Newtonian Dynamics
When we seek what is most fundamental — an idea that governs the whole of mechanics — we meet the concept of force, as the motive of the motion of a particle. In this respect, the laws formulated by Isaac Newton are essential [Newton:1687]. Even so, as we shall see, there exist further formulations, resting on other axioms, from which Newton's laws can be derived (Lagrangian Mechanics and Hamiltonian Mechanics).
A particle can translate; and if it has a volume — that is, if it is a body — it can also rotate. We shall introduce the axioms and definitions that allow us to understand the mechanics of a particle and of a body.
Inertia and mass
Inertia
Consider a particle. On what does its kinematics depend when nature interacts with it? We may suppose a property of the particle that plays a role in the motive of its motion, simply by virtue of its being in nature. We postulate:
If a particle moves in uniform rectilinear motion with respect to some inertial frame of reference, it will continue to do so unless it interacts with nature. Rests on Definitions 18.3 and 18.8.
This is a property possessed by every particle, known as inertia.
Interaction
An interaction is the motive by which a particle changes its position \(\vect{x}(t)=\vect{0}\ \forall t\) to another position \(\vect{x}\,'\). Rests on Definition 18.3 and Postulate 19.1.
Mass
One particle can have more inertia than another.
The mass is the scalar quantity that quantifies inertia, in the following sense: the mass of a particle is proportional to how much it costs nature to move it. Rests on Postulate 19.1 and Definition 18.3.
The Fundamentos formulary accordingly sharpens the notion of a particle introduced in Definition 18.3: a particle is a point of space to which a mass \(m\) is assigned.
Newton's laws and the Newtonian formulation: translation
Linear momentum and Newton's first law
Consider the following fact of nature. When a very heavy truck moves at great speed, we feel that it carries great power, implacable in any collision (let us not forget that if the truck moves at great speed it is because our frame of reference is outside the truck; if we were inside it and did not notice what lies outside, for us it would be at rest). Now, if we have a light piece of wood that moves very slowly, we feel that it carries less power than the truck.
On the other hand, we may consider the crossed case. What would happen if the truck moved very slowly while the piece of wood travelled at extravagant speeds close to the speed of light? Clearly we must introduce a quantity that manifests the power of which we speak — one that unifies the mass and the velocity associated with a given motion.
The linear momentum of a particle of mass \(m\) moving with velocity \(\vect{v}\) with respect to some frame of reference is
![[figure: momentum.pdf]](/images/treatise/de/figure/momentum.svg)
Linear momentum: the quantity unifying the mass of a particle with the velocity of its motion.
We enunciate as a fundamental principle:
The linear momentum of a free particle is constant, with respect to any inertial frame of reference. Rests on Definitions 18.8 and 19.5.
A body on which nothing acts keeps its velocity: neither its speed nor its direction changes, and no agent whatever is needed to maintain the motion. The fact is never directly visible on Earth, where friction and air resistance are never absent; it is exposed by removing them by degrees — a ball on a horizontal plane runs the farther the smoother the plane, which is Galileo's argument [Galilei:1638] — and it is seen outright in a coasting spacecraft. The contrary Aristotelian expectation, that motion requires a mover and ceases with it, is the oldest falsified theory in mechanics. Rests on Postulate 19.6 and Equation (19.1).
Derivation. Derives Phenomenon 19.7. By Postulate 19.6 the linear momentum of a free particle is constant. If in addition the mass of the particle is constant, then from Equation (19.1)
is constant as well, so the velocity keeps both magnitude and direction, and the trajectory is a straight line traversed uniformly. The observation therefore tests two things at once, and separating them matters: the constancy of \(\vect{p}\), which is the postulate, and the constancy of \(m\), which fails for a rocket and for a body accumulating or shedding matter. It is only with both that Equation (19.3) collapses to Equation (19.4) and the familiar statement follows.
∎Force and Newton's second law
Until now we have used the concept of an interaction of nature with particles. But what, at bottom, is an interaction? We must quantify this concept in order to introduce fundamental equations that govern the entire dynamics of a system. An interaction certainly perturbs the inertia of a particle, and so we introduce as a fundamental postulate:
The sum of all the interactions upon a particle is the measure of how much its linear momentum fails to be conserved. Rests on Definition 19.5 and Postulate 19.6.
We shall call the total force upon a particle the sum of all the interactions upon it, and we shall denote it by \(\vect{F}\). Newton's second law is then mathematically expressed by
Newton's second law is fundamental because it provides the equations of motion. That is, given the sum of the forces acting upon a particle, we can compute its trajectory. Moreover, we can finally conclude the following: a particle will continue its rectilinear motion unless a nonvanishing total force acts upon it — this with respect to an inertial frame of reference. The figure illustrates a particle that moves along a straight line and deviates from its trajectory.
![[figure: force.pdf]](/images/treatise/de/figure/force.svg)
A particle moving in a straight line is deflected from its trajectory by a nonvanishing total force.
From Equation (19.1) we see that
If the mass of the particle is constant, then from Equation (19.3) we see that
The Fundamentos formulary states the second law in the summary form: the change with respect to time of the linear momentum of a particle is proportional to its mass and to its acceleration, \(\vect{F}=\dv{\vect{p}}{t}=m\vect{a}\).
For a body of fixed mass the acceleration produced is proportional to the applied force and parallel to it; for a fixed applied force it is inversely proportional to the mass. The classical demonstration is Atwood's machine, in which two masses \(m_1>m_2\) hang from the ends of a cord over a light pulley and the pair is observed to move with a uniform acceleration smaller than that of free fall, in the ratio \((m_1-m_2)/(m_1+m_2)\). Bringing the two masses close in value makes the acceleration as small as the timing apparatus requires without changing its constancy — the same dilution trick as Galileo's inclined plane [Galilei:1638], and the arrangement by which the second law was first checked quantitatively in the laboratory [Newton:1687]. Rests on Equation (19.4) and Postulate 19.8.
Derivation. Derives Phenomenon 19.10. Idealize the cord as inextensible and massless and the pulley as light and frictionless, so that the tension \(T\) is the same throughout and the two masses have accelerations of equal magnitude \(a\) and opposite sense. Applying Equation (19.4) to each mass along the vertical, with the descending mass taken positive,
Adding the two eliminates the unknown tension, which is the point of the arrangement, and gives
The acceleration is constant, so the descent obeys the itinerary equation for constant acceleration of Kinematics and can be timed over a measured distance. Two limits check the result: for \(m_2=0\) it returns \(a=g\), free fall; and for \(m_1=m_2\) it returns \(a=0\), equilibrium. What the experiment establishes is that a single number \(m\) suffices — the same \(m\) that measures the weight \(mg\) also measures the resistance to acceleration in Equation (19.4) — which is the assumption whose exactness is tested in The Equivalence Principle and Classical Tests.
∎Covariance
A law of motion earns the name only if it says the same thing in every inertial frame; were it otherwise, Postulate 18.16 would fail, and an experiment performed inside a closed cabin could disclose the cabin's velocity. That Equation (19.4) passes this test is the content of the present heading. It is worth doing carefully, because the demonstration constrains the admissible forces a good deal more than it constrains the equation.
The transformation itself belongs to kinematics and is stated there: Definition 18.18 defines the Galilean transformation
with \(R\in\SO(3)\), \(\vect{V}\) a relative velocity in \(\mathrm{m}/\mathrm{s}\), \(\vect{a}\) a displacement in \(\mathrm{m}\) and \(b\) a time offset in \(\mathrm{s}\), and Proposition 18.20 proves that the ten of them close into the Galilei group with the composition law Equation (18.12). Both are used below without being restated.
The mass of a particle is the same number in every inertial frame, \(m'=m\): mass is a scalar by Definition 19.3, and nothing in Definition 18.18 acts on it. This is an assumption about what mass is, not a theorem, and it is exactly the assumption relativity abandons — see Relativistic Dynamics, where the inertia of a body depends on its speed. Rests on Definitions 18.18 and 19.3.
Let the mass of a particle be constant and let the total force upon it transform as \(\vect{F}\,'=R\vect{F}\) under Equation (18.11). Then Equation (19.4) holds in \(K\) if and only if \(\vect{F}\,'=m\vect{a}\,'\) holds in \(K'\). Rests on Definition 18.18, Equation (19.4) and Postulate 18.14.
