The Two-Body Problem and Reduced Mass

Derivation

The Two-Body Problem and Reduced Mass

D-028 Home PU-101 Threads force · matter Depends on The Center-of-Mass Theorem, Momentum Conservation from Newton's Third Law

Statement

For two point masses \(m_1\) and \(m_2\) interacting through a potential \(V\) that depends only on their separation \(\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2\), the change of variables to the center-of-mass position \(\mathbf{R}\) and the relative position \(\mathbf{r}\) decouples the dynamics exactly: the center of mass moves as a free particle of mass \(M=m_1+m_2\), while the relative coordinate obeys the equation of motion of a single fictitious particle of reduced mass \(\mu = \dfrac{m_1 m_2}{m_1+m_2}\) moving in the fixed external potential \(V(\mathbf{r})\).

Why it matters

The two-body problem is the first genuinely interacting system in mechanics, and naively it has six coupled degrees of freedom. Reduced mass is the trick that collapses it to a one-body problem we already know how to solve — Kepler orbits, the hydrogen atom, diatomic vibrations, and Rutherford scattering are all the same reduced-mass equation with different \(V\).

It also fixes a subtle error that appears everywhere in practice: treating the Sun, the proton, or a heavy nucleus as infinitely massive. Replacing \(m\) by \(\mu\) is the exact correction, and it shifts spectral lines, orbital periods, and vibrational frequencies by measurable amounts (the hydrogen/deuterium isotope shift is precisely this effect).

Assumptions

The two bodies are point masses (or rigid spheres with spherically symmetric mass).If either body has internal structure or a non-spherical mass distribution, tidal and multipole couplings enter and the internal coordinates no longer separate from \(\mathbf{r}\).

The interaction depends only on the relative separation, \(V=V(\mathbf{r}_1-\mathbf{r}_2)\), with equal and opposite forces (Newton's third law).If \(V\) depends on \(\mathbf{r}_1\) and \(\mathbf{r}_2\) separately (e.g. an external field pinning one body), the center of mass is no longer free and the two motions stay coupled.

The masses are constant and the dynamics are non-relativistic.If mass varies (rockets) or speeds approach \(c\), kinetic energy is no longer \(\tfrac12 m v^2\), momentum is not \(m\mathbf{v}\), and the clean split into \(M\) and \(\mu\) fails.

An inertial frame is used.In a rotating or accelerating frame, fictitious forces act on each body in proportion to its own mass and do not cancel in the relative equation, spoiling the reduction.

Derivation

\[ M\mathbf{R} = m_1\mathbf{r}_1 + m_2\mathbf{r}_2, \qquad \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2, \qquad M = m_1+m_2 \]

Define the two new coordinates. This is an invertible linear change of variables (Jacobian nonzero), so no information is lost A

\[ \mathbf{r}_1 = \mathbf{R} + \frac{m_2}{M}\,\mathbf{r}, \qquad \mathbf{r}_2 = \mathbf{R} - \frac{m_1}{M}\,\mathbf{r} \]

Invert the linear map: solve the two relations of Step 1 simultaneously for \(\mathbf{r}_1,\mathbf{r}_2\) A

\[ m_1\ddot{\mathbf{r}}_1 = \mathbf{F}_{12} = -\nabla_{\mathbf{r}_1} V, \qquad m_2\ddot{\mathbf{r}}_2 = \mathbf{F}_{21} = -\nabla_{\mathbf{r}_2} V \]

Write Newton's second law for each body; \(\mathbf{F}_{12}\) is the force on body 1 from body 2 A

\[ \nabla_{\mathbf{r}_1}V = \nabla_{\mathbf{r}}V, \qquad \nabla_{\mathbf{r}_2}V = -\nabla_{\mathbf{r}}V \]

