Part I · The Action Principle — Chapter 1.2

Stationary Action

One scalar function, one condition on it, and every equation of motion in this book.

Tools you'll need  — Chapter 0.1: the derivative as the coefficient of the best linear approximation, and the $o(h)$ bookkeeping. This chapter is that definition with $h$ replaced by an entire function. Nothing else changes, which is the whole point of having defined it that way. Chapter 0.2: integration by parts. Section 3 is one application of it, and the boundary term matters. Chapter 0.6: the multivariable chain rule, which is all that §7 is. Also stationary points and the Hessian, for §5, and Lagrange multipliers, for §6.4, where the multiplier is a physical force. Chapter 1.1: Fermat's principle, and the catalogue of things the force picture cannot do.

1 · Functionals

Let's start with the object this whole chapter is about, because it has a plain definition and an unfamiliar name. A function eats a number and returns a number. A functional eats a function and returns a number. That is the entire definition. The only reason it feels exotic is that you have spent your life with the first kind.

You have in fact been using functionals for years without the word. Here are three of them, and all three will be doing work before the chapter is out.

Arc length. Given a curve $y(x)$ between $x=a$ and $x=b$, we get its length by adding up little hypotenuses $\dd s = \sqrt{\dd x^{2}+\dd y^{2}}$:

Fermat's travel time. This is the same integral, weighted by the local slowness. If light moves at speed $c/n(x,y)$, where $n$ is the refractive index, the time to traverse the curve is

That is exactly the quantity Chapter 1.1 minimised, written out in general rather than for two flat slabs.

The action. Take a trajectory $x(t)$ between times $t_{1}$ and $t_{2}$. Take also some function $L(x,\dot x,t)$ of position, velocity and time. Then

We have not yet said what $L$ is, and that is deliberate. Section 5 is an honest discussion of where it comes from, and the answer is not what a textbook usually implies. For now, $L$ is any smooth function of three arguments, and $S$ is the functional it defines.

Notice the notation, because it is doing real work: square brackets. $S[x]$ is not "$S$ evaluated at the number $x$". It is "$S$ evaluated at the function $x$". You must hand over the entire path, every $x(t)$ for every $t$ in the interval, before $S$ will give you a number back.

2 · The variational derivative

This section is the conceptual centre of the chapter. Everything after it is consequences.

What we want is to say what it means for the functional $S$ to be stationary at a particular path. There is an honest difficulty in the way, and it is worth naming before we go round it.

We cannot write $\lim_{h\to0}\big(S[x+h]-S[x]\big)/h$. The displacement here is not a number $h$ that we can divide by. It is an entire function, and dividing by a function is not a meaningful operation.

The way out is the one Chapter 0.1 insisted on, at some cost in apparent pedantry, on its first page. The derivative is not a slope. It is the coefficient of the best linear approximation. That definition never mentioned division, and it never mentioned one-dimensionality. What it said was this: nudge the input, and identify the part of the output's response that is linear in the nudge. Nothing about that instruction breaks when the nudge is a function.

2.1 · Reduce to one variable

Fix the path $x(t)$ we want to test. Now fix a second function $\eta(t)$, which is the shape of the nudge. There is one restriction on it:

Such an $\eta$ is called an admissible variation. The restriction says that the nudge does not move the endpoints. Every path we compare starts at the same place at $t_{1}$ and finishes at the same place at $t_{2}$.

That is not a technicality. It is the boundary condition of the physical problem, and in §3 it is precisely what kills a term we would otherwise be stuck with.

Now build a one-parameter family of paths by scaling the nudge:

At $\epsilon=0$ this is the path we are testing. As $\epsilon$ grows, the path bulges away from it in the shape $\eta$, always pinned at both ends. Feed the family into the functional:

Let's pause and look carefully at what (1.2.6) is. $\Phi$ eats a single real number $\epsilon$ and returns a single real number. It is an ordinary function of one real variable. All of Chapter 0.1 applies to it verbatim, with no generalisation required at all.

So we have converted an infinite-dimensional problem into a one-dimensional one. The price we paid was agreeing, for the moment, to look in only one direction.

2.2 · The definition

Say that $S$ is stationary at the path $x$ if

The clause in italics is the entire content, so let's be careful about it. Any single $\eta$ probes one direction in the space of paths, and a function can easily be stationary along one direction and not along others. A saddle is flat along two lines and steep everywhere else.

Demanding (1.2.7) for all $\eta$ is demanding flatness in every one of the infinitely many available directions at once. That is a very strong condition, and it is strong enough to determine the path.

The quantity in (1.2.7) deserves a name and a symbol. Write

and call it the first variation of $S$ at $x$ in the direction $\eta$. Unwind (1.2.8) through Chapter 0.1's definition of a derivative and it says exactly this:

2.3 · The functional derivative

One more piece of notation before we compute anything. Physics uses it constantly and it is easy to be mystified by it, so it is worth ten lines here.

In finite dimensions, Chapter 0.6 wrote the derivative of $f:\R^{n}\to\R$ in the direction $\vv h$ as a sum over components,

The gradient components $\partial f/\partial x^{i}$ are the coefficients of the linear map, one per direction. Here the directions are labelled not by an index $i$ running over $1,\dots,n$ but by a time $t$ running over $[t_{1},t_{2}]$. So the sum over the label becomes an integral over it.

That tells us what to define. The functional derivative $\delta S/\delta x(t)$ is the coefficient function appearing in

Set (1.2.11) beside (1.2.10) and the translation is index $\to$ time, sum $\to$ integral, gradient component $\to$ functional derivative. It is a continuum gradient.

Section 3 computes it, and the answer will be the Euler–Lagrange expression. The condition $\delta S=0$ for all $\eta$ then becomes the statement that this continuum gradient vanishes at every $t$, which is one equation per instant, as promised.

3 · Euler–Lagrange, derived

Now for the calculation itself. Take $S[x]=\int_{t_{1}}^{t_{2}}L(x,\dot x,t)\,\dd t$ with $L$ smooth, and compute $\delta S$. There are exactly four moves, and you already have all four.

3.1 · Move one: differentiate under the integral sign

Write out $\Phi(\epsilon)$. Since $x_{\epsilon}=x+\epsilon\eta$ implies $\dot x_{\epsilon}=\dot x+\epsilon\dot\eta$,

What we want is $\Phi'(0)$. The integrand depends on $\epsilon$ and the limits do not, so assume for the moment that we may move the $\epsilon$-derivative inside:

⚑ Quoted. The interchange in (1.2.13) is the one step in this derivation we are not proving. The standard sufficient condition runs like this. If $L$ and its $\epsilon$-partial derivative are continuous on the rectangle $[t_{1},t_{2}]\times[-\epsilon_{0},\epsilon_{0}]$, then $\Phi$ is differentiable and (1.2.13) holds. We are assuming that $L$ is a smooth function of its three arguments, and that $x,\eta$ are smooth on a closed bounded interval, so the condition is met. The grind box below says what can go wrong when it is not.

