Lennard Jones, 3 particles, partition function

[SchroedingersLion]

TL;DR

Calculation of canonical integrals for 3 Lennard Jones particles.

Greetings,

similar to my previous thread
(https://www.physicsforums.com/threa...ce-between-two-particles.990055/#post-6355442),
I am trying to calculate the average inter-particle distance of particles that interact via Lennard Jones potentials. Now, however, I am dealing with 3 particles, and I fail to write down the partition function in a way that allows (numerical) computation.

The total potential is given by $\hat{U}(\mathbf{q_1}, \mathbf{q_2}, \mathbf{q_3})=U(r_{1,2})+U(r_{1,3})+U(r_{2,3})$ with $\mathbf{q_i}$ the position of particle $i$, $r_{i,j}$ the distance between particles $i$ and $j$, and $U(r)$ the usual Lennard Jones potential.

The partition function can then be written as $$Z=\int_{R³}\int_{R³}\int_{R³}e^{-\beta U(r_{1,2})}e^{-\beta U(r_{1,3})}e^{-\beta U(r_{1,3})}d³\mathbf{q_1}d³\mathbf{q_2}d³\mathbf{q_3}$$ I assume there is again a convenient way to transform this to a simpler integral as the potential only depends on the three inter-particle distances.
Can I again simply go to the coordinates $\mathbf{R}=\frac 1 3 (\mathbf{q_1}+\mathbf{q_2}+\mathbf{q_3})$ and $\mathbf{r_{i,j}}=\mathbf{r_i}-\mathbf{r_j}$?
(@vanhees71 's approach in my previous thread to two particles where it worked fine).

But this does not seem to do the trick, as I would arrive at a 12-dimensional integral, instead of a 9-dimensional integral.SL

 

[jambaugh]

So far as the calculus is concerned your new coordinates (plus R) will have an additional vector constraint. Defining the COM relative vector positions as: \mathbf{r}_k = \mathbf{q}_k -\mathbf{R} means they will satisfy:
\sum_{k=1}^3 \mathbf{r}_k = 0
So for example you would, say, integrate over \mathbf{R}, \mathbf{r}_1, and \mathbf{r}_2 and express the third position as a function of the two you chose to be independent.

As to the physics. The integration over \mathbf{R} would decouple but yield an infinite volume integral. But that's just the failure to recognize that in a practical setting one would be confining the entire system within a finite volume to regularize it. One can factor out this contribution, I believe, and simply renormalize the partition function.

 

[SchroedingersLion]

So far as the calculus is concerned your new coordinates (plus R) will have an additional vector constraint. Defining the COM relative vector positions as: \mathbf{r}_k = \mathbf{q}_k -\mathbf{R} means they will satisfy:
\sum_{k=1}^3 \mathbf{r}_k = 0
So for example you would, say, integrate over \mathbf{R}, \mathbf{r}_1, and \mathbf{r}_2 and express the third position as a function of the two you chose to be independent.

Thank you! While I understand your substitution, I am not sure how that leads to an integration over the $r_{i,j}$ from my OP. Then again, I am not sure that it is even possible to rewrite the integral in my desired form with more than two particles.

 

[jambaugh]

My inclination is to work relative to one of the particles, say the first and integrate over \mathbf{r}_{12} and \mathbf{r}_{13} using \mathbf{r}_{23} = \mathbf{r}_{13}-\mathbf{r}_{12}.

Still, I'm inclined to think that the integral is going to be nasty (intractable) due to the dependence of the potential on \lVert \mathbf{r}_{23}\rVert.

 

[SchroedingersLion]

Cheers,

I will probably not spend further time on this then, and instead "solve" the problem by using a particle simulation with very small step size.