Reduced Mass

Section 21.4 Reduced Mass

So far, we have found one decoupled equation to replace (21.2.1). What about the other \(n-1\) equations? It turns out that, in general, there is no way to decouple and solve the other equations. Physicists often say, “The \(n\)-body problem can not be solved in general.” Whenever you are stuck trying to solve a general problem, it often pays to start with simpler examples to build up your intuition. We will make two simplifying assumptions.

Assume that there are no external forces acting.
Assume that there are only two masses.

The system of equations (21.2.1) reduces to

\begin{align} m_1 \frac{d^2\rr_1}{dt^2} \amp= -\ff_{21}\notag\\ m_2 \frac{d^2\rr_2}{dt^2} \amp= +\ff_{21}\tag{21.4.1} \end{align}

Because we added the two equations of motion to find the equation of motion for the center-of-mass, we are led now to consider subtracting the equations so as to get \(\rr=\rr_2-\rr_1\text{.}\) We see that before we subtract, we should multiply the first equation in (21.4.1) by \(m_2\) and the second equation by \(m_1\) so that the factors in front of the second derivative are the same. Subtracting the first equation from the second and regrouping, we obtain:

\begin{align} m_1 m_2 \frac{d^2}{dt^2}(\rr_2-\rr_1) \amp = m_1 m_2 \frac{d^2}{dt^2}(\rr)\tag{21.4.2}\\ \amp = (m_1+m_2) \ff_{21}\tag{21.4.3} \end{align}

or rearranging:

\begin{equation} \frac{m_1m_2}{m_1+m_2} \frac{d^2\rr}{dt^2} = \ff_{21}\tag{21.4.4} \end{equation}

Definition 21.6. Reduced Mass.

The combination of masses

\begin{equation} \mu = \frac{m_1m_2}{m_1+m_2} \tag{21.4.5} \end{equation}

is called the reduced mass.

Plugging in \(\mu\) for the reduced mass, (21.4.4) becomes

\begin{equation} \mu \frac{d^2\rr}{dt^2} = \ff_{21}\text{.}\tag{21.4.6} \end{equation}

This equation has the same form as Newton’s law for a single fictitious mass \(\mu\text{,}\) with position vector \(\rr\text{,}\) moving subject to the force \(\ff_{21}\text{.}\) For the rest of this chapter, when we talk about “the mass”, we mean this fictitious particle subject to equation (21.4.6).

Note that to solve the original two mass problem we started with, we will need to transform the solutions for \(\rr\) back to \(\rr_1\) and \(\rr_2\text{.}\)

Figure 21.7. The position vectors \(\rr_1\) and \(\rr_2\) for masses \(m_1\) and \(m_2\) and the displacement vector \(\rr=\rr_2-\rr_1\) between them.

You can use Figure 21.7 to help you undo the change of coordinates, writing \(\rr_1\) and \(\rr_2\) in terms of \(\RRv\) and \(\rr\text{.}\)

Activity 21.1. Undo Formulas for Center of Mass (Algebra).

Find the positions of the two masses in terms of the position of the center of mass and the relative position, i.e. solve for:

\begin{align} \vec{r}_1\amp =\notag\\ \vec{r}_2\amp =\notag \end{align}

Hint.

The system of equations (21.3.3) and (21.2.2) is linear, i.e. each variable is to the first power, even though the variables are vectors. In this case, you can use all of the methods you learned for solving systems of equations while keeping the variables vector valued, i.e. you can safely ignore the fact that the \(\vec{r}\)s are vectors while you are doing the algebra as long as you don’t divide by a vector.

Activity 21.2. Undo Formulas for Center of Mass (Geometry).

The figure below shows the position vector \(\vec r\) and the orbit of a “fictitious” reduced mass \(\mu\text{.}\)

Figure 21.8. The position vectors for \(\mu\text{.}\)

Suppose \(m_1=m_2\text{,}\) sketch the position vectors and orbits for \(m_1\) and \(m_2\) corresponding to \(\rr\text{.}\) Describe a common physics example of central force motion for which \(m_1=m_2\text{.}\)
Repeat for \(m_2 > m_1\text{,}\) where the mass difference is neither very large nor very small.

Hint.

Think about what you expect in the limiting case \(m_2 >> m_1\text{.}\)

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