3.2 · Move two: linearise the integrand

The $\epsilon$-derivative inside is the multivariable chain rule of Chapter 0.6, applied to $L$ as a function of its first two slots. Write $\partial L/\partial x$ and $\partial L/\partial\dot x$ for the partial derivatives of $L$ with respect to its first and second arguments. As $\epsilon$ increases, the first argument moves at rate $\eta$ and the second at rate $\dot\eta$, so

where after setting $\epsilon=0$ the partials are evaluated on the original path $\big(x(t),\dot x(t),t\big)$. That gives

Now we are stuck, and it is worth being precise about why.

To conclude anything about the integrand, we need $\eta$ to appear as an overall factor, so that "this holds for every $\eta$" has some bite. It does appear that way in the first term. In the second term what appears is $\dot\eta$, and $\dot\eta$ is not independent of $\eta$. It is determined by it. We cannot choose the two separately, so we cannot argue term by term.

3.3 · Move three: integrate by parts

This is exactly the situation Chapter 0.2 built integration by parts for. That rule says you may move a derivative from one factor onto the other, at the cost of a minus sign and a boundary term. Apply it to the second term with $u = \partial L/\partial\dot x$ and $v'=\dot\eta$, so that $v=\eta$:

Now look at the boundary term. It is $\big(\partial L/\partial\dot x\big)\big|_{t_{2}}\eta(t_{2}) - \big(\partial L/\partial\dot x\big)\big|_{t_{1}}\eta(t_{1})$. Admissible variations vanish at the endpoints, by (1.2.4), so both $\eta$'s are zero. The boundary term dies.

It dies because we fixed the endpoints, and not because of any property of $L$. That is worth remembering rather than deriving twice.

The reason it is worth remembering is that you will eventually stop fixing the endpoints. It happens in Chapter 1.3, in Chapter 3.6 for gravity, and everywhere in field theory where the "boundary" is at spatial infinity. When it does, the surviving boundary term is not litter. It is where the momentum lives, and where the black-hole entropy will come from.

Substitute (1.2.16) into (1.2.15) and collect the common factor $\eta$:

Compare that with (1.2.11) and you can read off the answer to §2.3. The functional derivative of the action is the bracket:

3.4 · Move four: the fundamental lemma

Stationarity says that (1.2.17) vanishes for every admissible $\eta$, and what we want to conclude is that the bracket itself vanishes.

That step is not free. It is a genuine theorem, and skipping it means skipping the move that converts "the integral is zero" into "the integrand is zero". The same move recurs in Chapter 5.2 to produce field equations, and in Chapter 3.6 to produce Einstein's. So we prove it once, here, properly.

The theorem is not obvious and it is not trivially true. An integral can vanish through cancellation. Take $\int_{-1}^{1}t\,\dd t=0$, where the integrand is zero nowhere but the origin.

What defeats the cancellation here is that the hypothesis lets you choose $\eta$. If $f$ is positive somewhere, pick an $\eta$ that is supported only there, and there is nothing left for it to cancel against. Making that idea airtight means exhibiting such an $\eta$, and that is the content of the proof.

Proof. Suppose, for contradiction, that $f$ is not identically zero. Then there is a point $t_{0}$ with $f(t_{0})\neq0$. Replacing $f$ by $-f$ if necessary changes nothing, since the hypothesis is preserved, so assume $f(t_{0}) \gt 0$. If $t_{0}$ is an endpoint, continuity gives a nearby interior point where $f$ is still positive, so assume $t_{0}$ is interior.

Set $c=\tfrac12 f(t_{0}) \gt 0$. By continuity of $f$ at $t_{0}$, there is a $\delta \gt 0$ with $[t_{0}-\delta,\,t_{0}+\delta]\subset(t_{1},t_{2})$ and

That pins $f$ above a fixed positive number across a whole interval, rather than at the single point $t_{0}$. Now we need a probe that lives only on that interval, so that the integral in the hypothesis sees the positive part of $f$ and has nothing anywhere else to cancel against. Here is one, built by hand. Define

Three things about (1.2.20), each checkable by hand.

• It is non-negative everywhere, and strictly positive on the open interval $\abs{t-t_{0}} \lt \delta$.
• It vanishes at $t_{1}$ and $t_{2}$, since it vanishes outside a subinterval strictly inside $[t_{1},t_{2}]$.
• It is continuously differentiable. Away from the two junction points that is plain, and at $t=t_{0}\pm\delta$ both the value $[\delta^{2}-(t-t_{0})^{2}]^{2}$ and its derivative $-4(t-t_{0})\big[\delta^{2}-(t-t_{0})^{2}\big]$ approach zero from inside, matching the zero function outside.

So $\eta$ is admissible.

Feed it in. The integrand vanishes outside the bump, and inside it both factors are positive:

That contradicts the hypothesis that the integral vanishes for every admissible $\eta$. Hence no such $t_{0}$ exists and $f\equiv0$. $\blacksquare$

Grind box — the bump's integral, smoother bumps, and how much continuity you really need

The number $16\delta^{5}/15$. Substitute $u=t-t_{0}$ and expand:

$$\int_{-\delta}^{\delta}\big(\delta^{2}-u^{2}\big)^{2}\dd u = \int_{-\delta}^{\delta}\big(\delta^{4}-2\delta^{2}u^{2}+u^{4}\big)\dd u = 2\left(\delta^{5}-\frac{2\delta^{5}}{3}+\frac{\delta^{5}}{5}\right) = \frac{16\delta^{5}}{15}.$$

The value is irrelevant to the proof. All that matters is that it is strictly positive. It is worth computing once so that the inequality in (1.2.21) is not taken on faith.

A smoother bump. Our $\eta$ is $C^{1}$ but not $C^{2}$: differentiate twice and at $t=t_{0}\pm\delta$ you get $8\delta^{2}$ from inside and $0$ from outside. $C^{1}$ is all the lemma needs, because the Lagrangian depends on $x$ and $\dot x$ and no higher derivative. You will want a variation that is infinitely differentiable in field theory, where Lagrangians with second derivatives occur. For that, use

$$\eta(t) = \exp\!\left(\frac{-1}{\delta^{2}-(t-t_{0})^{2}}\right)\ \text{ for }\abs{t-t_{0}}\lt\delta,\qquad \eta=0 \text{ otherwise.}$$

Every derivative of this vanishes at the junctions, because $\ee^{-1/s}$ and all its derivatives tend to zero as $s\to0^{+}$ faster than any power. That is Chapter 0.3's remark that this function has a Taylor series identically zero and yet is not zero, the standard counterexample to "smooth implies analytic". Restricting to $C^{\infty}$ probes only strengthens the hypothesis of the lemma, so the conclusion still follows. You have fewer probes and still enough.

