Part II · Special Relativity — Chapter 2.4

Tensors, Honestly

An index is not decoration. It is a promise about what happens when someone else picks different coordinates.

Tools you'll need.  Chapter 0.4 §4: change of basis, the fact that components transform with $P^{-1}$ while basis vectors transform with $P$, and $A'=P^{-1}AP$. That section is the entire skeleton of this one. Chapter 0.6 §4: the gradient is really a one-form, $g_{ij}$, and $(\nabla f)^{i}=g^{ij}\partial_{j}f$. This chapter is the promised payoff of that section, and we pick the thread up by name in §3. Also 0.6 §5, the multivariable chain rule, which does almost all of the work below. Chapter 0.7 §4.3, the unique split of a matrix into symmetric plus antisymmetric parts, and §6, the continuity equation. Chapter 2.2 for the boost and rapidity. Chapter 2.3 for the interval, $\eta_{\mu\nu}$, and the statement that a four-vector is something transforming like $\dd x^{\mu}$, which is the statement we are about to turn into a definition.

1 · The problem, stated sharply

Start with the demand that the whole chapter answers to. Physics is not allowed to depend on the coordinates a human being chose. Nothing about the electron knows whether you set up your axes pointing north or north-north-east, or whether you are drifting past the laboratory at $0.6c$. Whatever is real must be the same for everyone.

But components are not the same for everyone, and nothing could make them so. That is what components are: numbers you get by comparing a physical thing to a set of reference directions you invented. Chapter 0.4 §4 made this precise for an ordinary vector. Change the basis by $P$ and the components change by $P^{-1}$. Chapter 2.2 made it precise for spacetime, where changing to a moving frame reshuffles the four numbers $(ct,x,y,z)$ by a boost.

So we have a tension. The things we can write down are components, and those are not invariant. The things we want to assert are laws, and those must be. The resolution is not to abandon components. It is to insist on a particular class of objects.

That is a design constraint, and like all good design constraints it rules things out. Let's watch it rule something out, because here the failure teaches more than the success would.

1.1 · Four numbers that are not a four-vector

Here is an object that has an index and is not a tensor. Define

Nothing in that definition is ill-formed. It is a perfectly good rule: hand me a frame, I hand you four numbers. You could imagine it standing for "the direction of time". You could also imagine it standing for the rest frame of the ether, which is what people historically did, and which is exactly the mistake Chapter 2.1 dismantled. It has an upper index, it has four components, and it looks like everything else we write. Watch what happens.

Take a particle with four-momentum $p^{\mu}=(E/c,\,p^{x},\,p^{y},\,p^{z})$, a genuine four-vector. Build the number

It looks like a scalar, with the indices all matched and summed and nothing left dangling. So let's evaluate it in two frames using specific numbers. Let $p^{\mu}=(5,\,3,\,0,\,0)$ in units of $\mathrm{MeV}/c$, and let $S'$ move at $\beta=3/5$, so $\gamma=5/4$. Chapter 2.2's boost gives $p'^{\mu}=(4,\,0,\,0,\,0)$. We will grind that arithmetic out in §9, and in the meantime you can check it in your head: $\tfrac54(5-\tfrac35\cdot3)=\tfrac54\cdot\tfrac{16}{5}=4$. Now feed both sets of components into the recipe.

Two observers, one recipe, two answers. Look carefully at what did not go wrong. No algebra was botched and no sign was dropped. The recipe $\eta_{\mu\nu}N^{\mu}p^{\nu}$ is unambiguous, and both computations are correct. What failed is the claim that the recipe defines a number. It defines a number per frame, which is to say it defines nothing at all. You cannot write "$\mathcal{E}=5$" and mean anything by it.

The damage gets worse if you write an equation rather than a single number. Consider what happens to this one.

In frame $S'$, where our particle is at rest, that equation is true: $p'^{\mu}=(4,0,0,0)$ and $mc=4$. Now suppose $N^{\mu}$ were a four-vector. The theorem of §6 would then let us conclude that (2.4.4) holds in every frame. And since $N^{\mu}$ has been declared to be $(1,0,0,0)$ everywhere, that would say $p^{\mu}=(mc,0,0,0)$ everywhere. This particle would be at rest for every observer, including the one watching it go past at $0.6c$. The conclusion is false, so one of the premises must be false. It is not the theorem. It is the assumption that $N^{\mu}$ is a four-vector, and the theorem's one hypothesis is exactly the one $N^{\mu}$ fails.

So here is the slogan for the chapter. Read it as a definition rather than as a warning.

2 · Vectors, defined by how they transform

Now let's build the definition. What we need first is a prototype: one object whose transformation law we can compute rather than choose. There is exactly one natural candidate, and Chapter 2.3 already leaned on it.

Let $x^{\mu}$ be one set of coordinates and $x'^{\mu}$ another, related by four functions $x'^{\mu}=x'^{\mu}(x^{0},x^{1},x^{2},x^{3})$ that are smooth and invertible. Take two nearby events separated by an infinitesimal displacement $\dd x^{\mu}$, and ask what that separation is in primed coordinates. The answer is not a matter of convention. It is the multivariable chain rule of Chapter 0.6 §5, applied to each of the four functions $x'^{\mu}$ in turn.

summed over $\nu=0,1,2,3$. Nothing was assumed there. Both sides are the same physical displacement between the same two events, described twice over. That array of partial derivatives is going to appear on every page from here on, so let's give it a name.

This is the Jacobian matrix of Chapter 0.6 §8, wearing index notation. The upper index labels the row and the new coordinate. The lower index labels the column and the old one. Keep the horizontal offset in $\Lambda^{\mu}{}_{\nu}$, because it records which slot comes first, and by §4 that will start to matter.