Since \(V\) depends only on \(\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2\), the chain rule gives \(\partial\mathbf{r}/\partial\mathbf{r}_1=+1\) and \(\partial\mathbf{r}/\partial\mathbf{r}_2=-1\); this encodes \(\mathbf{F}_{21}=-\mathbf{F}_{12}\) B

\[ m_1\ddot{\mathbf{r}}_1 + m_2\ddot{\mathbf{r}}_2 = -\nabla_{\mathbf{r}}V + \nabla_{\mathbf{r}}V = \mathbf{0} \;\Longrightarrow\; M\ddot{\mathbf{R}} = \mathbf{0} \]

Add the two equations of Step 3 and use Step 4; the left side is \(\frac{d}{dt}(m_1\dot{\mathbf{r}}_1+m_2\dot{\mathbf{r}}_2)=M\ddot{\mathbf{R}}\). This is the center-of-mass theorem B

\[ \ddot{\mathbf{r}} = \ddot{\mathbf{r}}_1 - \ddot{\mathbf{r}}_2 = \frac{\mathbf{F}_{12}}{m_1} - \frac{\mathbf{F}_{21}}{m_2} \]

Subtract the two equations of Step 3 after dividing each by its mass; this isolates the relative acceleration A

\[ \ddot{\mathbf{r}} = \frac{\mathbf{F}_{12}}{m_1} - \frac{-\mathbf{F}_{12}}{m_2} = \left(\frac{1}{m_1}+\frac{1}{m_2}\right)\mathbf{F}_{12} \]

Substitute \(\mathbf{F}_{21}=-\mathbf{F}_{12}\) (Step 4) and factor out the common force B

\[ \frac{1}{\mu} \equiv \frac{1}{m_1}+\frac{1}{m_2} = \frac{m_1+m_2}{m_1 m_2} \;\Longrightarrow\; \mu = \frac{m_1 m_2}{m_1+m_2} \]

Define \(\mu\) as the reciprocal-sum combination that appeared naturally; this is a definition, chosen to make the next line read like Newton's law A

\[ \mu\,\ddot{\mathbf{r}} = \mathbf{F}_{12} = -\nabla_{\mathbf{r}} V(\mathbf{r}) \]

Multiply Step 7 through by \(\mu\) and use Step 4. This is Newton's law for a single particle of mass \(\mu\) at position \(\mathbf{r}\) in the fixed potential \(V\) B

\[ E = \tfrac12 M\dot{\mathbf{R}}^2 + \tfrac12 \mu\,\dot{\mathbf{r}}^2 + V(\mathbf{r}), \qquad \mathbf{L} = M\,\mathbf{R}\times\dot{\mathbf{R}} + \mu\,\mathbf{r}\times\dot{\mathbf{r}} \]

Substitute Step 2 into \(T=\tfrac12 m_1\dot{\mathbf{r}}_1^2+\tfrac12 m_2\dot{\mathbf{r}}_2^2\); the cross term \(\propto\left(m_1\tfrac{m_2}{M}-m_2\tfrac{m_1}{M}\right)=0\) cancels, so energy and angular momentum split cleanly into COM and relative parts C

Result

\[ M\ddot{\mathbf{R}} = \mathbf{0}, \qquad \mu\,\ddot{\mathbf{r}} = -\nabla_{\mathbf{r}}V(\mathbf{r}), \qquad \mu=\frac{m_1 m_2}{m_1+m_2} \]

Reading. The six-dimensional two-body problem factorizes into two independent three-dimensional problems: a free center of mass (trivially solved, \(\mathbf{R}(t)=\mathbf{R}_0+\mathbf{V}t\)) and one fictitious particle of mass \(\mu\) in the potential \(V\). Every result about a single particle in a central field transfers verbatim once \(m\) is replaced by \(\mu\). Note \(\mu \le \min(m_1,m_2)\), and \(\mu\to m_1\) as \(m_2\to\infty\).

Units check. \(\mu=\dfrac{m_1 m_2}{m_1+m_2}\) has units \(\dfrac{[\text{kg}][\text{kg}]}{[\text{kg}]}=[\text{kg}]\), a mass, as required. Then \(\mu\ddot{\mathbf{r}}\) has units \(\text{kg}\cdot\text{m s}^{-2}=\text{N}\), matching \(-\nabla V\) with \(V\) in joules and \(\nabla\) carrying \(\text{m}^{-1}\).

