Part I · The Action Principle — Chapter 1.1

What's Wrong With Forces

Newton's physics is right. Newton's formulation is a dead end, and this chapter shows you exactly where the road stops.

Tools you'll need  — Chapter 0.1: the derivative as a linear approximation, and $\dot x,\ddot x$ notation. Chapter 0.2: the Fundamental Theorem, and substitution. Chapter 0.6: partial derivatives, the gradient, and the chain rule along a curve. Chapter 0.7 §2 is the load-bearing one, covering line integrals and the four equivalent characterisations of a conservative field. This chapter is where that section gets spent. Chapter 0.8: what a second-order ODE is, and the Picard–Lindelöf theorem, which is what turns $\vv F=m\vv a$ into a statement about determinism.

1 · Newton's laws, stated precisely

You have seen these three laws before. What you have probably not seen is the fine print, and the fine print is where the interesting failures live. So let's state them carefully, and pay attention to the exact wording each time.

1.1 · The first law is a definition, not a special case

First law. There exist frames of reference, called inertial frames, in which a body subject to no net force moves with constant velocity.

The usual textbook aside is that the first law is nothing more than $\vv F=m\vv a$ with $\vv F=\vv 0$. That reading throws away the entire content, and it is worth seeing why.

Read as a statement about one particular body, "no force implies no acceleration" really is a corollary of the second law, and it says nothing new. Read as an existence claim about frames, it is a separate and much stronger assertion. It says that the class of reference frames in which the second law takes its simple form is non-empty.

That matters because the second law is false in most frames. Stand in a braking train and every object around you accelerates backwards with no visible cause. Stand on the rotating Earth and the swing plane of a long pendulum creeps slowly around. The second law does not hold in either of those frames. If you insist on writing it there anyway, you have to invent forces, the "fictitious" ones, whose only job is to absorb the discrepancy.

The first law exists to say something quite specific: there is a class of frames where you don't have to do that, and the second law is a claim about those frames only.

1.2 · The second law, and what determinism means

Second law. In an inertial frame, the rate of change of a body's momentum equals the net force on it:

Let's read (1.1.1) as a piece of mathematics rather than as a slogan. It is a second-order ordinary differential equation for the function $\vv r(t)$, and it is exactly the type of equation Chapter 0.8 was built to handle. To bring that machinery to bear, we rewrite it as a first-order system in the six-dimensional variable $(\vv r,\vv v)$:

and now Picard–Lindelöf applies word for word. If $\vv F$ is continuous in $t$ and Lipschitz in $(\vv r,\vv v)$, then specifying $\vv r$ and $\vv v$ at a single instant determines the entire trajectory uniquely, forwards and backwards in time.

1.3 · The third law, in two strengths

Third law (weak form). If body $j$ exerts a force $\vv F_{ij}$ on body $i$, then body $i$ exerts $\vv F_{ji}=-\vv F_{ij}$ on body $j$.

Third law (strong form). In addition, $\vv F_{ij}$ is directed along the line joining the two bodies.

Gravity and the electrostatic Coulomb force satisfy both forms. Almost every textbook states only the weak one and then quietly uses the strong one, because the weak form is what gives you momentum conservation, while the strong form is what you need for angular momentum. Section 3 derives both of them, and then breaks both of them.

1.4 · What the second law does not do

Here is the omission that is almost never said out loud. (1.1.1) does not tell you what $\vv F$ is. It is an equation with an undetermined symbol sitting in it. Give it a force law and it predicts a motion. Give it none and it predicts nothing at all. Worse, on its own it cannot even be tested, because any observed motion whatsoever can be accommodated by defining $\vv F \equiv m\ddot{\vv r}$.

So Newtonian mechanics is not a theory. It is a framework awaiting a force law. The physics lives entirely in the forces, and the forces have to be supplied from outside: by Newton's inverse-square gravitation, by Hooke's law, by Coulomb's law, by an empirical drag coefficient. Each of those is a separate empirical input, carried in by hand and justified by fitting data.

Hold on to that thought, because it is the sharpest single argument for what comes next. Chapter 1.2's formulation has the same structure, in that you must supply one function by hand. The difference is that the function is a scalar. By Chapter 1.4, the requirement that this scalar respect a symmetry will constrain it so tightly that in Part VI the force laws stop being inputs and start being outputs.