Now make the leap that organises everything after it. We have one object, $\dd x^{\mu}$, that we know transforms by (2.4.6). We define the class of objects that behave the same way.

Two things should feel unsatisfying about that definition, and both are worth confronting before we go on.

"It's circular. You defined vectors by copying $\dd x$." Yes, deliberately. The definition is ostensive: it points at a known object and says "these ones". What makes it non-vacuous is that many other objects turn out to belong to the class, and their membership is a theorem rather than a stipulation. Take three of them in turn. The four-velocity $u^{\mu}=\dd x^{\mu}/\dd\tau$ belongs, because $\dd\tau$ is a scalar (Chapter 2.3), and dividing a vector by a scalar leaves the law (2.4.6) intact. The four-momentum $p^{\mu}=mu^{\mu}$ belongs, since $m$ is a scalar by construction. In this book $m$ always means rest mass, the number every observer computes from $p_{\mu}p^{\mu}=m^{2}c^{2}$. The current $J^{\mu}=(\rho c,\vv J)$ belongs as well, and §9 shows what that buys. Each of those is a claim about physics, checkable and falsifiable, and each of them fails for $N^{\mu}$ of (2.4.1).

"Where did the arrow go?" Nowhere. It is still available, and Chapter 0.4 tells you exactly how to recover it. There, a vector was an element of a vector space and its components were $x_{j}$ in a basis $\{e_{j}\}$. Changing basis by $e'_{j}=P_{ij}e_{i}$ sent the components to $\vv x'=P^{-1}\vv x$. Compare that with the definition above and read off the dictionary between the two accounts.

The components transform with $\Lambda$ and the basis vectors transform with $\Lambda^{-1}=P$, so their product, which is the actual arrow, does not move at all. Chapter 0.4 §4 put it in one sentence: make your ruler twice as long and every measurement in rulers is halved. "Contra-variant" is literally that. The components vary contrary to the basis. So the arrow is exactly as real as it ever was. What has changed is which end of the relationship we take as the definition. We take the components' end, because that is the end that still makes sense when there is no global vector space to draw arrows in. From Chapter 3.2 onward, that is always the situation.

2.1 · What $\Lambda$ is for a boost

Before going further, let's see what all of this looks like for the one transformation we already know. Specialise to the boost of Chapter 2.2, with $S'$ moving along $+x$ at speed $v=\beta c$ and $x^{\mu}=(ct,x,y,z)$. Chapter 2.2 gave us the coordinate change itself.

What we want from it is $\Lambda$, and getting it means differentiating each new coordinate with respect to each old one. The coefficients here are constants, so the partial derivatives are just those constants.

That is the matrix Chapter 2.3 wrote down. It is no longer a postulate. It is a computed Jacobian, and every entry of it is independent of position. That constancy is a special feature of Lorentz transformations rather than a general fact about coordinate changes. Hold onto the observation. It is the hinge on which §9's second worked example turns, and it is the single technical difference between this chapter and Chapter 3.3.

Grind box — the inverse boost, and why rapidity is the sane variable

Invert (2.4.8) by solving for $ct$ and $x$. Multiply the first by $\gamma$ and the second by $\gamma\beta$ and add:

$$\gamma\,ct' + \gamma\beta\,x' = \gamma^{2}(1-\beta^{2})\,ct = ct,$$

using $\gamma^{2}(1-\beta^{2})=1$, which is the definition of $\gamma$ rearranged. The same manoeuvre gives $x=\gamma(x'+\beta ct')$. So

$$(\Lambda^{-1})^{\mu}{}_{\nu} \;=\; \pdv{x^{\mu}}{x'^{\nu}} \;=\; \begin{pmatrix} \gamma & +\gamma\beta & 0&0\\ +\gamma\beta & \gamma & 0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix} \;=\; \Lambda(-\beta),$$

which is the physically unsurprising statement that if $S'$ moves at $+v$ relative to $S$ then $S$ moves at $-v$ relative to $S'$. Here that statement is derived rather than assumed.

Chapter 2.2 introduced the rapidity $\phi$ with $\tanh\phi=\beta$, whence $\gamma=\cosh\phi$ and $\gamma\beta=\sinh\phi$ (divide $\cosh^{2}-\sinh^{2}=1$ by $\cosh^{2}$ to see the first). In that variable

$$\Lambda(\phi) = \begin{pmatrix}\cosh\phi & -\sinh\phi & 0&0\\ -\sinh\phi & \cosh\phi & 0&0\\ 0&0&1&0\\0&0&0&1\end{pmatrix},$$

and the hyperbolic addition formulas give, in two lines of matrix multiplication,

$$\Lambda(\phi_{2})\,\Lambda(\phi_{1}) \;=\; \Lambda(\phi_{1}+\phi_{2}).$$

Rapidities add. That is why $\phi$ is the right variable and $\beta$ is not. It is also a consistency check on the whole framework, because the chain rule guarantees that composing coordinate changes composes their Jacobians,

$$\pdv{x''^{\mu}}{x^{\nu}} \;=\; \pdv{x''^{\mu}}{x'^{\rho}}\;\pdv{x'^{\rho}}{x^{\nu}},$$

so the transformation law of §2 is automatically consistent under composition. Nothing extra had to be imposed. Chapter 6.1 will call this "the transformations form a group and the components carry a representation of it". Here it is nothing more than the chain rule.

3 · Covectors — the other kind of index

Chapter 0.6 §4 ended with a promise: differentiation naturally produces lower indices, vectors naturally carry upper ones, the metric converts between them, and when the metric is the identity you cannot see the conversion happening. It said that Chapter 2.4 would make this a formal convention. Here we are. This section builds the second species of index, and it builds it without using a metric at all. The metric does not arrive until §4.

