Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My (non-expert) impression is that many physically important equations of motion can be obtained as Euler-Lagrange equations. For example in quantum fields theories and in quantum mechanics quantum equations of motion are obtained from the classical ones only if a Lagrangian (or Hamiltonian) is known for the classical case. Is my understanding oversimplified too much?

Are there examples of physically important equations which are not Euler-Lagrange for any Lagrangian?

More specifically, let us consider the classical motion of a particle in $\mathbb{R}^3$ with friction: $$\overset{\cdot\cdot}{\vec x}=-\alpha \overset{\cdot}{\vec x},\, \, \alpha>0,$$ namely acceleration is proportional to velocity with negative coefficient.

Is this equation Euler-Lagrange for an appropriate Lagrangian? Is there a quantum mechanical version of it?

share|improve this question
3  
mathoverflow.net/questions/101395/… might be relevant –  Thomas Rot Oct 27 '13 at 16:41
    
@ Thomas Rot: Thank you. This seems to be very relevant, I have to think on it. –  semyon alesker Oct 28 '13 at 7:49
2  
Uh, how is this Research-level? –  Dimensio1n0 Oct 28 '13 at 11:14

4 Answers 4

  1. Well, there is always the trivially enforced solution $$\tag{1} S[x,\lambda]~=~\int\! dt \sum_{i=1}^3\lambda_i(t) \left(\ddot{x}^i(t)+\alpha \dot{x}^i(t) \right),$$ where $\lambda_i(t)$ are three Lagrange multiplier variables. From now on we assume that we are not allowed to use other variables than $x^i(t)$.

  2. Whether the 3D ODE $\ddot{\bf x}+\alpha\dot{\bf x}=0$ has a Lagrangian formulation or not is in general a hard problem, see e.g. Douglas' theorem and the Helmholtz conditions on this Wikipedia page. (We are unaware if these conditions have been completely analyzed in 3D. On the other hand, it is easy to see that the corresponding 1D problem has a Hamiltonian action formulation, at least locally, cf. e.g. this Phys.SE answer.)

  3. The 3D problem simplifies enormously if we assume that the form of the kinetic part $T$ of the Lagrangian $L=T-U$ is just $T=\frac{1}{2}\dot{\bf x}^2$. (This assumption is of course heavily motivated from physics, and is undoubtedly necessary if we would like to make quantum mechanical sense of the system.) Then OP's question essentially boils down to if there exists a velocity-dependent potential $U({\bf x}, \dot{\bf x})$ for the friction force ${\bf F}=-\alpha \dot{\bf x}$. This is equivalent to asking if the Helmholtz conditions [1] $$\tag{2} \frac{\partial F_i}{\partial x^j} -\frac{1}{2}\frac{d}{dt}\frac{\partial F_i}{\partial \dot{x}^j} ~=~[i \longleftrightarrow j], \qquad \frac{\partial F_i}{\partial \dot{x}^j}~=~-[i \longleftrightarrow j].$$ are satisfied. It is a straightforward to see that this is not the case. (Already the second condition (2b) is not met.) See also this related Phys.SE post. For a general discussion of the notion of conservative force for velocity-dependent forces, see e.g. this Phys.SE answer.

References:

  1. H. Helmholtz, Ueber die physikalische Bedeutung des Prinzips der kleinsten Wirkung, J. für die reine u. angewandte Math. 100 (1887) 137.
share|improve this answer
    
Thanks very much for the answer, in particular for the reference to Douglas' theorem. I think that it does imply that the equation $\overset{\cdot\cdot}{\vec x}+\alpha\overset{\cdot}{\vec x}=0$ has a Lagrangian formulation. I think the Helmholtz conditions are satisfied by the matrix $g_{ij}(t,x,\overset{\cdot}{\vec x})=e^{-\alpha t/2}\delta_{ij}$. However it is not yet clear to me for the moment how to write a Lagrangian down explicitly, even whether it can be chosen to be time independent. –  semyon alesker Nov 30 '13 at 15:26

strictly speaking, the answer to your question is "no"; but if you insist, you can modify the Euler-Lagrange equation by adding a nonzero right-hand-side to incorporate friction:

$$\frac d{dt} \left( \frac{\partial L}{\partial \dot x_i} \right) - \frac{\partial L}{\partial x_i} = - \frac{\partial F}{\partial \dot x_i}$$

the function $F$ is called the dissipation function, in the example you give it is quadratic in the velocity, $F=\frac{1}{2}m\alpha \dot{x}^2$.