Continuity of $f$ is essential. Drop it and the lemma is false in the pointwise sense: a function that is $1$ at a single point and $0$ elsewhere has $\int f\eta=0$ for every $\eta$, yet is not identically zero. The correct statement without continuity is "$f=0$ almost everywhere". Physically this says that the equation of motion holds at every instant except possibly on a set of measure zero, and that is exactly the loophole permitting collisions, shocks, and the kinked worldlines of Chapter 5.6's path integral.

The interchange, and when it fails. The condition quoted in §3.1 is sufficient, not necessary, and it is a genuine condition. The standard failure is $\Phi(\epsilon)=\int_{0}^{1}\dfrac{\epsilon}{\epsilon^{2}+t^{2}}\dd t = \arctan(1/\epsilon)$ for $\epsilon\ne0$, which tends to $\pi/2$ as $\epsilon\to0^{+}$ and to $-\pi/2$ from below, while the pointwise limit of the integrand is $0$ for every $t\ne0$. The integrand's $\epsilon$-derivative is not continuous at $(\epsilon,t)=(0,0)$, and the interchange is invalid. In mechanics, with $L$ smooth and the interval compact, this never bites; in field theory with fields that do not decay it sometimes does, and the symptom is a divergent action.

3.5 · The Euler–Lagrange equation

Apply the lemma to (1.2.17). The bracket is continuous, since we assumed both $L$ and the path smooth, and the integral vanishes for every admissible $\eta$. So the bracket vanishes identically, and we have

for all $t\in[t_{1},t_{2}]$. This is the Euler–Lagrange equation.

Let's pause on what kind of equation it is. It is a second-order ordinary differential equation for $x(t)$. It is second order because $\partial L/\partial\dot x$ depends on $\dot x$, and differentiating that in time produces $\ddot x$. Two constants of integration, fixed by the two endpoint conditions. Everything is consistent.

Nothing in the derivation used the fact that there was only one coordinate. Suppose the configuration is described by $n$ numbers $q^{1},\dots,q^{n}$ and $L=L(q^{1},\dots,q^{n},\dot q^{1},\dots,\dot q^{n},t)$. Vary each of them independently: take $\eta=(\eta^{1},\dots,\eta^{n})$ and run §3.1–§3.4 with sums over $i$ throughout.

Move two produces $\sum_{i}\big[\partial_{q^{i}}L\,\eta^{i} + \partial_{\dot q^{i}}L\,\dot\eta^{i}\big]$. Move three integrates each term by parts separately. Move four is applied $n$ times, once with $\eta^{1}$ arbitrary and all the other $\eta^{i}=0$, then once with $\eta^{2}$ arbitrary, and so on. The result is

One equation per degree of freedom, all of them from the single scalar $L$. That is a real compression: $n$ coupled second-order differential equations packaged into one function. It is the first practical reason to care, and by §6 it will look like a bargain.

Grind box — what happens if the Lagrangian contains $\ddot q$

Nothing in §3 required $L$ to stop at first derivatives. Suppose $S=\int_{t_{1}}^{t_{2}}F(q,\dot q,\ddot q,t)\,\dd t$. Moves one and two are unchanged and give

$$\delta S = \int_{t_{1}}^{t_{2}}\left[\pdv{F}{q}\,\eta + \pdv{F}{\dot q}\,\dot\eta + \pdv{F}{\ddot q}\,\ddot\eta\right]\dd t.$$

Move three now has to be run twice on the last term:

$$\begin{aligned} \int\pdv{F}{\ddot q}\,\ddot\eta\;\dd t &= \left[\pdv{F}{\ddot q}\,\dot\eta\right]_{t_{1}}^{t_{2}} - \int\dv{}{t}\!\left(\pdv{F}{\ddot q}\right)\dot\eta\;\dd t\\[3pt] &= \left[\pdv{F}{\ddot q}\,\dot\eta - \dv{}{t}\!\left(\pdv{F}{\ddot q}\right)\eta\right]_{t_{1}}^{t_{2}} + \int\dv{^{2}}{t^{2}}\!\left(\pdv{F}{\ddot q}\right)\eta\;\dd t. \end{aligned}$$

There are now two boundary terms, and killing them needs both $\eta$ and $\dot\eta$ to vanish at the ends. That is to say the variational problem is only well posed if you fix the position and the velocity at each endpoint. Given that, the fundamental lemma applies verbatim and delivers

$$\pdv{F}{q} - \dv{}{t}\!\left(\pdv{F}{\dot q}\right) + \dv{^{2}}{t^{2}}\!\left(\pdv{F}{\ddot q}\right) = 0,$$

a fourth-order equation, and the obvious pattern continues with alternating signs for still higher derivatives. This is the version the Familiar Ground callout in §7 needs: with $F=\lambda f''(x)^{2}$ the equation reduces to $f''''=0$, whose solutions are the cubics.

⚑ One quoted warning, because it explains an omission you will otherwise notice. Ostrogradsky proved that a Lagrangian genuinely depending on $\ddot q$ has a conserved energy that is linear in one of its momenta and therefore unbounded below, so such theories are generically unstable. That is why every fundamental action in §8.1's table contains at most first derivatives of its fields. The Einstein–Hilbert action looks like an exception, since $R$ contains second derivatives of the metric. It escapes because those second derivatives enter only inside a total divergence, as Chapter 3.6 will show explicitly.

4 · It reproduces Newton

Time to check that this machinery is not describing some other universe. Take a single particle of mass $m$ moving in one dimension in a potential $V(x)$, and make the choice

Where that choice comes from is §5's problem. Right now we only want its consequences.

Compute the two partial derivatives, treating $x$ and $\dot x$ as independent slots of the function $L$. That is legitimate, for the reason spelled out in the warning callout below.

Those two partials are the only ingredients (1.2.22) asks for. We want to know what it says about this particular $L$, so substitute them in:

The right-hand side is the force, by Chapter 1.1's definition of a conservative force as minus the gradient of its potential. So (1.2.26) is $m\ddot x = F$.

In three dimensions nothing changes except the number of equations. With $L=\half m\,\dot{\vv x}\cdot\dot{\vv x} - V(\vv x)$ and $q^{i}=x^{i}$,

Now read (1.2.27) again, slowly, without substituting anything. The object $\partial L/\partial\dot x^{i}$ came out as $m\dot x^{i}$, which is the momentum. The object $\partial L/\partial x^{i}$ came out as $-\partial V/\partial x^{i}$, which is the force. So the Euler–Lagrange equation, in Cartesian coordinates with $L=T-V$, reads

Newton's second law is not a separate law that happens to agree with the action principle. It is the Euler–Lagrange equation, written in Cartesian coordinates. That is the entire relationship between Chapter 1.1 and this one.

That also explains why the notation generalises the way it does. For an arbitrary Lagrangian in arbitrary coordinates we define

and then (1.2.23) always reads $\dot p_{i} = \partial L/\partial q^{i}$. The definition is not a whim. It is chosen so that the Euler–Lagrange equation keeps the shape of Newton's law.