Take a scalar field $f$, meaning a function on spacetime whose value at an event is the same number for everyone, so that $f'(x')=f(x)$. Temperature at a point, say, or the proper time elapsed along some fixed worldline. Its four partial derivatives

are perfectly good components of something. The question is what. We want their transformation law, so apply the chain rule to $f$ regarded as a function of the primed coordinates.

Let's look at what appeared. Not $\pdv{x'^{\mu}}{x^{\nu}}$, but $\pdv{x^{\nu}}{x'^{\mu}}$, the derivatives of the old coordinates with respect to the new. That is the inverse Jacobian. Calling it the inverse is a claim about matrices, so let's establish it. We can do that without computing anything, again by the chain rule.

since the primed coordinates are independent of one another: $\partial x'^{0}/\partial x'^{0}=1$ and $\partial x'^{0}/\partial x'^{1}=0$, and so on. So write $(\Lambda^{-1})^{\nu}{}_{\mu}\equiv\pdv{x^{\nu}}{x'^{\mu}}$. The notation is honest. That array really is the matrix inverse of (2.4.6), and (2.4.12) reads $\Lambda\Lambda^{-1}=I$.

So $\partial_{\mu}f$ transforms with the inverse of $\Lambda$, while $\dd x^{\mu}$ transforms with $\Lambda$ itself. These are different laws. They give different numbers. Objects that obey the inverse law deserve their own name and their own index position.

3.1 · Why the two species exist: the pairing

Now for the structural fact that justifies having invented a second kind of object at all. Take a vector $V^{\mu}$ and a covector $\omega_{\mu}$ and contract them, which means summing over the repeated index with one copy up and one down. Watch what the two transformation laws do to each other.

The two Jacobians met and collapsed to a Kronecker delta, and the delta then did nothing except rename an index. So $V^{\mu}\omega_{\mu}$ takes the same value in every coordinate system. It is a scalar, meaning a real number rather than a number-per-frame. It is the first genuinely invariant thing we have built, and building it took no metric.

Let's read the two definitions again in the light of that calculation. The vector law and the covector law are not two arbitrary conventions that happen to be inverse to one another. Covectors are defined to be the objects that pair with vectors to give invariants. That is what they are for, and the inverse Jacobian is forced on them by that job. In Chapter 0.4's language, the covectors at a point form the dual space $V^{*}$, which is the space of linear maps from vectors to numbers.

3.2 · The picture: arrows and stacks

There is a genuine geometric image behind all of this, and it is not the one you would guess.

A vector is an arrow. A covector is not an arrow. A covector $\omega$ is a machine that eats an arrow and returns a number, linearly, and the honest picture of such a machine is a family of evenly spaced parallel surfaces. They are the level sets $\omega=0,1,2,3,\ldots$ of the linear function it computes. In two dimensions they are parallel lines. In four they are parallel hyperplanes. A "big" covector has closely spaced surfaces and a "small" one has widely spaced surfaces, so the covector's magnitude lives in the density of the stack. That is why doubling $\omega$ halves the spacing.

With that picture in hand, the contraction $V^{\mu}\omega_{\mu}$ acquires a completely concrete meaning.

For the gradient, this is a picture you have already seen without knowing what you were looking at. The level sets of $\dd f$ are the contours of $f$, and $V^{\mu}\partial_{\mu}f$ is the number of contour lines you cross walking along $V$, which is the change in $f$. Chapter 0.6's contour figure was a picture of a one-form all along.

skew ∠(e₁,e₂)

90.0°

|e₁|

1.00

|e₂|

1.00

show the wrong pairing

reset to Cartesian

V¹ = 2.60000 V² = 1.50000

ω₁ = 1.00000 ω₂ = 0.60000

Vⁱωᵢ = 3.50000000 (drift 0.0e+0)

lines pierced = 3 (tip sits at ω = 3.50)

3.3 · Why nobody told you this before

Because in Euclidean space with Cartesian coordinates the two species have numerically identical components, which makes the distinction invisible. Chapter 0.6 §4 showed why. Converting between them requires the metric, and in Cartesian coordinates on flat Euclidean space $g_{ij}=\delta_{ij}$, so the conversion multiplies everything by $1$. Every dot product you have ever computed as $\sum_{i}a_{i}b_{i}$ was silently a contraction of a vector with a covector, and you were spared the bookkeeping because the metric was doing nothing.

The invisibility ends the moment either of two things happens, and in relativity both of them do. Use non-Cartesian coordinates and $g_{ij}$ stops being $\delta_{ij}$, which is where Chapter 0.6 §4 got the polar gradient's notorious $1/r$. Or use a metric that is not positive-definite, such as $\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$, and the conversion starts flipping signs. That second case is what §4 is about.

4 · The metric as the dictionary

So far there has been no metric. Vectors, covectors and the pairing between them all exist without one. Now we add the object that Chapter 2.3 built out of the invariance of the interval, and look at exactly what it buys.

Chapter 2.3's central claim was that the quantity

is the same for all inertial observers. Let's write that out as an equation between two computations of the same thing, using (2.4.6) on the primed side.

Since $\dd x^{\rho}$ is arbitrary, the coefficients at the two ends must agree. (They must agree after symmetrising, and $\eta$ is already symmetric.) That gives us the defining property of a Lorentz transformation.

