Two bodies replaced by one that does not exist

Every closed-form result in celestial mechanics is a result about one body moving in a fixed field. The ellipse, the harmonic law, the vis-viva relation, Kepler’s equation — all of them are derived for a single mass moving about a centre that does not move, and the conic family with them.

No such system exists. Both bodies move; the Sun is displaced by Jupiter and so is every star by its planets. So the derivations ought to be answers to a question nobody asked.

They are not, because of a substitution that is exact rather than approximate. The two-body problem can be rewritten, with no loss and no error, as a one-body problem — and the body it produces has a mass that belongs to neither of the two real objects.

The rewriting, in three lines #

Two bodies attract each other. Newton’s second law for each, with $\mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1$ the separation vector:

$$m_1\ddot{\mathbf{r}}_1 = \frac{Gm_1m_2}{r^3}\mathbf{r}, \qquad m_2\ddot{\mathbf{r}}_2 = -\frac{Gm_1m_2}{r^3}\mathbf{r}.$$

Divide each by its own mass and subtract the first from the second:

$$\ddot{\mathbf{r}} = -\frac{G(m_1+m_2)}{r^3}\mathbf{r}.$$

That is the whole derivation. The separation vector obeys the equation of a single body moving about a fixed point of mass $m_1 + m_2$. Nothing was neglected, no ratio was assumed small, no term was dropped. It is an identity.

The consequences arrive immediately. The relative orbit is a conic, with the fixed centre at its focus. Kepler’s third law applies to the relative orbit with $M = m_1 + m_2$ in place of the central mass — which is why the harmonic law for a binary star measures the sum of the masses and not either one. The vis-viva relation holds for the relative speed. Every two-body result is a result about $\mathbf{r}$.

Where the reduced mass comes in #

The rewriting above never mentions $\mu$. It appears when the energy and angular momentum are wanted, rather than the trajectory.

Write the total kinetic energy in terms of the barycentre motion and the relative motion. The cross terms cancel — that is the point of using the barycentre — and what is left is

$$T = \tfrac{1}{2}(m_1+m_2)\dot{\mathbf{R}}^2 + \tfrac{1}{2}\mu\dot{\mathbf{r}}^2, \qquad \mu = \frac{m_1m_2}{m_1+m_2}.$$

The first term is the whole system drifting; in the barycentric frame it is zero and stays zero. The second is a single particle of mass $\mu$ moving with the relative velocity. The same combination appears in the angular momentum, $\mathbf{L} = \mu\,\mathbf{r}\times\dot{\mathbf{r}}$, and in the potential energy written as $-GM\mu/r$.

So the complete statement is: the two-body problem is exactly a one-body problem in which a particle of mass $\mu$ moves in the field of a fixed mass $M = m_1 + m_2$. The trajectory is governed by $M$; the energy and momentum are scaled by $\mu$.

Neither number belongs to an object. $M$ is the mass of nothing — no single body in the system has it — and $\mu$ is smaller than either mass, always. For equal masses $\mu = m/2$; for a very unequal pair $\mu \to m_2$, the smaller mass, which is the sense in which the familiar one-body picture is the limit of the exact one.

What the substitution keeps, and what it loses #

The one-body picture is exact for everything that depends on the separation. It is silent about everything that does not, and the distinction matters when a real observation is being interpreted.

Preserved. The shape of the relative orbit, its eccentricity, its orientation, its period, the timing of the bodies along it, the total energy, the total angular momentum, and every conserved quantity of the pair.

Not preserved. Where either body is. The relative orbit says the two are 4.2 AU apart at a particular date; it does not say where either of them is. Recovering that needs the mass ratio, which the relative orbit does not contain:

$$\mathbf{r}_1 = -\frac{m_2}{m_1+m_2}\mathbf{r}, \qquad \mathbf{r}_2 = \frac{m_1}{m_1+m_2}\mathbf{r}.$$

Two numbers come out of the relative orbit — the total mass, from the period and the semi-major axis — and a third is needed to split it. That structural fact is the reason binary-star astronomy is organised the way it is.

Why the harmonic law measures a sum #

The most-used consequence of the substitution is one that is easy to state wrongly, and the wrong version is in a great many textbooks.

Kepler’s third law for a planet about the Sun is usually written $P^2 = a^3$ in years and AU, with the Sun’s mass absorbed into the units. The exact statement, from the relative equation, is

$$P^2 = \frac{4\pi^2 a^3}{G(m_1 + m_2)},$$

where $a$ is the semi-major axis of the relative orbit — the separation’s ellipse, not either body’s own. Both of those refinements matter at different times.

For the solar system the mass correction is tiny and the distinction between the relative and heliocentric orbits is tinier still, because $m_2/m_1$ is at most $10^{-3}$. Even so it is measurable: Jupiter’s period is 0.096% longer than the naive law predicts, which over the two hundred years of modern observation is more than a week, and the discrepancy is the standard way of stating that the law needs the sum.

