The point that keeps moving as if nothing had happened

A firework shell rises, bursts, and throws two hundred fragments in every direction. Somewhere in that expanding cloud is a location — occupied by nothing at all — carrying on along the parabola the unburst shell was following, at the speed it had, as though the explosion had not occurred. It does so until the first fragment touches the ground.

That location is the mass-weighted average of the fragments’ positions, and its obedience is not a coincidence. It is the only part of a complicated system that answers to an elementary law.

The forces that cancel in pairs #

Take any collection of bodies — two, or two hundred, or the $10^{23}$ molecules in a room — and write down every force acting on every one of them. The list divides in two: forces exerted by something inside the collection, and forces exerted from outside it.

Newton’s third law says each force in the first group has a partner in the same group, equal in size and opposite in direction, acting at the same instant. Molecule 41 pulls on molecule 12 with exactly the force molecule 12 pulls back with, reversed. So when the whole list is summed, the internal forces annihilate one another in pairs and what survives is the external forces alone:

$$\sum \mathbf{F} = \sum \mathbf{F}_{\text{ext}}.$$

This is stronger than it looks. The internal forces may be of any size, duration and complexity — contact between deforming solids whose shape nobody knows, electrostatic forces between molecules no calculation will enumerate — and none of it enters, for the same reason that a collision can be solved without knowing the force curve inside it: the cancellation depends on the antisymmetry of the interaction and on nothing else.

The one-body version of that bookkeeping is the free-body diagram: draw a boundary, keep every arrow that crosses it, and discard the rest. A block on a slope has its weight, the normal force, and the weight resolved along the surface — at 22° that is $0.37mg$ — and the whole art is choosing the boundary and the axes. For a many-body system nothing changes except the size of the boundary. It is drawn once, around the whole collection, and every arrow that begins and ends inside it is thrown away before anything is computed.

What is left has to be attached to something. Newton’s second law applies to a particle, and two hundred fragments are not a particle. Summing the second law over the members answers it: the left-hand side turns out to be the acceleration of

$$\mathbf{R} = \frac{\sum m_i \mathbf{r}_i}{\sum m_i},$$

and the result is the centre-of-mass theorem: $\sum \mathbf{F}_{\text{ext}} = M\ddot{\mathbf{R}}$, with $M$ the total mass. One vector equation, carrying no information about the interior, true for a spinning, tumbling, exploding system exactly as for a brick.

The same cancellation runs through the torque sum, with one extra condition. A torque is a force times a distance from a pivot, so an internal pair contributes the force one way at one body and the other way at the other — and those cancel only if the pair acts along the line joining the two bodies, which is the strong form of the third law. Granted that, the external torques alone govern the rotation about the centre of mass, and the system’s spin is as free of its own internal history as its translation is.

A weighted mean that need not be inside anything #

The definition is an average, and averages are permissive about where they land. A hoop’s mass-weighted mean is at its centre, where there is no metal; a boomerang’s is in the air off its elbow. The Earth–Moon mean is 4,671 km from the Earth’s centre, 1,700 km below the surface, and the planet swings around it monthly. For Pluto and Charon it is 2,126 km out against Pluto’s radius of 1,188 km, so it sits in vacuum and both bodies orbit it.

None of it troubles the theorem, which does not say a piece of matter behaves well. It says a coordinate does.

None of this depends on the projectile being a projectile. A launch at 30° and a launch at 60° at the same speed land in the same place, short of the 45° range — a statement about a single particle that the centre of mass of any object inherits unchanged, whatever the object then does to itself in flight. A shell that bursts, a diver that tucks, a cat that rights itself: the mass-weighted mean of each of them travels the parabola its launch prescribed.

Which is where the shell comes back. Between launch and burst the only external force is gravity, and $Mg$ divided by $M$ is $g$ however the mass is arranged, so the whole object follows the parabola of a single particle. After the burst the fragments exert enormous forces on one another, and every one of them is internal. The external sum is still $Mg$, the centre of mass still accelerates at $g$, the parabola continues. Momentum conservation is not only obeyed by the wreckage — it is visible in it, as a point keeping to a curve nothing is travelling along any more.

The picture ends when the first fragment lands, because the ground is external and its force is in no cancelling pair. Until then, a photograph of the burst taken from the side contains a parabola nobody drew.

The horizontal half of that is worth separating, because it is where the cancellation is easiest to see. Gravity has no horizontal component, so the horizontal momentum of the whole system is untouched from launch to landing: whatever the fragments do to each other, their mass-weighted horizontal position advances at the constant rate it had before the burst. The vertical component passes through zero at the top and reverses. Neither is affected by anything the object does internally, because neither is what the internal forces sum to.