Derives Theorem 19.12. By Postulate 18.14 the two frames measure the same time increments, \(\dd t\,'=\dd t\), so differentiating with respect to \(t\,'\) is differentiating with respect to \(t\). Differentiating the second of Equation (18.11) once, \(R\), \(\vect{V}\) and \(\vect{b}\) being constant,
and a second time,
Multiplying Equation (19.7) by the constant mass and subtracting the assumed transformation of the force,
and \(R\) is invertible, so the left-hand side vanishes exactly when the right-hand side does.
∎The hypothesis on the force is not a formality, and it is the half of the statement the source note omits: Equation (19.4) is covariant provided the force is the kind of object that transforms as a vector and does not depend on the absolute position, the absolute time or the absolute velocity of the particle. That is a genuine restriction on the interactions nature is allowed to have, and it is satisfied by the ones it does have.
Let the force on particle \(s\) of a system be a sum of pair terms
with \(\vect{f}\) equivariant under rotations, \(\vect{f}\left(R\vect{u},R\vect{w}\right) =R\,\vect{f}\left(\vect{u},\vect{w}\right)\) for every \(R\in\SO(3)\). Then \(\vect{F}'_{s}=R\vect{F}_{s}\), so Theorem 19.12 applies to every particle of the system. Rests on Definition 18.18 and Theorem 19.12.
Derives Proposition 19.13. The displacement \(\vect{b}\) and the term \(\vect{V}t\) cancel in a difference of positions, and \(\vect{V}\) cancels in a difference of velocities: by Equations (18.11) and (19.6),
Equivariance then returns each pair term as \(R\) times its unprimed value, and the finite sum Equation (19.8) with it.
∎A pair force of the form \(\vect{f}\left(\vect{u}\right) =f\!\left(\abs{\vect{u}}\right)\vect{u}/\abs{\vect{u}}\) satisfies the hypothesis of Proposition 19.13, a rotation preserving both \(\abs{\vect{u}}\) and the direction of \(\vect{u}\) relative to itself. Newtonian gravitation and the Coulomb force are of this kind, and so is a Hooke spring joining two particles; the central case is developed in Central Forces and Statics. Rests on Proposition 19.13.
What the two frames share is the form of the law, not the values of the quantities in it. Under a pure boost the momentum of Equation (19.1) changes by \(-m\vect{V}\) and the kinetic energy of Equation (19.18) becomes
neither of which is unchanged. What is boost-invariant is the acceleration Equation (19.7), together with the separations between particles and their relative velocities — exactly the quantities Equation (19.8) is built from. The distinction returns in Section 19.5.5: a quantity can be conserved in every inertial frame without taking the same value in any two of them.
Theorem 19.12 was stated for constant mass, and the restriction cannot be lifted. Allowing \(m\) to depend on time and transforming under a pure boost,
so the two frames disagree by \(-\dot m\,\vect{V}\). No transformation law for the force can absorb that term, because it depends on the arbitrary boost velocity \(\vect{V}\) and not on anything physical. Equation (19.2) is therefore Galilean covariant only for a system whose mass is constant. A rocket is not a counterexample but a lesson in bookkeeping: the closed system is the rocket together with the exhaust it has already ejected, its total mass is constant, and Equation (19.2) applies to that. The same caveat was recorded in the derivation of Phenomenon 19.7.
The proof of Theorem 19.12 leaned on Postulate 18.14 at the very first step: only because \(t\,'\) and \(t\) differ by a constant does the second derivative in \(K'\) agree with the second derivative in \(K\). If instead the time of \(K'\) is allowed to depend on position — as the Lorentz transformations of Lorentz Transformations require it to — then Equation (19.7) is false, and no redefinition of the force restores Equation (19.4). The mechanics built on that other group is Relativistic Dynamics, and the disagreement between the two is of relative order \(v^{2}/c^{2}\), which Equation (18.8) bounds.
The observational counterpart of Theorem 19.12 is Phenomenon 18.17: mechanical experiments in a closed cabin do not disclose its uniform velocity. Nothing in the theorem is tested by that observation alone, however, since the covariance was built into the postulates; what is tested is the conjunction of Postulate 18.14 with the assumption that the real forces are of the admissible form Equation (19.8).
Newton's laws and the Newtonian formulation: rotation
Angular momentum
Let us note the following fact of nature. When we try to push a door, we can clearly observe that it costs much less to push it from a point far from the hinge (for example, the handle) than from a point closer to it. What is happening? We are applying the same force, yet we do not achieve the same motion. In other words, to succeed in opening the door we must apply more or less force depending on where we do it.
This suggests the following. To study the motion of a body we must introduce an analogue of the translational Newton laws, but with a quantity that carries information about the location at which there is a surge of momentum — that is, a quantity that is a function of the position and the momentum of a particle.
Thus we define:
The angular momentum of a particle of position \(\vect{x}\) and momentum \(\vect{p}\) is
with units of \(\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}\). Rests on Definitions 18.3 and 19.5.
The principle the source manuscript was reaching for at this point is the rotational counterpart of Postulate 19.6.
The angular momentum of a free particle, taken about any point at rest in an inertial frame, is constant. Rests on Definition 19.18 and Postulate 19.6.
Postulate 19.19 follows from Postulate 19.6 and Definition 19.18; it is therefore a theorem for a single particle, and only becomes an independent statement for a system of interacting particles (Section 19.5.4). Rests on Postulate 19.6 and Definition 19.18.
Derives Proposition 19.20. Let \(\vect{x}\) be measured from the fixed point. Differentiating Equation (19.11) and using the product rule for the cross product,
The first term vanishes because \(\vect{p}=m\vect{v}\) is parallel to \(\vect{v}\) by Equation (19.1), and the second vanishes because \(\vect{p}\) is constant for a free particle by Postulate 19.6. Hence \(\vect{L}\) is constant. The argument uses only that the point about which \(\vect{L}\) is taken does not move; about a point in arbitrary motion the statement is false, since then \(\dd\vect{x}/\dd t\) is no longer \(\vect{v}\).
∎For a system the situation is different and the postulate does real work: the internal forces of a system need not cancel in the torque sum merely because they cancel in the force sum, and what is required of them is stated as Postulate 19.61 below.
A body moving under a force always directed towards one fixed centre sweeps out equal areas about that centre in equal times: it moves quickest where it is nearest and slowest where it is farthest, in exact compensation. This is Kepler's second law, proved from the laws of motion as the first proposition of the Principia [Newton:1687], and it holds for every central force whatever, not only for the inverse square. The Earth's orbit is eccentric enough for the effect to be conspicuous: Newcomb's reduction of the inner-planet observations gives its eccentricity as \(e=0.01675\) at the epoch 1900.0, slowly decreasing [Newcomb:1895]. With a semi-major axis of one astronomical unit, \(a=1.495978707\times 10^{11}\,\mathrm{m}\) by definition [BIPM:2019], the perihelion and aphelion distances differ by \(2ae\approx5.0\times 10^{9}\,\mathrm{m}\), and the Earth runs correspondingly faster in January than in July. Rests on Postulate 19.19.
Derivation. Derives Phenomenon 19.21. Let the force be central about the origin, \(\vect{F}=F(\vect{x})\, \vect{x}/\abs{\vect{x}}\). Then \(\vect{x}\times\vect{F}=\vect{0}\), so by Equations (19.14) and (19.15) the angular momentum about that centre is constant. The area swept by the radius vector in a time \(\dd t\) is that of the triangle spanned by \(\vect{x}\) and \(\dd\vect{x}\),
of SI unit \(\mathrm{m}^{2}/\mathrm{s}\), which is constant with \(\abs{\vect{L}}\). Since \(\vect{L}\) is moreover fixed in direction, the motion stays in the plane through the centre perpendicular to it. At the two apses the velocity is perpendicular to the radius, so \(\abs{\vect{L}}=m r v\) there and the speeds are in the inverse ratio of the distances,
which for the eccentricity quoted above is \(1.03407\). That figure is a prediction computed from \(e\), not a second measurement: the independent test of the areal law, and of the inverse-square force behind the Earth's particular value of \(e\), belongs to Central Forces and Statics. The derivation assumed only that the force points along \(\vect{x}\); it says nothing at all about how the force depends on distance, which is why the areal law survived every revision of the force law between Kepler and Newton.