In matrix form that reads $\Lambda^{\mathsf T}\eta\,\Lambda=\eta$. Let's pause on what it is saying. The Lorentz transformations are exactly the linear maps that preserve $\eta$, in the same way that rotations are exactly the linear maps preserving $\delta_{ij}$, which is the condition $R^{\mathsf T}R=I$. Chapter 2.3 derived the boost from the two postulates. Equation (2.4.16) is that same result restated as an algebraic condition, and it is the form Chapter 6.1 generalises when it names the group $\mathrm{SO}(1,3)$.

4.1 · Lowering

Define, for any vector $V^{\mu}$,

The name for that operation is "lowering the index". Writing the result with a down index carries an implicit claim, which is that $V_{\mu}$ is a genuine covector obeying the law of §3. That is a claim, so let's prove it. Start from the definition in the primed frame and push.

We want that to equal $(\Lambda^{-1})^{\sigma}{}_{\mu}V_{\sigma}=(\Lambda^{-1})^{\sigma}{}_{\mu} \eta_{\sigma\rho}V^{\rho}$. So the whole question is whether the following identity holds.

Take (2.4.16) and contract it with $(\Lambda^{-1})^{\sigma}{}_{\alpha}$ on the index $\sigma$:

which is (2.4.19) with the free indices renamed and the symmetry $\eta_{\mu\nu}=\eta_{\nu\mu}$ used on both sides. So yes: $V_{\mu}$ is a covector, and (2.4.17) is a legitimate map from one species to the other. Note precisely what made it work. It was (2.4.16), the invariance of $\eta$. Lowering with an arbitrary array of numbers would not have produced a covector.

4.2 · Raising

Going the other way needs the inverse of $\eta$. Since $\eta$ is a non-degenerate matrix it has one. Call its entries $\eta^{\mu\nu}$, defined by

For our signature the inverse is easy to find, because $\mathrm{diag}(1,-1,-1,-1)$ squares to the identity. So $\eta^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ as an array of numbers, the same four numbers down the diagonal. That coincidence is a property of this particular metric and not a general fact. In Chapter 3.3 the inverse metric $g^{\mu\nu}$ is a genuinely different array from $g_{\mu\nu}$, and confusing the two is a classic and expensive error. With the inverse in hand, we can raise an index.

That is consistent, since $\eta^{\mu\nu}V_{\nu}=\eta^{\mu\nu}\eta_{\nu\rho}V^{\rho}=\delta^{\mu}{}_{\rho}V^{\rho}=V^{\mu}$. Raising after lowering returns you to where you started, which is the least one can ask of a dictionary.

4.3 · What it actually does to the numbers

Let's see what lowering does to actual numbers. Write (2.4.17) out component by component. Since $\eta$ is diagonal, only one term survives in each sum.

So with signature $(+,-,-,-)$, lowering flips the sign of the three spatial components and leaves the time component alone. Take the four-momentum of §1 as a concrete case,

and the invariant length is then the contraction of the two arrays against each other,

so $m=4\ \mathrm{MeV}/c^{2}$. Note that $p_{\mu}p^{\mu}$ is a contraction of a covector with a vector, so by (2.4.13) it is invariant with no further argument needed. Section 9 checks that by brute force in three different frames.

Grind box — index gymnastics, and a bonus identity

Three working rules, each a one-line consequence of (2.4.21).

(i) You may slide a contracted pair up and down together.

$$V_{\mu}W^{\mu} = \eta_{\mu\nu}V^{\nu}W^{\mu} = V^{\nu}\big(\eta_{\nu\mu}W^{\mu}\big) = V^{\nu}W_{\nu}.$$

So $V_{\mu}W^{\mu}=V^{\mu}W_{\mu}$: it does not matter which factor carries the lowered index, only that exactly one does.

(ii) $\eta$ with one index up and one down is the Kronecker delta. By (2.4.21), $\eta^{\mu}{}_{\nu}=\eta^{\mu\rho}\eta_{\rho\nu}=\delta^{\mu}{}_{\nu}$. So you never need to write $\eta^{\mu}{}_{\nu}$ at all. It is $\delta$.

(iii) The metric knows the inverse transformation. Define $\Lambda_{\mu}{}^{\nu}$ the only way the rules allow, by lowering the first index and raising the second: $\Lambda_{\mu}{}^{\nu}\equiv\eta_{\mu\alpha}\,\eta^{\nu\beta}\,\Lambda^{\alpha}{}_{\beta}$. Now contract the identity (2.4.20), namely $\eta_{\mu\alpha}\Lambda^{\mu}{}_{\rho}=\eta_{\rho\sigma}(\Lambda^{-1})^{\sigma}{}_{\alpha}$, with $\eta^{\rho\tau}$.

$$\begin{aligned} \text{left} &= \eta^{\rho\tau}\eta_{\mu\alpha}\Lambda^{\mu}{}_{\rho} = \Lambda_{\alpha}{}^{\tau},\\[3pt] \text{right} &= \eta^{\rho\tau}\eta_{\rho\sigma}(\Lambda^{-1})^{\sigma}{}_{\alpha} = \delta^{\tau}{}_{\sigma}(\Lambda^{-1})^{\sigma}{}_{\alpha} = (\Lambda^{-1})^{\tau}{}_{\alpha}. \end{aligned}$$

Hence $\Lambda_{\alpha}{}^{\tau}=(\Lambda^{-1})^{\tau}{}_{\alpha}$. Lowering and raising the indices of $\Lambda$ produces its inverse. This is why most books never write $\Lambda^{-1}$ at all. They write $\Lambda_{\mu}{}^{\nu}$ and let the index positions carry the information. We have kept $\Lambda^{-1}$ explicit up to this point, because the covector law of §3 is true with no metric in sight and it would have been dishonest to define it using one.

