# Stat mech: probability density - 2 interacting particles

1. May 28, 2015

### sltungle

1. The problem statement, all
variables and given/known data

Show that the probability density as a function of seperation, r, of two atoms interacting via a potential U(r) (e.g. a function of seperation only, such as a Coulombic interaction), is given by

$$\rho(r) = Cr^2e^{-\beta U(r)}$$

where C is some constant. For simplicity consider just 2 atoms in 3D and discard all other atoms.

Hint: Start by writing down the probability density of r in terms of configurational integrals for the two atoms,

$$\rho(r) = \dfrac{\int dq_{1}^3\int dq_{2}^3\delta(|{\bf{q_2}}-{\bf{q_2}}| - r)e^{-\beta U({\bf{q_2}},{\bf{q_2}})}}{Z_2}$$

and change coordinate systems so that one atom is at the origin and the second is at some relative position vector, then convert to spherical coordinates.

2. Relevant equations

3. The attempt at a solution

Setting particle 1 to be at the origin the distance given by $$|{\bf{q_2}}-{\bf{q_2}}|$$ is just given by the position of particle two, which I'm just calling $${\bf{q^{\prime}_{2}}}$$ for the time being, so:

$$\delta({|\bf{q^{\prime}_{2}}}| - r)$$

Which if I'm not mistaken should look something similar to (I feel that r was a poor choice of notation to use prior to having us swap to spherical polar, but whatever),

$$\delta(x_2 - r_x)\delta(y_2 - r_y)\delta(z_2 - r_z)$$

The potential energy now also only becomes a function of the separation of the two particles, that is to say:

$$e^{-\beta U({\bf{q_2}},{\bf{q_2}})} = e^{-\beta U({|\bf{q^{\prime}_{2}}}|)} = e^{-\beta U({q^{\prime}_{2}})}$$

Converting the Dirac Delta function to spherical polar coordinates, I'm led to believe (my physics degree was shockingly sparse on mathematics courses so I'm somewhat ignorant on a lot of the formalism behind these kinds of things, I apologise; I'm slowly but surely trying to make amends for that), requires one to divide by the Jacobian, that is the above becomes:

$$\dfrac{1}{|J|}\delta(r_2 - r)\delta(\theta_2 - \theta)\delta(\phi_2 - \phi)$$

Or

$$\dfrac{1}{r_2^2sin(\theta_2)}\delta(r_2 - r)\delta(\theta_2 - \theta)\delta(\phi_2 - \phi)$$

Where now 'r' on its own is literally just the radial separation of the two particles. Converting the infinitesimal volume element to spherical polar also picks me up a Jacobian term:

$$dq_{1}^3 = dx_1dy_1dz_1 = |J|dr_1d\theta_1 d\phi_1 = r_1^2sin(\theta_1)dr_1d\theta_1 d\phi_1$$

(And the same for particle 2 with a 2 in place of the subscript 1 of course)

However when I substitute this all back into the original equation I don't quite see where I'm going to pick up my r^2 term from.

$$\rho(r) = \dfrac{1}{Z_2}\int r_1^2sin(\theta_1)dr_1d\theta_1 d\phi_1 \int dr_2d\theta_2 d\phi_2 \delta(r_2 - r)\delta(\theta_2 - \theta)\delta(\phi_2 - \phi)e^{-\beta U(r_2)}$$

Assuming I've done everything right (which I've probably not or I'd have the right answer unless something in the following line of reasoning is incorrect) the only remaining term of interest in the second integral would be the exponential term in which the distance 'r' would be picked out from the delta function. $$e^{-\beta U(r)}$$ The first integral would yield $$4\pi$$ over the angular variables, however I'm still missing an r^2 term. Even if I could integrate over $$dr_1$$ (which I can't because the radius can be infinitely large) I'd still pick up an extra power in r that I don't want. The only way I can see this working is if the delta function didn't pick up a Jacobian term because then I'd still have an $$r{_2}^2$$ term in the second integral which when evaluated with the delta function would come out to be r^2 (although if the Jacobians didn't cancel out then I'd also have a surviving angular term which doesn't particularly make much sense as the potential cares not for the angle).

It's late, and I'm unsure where to proceed from here, so I'm probably going to give in for the night. I'm not expecting the answer, but some guidance would be much appreciated (my professor is away sick at the moment, so the internet will have to suffice for now!).

2. May 28, 2015

### Orodruin

Staff Emeritus
You are mistaken here. This is a one dimensional delta function, not a three dimensional one.

3. May 28, 2015

### sltungle

Is this because the argument of the delta function is the magnitude of separation which is the same regardless of coordinate system? That was my first attempt to solve the problem about 3 day ago but one of the other professors told me he was concerned that I hadn't treated my change of coordinate systems properly in terms of the delta function.

4. May 28, 2015

### Orodruin

Staff Emeritus
It is a one-dimensional delta because you are only interested in integrating over it in the region where the distance between the particles is r. Taking away the CoM motion, this is a sphere and so you need to integrate over the sphere's surface, which is two-dimensional - thereby reducing the three dimensional integral to a two-dimensional one.

A more straight forward way is to simply not introduce the delta and see what the distribution function becomes when you leave only the r integral.

5. May 28, 2015

### sltungle

Ah, of course! Thank you very much for the help :)