∎Torque
Analogously to Newton's second law in the translational case, we introduce as a fundamental postulate:
The sum of all the interactions upon a body at a given point is the measure of how much its angular momentum fails to be conserved. Rests on Definition 19.18 and Postulate 19.8.
We shall call the total torque, or moment of force, upon a body the sum of all the forces applied at a given point, and we shall denote it by \(\vect{\tau}\). Newton's second law in the rotational case is then expressed as
with units of \(\mathrm{N}\,\mathrm{m}=\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}^{2}\), dimensionally those of an energy.
Substituting the definition of angular momentum into Equation (19.14), we have
where the last step uses Equation (19.2); and since the velocity of a particle is always parallel to its linear momentum, we find the relation
This is a very important relation, for it yields the following conclusion: the torque is a quantity that manifests the total force applied at the place where the particle is. We can use this equation to describe the motion of a body when we apply a force at a specific point of it, as in the example of the door: it costs us more or less to move it depending on where we apply the force.
The same force applied to a door opens it easily at the handle and scarcely at all near the hinge, and a rigid bar pivoted at a point balances two weights when, and only when, the products of weight and distance from the pivot agree on the two sides — Archimedes' law of the lever, the oldest quantitative law in mechanics and the principle of every balance, crowbar and wheelbarrow. What governs rotation is therefore not the force alone but the force together with its point of application, and specifically the product of the force with the perpendicular distance from the axis to its line of action: a force directed straight at the axis has no such distance and never turns anything, which is why pulling a door towards its hinge is futile [Newton:1687]. Rests on Equations (19.14) and (19.15).
Derivation. Derives Phenomenon 19.23. Equation (19.15) gives \(\vect{\tau}=\vect{x}\times\vect{F}\). Let \(\vect{x}\) be the position of the point of application relative to the axis and \(\theta\) the angle between \(\vect{x}\) and \(\vect{F}\). The magnitude of the torque is then
where \(d\) is the perpendicular distance from the axis to the line of action — the lever arm. It grows in proportion to the distance at which the force is applied, which is the door; and it vanishes when \(\vect{F}\) is parallel or antiparallel to \(\vect{x}\), so that \(\sin\theta=0\), which is the futile pull towards the hinge. For the balanced bar, the angular momentum about the pivot does not change, so by Equation (19.14) the torques must cancel; with the two weights applied on opposite sides and perpendicular to the bar this reads \(d_1 F_1=d_2 F_2\), the law of the lever. Note that the derivation used Equation (19.15), which was itself obtained by discarding \(\vect{v}\times\vect{p}\); that term vanishes identically, so nothing was approximated.
∎Covariance
Theorem 19.12 settled the translational law. The rotational law raises a question the translational one does not, because \(\vect{L}\) and \(\vect{\tau}\) are both defined relative to a chosen point, and a change of frame moves that point.
The argument below uses the fact that the cross product of two vectors transforms under an orthogonal change of Cartesian axes \(a_{ij}\) as
that is, as a vector for a rotation and with an extra sign for a reflection — the reason \(\vect{L}\) and \(\vect{\tau}\) are called axial, or pseudo-, vectors. Differentiable Manifolds, Tensors, and Curvature defines the cross product in the index form \(\left(\vect{u}\times\vect{w}\right)_{i} =\varepsilon_{ijk}u_{j}w_{k}\) of Equation (13.225) and names it a pseudo-vector, but does not state Equation (19.17); the statement belongs there rather than here. It follows in one line from the Leibniz expansion Equation (5.19) of a determinant, which gives \(\varepsilon_{lmn}\det a=\varepsilon_{ijk}a_{il}a_{jm}a_{kn}\), combined with the transformation rule Equation (13.21) for Cartesian tensors. For \(R\in\SO(3)\), where \(\det R=+1\), Equation (19.17) reads \(\left(R\vect{u}\right)\times\left(R\vect{w}\right) =R\left(\vect{u}\times\vect{w}\right)\), and that is the form used below.
Let a particle of constant mass be subject to a force transforming as \(\vect{F}\,'=R\vect{F}\) under Equation (18.11), and let \(\vect{L}\,'\) and \(\vect{\tau}\,'\) be formed from the primed position and momentum, that is, about the origin of \(K'\). Then Equation (19.14) holds in \(K\) if and only if \(\vect{\tau}\,'=\dd\vect{L}\,'/\dd t\,'\) holds in \(K'\), even though neither \(\vect{L}\) nor \(\vect{\tau}\) is invariant. Rests on Equation (19.14), Definition 18.18 and Theorem 19.12.
Derives Theorem 19.25. Write \(\vect{c}(t)=\vect{V}t+\vect{b}\) for the displacement of the origin of \(K'\), so that Equations (18.11) and (19.6) read \(\vect{x}\,'=R\vect{x}-\vect{c}\) and \(\vect{v}\,'=R\vect{v}-\vect{V}\) with \(\dd\vect{c}/\dd t=\vect{V}\). Then
using Equation (19.15) in the last line. For the angular momentum, \(\vect{p}\,'=m\vect{v}\,'\) by Equation (19.1) and Definition 18.18, so
Differentiating with respect to \(t\), which by Postulate 18.14 is the same as differentiating with respect to \(t\,'\), and using \(\dd\vect{c}/\dd t=\vect{V}\) together with \(\vect{V}\times\vect{V}=\vect{0}\),
because the two middle terms are equal and opposite by the antisymmetry of the cross product, and \(m\,\dd\vect{v}/\dd t=\vect{F}\) by Equation (19.4). The two computations agree, so \(\dd\vect{L}\,'/\dd t\,'=\vect{\tau}\,'\); and running the argument with \(R^{-1}\), \(-R^{-1}\vect{V}\) and \(-R^{-1}\vect{b}\) gives the converse.
∎The displayed expression for \(\vect{L}\,'\) shows that angular momentum is frame-dependent in a way linear momentum is not: it depends on where the origin is put, on how fast that origin moves, and — through the term \(m\,\vect{c}\times R\vect{v}\) — on the time. A body at rest at \(\vect{x}=\vect{0}\) has \(\vect{L}=\vect{0}\) in \(K\) and a nonzero, time-varying \(\vect{L}\,'\) in any frame whose origin is elsewhere and moving. What survives the change of frame is the balance Equation (19.14) between the two sides, which is all a law of motion is ever asked to do.
The proof of Theorem 19.25 was written for \(R\in\SO(3)\), but it does not need \(\det R=+1\). For a reflection, Equation (19.17) carries an extra minus sign, so the same two computations return \(\vect{\tau}\,'=-R\vect{\tau} -\vect{c}\times R\vect{F}\) and \(\dd\vect{L}\,'/\dd t =-R\vect{\tau}-\vect{c}\times R\vect{F}\): the sign appears on both sides and cancels between them, and the law is form-invariant under the full orthogonal group \(\Ogrp(3)\). Newtonian mechanics is therefore exactly parity-symmetric, and the mirror image of a mechanical process is again a mechanical process. That this is a property of mechanics rather than of nature was established only in 1957, when the beta decay of oriented cobalt-60 nuclei was found to distinguish left from right [Wu:1957]; the experiment is reported in Experiment: Parity Violation.
Theorems 19.12 and 19.25 assert that no experiment can detect the orientation, position, epoch or uniform velocity of a laboratory. Each of the ten parameters of Equation (18.11) therefore carries an experimental bound, and the bounds are severe: the compilation of limits on violations of rotation and boost invariance [Kostelecky:2011] collects them across atomic, nuclear and particle physics. They are quoted there for the Lorentz group rather than the Galilei group, which is the correct modern setting; the Galilean statements of this chapter inherit whatever bound applies at speeds where Equation (18.8) is negligible.
Energy
Kinetic energy
The kinetic energy of a particle moving with velocity \(\vect{v}\) with respect to a frame of reference \(K\) is
with units of \(\mathrm{J}\). Rests on Definitions 18.10 and 19.3.