Nothing like that is available here. You cannot ask "what is the most general force consistent with rotational symmetry?" and get the Standard Model out. You can ask that question about scalars.

2 · Energy

Two quantities survive the transition out of the force picture, and they are energy and momentum. Both of them come out of (1.1.1) by a single manipulation, and both of them will be redefined in Chapter 1.4 as consequences of symmetries. Here we get them the ordinary way, partly because we need them and partly because the derivation shows exactly what they cost.

2.1 · The work–energy theorem

We want a scalar statement out of a vector equation, so start from $\vv F = m\dot{\vv v}$ and dot both sides with $\vv v$. There is nothing clever about that move. Dotting with the velocity is the only way to turn a vector equation into a scalar one without picking some direction by hand.

The right-hand side is a total derivative in disguise, and the product rule applied to a dot product shows it:

Notice what has happened. The combination $\half mv^{2}$ appeared on its own, without anyone naming it in advance. Something that shows up unbidden has earned a name, so let's give it one:

With that name in hand, (1.1.3) reads $\vv F\cdot\vv v = \dd T/\dd t$. We want a statement about a stretch of the motion rather than about one instant, so integrate both sides in time from $t_{1}$ to $t_{2}$ along the actual trajectory $\vv r(t)$. Since $\vv v\,\dd t=\dd\vv r$, the left-hand side is Chapter 0.7's parametrised line integral:

This is the work–energy theorem. Let's pause on how little it assumed, because that is where its value lies: it used no property of $\vv F$ whatsoever. It holds for friction, for magnetic forces, and for a force you invent on the spot. The work done along the actual path equals the change in kinetic energy, always.

What it does not give you is a conservation law. The left-hand side is a line integral along a path, and so far nothing obliges it to be expressible as a difference of anything. To get a conservation law we need one extra hypothesis, and Chapter 0.7 already told us precisely which one it is.

2.2 · Conservative forces — cashing Chapter 0.7

Chapter 0.7 §2.3 listed four conditions on a vector field $\vv F$ defined on a region $U$, and proved that on a simply connected domain all four are the same condition. Here they are again:

	Condition	Character
(A)	$\vv F = -\nabla V$ for a single-valued function $V$ on $U$	a potential exists
(B)	$\int_{C}\vv F\cdot\dd\vv r$ depends only on the endpoints of $C$	path-independence
(C)	$\oint_{C}\vv F\cdot\dd\vv r = 0$ for every closed loop	no free energy from a cycle
(D)	$\nabla\times\vv F = \vv 0$ throughout $U$	a local test

The minus sign in (A) is a convention, chosen so that $V$ decreases in the direction the force pushes. Chapter 0.7 wrote $\vv F=\nabla\phi$, so $V=-\phi$. A force satisfying these conditions is called conservative, and over the next half page that name gets earned rather than assumed.

The equivalence is the whole point, so let's not let it slide past. Condition (D) is three derivatives evaluated at a point, and it takes about ten seconds. Conditions (B) and (C) are statements about infinitely many paths, and you cannot check them directly at all. Condition (A) is the one you actually want, because it replaces a whole vector field by a single number at each point.

Being able to move freely between a ten-second local test and a global existence statement is the entire practical value of Chapter 0.7. It also deserves $2\pi$ of caution. The arrow (D)$\Rightarrow$(C) fails on a domain that is not simply connected, which is what the punctured-plane counterexample there was for.

Now suppose $\vv F$ is conservative. Then by (A) and Chapter 0.7's gradient theorem,

Now we have two expressions for the same work $W$. The work–energy theorem gave us one, and it says the work equals the change in kinetic energy. This one says the work equals minus the change in potential energy. Both hold along the actual path, so they have to agree. Set them equal, taking $\Delta T$ from (1.1.6):

That is energy conservation, derived, and derived in a way that shows exactly what it rests on. Let's take the three conditions one at a time. The quantity $V$ is called the potential energy, and it exists because of (A). The conservation itself holds because of (B). The reason you can verify any of it in practice is (D). Three faces of one condition, all three used, all three necessary.

