A.5 The reduced mass

Two-body systems, like the earth-moon system of celestial mechanics or the proton-electron hydrogen atom of quantum mechanics, can be analyzed more simply using reduced mass. In this note both a classical and a quantum derivation will be given. The quantum derivation will need to anticipate some results on multi-particle systems from chapter 5.1.

In two-body systems the two bodies move around their combined center of gravity. However, in examples such as the ones mentioned, one body is much more massive than the other. In that case the center of gravity almost coincides with the heavy body, (earth or proton). Therefore, in a naive first approximation it may be assumed that the heavy body is at rest and that the lighter one moves around it. It turns out that this naive approximation can be made exact by replacing the mass of the lighter body by an reduced mass. That simplifies the mathematics greatly by reducing the two-body problem to that of a single one. Also, it now produces the exact answer regardless of the ratio of masses involved.

The classical derivation is first. Let $m_1$ and ${\skew0\vec r}_1$ be the mass and position of the massive body (earth or proton), and $m_2$ and ${\skew0\vec r}_2$ those of the lighter one (moon or electron). Classically the force $\vec{F}$ between the masses will be a function of the difference ${\skew0\vec r}_{21}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2-{\skew0\vec r}_1$ in their positions. In the naive approach the heavy mass is assumed to be at rest at the origin. Then ${\skew0\vec r}_{21}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$, and so the naive equation of motion for the lighter mass is, according to Newton’s second law,

\begin{displaymath}
m_2 \ddot {\skew0\vec r}_{21} = \vec F({\skew0\vec r}_{21})
\end{displaymath}

Now consider the true motion. The center of gravity is defined as a mass-weighted average of the positions of the two masses:

\begin{displaymath}
{\skew0\vec r}_{\rm cg} = w_1 {\skew0\vec r}_1 + w_2 {\ske...
...qquad
w_1=\frac{m_1}{m_1+m_2} \quad w_2=\frac{m_2}{m_1+m_2}
\end{displaymath}

It is shown in basic physics that the net external force on the system equals the total mass times the acceleration of the center of gravity. Since in this case it will be assumed that there are no external forces, the center of gravity moves at a constant velocity. Therefore, the center of gravity can be taken as the origin of an inertial coordinate system. In that coordinate system, the positions of the two masses are given by

\begin{displaymath}
{\skew0\vec r}_1 = - w_2 {\skew0\vec r}_{21} \qquad {\skew0\vec r}_2 = w_1 {\skew0\vec r}_{21}
\end{displaymath}

because the position $w_1{\skew0\vec r}_1+w_2{\skew0\vec r}_2$ of the center of gravity must be zero in this system, and the difference ${\skew0\vec r}_2-{\skew0\vec r}_1$ must be ${\skew0\vec r}_{21}$. (Note that the sum of the two weight factors is one.) Solve these two equations for ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ and you get the result above.

The true equation of motion for the lighter body is $m_2\ddot{\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vec{F}({\skew0\vec r}_{21})$, or plugging in the above expression for ${\skew0\vec r}_2$ in the center of gravity system,

\begin{displaymath}
m_2 w_1 \ddot{\skew0\vec r}_{21} = \vec F({\skew0\vec r}_{21})
\end{displaymath}

That is exactly the naive equation of motion if you replace $m_2$ in it by the reduced mass $m_2w_1$, i.e. by
\begin{displaymath}
\fbox{$\displaystyle
m_{\rm red} = \frac{m_1m_2}{m_1+m_2}
$} %
\end{displaymath} (A.16)

The reduced mass is almost the same as the lighter mass if the difference between the masses is large, like it is in the cited examples, because then $m_2$ can be ignored compared to $m_1$ in the denominator.

The bottom line is that the motion of the two-body system consists of the motion of its center of gravity plus motion around its center of gravity. The motion around the center of gravity can be described in terms of a single reduced mass moving around a fixed center.

The next question is if this reduced mass idea is still valid in quantum mechanics. Quantum mechanics is in terms of a wave function $\psi$ that for a two-particle system is a function of both ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$. Also, quantum mechanics uses the potential $V({\skew0\vec r}_{21})$ instead of the force. The Hamiltonian eigenvalue problem for the two particles is:

\begin{displaymath}
H \psi = E \psi
\qquad
H = -\frac{\hbar^2}{2m_1}\nabla_1^2 - \frac{\hbar^2}{2m_2}\nabla_2^2
+ V({\skew0\vec r}_{21})
\end{displaymath}

where the two kinetic energy Laplacians in the Hamiltonian $H$ are with respect to the position coordinates of the two particles:

\begin{displaymath}
\nabla_1^2 \psi \equiv \sum_{j=1}^3 \frac{\partial^2\psi}{...
...\equiv \sum_{j=1}^3 \frac{\partial^2\psi}{\partial r_{2,j}^2}
\end{displaymath}