(iv) $\det\Lambda=\pm1$. Take determinants of $\Lambda^{\mathsf T}\eta\Lambda=\eta$. By Chapter 0.4 §5, $\det$ is multiplicative and $\det\Lambda^{\mathsf T}=\det\Lambda$, so $(\det\Lambda)^{2}\det\eta=\det\eta$, and $\det\eta=-1\neq0$ gives $(\det\Lambda)^{2}=1$. Boosts and rotations have $\det=+1$. A spatial reflection has $\det=-1$. That sign will matter exactly once in this chapter, in §8.

Two last remarks on $\eta$, both pointing forward.

It is what makes "length" mean anything. Without a metric there is no way to ask how long a four-vector is. You can pair a vector with a covector, but not a vector with a vector. $\eta$ supplies exactly the missing ingredient, and the price is that "length squared" can come out positive, negative or zero. That is the timelike, spacelike and null trichotomy of Chapter 2.3. The metric does not merely measure. It is what defines the causal structure.

It is about to become a field. Everything in this section used only that $\eta$ is a symmetric, invertible, position-independent array satisfying (2.4.16). Now drop the words "position-independent", and replace $\eta_{\mu\nu}$ by $g_{\mu\nu}(x)$, ten functions of position. Every formula above survives unchanged. What changes is that $\Lambda$ becomes position-dependent, and then exactly one thing breaks. What breaks is the derivative, as §9's second worked example shows. Repairing it is Chapter 3.3, and the repair is the gravitational field. That is the entire difference between special and general relativity, and it is a difference in one adjective.

5 · Tensors in general

Vectors carry one upper index and one factor of $\Lambda$. Covectors carry one lower index and one factor of $\Lambda^{-1}$. The generalisation writes itself, and it is a definition rather than a theorem.

Here is everything we have met so far, sorted by type.

Type	Object	Example
$(0,0)$	scalar	$m$, $\dd\tau$, $\dd s^{2}$, $p_{\mu}p^{\mu}$
$(1,0)$	vector	$\dd x^{\mu}$, $u^{\mu}$, $p^{\mu}$, $J^{\mu}$
$(0,1)$	covector	$\partial_{\mu}f$, $p_{\mu}$, $A_{\mu}$
$(0,2)$	rank-2, both down	$\eta_{\mu\nu}$, $g_{\mu\nu}$ (Ch 3.3)
$(1,1)$	mixed	$\delta^{\mu}{}_{\nu}$, $\partial_{\nu}V^{\mu}$
$(2,0)$	rank-2, both up	$F^{\mu\nu}$ (Ch 2.6), $T^{\mu\nu}$ (Ch 3.6)

A scalar has no indices and therefore no factors of $\Lambda$, so $\phi'(x')=\phi(x)$, the same number for everyone. That is what the general law says at $p=q=0$, and it is why "scalar" and "invariant" are the same word here.

5.1 · The three operations that preserve tensor character

Tensors are useful because you can compute with them without leaving the class. Exactly three basic moves are permitted, and the shortness of that list is itself worth knowing.

(1) Linear combination, same type. If $S$ and $T$ are both type $(p,q)$ and $a,b$ are scalars, then $aS+bT$ is type $(p,q)$. That follows at once from the linearity of the transformation law. Apply the law to each term and factor out the common string of $\Lambda$s. Note the restriction on the type. You cannot add a $(1,0)$ to a $(0,1)$, for the reason given in §3.3.

(2) Outer product. If $A^{\mu}$ is type $(1,0)$ and $B_{\nu}$ is type $(0,1)$, then the sixteen numbers $C^{\mu}{}_{\nu}\equiv A^{\mu}B_{\nu}$ form a type $(1,1)$ tensor. Proof: multiply the two transformation laws together and read off. In general, multiplying a $(p,q)$ by an $(r,s)$ component-by-component with all indices distinct gives a $(p+r,q+s)$.

(3) Contraction. Set one upper index equal to one lower index and sum. This is the interesting one, so we prove it.

Proof of the basic case. Transform, then set the indices equal:

It is the same collapse as (2.4.13), and it is the same collapse every time: a $\Lambda$ and a $\Lambda^{-1}$ sharing a summed index annihilate into a $\delta$. Once you have seen that, you have seen all of tensor algebra. The rest is bookkeeping about which indices are spectators.

Grind box — contraction in the general case, with spectators

Take a $(2,2)$ tensor $T^{\mu\nu}{}_{\rho\sigma}$ and contract $\nu$ against $\sigma$. Define $U^{\mu}{}_{\rho}\equiv T^{\mu\lambda}{}_{\rho\lambda}$ (sum on $\lambda$). Claim: $U$ is a $(1,1)$ tensor. Write the transformation law and set $\nu=\sigma$, calling the common value $\lambda$:

$$\begin{aligned} U'^{\mu}{}_{\rho} &= T'^{\mu\lambda}{}_{\rho\lambda}\\[2pt] &= \Lambda^{\mu}{}_{\alpha}\,\Lambda^{\lambda}{}_{\beta}\,(\Lambda^{-1})^{\gamma}{}_{\rho}\,(\Lambda^{-1})^{\delta}{}_{\lambda}\;T^{\alpha\beta}{}_{\gamma\delta}. \end{aligned}$$

Now the only factors carrying $\lambda$ are $\Lambda^{\lambda}{}_{\beta}$ and $(\Lambda^{-1})^{\delta}{}_{\lambda}$, and $\lambda$ is summed. Group them:

$$\Lambda^{\lambda}{}_{\beta}\,(\Lambda^{-1})^{\delta}{}_{\lambda} \;=\; \delta^{\delta}{}_{\beta},$$ $$U'^{\mu}{}_{\rho} \;=\; \Lambda^{\mu}{}_{\alpha}\,(\Lambda^{-1})^{\gamma}{}_{\rho}\;\delta^{\delta}{}_{\beta}\,T^{\alpha\beta}{}_{\gamma\delta} \;=\; \Lambda^{\mu}{}_{\alpha}\,(\Lambda^{-1})^{\gamma}{}_{\rho}\;U^{\alpha}{}_{\gamma},$$

which is exactly the $(1,1)$ law. The spectator indices $\mu$ and $\rho$ never participated. They carried their own $\Lambda$ and $\Lambda^{-1}$ through untouched. The argument is identical for any $(p,q)$ and any pair of indices, which is why nobody writes it out twice.