Two later chapters take that further. Chapter 1.3 treats $p_{i}$ as an independent variable in its own right and gets Hamiltonian mechanics. Chapter 1.4 shows that whenever $L$ fails to depend on some $q^{i}$ at all, the corresponding $p_{i}$ is conserved, which is Noether's theorem in its simplest guise, already visible in (1.2.23) for anyone who looks.

5 · Why $T-V$? An honest answer

You have just watched $L=T-V$ reproduce Newtonian mechanics exactly. The natural next question is where $T-V$ came from. The honest answer has three parts, and only the third of them is satisfying.

Part one: within classical mechanics, it is not derived. No argument starts from Newton's laws and ends at $L=T-V$ by necessity. What there is instead is the calculation of §4 run backwards. If you want a variational principle whose Euler–Lagrange equation is $m\ddot x=-V'(x)$, then $\half m\dot x^{2}-V(x)$ is a Lagrangian that does the job.

It is not the only one. Problem 4 in §10 shows that you may add any total time derivative without changing a thing, and multiplying $L$ by a nonzero constant also leaves the equations alone. So $T-V$ is a choice, justified after the fact by its working. Anyone who tells you the minus sign is intuitive is selling something.

Part two: "least action" is the wrong word. The condition we imposed, (1.2.7), is that the first variation vanishes. That is stationarity, exactly as in Chapter 0.6, and Chapter 0.6 was emphatic that a vanishing gradient signals a maximum, a minimum or a saddle, and that in more than one dimension saddles are the generic case.

The space of paths has infinitely many dimensions. Saddles are not the exception there. They are ordinary. The warning callout below makes this concrete with a system where the action is provably not a minimum.

Part three: the real reason, which is quantum. In Chapter 5.6 we will construct the Feynman path integral, in which a quantum particle going from $A$ to $B$ contributes an amplitude from every path, each weighted by the phase

The total amplitude is the sum over all paths. Now consider what that sum does.

Away from a stationary point of $S$, neighbouring paths have actions differing at first order in the separation, so their phases differ at first order too. And $\hbar$ is tiny, so $S/\hbar$ swings through many multiples of $2\pi$ over a microscopic change of path. The contributions from neighbouring paths therefore point in every direction on the complex plane, and they cancel.

Near a stationary point, by construction, $S$ changes only at second order. A whole neighbourhood of paths shares essentially the same phase, and their contributions add rather than cancel.

So the classical path is not selected because nature computes an integral and minimises it. Nature does not extremise anything. The classical path is where the quantum amplitudes fail to cancel, and nothing more than that.

The action principle is a statement about constructive interference that survives into the $\hbar\to0$ limit. The reason the classical path is the stationary one is that stationary phase is the only place a wildly oscillating sum can leave a residue.

That pushes the question of why $L=T-V$ and not something else back one level. $L$ is whatever sits in the exponent of the quantum theory, and that gets fixed by symmetry requirements. Lorentz invariance and locality do most of the work, in Chapter 5.2; the demand that the theory be unitary is Chapter 5.3's.

⚑ Part three is a promise, not a derivation. Nothing in it is proved here. It is stated so that you know a real explanation exists and where it is, and so that you do not spend the next six chapters believing that "nature is economical" is an explanation. It isn't. Chapter 5.6 redeems the promise, including the stationary-phase approximation that makes "wildly oscillating integrals localise on stationary points" a theorem rather than a story.

5.1 · Seeing stationarity

Before the abstractions pile up, let's look at the thing itself. The figure below takes the simplest system with a curved trajectory, a particle thrown straight up in uniform gravity, and lets you deform its path while watching the action.

Here is the setup. Mass $m=1\ \mathrm{kg}$, gravity $g=9.8\ \mathrm{m\,s^{-2}}$, and a path that starts at $y=0$ at $t=0$ and returns to $y=0$ at $t=T=1\ \mathrm{s}$. The Euler–Lagrange equation (1.2.26) with $V=mgy$ gives $\ddot y=-g$, and the solution with those endpoints is the parabola $y_{\ast}(t)=\tfrac12 g\,t(T-t)$, peaking at $1.225\ \mathrm{m}$. Its action, worked out directly, is

That number is the baseline every deformed path gets measured against. Now we need a way to deform the path. What we want is a family of shapes that moves the middle of the curve while leaving both endpoints fixed, because a path with different endpoints is not a competitor at all. So the trial path is $y(t)=y_{\ast}(t)+\eta(t)$ with

which vanishes at $t=0$ and $t=T$ for any $a_{k}$, so every trial path is admissible. The three sliders are the $a_{k}$.

The widget computes $\int T\,\dd t$ and $\int V\,\dd t$ by numerical quadrature along whatever path is on screen. It samples the curve, differences it to get the velocity, and sums. So the numbers you see are measurements of the drawn path rather than the output of a formula.

a₁

 0.2000

a₂

 0.0000

a₃

 0.0000

halve the deformation snap to true path

integrand: T − V

∫T dt =  5.3481 J s

∫V dt =  9.2511 J s

∫(T − V) dt : trial -3.9030 | true path -4.0017

Δ from true path = +0.098696

press "halve" twice to watch ΔS fall by a factor of four each time

Two of the facts the figure is showing can be established exactly, and it is worth doing that rather than trusting the pixels.

Write the trial path as $y_{\ast}+\eta$, with $\eta$ vanishing at both ends. Because $L$ is quadratic in $\dot y$ and linear in $y$, the expansion of $S$ terminates:

with no higher terms. Integrate the middle bracket's first piece by parts: $\int m\dot y_{\ast}\dot\eta\,\dd t = \big[m\dot y_{\ast}\eta\big]_{0}^{T} - \int m\ddot y_{\ast}\,\eta\,\dd t$. The boundary term vanishes because $\eta$ does, and $\ddot y_{\ast}=-g$, so the remaining integral is $+\int mg\,\eta\,\dd t$, which cancels the second piece exactly. The first variation is zero. That is stationarity, verified by hand, and what survives is

manifestly non-negative, and zero only for $\eta\equiv0$. So for this system the action is a genuine minimum and not merely stationary.

We can go one step further with the specific $\eta$ of (1.2.32). The sine modes are orthogonal on $[0,T]$ (Chapter 0.9), and what is needed here is one line of that argument rather than all of it, since after differentiating it is the cosines that must be orthogonal. That line is $\int_{0}^{T}\cos(j\pi t/T)\cos(k\pi t/T)\,\dd t = 0$ for $j\neq k$, which follows from the product-to-sum identity and one integration. So the cross terms integrate to zero and

This is purely quadratic in the amplitudes, which is why halving them quarters $\Delta S$. At $a_{1}=0.1$, $a_{2}=a_{3}=0$ it predicts $\Delta S=\pi^{2}(0.01)/4 = 0.024674$, and the widget's quadrature reports $0.024674$. The number on the screen is not decorative.