Limiting cases

Equal masses \(m_1=m_2=m\): \(\mu=m/2\). Both bodies orbit the common center of mass symmetrically, and the relative motion is that of a half-mass particle.
One mass dominant \(m_2\gg m_1\): \(\mu = m_1\big(1-\tfrac{m_1}{m_2}+\cdots\big)\to m_1\). The heavy body sits essentially at the center of mass and acts as a fixed force center — the textbook "planet around a fixed Sun" limit.
Extreme ratio \(m_2/m_1\to\infty\): \(\mu = m_1\) exactly in the limit; the finite-mass correction is first order in \(m_1/m_2\).
Atomic bound states: for positronium \(\mu=m_e/2\); for hydrogen \(\mu = m_e m_p/(m_e+m_p)\approx 0.99946\,m_e\). Small, but spectroscopically decisive.

Breaks when

Three or more interacting bodies. With a third mass no linear change of variables decouples all relative coordinates; the reduction to a single \(\mu\) is special to two bodies (the general three-body problem is non-integrable).
External or position-dependent field. If a fixed external potential \(V_{\text{ext}}(\mathbf{r}_1)\) acts — not depending only on the separation — the COM equation acquires a net force and \(\mathbf{R}\) no longer decouples from \(\mathbf{r}\).
Velocity-dependent or retarded interactions. Magnetic forces between moving charges, or the finite propagation delay of the field, break the instantaneous form of Newton's third law; momentum leaks into the field and the reduction is only approximate.
Radiating or relativistic systems. When the bodies radiate energy and momentum (gravitational or electromagnetic waves), the relative orbit decays and \(V(\mathbf{r})\) alone no longer governs \(\mathbf{r}\).

Failure modes

Using the total mass \(M\) as the orbital inertia. The inertia in \(\mu\ddot{\mathbf{r}}=\mathbf{F}\) is \(\mu\), not \(M\). \(M\) enters the gravitational force strength (Kepler's third law, \(\omega^2=GM/a^3\)); mixing which quantity is \(\mu\) and which is \(M\) is the most common error.
Adding masses instead of the reciprocal sum. Writing \(\mu=m_1+m_2\) rather than \(m_1m_2/(m_1+m_2)\); the reduced mass is smaller than either mass, never larger.
Dropping \(\mu\) in atomic spectra. Computing the Rydberg with \(m_e\) instead of \(\mu\) misses the isotope shift and the \(\sim 0.05\%\) hydrogen correction.
Double-counting kinetic energy. Writing \(T=\tfrac12 M\dot{\mathbf{R}}^2+\tfrac12 M\dot{\mathbf{r}}^2\); the relative term carries \(\mu\), not \(M\).
Assuming both bodies orbit a stationary heavy body. Even the Sun wobbles about the barycenter; ignoring it is exactly the \(\mu\to m_1\) approximation and fails for comparable masses (binary stars, Pluto–Charon).

Discussion

The deep reason the reduction works is symmetry. Translational invariance of \(V\) (it depends only on \(\mathbf{r}_1-\mathbf{r}_2\)) guarantees conservation of total momentum, and Noether's theorem ties that conserved quantity to the free, cyclic center-of-mass coordinate. Because \(\mathbf{R}\) does not appear in the Lagrangian \(L=\tfrac12 M\dot{\mathbf{R}}^2+\tfrac12\mu\dot{\mathbf{r}}^2-V(\mathbf{r})\), its conjugate momentum \(M\dot{\mathbf{R}}\) is constant and the \(\mathbf{R}\) motion factors out. Reduced mass is not an algebraic coincidence; it is the fingerprint of translational symmetry.

