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Is there any physics that cannot be expressed in terms of Lagrange equations?

+ 4 like - 0 dislike
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A lot of physics, such as classical mechanics, General Relativity, Quantum Mechanics etc can be expressed in terms of Lagrangian Mechanics and Hamiltonian Principles. But sometimes I just can't help wonder whether is it ever possible (in the future, maybe) to discover a physical law that can't be expressed in terms of Lagrangian Equations?

Or to put it in other words, can we list down for all the physical laws that can be expressed in terms of Lagrangian equations, what are the mathematical characteristics of them( such as it must not contain derivatives higher than 2, all the solutions must be linear etc)?


This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Graviton

asked Feb 7, 2011 in Theoretical Physics by Graviton (85 points) [ revision history ]
edited Jul 29, 2015 by Dilaton
All solutions must be linear? Why?

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user QGR
@QGR, I'm not saying that all solutions to physics law must obey linearity!! It's just an example I use to illustrate what I mean by "mathematical characteristics".

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Graviton

I'm also asking about what are the mathematical characteristics physical laws must obey if they are expressible in terms of Lagrangian mechanics.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Graviton

The issues in field theory are different, because you expect the Lagrangian to be a local density. For non-field-theories, you can allow a more general nonlocal notion of Lagrangian. The essential condition for this to be possible is symplectic formulation, so no dissipation. For field theories, there are more mundane examples which are nondissipative and are called non-Lagrangian, like IIB supergravity or some 2d models, which would not be considered non-Lagrangian here.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Ron Maimon

Here is a paper of Witten's that mentions theories that have no Lagrangians.

http://arxiv.org/pdf/0712.0157.pdf

4 Answers

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Hamilton's dynamics occurs on a phase space with an equal number of configuration and momentum variables $\{q_i,~p_i\}$, for $i~=~1,\dots n.$ The dynamics according to the symplectic two form ${\underline{\Omega}}~=~\Omega_{ab}dq^a\wedge dp^b$ is a Hamiltonian vector field $$ \frac{d\chi_a}{dt}~=~\Omega_{ab}\partial_b H, $$ with in the configuration and momentum variables $\chi_a~=~\{q_a,~p_a\}$ gives $$ {\dot q}_a~=~\frac{\partial H}{\partial p_a},~{\dot p}_a~=~-\frac{\partial H}{\partial q_a} $$ and the vector $\chi_a$ follows a unique trajectory in phase space, where that trajectory is often called a Hamiltonian flow.

For a system the bare action is $pdq$ ignoring sums. The Hamiltonian is found with imposition of Lagrangians as functions over configuration variables. This is defined then on half of the phase space, called configuration space. It is also a constraint, essentially a Lagrange multiplier. The cotangent bundle $T^*M$ on the configuration space $M$ defines the phase space. Once this is constructed a symplectic manifold is defined. Therefore Lagrangian dynamics on configuration space, or equivalently the cotangent bundle defines a symplectic manifold. This does not mean a symplectic manifold defines a cotangent bundle. The reason is that symplectic or canonical transformations mix the distinction between configuration and momentum variables.

As a result there are people who study bracket structures which have non-Lagrangian content. The RR sector on type IIB string is non-Lagrangian. The differential structure is tied to the Calabi-Yau three-fold, which defines a different dynamics.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Lawrence B. Crowell
answered Feb 7, 2011 by Lawrence B. Crowell (590 points) [ no revision ]
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If you have a classical theory specified by some partial differential equations, you can automatically come up with a Lagrangian by introducing a Lagrange multiplier for each PDE.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user QGR
answered Feb 7, 2011 by QGR (250 points) [ no revision ]
Not too sure what you mean here. Classical theory, as opposed to, quantum theory?

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Graviton
@Graviton: doesn't matter. You can use the classical Lagrangian to quantize your theory. Whether that works in a particular case is a different question though.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Marek
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You cannot model friction very well with Lagrangian Mechanics.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user ja72
answered Feb 7, 2011 by ja72 (0 points) [ no revision ]
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And why would that be?

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Marek
It is true that in introductory classes to classical mechanics usually only systems with energy conservation are considered. But it is possible to extend the framework to systems with dissipation, see e.g. Brogliato et. alt., “Dissipative Systems, Analysis and Control”, Springer, 2nd edition.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Tim van Beek
Looking at en.wikipedia.org/wiki/Lagrangian_mechanics how you would describe coulomb friction in a lagrangian?

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user ja72
@ja72 you could read this paper, which solves coulomb friction using a lagrangian: springerlink.com/content/t41m136288746gj0

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user spencer nelson
@Spencer Nelson - How is the new method validated? When it comes to friction modeling is very hard to ensure the results are realistic and accurate. Also, in my original statement I said it is difficult to model friction with LM, not impossible.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user ja72
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Why is this downvoted? This is exactly correct--- odd order differential equations, disspiative ones, cannot come out of an explicitly time independent Lagrangian, because they cannot conserve a time independent symplectic volume--- their time evolution takes any motion in phase space to a single point generically!

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Ron Maimon
Lagrange equations may model friction with the help of Rayleigh's dissipation function en.wikipedia.org/wiki/Lagrangian_mechanics#Dissipation_function The caveat is that Lagrange equations of frictional type are not variational, i.e., not stemming from an action principle.

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Qmechanic
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This question is rather generic in several different respects. For example it is not clear that even Schrodinger's equation is an equation that can be "expressed in Hamiltonian principles". Yes we have $\delta/\delta t \Psi=H \Psi$ but is this an expression in "Hamiltonian Principles"? How different must the equation become to not meet this requirement - assuming it does now, perhaps adding a non-linear term?

Furthermore "Hamilton's Principle" does not itself apply in the Quantum context, although the action paths that it introduces are used in Feynman Path Integrals. Hamilton's Principle being a classical concept.

Another generality is in the range of Physics. The whole area of Thermodynamics comes to mind. Now there are phase space explanations, but is that "Hamiltonian Principles?"

This post imported from StackExchange Physics at 2015-07-29 19:16 (UTC), posted by SE-user Roy Simpson
answered Feb 7, 2011 by Roy Simpson (165 points) [ no revision ]

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