Part 0 · The Toolkit — Chapter 0.6

Multivariable Calculus

One equation from Chapter 0.1, with a vector in it. Everything else is consequences.

Tools you'll need  — Chapter 0.1: the derivative as a linear approximation, the $o(h)$ bookkeeping, the chain rule as composition. Chapter 0.3: Taylor's theorem with remainder, used once in §6. Chapter 0.4: linear maps, their matrices in a basis, composition as matrix multiplication, and the determinant as a signed volume scaling factor. Chapter 0.5: inner products, Cauchy–Schwarz with its equality condition, orthogonal projection, and the spectral theorem for symmetric matrices. Chapters 0.4 and 0.5 are not decoration here. Sections 3, 6 and 7 each cash a specific theorem from them.

1 · Partial derivatives, and the trap

Let's start with the move everybody makes first. When a function has several inputs, freeze all but one of them and differentiate in the survivor. Let $f:\R^{n}\to\R$ and let $\vv e_{1},\ldots,\vv e_{n}$ be the standard basis. The partial derivative of $f$ with respect to $x^{i}$ at $\vv a$ is

Nothing new is happening here. Restricted to the line $\vv a + t\,\vv e_{i}$, the function $f$ is a function of the single variable $t$, and (0.6.1) is Chapter 0.1's definition applied to it. Every rule you already have applies verbatim: product, quotient, chain. The reason is that during the computation the other variables are literally constants. So $\partial_{x}(x^{2}y^{3}) = 2xy^{3}$ and $\partial_{y}(x^{2}y^{3}) = 3x^{2}y^{2}$, and there is no technique to learn.

The trap is that this feels like it ought to be the whole story, and it is not. Partial derivatives are a genuinely weaker notion than differentiability, and the gap between the two is not subtle. Here is the example that shows how wide it is.

Check the partials first. Along the $x$-axis, $f(x,0) = 0/(x^{2}) = 0$ for every $x$, so the function $t\mapsto f(t,0)$ is identically zero and its derivative at $t=0$ is $0$. Hence $\partial_{x}f(0,0)=0$, and by the symmetry $f(x,y)=f(y,x)$ also $\partial_{y}f(0,0)=0$. Both partials exist, are finite, and are perfectly well behaved.

Now let's do the thing the partial derivatives never do, which is to approach the origin along some other direction. Take the diagonal $y=x$. For $x\neq0$,

a constant. So $f\to\tfrac12$ along the diagonal and $f\to0$ along the axes. Two different limits mean there is no limit at all, so $f$ is not continuous at the origin. And a function that is not continuous at a point certainly does not have a good linear approximation there.

It is worth seeing why the function is so badly behaved, because the reason is geometrical rather than algebraic. We want the distance from the origin and the direction of approach to appear as separate quantities, so switch to polar coordinates. Write $x=r\cos\theta$ and $y=r\sin\theta$. Then

Let's look at what that line is actually saying. The radius has cancelled completely, so $f$ depends only on the direction of approach and never on the distance. Every value in $[-\tfrac12,\tfrac12]$ is attained arbitrarily close to the origin, on some ray or other.

Now notice where the two coordinate axes sit in that picture. They happen to be rays on which $f$ vanishes. The partial derivatives look along exactly those two rays, and so they see nothing wrong. They are two samples from a circle's worth of behaviour.

That is the honest indictment. Partial derivatives probe $n$ lines through a point. A function has infinitely many directions available to it, and it can misbehave in all the ones you didn't check.

So we need a definition that constrains the function's behaviour in every direction at once, and constrains it uniformly. Chapter 0.1 already wrote that definition down. We only have to let the displacement be a vector.

2 · The total derivative

Here is the definition the rest of the chapter runs on. Let $f:\R^{n}\to\R^{m}$.

Compare it with the one-variable equation letter by letter. The point $a$ became $\vv a$, and the displacement $h$ became $\vv h$. The number $f'(a)$ became the linear map $\mathrm{D}f(\vv a)$, and "multiply by $f'(a)$" became "apply the linear map". Multiplication by a number is exactly what a linear map $\R\to\R$ does, so the old definition is the case $n=m=1$ of the new one rather than an analogy to it.

Notice how much stronger this is than §1. The limit is over $\vv h\to\vv 0$ with no restriction on direction, so the approximation has to hold uniformly as $\vv h$ approaches zero any way at all. That includes along spirals, and along the diagonal that broke the counterexample.

One consequence is immediate. Differentiability at $\vv a$ forces continuity at $\vv a$: as $\vv h\to\vv 0$ the right-hand side of (0.6.4) tends to $f(\vv a)$, because a linear map sends $\vv 0$ to $\vv 0$ and is continuous.

2.1 · There is only one such map

The definition says "there exists". It is worth two lines to check that it cannot exist twice, because otherwise "the derivative" would be an abuse of language.

Suppose $A$ and $B$ both satisfy (0.6.4). Subtracting the two statements, the function values cancel and we are left with $(A-B)\vv h = o(\abs{\vv h})$. Our goal is to show that the map $A-B$ kills every vector. So fix any direction $\vv u$ with $\abs{\vv u}=1$, put $\vv h = t\vv u$ for small $t\gt0$, and see what the $o$ statement becomes. By linearity $(A-B)(t\vv u) = t\,(A-B)\vv u$, so

The left side must tend to $0$ as $t\to0$. The right side does not depend on $t$ at all. A constant that tends to zero is zero, so $(A-B)\vv u = \vv 0$ for every $\vv u$, and therefore $A=B$. The derivative, when it exists, is unique.

Note the mechanism, because we will reuse it. The linearity did all the work, by letting us pull the small parameter out in front and cancel it.

2.2 · Its matrix is the Jacobian

$\mathrm{D}f(\vv a)$ is a linear map, so by Chapter 0.4 it has a matrix once we fix bases. The recipe for that matrix is worth restating: the $j$-th column is the image of the $j$-th basis vector. So our goal is to find out what the map does to each $\vv e_{j}$, and the way to find out is to feed basis vectors into the definition. Put $\vv h = t\,\vv e_{j}$ in (0.6.4):

where we used $\mathrm{D}f(\vv a)(t\vv e_{j}) = t\,\mathrm{D}f(\vv a)\vv e_{j}$ by linearity. We want the column $\mathrm{D}f(\vv a)\vv e_{j}$ on its own, so rearrange and divide by $t$:

Look at the left-hand side. Componentwise, it is exactly the definition (0.6.1) of the partial derivative with respect to $x^{j}$. So the $j$-th column of the matrix of $\mathrm{D}f(\vv a)$ is the column of partials $\partial_{j}f^{1},\ldots,\partial_{j}f^{m}$. Collecting the columns side by side gives the matrix a name, the Jacobian:

Read the shape off it. There are $m$ rows, one per output component, and $n$ columns, one per input direction. A map $\R^{3}\to\R^{2}$ therefore has a $2\times3$ Jacobian. There is never any doubt about which way round it goes, because a matrix acting on a column vector in $\R^{n}$ must have $n$ columns.