The reduced-mass equation is universal precisely because it is blind to the nature of \(V\). Gravity gives Kepler ellipses; the Coulomb potential gives hydrogen and positronium; a harmonic \(V\) gives the vibrating diatomic molecule; a screened potential gives Rutherford-with-screening. In quantum mechanics the same substitution appears: the two-body Schrödinger equation separates identically, and the hydrogen energy levels carry \(\mu\), not \(m_e\). This is why the Rydberg "constant" is not truly constant across isotopes.

Physically, \(\mu\) measures how sluggishly the separation responds to the mutual force. When one body is heavy it barely moves, so the light body's inertia sets the pace and \(\mu\to m_{\text{light}}\). When the masses are equal, both share the motion and the separation responds as though it had half the mass. The reciprocal-sum form \(1/\mu=1/m_1+1/m_2\) is the same combination as springs in series or capacitors in series, and that is no accident: in each case two elements share a common force and their compliances add.

In the Hamiltonian and quantum settings the change of variables must be canonical: one defines the conjugate momenta \(\mathbf{P}=M\dot{\mathbf{R}}\) (total momentum) and \(\mathbf{p}=\mu\dot{\mathbf{r}}\) (relative momentum) and checks that the Poisson brackets (or commutators) are preserved, \(\{R_i,P_j\}=\delta_{ij}\), \(\{r_i,p_j\}=\delta_{ij}\), with all cross-brackets vanishing. Only then does \(H=\dfrac{\mathbf{P}^2}{2M}+\dfrac{\mathbf{p}^2}{2\mu}+V(\mathbf{r})\) legitimately separate the Hilbert space into a plane-wave COM factor and an internal factor. The cancellation of the cross term in Step 10 is exactly the statement that the new coordinates are orthogonal in the mass metric \(\operatorname{diag}(m_1,m_2)\).

Common misconceptions. "Reduced mass is the average of the two masses" — false; it is smaller than both. "Only the light particle matters" — only in the extreme-ratio limit; for comparable masses both wobble about the barycenter. "The center of mass sits at the heavy body" — only when \(m_2\gg m_1\); in general it lies between them, closer to the heavier one.

Worked examples

Earth–Moon reduced mass, given \(m_E=5.972\times10^{24}\,\text{kg}\) and \(m_M=7.342\times10^{22}\,\text{kg}\).

Set up the reduced-mass formula symbolically first A

\[ \mu = \frac{m_E m_M}{m_E+m_M} = \frac{(5.972\times10^{24})(7.342\times10^{22})}{5.972\times10^{24}+7.342\times10^{22}}\,\text{kg} \]

Substitute numbers; denominator \(=6.0454\times10^{24}\,\text{kg}\) A

\[ \mu = \frac{4.3846\times10^{47}}{6.0454\times10^{24}}\,\text{kg} = 7.253\times10^{22}\,\text{kg} \]

Evaluate; note \(\mu\) is just below \(m_M\), as expected since \(m_E\gg m_M\) A

\[ \mu_{\text{EM}} \approx 7.25\times10^{22}\,\text{kg} \approx 0.9878\,m_M \]

Reading. The reduced mass is about \(1.2\%\) below the Moon's mass, reflecting the Earth's small wobble about the barycenter (which in fact lies inside the Earth, \(\sim 4670\,\text{km}\) from its center).

Units check. \(\text{kg}^2/\text{kg}=\text{kg}\). Correct.

Hydrogen ground-state energy with finite proton mass. The rigid-nucleus Rydberg energy is \(E_\infty=13.6057\,\text{eV}\); \(m_e=9.109\times10^{-31}\,\text{kg}\), \(m_p=1.6726\times10^{-27}\,\text{kg}\).

