Seperating Free Energy into Potential and Kinetic Terms

We coarse-grain a Hamiltonian system $H (\vec{q}, \vec{p})$ with the function $f \in ℝ^{n} \to ℝ^{N}$ mapping microstates to macrostates $\vec{Q} = f (\vec{q})$ . The free energy of a macrostate can be calculated from the restricted partition function.

e^{- \frac{F (\vec{Q}, \vec{P})}{k_{B} T}} = \frac{1}{h^{n}} \int_{V} d \vec{q} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \prod_{i} δ (\frac{s}{c_{i}} (\frac{\partial F}{\partial P_{i}} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \frac{\partial H}{\partial p_{j}})) e^{- \frac{H (\vec{p}, \vec{q})}{k_{B} T}}

Here $h$ is Planks constant, $c_{i}$ are the unit constant with units of $Q_{i}$ , and $s$ is some the unit of time. They make the equations inside the Dirac delta functions unitless so dimensional analysis still works.

We assume a Hamiltonian in a quadratic form with respect to the momentum.

H (\vec{q}, \vec{p}) = \frac{1}{2} {\vec{p}}^{⊤} 𝐌 (q) \vec{p} + U (q) = \frac{1}{2} \sum_{k} \sum_{l} M_{k l} (q) p_{k} p_{l} + U (q)

The velocities can be caluclated as below

\frac{\partial H (\vec{q}, \vec{p})}{\partial p_{j}} = \sum_{k} S_{j k} (q) p_{k}

where

𝐒 (q) = \frac{1}{2} (𝐌^{⊤} (q) + 𝐌 (q))

is a symmetric matrix. Substituting for the velocities we get the below.

e^{- \frac{F (\vec{Q}, \vec{P})}{k_{B} T}} = \frac{1}{h^{n}} \int_{V} d \vec{q} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))) \int_{- \infty}^{\infty} d \vec{p} \prod_{i} δ (\frac{s}{c_{i}} (\frac{\partial F}{\partial P_{i}} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})) e^{- \frac{\frac{1}{2} \sum_{k} \sum_{l} {\vec{p}}^{⊤} 𝐌 (\vec{q}) \vec{p} + U (\vec{q})}{k_{B} T}}

The potential energy can be split out of the exponent and since it only depends on position it can be moved out of the momentum integral. And the quadratic part can be switched to use $𝐒$ because ${\vec{p}}^{⊤} 𝐒 \vec{p} = {\vec{p}}^{⊤} 𝐌 \vec{p}$

= \frac{1}{h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))) \int_{- \infty}^{\infty} d \vec{p} \prod_{i} δ (\frac{s}{c_{i}} (\frac{\partial F}{\partial P_{i}} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})) e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p})}

Let's assume that the free energy is also quadratic, and try to solve for $𝐑 (\vec{Q})$ and $V (\vec{Q})$ .

F (\vec{Q}, \vec{P}) = {\vec{P}}^{⊤} 𝐑 (\vec{Q}) \vec{P} + V (\vec{Q})

So we have the below equality.

e^{- β ({\vec{P}}^{⊤} 𝐑 (\vec{Q}) \vec{P} + V (\vec{Q}))} = \frac{1}{h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \prod_{i} δ (\frac{s}{c_{i}} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})) e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p})}

Substitute in the Fourier definition for the Dirac delta.

= \frac{1}{h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \prod_{i} \frac{1}{τ} \int_{- \infty}^{\infty} d k e^{i k \frac{s}{c_{i}} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p})}

Do change of variables on the $k_{i}$ 's to factor out the $\frac{s}{c_{i}}$ . So now $k_{i}$ has units of $\frac{s}{c_{i}}$

= \frac{1}{h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \prod_{i} \frac{1}{τ} \frac{c_{i}}{s} \int_{- \infty}^{\infty} d k e^{i k (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p})}

Expand all the products out. Also let $C = \prod_{i} c_{i}$ .