So the partials do reconstruct the derivative, once you know the derivative exists. That qualifier is the entire content of §1. The logic runs one way only:

and the converse is false, by (0.6.3). What rescues the converse is one extra hypothesis. This is the one theorem in the chapter we will quote in the main text rather than prove there:

Here is why it is safe to lean on that theorem. Every function you will differentiate in this book is built from polynomials, exponentials, logarithms and trigonometric functions by arithmetic and composition. The partials of such a thing are again of the same type, and they are continuous wherever they are defined. So in practice you compute the partials, observe that they are continuous, and conclude that the Jacobian really is the derivative.

That tells you how to read the counterexample of §1. It is not a warning about your daily calculations. It is a warning about what the definition has to be if the theorems are going to come out true. As for the proof, we have put it below rather than here because it needs the mean value theorem from Chapter 0.2 and adds nothing to the picture. It is written out in full, because this book does not say "it can be shown".

Grind box — the boundedness lemma, and continuous partials ⇒ differentiable

Lemma (linear maps are bounded). For any linear $A:\R^{n}\to\R^{m}$ there is a constant $C$ with $\abs{A\vv v}\le C\abs{\vv v}$ for all $\vv v$.

Proof. Write $\vv v = \sum_{i}v^{i}\vv e_{i}$, so $A\vv v = \sum_{i}v^{i}A\vv e_{i}$. By the triangle inequality and then Cauchy–Schwarz (Chapter 0.5) applied to the two lists of numbers $\abs{v^{i}}$ and $\abs{A\vv e_{i}}$,

$$\abs{A\vv v} \;\le\; \sum_{i}\abs{v^{i}}\,\abs{A\vv e_{i}} \;\le\; \left(\sum_{i}(v^{i})^{2}\right)^{1/2}\!\left(\sum_{i}\abs{A\vv e_{i}}^{2}\right)^{1/2} \;=\; C\abs{\vv v}$$

with $C=\big(\sum_{i}\abs{A\vv e_{i}}^{2}\big)^{1/2}$, a number depending only on $A$. ∎

This innocuous fact is used constantly and silently. It is what lets us say $\mathrm{D}f\big(o(\abs{\vv h})\big) = o(\abs{\vv h})$, meaning that a linear map does not turn a negligible displacement into a non-negligible one. It is also where finite-dimensionality enters. In infinite dimensions (Chapter 4.4) linear maps can be unbounded, $\dv{}{x}$ being the standard offender, and a great deal of quantum mechanics' technical difficulty descends from exactly that.

Theorem. If the partials of $f$ exist near $\vv a$ and are continuous at $\vv a$, then $f$ is differentiable at $\vv a$.

Proof for $n=2$, $m=1$. The general case is the same argument with more terms. Write $\vv a=(a,b)$ and $\vv h=(h,k)$. We want to reach a one-variable situation, where the mean value theorem is available, so split the total change into two moves, one along each axis, by adding and subtracting the corner value:

$$f(a+h,b+k)-f(a,b) = \underbrace{\big[f(a+h,b+k)-f(a,b+k)\big]}_{\text{move in }x} + \underbrace{\big[f(a,b+k)-f(a,b)\big]}_{\text{move in }y}.$$

Each bracket changes only one variable, so each is a one-variable increment and the mean value theorem of Chapter 0.2 applies to it. There exist $\theta_{1},\theta_{2}\in(0,1)$ with

$$f(a+h,b+k)-f(a,b) = h\,\partial_{x}f(a+\theta_{1}h,\,b+k) \;+\; k\,\partial_{y}f(a,\,b+\theta_{2}k).$$

Now use continuity of the partials at $(a,b)$: both evaluation points converge to $(a,b)$ as $(h,k)\to(0,0)$, so we may write $\partial_{x}f(a+\theta_{1}h,b+k) = \partial_{x}f(a,b)+\varepsilon_{1}$ and $\partial_{y}f(a,b+\theta_{2}k) = \partial_{y}f(a,b)+\varepsilon_{2}$ with $\varepsilon_{1},\varepsilon_{2}\to0$. Then

$$f(\vv a+\vv h)-f(\vv a) = \underbrace{h\,\partial_{x}f(\vv a) + k\,\partial_{y}f(\vv a)}_{\text{linear in }\vv h} \;+\; \underbrace{h\varepsilon_{1}+k\varepsilon_{2}}_{\text{remainder}},$$

and since $\abs{h}\le\abs{\vv h}$ and $\abs{k}\le\abs{\vv h}$,

$$\frac{\abs{h\varepsilon_{1}+k\varepsilon_{2}}}{\abs{\vv h}} \;\le\; \abs{\varepsilon_{1}}+\abs{\varepsilon_{2}} \;\longrightarrow\; 0 .$$

That is exactly (0.6.4) with the Jacobian as the linear map. ∎

Now let's see where the counterexample of §1 fails this test. Its partials do exist away from the origin. But a short computation from (0.6.3) shows that they blow up like $1/r$ as you approach it, so they are nowhere near continuous at the origin. The hypothesis is not decorative.

3 · The gradient

Now let's specialise to a scalar-valued function, meaning $m=1$. This is the case physics uses most: a temperature field, a potential, an action, a log-likelihood. Here $\mathrm{D}f(\vv a):\R^{n}\to\R$ is a linear map into the reals, which is called a linear functional. In Chapter 0.4's terms it is a linear map whose target happens to be one-dimensional. Its matrix is therefore a single row, $\big(\partial_{1}f\;\cdots\;\partial_{n}f\big)$.

A row of numbers looks like a vector lying on its side, and the temptation to stand it up is irresistible. Give in to it. But notice the step as you take it, because §4 is about what it costs. Standing the row up means pairing it with a column, and pairing needs a dot product, so use the standard inner product of Chapter 0.5 to define the gradient $\nabla f(\vv a)$ as the unique vector satisfying

That says what $\nabla f$ does, and now we want its components. They follow by feeding in basis vectors, exactly as in §2.2. Taking $\vv h=\vv e_{i}$ gives $\nabla f\cdot\vv e_{i} = \partial_{i}f$, and the left side of that is the $i$-th component of $\nabla f$. So

3.1 · Steepest ascent, derived from Cauchy–Schwarz

Let's ask the natural question. In which direction does $f$ increase fastest? To answer it we first have to say what "rate of increase in a direction" means. For a unit vector $\hat u$, the directional derivative is the rate of change of $f$ along $\hat u$,

The middle equality there is (0.6.4) with $\vv h = t\hat u$, divided by $t$, in the limit. It is the same computation as (0.6.7), with an arbitrary direction in place of a basis vector.

