Lennard Jones Potential and the Average Distance Between Two Particles

[SchroedingersLion]

Greetings!

Suppose I have 2 particles that interact via a Lennard Jones potential $$U(\mathbf{q}_{1}, \mathbf{q}_{2}) = 4 \epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^6 \right] $$
with interparticle distance $r=|\mathbf{q}_{1} - \mathbf{q}_{2}|$.

The canonical density $\rho(\mathbf{q}_{1}, \mathbf{q}_{2})=\frac{1}{Z}e^{-\beta U}$ and I want to calculate the average distance between the particles in 2 dimensions. In a first attempt, I thought it was legit to simply write

$$ <r> = \int_{0}^{\infty} r \rho(r)\, dr .$$

But then I wondered: Is this even equivalent to the expression one needs to start with, i.e.
$$<r> = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} |\mathbf{q}_{1} - \mathbf{q}_{2}| \rho(\mathbf{q}_{1}, \mathbf{q}_{2}) \,d\mathbf{q_1} \, d\mathbf{q_{2}}$$

This is a volume integral and it feels like turning it into a 1D integral is wrong, as I lose the information about the number of dimensions in doing so.
So I could start by doing a proper substitution, say $r=|\mathbf{q}_{1} - \mathbf{q}_{2}|$ but how would I transform the differentials then?
SL

 

[anuttarasammyak]

Say
$$\mathbf{r}=\mathbf{q_2}-\mathbf{q_1}$$
, the last equation of OP is
$$<r>=\int \frac{1}{Z}\mathbf{dq_1} \int e^{-\beta U(r)} d\mathbf{r}=\frac{V}{Z} \int_0^\infty e^{-\beta U(r)} 4\pi r^2 dr$$
where volume integrals are mentioned single integral symbol and the potential dumps almost zero at size of the volume V.

 

[Delta2]

I on the contrary agree with your first approach, that involves only the interparticle distance $r$ and i think its a nice and neat way to find the average interparticle distance since everything is given as functions of $r$. Why did you abandon this approach?

Your second approach instead resolves to the individual distances $r_1,r_2$ and is not what you think it is. I worked that "double" integral (which is not a volume integral btw) in cartesian coordinates and i got for the x-coordinate of the average distance the result
$$x=\int\int \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}e^{-\beta U(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2})}dx_1dx_2 $$

which is simply meaningless as the coordinates $y_i,z_i$ remain as variables in the expression for x. Thats why i said this approach is not what you think it is.

EDIT: I interpreted your "double" integral having in mind standard vector calculus notation. For example for the differentials $dq_1=dx_1\hat i+dy_1\hat j+dz_1\hat k$. Perhaps you meant to write $d^3q_1d^3q_2$ there for the differentials and then it indeed becomes a double volume integral (6-ple integral).

 

[vanhees71]

Say
$$\mathbf{r}=\mathbf{q_2}-\mathbf{q_1}$$
, the last equation of OP is
$$<r>=\int \frac{1}{Z}\mathbf{dq_1} \int e^{-\beta U(r)} d\mathbf{r}=\frac{V}{Z} \int_0^\infty e^{-\beta U(r)} 4\pi r^2 dr$$
where volume integrals are mentioned single integral symbol and the potential dumps almost zero at size of the volume V.

Here, somehow the factor $|\vec{q}_1-\vec{q}_2|$ seems to be missing.

I think the right thing is given in #1 already. With
$$\rho(\vec{q}_1,\vec{q}_2)=\frac{1}{Z} \exp[-V(|\vec{q}_1-\vec{q}_2|)], \quad Z=\int_V \mathrm{d}^3 q_1 \int_V \mathrm{d}^3 q_2 \exp[-V(|\vec{q}_1-\vec{q}_2|)].$$
First we introduce the center-of-mass position of two particles and their relative position vector
$$\vec{R}=\frac{1}{2} (\vec{q}_1+\vec{q}_2), \quad \vec{r}=\vec{q}_1-\vec{q}_2.$$
The Jacobian for the 6D integral is
$$\mathrm{d}^3 R \mathrm{d}^3 r=\mathrm{d}^3 q_1 \mathrm{d}^3 q_2.$$
From this you get (assuming that $V$ is a macroscopic volume, so that you can integrate over all $\vec{r} \in \mathbb{R}^3$)
$$Z= \int_V \mathrm{d}^3 R \int_{\mathbb{R}^3} \exp[-V(r)]= 4 \pi V \int_{0}^{\infty} \mathrm{d} r r^2 \exp[-V(r)].$$
For the average $r$ you need the integral
$$\int_V \mathrm{d}^3 R \int_{\mathbb{R}}^3 r \exp[-V(r)]=4 \pi V \int_0^{\infty} \mathrm{d} r r^3 \exp[-V(r)].$$
So finally
$$\langle r \rangle=\frac{\int_0^{\infty} \mathrm{d} r r^3 \exp[-V(r)]}{\int_0^{\infty} \mathrm{d} r r^2\exp[-V(r)]}.$$

 

[SchroedingersLion]

Here, somehow the factor $|\vec{q}_1-\vec{q}_2|$ seems to be missing.