Now run the same expansion on $\tilde S=\int(T+V)\dd t$, and the cancellation fails. The sign flip turns $-\int mg\eta$ into $+\int mg\eta$, so instead of cancelling the by-parts term it doubles it, leaving a first variation $\delta\tilde S = 2mg\int_{0}^{T}\eta\,\dd t$. For the fundamental mode that is $2mgT\cdot(2/\pi)a_{1} = 4mgT a_{1}/\pi$, a slope of $12.48$ per metre of amplitude, which is the $\pm0.125$ the caption quotes for $a_{1}=\pm0.01$. Not stationary, and not by a little.

That callout set two facts side by side: a second path through the same endpoints, and a direction of strictly smaller action. You can watch them arrive together. The figure runs three separate computations and hands you the one quantity all three depend on.

trip duration $\mathcal{T}$

T = 2.19911 = 0.700 π/ω

amplitude A

A = +0.6000 (mode n = 1)

mode 1 mode 2 mode 3

jump to the first conjugate point

the second path

step past the next conjugate point

reset

ΔS quadrature, 20000 cells: +0.205998719 modal identity: +0.205998718 difference 3.9e-10

Jacobi zeros, RK4 + bisection: 3.1415927 6.2831853 9.4247780 largest gap from nπ/ω: 1.1e-12

fan of extremals: spread at t = T is 6.877e-1 — max |x| over 21 RK4 solutions

negative directions k = 0 Jacobi zeros inside (0, T): 0 λ1 = +1.040816

E–L residual along the purple path, max |x″ + ω²x| = 6.2449e-1 A·|λ1| = 6.2449e-1

every eigenvalue positive — no conjugate point yet, and the action really is a minimum here.

Grind box — the second variation in general, and the one condition you can check by eye

The oscillator's $\Delta S$ was exact because its $L$ was quadratic. In general it is only the leading correction, and getting it means applying Chapter 0.6's second-order Taylor expansion to $\Phi(\epsilon)=S[x+\epsilon\eta]$. Differentiate (1.2.15) once more in $\epsilon$ and set $\epsilon=0$:

$$\delta^{2}S \;\equiv\; \Phi''(0) \;=\; \int_{t_{1}}^{t_{2}}\!\left[\pdv{^{2}L}{x^{2}}\eta^{2} + 2\,\frac{\partial^{2}L}{\partial x\,\partial\dot x}\,\eta\dot\eta + \pdv{^{2}L}{\dot x^{2}}\dot\eta^{2}\right]\dd t,$$

so that $S[x+\epsilon\eta] = S[x] + \epsilon\,\delta S + \tfrac12\epsilon^{2}\,\delta^{2}S + o(\epsilon^{2})$. This is exactly Chapter 0.6's Hessian, with the matrix replaced by an integral kernel. Tidy it using $2\eta\dot\eta = \dv{}{t}(\eta^{2})$ and one integration by parts, whose boundary term dies as usual:

$$\delta^{2}S = \int_{t_{1}}^{t_{2}}\Big[P(t)\,\dot\eta^{2} + Q(t)\,\eta^{2}\Big]\dd t, \qquad P = \pdv{^{2}L}{\dot x^{2}}, \quad Q = \pdv{^{2}L}{x^{2}} - \dv{}{t}\!\left(\frac{\partial^{2}L}{\partial x\,\partial \dot x}\right).$$

The condition you can check by eye. A minimum requires $P(t)\ge0$ everywhere along the path. The argument is the reason fast wiggles matter. Take $\eta$ to be a tiny bump of height $A$ concentrated in a window of width $w$ around a point where $P \lt 0$. Then $\int Q\eta^{2}\sim A^{2}w$ while $\int P\dot\eta^{2}\sim P A^{2}/w$, so shrinking $w$ makes the $P$ term dominate with whatever sign $P$ has. If $P \lt 0$ anywhere, $\delta^{2}S$ can be made as negative as you like, and there is no minimum. This is the Legendre condition, and for $L=T-V$ it reads $P=m \gt 0$, satisfied automatically.

It is also worth checking on the one action in §8.1's table you can already differentiate. Chapter 2.5's free relativistic particle has $L=-mc^{2}\sqrt{1-\dot x^{2}/c^{2}}$, giving $\partial L/\partial\dot x = \gamma m\dot x$ and $P = \partial^{2}L/\partial\dot x^{2} = m\gamma^{3} \gt 0$. So $S$ is minimised. And since $S=-mc^{2}\!\int\dd\tau$ is minus the proper time, minimising $S$ means maximising the proper time along the worldline. The minus sign in front of $mc^{2}$ is there precisely to arrange that, and it is why the travelling twin, who takes a non-extremal path, comes back younger.

⚑ Legendre's condition is necessary, not sufficient. The sufficient statement is Jacobi's theorem, that $\delta^{2}S \gt 0$ for all nonzero admissible $\eta$ precisely when $P \gt 0$ and the interval contains no conjugate point. It is quoted here and not proved. It is a boundary-value problem for the equation $\dv{}{t}(P\dot u) - Qu = 0$, whose zeros are the conjugate points. For the oscillator, $P=m$, $Q=-m\omega^{2}$, that equation is $\ddot u = -\omega^{2}u$, its solution starting at $u(0)=0$ is $\sin\omega t$, and its next zero is at $t=\pi/\omega$. That is exactly the half-period the callout found by exhibiting a variation. The general theory reproduces the special case, which is the least one can ask of it.

6 · Constraints become trivial — the payoff

So far the action principle has cost effort and returned Newton's laws, which we already had. This section is where it starts paying.

6.1 · Generalised coordinates

A generalised coordinate system is any set of $n$ numbers $q^{1},\dots,q^{n}$ that fixes the configuration of the system, where $n$ is the number of independent ways the system can move. The $q$'s need not be lengths. They can be angles, areas, arc lengths along a wire, the amplitude of a normal mode, or anything else, and they need not even have the same dimensions as each other.

Why that is allowed is the whole point of §3. The derivation of (1.2.23) never assumed anything about what the $q^{i}$ meant. It assumed only that they could be varied independently. So the recipe is:

1. Choose coordinates that automatically satisfy the constraints, one coordinate per genuine degree of freedom, so that every value of the $q^{i}$ describes an allowed configuration and every allowed configuration corresponds to some $q^{i}$.
2. Write $T$ and $V$ in terms of them.
3. Turn the crank (1.2.23).

Constraint forces never appear, because in step 1 the constraint stopped being a force and became a coordinate.

6.2 · The pendulum, in three lines

Take a bob of mass $m$ on a rigid massless rod of length $\ell$, swinging in a vertical plane about a fixed pivot. There are two Cartesian coordinates and one rigid constraint, so $n=1$. The coordinate to use is $\theta$, the angle from the downward vertical.