So the question has become a question about a dot product: over all unit $\hat u$, which one maximises $\nabla f\cdot\hat u$? That is precisely the question Cauchy–Schwarz answers, and we proved it in Chapter 0.5. For any two vectors,

We want a bound on $\nabla f\cdot\hat u$, so apply that with $\vv p = \nabla f$ and $\vv q = \hat u$, remembering that $\abs{\hat u}=1$ makes the right-hand side collapse:

A bound is only useful if something reaches it, and the equality clause says both of these are reached. The upper bound is attained at $\hat u = \nabla f/\abs{\nabla f}$, where $\nabla f\cdot\hat u = \abs{\nabla f}^{2}/\abs{\nabla f} = \abs{\nabla f}$. The lower bound is attained at $\hat u = -\nabla f/\abs{\nabla f}$. So, with nothing assumed and nothing waved at:

This is the cleanest single payoff for having proved Cauchy–Schwarz. Now look at the other extreme of the same inequality. If $\nabla f\cdot\hat u = 0$ then $f$ does not change at all, to first order, along $\hat u$. Those directions are worth a name, and they have a geometric meaning.

3.2 · The gradient is perpendicular to level sets

A level set of $f$ is a set $\{\vv x : f(\vv x)=c\}$. Think of an isotherm, an equipotential, or a contour line on a map. Let $\vv x(t)$ be any differentiable curve that lies inside one of them, so that $f(\vv x(t)) = c$ for all $t$. Our plan is to differentiate that identity, since a constant has derivative zero and that will hand us a statement about $\nabla f$.

We can do the differentiation straight from the definition, which has the side benefit of proving the special case of the chain rule we will need repeatedly. Expand $\vv x$ to first order as $\vv x(t+s) = \vv x(t) + s\,\dot{\vv x}(t) + o(s)$, and feed the result into (0.6.4):

The boundedness lemma of §2 was used there to absorb $\mathrm{D}f\big[o(s)\big]$ into $o(s)$. Now compare that last line with the one-variable definition of a derivative in $s$. The coefficient of $s$ is the derivative, so we can read it straight off:

That holds for any curve at all. Now impose the extra condition that our curve lies in a level set. Then $f(\vv x(t))$ is the constant $c$, so the left side is $\dv{c}{t}=0$, and we are left with

The velocity $\dot{\vv x}$ of such a curve is exactly what we mean by a vector tangent to the level set. So $\nabla f$ is orthogonal to every tangent direction of the level set through $\vv a$. The gradient is perpendicular to the level surface, and points across it in the direction of increase.

The two facts fit together into one picture. Contours are the lines along which nothing changes. The gradient is perpendicular to them. And it is longest where the contours are closest together, because that is where a small step changes $f$ the most. The figure below is that picture, with the arithmetic exposed so you can check it rather than believe it.

point x

point y

show constraint (§7)

∇f = (2.80, 5.60), |∇f| = 6.261

∇f · t̂ = -2.8e-12 (zero)

4 · The gradient is really a one-form — and why you must care

Go back to (0.6.10) and look at what was actually done there. We had a perfectly good object, the linear map $\mathrm{D}f(\vv a)$, which eats a displacement vector and returns a number. We converted it into a vector, $\nabla f$, by demanding that dotting with that vector reproduce the map. The conversion required a dot product. It could not have been done without one.

So let's separate the two objects and take them one at a time. First the one that came for free. It deserves its usual name, so write

with a sum over $i$. The object $\dd f$ is a one-form, also called a covector. It is a linear machine that eats a vector and returns a number. In Chapter 0.4's language it is nothing but a linear map whose target happens to be one-dimensional. Such maps can be added and scaled, and the result is another one, so they form a vector space in their own right. That space is called the dual space $V^{*}$, and Chapter 2.4 will treat it properly.

What matters here is that $\dd f$ is defined by $f$ alone. No inner product, no notion of length and no notion of angle went into (0.6.18). All that went in was the pairing of a displacement with a rate of change, which is exactly what a physical derivative is. This is also the honest definition of the $\dd x$ you have been cancelling since school, and Chapter 0.1's warning callout promised you it would get one here.

The gradient, by contrast, is defined by two things: $f$ and an inner product. Look at the index positions, which are about to earn their keep. In (0.6.18) the components $h^{i}$ of a vector carry an upper index. The components $\partial_{i}f$ of the one-form carry a lower one. And the two are summed against each other to make a number.

Now let's make the inner product explicit, so that we can see it appear in the formula for $\nabla f$ instead of hiding in it. Given a basis, bilinearity determines an inner product completely by what it does to basis vectors. Write $\vv p = p^{i}\vv e_{i}$ and $\vv q = q^{j}\vv e_{j}$ and expand:

The array $g_{ij}$ is called the metric in that basis. It is symmetric because the inner product is. It is also invertible, because the inner product is non-degenerate: if some $\vv v$ had $g_{ij}v^{i}=0$ for all $j$, then $\vv v$ would be orthogonal to everything, including itself, and so $\vv v=\vv 0$. Both index positions are down, which matches the fact that the metric eats two vectors.

We now have everything we need to see the cost of the gradient. Our goal is to solve (0.6.10) for the components of $\nabla f$, so write that defining relation out in these components:

Both sides are sums over $j$, and the statement holds for every $\vv h$, so the coefficients of each $h^{j}$ must agree: $g_{ij}(\nabla f)^{i} = \partial_{j}f$. That is a linear system for $(\nabla f)^{i}$, so solve it by multiplying through by the inverse matrix $g^{ij}$, which exists because the inner product is non-degenerate:

There it is, in the open. Converting the one-form $\partial_{j}f$ into the vector $(\nabla f)^{i}$ is exactly the act of contracting with the inverse metric. It is called raising an index, and it is not free.