Two consequences worth stating. Contracting a vector with a covector is the case $(1,0)\otimes(0,1)\to(0,0)$, which is (2.4.13) again. And $p_{\mu}p^{\mu}=\eta_{\mu\nu}p^{\mu}p^{\nu}$ is $(0,2)\otimes(1,0)\otimes(1,0)$ contracted twice down to $(0,0)$. That is why it is a scalar. Not because someone declared mass invariant, but because of the index structure.

Grind box — the quotient theorem, or how to prove something is a tensor without transforming it

Often you meet an array and want to know whether it is a tensor, and transforming it directly is painful. There is a shortcut, and it is used constantly.

Claim. Suppose $K_{\mu\nu}$ is an array of sixteen numbers per frame, and suppose that for every vector $V^{\nu}$ the sixteen-fold sum $K_{\mu\nu}V^{\nu}$ is a covector. Then $K_{\mu\nu}$ is a $(0,2)$ tensor.

Proof. By hypothesis $W_{\mu}\equiv K_{\mu\nu}V^{\nu}$ is a covector, so

$$K'_{\mu\nu}V'^{\nu} \;=\; W'_{\mu} \;=\; (\Lambda^{-1})^{\alpha}{}_{\mu}W_{\alpha} \;=\; (\Lambda^{-1})^{\alpha}{}_{\mu}K_{\alpha\beta}V^{\beta}.$$

On the left substitute $V'^{\nu}=\Lambda^{\nu}{}_{\beta}V^{\beta}$:

$$\Big[K'_{\mu\nu}\Lambda^{\nu}{}_{\beta} - (\Lambda^{-1})^{\alpha}{}_{\mu}K_{\alpha\beta}\Big]V^{\beta} \;=\; 0 \qquad\text{for every }V^{\beta}.$$

A bracket that annihilates every vector is zero, as you can see by taking $V$ to be each of the four basis vectors in turn. Contract the resulting identity with $(\Lambda^{-1})^{\beta}{}_{\sigma}$ to free the index $\nu$, and you get

$$K'_{\mu\sigma} \;=\; (\Lambda^{-1})^{\alpha}{}_{\mu}(\Lambda^{-1})^{\beta}{}_{\sigma}K_{\alpha\beta},$$

the $(0,2)$ law. $\blacksquare$

The same argument works for any index structure. Why it matters: this is how you certify that the objects physics hands you are tensors. In Chapter 2.6 the field $F^{\mu\nu}$ is identified by the requirement that $f^{\mu}=qF^{\mu\nu}u_{\nu}/c$ be a four-vector for every four-velocity $u_{\nu}$. The quotient theorem then forces $F^{\mu\nu}$ to be a tensor, which is most of the work of that chapter done in advance.

One caution, and it is the whole content of the phrase "for every": the hypothesis must hold for all vectors, not merely for the one you happen to care about. In §1, $\eta_{\mu\nu}N^{\mu}p^{\nu}$ was a number for each particular $p$, and that never made $N^{\mu}$ a vector.

6 · The whole point, in one theorem

Everything so far has been apparatus. Here is what it was for.

Proof. Define $D\equiv T-S$. By operation (1) of §5.1, $D$ is a tensor of type $(p,q)$. The hypothesis says every component of $D$ vanishes in the unprimed frame. Now apply the transformation law. Each primed component of $D$ is a sum of terms, and every term contains a factor $D^{\alpha\beta\ldots}{}_{\gamma\delta\ldots}=0$. A sum of terms each containing a factor of zero is zero. Hence $D'=0$ in the primed frame too, which is to say $T'=S'$. Since the primed frame was arbitrary, the equation holds everywhere. $\blacksquare$

That proof is four lines long and it is not deep. The transformation law is linear and homogeneous, meaning that every term has exactly one factor of $D$ and there is no constant term, so the law maps zero to zero. That is the entire mechanism. What is remarkable is not the proof but the consequence.

6.1 · What this buys

Consider what you would otherwise have to do. You have a candidate law of physics, and you want to know whether it is consistent with relativity, which means asking whether an observer sailing past at $0.9c$ would write the same law. The honest procedure is to express everything in the moving observer's coordinates, substitute, grind, and see whether the result has the same form. That is a long calculation, it must be redone for every law, and it is easy to get wrong.

The theorem replaces all of that with an inspection. If both sides of your equation are tensors of the same type, meaning the free indices match up with up and down with down on both sides, then you are done. There is no calculation to do. The frame-independence is visible in the shape of the equation. This property has a name, manifest covariance, and "manifest" is the operative word. The law is not merely true. It is true in a way you can see.

So the rest of this book has a writing convention that is also a physical principle:

6.2 · The scoreboard

$\vv F = m\vv a$ is not a tensor equation. Chapter 1.1 §4.2 already showed the damage from a different direction. Newton's second law changes form under an innocent change of spatial coordinates, sprouting terms that look like forces and are not. Relativity makes the failure sharper. The left side is a three-component object. The right side involves $\dd^{2}\vv x/\dd t^{2}$, in which $t$ is a coordinate rather than a scalar, and Chapter 2.2 established that $t$ is not shared between frames. There is no way to read (2.4.9) as acting on this equation, because the equation is not built from objects the matrix knows how to act on. The trouble is not that $\vv F=m\vv a$ is false. At $10^{-6}c$ it is superb. The trouble is that its content changes when you change frames, and that disqualifies it as a statement about nature rather than about a laboratory.

