considering the Hamiltonian associated with propagation equation Eq. (\[partial\]), $$\mathcal{H}=\iint \left(|\nabla \psi|^2-\frac{1}{2}|\psi|^4-V|\psi|^2 \right)~\mathrm{d}x ~\mathrm{d}y,
\label{hamiltonian}$$ which is a conserved quantity upon propagation. Depending on their orientation, stationary dipoles ($B=0$) have a slightly different value of $\mathcal{H}$ in nonlinear regime. If we take our diagonal (w.r.t. waveguide cross-section) dipoles $D_1$ and $D_2$ from above, we can construct a parallel dipole $D_p$ in the following way: $$D_p=\frac{D_1+D_2}{\displaystyle \sqrt{\iint |D_1+D_2|^2 ~\mathrm{d}x ~\mathrm{d}y}}.$$ The intensity distribution of this dipole $D_p$ is very close to that shown in the snapshot of the rotating azimuthon shown in the right panel in Fig. \[square\_r\_evolution\]. One can show using the definition of the Hamiltonian Eq. (\[hamiltonian\]) that $\mathcal{H}(D_p)>\mathcal{H}(D_1)=\mathcal{H}(D_2)$. In fact, it turns out that those two dipole orientations (parallel and diagonal) correspond to the extremal values of the Hamiltonian. Now, if the azimuthon rotates it has to pass through all possible orientations, i.e., its value of $\mathcal{H}$ has to be larger than $\mathcal{H}(D_p)$. We can use this reasoning to estimate the value of $B_{\rm cr}$: The left panel of Fig. \[angularmomentumcurve\] depicts the dependence of the Hamiltonian of superposed diagonal dipoles as a function of $B$ (red line) for constant power $P$. The blue line represents the value 