In Cartesian coordinates on Euclidean space $g_{ij}=\delta_{ij}$, so $g^{ij}=\delta^{ij}$ and (0.6.21) reads $(\nabla f)^{i}=\partial_{i}f$. The components are then numerically identical and the distinction is invisible. That invisibility is why the gradient is usually taught as a vector, and why the whole issue can seem like pedantry. It is not pedantry. Here are three places where it stops being invisible:

• Non-Cartesian coordinates, immediately. In plane polar coordinates the metric is $g_{ij}=\mathrm{diag}(1,r^{2})$, so $g^{ij}=\mathrm{diag}(1,r^{-2})$ and the gradient's angular component is $r^{-2}\partial_{\theta}f$ rather than $\partial_{\theta}f$. If you have ever wondered where the stray $1/r$ in the polar gradient formula comes from, it comes from here. The grind box does the calculation.
• Relativity, permanently. The Minkowski metric has signature $(+,-,-,-)$ (Chapter 2.3), so raising a spatial index flips its sign: $(\nabla f)^{i} = -\partial_{i}f$. No coordinate choice makes this go away, because no coordinate choice makes the metric the identity. From Chapter 2.4 onward, upper and lower indices are different objects, and mixing them is an error rather than a style preference.
• Curved space, structurally. On a manifold (Chapter 3.2) there is no inner product at all until you supply one, and supplying one is the gravitational field (Chapter 3.3). So on a manifold $\dd f$ exists always, and $\nabla f$ exists only once gravity has been specified. That is a statement about physics rather than about notation.

You have just met your first covector, and it was hiding inside the most familiar object in vector calculus. The rule to carry forward is short. Differentiation naturally produces lower indices. Vectors naturally carry upper ones. The metric is what converts between them, and when the metric is the identity you cannot see the conversion happening. Chapter 2.4 makes this a formal convention, and Chapter 3.2 makes it unavoidable. Nothing more is needed here.

Grind box — the polar gradient, derived rather than memorised

In plane polar coordinates $(r,\theta)$ the squared length of a small displacement is, by Pythagoras on a radial step $\dd r$ and a perpendicular arc $r\,\dd\theta$,

$$\dd s^{2} = \dd r^{2} + r^{2}\dd\theta^{2} \qquad\Longrightarrow\qquad g_{ij} = \begin{pmatrix}1&0\\0&r^{2}\end{pmatrix}.$$

Inverting a diagonal matrix is easy: $g^{ij}=\mathrm{diag}(1,\,r^{-2})$. So by (0.6.21) the gradient's components in the coordinate basis $\{\partial_{r},\partial_{\theta}\}$ are

$$(\nabla f)^{r} = \partial_{r}f, \qquad (\nabla f)^{\theta} = \frac{1}{r^{2}}\,\partial_{\theta}f.$$

Those are components in the coordinate basis, and the formula in the textbooks is written in the unit basis, so one more step is needed. The coordinate basis vector $\partial_{\theta}$ is not a unit vector, because moving one unit of $\theta$ moves you a distance $r$. Its normalised version is therefore $\hat{\boldsymbol\theta} = r^{-1}\partial_{\theta}$, which we can rearrange to $\partial_\theta = r\,\hat{\boldsymbol\theta}$. Substituting that in,

$$\nabla f = (\nabla f)^{r}\,\hat{\vv r} + (\nabla f)^{\theta}\,r\,\hat{\boldsymbol\theta} = \pdv{f}{r}\,\hat{\vv r} + \frac{1}{r}\pdv{f}{\theta}\,\hat{\boldsymbol\theta}.$$

That is the formula printed on the inside cover of every vector-calculus text, and you have just derived it. There are two morals. First, the notorious $1/r$ is not a quirk of polar coordinates. It is one factor of $r^{-2}$ from raising the index and one factor of $r$ from normalising the basis vector. Second, and this is the part that matters later, the "$h_{i}$ scale factors" that vector-calculus books ask you to memorise for cylindrical and spherical coordinates are nothing but the diagonal entries $\sqrt{g_{ii}}$ of the metric. Chapter 3.3 replaces the table with the metric and never mentions scale factors again.

Two sanity checks. For $f=r$ we get $\nabla f = \hat{\vv r}$, which is right: the fastest way to increase your distance from the origin is to move directly away from it, at unit rate. For $f=\theta$ we get $\nabla f = \hat{\boldsymbol\theta}/r$, which correctly says that far from the origin you must travel a long way to change your bearing.

5 · The chain rule in several variables

Chapter 0.1 derived the chain rule by composing two linear approximations and observing that linear maps compose by multiplication. In one dimension "multiplication" meant multiplying two numbers. Here it means composing two linear maps, which by Chapter 0.4 means multiplying two matrices. The derivation is word-for-word the same, so let's do it and watch the matrices appear.

Let $g:\R^{n}\to\R^{p}$ be differentiable at $\vv a$ and $f:\R^{p}\to\R^{m}$ differentiable at $\vv b = g(\vv a)$. We want the derivative of the composition, so nudge the input by $\vv h$ and follow the nudge through both maps in turn. First, $g$ moves its output by

That displacement $\vv k$ is what arrives at the input of $f$, so the next step is to feed it in there, using $f$'s own linear approximation at $\vv b$ and then substituting for $\vv k$:

Two small steps hide in that last line, and both are the boundedness lemma. First, $\mathrm{D}f(\vv b)$ applied to something $o(\abs{\vv h})$ is again $o(\abs{\vv h})$. Second, $\abs{\vv k}\le C\abs{\vv h}$ for small $\vv h$ by (0.6.22), so anything that is $o(\abs{\vv k})$ is also $o(\abs{\vv h})$.

The last line now has the exact shape of the definition (0.6.4), with a linear map sitting where the derivative belongs. Uniqueness (§2.1) says that map has no choice but to be the derivative, so we can read the coefficient off:

In matrices that reads: an $(m\times p)$ Jacobian times a $(p\times n)$ Jacobian gives an $(m\times n)$ Jacobian. The inner dimensions match automatically, because the output space of $g$ is the input space of $f$, so the shapes cannot be wrong. This is one of the reasons to think of the derivative as a map rather than as a table of numbers. The bookkeeping is enforced by the structure.

5.1 · The case you will use every day

Let $\vv x(t)$ be a curve in $\R^{n}$ and $f$ a scalar field. Then $f\circ\vv x$ is a function of one variable, $\mathrm{D}\vv x(t)$ is the column $\dot{\vv x}$, and $\mathrm{D}f$ is the row of partials, so (0.6.24) reduces to a row times a column:

which is (0.6.16), now obtained as a special case rather than from scratch. This is the workhorse of the chapter. It is how you compute the rate of change of a temperature felt by a moving thermometer, or of a potential felt by a moving charge, or of a Lagrangian along a trajectory. Chapter 1.2 opens with it.