The same permissiveness has been engineered deliberately, in a stadium. A high jumper’s centre of mass rises by an amount set by the athlete’s power, and no technique alters that arithmetic; what technique alters is where the bar sits relative to that point. In the Fosbury flop the body arches backwards so that head, torso and legs cross in succession while the mean of the body’s mass tracks a lower curve — low enough, in a well-executed jump, to pass below the bar the body has cleared. The quantity gravity charges for is a weighted average, and an average can be sent under an obstacle its parts go over.

The frame in which the total momentum is zero #

If a single point travels in a predictable straight line, the obvious move is to ride along with it — and doing so turns the whole family of collision results into one sentence.

In that frame the total momentum is zero by construction; that is what the frame is. Two bodies therefore approach with equal and opposite momenta and must leave with equal and opposite momenta. For an elastic collision, kinetic energy being unchanged as well, one possibility remains: each body keeps its own speed and reverses direction.

The hero figure follows from the same reversal, scaled. Its centre of mass travels at 1.00 m/s, so in that frame the bodies arrive at $+1.00$ and $-3.00$ m/s. A restitution of 0.4 reverses both and shrinks both by 0.4, giving $-0.40$ and $+1.20$; adding the 1.00 m/s back returns 0.60 and 2.20, which is what the figure prints. The mass ratio is never thought about at all.

Which frame that arithmetic is done in matters less than it seems, and the way it matters is worth drawing.

The zero-momentum frame is therefore not a privileged place where the true energy lives. It is the one frame in which the untouchable share $\tfrac12 MV^2$ is zero, so that everything left is the part a collision can spend.

The energy becomes equally transparent. Of the hero figure’s 8.00 J, exactly $\tfrac12 M V^2 = \tfrac12 \times 4 \times 1.00^2 = 2.00$ J belongs to the centre of mass’s own motion, and no internal process can touch it. The remaining 6.00 J is internal, and reversing both velocities while scaling them by $e$ multiplies it by $e^2$: $0.16 \times 6.00 = 0.96$ J, which added to the untouchable 2.00 J gives 2.96 J — the figure’s number, with no collision formula in sight.

The maximum absorbable fraction follows at once, and is fixed by the masses rather than the materials: the surviving sixth is $m_1/(m_1+m_2)$, so five sixths of the incoming energy is available to become deformation, heat and sound. Reverse the masses and only one sixth is.

What splits without a cross term, and the two bodies that become one #

The decomposition generalises past momentum, on the strength of a piece of algebra that could easily have come out differently.

Write each body’s position as the centre of mass plus a displacement from it, $\mathbf{r}_i = \mathbf{R} + \mathbf{r}_i'$, and each velocity likewise. Squaring and summing for the kinetic energy gives three groups of terms: one in $V^2$, one in $v_i'^2$, and a cross term proportional to $\mathbf{V}\cdot\sum m_i \mathbf{v}_i'$. That last sum is the total momentum in the centre-of-mass frame, which is zero by definition. The cross term vanishes identically, and

$$T = \tfrac12 M V^2 + \sum \tfrac12 m_i v_i'^2.$$

This is König’s theorem, and the absence of the cross term is the whole of its value: the motion of the centre of mass and the motion about it can be tallied separately and added, with no interference term between them. The same construction applied to angular momentum gives $\mathbf{L} = \mathbf{R}\times\mathbf{P} + \mathbf{L}'$, again with nothing in between.

Height and speed trade against each other along the flight, and the trade is a statement about the centre of mass and about nothing else. For a tumbling object those two columns are the whole of what this accounting describes: the energy of the spin about that point sits outside them, in the second term, and can be changed by internal forces at any moment without disturbing the first at all. A skater pulling her arms in changes the second and leaves the first exactly where it was.

Everyday mechanics leans on this without saying so. A rolling body’s energy is quoted as $\tfrac12 mv^2 + \tfrac12 I\omega^2$, and that expression is legitimate only because the cross term is absent. Had it survived, a wheel’s energy would contain a term mixing speed with spin, and no division into shares would exist.

Where the released energy goes is fixed by the shape and by nothing else. A solid cylinder puts 67 per cent of a metre’s drop into translation and 33 into rotation, a sphere 71 and 29, and the split depends only on $k = I/mR^2$ — 0.50 and 0.40 — and not on mass, radius or the angle of the slope. That is what decides the race, and it is legible only because the two pots do not mix.

Only the first pot is speed, which is what makes the split visible from the bottom of the slope. After a metre’s drop the sphere reaches 3.74 m/s where a point particle sliding without friction would reach $\sqrt{2gh} = 4.43$ m/s, and the missing energy is not lost: it is turning.

