model H

For simplicity, we first look at the continuum equations of motion for $(\phi, \pi_T)$ without dissipation or noise:

$$\begin{aligned} \dot{\phi} & = - \nabla_\mu ( \phi , \pi_{T,\mu} ) = - \pi_{T,\mu} \nabla_\mu \phi, \\ \dot{\pi}_{T,i} & = - \hat{{\cal T}}_{ij} \nabla_k ( \pi_{T,k} \pi_{T,j} + \nabla_k \phi \nabla_j \phi ) \\ & = - \hat{{\cal T}}_{ij} ( \nabla_k \left(\pi_{T,k} \pi_{T,j} \right) + \nabla_j \phi \nabla^2 \phi ) \end{aligned}$$

The time derivative of the Hamiltonian is (assumed $\rho = 1$), $$\dot{H} = \int d^3x [ - \dot{\phi} \nabla^2 \phi + \pi_{T,i} \dot{\pi}_{T,i} + V'(\phi) \dot{\phi} ] $$

Substituting

$$\begin{aligned} \dot{{\cal H}} = - \int d^3x \ \left[ - \nabla^2\phi \\ \vec{\pi}_T \cdot \vec{\nabla} \phi % + \vec{\pi}_T \cdot \left( \vec{\pi}_T \cdot \vec{\nabla} \right) \vec{\pi}_T % + \nabla^2\phi \ \vec{\pi}_T \cdot \vec{\nabla} \phi % + \vec{\nabla} \cdot \left( V(\phi) \vec{\pi}_T \right) \right]. \end{aligned}$$

The first and third terms cancel each other. In the continuum, the second term can be written as a divergence,

$$\begin{aligned} \vec{\pi}_T \cdot \left( \vec{\pi}_T \cdot \vec{\nabla} \right) \vec{\pi}_T = \vec{\nabla} \cdot \left( \vec{\pi}_T \ \frac{\vec{\pi}_T^2}{2} \right). \end{aligned}$$

However, this is not guaranteed in the discretised version if we use the divergence type term ∇_k(π_T, kπ_T, j). Morinishi et al suggests we use a skew-symmetric form to take care of this:

$$\begin{aligned} \nabla_k \left(\pi_{T,k} \pi_{T,j} \right) = \frac{1}{2} \ \nabla_k \left(\pi_{T,k} \pi_{T,j} \right) + \frac{1}{2} \ \pi_{T,k} \ \nabla_k \pi_{T,j}. \end{aligned}$$

Also, in the discretised version we expect the first and third terms to cancel each other if we use the central difference scheme throughout, i.e, for (∇_jϕ∇²ϕ) term and also the kinetic term, (∇ϕ)², in the Hamiltonian:

$$\begin{aligned} \nabla_j \phi &\to \frac{\phi(\vec{x}+\hat{\nu}_j) - \phi(\vec{x} - \hat{\nu}_j)}{2}, \nonumber \\ % \nabla^2\phi &\to \sum_\nu \frac{1}{4} \ \left(\phi(\vec{x} + 2 \hat{\nu}) + \phi(\vec{x} - 2 \hat{\nu}) - 2 \ \phi(\vec{x}) \right) \end{aligned}$$

We note that the central difference scheme satisfies ∫d³x(∇ϕ)² →  − ∫d³x ϕ∇²ϕ in Hamiltonian:

$$\begin{aligned} {\cal H} &= \frac{1}{2} \ \sum_i \ \left(\frac{\phi_{i+1} - \phi_{i-1}}{2}\right)^2 = \frac{1}{2} \ \frac{1}{4} \sum_{i} \left( \phi_{i+1}^2 + \phi_{i-1}^2 - 2 \ \phi_{i+1} \phi_{i-1} \right), \nonumber \\ % & = \frac{1}{2} \ \frac{1}{4} \ \sum_i \left( 2 \phi_i^2 - \phi_i \ \phi_{i+2} \ - \phi_i \phi_{i-2} \right) = - \frac{1}{2} \ \sum_i \ \phi_i \left( \frac{\phi_{i+2} + \phi_{i-2} - 2 \phi_i}{4} \right) \end{aligned}$$

If we discretize the spatial derivatives we get (leaving aside the projector and using the skew symmetric form of Morinishi et al for ∇_k(π_T, kπ_T, i)),

$$\begin{aligned} \dot{\phi} &= - \pi_{T,\mu}(x) \ \nabla^c_{\mu} \phi \\ % \dot{\pi}_{T,\mu} &= - \Bigg[ \frac{1}{2} \nabla^c_{\nu} \left( \pi_{T,\nu} \pi_{T,\mu} \right) + \frac{1}{2} \ \pi_{T,\nu} \ \nabla^c_{\nu} \pi_{T, \mu} \nonumber \\ % & + \nabla^c_\mu \phi \ \sum_\nu \ \frac{ \left(\phi(\vec{x} + 2 \hat{\nu}) + \phi(\vec{x} - 2 \hat{\nu}) - 2 \ \phi(\vec{x}) \right)}{4} \Bigg], \end{aligned}$$

where ∇_μ^cψ(x⃗) = (ψ(x⃗+μ̂)−ψ(x⃗−μ̂))/2. For the projector we use

$$\begin{aligned} P_{\mu\nu} = \delta_{\mu\nu} - \frac{\tilde{k}_\mu \ \tilde{k}_\nu}{\tilde{k}^2}, \end{aligned}$$

with k̃_μ = sin (k_μ), and k_μ = (2π/L)n̂_k, μ. The time derivatives can be solved using Runge-Kutta methods.

Writing the evolution equations as,

$$\begin{aligned} \dot{\phi}_\mu = {\cal F}_\mu (\phi_\mu), \end{aligned}$$

the ‘third-order’ RK scheme gives,

$$\begin{aligned} \phi^{n+1}_\mu = \phi^n_{\mu} + \Delta t \ \left( \frac{1}{6} \ k_{\mu,1} + \frac{1}{6} \ k_{\mu,2} + \frac{2}{3} \ k_{\mu,3} \right), \end{aligned}$$

where k_μ, i are given by

$$\begin{aligned} k_{\mu,1} &= {\cal F}_\mu (\phi^n_\mu), \\ % k_{\mu, 2} &= {\cal F}_\mu \left( \phi^n_\mu + \Delta t \ k_{\mu,1} \right) \\ % k_{\mu,3} &= {\cal F}_\mu \left( \phi^n_\mu + \frac{\Delta t}{4} \ \left( k_{\mu,1} + k_{\mu,2} \right) \right). \end{aligned}$$