Grind box · where friction's energy went, and a first look at coarse-graining

Take a projectile with quadratic drag, $\vv F = -mg\hat{\vv y} - k\abs{\vv v}\vv v$. The drag term is not conservative: its work depends on the path, and a closed loop returns negative work, which violates (C). Integrate the motion numerically ($m=1.7$, $k=0.42$, $g=9.81$, launched from the origin at $\vv v_{0}=(9,14)$, run to $t=2.4$) and check the two statements separately.

Quantity	Value
$\int\vv F\cdot\vv v\,\dd t$ (all forces)	$-202.4463642316$
$\Delta T$	$-202.4463642317$
$\Delta(T+mgy)$	$-282.6869372$
work done by drag alone	$-282.6869372$

The work–energy theorem (1.1.6) holds to eleven digits, as it must, since it assumed nothing. But $T+V$ is not conserved, and the deficit is exactly the work done by the non-conservative force. (1.1.8) failed, at precisely the step that used (A).

Now the physical question. Is energy conservation actually violated? No. The missing $282.69\,\mathrm{J}$ sits in the air and the projectile as thermal energy, which is the kinetic energy of $\sim10^{25}$ molecules whose individual positions and velocities you declined to track. Every one of those microscopic degrees of freedom obeys conservative electromagnetic forces. Friction is not a violation of energy conservation. It is a bookkeeping choice: you threw away $10^{25}$ coordinates and the energy went with them.

State that as a principle, because it recurs at every scale of this book. Non-conservative forces are what conservative forces look like after you stop tracking some of the degrees of freedom. The technical name for declining to track them is coarse-graining. It introduces an asymmetry, in that you can predict where the energy goes but you cannot get it back, and that asymmetry is the origin of the second law of thermodynamics. The second law is therefore a statement about the coarse-graining, not about the microscopic dynamics, all of which is time-reversible. The same move reappears in Chapter 5.11 under the name renormalisation group: integrate out the degrees of freedom you cannot resolve, and the effective description of what is left acquires properties the microscopic one did not have.

3 · Momentum, angular momentum — and a crack

3.1 · Total momentum, from the weak third law

Let's do momentum first, and take $N$ particles so that the bookkeeping is honest. Particle $i$ feels an external force $\vv F_{i}^{\text{ext}}$, plus internal forces $\vv F_{ij}$ from each of the others, so (1.1.1) reads $\dot{\vv p}_{i} = \vv F_{i}^{\text{ext}} + \sum_{j\neq i}\vv F_{ij}$. We want a statement about the system as a whole, so define the total momentum $\vv P = \sum_{i}\vv p_{i}$ and add the equations up:

Now look at the double sum. It runs over ordered pairs, so each unordered pair $\{i,j\}$ turns up twice, once as $\vv F_{ij}$ and once as $\vv F_{ji}$. By the weak third law those two cancel. Every pair cancels in the same way, so the whole double sum is zero, and

Internal forces, however violent, cannot move the total momentum. This is why a rocket works, and why the centre of mass of an exploding shell carries on along its parabola.

3.2 · Angular momentum, from the strong third law

Angular momentum goes the same way, with one extra demand at the end. Define $\vv L = \sum_{i}\vv r_{i}\times\vv p_{i}$ and differentiate. The product rule gives two kinds of term, and one kind dies immediately: $\dot{\vv r}_{i}\times\vv p_{i} = \vv v_{i}\times m_{i}\vv v_{i}$ vanishes, because a vector's cross product with itself is zero. That leaves

Pair up the internal terms again. Using $\vv F_{ji}=-\vv F_{ij}$, the pair $\{i,j\}$ contributes

which vanishes if and only if $\vv F_{ij}$ is parallel to $\vv r_{i}-\vv r_{j}$, the line joining the two particles. That condition is exactly the strong form. With it we get $\dd\vv L/\dd t = \boldsymbol\tau^{\text{ext}}$, so an isolated system conserves angular momentum. Without it, the internal forces can spin the system up out of nothing.

Grind box · why the weak form is not enough, with a concrete counterexample

Take two particles at $\vv r_{1}=(0,0,0)$ and $\vv r_{2}=(d,0,0)$, and suppose they exert on each other forces that are equal and opposite but not along the line joining them, say $\vv F_{12} = f\hat{\vv y}$ and $\vv F_{21} = -f\hat{\vv y}$. The weak third law holds exactly, so by (1.1.10) the total momentum is conserved.