The bob moves on a circle of radius $\ell$, so its speed is $\ell\dot\theta$ and its height relative to the pivot is $-\ell\cos\theta$. Hence

That completes step two of the recipe, and $L$ is now the entire physical input to the problem. Turning the crank needs two partial derivatives of it, one in the coordinate and one in its rate of change. Take them:

What we want out of this is the equation of motion for $\theta$, and (1.2.22) is what converts a pair of partials into one. So feed them in:

That is the whole derivation. No forces were resolved and no free-body diagram was drawn. And, which is the point, the tension in the rod never appeared.

Chapter 1.1 got the same equation by resolving forces along and perpendicular to the rod. That required introducing the unknown tension $\mathcal{F}$, writing two component equations, and eliminating $\mathcal{F}$ between them. Worked example 1 does the two side by side so you can count the lines.

Let's pause on why the tension is absent here. The rod's tension is always perpendicular to the bob's motion, so it does no work. Energies are what $L$ is built from, and a force that does no work contributes no energy, so it cannot appear in $L$.

Meanwhile the geometric job the tension was doing, which was keeping the bob at distance $\ell$, has been absorbed into the choice of $\theta$ as the coordinate. Every value of $\theta$ automatically puts the bob at radius $\ell$.

So the constraint has been converted from a force into a coordinate, and that trade is the central practical advantage of the Lagrangian formulation.

6.3 · A harder one, to show the method scales

Now the double pendulum. There is a rod of length $\ell_{1}$ from a fixed pivot to mass $m_{1}$, and a second rod of length $\ell_{2}$ from $m_{1}$ to $m_{2}$. Four Cartesian coordinates, two rigid constraints, so $n=2$. Take $\theta_{1}$ and $\theta_{2}$, each measured from the downward vertical.

Positions, with $y$ measured downward from the pivot:

Differentiate and square. For the second mass, $\dot x_{2}^{2}+\dot y_{2}^{2} = \ell_{1}^{2}\dot\theta_{1}^{2} + \ell_{2}^{2}\dot\theta_{2}^{2} + 2\ell_{1}\ell_{2}\dot\theta_{1}\dot\theta_{2}\big(\cos\theta_{1}\cos\theta_{2} + \sin\theta_{1}\sin\theta_{2}\big)$, and the bracket collapses by the angle-subtraction identity. Heights are $-y$, so

Two lines, and the dynamics is completely specified. Apply (1.2.23) twice and you have the equations of motion, coupling and all.

Set the two routes beside each other.

• The Newtonian route requires four unknown constraint-force components, since there are two rods and each exerts a force with two components on each of its ends. That means four vector equations, and the elimination of all four unknowns.
• The Lagrangian route needed no unknowns at all, because the coordinates $(\theta_{1},\theta_{2})$ already encode both rigid rods.

Now the scaling argument. Adding a third link adds one coordinate, one term to $T$, one term to $V$, and one Euler–Lagrange equation. In the force picture it adds two more unknown constraint forces to be eliminated.

So the Lagrangian method grows linearly in the number of degrees of freedom. The force method grows with the number of constraints, which is exactly the quantity you were hoping to make go away.

Grind box — the double pendulum's kinetic energy, every step

Differentiate (1.2.39):

$$\begin{aligned} \dot x_{1} &= \ell_{1}\dot\theta_{1}\cos\theta_{1}, & \dot y_{1} &= -\ell_{1}\dot\theta_{1}\sin\theta_{1},\\ \dot x_{2} &= \ell_{1}\dot\theta_{1}\cos\theta_{1}+\ell_{2}\dot\theta_{2}\cos\theta_{2}, & \dot y_{2} &= -\ell_{1}\dot\theta_{1}\sin\theta_{1}-\ell_{2}\dot\theta_{2}\sin\theta_{2}. \end{aligned}$$

The first mass is easy: $\dot x_{1}^{2}+\dot y_{1}^{2} = \ell_{1}^{2}\dot\theta_{1}^{2} (\cos^{2}\theta_{1}+\sin^{2}\theta_{1}) = \ell_{1}^{2}\dot\theta_{1}^{2}$. For the second, square both and add:

$$\begin{aligned} \dot x_{2}^{2} &= \ell_{1}^{2}\dot\theta_{1}^{2}\cos^{2}\theta_{1} + \ell_{2}^{2}\dot\theta_{2}^{2}\cos^{2}\theta_{2} + 2\ell_{1}\ell_{2}\dot\theta_{1}\dot\theta_{2}\cos\theta_{1}\cos\theta_{2},\\ \dot y_{2}^{2} &= \ell_{1}^{2}\dot\theta_{1}^{2}\sin^{2}\theta_{1} + \ell_{2}^{2}\dot\theta_{2}^{2}\sin^{2}\theta_{2} + 2\ell_{1}\ell_{2}\dot\theta_{1}\dot\theta_{2}\sin\theta_{1}\sin\theta_{2}. \end{aligned}$$

The Pythagorean identity collapses the first two columns, and the cross terms combine through $\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2} = \cos(\theta_{1}-\theta_{2})$:

$$\dot x_{2}^{2}+\dot y_{2}^{2} = \ell_{1}^{2}\dot\theta_{1}^{2}+\ell_{2}^{2}\dot\theta_{2}^{2}+2\ell_{1}\ell_{2}\dot\theta_{1}\dot\theta_{2}\cos(\theta_{1}-\theta_{2}).$$

Then $T=\half m_{1}\ell_{1}^{2}\dot\theta_{1}^{2} + \half m_{2}\big(\dot x_{2}^{2}+\dot y_{2}^{2}\big)$, which on collecting the $\dot\theta_{1}^{2}$ terms gives (1.2.40). With $y$ measured downward, the height of mass $j$ above the pivot is $-y_{j}$, so $V = -m_{1}gy_{1}-m_{2}gy_{2}$, giving the two potential terms with the signs shown.

A structural remark you will need in Chapter 1.3. The kinetic energy came out as a quadratic form in the generalised velocities, $T=\half\sum_{ij}M_{ij}(\theta)\,\dot\theta^{i}\dot\theta^{j}$ with

$$M(\theta) = \begin{pmatrix} (m_{1}+m_{2})\ell_{1}^{2} & m_{2}\ell_{1}\ell_{2}\cos(\theta_{1}-\theta_{2})\\ m_{2}\ell_{1}\ell_{2}\cos(\theta_{1}-\theta_{2}) & m_{2}\ell_{2}^{2}\end{pmatrix},$$

symmetric and positive definite, the "mass matrix". That is not an accident of this example. Whenever the coordinate change from Cartesians is time-independent, $T$ is always a positive-definite quadratic form in $\dot q$, because it is $\half\sum m_{a}\abs{\dot{\vv x}_{a}}^{2}$ pushed through a linear map (Chapter 0.5). Positive-definiteness is what makes $p=\partial L/\partial\dot q$ invertible for $\dot q$, which is precisely the condition Chapter 1.3 needs for the Legendre transform to exist. And a metric-like symmetric positive-definite matrix that depends on position is exactly the object Chapter 3.3 calls a metric. The kinetic energy is a metric on configuration space, and free motion is a geodesic in it.