Now make a change of variables from ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ to ${\skew0\vec r}_{\rm {cg}}$ and ${\skew0\vec r}_{21}$ where

\begin{displaymath}
{\skew0\vec r}_{\rm cg} = w_1 {\skew0\vec r}_1 + w_2 {\ske...
... \qquad {\skew0\vec r}_{21}={\skew0\vec r}_2-{\skew0\vec r}_1
\end{displaymath}

The derivatives of $\psi$ can be converted using the chain rule of differentiation:

\begin{displaymath}
\frac{\partial\psi}{\partial r_{1,j}} =
\frac{\partial\p...
..._{{\rm cg},j}} w_1
- \frac{\partial\psi}{\partial r_{21,j}}
\end{displaymath}

or differentiating once more and summing

\begin{displaymath}
\nabla_1^2 \psi
= \sum_{j=1}^3 \frac{\partial^2\psi}{\pa...
...}
+ \sum_{j=1}^3 \frac{\partial^2\psi}{\partial r_{21,j}^2}
\end{displaymath}

and a similar expression for $\nabla_2^2\psi$, but with $w_2$ instead of $w_1$ and a plus sign instead of the minus sign. Combining them together in the Hamiltonian, and substituting for $w_1$ and $w_2$, the mixed derivatives drop out against each other and what is left is

\begin{displaymath}
H = - \frac{\hbar^2}{2(m_1+m_2)} \nabla^2_{\rm cg}
- \fr...
...hbar^2}{2 m_{\rm red}} \nabla^2_{21} + V({\skew0\vec r}_{21})
\end{displaymath}

The first term is the kinetic energy that the total mass would have if it was at the center of gravity; the next two terms are kinetic and potential energy around the center of gravity, in terms of the distance between the masses and the reduced mass.

The Hamiltonian eigenvalue problem $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ has separation of variables solutions of the form

\begin{displaymath}
\psi=\psi_{\rm cg}({\skew0\vec r}_{\rm cg})\psi_{21}({\skew0\vec r}_{21})
\end{displaymath}

Substituting this and the Hamiltonian above into $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ and dividing by $\psi_{\rm {cg}}\psi_{21}$ produces

\begin{displaymath}
- \frac{\hbar^2}{2(m_1+m_2)} \frac{1}{\psi_{\rm cg}}
\na...
...bar^2}{2 m_{\rm red}} \nabla^2_{21} + V\right]\psi_{21}
= E
\end{displaymath}

Call the first term in the left hand side $E_{\rm {cg}}$ and the second $E_{21}$. By that definition, $E_{\rm {cg}}$ would normally be a function of ${\skew0\vec r}_{\rm {cg}}$, because $\psi_{\rm {cg}}$ is, but since it is equal to $E-E_{21}$ and those do not depend on ${\skew0\vec r}_{\rm {cg}}$, $E_{\rm {cg}}$ cannot either, and must be a constant. By similar reasoning, $E_{21}$ cannot depend on ${\skew0\vec r}_{21}$ and must be a constant too. Therefore, rewriting the definitions of $E_{\rm {cg}}$ and $E_{21}$, two separate eigenvalue problems are obtained:

\begin{displaymath}
- \frac{\hbar^2}{2(m_1+m_2)}\nabla^2_{\rm cg}\psi_{\rm cg}...
... red}} \nabla^2_{21} + V\right]\psi_{21}
= E_{21} \psi_{21}
\end{displaymath}

The first describes the quantum mechanics of an imaginary total mass $m_1{+}m_2$ located at the center of gravity. The second describes an imaginary reduced mass $m_{\rm {red}}$ at a location ${\skew0\vec r}_{21}$ away from a fixed center that experiences a potential $V({\skew0\vec r}_{21})$.

For the hydrogen atom, it means that if the problem with a stationary proton is solved using an reduced electron mass $m_{\rm p}m_{\rm e}$$\raisebox{.5pt}{$/$}$$(m_{\rm p}+m_{\rm e})$, it solves the true problem in which the proton moves a bit too. Like in the classical analysis, the quantum analysis shows that in addition the atom can move as a unit, with a motion described in terms of its center of gravity.

It can also be concluded, from a slight generalization of the quantum analysis, that a constant external gravity field, like that of the sun on the earth-moon system, or of the earth on a hydrogen atom, causes the center of gravity to accelerate correspondingly, but does not affect the motion around the center of gravity at all. That reflects a basic tenet of general relativity.