There is a second refinement hiding in the same formula, and it is the one that trips up satellite work. The $a$ in the law is the semi-major axis of the relative orbit, and for a satellite about the Earth that is the satellite’s distance from the Earth’s centre — not from the barycentre of the pair, and not from the ground. The distinction is negligible for a spacecraft and is not for the Moon: the Earth–Moon barycentre sits 4,670 km from the Earth’s centre, three-quarters of the way to the surface, and the Moon’s orbit about that point has a semi-major axis 1.2% smaller than its orbit about the Earth’s centre. Using the wrong one puts the sidereal month out by half a day.

For a binary star the correction is the entire content. Two solar-mass stars in a 1 AU relative orbit have a period of $1/\sqrt{2} = 0.707$ years, not one year. Nothing about the system can be inferred without treating the two masses symmetrically, and the quantity the observation delivers is irreducibly their sum.

What was actually measured #

A binary star’s orbit is measured in one of three ways, and each gives a different combination of the masses. Which combination is available is the whole of what can be said about the system.

A visual binary — one resolved into two points of light — gives the relative orbit directly: the separation and position angle over decades, fitted to an ellipse. With the parallax to convert arcseconds into AU, Kepler’s third law gives

$$m_1 + m_2 = \frac{a^3}{P^2}$$

in solar masses, AU and years. The sum, and only the sum. Splitting it requires measuring the barycentre against background stars, which is a far harder astrometric problem, and it has been done for a few dozen systems.

A spectroscopic binary — unresolved, detected by the Doppler wobble — gives the radial velocity of one or both components. If only one spectrum is visible, what comes out is the mass function

$$f(m) = \frac{(m_2\sin i)^3}{(m_1+m_2)^2} = \frac{P K_1^3 (1-e^2)^{3/2}}{2\pi G},$$

a single number combining both masses and the unknown inclination. It is a lower bound on $m_2$ and nothing more. If both spectra are visible, the ratio of the two velocity amplitudes gives $m_1/m_2$ immediately and exactly — no distance, no inclination needed — because the barycentre condition is a ratio.

An eclipsing binary fixes the inclination near 90°, removing the $\sin i$ ambiguity. A system that is both eclipsing and double-lined spectroscopic gives every mass and radius in it, in absolute units, with no distance required at all. That is why such systems are the fundamental calibrators of stellar astrophysics: about two hundred of them are known to better than 3% in mass, and the mass–luminosity relation rests on them. The sharpest test of the whole framework is not a star but a pulsar. The Hulse–Taylor binary pulsar PSR B1913+16 has a pulse period known to fourteen significant figures, so its orbit is measured with a precision no optical technique approaches: the orbital period is 27,906.98163 seconds, the eccentricity 0.6171334, and the two neutron-star masses are 1.4398 and 1.3886 solar masses, each to about 0.0002. The two-body framework, with relativistic corrections, reproduces the observed orbital decay to within 0.2% — and that decay, 76.5 microseconds per year, is the first evidence for gravitational radiation.

The barycentre is an observable, and it is not at the centre of the Sun #

The fictitious fixed centre of the one-body problem is not the barycentre — it is a point of mass $m_1 + m_2$ at the origin of the separation vector, which is the position of one of the bodies. The barycentre is a different point, and unlike the fictions it is real, in the sense that something can be measured against it.

For the Sun and Jupiter the barycentre sits at

$$\frac{m_J}{m_\odot + m_J} \times 5.204\ \text{AU} = 0.00496\ \text{AU} = 742{,}000\ \text{km},$$

which is 1.07 solar radii — just outside the Sun’s surface. The Sun does not wobble about a point inside itself; it orbits a point in space above its own photosphere, once every 11.86 years, at 12.4 metres per second.

That last number is the one that mattered. A star’s reflex motion of a few metres per second is precisely what a Doppler shift can detect, and the first exoplanet around a Sun-like star was found in 1995 by measuring 51 Pegasi moving at 59 m/s. An observer fifty light years away, watching the Sun with the same instruments, would have found Jupiter. The full solar system’s barycentre is the sum over all the planets, and it wanders in and out of the Sun’s body on a timescale set by the beat between Jupiter’s and Saturn’s periods. It has been outside the photosphere for most of the last century. Every ephemeris in use is referred to it rather than to the Sun’s centre, because it is the point that moves uniformly and the Sun’s centre is not.

The generalisation: any pair of anything #

The reduced mass is not a fact about gravity. It is a fact about two bodies interacting through a force that depends only on their separation, and it turns up wherever that happens.

A diatomic molecule vibrates at $\omega = \sqrt{k/\mu}$, with $\mu$ the reduced mass of the two nuclei — which is why replacing hydrogen with deuterium shifts an infrared band by very nearly $\sqrt{2}$ and why isotopic substitution is a standard tool for assigning spectra.