6.4 · When you want the constraint force: Lagrange multipliers

Sometimes the constraint force is the answer you need. You are designing the rod and you want to know whether it will snap. Eliminating the tension was convenient, and now we want it back. Chapter 0.6 already built the tool.

There, the problem was to make $f(\vv x)$ stationary subject to $g(\vv x)=0$. The resolution was to make the unconstrained function $f - \lambda(g-c)$ stationary in both $\vv x$ and $\lambda$, trading a restriction on the search space for one extra equation and one extra unknown. We do exactly the same thing here, with the search space now a space of paths.

Suppose we use redundant coordinates $q^{1},\dots,q^{N}$, with $N$ larger than the number of degrees of freedom, subject to a constraint $g(q,t)=0$ that must hold at every instant. Consider the modified action

Here $\lambda(t)$ is one more variable to be varied over. Note that it is a function of time, one multiplier per instant, because the constraint must hold at every instant.

Varying with respect to $\lambda$ is easy. It appears without its derivative, so (1.2.22) with $\partial\tilde L/\partial\dot\lambda=0$ gives $\partial\tilde L/\partial\lambda = g(q,t) = 0$. The constraint has been recovered as an equation of motion.

Varying with respect to $q^{i}$ gives

Compare that with (1.2.23). The left-hand side is the free Euler–Lagrange expression, which in Cartesian coordinates is $\dot p_{i}-F_{i}$. So the right-hand side is an extra force, and $\lambda\,\partial g/\partial q^{i}$ is precisely the force that enforces the constraint. Two features are worth naming:

• The constraint force points along $\nabla g$, perpendicular to the constraint surface, exactly as Chapter 0.6 showed the gradient is normal to a level set. That is the mathematical content of "the constraint force does no work". Motion allowed by the constraint is tangent to the surface, and a force normal to it has zero dot product with the motion.
• The multiplier is the magnitude of that force, up to the normalisation of $g$, which matches Chapter 0.6's finding. There is a sign convention worth stating rather than stumbling over. Chapter 0.6 wrote $\mathcal L = f-\lambda(g-c)$ and this section writes $L+\lambda g$, so the two $\lambda$'s are negatives of one another. Either way the multiplier is the sensitivity of the optimum to relaxing the constraint. Relax a rod's length and the work you must do is set by its tension. Same $\lambda$, same meaning.

Worked example 1 runs this on the pendulum and recovers the tension explicitly. The same trade, which is either to restrict the space or else to enlarge it and add a multiplier, reappears twice more. It is in Chapter 3.6, where the metric is varied subject to nothing, and yet the analogous structure produces constraint equations. And it is in Chapter 6.3, where the "constraint" being imposed by a multiplier is a gauge condition.

7 · Form invariance — the property that survives to general relativity

Here is the structural property that makes the Lagrangian formulation not merely convenient but necessary for everything after Chapter 2.

Claim. Let $q^{i}=q^{i}(Q^{1},\dots,Q^{n},t)$ be any invertible, smooth change of coordinates (possibly time-dependent), and define the Lagrangian in the new coordinates by substitution,

Then $Q^{a}(t)$ satisfies the Euler–Lagrange equations of $\tilde L$ if and only if $q^{i}(t)$ satisfies the Euler–Lagrange equations of $L$.

The one-line reason. $S$ is a number attached to a path. Relabelling which numbers you use to name the points of the path cannot change that number, because you fed in the same physical path and got the same integral back.

So the function $\Phi(\epsilon)=S[\text{path deformed by }\epsilon]$ is identical in both descriptions, and so is the statement $\Phi'(0)=0$. Stationarity is a property of the path, and not of the names.

That argument is correct and it is the reason. It is worth also seeing the derivative-level version, because the two lemmas it uses recur constantly.

Lemma 1 (cancellation of the dots). Differentiating $q^{i}(Q(t),t)$ along the path,

The right-hand side is linear in $\dot Q^{a}$ with coefficients depending only on $Q$ and $t$. Reading off the coefficient,

Lemma 2 (partial and total derivatives commute). Differentiate (1.2.44) with respect to $Q^{a}$ at fixed $\dot Q$. Separately, take the total time derivative of $\partial q^{i}/\partial Q^{a}$ using the chain rule. Both give the same thing:

where the middle step used the equality of mixed partials (Chapter 0.6). Now the computation. From (1.2.43) and the chain rule, using Lemma 1,

and, since $\tilde L$ depends on $Q^{a}$ both directly and through $\dot q$,

Take $\dd/\dd t$ of (1.2.47) with the product rule and subtract (1.2.48):

By Lemma 2 the second and fourth terms are equal and cancel. What is left factorises:

The Euler–Lagrange expressions transform into each other by multiplication by the Jacobian matrix $\partial q^{i}/\partial Q^{a}$, which is invertible because the coordinate change is. So one set vanishes exactly when the other does. $\blacksquare$

7.1 · Why this matters more than it looks

Now the contrast. $F=ma$ is not form invariant. Write Newton's law in plane polar coordinates and it does not read $m\ddot r = F_{r}$. The acceleration of a particle in polar coordinates has extra pieces:

The $-r\dot\theta^{2}$ is the centrifugal term and the $2\dot r\dot\theta$ is the Coriolis term. They arise because the basis vectors $\hat{\vv r},\hat{\boldsymbol\theta}$ themselves rotate as the particle moves, so differentiating a vector twice requires differentiating the basis too. Getting (1.2.51) honestly takes half a page of unit-vector derivatives.

Now get the same thing from the Lagrangian. In polar coordinates $T=\half m(\dot r^{2}+r^{2}\dot\theta^{2})$, which is Pythagoras applied to the radial and tangential velocity components. So with a potential $V(r,\theta)$,

Turn the crank. For $q=r$ we have $\partial L/\partial\dot r = m\dot r$ and $\partial L/\partial r = mr\dot\theta^{2} - \partial V/\partial r$, so

For $q=\theta$ we have $\partial L/\partial\dot\theta = mr^{2}\dot\theta$ and $\partial L/\partial\theta = -\partial V/\partial\theta$, so

Both the centrifugal and the Coriolis terms appeared automatically, out of differentiating a scalar. No unit vectors were differentiated, and no fictitious forces had to be remembered.

And (1.2.54) hands you a bonus. If $V$ does not depend on $\theta$, the left side says that $mr^{2}\dot\theta$ is constant. That is angular momentum conservation, for free, because $\theta$ was missing from $L$.

8 · One equation to rule them all — a preview

Everything so far has had a path $x(t)$ at the centre of it, meaning one independent variable, $t$, and finitely many dependent ones. Physics after Chapter 5 has fields instead. A field is a quantity $\phi$ defined at every point of spacetime, so the dependent variable is indexed by four independent ones, $x^{\mu}=(t,x,y,z)$.

