Geometry Rescaling

One of the major issues with the stability of the timemachine is whether or not your Hessians are PSD. A non-PSD Hessian induces a transition matrix with a spectral radius $p(r) > 1$, causing a divergent trajectory.

Recall Brownian dynamics: $x_t = x_{t-1} - k G(x, \theta) + noise$

Assuming a zero mixed partial, we can differentiate with respect to $\theta$ $x'_t = x'_{t-1} - k H(x_{t-1}) x'_{t-1}$ $x'_t = (I - k H) x'_{t-1}$

G and H are the Gradients and Hessians, respectively. Since $H$ is rarely PSD in Cartesian coordinates, how can we fix this?

By playing dirty: consider an alternative dynamical equation where we shrink the geometry by a small amount: $x_t = (1 - k c) x_{t-1} - k G(x, \theta) + noise$

Which has derivative: $x'_t = (1 - k c) x'_{t-1} - k H(x_{t-1}) x'_{t-1}$ $x'_t = (I - k c I - k H) x'_{t-1}$ $x'_t = (I - k (c I + H) ) x'_{t-1}$

Which is equivalent to adding a diagonal to the Hessian. So long as $c$ is bigger than the sum of each row/column, our Hessian is now PSD.

The downside to all of us? No one really knows what statistical ensemble this belongs to anymore.

Derivatives and Internal Coordinates

Brownian Dynamics has the following equations of motion, where we assume that the geometries $x$ is dependent on the forcefield parameters $\theta$ $x_t = x_{t-1} - k(g(x_{t-1}(\theta), \theta) + noise)$

We can differentiate the above w.r.t. $\theta$ $\dfrac{\partial x_t}{\partial \theta} = \dfrac{\partial x_{t-1}}{\partial \theta} - k( \dfrac{\partial g}{\partial x}(x_{t-1})\dfrac{\partial x_{t-1}}{\partial \theta} + \dfrac{\partial g}{\partial \theta})$

For the purposes of stability analysis and without loss of generality we can assume: $\dfrac{\partial g}{\partial \theta} = 0$

So we can proceed to simplify into: $\dfrac{\partial x_t}{\partial \theta} = (I-k \dfrac{\partial g}{\partial x}) \dfrac{\partial x_{t-1}}{\partial \theta}$

It can be shown that this converges if and only if the Hessian $\dfrac{\partial g}{\partial x}$ is PSD and $k$ is sufficiently small.

The issue is that the Hessian is not PSD in standard equations of motion due to “contamination” of the translational and rotational degrees of freedom even for the simplest of systems.

What we’d really like is something that looks like: $\dfrac{\partial x_t}{\partial \theta} = (I-k P^T \dfrac{\partial g}{\partial x} P) \dfrac{\partial x_{t-1}}{\partial \theta}$

We can back compute the following anti-derivative : $x_t = x_{t-1} - k P^T (g(P x, \theta) + noise)$

But this is invalid since we’d be modifying the positions in two different reference frames, so how would we modify the equations of motion to give us Hessians are that always PSD?

Pure internal coordinates

We can trivially implement Brownian dynamics in either Cartesian Coordinates or Internal Coordinates, since the kinetic energy is irrelevant. Suppose we have a two particle harmonic oscillator: $U(r(x), b) = (r(x)-b)^2$

Where $r$ is the distance between the two particles.

We have two equivalent Brownian equations of motion we can use: $x_t = x_{t-1} - k(g(x_{t-1}(\theta), \theta) + noise_{cart})$ $r_t = r_{t-1} - k(g(r_{t-1}(\theta), \theta) + noise_{int})$ $noise_{int} = r(noise_{cart})$

They have the following derivatives w.r.t. $\theta$ if we again assume the mixed partial is zero. $x'_t = x'_{t-1} - k(H_x(x_{t-1})x'_{t-1})$ $r'_t = r'_{t-1} - k(H_r(r_{t-1})r'_{t-1})$

Where $H_x$ is the Hessian with respect to the coordinate system $x$

The same same equations of motion are generated, except the former is divergent if we sample geometries where $r < b$. What?

Asymmetric Hessians

I encountered a really counter-intuitive result this past week regarding the Hessians of a simple 2D Harmonic Oscillator: $U(r(x_0, y_0, x_1, y_1), b) = (r-b)^2/2 =(\sqrt{(x_0-x_1)^2+(y_0-y_1)^2}-b)^2/2$

It is clear that the energies for $r+\delta$ and $r-\delta$ are the same, and the forces are identical in magnitude except opposite in direction.

However, while the internal Hessian $\dfrac{\partial^2 U}{\partial r^2} = 1$ and therefore positive semi-definite (PSD) every where w.r.t. $r$, this is not the case for the 4×4 Cartesian Hessian $\dfrac{\partial^2 U}{\partial x_i \partial x_j}$. In fact, the Cartesian Hessian is PSD only if $r \geq b$, that is, it has only positive eigenvalues or zero. We can actually analytically show that the eigenvalues of the Cartesian Hessian are $0, 0, 2$ and $\lambda$ where: $\lambda= 2(1-\dfrac{b}{r^2})$

So if $r < b$ then the eigenvalue is negative (no longer PSD), and if $r>b$ then the eigenvalue is positive (PSD).

To see this for yourself in sympy, the code is:

import sympy as sp
from sympy.functions import Abs, sqrt
from IPython.display import display, Math
x0, x1, y0, y1, b = sp.symbols('x0 x1 y0 y1 b', real=True)
U = ((sqrt((x0-x1)**2+(y0-y1)**2)-b)**2)/2
sp.init_printing()
args = (x0, y0, x1, y1)
H = sp.Matrix(list(sp.derive_by_array(sp.derive_by_array(U, args), args)))
H = H.reshape(4,4)
H = sp.simplify(H)
H.eigenvals()

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