Is there a way to know if a system described by a known equation of motion admits a Hamiltonian function? Take for example $$ \dot \vartheta_i = \omega_i + J\sum_j \sin(\vartheta_j-\vartheta_i)$$ where $\omega_i$ are constants. How can I know if there exist suitable momenta and a Hamiltonian?

  • 1
    $\begingroup$ Very long shot but you could use that $\dot{\vartheta}_i=\{\vartheta_i,H\}$ and write a very ugly system of partial differential equations for $H$. Nice question, however. $\endgroup$ – Jorge Lavín Jan 18 '13 at 8:50
  • 1
    $\begingroup$ @Burzum: I agree that the $\vartheta_i$ and $\omega_i$ are dimensionless, so on the LHS OP should put a characteristic time to compensate for the time derivative. $\endgroup$ – Vibert Jan 18 '13 at 12:49
  • 1
    $\begingroup$ Sorry, but shouldn't the equations be second-order for a hamiltonian formalism to exist? The other option would be to group half the $\vartheta_i$ into positions and the other half into momenta, but that is very violent on the initial symmetries. $\endgroup$ – Emilio Pisanty Jan 18 '13 at 17:10
  • 1
    $\begingroup$ If the number of angles $\vartheta_i$ is odd though, then it is impossible to formulate the system, as it is, in hamiltonian shape. You might be able to introduce (superfluous) suitably-evolving momenta to make it work in that case, though. $\endgroup$ – Emilio Pisanty Jan 18 '13 at 19:49
  • 1
    $\begingroup$ There is a research article on the Kuramoto model uncovered in a class of Hamiltonian systems $\endgroup$ – Marc Jul 1 '16 at 15:17

In general, it can be hard to tell if a given set of equations of motion (eom) are part of a (possibly larger) set of eom that can be put on Hamiltonian (or on Lagrangian) form.

Specifically, OP asks about the Kuramoto model with eom

$$\tag{1} \dot{\theta}_j -\omega_j ~=~\frac{K}{N}\sum_{k=1}^N\sin(\theta_k-\theta_j) ~\equiv~ K ~{\rm Im} \left( e^{-i\theta_j} \frac{1}{N}\sum_{k=1}^N e^{i\theta_k} \right).$$

We did not find a Hamiltonian formulation of eq. (1). Nevertheless, we hope that OP would still find the following considerations for interesting.

As explained in the Wikipedia page, the Kuramoto model describes $N$ oscillators with eigen-frequences $\omega_1, \ldots, \omega_N$, which couple

$$\tag{2} \dot{\theta}_j -\omega_j ~=~KR\sin(\Theta-\theta_j) ~\equiv~ K ~{\rm Im} \left( e^{-i\theta_j} \Phi \right),$$

with coupling constant $K$, via a complex order parameter

$$\tag{3} Re^{i\Theta}~\equiv~\Phi~=~\frac{1}{N}\sum_{k=1}^N e^{i\theta_k}, $$

which can be taken to be constant for $N$ large for statistical reasons.

Consider from now on the version of the Kuramoto model that is described by eq. (2) but without eq. (3), so the complex order parameter $\Phi$ is treated as just a free external complex parameter.

We now introduce complex fields with polar decomposition

$$\tag{4} \phi_j~:=~r_je^{i\theta_j}, \qquad j~\in~\{1,\ldots,N\}.$$

Consider next the Hamiltonian action

$$\tag{5} S_H ~:=~ \int\!dt~L_H, $$

with Hamiltonian Lagrangian

$$\tag{6} L_H~:=~ -\frac{i}{2} \sum_{j=1}^N \phi_j^{*}\dot{\phi}_j -H, $$

and with Hamiltonian

$$\tag{7} H ~:=~ \sum_{j=1}^N H_j , $$


$$\tag{8} H_j ~:=~ \frac{1}{2}\omega_j \phi_j^{*}\phi_j + K ~{\rm Im} (\phi_j^{*}\Phi ). $$

The Euler-Lagrange equations reads

$$\tag{9} -\frac{i}{2} \dot{\phi}_j -\frac{1}{2}\omega_j \phi_j~=~ \frac{K\Phi}{2i} , \qquad j~\in~\{1,\ldots,N\} ,$$

or in polar coordinates

$$\tag{10} \dot{\theta}_j -\omega_j r_j^2 ~=~KRr_j\sin(\Theta-\theta_j) ~\equiv~ K r_j~{\rm Im} \left( e^{-i\theta_j} \Phi \right),$$


$$\tag{11} \dot{r}_j ~=~KRr_j^2\cos(\Theta-\theta_j) ~\equiv~ K r_j^2~{\rm Re} \left( e^{-i\theta_j} \Phi \right).$$

Note that eq. (10) would reduces to the Kuramoto model eq. (2) if it was allowed to set $r_j=1$.