The generalisation writes itself. Replace the Lagrangian by a Lagrangian density $\mathcal{L}\big(\phi,\partial_{\mu}\phi\big)$ and the time integral by an integral over a region $\Omega$ of spacetime:

Now repeat §3 verbatim. Vary $\phi\to\phi+\epsilon\eta$, with $\eta$ vanishing on the boundary $\partial\Omega$ instead of at two endpoints. Differentiate under the integral sign. Linearise the integrand by the chain rule:

Then integrate by parts to move the derivative off $\partial_{\mu}\eta$. In four dimensions, "by parts" means the divergence theorem of Chapter 0.7, which converts $\int_{\Omega}\partial_{\mu}(\cdots)^{\mu}\dd^{4}x$ into a surface integral over $\partial\Omega$. That surface term dies because $\eta$ vanishes there, which is the same reason and the same sentence as before. Apply the fundamental lemma, whose proof needs only the four-dimensional bump function you would expect. Out comes

Put (1.2.57) beside (1.2.22). The dictionary is $t\to x^{\mu}$, $\dd t\to\dd^{4}x$, $\dv{}{t}\to\partial_{\mu}$, $\dot x\to\partial_{\mu}\phi$, and nothing else. The derivation you did in §3 is the derivation that produces every field equation in physics.

Chapter 5.2 develops this properly, including what happens when $\phi$ carries indices and what Lorentz invariance demands of $\mathcal{L}$, so there is no point duplicating it here. What matters now is that there is nothing new to learn. It is the same four moves in more dimensions.

8.1 · The actions this book will write down

⚑ Everything in this table is quoted forward. Not one line of it is derived here. Each is constructed, from symmetry principles, in the chapter named. It is on this page because seeing the list once, early, tells you what kind of book this is. After Chapter 1.2, physics consists of choosing a scalar and varying it.

System	Action	Chapter
Free particle	$S=\int \half m\,\dot{\vv x}^{2}\,\dd t$	§4
Relativistic particle	$S=-mc^{2}\!\int\!\dd\tau$	2.5
Electromagnetic field	$S=-\tfrac14\!\int\! F_{\mu\nu}F^{\mu\nu}\,\dd^{4}x$	2.6
Gravity (Einstein–Hilbert)	$S=-\dfrac{1}{2\kappa}\!\int\! R\,\sqrt{-g}\;\dd^{4}x$	3.6
Scalar field (Klein–Gordon)	$S=\!\int\!\big(\half\partial_{\mu}\phi\,\partial^{\mu}\phi-\half m^{2}\phi^{2}\big)\dd^{4}x$	5.3
Spin-½ field (Dirac)	$S=\!\int\!\bar\psi\big(\ii\gamma^{\mu}\partial_{\mu}-m\big)\psi\;\dd^{4}x$	5.5
Non-abelian gauge field (Yang–Mills)	$S=-\half\!\int\!\operatorname{tr}\big(F_{\mu\nu}F^{\mu\nu}\big)\dd^{4}x$	6.4
Relativistic string (Polyakov)	$S=-\dfrac{T}{2}\!\int\!\sqrt{-h}\;h^{ab}\,\partial_{a}X^{\mu}\partial_{b}X_{\mu}\;\dd^{2}\sigma$	7.2

The overall sign of the gravitational entry depends on the metric signature, and with this book's $(+,-,-,-)$ it is negative. Chapter 3.6 derives it rather than adopting it, and says in place why texts using the opposite signature write it the other way round.

Eight lines. Between them they contain classical mechanics, special relativity, electromagnetism, gravitation, quantum field theory, the strong and electroweak interactions, and string theory. Every one is fed to (1.2.22) or (1.2.57), and the output is the equations of motion of that theory.

To a first approximation, the rest of this book is the story of how each of those eight lines gets chosen. And that is always the same story: pick a symmetry, and write the most general scalar invariant under it.

9 · Worked examples

Grind box — integrating the sphere's geodesic equation in the general case

Solve $\sin^{2}\theta\,\phi'/\sqrt{1+\sin^{2}\theta\,\phi'^{2}} = k$ for $\phi'$. Square and cross-multiply:

$$\sin^{4}\theta\,\phi'^{2} = k^{2}\big(1+\sin^{2}\theta\,\phi'^{2}\big) \;\Longrightarrow\; \phi'^{2}\sin^{2}\theta\big(\sin^{2}\theta-k^{2}\big) = k^{2} \;\Longrightarrow\; \phi' = \frac{k}{\sin\theta\sqrt{\sin^{2}\theta-k^{2}}}.$$

Substitute $u=\cot\theta$, so $\dd u = -\csc^{2}\theta\,\dd\theta$ and $\sin^{2}\theta - k^{2} = \sin^{2}\theta\big(1-k^{2}(1+u^{2})\big)$. The integral collapses:

$$\phi-\phi_{0} = \int\frac{k\,\dd\theta}{\sin^{2}\theta\sqrt{1-k^{2}-k^{2}u^{2}}} = -\int\frac{k\,\dd u}{\sqrt{1-k^{2}-k^{2}u^{2}}} = -\int\frac{\dd u}{\sqrt{A^{2}-u^{2}}}, \quad A^{2}\equiv\frac{1-k^{2}}{k^{2}}.$$

Hence $\phi-\phi_{0} = -\arcsin(u/A)$, that is, $\cot\theta = -A\sin(\phi-\phi_{0})$. Multiply through by $\sin\theta$ and expand the sine:

$$\cos\theta = -A\sin\theta\big(\sin\phi\cos\phi_{0}-\cos\phi\sin\phi_{0}\big).$$

Now recognise the Cartesian coordinates of a point on the unit sphere: $X=\sin\theta\cos\phi$, $Y=\sin\theta\sin\phi$, $Z=\cos\theta$. The equation becomes

$$\big(A\sin\phi_{0}\big)X \;-\; \big(A\cos\phi_{0}\big)Y \;-\; Z \;=\; 0,$$

which is a plane through the origin. A plane through the centre of a sphere cuts it in a great circle. So every solution, for every $k$, is a great circle, with the $k=0$ case being the limit in which the plane contains the polar axis. (Verified numerically to $7\times10^{-16}$ for $A=0.8$, $\phi_{0}=0.4$.)

Notice that $k=\partial F/\partial\phi'$ is conserved because $\phi$ is absent from $F$. That is the same structure that gave angular momentum conservation in (1.2.54). On a sphere it is the statement that a geodesic's bearing satisfies $\sin\theta\cos(\text{bearing})=$ constant, which navigators call Clairaut's relation and which is the reason a great-circle route from London to Tokyo goes over the Arctic. It is Noether's theorem (Chapter 1.4), arriving three chapters early disguised as cartography.

10 · Your turn