## Tuesday, August 6, 2019

### What is the mass of a molecular vibration?

This is a basic question that I have been puzzling about. I welcome solutions.

Consider a diatomic molecule containing atoms with mass m1 and m2. It has a stretch vibration that can be described by a harmonic oscillator with a reduced mass mu given by
Now consider a polyatomic molecule containing N atoms.
It will have 3N-6 normal modes of vibration.
[The 6 is due to the fact that there are 6 zero-frequency modes: 3 rigid translations and 3 rotations of the whole molecule].
In the harmonic limit, the normal mode problem is solved below.
[I follow the classic text Wilson et al., Molecular Vibrations].
The problem is also solved in matrix form in Chapter 6 of Goldstein, Classical Mechanics].

One now has a collection of non-interacting harmonic oscillators. All have mass = 1. This is because the normal mode co-ordinates have units of length * sqrt(mass).

The quantum chemistry package Gaussian does more. It calculates a reduced mass mu_i for each normal mode i using the formula below.
This is discussed in these notes on the Gaussian web site. From mu_i and the normal mode frequency_i it then calculates the spring constant for each normal mode.

I have searched endlessly, and tried myself, but have not been able to answer the following basic questions:

1. How do you derive this expression for the reduced mass?
2. Is this reduced mass physical, i.e. a measurable quantity?

Similar issues must also arise with phonons in crystals.

Any recommendations?

1. Ehm... It's elementary physics. If you have the kinetic energy of two bodys that have masses m1 & M2, and velocities v1 & V2, T=½m1•v1²+½m2•v2². The centre of mass has velocity (m1•v1+m2•v2)/(m1+m2). If you want a mass asociated with v1-v2, preserving the same kinetic energy, where the other mass is m1+M2 with velocity of centre of mass, the mass you get is the reduced mass. Ir works the same for non-relativistic Quantum mechanics (when you change vi for -i\frac{\hbar\nabla}{mi}).

2. Once you have a total speed V, you can define a reduced mass μ so that the kinetic energy is T = ½μV². I think the crux is how to add up the particle speeds.

In the two particle case, say that one particle has mass m and is going left with speed v₁=Mu, and the other has mass M and is going right with speed v₂=mu. (I suggest readers get a pen and draw that.) The centre of mass is stationary.

Then T = ½m·(Mu)² + ½M·(mu)² = ½ mM/(M+m)·(mu+Mu)² = ½μ(v₁+v₂)². In the simplest case, you just add up the particle speeds.

So this might be a start. Pick a time when the kinetic energy of the classical vibrational mode is a maximum. The particles have masses m_i and speeds v_i. Choose μ such that ∑_i m_i·v_i² = μ·(∑_i v_i)².

3. For a phonon with dispersion relation ω(k), there's a kind of group dispersion. Define the mass μ of the mode with wave number k such that ω(k+k') = ω(k) + v_g·k' + (ℏ/2μ)·k'² + O(k'³). I don't know how that relates to the classical version.