5.2 · Changing coordinates, and a convention appearing on its own

Now the case that runs all of tensor analysis. Suppose the $x^{i}$ are themselves functions of new coordinates $u^{a}$. Think of polar coordinates written in terms of Cartesian ones, or of one observer's coordinates written in terms of another's. We want to know how a rate of change looks in the new coordinates, so take (0.6.24) and write it out entry by entry:

Three things about that equation deserve to be noticed now rather than in Chapter 2.4.

It is a matrix product. The object $\partial x^{i}/\partial u^{a}$ is the Jacobian of the coordinate change, and (0.6.26) says: to convert the components of $\dd f$ from one coordinate system to another, hit them with that Jacobian. That single sentence, promoted to a definition, is what a tensor is.

The summation happens by itself. The index $i$ appears twice on the right and is summed over. The index $a$ appears once on each side and is not summed. That pattern is not a coincidence of this particular formula. It is what always happens when you compose linear maps, because matrix multiplication sums over the shared index. Chapter 2.4 will therefore drop the $\sum$ entirely and adopt the Einstein summation convention, under which a repeated index is summed. The convention is possible because the pattern is universal, and the pattern is universal because everything in sight is a composition of linear maps.

The index heights are already right. Look at where the repeated index $i$ sits. It is upstairs in $\partial x^{i}/\partial u^{a}$, in the numerator, and downstairs in $\partial f/\partial x^{i}$, in the denominator. The rule of §4 was that differentiation lowers and coordinates raise, so one of the pair is up and the other is down. This is not luck. It is (0.6.21) and (0.6.18) telling you that the only combinations that can be summed to give a coordinate-independent number are one-up-one-down. In Chapter 2.4 that becomes a rule you can use to check your algebra. If you have written down a formula with two indices at the same height summed against each other, you have made a mistake.

6 · Second derivatives, the Hessian, and what a critical point looks like

So far everything has been first order. Let's go one order further, because that is where the shape of a function near a point lives. Differentiate a partial derivative again and you get a second partial, written $\partial_{i}\partial_{j}f$ or $\partial^{2}f/\partial x^{i}\partial x^{j}$. There are $n^{2}$ of them, and the first question to settle is whether the order of differentiation matters.

6.1 · Clairaut's theorem, with the hypothesis said out loud

The continuity hypothesis is real. The grind box below both proves the theorem and exhibits a function for which the two mixed partials at the origin come out as $+1$ and $-1$. But for every function you will meet in this book the hypothesis holds, so second partials commute and the matrix of them is symmetric. That matrix is the one we will work with from here on, and it has a name:

It is the Hessian. Note what it is structurally. The gradient $\nabla f$ is a map $\R^{n}\to\R^{n}$, and $H$ is its Jacobian. So the second derivative of a scalar field is the derivative of a vector field.

Grind box — proof of Clairaut, and the counterexample when continuity fails

Proof. Work in the two relevant variables and call them $x,y$. Consider the double difference across a small rectangle of sides $h,k$:

$$\Delta \;=\; f(a+h,b+k) - f(a+h,b) - f(a,b+k) + f(a,b).$$

It is symmetric under swapping the roles of the two variables, and we will evaluate it in the two possible orders. First group it as a difference in $x$ of the function $\varphi(x) = f(x,b+k)-f(x,b)$, so $\Delta=\varphi(a+h)-\varphi(a)$. By the mean value theorem (Chapter 0.2) there is $\theta\in(0,1)$ with $\Delta = h\,\varphi'(a+\theta h)$, i.e.

$$\Delta = h\Big[\partial_{x}f(a+\theta h,\,b+k) - \partial_{x}f(a+\theta h,\,b)\Big].$$

We now want a second derivative out of that, so look at the bracket. It is a difference in $y$ of the function $\partial_{x}f(a+\theta h,\cdot)$, so the mean value theorem applies to it as well, giving $\theta'\in(0,1)$ with

$$\Delta = h\,k\;\partial_{y}\partial_{x}f\big(a+\theta h,\;b+\theta' k\big).$$

Now group $\Delta$ the other way, as a difference in $y$ first. The identical argument gives $\sigma,\sigma'\in(0,1)$ with $\Delta = h\,k\;\partial_{x}\partial_{y}f(a+\sigma h, b+\sigma' k)$. Equate the two, divide by $hk\neq0$, and let $(h,k)\to(0,0)$. Both evaluation points converge to $(a,b)$, and by the assumed continuity of the two second partials at $(a,b)$ both sides converge to their values there. Hence $\partial_{y}\partial_{x}f(a,b)=\partial_{x}\partial_{y}f(a,b)$. ∎

The counterexample. Continuity was used in the very last step of that argument, so removing it ought to break the theorem, and it does. Take

$$f(x,y) = \frac{xy\,(x^{2}-y^{2})}{x^{2}+y^{2}} \ \ \text{for }(x,y)\neq(0,0), \qquad f(0,0)=0.$$

This one is continuous, and differentiable, and has second partials everywhere. Compute $\partial_{x}f$ in general and then restrict it to the $y$-axis: a short calculation gives $\partial_{x}f(0,y) = -y$. Restricting $\partial_{y}f$ to the $x$-axis gives $\partial_{y}f(x,0) = +x$. Therefore

$$\partial_{y}\partial_{x}f(0,0) = \dv{}{y}(-y) = -1, \qquad \partial_{x}\partial_{y}f(0,0) = \dv{}{x}(+x) = +1.$$

The mixed partials at the origin differ. What fails is precisely the hypothesis. Away from the origin the second mixed partial is a bounded but direction-dependent function of $\theta$, which is the same pathology as §1's counterexample one derivative further up. So it has no limit at the origin and is not continuous there. The theorem is not true for free. It is true under a condition, and the condition is exactly what stops direction-dependence from surviving to the limit.

6.2 · Taylor to second order

We already have all the machinery for this, and the trick is to turn the several-variable problem into a one-variable one. Fix $\vv a$ and $\vv h$ and define the one-variable function $\varphi(t) = f(\vv a + t\vv h)$, which traces $f$ along the straight line through $\vv a$ in the direction $\vv h$. Our goal is $\varphi$'s first two derivatives, since Chapter 0.3's Taylor theorem will then do the rest. Differentiate it with the chain rule (0.6.25), noting that the "velocity" of the line is the constant vector $\vv h$:

That is the first derivative, and we want the second. Each $\partial_{i}f$ in it is itself a scalar field evaluated along the same line, so apply the same chain rule to each of them in turn:

We have $\varphi(0)=f(\vv a)$ and both derivatives at $t=0$, which is everything Taylor needs. So apply the one-variable Taylor theorem of Chapter 0.3 to $\varphi$ about $t=0$, and then evaluate the result at $t=1$, which lands exactly on $f(\vv a+\vv h)$:

Look at what that boxed line contains. Three terms: a constant, a linear form, and a quadratic form. That is the general shape of a smooth function near a point. It is also why quadratic forms, which were Chapter 0.5's whole subject, are not an algebraic curiosity. They are the universal local model of everything.