The most economical use of the splitting is the two-body problem. Two bodies interacting only with each other have six coordinates between them; changing to the centre-of-mass position $\mathbf{R}$ and the separation $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$ divides them. $\mathbf{R}$ moves at constant velocity and can be forgotten, leaving one equation,

$$\mu \ddot{\mathbf{r}} = \mathbf{F}(\mathbf{r}), \qquad \mu = \frac{m_1 m_2}{m_1 + m_2},$$

which is one body of mass $\mu$ in a fixed field: a two-body problem reduced to a one-body problem exactly, with no approximation anywhere. For the hero figure $\mu = 3\times1/4 = 0.75$ kg and the approach speed is 4.00 m/s, so the internal energy is $\tfrac12\mu v_{\text{rel}}^2 = 6.00$ J — the same 6.00 J the frame argument produced, reached from the other end.

What it costs #

The theorem gives one equation about one point, and everything it does not mention is the price.

A crash structure is designed against the internal energy, not the total. The centre-of-mass share cannot be absorbed, so a crumple zone can only take what is left. Two identical cars meeting head-on at equal speeds have zero total momentum and the whole energy is available; the same car striking a parked one has half of it locked into the pair’s centre-of-mass motion. So a head-on collision at 50 km/h each is equivalent to hitting a rigid wall at 50 km/h and not at 100, and the factor of two intuition supplies is exactly that locked share.

The point can be tracked while the body cannot. A gymnast’s rotation, a diver’s tuck and a cat’s fall are internal rearrangements, forbidden by the theorem from shifting the mean position’s trajectory. A diver in the air cannot change where the centre of mass will enter the water; every available correction is a redistribution about it. Motion-capture software exploits the reverse, fitting a parabola and treating deviations from it as measurement error, because no real one is permitted.

The equation withholds everything about the interior. A brick and a bomb of the same mass and velocity have the same centre-of-mass motion. Stress, spin, fracture and internal temperature live in the discarded terms, which is why external forces alone can never say whether a block on a slope slides or topples — a question a free-body diagram cannot answer either.

Drawing the boundary is a judgement, and the arithmetic does not check it. The theorem is exact for whatever collection is named, and naming the wrong one gives a confident wrong answer with nothing in it to indicate the mistake.

Where the model stops #

Three assumptions are doing real work, and each fails at a size that can be quoted.

The external field must be uniform across the body. The theorem attaches the external force to the centre of mass; gravity acts on each element with a strength that varies over the body, and its resultant acts at the centre of gravity. For a body of length $L$ at distance $r$ from the source the two separate by about $L^2/6r$, the centre of gravity being nearer the source: 85 nanometres for a 1.8 m person on the Earth’s surface, 2.4 micrometres for a ten-metre satellite at 500 km altitude. That second offset, acting on a spacecraft’s weight of some kilonewtons, supplies a couple of order $10^{-2}$ N·m which always turns the long axis towards the planet, and gravity-gradient stabilisation is a boom extended to make $L$ large enough for a torque that never runs out of fuel. One exemption is exact: a uniform sphere’s two centres coincide precisely, because the average of $1/x$ over a uniform sphere is exactly $1/r$. Why a uniform field cannot be told from an acceleration is this same condition, and tidal effects are what remains once it is removed.

The mass must be constant. $\sum \mathbf{F}_{\text{ext}} = M\ddot{\mathbf{R}}$ was derived with $M$ fixed, and a rocket’s mass is not. Redone with the exhaust inside the system it holds again, and says something remarkable: the centre of mass of vehicle-plus-exhaust accelerates downward at $g$ alone while the vehicle climbs. A Saturn V first stage threw about 13,000 kg of propellant per second rearwards at 2,600 m/s — the 34 MN of thrust — and lost two thirds of its 2,970-tonne launch mass in 168 seconds. Written with $M$ as the vehicle’s current mass the description is wrong by the full size of the thrust.

The third law describes simultaneous action, which relativity does not permit. Equal and opposite at the same instant presupposes an instant every observer agrees on, and simultaneity is a choice of slicing. Two charges in general motion do not exert equal and opposite forces on each other: the magnetic parts fail to match, by a fraction of order $(v/c)^2$, about $10^{-4}$ at one per cent of light speed. The theorem survives anyway, because the field carries momentum at a density $\varepsilon_0\mathbf{E}\times\mathbf{B}$ and the missing amount is in the space between the charges in the interval — magnetism read as electricity seen sideways, from the other side. The conserved quantity is matter plus field, and what keeps to a straight line is the centre of energy: a weighted mean over positions taken at one time, and so dependent on which slicing does the weighting.

The same theorem where it was not expected #

The reduction to a body of mass $\mu$ outlived its subject.