Potential energy and conservative forces
Consider a force \(\vect{F}_V\) upon a particle. This force is conservative if and only if there exists a real function \(V\), which we shall call the potential energy, such that
Rests on Postulate 19.8 and Equation (19.2).
Let us see that if \(\vect{F}\) is conservative one has
and therefore
We may consider that a free particle is subject to a constant potential energy \(V_0\), since the force \(\vect{F}_V\) upon it would still be zero.
On occasion, instead of saying that a particle is subject to a conservative force \(\vect{F}_V\), one says that it is subject to a potential \(V\).
Energy
Consider a particle subject to a potential \(V\). The total energy of the particle is
The law of conservation of energy
The source manuscript reserves this heading with the instruction to demonstrate that \(\dd E/\dd t=0\) for every inertial frame. Carried out, the instruction turns out to be right for a closed system and wrong for a single particle in an externally imposed field, and the difference is worth stating precisely.
Let the total force upon a particle derive from a potential \(V\) that carries no explicit dependence on time. Then the energy \(E=T+V\) satisfies
with respect to the inertial frame in which \(V\) is time-independent, but not in general with respect to any other. If instead the system is a closed one of \(N\) particles of constant masses, whose internal forces derive from a potential \(U\) depending only on the mutual distances \(\abs{\vect{x}_{s}-\vect{x}_{r}}\), then Equation (19.22) holds for \(E=\sum_{s}\tfrac{1}{2}m_{s}\abs{\vect{v}_{s}}^{2}+U\) with respect to every inertial frame. Rests on Definition 19.31, Equation (19.19) and Equation (19.4).
Derives Theorem 19.32. One particle, one frame. Differentiating Equation (19.18) along the motion and using Equation (19.4), \(\dd T/\dd t=\vect{F}\cdot\vect{v}\); and since \(V\) carries no explicit time dependence, the chain rule gives \(\dd V/\dd t=\vect{\nabla}V\cdot\vect{v}=-\vect{F}\cdot\vect{v}\) by Equation (19.19). The two cancel, which is Equation (19.22).
One particle, another frame. Boost to a frame \(K'\) moving with constant velocity \(\vect{V}\), so that \(\vect{x}=\vect{x}\,'+\vect{V}t\) and \(\vect{v}\,'=\vect{v}-\vect{V}\) by Equation (19.6). The potential seen there is \(V\left(\vect{x}\,'+\vect{V}t\right)\), which does depend explicitly on the time, and
which vanishes only when the force is perpendicular to the boost. The reading is physical rather than paradoxical: in \(K'\) the agent maintaining the external field is itself in motion, and it does work at the rate \(-\vect{F}\cdot\vect{V}\). An external static potential is always an idealization of a second body held fixed, and holding it fixed is what the second frame sees being paid for.
The closed system. Here there is no external agent. Under the general transformation Equation (18.11) the mutual distances are unchanged, since \(\vect{x}'_{s}-\vect{x}'_{r}=R\left(\vect{x}_{s}-\vect{x}_{r}\right)\) and \(R\) preserves lengths, so \(U'=U\). The kinetic energy transforms by Equation (19.6) as
with \(\vect{P}=\sum_{s}\vect{p}_{s}\) the total momentum and \(M\) the total mass Equation (19.32). Hence \(E'=E-\vect{V}\cdot R\vect{P}+\tfrac{1}{2}MV^{2}\). Each of the three terms on the right is constant in time — \(E\) by the first part of this proof applied to the system, \(\vect{P}\) by Theorem 19.67, and \(M\) and \(\vect{V}\) by hypothesis — so \(E'\) is constant too. Note what this does not say: \(E'\neq E\), and the two frames disagree about how much energy the system has. They agree only that the number does not change.
∎A pendulum bob released from rest at a given height rises on the far side to that same height, and continues to do so when a nail placed beneath the point of suspension catches the cord in mid-swing and forces the bob onto a shorter arc of a different radius: what the descent confers is fixed by the vertical drop alone and not by the path taken. Galileo reports the interrupted-pendulum experiment for exactly this purpose [Galilei:1638]. The same law governs a ball rolling in a bowl, a mass bouncing on a spring, and a projectile exchanging height for speed; and it fails visibly, as a decaying amplitude, precisely when friction is present — that is, precisely when the total force does not derive from a potential. Rests on Theorem 19.32 and Equation (19.19).
Derivation. Derives Phenomenon 19.33. Let the total force on a particle of constant mass derive from a potential that does not depend explicitly on time, \(\vect{F}=-\vect{\nabla}V\) as in Equation (19.19). Differentiating the kinetic energy Equation (19.18) along the motion and using Equation (19.4),
while the potential energy evaluated along the same motion changes, by the chain rule, at the rate
Adding the two, the energy Equation (19.21) obeys \(\dv{E}{t}=0\), which is Equation (19.22). For the pendulum, \(V=mgh\) and \(T=\tfrac{1}{2}mv^2\), so a bob released from rest at height \(h_0\) carries \(E=mgh_0\) throughout and can be momentarily at rest only where \(h=h_0\) — whatever the shape of the arc it has travelled, which is the interrupted-pendulum observation. Two limitations are worth recording. The frame-dependence is real and is settled in Theorem 19.32: for the bob taken alone, in a frame moving horizontally, the energy is not constant by Equation (19.23), and constancy is recovered only when the Earth is included and the system is closed. And the argument fails as soon as \(V\) depends explicitly on time or the force has a non-conservative part, which is why Equation (19.20) is the criterion to test before invoking the result — the decaying amplitude of a real pendulum is that criterion failing.
∎Work
The work performed by a particle along a curve \(C\), whose trajectory is due to a total force \(\vect{F}\), is
with units of \(\mathrm{J}\). Rests on Postulate 19.8 and Definition 18.3.
Let us see that if \(\vect{F}\) is conservative and \(C\) is a closed curve, one has
Therefore, the work performed by a particle under a conservative total force along a closed curve is null.
The differential of work performed by a particle in a displacement \(\dd\vect{x}\), due to a force \(\vect{F}\), is given by
Now, if the total force upon the particle comes from a potential \(V\), one has
and, integrating, we see that
Power
The power is the rate at which work is performed,
with units of \(\mathrm{W}=\mathrm{J}/\mathrm{s}\). Rests on Definition 19.34.
Work for the magnitude of a torque
The formulary asserts, without derivation, that the differential of work associated with the magnitude of a torque \(\tau\) through an angle \(\phi\) is
The relation is true under a hypothesis the formulary leaves unsaid, and stating it is the whole of the derivation.
Let a body turn by an infinitesimal rotation \(\dd\vect{\phi} =\hat{\vect{n}}\,\dd\phi\) about an axis \(\hat{\vect{n}}\) through the origin, and let a force \(\vect{F}\) be applied at the point \(\vect{x}\). Then the work done is
in \(\mathrm{J}\), with \(\vect{\tau}\) the torque Equation (19.15) about the origin. The scalar form Equation (19.28) is the special case in which the torque is parallel to the axis of rotation. Rests on Definition 19.34 and Equation (19.15).
Derives Proposition 19.36. Under a rotation by \(\dd\vect{\phi}\) about an axis through the origin the point of application moves by \(\dd\vect{x}=\dd\vect{\phi}\times\vect{x}\). This is the generator of \(\mathfrak{so}(3)\) acting on a vector: with the basis Equation (14.32) of Proposition 14.27, whose entries are \(\left(L_{k}\right)_{ij}=-\epsilon_{kij}\), one has \(\left(\dd\phi^{k}L_{k}\vect{x}\right)_{i} =\epsilon_{ikj}\dd\phi^{k}x_{j} =\left(\dd\vect{\phi}\times\vect{x}\right)_{i}\). Then, by Definition 19.34,
the third line by the cyclic invariance of the scalar triple product, which in the index notation Equation (13.225) is the cyclic symmetry of \(\varepsilon_{ijk}\), and the fourth by Equation (19.15). Writing \(\dd\vect{\phi}=\hat{\vect{n}}\,\dd\phi\) gives Equation (19.29). The component of \(\vect{\tau}\) perpendicular to \(\hat{\vect{n}}\) does no work: it is taken up by whatever constrains the axis, which is why a door hinge does not absorb energy while the door swings. Consequently Equation (19.28) as the formulary states it holds when, and only when, \(\vect{\tau}\) is parallel to \(\hat{\vect{n}}\) — the case of a body free to turn about a single fixed axis with the force applied in the plane perpendicular to it.
