Question

MC simulation: phase space being 3N-dimensional vs phase being 6N-dimensional

Answer 1

How can a Monte Carlo simulation be used to calculate thermodynamic quantities, given that phase space is 6N-dimensional, but we sample in 3N-dimensional (in what I expect to be the so called "conformational space"

Answer 2

Elaboration:
In most literature I read from the general learning books they write that the normal phase space is 6N dimensional, with N being the amount of particles in the system. 1 positional vector and 1 vector of momentum for each particle with the total energy given by the Hamiltonian.
In Monte Carlo I work with people talk about the conformational space. So I was thinking to unify the two terminologies:
The Monte Carlo phase space is 3N as we only got positional vectors, further more the the energy difference in the 3N dimensional phase, the energy difference in each state must from the potential energy function, while the kinetic energy contribution remains constant (so it doesn't violent the quantum mechanical theory of a particle in rest)?

To put a little math to it:
Lets assume we work in a system with a fixed amount of particles, volume, and temperature the canonical ensemble partition function \(Q\), with the system containing \(N\) identical particles of the mass \(m\), the partition function \(Q\) is given by:
\[\large Q_{NVT}=\frac{ 1 }{ N!~ h^{3N} }\int\limits_{}^{}\int\limits_{}^{} d \textbf{p}^{N}d\textbf{r}^N \exp \left( -\frac{ H \left( \textbf{p}^{N},\textbf{r}^N \right) }{ k_BT } \right) \]
With the Hamiltonian \(H\) equal to:
\[\large H(\textbf{p}^{N},\textbf{r}^N)=\sum_{i=1}^{N}\frac{ \left| \textbf{p}_i \right|^{2} }{ 2m }+V \left( \textbf{r}^N \right)\]
With \(V(\textbf{r}^N)\) being the potential energy function.
@Kainui

Answer 3

I was personally (in terms of math) thinking about separating the double integral as one only deals with positions and one with momentum. (I guess it is a fair assumption to say that the potential energy function does not depend upon the momentum of a particle)

Answer 4

I guess I need more clarification on the dimensions, when you say 6N dimensional this is 6 particles? I was envisioning this as being 3 components of position and 3 components of velocity for a single particle. Also, is it possible that the 3N dimensions are complex dimensions so that each one actually contains 2 dimensions inside of it?

Also, I'm not entirely sure about separating the potential energy and momentum, since they are both related by the total energy. It seems like the momentum, specifically the velocity, depends on the potential energy but maybe I'm wrong here, I'm not entirely sure. I am not entirely sure how the MC dimensions are working at all really, could you perhaps draw a picture of any of this? As far as I know MC is a method for approximating integrals with random numbers.

Answer 5

What I mean is that N being the amount of particles in the system. The phase space (all information in a dynamic system), can then be found to contain 6N dimensions. 3 from the position (x,y,z coordinates) and the momentum vector (p_x, p_y, p_z) so to say.

Answer 6

What I know from Monte Carlo simulations which is what I work with is that the we sample in the "conformational space" that being all the atomic positions and all "related information". with related information I mean all parameters that depends on the atomic positions is contained within the conformational space like torsion angles.
An example would be if you drew the potential energy vs the torsion angle:

The problem for me is that, in order to calculate the total energy of system we need to have both the momentum and the atomic positions. But somehow we evade that problem in monte carlo simulations