks very much. – Spiro Karigiannis Nov 25 '11 at 14:14
Dear Igor Khavkine, your answer is fantastic, and even more is such your work at the nLab about the phase space. This is much more than I would have expected, thank you. – Giuseppe Tortorella Nov 25 '11 at 16:28

Have you tried reading Arnold's book? Other possibilities include the several books on the geometry of classical mechanics by Jerry Marsden (with various coauthors). You will probably find the answers to your questions in here.

"Mechanics and Symmetry" by Marsden and Ratiu

"Foundations of Mechanics" by Abraham and Marsden

"Lectures on Mechanics" by Marsden

-
In case you read french, you can also have a look at this set of lecture notes by Colin de Verdiere: www-fourier.ujf-grenoble.fr/~ycolver/All-Articles/93b.pdf (chapter 1, section 4). – DamienC Nov 24 '11 at 16:32
Dear Spiro Karigiannis, thanks for the references, indeed the works of Arnold and Marsden(with his co-authors) are the sources from which now I'm trying to learn more. – Giuseppe Tortorella Nov 25 '11 at 16:10

I'm just going to make a basic point, so apologies if this is completely obvious to you (it's also contained in Igor Khavkine's much more thorough answer).

Let $Q$ be the configuration manifold of the system. The corresponding cotangent bundle $T^*Q$ has an intrinsic symplectic form $\omega=-d\Theta$, where $\Theta$ is the tautalogical one-form on $T^*Q$. For a Hamiltonian $H:T^*Q\rightarrow \mathbb{R}$, Hamilton's equations can be expressed in terms of the Hamiltonian vector field $X_H$ (defined by $i_{X_H}\omega=dH$). Note $\omega$ is intrinsic to the phase space $T^*Q$, and doesn't depend on the Hamiltonian $H$.

Now given a Lagrangian $L:TQ\rightarrow\mathbb{R}$, and corresponding Legendre transform $\mathbb{F}L:TQ\rightarrow T^*Q$, and assuming here for simplicity that $\mathbb{F}L$ is a diffeomorphism ("L is hyperregular"), one can use $\mathbb{F}L$ to pull everything back to $TQ$. $TQ$ becomes a symplectic manifold, with symplectic form $\omega_L=(\mathbb{F}L)^*\omega$, and the Euler-Lagrange equations are just the equations for the flow of the Hamiltonian vector field $X_E$ defined by $i_{X_E}\omega_L = dE$, where $E=(\mathbb{F}L)^*H = H\circ\mathbb{F}L$ is the energy function on $TQ$. I guess the main point though is that $TQ$ is not intrinsically a symplectic manifold. The symplectic form $\omega_L$ depends also on the choice of Lagrangian. This is one possible answer to why the geometric formulation is more common on the Hamiltonian side, and appears to be hidden' on the Lagrangian side: the geometry of $TQ$ (as it pertains to the E-L eqns) is tied up with the particular Lagrangian $L$, whereas the geometry of $T^*Q$ is independent of the particular Hamiltonian $H$.

I'd also recommend any of the books by Jerry Marsden mentioned in Spiro Karigiannis' answer as the best place to learn this (my notation is consisent with his).

Edit: I should clarify that by Legendre transform' above I mean (in coordinates) the map $p_i(x, \dot{x}) = \frac{\partial L}{\partial \dot{x}^i}(x, \dot{x})$. This is standard in the literature, but it differs from the classical meaning $H(x, p) = p\dot{x}-L(x, \dot{x})$ (where $p_i = \frac{\partial L}{\partial \dot{x}^i}$).

-