Now compute the internal torque from (1.1.12):

$$\big(\vv r_{1}-\vv r_{2}\big)\times\vv F_{12} \;=\; (-d\hat{\vv x})\times(f\hat{\vv y}) \;=\; -df\,\hat{\vv z} \;\neq\; \vv 0.$$

The pair develops angular momentum with no external agent. A force like that would be a reactionless flywheel: the system spins faster and faster, and since rotational kinetic energy grows as $L^{2}$, you could extract unbounded work from it. The strong third law is what forbids this. It is a genuinely separate assumption, and it does not follow from the weak one, as this example shows in three lines.

Note also which symmetry each form is standing in for. The weak form does the job that Chapter 1.4 will assign to translational invariance, and the strong form does the job of rotational invariance. In the Lagrangian formulation you assume the symmetries and the conservation laws follow. Here you assume the conservation-law-shaped force conditions and hope they hold. The next section shows that one of them does not.

3.3 · The crack: the third law is false

Everything in §3.1 and §3.2 rests on the third law, and the third law is an assumption about forces rather than a theorem. So it is worth asking whether any real force violates it. One does, and it is not exotic at all. It is the magnetic force between two moving charges.

We need two ingredients. This book derives both of them in Chapter 2.6, and quotes both of them here.

Now for the configuration, which is chosen to make the failure as stark as possible. Put charge $q_{1}$ at the origin, moving with $\vv v_{1}=v\hat{\vv x}$. Put charge $q_{2}$ at $\vv r_{2}=d\hat{\vv x}$, directly ahead of the first, and let it move perpendicular to that direction, with $\vv v_{2}=v\hat{\vv y}$.

Force on 2 from 1. The displacement from 1 to 2 is $d\hat{\vv x}$, so $\hat{\vv r}=\hat{\vv x}$, and

because the cross product of $\hat{\vv x}$ with itself is zero. Charge 1 is moving straight at charge 2, and a charge produces no magnetic field directly ahead of itself. So $\vv F_{21}^{\text{mag}} = q_{2}\vv v_{2}\times\vv B_{1} = \vv 0$, and charge 2 feels no magnetic force at all.

Force on 1 from 2. The displacement from 2 to 1 is $-d\hat{\vv x}$, so now $\hat{\vv r}=-\hat{\vv x}$ and

using $\hat{\vv y}\times\hat{\vv x}=-\hat{\vv z}$. This is not zero, and the magnetic force on charge 1 is

The electric forces between the two charges are equal and opposite along the line joining them, so they cancel in the sum. What is left is the magnetic residue:

The third law fails. Not approximately, and not in some limit: one particle is pushed and the other is not. The strong form fails twice over, because the leftover force is perpendicular to the line joining the charges, so (1.1.12) delivers a net internal torque as well.

Now feed that into (1.1.10). There is nothing outside this system, which is two charges alone in the universe, and yet

The total mechanical momentum of an isolated system changes. Momentum is appearing out of nowhere, at a rate you can compute. Before deciding how seriously to take that, let's be honest about how big the effect is. Compare (1.1.16) to the Coulomb force $q_{1}q_{2}/4\pi\varepsilon_{0}d^{2}$, and use $\mu_{0}\varepsilon_{0}=1/c^{2}$:

which for $v=10^{6}\,\mathrm{m\,s^{-1}}$ comes to $1.11\times10^{-5}$. Small, but not zero, and not subtle at all inside a particle accelerator. (Both the ratio and the vector directions above were checked numerically. The computed ratio is $1.112650\times10^{-5}$, against $v^{2}/c^{2}=1.112650\times10^{-5}$.)

4 · Where the force picture actually breaks

There are four failures to see, and we will take them one at a time. None of them is a matter of taste, and each one is demonstrated rather than asserted.

4.1 · Failure one: constraints

Consider a plane pendulum: a bob of mass $m$ hanging on a string of fixed length $\ell$ from a pivot, swinging in a vertical plane. This is the simplest constrained system in physics. Let's solve it with Newton, in Cartesian coordinates, without cheating anywhere.

Put the origin at the pivot, with $\hat{\vv x}$ horizontal and $\hat{\vv y}$ vertically upward. The bob sits at $(x,y)$, and two forces act on it. One is gravity, $(0,-mg)$. The other is the string tension, directed from the bob toward the pivot, that is, along $-(x,y)/\ell$. Call its magnitude $\mathcal{F}$, since the letter $T$ is already spoken for by (1.1.5).