I think the right thing is given in #1 already. With
$$\rho(\vec{q}_1,\vec{q}_2)=\frac{1}{Z} \exp[-V(|\vec{q}_1-\vec{q}_2|)], \quad Z=\int_V \mathrm{d}^3 q_1 \int_V \mathrm{d}^3 q_2 \exp[-V(|\vec{q}_1-\vec{q}_2|)].$$
First we introduce the center-of-mass position of two particles and their relative position vector
$$\vec{R}=\frac{1}{2} (\vec{q}_1+\vec{q}_2), \quad \vec{r}=\vec{q}_1-\vec{q}_2.$$
The Jacobian for the 6D integral is
$$\mathrm{d}^3 R \mathrm{d}^3 r=\mathrm{d}^3 q_1 \mathrm{d}^3 q_2.$$
From this you get (assuming that $V$ is a macroscopic volume, so that you can integrate over all $\vec{r} \in \mathbb{R}^3$)
$$Z= \int_V \mathrm{d}^3 R \int_{\mathbb{R}^3} \exp[-V(r)]= 4 \pi V \int_{0}^{\infty} \mathrm{d} r r^2 \exp[-V(r)].$$
For the average $r$ you need the integral
$$\int_V \mathrm{d}^3 R \int_{\mathbb{R}}^3 r \exp[-V(r)]=4 \pi V \int_0^{\infty} \mathrm{d} r r^3 \exp[-V(r)].$$
So finally
$$\langle r \rangle=\frac{\int_0^{\infty} \mathrm{d} r r^3 \exp[-V(r)]}{\int_0^{\infty} \mathrm{d} r r^2\exp[-V(r)]}.$$

Thank you for this outline. But it differs from my first approach which was:

$$
<r> = \frac{\int_{0}^{\infty} re^{-U(r)} \mathrm{d}r}{\int_{0}^{\infty} e^{-U(r)} \mathrm{d}r}
$$
This would only be the same to your result if one could cancel the extra $r$. Now, my simulation approaches a value that, within the limits of the statistics, resembles my analytical result.

Is it legit to treat the new variables $\mathbf{R}(\mathbf{q_1}, \mathbf{q_2})$ and $\mathbf{r}(\mathbf{q_1}, \mathbf{q_2})$ as independent from one another?

As I write this, I notice that I, in my simulation, treat the 1D case. In this case, your calculation reduces to my approach. So the formulas look different in 1, 2, or 3 dimensions. I find this to be a bit strange, as the potential is isotropic. How could you intuitively explain different average distances then?

 

[mastrofoffi]

As I write this, I notice that I, in my simulation, treat the 1D case. In this case, your calculation reduces to my approach. So the formulas look different in 1, 2, or 3 dimensions. I find this to be a bit strange, as the potential is isotropic. How could you intuitively explain different average distances then?

In D=1 you are averaging over a line, in D=2 over a circumference, in D=3 over a sphere. That should make some difference.
You are right that the symmetry of your LJ potential makes the problem 'effectively' 1-dimensional, but we might say that it is to be intended in a somewhat different sense from a really 1-dimensional problem.

Concretely, you can interpret the 1-dimensional mean as a mean over points grouped in bins of size d$r$; instead in 3 dimensions your "bins" will be spherical shells with surface area $4\pi r^2$ and thickness d$r$. That's how I am used to see it, but maybe someone comes up with a better picture.

Unless you are explicitly working with particles living in a one-dimensional space, vanhees71's answer is indeed the right one. Just make sure that you understand what does it actually mean to make these change of variables, they are not "just" abstract mathematical steps

 

[SchroedingersLion]

Well, I assume the substitutions are to decouple the two integrals. The potential $U(r)$ depends on both position variables. However, it does only depend on their relative position w.r.t. each other, not their absolute position. Therefore, splitting the dynamics into the movement of the center of mass (variable $\mathbf{R}$) and an orientation vector (variable $\mathbf{r}$) allows for a simpler integration.

To your comments to dimensionality: I understand that his formula is correct. I was just mentioning that, as I am treating the 1D case (i.e. particles moving along a line) in my simulation, his formula reduces to my formula - only that, originally, I thought my formula would also hold for higher dimensions.

Intuitively, I still don't see why I should get a different average distance in 1D or 2D. How would you explain the physical fact that you get different distances to a pupil with no idea of the maths?

 

[hutchphd]

I would discuss the difference in the result of a 1 dimensional and 3 dimensional random walk with equal probability on each axis. I think that becomes obvious with proper explanation.

 

[vanhees71]

Thank you for this outline. But it differs from my first approach which was:

$$
<r> = \frac{\int_{0}^{\infty} re^{-U(r)} \mathrm{d}r}{\int_{0}^{\infty} e^{-U(r)} \mathrm{d}r}
$$
This would only be the same to your result if one could cancel the extra $r$. Now, my simulation approaches a value that, within the limits of the statistics, resembles my analytical result.

Is it legit to treat the new variables $\mathbf{R}(\mathbf{q_1}, \mathbf{q_2})$ and $\mathbf{r}(\mathbf{q_1}, \mathbf{q_2})$ as independent from one another?

As I write this, I notice that I, in my simulation, treat the 1D case. In this case, your calculation reduces to my approach. So the formulas look different in 1, 2, or 3 dimensions. I find this to be a bit strange, as the potential is isotropic. How could you intuitively explain different average distances then?

You have to use the Jacobian between spherical coordinates and Cartesian coordinates. For 1D there's no transformation, so there $J=1$. For 2D you have polar coordinates in the plane. There $J=r$. For 3D you have $J=r^2 \sin \vartheta$.

 

[HomogenousCow]

The integral doesn't converge without a volume regularisation by the way, that's always the case with potentials that approach a constant at infinity.