All-atom simulation and coarse-grained simulations

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SUMMARY

This discussion focuses on the rigorous bridge between all-atom and coarse-grained (CG) simulations as outlined in the paper by Noid et al. The key equations discussed include the Hamiltonians for both atomistic and CG models, specifically the evaluation of forces on the CG model as presented in equations 22-26. The confusion arises around the differentiation of the delta function in equation 23 and the integration by parts leading to equation 26. Clarification is sought on the mathematical manipulations involved in these equations.

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Sat D
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TL;DR
I have been studying molecular dynamics and simulation, and I want to learn and perform coarse-grained simulations. I have been reading about these, and I have a question about the math.
I am currently reading this [paper by Noid et. al.](https://doi.org/10.1063/1.2938860) on the rigorous bridge between atomistic and coarse-grained simulations.

In the paper, he defined a linear map from the atomistic coordinates and momenta $$\mathbf{r}^n, \mathbf{p}^n$$ to the coarse-grained coordinates $$\mathbf{R}^N, \mathbf{P}^N$$ He then defined the Hamiltonian for both frames of reference, $$h\, (\text{all-atom}),H \,(\text{CG})$$

The part of the paper I don't understand is when they evaluate the forces on the CG model (equations 22-26).
They write,
$$ \mathbf{F}_I(\mathbf{R}^N) = -\frac{\partial U (\mathbf{R}^N)}{\partial \mathbf{R}_I} = \frac{k_BT}{z(\mathbf{R}^N)}\frac{\partial z(\mathbf{R}^n)}{\partial \mathbf{R}_I} = \frac{k_BT}{z(\mathbf{R}^N)} \int d\mathbf{r}^n \exp(-u(\mathbf{r}^n)/k_BT)\prod_{J\neq I} \delta (M_{RJ}(r^n)-\mathbf{R}_J)\frac{\partial}{\partial \mathbf{R}_I}\delta \left( \sum _{i\in \mathcal{I}_I } c_{Ii}\mathbf{r}_i-\mathbf{R}_I\right)$$

This is the part that confuses me. I know $$\mathbf{R}_I = \sum _{i\in \mathcal{I}_I} c_{Ii}\mathbf{r}_i$$
So shouldn't $$\frac{\partial X}{\partial \mathbf{R}_I} = \sum _{i\in \mathcal{I}_I} \frac{\partial X}{\partial \mathbf{r}_i} \frac{\partial \mathbf{r}_i}{\mathbf{R}_I} = \sum _{i\in \mathcal{I}_I}\frac{\partial X}{\partial \mathbf{r}_i}\frac{1}{c_{Ii}}$$
be the case?
However, equation 23 simply states that
$$\frac{\partial}{\partial \mathbf{R}_I} \delta \left( \sum _{i\in \mathcal{I}_I} c_{Ii}\mathbf{r}_i - \mathbf{R}_I \right) = -\frac{1}{c_{Ik}} \frac{\partial }{\partial \mathbf{r}_k} \delta \left(\sum _{i\in \mathcal{I}_I} c_{Ii}\mathbf{r}_i - \mathbf{R}_I\right)$$

How does this equation work?

Furthermore, I don't understand how they perform integration by parts on the higher-dimensional integral they have here, and arrive at the equation that they do. I would greatly appreciate it if someone could help me reach equation 26 from equation 22 in the paper.

I appreciate any advice that you may have!
 
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Sat D said:
However, equation 23 simply states that
$$\frac{\partial}{\partial \mathbf{R}_I} \delta \left( \sum _{i\in \mathcal{I}_I} c_{Ii}\mathbf{r}_i - \mathbf{R}_I \right) = -\frac{1}{c_{Ik}} \frac{\partial }{\partial \mathbf{r}_k} \delta \left(\sum _{i\in \mathcal{I}_I} c_{Ii}\mathbf{r}_i - \mathbf{R}_I\right)$$

How does this equation work?

I'm unfamiliar with molecular dynamics, but that equation is saying something purely mathematical. It's just saying that
$$\frac{\partial}{\partial x} \delta \left( x-y \right) = -\frac{\partial}{\partial y} \delta \left( x-y \right)$$
to re-write it in a suggestive form, let ##z = -y##, then
$$\frac{\partial}{\partial x} \delta \left( x+z \right) = \frac{\partial}{\partial z} \delta \left( x+z \right)$$
which should be easy to see because the delta function is symmetric in x and z
 

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