A All-atom simulation and coarse-grained simulations

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The discussion focuses on understanding the transition from all-atom to coarse-grained simulations as outlined in Noid et al.'s paper. Key points include the definition of a linear map between atomistic and coarse-grained coordinates, and the formulation of the Hamiltonians for both models. The user expresses confusion regarding the evaluation of forces on the coarse-grained model and the implications of certain equations, particularly the relationship between derivatives of delta functions. There is also a request for clarification on the integration by parts used to derive specific equations in the paper. Overall, the conversation emphasizes the mathematical intricacies involved in bridging these two simulation approaches.
Sat D
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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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