2.4. Reduced mass

Here, we derive the reduced mass for a system of two point-like masses connected by a force that obeys Hooke’s law.

We start out with the two point-like masses \(m_1\) and \(m_2\) at positions \(x_1\) and \(x_2\), assuming \(x_2>x_1\). The bond length is the difference \(r = x_2 - x_1\). The equilibrium bond length is \(r_\mathrm{e}\).

If the two masses are connected by a spring that follows Hooke’s law with force constant \(k\), the forces acting on the two are

(2.74)\[ \begin{align}\begin{aligned}F_1 = m_1\frac{\mathrm{d}^2}{\mathrm{d}t^2}x_1 = +k (r-r_\mathrm{e})\\F_2 = m_2\frac{\mathrm{d}^2}{\mathrm{d}t^2}x_2 = -k (r-r_\mathrm{e})\end{aligned}\end{align} \]

Dividing the first equation by \(m_1\) and the second equation by \(m_2\), and then subtracting the first from the second equation gives

(2.75)\[\frac{\mathrm{d}^2}{\mathrm{d}t^2}(x_2-x_1) = \frac{\mathrm{d}^2}{\mathrm{d}t^2}r = -k\left(\frac{1}{m_1}+\frac{1}{m_2}\right)(r-r_\mathrm{e})\]

The expression in parentheses can be abbreviated by defining the reduced mass \(\mu\)

(2.76)\[\frac{1}{\mu} = \frac{1}{m_1}+\frac{1}{m_2} \qquad\qquad \mu = \frac{m_1m_2}{m_1+m_2}\]

This gives finally

(2.77)\[\mu\frac{\mathrm{d}^2}{\mathrm{d}t^2}r = -k (r-r_\mathrm{e})\]

This equation is Hooke’s law for a single effective particle of mass \(\mu\) on a spring with force constant \(k\) and equilibrium length \(r_\mathrm{e}\).

This shows that the two spring-connected masses can be modeled as a single mass connected to a spring of the same force constant.