Now note the critical fact. $\mathcal{F}$ is unknown. No force law hands it to us. A string pulls with whatever tension is required to keep its length fixed, and how much that is depends on the motion, which is the very thing we are trying to find.

Newton's second law in components:

Count what we have: two equations, and three unknown functions $x(t)$, $y(t)$ and $\mathcal{F}(t)$. The system is underdetermined, and Picard–Lindelöf does not apply to it. The missing information is the constraint

which is an algebraic relation rather than a differential equation, so we cannot append it to (1.1.19) as it stands. To use it, we have to differentiate it into a compatible form. Once:

which says that the velocity is perpendicular to the string. True, and not yet enough, because it involves no accelerations at all. So differentiate again, using the product rule:

Now we have a relation among the accelerations, which is something we can combine with (1.1.19). Multiply the first equation of (1.1.19) by $x$, the second by $y$, and add them:

using (1.1.20) in the last step. Substituting (1.1.22) on the left and solving for the tension:

We have now determined the unknown force, but only in terms of the motion, which we still do not know. So put (1.1.24) back into (1.1.19) to close the system:

Let's stop and look at what we have. It is a pair of coupled nonlinear second-order ODEs in two variables, describing a system that visibly has one degree of freedom. Nothing about (1.1.25) suggests a pendulum. You cannot see the period in it. You cannot see the small-oscillation limit. You cannot integrate it in this form.

It is numerically treacherous too, because the constraint (1.1.20) is only enforced through its second derivative, so any integration error accumulates and the bob slowly drifts off its circle.

So let's do the thing we should have done at the start, and introduce the angle $\theta$ measured from the downward vertical:

which satisfies (1.1.20) automatically. That is already the gain, because the constraint is now built into the coordinate rather than enforced alongside it. What we want next is (1.1.25) written in $\theta$, and it is built out of Cartesian velocities and accelerations. So differentiate twice to get them:

Then $\dot x^{2}+\dot y^{2} = \ell^{2}\dot\theta^{2}(\cos^{2}\theta+\sin^{2}\theta) = \ell^{2}\dot\theta^{2}$, and $-gy = g\ell\cos\theta$, so the bracket in (1.1.25) is $\ell^{2}\dot\theta^{2}+g\ell\cos\theta$. Substituting into the first line of (1.1.25):

The $\ell\sin\theta\,\dot\theta^{2}$ terms cancel on both sides, leaving $\ell\cos\theta\,\ddot\theta = -g\sin\theta\cos\theta$, and dividing by $\ell\cos\theta$ (legitimate whenever $\cos\theta\neq0$):

And the tension, from (1.1.24), comes out as $\mathcal{F} = m\big(\ell\dot\theta^{2}+g\cos\theta\big)$, which is the centripetal requirement plus the component of the weight along the string.

An experienced hand would not have gone this way at all. Instead of Cartesian components, you resolve the two forces along and perpendicular to the string, which are the directions the geometry actually prefers. Perpendicular to the string, meaning tangentially, the tension contributes nothing, gravity contributes $-mg\sin\theta$, and the tangential acceleration is $\ell\ddot\theta$. Along the string, meaning radially inward, the acceleration is the centripetal $\ell\dot\theta^{2}$, and the inward forces are $\mathcal{F}$ less the inward component $mg\cos\theta$ of the weight:

That gives two component equations and two unknowns, $(\theta,\mathcal{F})$, and the tension is eliminated by the expedient of ignoring the second equation, which exists only to determine a quantity we did not want and are about to discard. It is quicker, and it is what anyone competent actually does.

But look at what it presupposed. You had to know in advance that the acceleration in polar coordinates splits into a tangential $\ell\ddot\theta$ and a centripetal $\ell\dot\theta^{2}$. That is not a guess. It is (1.1.35), which §4.2 spends half a page deriving. The Cartesian route above is longer precisely because it assumes nothing. Either way, both routes carry an unknown force through to the end and then throw it away. (Both results were checked by integrating (1.1.25) and (1.1.29) numerically from the same initial condition. The trajectories agree to $3\times10^{-11}$ over seven seconds, and a bob released from horizontal has $\mathcal{F}=3.000\,mg$ at the bottom of its swing.)