$\dfrac{\dd p^{\mu}}{\dd\tau}=f^{\mu}$ is one. Both sides are $(1,0)$. Here $p^{\mu}$ is a vector and $\dd\tau$ is a scalar, so the derivative is a vector, and $f^{\mu}$ is defined to be whatever four-vector sits on the right. Chapter 2.5 builds it and shows that its spatial part reduces to $\vv F$ at low speed, so nothing is lost. The old law is recovered as a limit, exactly as $\tfrac12mv^{2}$ was recovered in Chapter 0.1.

$\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}$ is one. Both sides are $(1,0)$, the left-hand one after its contraction. That single equation contains two of Maxwell's four equations. Chapter 2.6's punchline is that writing electromagnetism this way is not a repackaging but an explanation. The theory was relativistic before relativity was invented, which is why it broke Galilean invariance in Chapter 2.1.

$G_{\mu\nu}=\dfrac{8\pi G}{c^{4}}T_{\mu\nu}$ is one. Both sides are $(0,2)$ and symmetric. Chapter 3.6 derives it. The reason it can be guessed before it is derived is this chapter. Demand a $(0,2)$ symmetric tensor built from the metric and its first two derivatives, and there are almost no candidates. Tensor structure does not merely check laws. It drastically narrows the field of possible ones, and that is the method the whole second half of this book runs on.

7 · Symmetry, antisymmetry, and a promise

Chapter 0.7 §4.3 proved that any square matrix splits uniquely into a symmetric and an antisymmetric part, and then spent that result. Applied to the Jacobian of a velocity field, the symmetric part is strain and the antisymmetric part is rotation, at twice the local angular velocity. The same split applies to any rank-2 tensor, word for word.

with round brackets denoting the symmetric part and square brackets the antisymmetric part. That is standard notation, and it is worth learning now because Chapter 3.4 uses it heavily. Existence and uniqueness are Chapter 0.7's argument unchanged. Nothing in that argument used three dimensions or a metric.

7.1 · The split is frame-independent

Here is what is new, and it is the reason the split is physics rather than bookkeeping. Suppose $T^{\mu\nu}=T^{\nu\mu}$ in one frame, and ask what every other frame sees. Transform both indices and compare.

The first equality is the $(2,0)$ law with the two dummy indices named $\beta,\alpha$ instead of $\alpha,\beta$, which is a free choice by §8. The second equality uses only that numbers commute and that $T^{\beta\alpha}=T^{\alpha\beta}$ by hypothesis. So symmetry survives the transformation. The identical computation with one minus sign inserted shows that antisymmetry survives too.

7.2 · Counting, and what the counts are hiding

How many independent components does each piece have in $n$ dimensions? A symmetric array is fixed by its diagonal ($n$ entries) plus its strictly upper triangle ($\tfrac12 n(n-1)$ entries):

the antisymmetric count being the strictly upper triangle alone, since the diagonal is forced to zero by $A^{\mu\mu}=-A^{\mu\mu}$ for each fixed $\mu$ (no sum). The two add to $n^{2}$, as they must.

Now put numbers in. A coincidence is about to appear, and it has caused a great deal of confusion.

In three dimensions, $\tfrac12\cdot3\cdot2=3$. An antisymmetric $3\times3$ array has exactly three independent components, which is the same number as a vector. So it can masquerade as a vector, and in three-dimensional physics it invariably does. Chapter 0.7 §4.4 showed you the mechanism. The curl is really the antisymmetric part of a Jacobian, and it is only because $3=\tfrac12\cdot3\cdot2$ that we can package it as an arrow at all. The magnetic field is the same trick. $\vv B$ is not a vector in the way $\vv v$ is. It is an antisymmetric rank-2 object wearing a disguise that fits only in three dimensions. That is why it behaves oddly under reflection, and why the right-hand rule needs a convention where the laws of physics should not.

In four dimensions, $\tfrac12\cdot4\cdot3=6$. Six, not four, so the disguise fails. An antisymmetric rank-2 tensor in spacetime cannot possibly be a four-vector. It is its own kind of thing, with six slots.

Six independent numbers, sitting in the most natural rank-2 object spacetime admits. If you have been kept waiting since Chapter 0.7 §4.4, you should now be doing the arithmetic: three components of $\vv E$, three components of $\vv B$, six in total.

One more count before we leave the section: $\tfrac12\cdot4\cdot5=10$. A symmetric rank-2 tensor in four dimensions has ten independent components. That is the metric $g_{\mu\nu}$ of Chapter 3.3, and it is why "the gravitational field" turns out to be ten functions rather than one. Problem 4 asks you to do these counts yourself, because doing them is how they stick.

8 · Notation hygiene

This is a short section with a high yield. These rules are not style preferences. Each one is a consequence of something already proved, and each violation is a detectable error.

Free indices must match on both sides. A free index is one that appears exactly once in a term. It labels which component you are talking about, so both sides must be labelled the same way, with the same letters in the same positions and the same order of appearance. $A^{\mu}=B^{\mu}$ is fine. $A^{\mu}=B_{\mu}$ is not (different species, §3.3). $A^{\mu}=B^{\mu\nu}$ is not (the right side is sixteen numbers per value of $\mu$). Every term in a sum must carry the same free indices too, so $A^{\mu}+B^{\nu}$ is meaningless.

