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Is there a valid Lagrangian formulation for all classical systems?

+ 4 like - 0 dislike
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Can one use the Lagrangian formalism for all classical systems, i.e. systems with a set of trajectories $\vec{x}_i(t)$ describing paths?

On the wikipedia page of Lagrangian mechanics, there is an advertisement, which says that it also works for systems for which energy and momentum is not conserved. It's unreferenced but it sounds nice, but I wonder if there are other problems one might encounter. Does this statement already mean that all systems can be described by a Lagrangian?

I have found that, at least in some dissipative systems, you have to introduce non-standard Lagrangians, which are not of the form $L=T-V$ and so there is no clear kinetic and potential term. However, from a Newtonian point of view, there is still the $T:=\sum \frac{m}{2}\vec{x}_i'(t)^2$ term. Does that mean that you have a kinetic term, but it just isn't part of the Lagrangian?

Also, if it's possible to write down the Lagrangian (even if there is not such thing as a conserved energy), what is the meaning of the Legendre transform of this Lagrangian? Usually, it would be the Hamiltonian, but now, is it just any random function without any use?

I don't mind some differential geometry speak btw.

Lastly, are there extension of this, i.e. variational principles, which are far away from the Lagrangian ideas?

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user NikolajK
asked Jan 29, 2012 in Theoretical Physics by NikolajK (195 points) [ no revision ]
retagged Jul 29, 2015
I'm pretty sure if you read the 1st chapter of Goldstein's classical mechanics carefully you will find the answer to this question. OFf the top of my head, I can't really say for sure because I didn't really pay attention to the stuff about the dissipation function. Generally, I'm assuming anything classical can be modeled with classical mechanics I mean isn't that the whole point lol?

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user Timtam

3 Answers

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Roughly speaking, physicists thought a lot about this right before the revolutions of relativity and quantum theory. Heinrich Hertz reduced all of classical mechanics to a kind of Lagrangian and Hamiltonian framework and a new principle of least curvature. See Hertz, The Principles of Mechanics, out of copyright, http://www.archive.org/details/principlesofmech00hertuoft and Whittaker, Analytical Dynamics, pp. 254ff. Their thoughts turned out to be very helpful for general relativity, Wave Mechanics, and Quantum Field Theory.

Hertz's ideas of least curvature are very close to Lagrange's ideas.

All classical mechanics can be put into the Lagrangian framework: if energy is not conserved (for example, if the system is an open system, if friction is present, etc.) then merely one adjusts to allowing a time-varying Lagrangian.

But the practical utility of this formulation is sometimes low: questions about Statistical Physics require a different way of looking at the phase space and the system: its laws of motion are almost irrelevant and the kind of information about the trajectories of the parts of the system which the Lagrangian equations gives you are almost useless, one instead wants to know things such as their auto-correlation functions, which are almost independent of the particular trajectory or initial condition chosen.

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user joseph f. johnson
answered Jan 30, 2012 by joseph f. johnson (490 points) [ no revision ]
+ 0 like - 0 dislike

As far as I know, the hamiltonian formulation is even more general than the lagrangian one, in the sense that you may not be able to find a lagrangian description for a particular system, which nonetheless can be treated in a hamiltonian framework. Remember how the hamiltonian formalism can be introduced: we define generalized momenta $p_k=\partial L/\partial \dot q_k$ and notice that

\begin{equation} \frac{\partial L}{\partial \dot {q_k}}=\frac{\partial T}{\partial \dot{q_k}}=\frac{\partial}{\partial \dot{q_k}}\left(\frac{1}{2}a_{rs}(q,t)\dot{q_r}\dot{q_s}+b_r(q,t)\dot{q_r}+c(q,t)\right)=a_{ks}\dot{q_s}+b_k \end{equation}

, where there is a sum on the indices $r,s$ and where we have decomposed the kinetic energy as the sum of a term quadratic in the generalized velocities, one linear and a constant (as it can always be done, given holonomic constraints). The symmetric matrix $\{a_{ks}\}$ is invertible so $\dot{q_s}=\phi_s(q,p,t)$. All this to say that Lagrange equations can always be put in normal form:

\begin{equation} \dot{q_s}=\phi(q,p,t) \end{equation}

\begin{equation} \dot{p_s}=\frac{\partial L}{\partial \dot{q_s}} \end{equation}

We can define the hamiltonian $H(q,p,t)$ via the usual Legendre transformations and derive Hamilton's equations of motion. Once we have developed the hamiltonian formalism, we can forget how we get there and treat the $q$'s and the $p$'s as independent variables. Is it possible to get back to Lagrange equations and prove the two formalisms are equivalent? Yes, but only under a very general condition: given the hamiltonian and hamilton's equations, it must be possible to express the $\dot{q}$'s as functions of the canonical coordinates. If it is possible, define $L=p_k\partial h/\partial p_k-H$, where it is understood that now $L$ is thought of as a function of $(q,\dot q,t)$. It can be proven from here that then Lagrange equations must also hold. So, no, not all mechanical systems have a lagrangian description since you may start from a hamiltonian and find out that the relations that give the $\dot q$'s in terms of $(q,p)$ are not invertible. The hamiltonian can be a very general function, not necessarily decomposable in a kinetic and a potential term. By the way,the total derivative of $H$ is equal to its partial derivative with respect to time; so $H$ is the energy only if $H=H(q,p)$, that is the constraints do not depend on time.

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user quark1245
answered Feb 3, 2012 by quark1245 (0 points) [ no revision ]
Comment to the answer(v1): A Legendre transformation may be singular in both directions Lagrangian formalism $\leftrightarrow$ Hamiltonian formalism.

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user Qmechanic
+ 0 like - 0 dislike

There are some classical systems, which can not be described in Lagrangian formalism, e.g. particles with spin or polarization. But for these systems there exists a valid Hamiltonian!

Furthermore, one can go from every Lagrangian to the Hamiltonian formalism (H could then be a multivalued function). So the last one is the "more fundamental" one.

(For details see Souriau: Structure of dynamical systems: a symplectic view of physics, site XXI, eg. http://books.google.de/books?id=4tBrbryIKQAC&printsec=frontcover#v=onepage&q&f=false)

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user Tobias Diez
answered Feb 4, 2012 by Tobias Diez (85 points) [ no revision ]
Comment to the answer(v1): If one has an action principle on Hamiltonian form, i.e., schematically $S=\int dt(p\dot{q}-H)$, one can always reinterpret it as an action principle on Lagrangian form with twice as many variables. In this sense, the Lagrangian formalism is more general than the Hamiltonian formalism. Souriau seems to make the opposite claim. Unfortunately, Google does not provide free access to all relevant parts of Souriau's book. Could you please include in your answer how Souriau defines Lagrangian and Hamiltonian formalism?

This post imported from StackExchange Physics at 2015-07-29 19:14 (UTC), posted by SE-user Qmechanic

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