Categories: Physics.

# Reduced mass

Problems with two interacting objects can be simplified by combining them into a pseudo-object with reduced mass $\mu$, whose position equals the relative position of the objects. For bodies 1 and 2 with respective masses $m_1$ and $m_2$:

\begin{aligned} \boxed{ \mu \equiv \frac{m_1 m_2}{m_1 + m_2} } \end{aligned}

If $\va{x}_1$ and $\va{x}_2$ are the objects’ respective positions, then we define the relative position $\va{x}_r$, the relative velocity $\va{v}_r$, and the relative acceleration $\va{a}_r$:

\begin{aligned} \va{x}_r \equiv \va{x}_1 - \va{x}_2 \qquad \va{v}_r \equiv \va{v}_1 - \va{v}_2 = \dv{\va{x}_r}{t} \qquad \quad \va{a}_r \equiv \va{a}_1 - \va{a}_2 = \dvn{2}{\va{x}_r}{t} \end{aligned}

We now choose the coordinate system’s origin to be the center of mass of both objects:

\begin{aligned} m_1 \va{x}_1 + m_2 \va{x}_2 = 0 \end{aligned}

Rearranging and differentiating then yields the following useful equations:

\begin{aligned} \va{x}_2 = - \frac{m_1}{m_2} \va{x}_1 \qquad \quad \va{v}_2 = - \frac{m_1}{m_2} \va{v}_1 \qquad \quad \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1 \end{aligned}

Using these relations, we can rewrite the relative quantities we defined earlier:

\begin{aligned} \va{x}_r = \Big( 1 + \frac{m_1}{m_2} \Big) \va{x}_1 = \frac{m_1 + m_2}{m_2} \va{x}_1 \qquad \va{v}_r = \frac{m_1 + m_2}{m_2} \va{v}_1 \qquad \va{a}_r = \frac{m_1 + m_2}{m_2} \va{a}_1 \end{aligned}

Meanwhile, Newton’s third law states that if object 1 experiences a force $\va{F}_1 = m_1 \va{a}_1$ caused by object 2, then object 2 experiences an opposite and equal force $\va{F}_2 = - \va{F}_1$. In fact, our earlier relation between $\va{a}_1$ and $\va{a}_1$ boils down to Newton’s third law:

\begin{aligned} \va{F}_2 = m_2 \va{a}_2 = - m_1 \va{a}_1 = - \va{F}_1 \quad \implies \quad \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1 \end{aligned}

With all that in mind, let us take a closer look at the relative acceleration $\va{a}_r$:

\begin{aligned} \va{a}_r = \frac{m_1 + m_2}{m_2} \Big( \frac{m_1}{m_1} \Big) \va{a}_1 = \frac{m_1 + m_2}{m_1 m_2} \big( m_1 \va{a}_1 \big) = \frac{\va{F}_1}{\mu} = - \frac{\va{F}_2}{\mu} \end{aligned}

Where $\mu$ is the reduced mass, as defined above. In other words, the relative acceleration $\va{a}_r$ is just $\va{a}_1 = \va{F}_1 / m_1$ multiplied by $m_1 / \mu$. This can be regarded as focusing on the dynamics of body 1, while correcting for the effects of body 2.

This also suggests the following way to recover the original positions $\va{x}_1$ and $\va{x}_2$ from $\va{x}_r$, which you can easily verify for yourself:

\begin{aligned} \va{x}_1 = \frac{\mu}{m_1} \va{x}_r = \frac{m_2}{m_1 + m_2} \va{x}_r \qquad \quad \va{x}_2 = \frac{\mu}{m_2} \va{x}_r = - \frac{m_1}{m_1 + m_2} \va{x}_r \end{aligned}

With this, we can rewrite the total kinetic energy $T$ in an elegant way:

\begin{aligned} T &= \frac{1}{2} m_1 \va{v}_1^2 + \frac{1}{2} m_2 \va{v}_2^2 = \frac{1}{2} m_1 \Big( \frac{\mu}{m_1} \va{v}_r \Big)^2 + \frac{1}{2} m_2 \Big( \frac{\mu}{m_2} \va{v}_r \Big)^2 \\ &= \frac{1}{2} \frac{\mu^2}{m_1} \va{v}_r^2 + \frac{1}{2} \frac{\mu^2}{m_2} \va{v}_r^2 = \frac{1}{2} \Big( \frac{m_2 \mu^2}{m_1 m_2} + \frac{m_1 \mu^2}{m_1 m_2} \Big) \va{v}_r^2 \\ &= \frac{1}{2} \frac{(m_1 + m_2) \mu^2}{m_1 m_2} \va{v}_r^2 = \frac{1}{2} \frac{\mu^2}{\mu} \va{v}_r^2 = \frac{1}{2} \mu \va{v}_r^2 \end{aligned}

Then, assuming that the system’s potential energy $V$ only depends on the distance between the two objects, i.e. $V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$, we just showed that we can rewrite both $T$ and $V$ to contain only $\mu$ and relative quantities. This is relevant for both Lagrangian mechanics and Hamiltonian mechanics, where $L = T - V$ and $H = T + V$ respectively.