Grind box · the second component equation, and what its redundancy is telling you

We used only the $x$-equation of (1.1.25). Honesty requires checking the $y$-equation too. Substituting (1.1.27):

$$\begin{aligned} \ell\sin\theta\,\ddot\theta + \ell\cos\theta\,\dot\theta^{2} \;&=\; \frac{\ell\cos\theta}{\ell^{2}}\Big(\ell^{2}\dot\theta^{2}+g\ell\cos\theta\Big) - g\\[3pt] &=\; \ell\cos\theta\,\dot\theta^{2} + g\cos^{2}\theta - g \;=\; \ell\cos\theta\,\dot\theta^{2} - g\sin^{2}\theta, \end{aligned}$$

using $\cos^{2}\theta-1=-\sin^{2}\theta$. The $\ell\cos\theta\,\dot\theta^{2}$ terms cancel and we are left with $\ell\sin\theta\,\ddot\theta = -g\sin^{2}\theta$, i.e. $\ddot\theta=-(g/\ell)\sin\theta$ again, valid whenever $\sin\theta\neq0$.

The two equations are not independent. Each degenerates to $0=0$ somewhere, the $x$-equation at $\theta=\pm\pi/2$ and the $y$-equation at $\theta=0,\pi$, and between them they always deliver the same single equation of motion. That redundancy is the constraint speaking. Two Cartesian equations minus one constraint equals one genuine equation, and the system was one-dimensional from the start. We paid for two coordinates, carried an unknown force through five steps, and then had to discover that we had been solving one equation all along.

There is a slicker elimination, worth seeing because it shows how much depends on spotting the right trick. Multiply the first equation of (1.1.19) by $y$, the second by $x$, and subtract. The tension terms are $-\mathcal{F}xy/\ell$ in both, so they cancel without ever being evaluated:

$$m\big(y\ddot x - x\ddot y\big) \;=\; mgx \qquad\Longrightarrow\qquad y\ddot x - x\ddot y = gx.$$

Substituting (1.1.27), the left side collapses to $-\ell^{2}\ddot\theta$ (the $\dot\theta^{2}$ terms cancel identically) and the right to $g\ell\sin\theta$, which gives (1.1.29) in one line. Elegant, and entirely dependent on noticing that $y\times$(first)$-x\times$(second) kills the unknown. That combination is, not coincidentally, the $z$-component of the torque about the pivot. You have rediscovered angular momentum by hand, because Cartesian components hid it. Chapter 1.4 makes that systematic.

One more piece of fine print. Differentiating (1.1.20) twice enlarges the solution set, because (1.1.22) is implied by the constraint but does not imply it. The general solution of the differentiated system has $x^{2}+y^{2}=\ell^{2}+at+b$ for constants fixed by the initial data, so you must impose (1.1.20) and (1.1.21) at $t=0$ to kill $a$ and $b$. Forget either one and you are integrating the wrong problem. This is the standard hazard of index reduction in constrained dynamics, and a real source of bugs in physics engines. Chapter 1.2 never meets it, because there is no constraint left to differentiate.

4.2 · Failure two: coordinates, and the loss of form invariance

The pendulum hinted at this one. Now let's make it precise. Take $\vv F=m\vv a$ in the plane and write it out in polar coordinates $(r,\theta)$, which are the natural choice for any central force. The result is famous, and it ought to be disturbing.

Everything hinges on the fact that the polar basis vectors point in different directions at different places:

Along a trajectory $\theta$ changes with time, so these vectors are themselves functions of time and have to be differentiated as we go. By the chain rule,

Now differentiate the position $\vv r = r\,\hat{\vv r}$ twice, using the product rule and (1.1.32) at each step:

That is the velocity. Newton's law is a statement about acceleration, so we need one more derivative. Differentiate again, and use (1.1.32) once more, this time on both of the basis vectors:

Collect components. Newton's law $\vv F=m\vv a$ in polar coordinates therefore reads

Let's pause and compare that with what we started from. In Cartesian coordinates the law was $m\ddot x = F_{x}$ and $m\ddot y = F_{y}$: mass times second derivative equals force, one term on each side. In polar coordinates two extra terms have materialised. One is $-mr\dot\theta^{2}$, the centrifugal term. The other is $2m\dot r\dot\theta$, the Coriolis term.