Hydrogen is a two-body problem, and its energy levels contain $\mu$ rather than the electron mass. Since $m_p/m_e = 1836$ the correction is a factor 0.99946 — 0.054 per cent, small and consequential. Deuterium’s nucleus is twice as heavy, its reduced mass differs from hydrogen’s by 0.027 per cent, and every spectral line shifts by that fraction: Balmer-alpha moves from 656.28 nm to 656.10 nm. Urey found deuterium in 1931 by photographing that 0.18 nm gap — an isotope discovered because a mass-weighted average had been defined correctly.

A diatomic molecule’s vibration is the same reduction again, with $\omega = \sqrt{k/\mu}$. Hydrogen chloride absorbs at 2,886 cm⁻¹; swapping the hydrogen for deuterium changes $\mu$ from 0.97 to 1.89 atomic mass units and multiplies the frequency by $\sqrt{0.97/1.89} = 0.72$, predicting 2,091 cm⁻¹, which is what deuterium chloride does. The spring constant belongs to the electrons and never moved.

Astronomy uses the barycentre as an instrument. The Sun–Jupiter centre of mass is 742,000 km from the Sun’s centre and the Sun’s radius is 696,000 km, so the Sun orbits a point outside itself every twelve years at 12.7 m/s — and every radial-velocity exoplanet detection is that motion, measured about an invisible partner.

There is a deeper reading. Each conservation law belongs to a symmetry, and this theorem belongs to one easy to overlook: the physics is unchanged when viewed from a frame moving at constant velocity. The conserved quantity is $M\mathbf{R} - \mathbf{P}t$, whose constancy is the statement that the centre of mass moves uniformly. Momentum belongs to translation in space, angular momentum to rotation, energy to translation in time; the fourth and least discussed member of the family is this one. The relativistic version survives with energy replacing mass in the weighting, which is one more reason mass is a form of energy rather than a separate ledger.

Euler’s principle, and a corollary Newton had already written #

The theorem is in the Principia of 1687, as Corollary IV to the laws of motion: the common centre of gravity of two or more bodies does not alter its state of motion or rest through the actions of the bodies among themselves. Newton put it among the corollaries rather than the propositions, so it reads as a consequence rather than a discovery — which is how it has been treated ever since, to its cost.

The separation into translation and rotation was made properly by Euler, in a 1750 paper whose title claims a new principle of mechanics and whose content is that a rigid body’s motion is fully described by two equations: the external force sum governing the centre of mass, and the external torque sum governing the rotation about it. Sixty-three years after the corollary, and the harder half.

König’s theorem followed in 1751, and the reduced mass reached its modern form through Lagrange’s work on the two- and three-body problems in the 1770s. The order is worth noticing — conservation statement first, decomposition afterwards, algebraic reduction last — because each step made calculable a class of problem the previous one had only made consistent. Dick Fosbury, clearing 2.24 m in Mexico City in 1968, used the first of them and none of the algebra.

What the picture cannot show #

The centre of mass is not drawn in any of the ten figures. It cannot be: it is an average, and the figures show the things averaged. Every claim above about it is arithmetic performed on the numbers the figures print, which is honest but is not the same as seeing it.

The collision figures are one-dimensional and show no rotation. An off-centre hit at the same speeds gives an identical centre-of-mass motion and a different outcome, with energy diverted into spin and the bodies leaving at angles a row of blocks cannot represent. The pictures are consistent with a family of collisions they do not distinguish between — and so is the theorem, which is both the point being made and a warning about it.

The absence of the cross term cannot be drawn either. The rolling figures show two shares as percentages, which is what the absence permits, but two bars give no hint that a third term was possible and cancelled. That cancellation is König’s theorem, and it is invisible in every drawing of its consequences.

Nor can the field momentum appear. Two bodies and two arrows has already asserted that the space between them holds nothing that counts, when in the relativistic case the missing momentum is exactly there. Field lines are a choice that makes the momentum in a field particularly hard to see.

The ladder from here #

Later rungs on this anchor: two-dimensional collisions and the impact parameter, where three equations meet four unknowns; the rocket equation, derived by putting the exhaust inside the system; angular momentum conservation, and the collisions that put energy into spin; the moment-of-inertia tensor, and a body that tumbles because its axes disagree; and the relativistic centre of energy, where hidden momentum appears in systems that look static.

The neighbouring ladders are collisions and the law underneath them, where this argument starts; rotation and the moment of inertia, which are the motion about the point rather than of it; energy and the forces that give it back, which the splitting theorem divides in two; and pressure as a rate of arrival, where $10^{23}$ members have a centre of mass that sits still while every one of them travels at hundreds of metres per second.