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} \prod_{i} e^{i k_{i} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p})}

A product of exponents becomes an exponent of sums.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i \sum_{i} k_{i} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p})}

Combine the exponents.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- β (\frac{1}{2} {\vec{p}}^{⊤} 𝐒 (\vec{q}) \vec{p}) + i \sum_{i} k_{i} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})}

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- β \frac{1}{2} p^{⊤} 𝐒 p + i k^{⊤} (𝐑 P - 𝐉_{f} 𝐒 p)}

Swap the order of integration.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} \int_{- \infty}^{\infty} d \vec{p} e^{- β \frac{1}{2} p^{⊤} 𝐒 p + i k^{⊤} (𝐑 P - 𝐉_{f} 𝐒 p)}

The part the doesn't depend on the $k$ 's can be moved out.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} \int_{- \infty}^{\infty} d \vec{p} e^{- β \frac{1}{2} p^{⊤} 𝐒 p - i k^{⊤} 𝐉_{f} 𝐒 p}

Do complete the squares on the exponent

- β \frac{1}{2} p^{⊤} 𝐒 p - i k^{⊤} 𝐉_{f} 𝐒 p

= - \frac{1}{2} p^{⊤} (β 𝐒) p + (- i 𝐒 𝐉_{f}^{⊤} k)^{⊤} p

And assume $𝐒$ is positive definite then do complete the squares

= - \frac{1}{2} (p - i (\frac{1}{β} 𝐉_{f}^{⊤} k))^{⊤} (β 𝐒) (p - i (\frac{1}{β} 𝐉_{f}^{⊤} k)) - \frac{1}{2} \frac{1}{β} (𝐉_{f}^{⊤} k)^{⊤} 𝐒 (𝐉_{f}^{⊤} k)

Then substitute it back in to get

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} \int_{- \infty}^{\infty} d \vec{p} e^{- \frac{1}{2} (p - i (\frac{1}{β} 𝐉_{f}^{⊤} k))^{⊤} (β 𝐒) (p - i (\frac{1}{β} 𝐉_{f}^{⊤} k)) - \frac{1}{2} \frac{1}{β} (𝐉_{f}^{⊤} k)^{⊤} 𝐒 (𝐉_{f}^{⊤} k)}

The part that doesn't depend on momentum can be moved out.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} e^{- \frac{1}{2} \frac{1}{β} (𝐉_{f}^{⊤} k)^{⊤} 𝐒 (𝐉_{f}^{⊤} k)} \int_{- \infty}^{\infty} d \vec{p} e^{- \frac{1}{2} (p - i (\frac{1}{β} 𝐉_{f}^{⊤} k))^{⊤} (β 𝐒) (p - i (\frac{1}{β} 𝐉_{f}^{⊤} k))}

for an arbitrary multidimensional gaussian and symmetric positive definite matrix $A$ .

\int d \vec{x} e^{\frac{1}{2} (x - a)^{⊤} 𝐌 (x - a)} = \sqrt{\frac{τ^{n}}{det 𝐌}}

You can show it works even if the offset $a$ is an complex vector.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} e^{- \frac{1}{2} \frac{1}{β} (𝐉_{f}^{⊤} k)^{⊤} 𝐒 (𝐉_{f}^{⊤} k)} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

Now the expression is independent of the fine grain momentum. We need to do complete the squares again.

i k^{⊤} 𝐑 P - \frac{1}{2} \frac{1}{β} (𝐉_{f}^{⊤} k)^{⊤} 𝐒 (𝐉_{f}^{⊤} k)

Rearrange a bit.

= - \frac{1}{2} \frac{1}{β} k^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤}) k + i (𝐑 P)^{⊤} k

Completing the squares yields the below.

= - \frac{1}{2} \frac{1}{β} (k + i β (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P)^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤}) (k + i β (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P) - β P 𝐑^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P

Note this requires that $𝐉_{f} 𝐒 𝐉_{f}^{⊤}$ is positive definite. Substitute it back in.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- \frac{1}{2} \frac{1}{β} (k + i β (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P)^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤}) (k + i β (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P) - β P 𝐑^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) e^{- β P 𝐑^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- \frac{1}{2} \frac{1}{β} (k + i β (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P)^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤}) (k + i β (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P)} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

Then do the same thing again with the Gaussian.