The binding energy scales linearly with the reduced mass: \(E=\dfrac{\mu}{m_e}E_\infty\) B

\[ \frac{\mu}{m_e} = \frac{1}{1+m_e/m_p} = \frac{1}{1+(9.109\times10^{-31})/(1.6726\times10^{-27})} \]

Rewrite \(\mu=\dfrac{m_e m_p}{m_e+m_p}\) as \(m_e/(1+m_e/m_p)\) to expose the small correction B

\[ \frac{m_e}{m_p} = 5.446\times10^{-4} \;\Longrightarrow\; \frac{\mu}{m_e} = \frac{1}{1.0005446} = 0.999456 \]

Evaluate the mass ratio and the correction factor A

\[ E = 0.999456 \times 13.6057\,\text{eV} = 13.598\,\text{eV} \]

Multiply through; the shift is \(-0.0074\,\text{eV}\) relative to the infinite-mass value A

\[ E_{1}(\text{H}) = 13.598\,\text{eV}, \qquad \Delta E \approx 7.4\,\text{meV} \]

Reading. Using \(\mu\) instead of \(m_e\) lowers the binding energy by about \(0.054\%\). The corresponding difference between hydrogen and deuterium (whose nucleus is roughly twice as heavy) is the measurable isotope shift Urey used to discover deuterium.

Units check. \(\mu/m_e\) is dimensionless, so \(E\) stays in eV. Correct.

Problems

Two equal masses \(m=3.0\,\text{kg}\) are connected by a spring. Compute the reduced mass and state the effective mass entering the vibration frequency \(\omega=\sqrt{k/\mu}\).
Solution
\(\mu=\dfrac{m\cdot m}{2m}=\dfrac{m}{2}=1.5\,\text{kg}\). The oscillation frequency uses \(\mu=1.5\,\text{kg}\), so \(\omega=\sqrt{k/1.5}\) — a factor \(\sqrt2\) higher than a single mass \(m\) on the same spring against a fixed wall.
The CO molecule has \(m_C=12.0\,\text{u}\), \(m_O=16.0\,\text{u}\) (\(1\,\text{u}=1.6605\times10^{-27}\,\text{kg}\)). Find \(\mu\) in kg.
Solution
\(\mu=\dfrac{(12)(16)}{12+16}\,\text{u}=\dfrac{192}{28}\,\text{u}=6.857\,\text{u}\). In SI: \(6.857\times1.6605\times10^{-27}=1.139\times10^{-26}\,\text{kg}\).
A binary star has components \(m_1=2M_\odot\) and \(m_2=6M_\odot\). What fraction of the total mass is the reduced mass, and how far from \(m_2\) does the barycenter lie if the separation is \(a\)?
Solution
\(\mu=\dfrac{(2)(6)}{8}M_\odot=1.5M_\odot\), so \(\mu/M=1.5/8=0.1875\). Barycenter distances scale inversely with mass: distance from \(m_2\) is \(\dfrac{m_1}{M}a=\dfrac{2}{8}a=0.25\,a\); from \(m_1\) it is \(0.75\,a\).
Positronium is a bound electron–positron pair, each of mass \(m_e\). Show its ground-state binding energy is half the infinite-mass hydrogen value, \(6.80\,\text{eV}\).
Solution
\(\mu=\dfrac{m_e m_e}{2m_e}=\dfrac{m_e}{2}\). Since \(E\propto\mu\), \(E=\dfrac{\mu}{m_e}E_\infty=\dfrac12(13.6057)=6.80\,\text{eV}\). The Coulomb charge product is unchanged (\(+e\) and \(-e\)).
Two bodies \(m_1=1.0\,\text{kg}\) and \(m_2=4.0\,\text{kg}\) start at rest and attract each other. When \(m_1\) has moved \(0.80\,\text{m}\), how far has \(m_2\) moved, and what is \(\mu\)? Use momentum conservation.
Solution
The center of mass stays fixed (zero total momentum, from Newton's third law), so \(m_1\,\Delta x_1=m_2\,\Delta x_2\): \(\Delta x_2=\dfrac{m_1}{m_2}\Delta x_1=\dfrac{1}{4}(0.80)=0.20\,\text{m}\). Reduced mass \(\mu=\dfrac{(1)(4)}{5}=0.80\,\text{kg}\). The separation changed by \(0.80+0.20=1.0\,\text{m}\), and \(\mu\ddot{\mathbf r}=\mathbf F\) governs that relative coordinate.