# Stat mech: probability density - 2 interacting particles

• sltungle
In summary, the conversation discusses deriving the probability density of two atoms interacting via a potential U(r) in terms of separation, r. The derived formula is given by ρ(r) = Cr^2e^-βU(r), where C is a constant. The conversation also discusses converting to spherical coordinates and integrating over the two atoms, with the potential energy now only being a function of the separation of the particles. The mistake of using a three-dimensional delta function is also addressed and the correct approach is suggested.
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 separation 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.

## The Attempt at a Solution

[/B]
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!).

sltungle said:
Which if I'm not mistaken should look something similar to
You are mistaken here. This is a one dimensional delta function, not a three dimensional one.

Orodruin said:
You are mistaken here. This is a one dimensional delta function, not a three dimensional one.
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.

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.

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

## 1. What is statistical mechanics?

Statistical mechanics is a branch of physics that uses statistical methods to explain the behavior of a large number of particles in a system. It helps us understand the macroscopic properties of a system by studying the microscopic behavior of its individual particles.

## 2. What is the probability density function in statistical mechanics?

The probability density function in statistical mechanics is a mathematical function that describes the probability of finding a particle at a given position and momentum in a system. It is used to calculate the likelihood of a particular configuration of particles in a system.

## 3. How do you calculate the probability density function for two interacting particles?

The probability density function for two interacting particles can be calculated using the partition function, which takes into account the energy levels and interactions between the particles. By integrating the partition function with respect to position and momentum, we can obtain the probability density function for the system.

## 4. How does statistical mechanics explain the behavior of interacting particles?

Statistical mechanics uses the laws of thermodynamics and probability to explain the behavior of interacting particles in a system. It considers factors such as temperature, energy, and entropy to describe the overall behavior of the particles and how they interact with each other.

## 5. What are the limitations of statistical mechanics in studying interacting particles?

One limitation of statistical mechanics is that it assumes that all particles in a system are identical and interchangeable, which may not always be the case in real systems. It also does not take into account quantum effects, which can have a significant impact on the behavior of particles at small scales.

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