∎Dividing Equation (19.29) by \(\dd t\) and using Equation (19.27),
with \(\vect{\omega}=\dd\vect{\phi}/\dd t\) the angular velocity in \(\mathrm{rad}/\mathrm{s}\) and \(P\) in \(\mathrm{W}\). This is the relation by which the shaft power of an engine is computed from a dynamometer reading of the torque and a tachometer reading of the rotation rate. Rests on Proposition 19.36 and Equation (19.27).
The work–energy theorem
The work performed by a particle equals the variation of its kinetic energy,
Rests on Definition 19.34, Equation (19.4) and Definition 19.29.
Derivation. Derives Theorem 19.38. Consider a particle under a total force \(\vect{F}\) that performs a work \(W\). From the definition of work and Equation (19.4),
Therefore, from Equation (19.18), \(W=\Delta T\).
∎This result is known as the work–energy theorem, or the theorem of the vires vivae (living forces).
Systems of particles
In the preceding sections we introduced all the fundamental principles that govern the dynamics of a particle and of a body. We began with particles because Newton's laws take their simplest form there. We did not do the same with bodies of definite volume, since in essence it is not necessary: to speak of particles is not more general, but it is more canonical.
We shall now introduce the concepts, and extend the definitions in the appropriate manner, for systems of many particles and bodies.
Inertia and mass of a system
Body as a system of particles
A body is a set of \(N\) particles occupying a region \(V\) of space. This region of space is called the volume of the body. Rests on Definitions 18.4 and 19.3.
Total mass of the system
Consider a closed system of \(N\) particles of masses \(m_s\), with \(s=1,\ldots,N\).
The total mass of the system is simply the sum of all the masses,
Because we have considered the system to be closed, \(M\) is constant.
![[figure: mass.pdf]](/images/treatise/de/figure/mass.svg)
A closed system of \(N\) particles: the total mass is the sum of the individual masses.
Moment of inertia of the system
Definition 19.3 makes the mass the measure of how much a particle resists being translated. A body resists being turned as well, and by an amount that depends on how its mass is distributed about the axis of the turn: this is the everyday fact that a long pole is harder to spin about its middle than a compact one of the same weight.
Let a system of \(N\) particles of masses \(m_{s}\) have positions \(\vect{x}_{s}\) measured from a point \(O\), and let \(\hat{\vect{n}}\) be a unit vector through \(O\). The moment of inertia of the system about the axis \((O,\hat{\vect{n}})\) is
with \(d_{s}\) the perpendicular distance from the particle to the axis, of SI unit \(\mathrm{kg}\,\mathrm{m}^{2}\). For a body of mass density \(\rho\) the sum is replaced by \(I=\int_{V}\rho\,d^{2}\,\dd V\). Rests on Definitions 19.3 and 19.39.
The dependence on the axis is quadratic and can be separated out.
The inertia tensor of the system about the point \(O\) is the array
of SI unit \(\mathrm{kg}\,\mathrm{m}^{2}\). Rests on Definitions 13.5 and 19.41.
For every unit vector \(\hat{\vect{n}}\),
and \(I_{ij}\) is symmetric, is a Cartesian tensor of rank two in the sense of Definition 13.5, and is positive semidefinite, vanishing on \(\hat{\vect{n}}\) only when every particle of the system lies on the axis. Rests on Definitions 13.5 and 19.42.
Derives Proposition 19.43. Since \(\hat{\vect{n}}\) is a unit vector, \(d_{s}^{2}=\abs{\hat{\vect{n}}\times\vect{x}_{s}}^{2} =\abs{\vect{x}_{s}}^{2} -\left(\hat{\vect{n}}\cdot\vect{x}_{s}\right)^{2}\), which is Pythagoras' theorem applied to the decomposition of \(\vect{x}_{s}\) into its components along and across the axis. Substituting into Equation (19.33) and writing \(\abs{\vect{x}_{s}}^{2}=\hat{n}_{i}\delta_{ij}\hat{n}_{j} \abs{\vect{x}_{s}}^{2}\) and \(\left(\hat{\vect{n}}\cdot\vect{x}_{s}\right)^{2} =\hat{n}_{i}x_{s,i}x_{s,j}\hat{n}_{j}\) gives Equation (19.35) with \(I_{ij}\) as in Equation (19.34). Symmetry is manifest. Each summand of Equation (19.34) is built from \(\delta_{ij}\), which is invariant, and from the outer product \(x_{s,i}x_{s,j}\) of a vector with itself, which obeys Equation (13.22) because each factor obeys the vector rule; the finite sum of tensors is a tensor, so \(I_{ij}\) satisfies Definition 13.5. Finally Equation (19.33) exhibits \(I\) as a sum of terms \(m_{s}d_{s}^{2}\geq0\), so \(I\geq0\), with equality precisely when every \(d_{s}=0\).
∎Let \(I_{\text{CM}}\) be the moment of inertia of the system about an axis of direction \(\hat{\vect{n}}\) through its centre of mass (Definition 19.47, below), and let \(I\) be the moment of inertia about the parallel axis of the same direction through a point at perpendicular distance \(D\) from it. Then
with \(M\) the total mass Equation (19.32). Rests on Definition 19.41, Equation (19.38) and Equation (19.32).
Derives Theorem 19.44. Take the origin at the centre of mass and let the second axis pass through the point \(\vect{D}\), chosen perpendicular to \(\hat{\vect{n}}\) without loss of generality, so that \(\abs{\vect{D}}=D\). Write \(\vect{x}_{s}^{\perp}\) for the component of \(\vect{x}_{s}\) perpendicular to \(\hat{\vect{n}}\), so that \(d_{s}=\abs{\vect{x}_{s}^{\perp}}\) for the first axis and \(\abs{\vect{x}_{s}^{\perp}-\vect{D}}\) for the second. Then
because the middle sum is \(M\) times the perpendicular component of the position of the centre of mass Equation (19.38), which vanishes when the origin is the centre of mass itself. The theorem therefore singles the centre of mass out: among all parallel axes of a given direction, the one through the centre of mass has the smallest moment of inertia, and Equation (19.36) says by how much any other exceeds it.
∎For a homogeneous sphere of mass \(M\) and radius \(R\) the symmetry forces \(I_{ij}=I\delta_{ij}\), so every axis through the centre gives the same moment of inertia and it may be extracted from the trace of Equation (19.34), which is \(I_{ii}=2\int_{V}\rho\abs{\vect{x}}^{2}\dd V=3I\). With \(\rho=3M/\left(4\pi R^{3}\right)\) constant,
whence
A measured moment of inertia smaller than \(\tfrac{2}{5}MR^{2}\) is therefore direct evidence that a body's mass is concentrated towards its centre, which is how the internal structure of a planet is read off its precession. Rests on Definition 19.42 and Equation (19.33).
Because \(I_{ij}\) is real and symmetric, there is an orthonormal basis of \(\R^{3}\) in which it is diagonal, \(I_{ij} =\diag\left(I_{1},I_{2},I_{3}\right)\) with \(I_{k}\geq0\); the basis directions are the principal axes and the \(I_{k}\) the principal moments. Theorem 5.75 proves the spectral theorem for a self-adjoint operator on a complex inner-product space, which gives the reality of the \(I_{k}\) at once; the statement actually used here is its real form, that a real symmetric matrix is diagonalized by a real orthogonal change of basis, and Part II does not state it separately. It is a standard corollary — the eigenvalues being real, the eigenvectors may be taken real — but it is mathematics and belongs in Linear Algebra and Representation Theory, not here. Everything that follows from the principal axes, from Euler's equations to the free precession of a symmetric top, is the subject of Rigid Bodies and Rotating Frames [Euler:1765] [Poinsot:1834].
