- #1
Sat D
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- TL;DR Summary
- 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!
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!