= \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) e^{- β P^{⊤} 𝐑^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P} \sqrt{\frac{β^{N} τ^{N}}{det (𝐉_{f} 𝐒 𝐉_{f}^{⊤})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

This almost gives a solution for seperating the free energy, but not quite there yet.

e^{- β (P^{⊤} 𝐑 (Q) P + V (Q))} = \frac{C}{τ^{N} s^{N} h^{n}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q}))) e^{- β P^{⊤} 𝐑^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P} \sqrt{\frac{β^{N} τ^{N}}{det (𝐉_{f} 𝐒 𝐉_{f}^{⊤})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

This gives a final, exact, expression of (with $O$ being whatever units make the rightmost sqrt dimensionless)

\exp (- β (P^{⊤} 𝐑 (Q) P + V (Q))) = \frac{O}{τ^{N} h^{n}} \int_{V} d \vec{q} \exp (- β U (q) + \ln (\frac{C}{s^{N} O} \sqrt{\frac{β^{N} τ^{N}}{det (𝐉_{f} 𝐒 𝐉_{f}^{⊤})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \exp (- β P^{⊤} 𝐑^{⊤} (𝐉_{f} 𝐒 𝐉_{f}^{⊤})^{- 1} 𝐑 P) \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))

Seperating with additional asumptions

Let $𝐋 = 𝐉_{f} (\vec{q}) 𝐒 (\vec{q}) 𝐉_{f}^{⊤} (\vec{q})$ . If $𝐋$ can be written solely as a function of $\vec{Q}$ instead of $\vec{q}$ , so it only depends on the coarse grain positions, then we can move it out of the integral. We can do that because it is practically constant in the integral since any time it differs the Dirac delta function removes it's contributions

\exp (- β (P^{⊤} 𝐑 (Q) P + V (Q))) = \frac{O}{τ^{N} h^{n}} \exp (- β P^{⊤} 𝐑^{⊤} 𝐋^{- 1} (\vec{Q}) 𝐑 P) \int_{V} d \vec{q} \exp (- β U (q) + \ln (\frac{C}{s^{N} O} \sqrt{\frac{β^{N} τ^{N}}{det 𝐋 (\vec{Q})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))

Then a simple solution as $𝐑 (\vec{Q}) = 𝐋 (\vec{Q})$ , and in the exponent $𝐋^{- 1} (\vec{Q}) 𝐑 (\vec{Q})$ (or the transpose since $L$ symmetric.

\exp (- β (P^{⊤} 𝐑 (Q) P + V (Q))) = \frac{O}{τ^{N} h^{n}} \exp (- β P^{⊤} 𝐑 P) \int_{V} d \vec{q} \exp (- β U (q) + \ln (\frac{C}{s^{N} O} \sqrt{\frac{β^{N} τ^{N}}{det 𝐋 (\vec{Q})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))

And the momentum contribution can be divided out on both sides.

\exp (- β V (Q)) = \frac{O}{τ^{N} h^{n}} \int_{V} d \vec{q} \exp (- β U (q) + \ln (\frac{C}{s^{N} O} \sqrt{\frac{β^{N} τ^{N}}{det 𝐋 (\vec{Q})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))

Then just rearrange and basic algebra.

\exp (- β V (Q)) = \frac{O}{τ^{N} h^{n}} \frac{C}{s^{N}} \sqrt{\frac{β^{N} τ^{N}}{det 𝐋 (\vec{Q})}} \int_{V} d \vec{q} \exp (- β U (q) + \ln (\frac{1}{O} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))

V (Q) = - \frac{1}{β} (\ln (\frac{1}{τ^{N}}) + \ln (\frac{C}{s^{N}} \sqrt{\frac{β^{N} τ^{N}}{det 𝐋 (\vec{Q})}}) + \ln (\frac{O}{τ^{N} h^{n}} \int_{V} d \vec{q} \exp (- β U (q) + \ln (\frac{1}{O} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \prod_{i} δ (\frac{1}{c_{i}} (Q_{i} - f_{i} (\vec{q})))))