Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 344 / 2

Write ${\mu }_j=\delta F/\delta p_j$ and $\int ={\int }_{t_1}^{t_2}dt\int d\mathbf{r}$. Eliminating the noise gives $f=\dot{p}+\Gamma \mu$. The Onsager--Machlup path probability is thereforeThese are conditional path densities given the corresponding starting configuration. For a time-even vector order parameter, reversal reads the configurations backward, changes $\dot{p}$ to $-\dot{p}$, and keeps $\mu$ unchanged on each corresponding configuration.

The two Gaussian white noise measures have the same normalization. With a common midpoint discretization, the noise-to-path Jacobian also agrees under reversal by time-reversal invariance of a path Jacobian. The common factor $\mathcal{K}[p]$ may include that path-dependent Jacobian; it need not be a universal constant. This makes the shared-factor assertion precise while leaving the requested ratio unaffected.

The difference between the two squared actions is $|\dot{p}+\Gamma \mu {|}^2-|-\dot{p}+\Gamma \mu {|}^2=4\Gamma \dot{p}\cdot \mu$. The common factor cancels, givingThe last equality uses the functional chain rule with the same midpoint convention as the Onsager–Machlup functional.

At equilibrium, microscopic reversibility requires $e^{-F_1/(k_{B}T)}{P}_F=e^{-F_2/(k_{B}T)}{P}_B$, equivalently $P_F/P_B=e^{-(F_2-F_1)/(k_{B}T)}$. Comparing the coefficients of an arbitrary free-energy change gives the model A fluctuation-dissipation relation

The Fourier transform turns ${\nabla }^2$ into multiplication by $-q^2$. Each Cartesian component is an Ornstein-Uhlenbeck process:Since $r(q)>0$, the transient from the remote past vanishes, and the integrating-factor solution isUsing the stated Gaussian white noise covariance and assuming $t_1\ge t_2$,Interchanging the times gives the stationary two-time correlation functionAt equal times this is the equilibrium Gaussian field theory covariance, and the decay time is $r(q{)}^{-1}$.

Put $u_j={\dot{p}}_j+{\partial }_i{W}_{ij}$. The noises required by a joint trajectory are $f_j=u_j+\Gamma {\mu }_j$ and $N_{ij}=W_{ij}+M{\partial }_i{\mu }_j$. Their independent Gaussian white noise weights give the joint path probability of an order parameter and its flux:For the physically reversed path, $p$ is time-even and the transport flux $W$ is time-odd. Therefore $u\mapsto -u$, $W\mapsto -W$, andThe common factor includes the two noise normalizations and the midpoint Jacobian determinant. Its calculation must include both relaxation channels. On a finite spatial grid, let $\mathsf{D}$ and $\mathsf{G}$ represent the divergence and gradient, with $\Delta =\mathsf{DG}$ the discrete Laplacian, and let ${\mathcal{H}}_n=\partial \mu /\partial p$ be the Hessian matrix of $F$ at the midpoint of time step $n$. Differentiating the two required noises with respect to the next configuration and the interval flux gives the Jacobian matrixUsing the Schur complement, its Jacobian determinant, apart from the path-independent power of $\Delta t$, isReversal visits the same midpoint configurations in reverse order, so the product of these factors is unchanged. The joint time-reversal invariance of a path Jacobian therefore involves the complete relaxation operator $\Gamma I-M\Delta$.

Taking the action difference givesRepeated component indices are summed. The second squared norm sums over both flux indices. This is mixed conserved and nonconserved order-parameter dynamics; reversing the order parameter history while leaving the flux unreversed would give the wrong probability ratio.

Set the noise strengths to the model A fluctuation-dissipation relation and the fluctuation-dissipation relation for a conserved flux:The preceding logarithmic ratio becomesThe last two terms combine as ${\partial }_i({\mu }_j{W}_{ij})$. They integrate to zero under the periodic boundary conditions, by integration by parts. The remaining term is $F_2-F_1$ by the functional chain rule, henceAfter including the equilibrium initial weights, the forward and reversed histories have equal probability. Thus microscopic reversibility is satisfied by the two channels together, with each channel's own thermal noise strength.