Kinematics of the centre of mass
Consider the preceding system, in which the positions of the particles are \(\vect{x}_s\). How can we assign to the system a quantity that allows us to treat it as a single particle? We can define a point of space that takes an average of the positions, according to the amount of mass assigned to each particle.
The position of the centre of mass of a system of particles is
The centre of mass may itself be regarded as a point of space to which the mass \(M\) is assigned. By definition, the velocity of the centre of mass, with respect to a frame of reference in which its position is \(\vect{x}_{\text{CM}}\), is given by
and, in the same fashion, the acceleration of the centre of mass is given by
Internal dynamics of the system
Newton's third law
Let us note the following fact of nature. When we push a wall with a given force, directed towards it, it is as if the wall pushed us back with the same force, but in the opposite direction.
Considering again the aforementioned system of particles: what about the interaction among them? Well, since the particles can interact, one of them feels a total force due to the others, and it exerts a force upon the others. What is the relation between these forces? To give an answer, let us take two particles of the system, which we shall call particle 1 and particle 2.
We shall denote by \(\vect{F}_{12}\) the force felt by particle 1 due to particle 2, and by \(\vect{F}_{21}\) the force felt by particle 2 due to particle 1.
On the basis of the philosophical conclusions about action and reaction, we may intuit that the magnitudes of these forces must be equal; but we must in addition specify an orientation for these vectors. In any case, owing to the interaction between these two particles they approach or recede from each other; and we may also intuit that this trajectory of approach or recession is in principle a straight line, so that these forces would be parallel.
We then enunciate as a fundamental principle:
The force that one particle exerts upon another is equal to the force that the latter exerts upon the former, but with opposite sign,
Rests on Postulate 19.8 and Definition 19.2.
This is known by the name of Newton's third law, or the principle of action and reaction.
![[figure: actrec.pdf]](/images/treatise/de/figure/actrec.svg)
![[figure: actrec2.pdf]](/images/treatise/de/figure/actrec2.svg)
Action and reaction: the force one particle exerts upon another is equal and opposite to the force it receives from it.
Net internal force
There are two types of forces to which our system of particles is subject: the internal ones, \(\vect{F}_{\text{INT}}\), which are the interactions among the particles; and the external ones, \(\vect{F}_{\text{EXT}}\), which are those alien to the system — they affect it as a whole, not each particle separately. An example of this can be a gas of molecules inside a balloon: the particles interact with one another repulsively, forming the gas, while an external force could be the action of gravity upon the balloon. We shall assume:
The internal interactions of the system do not contribute to its motion as a whole,
Rests on Postulate 19.48.
since for the most part they cancel by Newton's third law. In our example applied to nature, this amounts to the following: the molecular interactions are of no great importance for the motion of the balloon as such.
When two bodies interact and nothing external acts on the pair, the vector sum of their linear momenta is the same after the interaction as before. A gun recoils as the shot leaves it; a swimmer moves forward by driving water backward; a rocket accelerates in vacuum with nothing to push against. In a collision on an air track the sum \(m_1\vect{v}_1+m_2\vect{v}_2\) is found unchanged whether the bodies rebound or stick together, and whether or not the kinetic energy is conserved — which it usually is not. Newton records the result as the third corollary to the laws of motion [Newton:1687]. Rests on Postulate 19.48 and Equation (19.2).
Derivation. Derives Phenomenon 19.50. Take two particles subject only to their mutual interaction. Applying Equation (19.2) to each,
and adding the two, Newton's third law Equation (19.41) makes the right-hand side vanish identically:
The total momentum is therefore constant throughout the interaction, however violent and however brief, and — this is the strength of the result — no property of the interaction law enters, so the same conclusion covers collisions whose forces are entirely unknown. For \(N\) particles the internal contributions cancel in the same way, pair by pair, which is the content of Postulate 19.49. Energy is not similarly protected: Equation (19.43) holds for an inelastic collision, in which kinetic energy is demonstrably lost to heat and deformation.
∎Translational Newton laws for systems of particles
Linear momentum of the centre of mass
Consider a (closed) system of particles, in which the \(s\)-th particle has position \(\vect{x}_s\) and linear momentum \(\vect{p}_s\).
The linear momentum of the centre of mass is
Completeness
Let us compute:
so that, by Equation (19.44),
Restricting to the case in which the mass of each particle is constant, we obtain the following completeness of the total linear momentum of the system:
that is, the sum of the linear momenta of all the particles of the system is equal to the linear momentum of the centre of mass.
Translational Newton's second law for the centre of mass
Let us consider the centre of mass as a particle. Denote the total force upon our system of particles by \(\vect{F}\). Recalling the discussion of the net internal force, one has
and, as we have assumed, \(\vect{F}_{\text{INT}}=\vect{0}\). Thus, applying Newton's second law to our centre of mass, we arrive at the expression
and, since \(M\) is constant, one has
The source flags at this point, without carrying it out, the demonstration that the interaction of the centre of mass with the remaining particles cancels, so that Newton's second law may legitimately be applied to it. That demonstration is the following, and its by-product is that Postulate 19.49 need not have been postulated at all.
Let a closed system of \(N\) particles of constant masses have internal forces obeying Equation (19.41). Then the net internal force vanishes,
and consequently Equation (19.46) holds: the centre of mass accelerates as a single particle of mass \(M\) acted on by the total external force alone. Rests on Postulate 19.48, Equation (19.45) and Equation (19.44).
Derives Theorem 19.52. Group the double sum in Equation (19.47) into unordered pairs. The pair \(\set{s,r}\) contributes \(\vect{F}_{sr}+\vect{F}_{rs}\), which is \(\vect{0}\) by Equation (19.41); every pair occurs exactly once, so the whole sum vanishes. Now differentiate the completeness relation Equation (19.45) and apply Equation (19.2) to each particle:
and since \(M\) is constant, \(\dd\vect{p}_{\text{CM}}/\dd t=M\vect{a}_{\text{CM}}\) by Equations (19.40) and (19.44), which is Equation (19.46).
∎Equation (19.47) is exactly the statement Postulate 19.49 assumes, so that assumption may be discharged: for a system whose particles interact in pairs obeying the third law, the vanishing of the net internal force is a consequence, not an independent principle. The hedge in the discussion following Postulate 19.49 — “for the most part they cancel” — is therefore too weak, and can be strengthened to an exact cancellation. What it does require is that the forces be pair forces: a genuine three-body interaction, one not expressible as a sum over pairs, is not covered by Equation (19.41) and would have to be examined on its own. No such force is known among the fundamental interactions.
A body that flies apart in mid-air does not thereby alter the path of its centre of mass: the fragments of a bursting firework, and the two halves of a rod that breaks in flight, continue to straddle the same smooth arc their common centre was already describing, and the arc shows no kink at the instant of the burst. A wrench tossed spinning across a room follows, in its centre of mass, exactly the path of a wrench that does not spin, however complicated the tumbling superposed on it. The observation is what licenses the everyday practice of treating an extended body as a point. Rests on Equation (19.46) and Postulate 19.49.
Derivation. Derives Phenomenon 19.54. By Equation (19.46) the acceleration of the centre of mass answers to \(\vect{F}_{\text{EXT}}\) alone; the internal forces have already dropped out through Postulate 19.49, cancelling pair by pair by Equation (19.41). An explosion, a rotation, or any other rearrangement of the parts is internal by definition, so it alters neither \(\vect{F}_{\text{EXT}}\) nor the total mass \(M\), and therefore leaves \(\vect{a}_{\text{CM}}\) untouched at the instant it occurs and afterwards. With the external force of a body near the ground, \(\vect{F}_{\text{EXT}}=M\vect{g}\), one has \(\vect{a}_{\text{CM}}=\vect{g}\) constant, so the centre of mass describes the parabola of Experiment: Free Fall and Projectile Motion whatever the fragments do individually. Two caveats belong with the result: the cancellation of the internal forces requires the third law in the form Equation (19.41), and \(M\) must be constant, which excludes bodies that expel matter altogether rather than merely redistributing it.
∎Rotational Newton laws for systems of particles
Angular momentum of the centre of mass
The angular momentum of the centre of mass is
Completeness
The translational case gave the clean identity Equation (19.45): the momentum of the centre of mass is the sum of the momenta of the particles, with nothing left over. The rotational case does not, and the difference is the single most consequential fact about the mechanics of extended bodies.