The hydrogen atom’s energy levels are proportional to $\mu$, and the small correction from the electron-to-proton reduced mass rather than the bare electron mass is measurable: it separates hydrogen from deuterium in the Balmer lines by 1.8 Å at H$\alpha$, and that splitting is how deuterium was discovered in 1931.

Two-particle scattering is universally reduced to a single particle of mass $\mu$ scattering off a fixed potential, and the difference between the laboratory frame and the centre-of-mass frame — a permanent source of confusion in nuclear physics — is exactly the barycentre term that the substitution sets aside.

In each case the same two things are true: the trajectory is governed by the total mass or the interaction, the energetics by the reduced mass, and the substitution is exact. It is worth stating plainly because “reduced mass” sounds like an approximation and is not one. Nothing is neglected in obtaining it.

The combination a merger measures #

There is one observation in which neither the total mass nor the reduced mass is what comes out, and a third combination of the two appears instead — which makes it the sharpest illustration of the essay’s point that the masses enter different questions differently.

Two compact objects spiralling together radiate gravitational waves, and the rate at which their orbit shrinks depends on how strongly they radiate, which depends on both masses. Working it through, the frequency of the emitted wave sweeps upward at a rate governed by

$$\mathcal{M} = \frac{(m_1 m_2)^{3/5}}{(m_1 + m_2)^{1/5}},$$

the chirp mass — which is $\mu^{3/5} M^{2/5}$, a weighted geometric mean of the two quantities this essay is about.

At leading order nothing else appears. The waveform of an inspiral is fixed by the chirp mass, the distance and the geometry, so a signal detected with modest sensitivity returns that one number precisely and says almost nothing about the individual masses. The first detected merger’s chirp mass was known to a few per cent while the two component masses were each uncertain by twenty.

Separating them requires the parts of the waveform where the leading-order description fails: the higher-order corrections to the inspiral, which depend on the mass ratio, and the merger and ringdown, which depend on the final object. Those are the faintest and briefest parts of the signal, which is why the individual masses are always the worse-determined quantities.

So the pattern holds in a third setting. A Kepler orbit’s period gives the total mass; the amplitude ratio of two velocity curves gives the mass ratio; and a chirping waveform gives a combination that is neither, because what is being measured is a rate of energy loss rather than a frequency of revolution.

Each question picks out its own function of $m_1$ and $m_2$, and knowing which one a measurement returns is most of understanding what it has reported.

It also explains a habit of the gravitational-wave literature that looks evasive and is not. A catalogue entry quotes the chirp mass with a small uncertainty and the component masses with large ones, and the two are not independent — the pair of masses is constrained to a narrow curve in the plane rather than to a box, and quoting the two numbers separately with their own error bars throws that away.

The same care is owed to any quantity derived from a combination rather than measured directly, and this essay’s subject is the oldest example of one.

Reading a catalogue that way takes one extra step and prevents the commonest misuse of it, which is to treat the two component masses as though each had been measured.

The same warning applies to the mass ratio inferred from an amplitude ratio, which is exact, and to the individual masses inferred from it together with an assumed inclination, which are not.

Where the model stops #

Two bodies exactly. The cancellation that produces the relative equation works because there are two terms to subtract. With three bodies there is no separation vector whose equation closes, and the whole method fails — not approximately, but structurally. That is the sharp boundary between the solvable and the unsolvable in this subject.

Point masses. Two extended bodies exert forces that depend on their orientations as well as their separation, so the relative equation acquires terms that do not reduce. For close binaries that deform each other, the two-body ellipse is not the orbit.

No dissipation. The Hulse–Taylor orbit shrinks. Once energy leaves the system the relative orbit is not a fixed conic, and the elements become slowly varying.

The figures show a snapshot of a construction, not a motion. Both panels of the hero figure draw complete orbits, which no observation ever sees: a visual binary is measured over decades and yields a few dozen points on an arc, and the ellipse is a fit. The picture is the conclusion of the measurement, drawn as though it were the data.

The ladder from here #

Later rungs on this anchor: the barycentric and relative coordinates derived as a canonical transformation. The mass function, and what a single-lined spectroscopic binary can and cannot say. Eclipsing binaries as absolute calibrators. Astrometric orbits and the barycentre against the background. The Hulse–Taylor system’s post-Keplerian parameters, and how five of them overdetermine two masses. Reduced mass in molecular vibration and in the Rydberg constant.

There is a small pleasure in the fact that the object at the centre of the exact solution — mass $m_1 + m_2$, sitting still — is not there, and the object orbiting it — mass $\mu$ — is not there either. The two-body problem is solved by replacing both bodies with two fictions, and the answer is correct to the last digit.