Dummy indices are summed and freely renameable. A dummy index appears exactly twice, once up and once down, and is summed. Its name is invisible to the result, so $V^{\mu}\omega_{\mu}=V^{\alpha}\omega_{\alpha}$. Use that freedom aggressively. In a long computation, renaming dummies to keep them distinct is the difference between an answer and a mess.

An index may not appear three times. If it does, either you meant two different indices and reused a letter, or you have a genuine error. There is no valid expression in which a letter appears three times in one term. This rule catches more mistakes than any other, precisely because it is purely mechanical: you can check it without understanding the expression at all.

The Kronecker delta is the mixed-index identity. $\delta^{\mu}{}_{\nu}=1$ if $\mu=\nu$ and $0$ otherwise. It has one index up and one down because it is a $(1,1)$ tensor. Problem 1 asks you to verify that, and the answer is more interesting than it sounds. Its two working properties are that it renames an index, $\delta^{\mu}{}_{\nu}V^{\nu}=V^{\mu}$, and that its trace counts dimensions.

Not $1$. Getting that wrong is a rite of passage. The number $4$ will appear all over Chapter 3.4 and, in the guise of $d$, throughout Chapter 5.10's dimensional regularisation.

$\partial_{\mu}$ carries a lower index. That is (2.4.11), and it is the most frequently forgotten fact in the subject, so here it is a second time. $\partial_{\mu}=\pdv{}{x^{\mu}}$ carries a lower index even though $x^{\mu}$ carries an upper one, because differentiating with respect to a thing inverts how it transforms. If you find yourself writing $\partial^{\mu}$, you must have raised it with the metric, and there is a minus sign on the spatial components waiting for you.

8.1 · The Levi-Civita symbol, and an honest caveat

One more array is worth naming. Define $\epsilon_{\mu\nu\rho\sigma}$ to be totally antisymmetric, meaning that it changes sign under the exchange of any two indices, and fix $\epsilon_{0123}=+1$. Total antisymmetry means it vanishes unless all four indices are different, so its $256$ entries are $\pm1$ on the $24$ permutations of $(0,1,2,3)$ and zero elsewhere.

How does it transform? We need one fact about any $4\times4$ matrix $M$ first, taken straight from Chapter 0.4 §5. The determinant is the alternating multilinear function of the columns normalised by $\det I=1$, so contracting a totally antisymmetric array with four copies of $M$ can only reproduce that array times the determinant.

Now ask what the $(0,4)$ tensor law would do to $\epsilon$. That law uses four factors of $\Lambda^{-1}$, so set $M=\Lambda^{-1}$ in (2.4.31):

By the grind box in §4, $\det\Lambda=\pm1$ for Lorentz transformations, so $1/\det\Lambda=\det\Lambda$ and the factor is $\pm1$. Under boosts and rotations, where $\det=+1$, the symbol is genuinely invariant and may be used exactly like a $(0,4)$ tensor. Under a spatial reflection, where $\det=-1$, it flips sign. That is where "pseudovector" and "pseudoscalar" come from, and Chapter 5.5 will find it decisive when the weak interaction turns out to care about the difference.

⚑ For a general coordinate change $\det\Lambda$ is neither $\pm1$ nor constant. Equation (2.4.32) then says that an $\epsilon$ with fixed entries is not a tensor at all but a tensor density, an object that picks up a power of the Jacobian determinant. The repair is to multiply by $\sqrt{-g}$, where $g=\det g_{\mu\nu}$, because that factor carries a compensating determinant. We quote the repair here and defer the derivation to Chapters 3.3 and 3.5, where the same factor is also what makes $\sqrt{-g}\,\dd^{4}x$ the invariant volume element. Within Part II, where $\det\Lambda=\pm1$, none of this bites.

9 · Worked examples

Grind box — where Step 1 breaks, and why Part III needs a new derivative

Step 1 above used one thing beyond the definitions: that $\Lambda^{\mu}{}_{\alpha}$ is constant, so $\partial_{\beta}$ passes through it. Suppose it is not constant, as it is not for a change to polar coordinates, or an accelerating frame, or any coordinate patch on a curved manifold. Then the product rule gives an extra term.

$$\partial'_{\nu}V'^{\mu} \;=\; (\Lambda^{-1})^{\beta}{}_{\nu}\Big[\Lambda^{\mu}{}_{\alpha}\,\partial_{\beta}V^{\alpha} \;+\; \underbrace{\big(\partial_{\beta}\Lambda^{\mu}{}_{\alpha}\big)V^{\alpha}}_{\text{inhomogeneous}}\Big].$$

The first term is the tensor law. The second is not proportional to $\partial V$ at all. It is proportional to $V$ itself, so the whole expression is not of the form (matrix)$\times$(components of $\partial V$). $\partial_{\nu}V^{\mu}$ is not a tensor under general coordinate changes.

Look closely at why this is fatal rather than annoying. The failure is an inhomogeneous term, and §6's theorem needed the transformation law to be homogeneous. With an inhomogeneous term, an expression that vanishes in one frame need not vanish in another. The proof of the central theorem collapses, and with it the guarantee that makes tensor equations worth writing.

The repair is to invent a new derivative $\nabla_{\mu}$ which differs from $\partial_{\mu}$ by a correction term chosen so that the two inhomogeneous pieces cancel. The correction is called the connection, it is built from derivatives of the metric, and constructing it is Chapter 3.3. Its failure to commute with itself is curvature, which is Chapter 3.4, which is gravity.

So the flat-space simplification that let this chapter be short is exactly the thing Part III gives up, and everything hard about general relativity is downstream of that one product rule.

10 · Your turn