6.3 · Critical points, classified by eigenvalues

A critical point is a point where $\nabla f=\vv 0$, meaning that no direction changes $f$ to first order. Every maximum, minimum and saddle is one. The reason is that if $\nabla f\neq\vv 0$ you can increase $f$ by stepping along it and decrease $f$ by stepping the other way, so you are not at an extremum. Now put $\nabla f=\vv 0$ into (0.6.30). The linear term drops out, and what is left is

So the local behaviour is decided by the sign of the quadratic form $\vv h^{\mathsf T}H\vv h$. A quadratic form with cross terms in it is hard to read a sign off, so we want a basis in which the cross terms disappear. Clairaut has already given us one: $H$ is symmetric, which is exactly the hypothesis of the spectral theorem of Chapter 0.5. That theorem supplies an orthonormal basis $\vv e_{1},\ldots,\vv e_{n}$ of eigenvectors of $H$, with real eigenvalues $\mu_{1},\ldots,\mu_{n}$. Expand the displacement in that basis as $\vv h=\sum_{k}c_{k}\vv e_{k}$, and the quadratic form collapses to a sum of squares:

All cross terms are gone. The classification now reads itself off:

• All $\mu_{k}\gt0$: then $\sum\mu_{k}c_{k}^{2}\ge\mu_{\min}\abs{\vv h}^{2}$ with $\mu_{\min}\gt0$, so the quadratic term is positive and of order $\abs{\vv h}^{2}$, while the remainder is $o(\abs{\vv h}^{2})$ and therefore eventually smaller. For small enough $\abs{\vv h}$ the difference in (0.6.31) is positive in every direction: a strict local minimum.
• All $\mu_{k}\lt0$: the same argument with the signs reversed, giving a strict local maximum.
• Mixed signs: a saddle. Along an eigenvector with $\mu_{k}\gt0$ the function rises, and along one with $\mu_{k}\lt0$ it falls. It is a maximum and a minimum at the same point, depending on which way you look.
• Some $\mu_{k}=0$: the test is silent. The quadratic term vanishes in that direction, so the $o(\abs{\vv h}^{2})$ remainder decides, and we have no control over it. You need higher order. Chapter 6.6's Higgs potential is exactly such a case at its minimum, where the potential is flat along the direction the vacuum is free to rotate in, and what happens in that flat direction is the entire physics. (At the symmetric point $\phi=0$ the Hessian is strictly negative, so that point is an ordinary maximum and the test is not silent there.)

The eigenvectors are not bystanders in this, and here is the calculation that shows what they mean. Take a displacement along one of them by setting $\vv h = t\vv e_{k}$ in (0.6.31). Since $\vv e_{k}^{\mathsf T}H\vv e_{k}=\mu_{k}$, we get

Along the $k$-th eigenvector, then, $f$ is a one-dimensional parabola whose second derivative is $\mu_{k}$. The eigenvalues of the Hessian are the curvatures of $f$ along the principal directions, and the eigenvectors are those directions.

Near a minimum this makes the level sets ellipsoids. Setting $\tfrac12\sum\mu_{k}c_{k}^{2}=\delta$ gives semi-axes $c_{k}=\sqrt{2\delta/\mu_{k}}$, so a small eigenvalue means a long axis, which is a shallow and extended valley in that direction. Hold onto the picture of a long thin valley. The next callout is about nothing else.

7 · Lagrange multipliers

Now the constrained problem. We want to extremise $f(\vv x)$ subject to $g(\vv x)=c$. The recipe for doing that is famous, and it is usually handed over as a recipe. We are going to derive it instead, in three sentences of geometry, and then extract something from it that the recipe hides.

Let $\vv a$ be a point on the constraint surface $S=\{g=c\}$ at which $f$ is extremal among points of $S$, and assume $\nabla g(\vv a)\neq\vv 0$. We need two facts, and both of them have already been proved.

Which directions are tangent to $S$. If $\vv x(t)$ is a curve lying in $S$ then $g(\vv x(t))=c$, so (0.6.16) gives $\nabla g\cdot\dot{\vv x}=0$. Every tangent vector to $S$ is therefore orthogonal to $\nabla g$. The converse says that every vector orthogonal to $\nabla g$ is the velocity of some curve in $S$, and that is the implicit function theorem, ⚑ which we quote. It is guaranteed by the assumption $\nabla g\neq\vv 0$, which is what says the constraint really does cut down the dimension by one near $\vv a$. So the tangent space to $S$ at $\vv a$ is exactly $(\mathrm{span}\,\nabla g)^{\perp}$.

What extremality says. If $\vv x(t)$ is a curve in $S$ through $\vv a$, then $t\mapsto f(\vv x(t))$ has an extremum at that point, so its derivative vanishes and $\nabla f\cdot\dot{\vv x}=0$. Moving along the constraint surface cannot change $f$ to first order, because that is what being extremal on $S$ means. So $\nabla f(\vv a)$ is orthogonal to every tangent vector of $S$.

Now let's put the two facts together. The plan is to show that $\nabla f$ has nothing left over once you remove its component along $\nabla g$, and Chapter 0.5's orthogonal projection is the tool for separating those two pieces. Write $\vv n = \nabla g(\vv a)$ and split $\nabla f$ into its component along $\vv n$ and whatever remains:

The residual is orthogonal to $\vv n$ by construction, since $\vv w\cdot\vv n = \nabla f\cdot\vv n - \lambda\abs{\vv n}^{2} = 0$. So by the first fact, $\vv w$ is a tangent vector to $S$. But the second fact says $\nabla f$ is orthogonal to every tangent vector, and that includes $\vv w$ itself. Dot the split with $\vv w$ and see what happens:

A vector of zero length is the zero vector, so $\vv w=\vv 0$. Feeding that back into the split leaves $\nabla f$ with only its component along $\vv n$, which is the result we were after:

Notice that the argument even handed us the multiplier explicitly, as $\lambda = \nabla f\cdot\nabla g/\abs{\nabla g}^{2}$. That is worth remembering when you want to check a solution rather than find one.

The geometric statement is the one to carry away. At a constrained extremum the level set of $f$ is tangent to the constraint surface, since both have the same normal direction. Suppose instead that they crossed. Then you could slide along the constraint from one side of the $f$-contour to the other and change $f$, so you were not at an extremum after all. Press the button in the §3 figure and this is exactly what you are looking at.