Where did they come from? Not from any force. We changed no physics, we added no interaction, and $\vv F$ is the same vector it always was. They came from (1.1.32), which is to say from the basis vectors turning as you move. They are artefacts of the coordinate system.

Here is the cleanest demonstration that they are artefacts. Take a completely free particle, $\vv F=\vv 0$, moving in a straight line: $x=v_{0}t$, $y=b$ with $b\neq0$. Then $r=\sqrt{b^{2}+v_{0}^{2}t^{2}}$, and differentiating twice gives

A particle with no force on it has a nonzero second derivative of its radial coordinate. At $v_{0}=b=1$ and $t=1$ the value is $0.35355$. It is cancelled exactly by $r\dot\theta^{2}$, which at that same instant is also $0.35355$, so (1.1.35) gives $F_{r}=0$ as it must. But "$m\ddot r = F_{r}$" would have been flatly wrong.

And that is the consequence deciding the structure of this entire book. In general relativity spacetime is curved, and a curved space admits no global Cartesian coordinate system at all. That is not a matter of difficulty. It is a theorem (Chapter 3.4: the obstruction is the Riemann tensor, and it is nonzero exactly when curvature is present). So there is no privileged frame in which the correction terms vanish everywhere, and therefore no frame in which $\vv F=m\vv a$ can be stated in its clean form. A law whose shape depends on using coordinates that do not exist cannot survive.

What we need instead is a formulation that is form invariant from the outset, one that produces the correct equations in any coordinates at all, including coordinates nobody has tabulated. Chapter 1.2 §7 proves that the Euler–Lagrange equations have exactly this property, for arbitrary smooth invertible coordinate changes. The reason fits in one line: the action is a number attached to a path, and relabelling the points of a path cannot change a number. Chapter 3.2 then does gravity in that language, and the centrifugal and Coriolis terms of (1.1.35) reappear there as Christoffel symbols, the same artefacts, finally given a proper name and a geometric meaning.

4.3 · Failure three: fields

Newtonian mechanics is a theory of $N$ particles. Its state is $6N$ numbers, its input is a list of pairwise forces, and its output is $3N$ coupled equations. Every part of that presupposes that $N$ is finite.

An electromagnetic field is not like that. It has a value at every point of space, six numbers there, the components of $\vv E$ and $\vv B$, and there are uncountably many points. So the configuration is not a list of numbers. It is a function, and the space of functions is infinite-dimensional. Chapter 0.4 built exactly that vector space, and Chapter 4.3 will take its infinite-dimensionality seriously.

So ask the Newtonian question of a field. What is the force on it? The question has no referent. There is no particle to apply $\vv F=m\vv a$ to, no mass to divide by, and no finite list to sum over. You can ask what force the field exerts on a charge, which is the Lorentz law. But the dynamics of the field itself, which is what Maxwell's equations describe, is not the kind of thing the second law can express at all. "Sum the forces on each particle and set the total equal to $m\vv a$" has no meaning when there are no particles and no $m$.

You have already met the transition. Chapter 0.8 §7.6 took a chain of $N$ masses coupled by springs, wrote down the $N$ coupled equations, and let $N\to\infty$ with the spacing going to zero. The $N$ ordinary differential equations became one partial differential equation, the wave equation, and the $N$ coordinates became a field $\phi(x,t)$. That limit is exactly where the particle description stops being useful and the field description takes over.

What we need instead is a formulation whose input is a single scalar function, one that does not care whether the thing it describes has three degrees of freedom or a continuum of them. That is a Lagrangian. The extension to fields, which replaces $L(q,\dot q)$ by a Lagrangian density $\mathcal{L}(\phi,\partial_{\mu}\phi)$ integrated over space, is Chapter 5.2. Maxwell's equations then come from a single scalar (Chapter 2.6), as do Einstein's (Chapter 3.6) and the Standard Model's (Chapter 6.8). Not one of those can be written as $\vv F=m\vv a$.

4.4 · Failure four: quantum mechanics

The last failure is the most abrupt. In quantum mechanics, force is essentially not a concept.

Take that as the practical argument, if the structural ones have not landed. The objects that survive into the rest of physics are the Lagrangian and the Hamiltonian. The force does not. Every hour spent on Chapters 1.2 and 1.3 is spent on machinery that reappears in Chapter 4.2, in Chapter 5.3 and in Chapter 6.8. Time spent perfecting free-body diagrams is not.