Throughout this subsection, a primed symbol denotes a quantity referred to the centre of mass:
The primes here have nothing to do with the change of inertial frame of Equation (18.11); the centre of mass need not move uniformly.
With the masses constant,
Derives Lemma 19.57. By Equations (19.38) and (19.49),
using Equation (19.32). Differentiating with respect to time, the masses being constant, gives the second identity.
∎The total angular momentum of the system about a fixed origin is
with \(\vect{L}_{\text{CM}}\) the angular momentum Equation (19.48) of the centre of mass and \(\vect{S}\) the internal, or spin, angular momentum, both of SI unit \(\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}\). Rests on Definition 19.55, Lemma 19.57 and Equation (19.44).
Derives Theorem 19.58. Substituting Equation (19.49) into the sum and expanding the four cross products,
The two middle sums vanish by Equation (19.50), and the first term is \(\vect{x}_{\text{CM}}\times M\vect{v}_{\text{CM}} =\vect{x}_{\text{CM}}\times\vect{p}_{\text{CM}} =\vect{L}_{\text{CM}}\) by Equations (19.44) and (19.48).
∎With the same notation,
Rests on Lemma 19.57 and Equation (19.18).
Derives Corollary 19.59. Expanding \(\abs{\vect{v}_{\text{CM}}+\vect{v}'_{s}}^{2}\) in \(T=\sum_{s}\tfrac{1}{2}m_{s}\abs{\vect{v}_{s}}^{2}\) gives three groups of terms; the cross term is \(\vect{v}_{\text{CM}}\cdot\sum_{s}\vect{p}'_{s}\), which vanishes by Equation (19.50).
∎Equation (19.51) is not the analogue of Equation (19.45): the angular momentum of the centre of mass is not the total angular momentum, and the defect \(\vect{S}\) does not vanish for any system that is turning. A flywheel spinning in place has \(\vect{L}_{\text{CM}}=\vect{0}\) and \(\vect{S}\neq\vect{0}\); the Earth in its orbit carries both, and they are independent — the one associated with the year, the other with the day. The reason the linear case is simpler is that translation has no internal counterpart: there is no way to redistribute the parts of a system so as to give it internal momentum, because Equation (19.50) forbids it, while \(\vect{x}'_{s}\times\vect{p}'_{s}\) can perfectly well sum to something nonzero. The two pieces obey separate equations of motion, which is the subject of the next heading, and it is that separation which lets an extended body be treated as a point for its translation while keeping a full rotational dynamics of its own (Rigid Bodies and Rotating Frames).
Rotational Newton's second law for the centre of mass
For the translational law it sufficed that the internal forces cancel in pairs, which is Postulate 19.49. For the rotational law that is not enough: two equal and opposite forces applied at different points cancel as forces but form a couple, and a couple turns things. What is needed in addition is that the pair force act along the line joining the two particles.
The force between two particles is directed along the line joining them,
in addition to satisfying Equation (19.41). Rests on Postulate 19.48 and Corollary 19.14.
Let the internal forces of a system satisfy Postulate 19.61 and let the masses be constant. Then, about any point at rest in an inertial frame,
where \(\vect{F}^{\text{EXT}}_{s}\) is the external force on the \(s\)-th particle. Only the external torques appear. Rests on Postulate 19.61, Equation (19.14) and Equation (19.2).
Derives Theorem 19.62. Split the force on particle \(s\) into external and internal parts, \(\vect{F}_{s}=\vect{F}^{\text{EXT}}_{s}+\sum_{r\neq s}\vect{F}_{sr}\), and differentiate the sum, the term \(\vect{v}_{s}\times\vect{p}_{s}\) vanishing for each \(s\) as in Proposition 19.20:
In the double sum group the terms in pairs. Using Equation (19.41) in the form \(\vect{F}_{rs}=-\vect{F}_{sr}\), the pair \((s,r)\) contributes
which vanishes precisely because Equation (19.53) makes the two factors parallel. Every pair vanishes, so the double sum does, leaving Equation (19.54).
∎Under the same hypotheses, and with \(\vect{S}\) the internal angular momentum of Equation (19.51),
and this holds even when the centre of mass is accelerating, so that its rest frame is not inertial. Rests on Theorem 19.58, Lemma 19.57 and Postulate 19.61.
Derives Theorem 19.63. Differentiating \(\vect{S}\) as defined in Equation (19.51), the term \(\vect{v}'_{s}\times m_{s}\vect{v}'_{s}\) vanishes for each \(s\), and by Equations (19.40) and (19.49)
using Equation (19.4) for the \(s\)-th particle. Hence
and the second term vanishes by Equation (19.50) — which is exactly why the accelerating centre of mass costs nothing here, and why no other point of the body enjoys the same privilege. The internal contributions to the first term cancel pair by pair as in Theorem 19.62, since \(\vect{x}'_{s}-\vect{x}'_{r}=\vect{x}_{s}-\vect{x}_{r}\) by Equation (19.49), leaving Equation (19.55).
∎Differentiating Equation (19.48) and using Equation (19.46),
so the centre of mass turns about a fixed origin exactly as a single particle of mass \(M\) at \(\vect{x}_{\text{CM}}\) would. Together with Equation (19.55) this splits the rotational dynamics of any system into an orbital equation and a spin equation that can be solved separately. Rests on Equations (19.46) and (19.48).
A skater turning on the spot spins faster the closer the arms are drawn to the axis, and slower as they are extended again, without any push from outside; a diver in the air completes a somersault by tucking; a falling cat that begins with no rotation at all can turn itself over and land on its feet, ending as it began with zero total internal angular momentum. In each case the body changes its own shape, hence its moment of inertia Equation (19.33), while nothing external exerts a torque about the centre of mass. Rests on Theorem 19.63.
Derivation. Derives Phenomenon 19.65. Gravity acting on a body near the ground contributes \(\sum_{s}\vect{x}'_{s}\times m_{s}\vect{g} =\left(\sum_{s}m_{s}\vect{x}'_{s}\right)\times\vect{g}=\vect{0}\) by Equation (19.50), so a uniform gravitational field exerts no torque about the centre of mass. With the skate pivot and the air resistance also contributing nothing appreciable, \(\vect{\tau}'_{\text{EXT}}=\vect{0}\) and Equation (19.55) gives \(\vect{S}\) constant. For a body turning about a fixed principal axis (Remark 19.46) the component of \(\vect{S}\) along the axis is \(I\omega\) with \(I\) the moment of inertia Equation (19.33) about it, so
and drawing the arms in, which lowers \(I\), raises \(\omega\) in the same proportion. Note what is not conserved: by Equation (19.52) the rotational kinetic energy is \(\tfrac{1}{2}I\omega^{2}=\tfrac{1}{2}S^{2}/I\), which rises as \(I\) falls — the skater's muscles do the work, and the energy budget is what distinguishes this from a free body. The cat is subtler, since its axis is not fixed and \(I_{ij}\) changes shape as well as scale; \(\vect{S}=\vect{0}\) throughout, and the reorientation is possible because the body is not rigid. Both cases are treated properly in Rigid Bodies and Rotating Frames [Euler:1765].
∎Postulate 19.61 holds for gravitation, for the Coulomb force and for elastic contact — for every force of the central kind of Corollary 19.14 — and for those Equation (19.54) is exact. It fails for the magnetic force between two moving charges, which is not directed along the line joining them and for which even the weak form Equation (19.41) can fail instantaneously. The angular momentum of the two charges alone is then not conserved; what is conserved is the total including the angular momentum stored in the electromagnetic field, as The Maxwell Equations shows. This is the characteristic pattern of every later repair to the Newtonian scheme: the conservation law is not abandoned but enlarged, by finding the carrier of what appeared to have been lost.
Conservation laws
Everything a closed system does leaves certain numbers unchanged. There are two ways to see which numbers those are, and this book uses both because they have different reach.
The first is direct: apply the laws already postulated and see what their time derivative turns out to be. This route asks nothing of the forces beyond Equation (19.41), so it survives friction, plastic deformation and every other dissipative process, and it is what justifies the everyday use of momentum conservation in collision problems where the interaction is entirely unknown.