The mechanical version everyone quotes follows immediately, and it comes from packaging the two conditions $\nabla f=\lambda\nabla g$ and $g=c$ as the critical-point equations of a single function. Define

and look for its unconstrained critical points. Setting $\partial\mathcal{L}/\partial x^{i}=0$ gives $\nabla f=\lambda\nabla g$. Setting $\partial\mathcal{L}/\partial\lambda=0$ gives back $g=c$. So the constraint has been converted from a restriction on the search space into one more equation to solve, at the price of one more unknown.

That trade is worth pausing on. You may restrict the space, or you may enlarge it and add a multiplier. It is one of the deepest moves in theoretical physics. It returns in Chapter 1.2 as the way to handle a bead on a wire, and again, much later, as the way gauge symmetry is imposed.

7.1 · The multiplier means something

$\lambda$ looks like a bookkeeping device you introduce and then discard. It is not, and here is how to see what it measures. Let $\vv x^{*}(c)$ be the constrained optimum as a function of the constraint level $c$, and let $f^{*}(c) = f(\vv x^{*}(c))$ be the optimal value. Assume $\vv x^{*}$ varies differentiably with $c$, which it does under the same non-degeneracy that made the multiplier exist. We want to know how the best achievable value responds to a change in $c$, so differentiate $f^{*}$ using the chain rule (0.6.25):

where the second step used (0.6.36). That has left us needing the quantity $\nabla g\cdot\dv{\vv x^{*}}{c}$, and the constraint itself will supply it. Differentiate $g(\vv x^{*}(c)) = c$ with respect to $c$, which is the same chain rule applied to an identity holding for all $c$:

The awkward factor is exactly $1$. So substitute (0.6.39) into (0.6.38), and everything on the right except the multiplier disappears:

So the multiplier is a sensitivity. It is the rate at which the best achievable value improves when you relax the constraint by one unit. Economists call it a shadow price. In mechanics it is the constraint force, and in thermodynamics, as Worked example 1 will show, it is the temperature. A large $\lambda$ says the constraint is expensive and worth negotiating, and $\lambda=0$ says the constraint is not binding at all.

Let's check that on a case small enough to do by hand, which is also the figure's setup. Minimise $f=x^{2}+2y^{2}$ subject to $x+y=c$. Then (0.6.36) reads $(2x,4y)=\lambda(1,1)$, so $x=\lambda/2$ and $y=\lambda/4$. Putting those into the constraint gives $3\lambda/4=c$, hence $\lambda=4c/3$ and $\vv x^{*}=(2c/3,\,c/3)$. The optimal value and its derivative are then

Take $c=3$ for a specific number. The optimum is $(2,1)$, with $f^{*}=6$ and $\lambda=4$. That $4$ is the number printed in the figure when the constraint is showing.

Grind box — when Lagrange fails, and why the fine print is the interesting part

The derivation used $\nabla g(\vv a)\neq\vv 0$ twice: to know the tangent space is a genuine hyperplane, and to invoke the implicit function theorem. Drop it and the method can be wrong. Minimise $f(x,y)=x$ subject to

$$g(x,y) = x^{3}-y^{2} = 0.$$

The constraint set is the curve $y^{2}=x^{3}$, a cusp at the origin, defined only for $x\ge0$. The minimum of $x$ on it is at the origin. But there $\nabla g = (3x^{2},-2y) = (0,0)$, while $\nabla f = (1,0)$ everywhere, and no $\lambda$ satisfies $(1,0)=\lambda(0,0)$. The method returns nothing, and the answer it failed to find was sitting at the origin all along.

The condition $\nabla g\neq\vv 0$ is called a constraint qualification. The moral is that the multiplier exists because the constraint surface is smooth and non-degenerate there, rather than because algebra says so. Where constraints are degenerate, the multiplier framework needs repair. That is not an exotic case. In gauge theories the constraints are degenerate by design, because a gauge symmetry means several different field configurations describe the same physics. Handling that properly is what Dirac's constrained-Hamiltonian machinery and the Faddeev–Popov procedure exist to do, and every one of those is a descendant of (0.6.36) plus an honest accounting of when it breaks.

Several constraints. With $g_{1}=c_{1},\ldots,g_{r}=c_{r}$ the argument is identical word for word, with $\vv n$ replaced by the subspace $W=\mathrm{span}\{\nabla g_{1},\ldots,\nabla g_{r}\}$ and the one-line split (0.6.34) replaced by Chapter 0.5's orthogonal projection onto $W$. Tangent vectors are those orthogonal to every $\nabla g_{\alpha}$. The residual $\vv w = \nabla f - P\nabla f$ is one of them, and extremality kills it exactly as before. What survives is

$$\nabla f = \sum_{\alpha=1}^{r}\lambda_{\alpha}\nabla g_{\alpha},$$

with the qualification now reading "the $\nabla g_{\alpha}$ are linearly independent at $\vv a$", which is Chapter 0.4's notion of independence doing real work. Worked example 1 uses two constraints and therefore two multipliers, one of which turns out to be the temperature.

8 · Change of variables and the Jacobian determinant

Here is the last consequence, and it is the one that pays a debt from Chapter 0.2. Let $\Phi:U\to\R^{n}$ be a differentiable, injective change of coordinates with $\det\mathrm{D}\Phi\neq0$. The question is: how does an integral transform under it?

That last condition earns a name, because two later chapters lean on it directly and it is worth having stated rather than assumed. If $\mathrm{D}\Phi(\vv a)$ is invertible then $\Phi$ itself is invertible on some neighbourhood of $\vv a$, with a differentiable inverse whose derivative there is $\big(\mathrm{D}\Phi(\vv a)\big)^{-1}$. That is the inverse function theorem, and it says the obvious thing: near a point, the map is its linear part, and a linear part you can undo is an undoable map. ⚑ The proof is a contraction-mapping argument that belongs to a real analysis course, so it is quoted here. It is equivalent to the implicit function theorem quoted in §7, and Chapter 1.3 spends it twice — once to ask when velocities can be traded for momenta, and once to identify the Hessian that decides it.

The answer is forced by two things we already have, so let's set them side by side. First, from §2, near any point $\vv u$ the map $\Phi$ is a linear map, up to an error that dies faster than the displacement:

Second, from Chapter 0.4, a linear map $A$ takes a region of volume $V$ to a region of volume $\abs{\det A}\,V$. That is what the determinant is: the signed factor by which a linear map scales volume. The absolute value is there because volume does not care about orientation.