5 · The alternative, glimpsed: Fermat's principle

Everything so far has been demolition. Now let's look at one small, complete, correct piece of physics that works in a completely different way. Once you have seen it, Chapter 1.2 is nothing more than the general theory of the same move.

It is about light, and it is older than Newton. Hero of Alexandria knew that a light ray reflecting off a mirror takes the shortest path. Fermat, in 1662, guessed the right generalisation:

Let's pause and read what kind of statement that is, because it is unlike anything in §1–§4. It says nothing about what happens to the ray at any particular place. It describes no local push. Instead it assigns a single number, the total travel time, to each entire candidate path, and then selects the path by a condition on that number. It is a statement about whole histories rather than about instants.

5.1 · Snell's law, derived

Imagine two media meeting at a flat horizontal interface. Light travels at speed $v_{1}=c/n_{1}$ above it and $v_{2}=c/n_{2}$ below it, where $n$ is the refractive index. Place the source $A$ at height $a$ above the interface, and the target $B$ at depth $b$ below it, with horizontal separation $d$ between them.

Within each medium the speed is constant, so within each medium the fastest route between two points is a straight line. There is nothing to trade off there. That means any candidate path is completely specified by one number: the horizontal position $x$ at which it crosses the interface. Set the origin directly below $A$. The two legs then have lengths $\sqrt{a^{2}+x^{2}}$ and $\sqrt{b^{2}+(d-x)^{2}}$ by Pythagoras, so the total time is

Look at what that did. An infinite-dimensional problem, the problem of choosing a whole path, has collapsed into an ordinary one-variable minimisation, which means Chapter 0.1 is enough. We want the stationary point, so differentiate, using the chain rule on each square root:

Now recognise the two fractions. Let $\theta_{1}$ be the angle the incident ray makes with the normal to the interface, and $\theta_{2}$ the same for the refracted ray. In the right triangle with vertical leg $a$ and horizontal leg $x$, the hypotenuse is the ray and the side opposite $\theta_{1}$ is $x$, so

These are not approximations. They are the definition of the sine, applied to the two triangles. Substitute them into (1.1.38) and impose stationarity, $T'(x)=0$:

Finally put $v_{i}=c/n_{i}$, so $1/v_{i}=n_{i}/c$, and cancel the $c$:

That is Snell's law, the empirical rule of refraction found by Ibn Sahl in 984 and rediscovered by Snellius in 1621. We derived it in half a page, from a statement about entire paths that mentions neither angles nor interfaces.

Two details before the figure. First, the stationary point is a genuine minimum here, and unique: differentiating (1.1.38) once more gives

everywhere, so $T$ is strictly convex and has exactly one stationary point. Second, notice what kind of statement (1.1.41) is. It is local, constraining the two angles at a single place, the crossing point. Fermat's principle was global, a statement about the whole path. Something has been converted along the way.

5.2 · Watch it happen

crossing point x

x =  0.5500 m

index n₁

1.00

index n₂

1.50

jump to the minimum of T

n₁ sin θ₁ =  0.48192

n₂ sin θ₂ =  1.23482

difference =  -0.75290

dT/dx =  -2.5114 ns/m

T = 12.619925 ns (minimum 11.571240 ns at x = 1.3627 m)

5.3 · The template

Let's step back from the optics and look at the shape of what happened, because that shape is the subject of Chapter 1.2.

Two features of the method deserve to be named, because they are precisely the two failures of §4.1 and §4.2.

It is coordinate-free. Nothing in Fermat's principle mentions $x$, $y$, or any axis at all. "The travel time along this path" is a property of the path and the media. You may compute it in whatever coordinates you like and you will get the same answer, because it is a number. Contrast that with §4.2, where the form of the law itself changed when we relabelled the plane. A principle stated as "make this number stationary" cannot suffer that, and Chapter 1.2 §7 turns the observation into a theorem.

It needs only one scalar function. The entire input was $n(\vv x)$, one number at each point of space, from which the travel time of any path follows by integration. There were no vectors, no components, and no free-body diagrams. Contrast that with §4.3, where the force picture had nothing at all to say about a system with a continuum of degrees of freedom. A formulation whose input is a single scalar does not care how many degrees of freedom there are.

6 · Worked examples

7 · Your turn