The second is by symmetry. Each of the ten parameters of the Galilei group Equation (18.11) that leaves a system's description unchanged yields one conserved quantity, by Noether's first theorem Theorem 16.84 [Noether:1918]. That route explains why these particular quantities and no others, but it requires the dynamics to come from an action, which excludes the dissipative cases the first route covers.
The three laws, directly
For a closed system of constant masses subject to no external force, the total momentum
of SI unit \(\mathrm{kg}\,\mathrm{m}/\mathrm{s}\), is constant in time. Rests on Theorem 19.52 and Equation (19.45).
Derives Theorem 19.67. By Equation (19.45) the sum is \(\vect{p}_{\text{CM}}\), and by Theorem 19.52 \(\dd\vect{P}/\dd t=\vect{F}_{\text{EXT}}\), which vanishes by hypothesis. Nothing was assumed about the internal forces except that they obey Equation (19.41); in particular they may be dissipative, which is why the total momentum is conserved in a perfectly inelastic collision, as recorded in Phenomenon 19.50, while the kinetic energy is not.
∎For a closed system of constant masses whose internal forces satisfy Postulate 19.61, and on which the external torque about a fixed point of an inertial frame vanishes, the total angular momentum about that point is constant. If the external torque about the centre of mass vanishes, the spin \(\vect{S}\) is constant separately. Rests on Theorems 19.62 and 19.63.
Derives Theorem 19.68. Immediate from Equation (19.54) and Equation (19.55) with the respective right-hand sides set to zero. The two statements are independent, because Equation (19.51) splits the total into two pieces obeying separate equations: a body may conserve its spin while its orbital angular momentum changes, which is the case of a spinning projectile in flight.
∎For a closed system of constant masses subject to no external force, the quantity
of SI unit \(\mathrm{kg}\,\mathrm{m}\), is constant in time. Equivalently, the centre of mass moves in a straight line with constant velocity. Rests on Theorem 19.67 and Equation (19.44).
Derives Theorem 19.69. Differentiating Equation (19.59),
using \(\dd\vect{P}/\dd t=\vect{0}\) from Theorem 19.67 and \(\vect{P}=M\vect{v}_{\text{CM}}\) from Equation (19.44). Solving Equation (19.59) for \(\vect{x}_{\text{CM}}\) gives \(\vect{x}_{\text{CM}}(t)=\vect{x}_{\text{CM}}(0) +\vect{v}_{\text{CM}}t\), the second form of the statement.
∎Theorem 19.69 looks like a restatement of Theorem 19.67 and is often passed over as one, but it is a separate conserved quantity: \(\vect{G}\) depends explicitly on the time, and knowing \(\vect{P}\) does not fix the constant \(\vect{x}_{\text{CM}}(0)\) that \(\vect{G}\) encodes. The three components of \(\vect{G}\) are what the three boost parameters of Equation (18.11) contribute, and without them the count in Table 19.1 would come to seven rather than ten. It is also the one of the four whose Noether derivation is not routine, for the reason given in Remark 19.72.
Energy is the remaining kind, and it was settled in Theorem 19.32: for a closed system whose internal forces derive from a potential in the mutual distances, \(E\) is constant with respect to every inertial frame, though it takes different values in different ones by Equation (19.24). Unlike the three above, it requires the internal forces to be conservative — exactly the condition Equation (19.20) tests.
| Symmetry | Parameters | Conserved quantity | SI unit |
|---|---|---|---|
| Time translation | 1 | Energy $E$ | \(\mathrm{J}\) |
| Space translation | 3 | Momentum $\vect{P}$ | \(\mathrm{kg}\,\mathrm{m}/\mathrm{s}\) |
| Rotation | 3 | Angular momentum $\vect{L}$ | \(\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}\) |
| Boost | 3 | $\vect{G}=\vect{P}t-M\vect{x}_{\text{CM}}$ | \(\mathrm{kg}\,\mathrm{m}\) |
The same laws from symmetry
Calculus of Variations proves both of Noether's theorems, and the first of them (Theorem 16.84) is stated exactly in the form needed here: one independent variable, several dependent ones. Its two standard corollaries are already the first two entries of Table 19.1 — Corollary 16.87 for energy from invariance under time translation, and Corollary 16.88 for momentum and angular momentum from invariance under space translation and rotation. Nothing of that needs repeating; what belongs here is the identification of the mechanical system to which it applies, and the one case the corollaries do not cover.
Let a closed system of \(N\) particles of constant masses have internal forces derived from a potential \(U\) depending only on the mutual distances. Then the equations of motion Equation (19.4) for all \(N\) particles are the Euler–Lagrange equations of the action Equation (16.86) built from
in \(\mathrm{J}\), and \(\Lag\) is invariant under all ten transformations Equation (18.11) up to a total time derivative. Rests on Theorem 16.97, Equation (19.4) and Definition 19.30.
Derives Proposition 19.71. Theorem 16.97 establishes the equivalence for one particle in a potential; the \(N\)-particle statement is the same computation carried out for each of the \(3N\) coordinates in turn, the Euler–Lagrange system Equation (16.30) being one equation per dependent variable. For the invariance: the kinetic term is a sum of squared speeds, unchanged by a rotation and by a time or space translation, and \(U\) depends only on the distances \(\abs{\vect{x}_{s}-\vect{x}_{r}}\), which Proposition 19.13 showed to be invariant under the whole group. The one transformation that does not leave \(\Lag\) strictly unchanged is the boost, and it changes it by a total derivative, as Remark 19.72 computes.
∎Under the one-parameter family of boosts \(\vect{x}_{s}\longmapsto\vect{x}_{s}+\varepsilon\vect{u}\,t\), with \(\vect{u}\) a fixed velocity, the Lagrangian Equation (19.60) is not invariant: to first order in \(\varepsilon\) it acquires
by Equation (19.38). This is precisely the case Theorem 16.84 admits with a nonzero divergence term \(\Lambda\), and taking \(\xi=0\), \(\phi_{s}=\vect{u}t\) and \(\Lambda=M\vect{x}_{\text{CM}}\cdot\vect{u}\) in the Noether charge Equation (16.76) gives
with \(\vect{G}\) the quantity Equation (19.59) of Theorem 19.69, and \(\vect{u}\) arbitrary. Had \(\Lambda\) been dropped, the charge would have come out as \(\vect{P}\cdot\vect{u}\,t\), which is not conserved. The term that cannot be dropped is the same object that appears in Lie Groups, Lie Algebras, and Fibre Bundles as the central charge of the Galilei algebra — the mass, in Example 14.80 — and in The Poisson Algebra and the Canonical Bridge to Quantum Mechanics as the constant defect in the Poisson brackets of a free particle (Proposition 25.14). Three descriptions of one fact: the free particle realizes not the Galilei algebra but its central extension by the mass.
The two derivations are not redundant, and neither contains the other.
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The direct route is stronger in scope. It needs only Equation (19.41) and yields Theorem 19.67 for internal forces of any kind, dissipative ones included. A perfectly inelastic collision has no Lagrangian description of the colliding bodies alone, yet conserves momentum exactly.
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The symmetry route is stronger in explanation. It says why there are ten conserved quantities and not eleven, ties each to a geometrical property of space and time, and generalizes: replacing the Galilei group by the Poincaré group repeats the argument unchanged and delivers the relativistic conservation laws of Relativistic Dynamics, while the field-theoretic form Theorem 16.90 delivers those of Generalized Classical Field Theory.
Both are used in what follows. The systematic development of the symmetry route for mechanical systems, with generalized coordinates and constraints, is Lagrangian Mechanics; the conserved quantities reappear there as first integrals and cyclic coordinates (Proposition 16.28), and again in Hamiltonian Mechanics as generators of the very transformations they are conserved under.
The Newtonian formulation presented in this chapter provides the equations of motion whenever all the forces are known explicitly. When the motion is restricted by constraints, however, the constraint forces are not known in advance, and the formulation must be recast in terms of generalized coordinates; this is the subject of Lagrangian Mechanics. The special cases of equilibrium and of central forces — the latter the gateway to celestial mechanics — are developed in Central Forces and Statics.