Now put the two together. Chop $U$ into tiny cubes of side $\epsilon$. The cube at $\vv u$ is a set of displacements $\vv h$ of size at most $\epsilon\sqrt n$, so (0.6.42) says its image under $\Phi$ is a parallelepiped, namely the image of the cube under the linear map $\mathrm{D}\Phi(\vv u)$, translated to $\Phi(\vv u)$, with a distortion that is small compared to $\epsilon$. Apply the second fact to that parallelepiped and its volume is

That is the statement about one cell, and an integral is a sum over cells. So sum $f(\Phi(\vv u))$ times that volume over all the cubes and let $\epsilon\to0$. Doing so turns the Riemann sums of Chapter 0.2 on one side into the Riemann sums on the other, and what comes out is

Check it against the one-dimensional case you already know. There $\det\mathrm{D}\Phi = \Phi'$, and the formula is the substitution rule of Chapter 0.2, with the absolute value doing the job that reversing the limits of integration does there. The factor $\abs{\det \mathrm{D}\Phi}$ is called the Jacobian determinant. It is the local volume-magnification of the coordinate change, and it is nothing more mysterious than that.

Grind box — what that argument does and doesn't establish

The chop-and-sum argument is the right picture and it is how every physicist thinks about the Jacobian, but it is not a proof, and it is worth being clear about where the gap is.

Two things were glossed. (i) Uniformity. The $o(1)$ in (0.6.43) must shrink at a rate that does not depend on which cube you are looking at. Otherwise summing a large number of small errors need not give a small total. This is delivered by continuity of $\mathrm{D}\Phi$ on a compact region, which is the usual hypothesis. (ii) The boundary. Cubes that straddle the edge of $U$ are only partly inside, and one must show their total contribution vanishes. It does, because the boundary of a reasonable region has $n$-dimensional volume zero. But "reasonable" needs defining, and doing that properly is where measure theory enters and where the theorem is genuinely a theorem.

What the argument does establish, and what you should take from it, is the mechanism. The $\abs{\det}$ appears because differentiability means local linearity, and because determinants are what linear maps do to volume. Change either fact and the formula changes with it.

8.1 · Polar coordinates — paying Chapter 0.2's debt

In Chapter 0.2, deriving the Gaussian integral, we needed the polar area element and obtained it geometrically, by differencing two circular sectors. It was an honest argument, but it was special to the plane and to circles, and the chapter explicitly said that Chapter 0.6 would produce the same factor mechanically. Here is that promise kept.

The plan is to write down the map, take its Jacobian, and take the determinant of that. So take $\Phi(r,\theta) = (r\cos\theta,\;r\sin\theta)$. Its Jacobian is a $2\times2$ matrix of partials, computed by the rules of §1, which is to say by differentiating each output with respect to each input:

What we want from that matrix is the volume factor, which means its determinant. Take it by the $2\times2$ rule $ad-bc$ of Chapter 0.4, and use $\cos^{2}\theta+\sin^{2}\theta=1$ at the end:

The general formula (0.6.44) asks for the absolute value of that, and $r\ge0$ makes the absolute value do nothing. So the area element in polar coordinates is

That is the factor Chapter 0.2 used on faith, now derived from the definition of the determinant. The debt is paid, and the derivation cost three lines instead of a picture.

Notice that it also explains the geometric argument rather than replacing it. The columns of (0.6.45) are the displacement produced by a unit change in $r$, which has length $1$ and points radially, and the displacement produced by a unit change in $\theta$, which has length $r$ and points tangentially. Those two are perpendicular, so the parallelogram they span has area $1\times r = r$. The determinant computed the area of the little polar cell, which is precisely what the sector argument computed by hand.

8.2 · Spherical coordinates

The same procedure works in three dimensions, and it is worth doing once because the answer is one you will use constantly. Take $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$ and $z=r\cos\theta$, with $\theta$ measured from the $z$-axis as physics always does. Differentiating each output with respect to each input gives

We want that matrix's determinant, and we may expand along whichever row or column is cheapest. The bottom row has a zero in it, so expanding along it costs only two cofactors instead of three. Start with the $\cos\theta$ entry, which multiplies

after using $\cos^{2}\phi+\sin^{2}\phi=1$. The other surviving entry is $-r\sin\theta$, and with the alternating sign $(-1)^{3+2}=-1$ it multiplies $-\big(r\sin^{2}\theta\big)$. That is both cofactors, so add the two contributions to get the determinant:

using $\cos^{2}\theta+\sin^{2}\theta=1$ once more. Feed that determinant into (0.6.44) and the volume element in spherical coordinates is

This is positive for $\theta\in(0,\pi)$, so once again the absolute value takes care of itself. Problem 1 uses this to compute the volume of a sphere, which is the cheapest possible check that the factor is right.

8.3 · The same factor, in general relativity

One forward glance, because the connection is exact rather than analogical. In curved spacetime the invariant volume element is written $\sqrt{-g}\;\dd^{4}x$, where $g=\det g_{\mu\nu}$ is the determinant of the metric. It is negative because of the Lorentzian signature, which is why there is a minus sign under the root. That $\sqrt{-g}$ is (0.6.44) wearing different clothes.

You can see that without any relativity at all, using the flat-space metrics of §4 and comparing them with the two determinants we just computed. In plane polar coordinates $g_{ij}=\mathrm{diag}(1,r^{2})$, so

which is exactly the factor in (0.6.47). Now try the same thing one dimension up. In spherical coordinates the metric is $\dd s^{2}=\dd r^{2}+r^{2}\dd\theta^{2}+r^{2}\sin^{2}\theta\,\dd\phi^{2}$, so $g_{ij}=\mathrm{diag}(1,\,r^{2},\,r^{2}\sin^{2}\theta)$, and its determinant gives

which is exactly (0.6.51). The agreement is not a coincidence of these two examples. The reason is structural, and Chapter 3.5 §6.2 will show it. Under a coordinate change the metric picks up two factors of the Jacobian matrix, so its determinant picks up $(\det J)^{2}$ and $\sqrt{\abs{g}}$ picks up $\abs{\det J}$. That is precisely what cancels the $\abs{\det J}^{-1}$ that $\dd^{n}x$ acquires. The combination $\sqrt{\abs{g}}\,\dd^{n}x$ is therefore the same in every coordinate system, which is what "invariant volume element" means. The Einstein–Hilbert action of Chapter 3.6 carries that factor for this reason and no other.

The same $\abs{\det}$ turns up every time you change variables in a path integral (Chapter 5.7). There it is no longer a nuisance factor but the whole story. The Faddeev–Popov ghosts of non-abelian gauge theory are the Jacobian determinant of a change of variables, promoted to a field.

9 · Worked examples

10 · Your turn