The Lagrangian formulation of mechanics without going through variational principles. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:52:19Z http://mathoverflow.net/feeds/question/81800 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81800/the-lagrangian-formulation-of-mechanics-without-going-through-variational-princip The Lagrangian formulation of mechanics without going through variational principles. Giuseppe Tortorella 2011-11-24T13:06:51Z 2011-11-25T09:35:09Z <p>In some texts on classical mechanics and not only, the Euler--Lagrange equations of motion are directly obtained as solution of variational problems.<br> 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.</p> <p>This observation suggested to myself the following question:</p> <p>Is the variational approach to the Euler--Lagrange equations the only one viable?<br> 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?</p> <blockquote> <p>I was searching for suggestion of reading for best tackle this question.</p> </blockquote> <p>As usual any feedback is welcome.</p> http://mathoverflow.net/questions/81800/the-lagrangian-formulation-of-mechanics-without-going-through-variational-princip/81819#81819 Answer by Spiro Karigiannis for The Lagrangian formulation of mechanics without going through variational principles. Spiro Karigiannis 2011-11-24T16:14:38Z 2011-11-24T16:14:38Z <p>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.</p> <p>Here are some links:</p> <p><A HREF="http://books.google.ca/books?id=I2gH9ZIs-3AC&amp;printsec=frontcover&amp;dq=marsden+mechanics&amp;hl=en&amp;ei=ZGzOTp7BLubg0QGrzOgw&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CC8Q6AEwAA#v=onepage&amp;q=marsden%20mechanics&amp;f=false" rel="nofollow">"Mechanics and Symmetry"</A> by Marsden and Ratiu</p> <p><A HREF="http://books.google.ca/books?id=f6DAVMTqtdQC&amp;printsec=frontcover&amp;dq=marsden+mechanics&amp;hl=en&amp;ei=ZGzOTp7BLubg0QGrzOgw&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=2&amp;ved=0CDUQ6AEwAQ#v=onepage&amp;q=marsden%20mechanics&amp;f=false" rel="nofollow">"Foundations of Mechanics"</A> by Abraham and Marsden</p> <p><A HREF="http://books.google.ca/books?id=3M07FsrNn9gC&amp;printsec=frontcover&amp;dq=marsden+mechanics&amp;hl=en&amp;ei=ZGzOTp7BLubg0QGrzOgw&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=4&amp;ved=0CEEQ6AEwAw#v=onepage&amp;q=marsden%20mechanics&amp;f=false" rel="nofollow">"Lectures on Mechanics"</A> by Marsden</p> <p><A HREF="http://books.google.ca/books?id=Pd8-s6rOt_cC&amp;printsec=frontcover&amp;dq=arnold+mechanics&amp;hl=en&amp;ei=K23OTv7-OYP20gHxso1M&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CDIQ6AEwAA#v=onepage&amp;q=arnold%20mechanics&amp;f=false" rel="nofollow">"Mathematical Methods of Classical Mechanics"</A> by Arnold</p> http://mathoverflow.net/questions/81800/the-lagrangian-formulation-of-mechanics-without-going-through-variational-princip/81846#81846 Answer by Igor Khavkine for The Lagrangian formulation of mechanics without going through variational principles. Igor Khavkine 2011-11-25T04:10:17Z 2011-11-25T04:10:17Z <p>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.</p> <h3>The Hamiltonian formalism is a special case of the Lagrangian formalism.</h3> <p>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 <em>auxiliary</em> 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 $L'(x,y(x))=L(x)$.</p> <p>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.</p> <p>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 (<a href="http://dx.doi.org/10.1103/PhysRevD.44.R939" rel="nofollow">PRD</a>, 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.</p> <h3>The phase space is the space of solutions of the Euler-Lagrange equations.</h3> <p>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.</p> <p>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.</p> <h3>Any Lagrangian defines a (pre)symplectic form (current).</h3> <p>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 <em>covariant phase space</em> method. A very nice (though not original) reference is Lee &amp; Wald (<a href="http://dx.doi.org/10.1103/PhysRevD.44.R939" rel="nofollow">JMP</a>, 1990). See also this <a href="http://ncatlab.org/nlab/show/phase+space#covariant_phase_space_8" rel="nofollow">nLab</a> page, which also has a more extensive <a href="http://ncatlab.org/nlab/show/phase+space#covariant_phase_space_8" rel="nofollow">reference list</a>.</p> <p>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.</p> <p>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 <em>current</em>). 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.</p> <h3>A (pre)symplectic form (current) defines a Lagrangian.</h3> <p>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.</p> <p>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.</p> <p>To my knowledge, the above observation first appeared in Henneaux (<a href="http://dx.doi.org/10.1016/0003-4916%2882%2990334-7" rel="nofollow">AnnPhys</a>, 1982) for ODEs and in Bridges, Hydon &amp; Lawson (<a href="http://dx.doi.org/10.1017/S0305004109990259" rel="nofollow">MathProcCPS</a>, 2010) for PDEs. The calculation demonstrating this observation is given in a bit more detail on this <a href="http://ncatlab.org/nlab/show/phase+space#InverseProblem" rel="nofollow">nLab page</a>.</p> <p>Another way to look at this result is to consider a conserved (pre)symplectic form as a certificate for the solution of the <a href="http://en.wikipedia.org/wiki/Inverse_problem_for_Lagrangian_mechanics" rel="nofollow">inverse problem of the calculus of variations</a>.</p> <hr> <p>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 (<a href="http://en.wikipedia.org/wiki/Darboux%27s_theorem" rel="nofollow">Darboux's theorem</a>). The Legendre transform identifies this choice of coordinates explicitly.</p> http://mathoverflow.net/questions/81800/the-lagrangian-formulation-of-mechanics-without-going-through-variational-princip/81857#81857 Answer by Paul Skerritt for The Lagrangian formulation of mechanics without going through variational principles. Paul Skerritt 2011-11-25T08:55:54Z 2011-11-25T09:35:09Z <p>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).</p> <p>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 $H$.</p> <p>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 <em>intrinsically</em> 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 $H$. </p> <p>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).</p> <p>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 $p_i = \frac{\partial L}{\partial \dot{x}^i}$).</p>