  • $\begingroup$ That's probably the best answer. It's very interesting. Actually, you were able to find a Lagrangian, though then the Euler's equations must still be coupled with (3). $\endgroup$ – Bzazz Oct 1 '13 at 8:31

As Michael Brown asked for an elaboration, here's a short introduction into the geometry of Hamiltonian mechanics:

While the traditional formulation works with special sets of coordinates (the 'canonical' ones), the differential-geometric approach de-emphasizes these and instead introduces the symplectic form as the characteristic structure of Hamiltonian phase space.

From a hands-on perspective, what this means is changing the minus that occurs in Hamilton's equations and the Poisson brackets into a full-blown antisymmetric tensor to make a coordinate-free formulation possible.

If you want to know if a given set of first-order equations can be treated (locally) as Hamiltonian, you'll have to look for such a tensor $\omega$ (more precisely: a non-degenerate closed antisymmetric bilinear differential form) that is invariant under the motion of the system.

The first order equations define a vector field $X$ and the invariance condition reads $\mathcal L_X\omega=0$ where $\mathcal L_X$ denotes the Lie-derivate, ie the difference in $\omega$ when moving along the flow of $X$.

From differential calculus and because $\mathrm d\omega=0$ by definition, this is equivalent to $$ \mathrm d\iota_X\omega=0 $$ where $\iota_X$ means contraction with $X$, ie $$ \iota_X\omega(\cdot)\equiv \omega(X,\cdot) $$ I did the calculation for Bzazz' equations in case $i=1\dots2$ (the dimension of phase space is necessarily even) on my way home, but the result wasn't particularly useful: So we do have a condition that tells us if a chosen symplectic form works, but in general no obvious way to find one (at least no way immediately obvious to me).

This ties in with Emilio Pisanty's comment: Classical (analytical) mechanics is normally concerned with second-order systems, which can be modelled geometrically in various ways:

  • If you write down the equations of motion in explicit form, you end up with a semi-spray, a section of the second tangent bundle bundle $\mathrm T\mathrm TM\to\mathrm TM$
  • If you introduce the concept of mass (an isomorphism $\mathrm TM\to\mathrm T^*M$) you end up with a velocity-dependent Newtonian force field, a section of $\mathrm T\mathrm T^*M\to\mathrm TM$
  • As $\mathrm T\mathrm T^*M$ and $\mathrm T^*\mathrm TM$ are naturally isomorphic, we can translate this to a section $\mathrm T^*\mathrm TM\to\mathrm TM$, which gives us the Lagrangian formulation - the differential $\mathrm dL$ of the Lagrange function is just a fancy way to write down a Newtonian force
  • Using the mass of the system (given by the fibre derivative $\mathbb FL$ in the Lagrangian formulation) we can move to the Hamiltonian picture: The differential $\mathrm dH$ of the Hamilton function gives us a section $\mathrm T^*\mathrm T^*M\to\mathrm T^*M$ which can be mapped to the Hamiltonian vector field $X$, a section $\mathrm T\mathrm T^*M\to\mathrm T^*M$, via the canonical symplectic form of the cotangent bundle.

This last equivalence between $\mathrm dH$ and $X$ is the geometric form of Hamilton's equations. For Hamiltonian mechanics, we can forget about the bundle structure of the tangent and cotangent spaces and use an arbitrary symplectic product, generalizing the formalism to a larger class of systems.

Normally, we do not start with first-order equations as in Bzazz' example (where no additional structure is present if they are not already in Hamiltonian form), but rather second-order ones (where the bundle structure is implicit). Then, we look for a Lagrange function, and check that it is hyper-regular (non-zero determinant of the Hessian matrix) - if this is the case, the mass $\mathbb FL$ (mapping velocity to canonical momentum) is an isomorphism and the system can be modelled in the Hamiltonian picture.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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