Friday 13 March 2015

complex systems - Hamiltonian or not?


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?



Answer



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


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


where


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


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


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



and


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


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


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