Quantum mechanical evolution is unitary, so dissipation and the associated irreversibility is not readily incorporated (however, see Lindblad equation).

share|improve this answer
    
Thank you. Do you have an explanation or reference why the answer is "no"? Regarding the "quantum" part of your answer, I missed if there is some convention to quantize motion with friction. Probably it is hard to state this question formally, but as a first guess one could try to write a Schroedinger equation with general (time dependent?) Hamiltonian such that a version of the Ehrenfest theorem would be compatible with the classical equation of motion with friction. –  semyon alesker Oct 28 '13 at 8:01
    
I think that the quantization concerns only non dissipative symplectic systems. What can you think as a nuclear or sub-nuclear dissipative quantum system? What would mean dissipate in this context? I'm not sure it makes sense. –  Patrick I-Z Nov 27 '13 at 22:51

There is another way around, starting with Lagrange, but bypassing his variational construction to reach immediately the presymplectic construction. Considering the motion of a point submitted to a force $F$, you can define the following $2$-form(${}^*$) $\omega$ on the space $Y$ of initial conditions $(x,v,t) \in {\bf R}^3 \times{\bf R}^3 \times {\bf R}$ $$ \omega(\delta y)(\delta'y) = \langle\delta v - F\delta t , \delta' x - v\delta' t \rangle - \langle \delta' v - F\delta' t , \delta x - v\delta t \rangle, $$ where $\delta y = (\delta x, \delta v, \delta t) \in T_yY$, idem for $\delta'y$. You can notice that the kernel of this 2-form is generated by the equations of motion: $$ \ker(\omega_y) = \{ \delta y \mid \delta x = v \delta t \quad \mbox{and} \quad \delta v = F \delta t\}. $$ In other words, the motions of the point submitted to the force $F$ are the integral curves of the vectorial distribution $y \mapsto \ker(\omega_y)$.

You can check that if $F$ does not depend on $v$ and if $F$ is derived from a potential $F = - dU$ then $\omega$ is closed, therefore presymplectic, and the space of characteristics of $\omega$, that is, the solutions of your dynamical system, is then equipped with a symplectic structure.

Now we can rephrase your question in the presymplectic framework: what are the condition on $F$, depending a priori from $y=(x,v,t)$, for $\omega$ to be closed?

First of all you can notice that the form $\omega$ writes $$ \omega = \omega_0 \oplus F \cdot dx \wedge dt, $$ where $\omega_0$ is the standard symplectic form on ${\bf R}^3 \times {\bf R}^3$. The condition for $\omega$ to be closed writes then $d[F\cdot dx] \wedge dt = 0$. In other words we need $d[F \cdot dx]$ to be proportional to $dt$. If you make the computation you'll find that the partial derivative $\partial F_i / \partial v^j$ must vanish and for all $t$, $d[x \mapsto F_t(x)] = 0$, that is, $F_t = -dU$, where $U$ depends on $(x,t)$.

Thus if you add a dissipative term in your force, you lose the symplectic structure on the space of solutions, and therefore all the conservation theorems associated.

(*) This construction is due to Cartan, Galissot, Souriau.

share|improve this answer

There are (at least) two methods where certain dissipative systems can be obtained from a Lagrangian. These methods are described (among other treatments of dissipation in mechanical systems) in the following lecture by Anthony Bloch.

One method is achived by means of the coupling of the (finite dimensional) mechanical system to an infinite dimensional one. The prototype of this construction is the Lamb model (discovered by H. Lamb in 1900) in which a harmonic oscillator is attached to an infinite string. This model equation of motion can be obtained from a the sum of the harmonic oscillator's and the elastic string actions subject to the boundary condition at the attachment point. After the insertion of the D'Alembert retarted solution of the string, the equation of motion of the Harmonic oscillator acquires a viscous dissipation term.

Another method is described in page 21 of Bloch's lectures, in which the dissipation can be achieved through a noncanonical symplectic structure. This symplectic structure can be obtained from a Lagrangian with a Wess-Zumino term, please see for example the following article by: Mendes, Neves, Oliveira, and Takakura in which radiation daming in two dimensions is described according to this model. The Hamiltonian formulation of this construction allows the quantization of this type of systems.

share|improve this answer
    
I'll look forward to understand what this constructions give exactly, but what I just get, by a rapid lecture of your second paper, is this sentence: "Since the system (1) is dissipative a straightforward Lagrangian description leading to a consistent canonical formalism is not available." This is what I believe for structural reason: symplectic means non-dissipative (intuitively $d\omega=0$ means: there is no leak). Thus if you want to make a dissipative system starting with a symplectic structure you have to use some trick to compensate on one hand what you lose on the other. –  Patrick I-Z Dec 1 '13 at 16:32
    
The authors provide a Lagrangian in equation (5), also the Dirac brackets in equation (13) are of the type referred to in the first reference. I guess i shouldn't have used the name symplectic for these brackets. –  David Bar Moshe Dec 1 '13 at 16:40
    
They have a Lagrangian the same way we can write always the solutions of the Newton equations as the kernel of an skew-symmetric 2-form on the space of initial conditions. The system is still a 2nd degree ODE. The point is that this form is not closed when the system is dissipative. I agree that there is something to understand here. Thanks BTW for the reference you give. –  Patrick I-Z Dec 1 '13 at 16:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.