Part III · General Relativity — Chapter 3.3

Metric and Connection

Two objects are missing from the arena: one that measures, and one that compares. This chapter builds both, and finds that the second of them is the gravitational field.

Tools you'll need.  Chapter 3.2 in its entirety: the tangent space, the coordinate basis $\partial_{\mu}$, the transformation law (3.2.16) for vector components, and above all §5, which is the problem this chapter solves. Chapter 2.4 §4: the metric as the dictionary between vectors and covectors, raising and lowering, and §6, the theorem that a tensor equation true in one chart is true in all of them. Chapter 0.5 §1 and §6: inner products defined by axioms, and the spectral theorem. Chapter 0.6 §5 (the multivariable chain rule, used constantly) and §6.1 (Clairaut's theorem on mixed partials, which does the essential work in §7). Chapter 1.2 §3 (the Euler–Lagrange equation) and its Worked example 2, which minimised path length in the plane and on a sphere and promised that this chapter would replace the flat line element with a curved one. Chapter 2.5 §5, the action $S=-mc^{2}\!\int\dd\tau$. Chapter 2.3 §5, proper time as the length of a worldline. Chapter 0.8 §1 for linear systems of ordinary differential equations, used once in §6.

1 · The metric: an inner product at every point

Chapter 0.5 defined an inner product on a vector space by axioms rather than by a formula: a symmetric bilinear map taking two vectors to a number. Chapter 2.4 §4 put one on Minkowski spacetime, with indefinite signature, and showed that it does two jobs at once. It measures intervals, and it converts vectors into covectors.

Both of those constructions apply to a single vector space. We now have one vector space per point, so we install one inner product per point. That is the whole idea, and the definition below is just careful bookkeeping about it.

Three remarks on that definition, each of which will be used.

(i) It is a tensor field. The map eats two vectors and returns a number, linearly in each. That is Chapter 3.2 §7's definition of a type $(0,2)$ tensor, word for word. So its components obey the tensor law with two factors of the inverse Jacobian. We can also derive that law directly, and it is worth doing once, because the derivation is three lines and it fixes the index placement.

What we want is a relation between the components in two charts, so we start from the one thing that must hold whichever chart we use. Expand the same two vectors both ways, and demand that $g(X,Y)$ come out as one number:

Our next goal is to get the right-hand side written in primed components, so that both sides can be compared term by term. Substitute the component law $X^{\alpha}=(\partial x^{\alpha}/\partial x'^{\mu})X'^{\mu}$, which is (3.2.16) read backwards, and do the same for $Y$:

Both sides now carry the same factor $X'^{\mu}Y'^{\nu}$, and $X'$ and $Y'$ were arbitrary. So the coefficients must agree index by index, and that gives the law we were after.

(ii) The signature is fixed. At each point $g_{\mu\nu}$ is a real symmetric matrix. The spectral theorem (Chapter 0.5 §6) therefore gives it real eigenvalues and an orthogonal eigenbasis, and non-degeneracy says that none of those eigenvalues is zero. The count of positive and negative eigenvalues is the signature. On a connected manifold that count cannot change from point to point, because changing it would need an eigenvalue to pass through zero on the way, and non-degeneracy forbids exactly that. ⚑ Making the continuity argument airtight takes Sylvester's law of inertia together with a connectedness argument. We quote the conclusion and use it.

For spacetime we impose signature $(+,-,-,-)$: three minus signs and one plus, everywhere. That is what makes each tangent space a copy of the Minkowski spacetime of Chapter 2.3. It is also the precise statement that special relativity holds locally.

(iii) The line element. The traditional notation for a metric is

and it deserves one sentence of honesty, because Chapter 3.2 §6 taught us what $\dd x^{\mu}$ really is. It is a basis covector, and the product of two covectors written side by side means their symmetric tensor product. So (3.3.4) is shorthand for the tensor $g=g_{\mu\nu}\,\dd x^{\mu}\otimes\dd x^{\nu}$, and feeding it the same tangent vector twice returns the squared length of that vector.

There is an older reading of the same symbols: "the squared length of an infinitesimal displacement". It gives the right answers, because it is the statement above with the vector left implicit. We use the notation freely from here on, having said once what licenses it.

1.1 · Raising, lowering, and the gradient that finally exists

Non-degeneracy says the matrix $g_{\mu\nu}$ is invertible at each point. Write the inverse $g^{\mu\nu}$, defined by

Everything Chapter 2.4 §4 built now transfers unchanged, because it was pointwise algebra: $V_{\mu}\equiv g_{\mu\nu}V^{\nu}$ lowers an index, $\omega^{\mu}\equiv g^{\mu\nu}\omega_{\nu}$ raises one, and the two are inverse operations by (3.3.5). The one alteration is the one Chapter 3.2 §7 identified: the dictionary is now different at every point.

The length of a vector and the angle between two are read off in the usual way, with one warning inherited from Part II. In Lorentzian signature $g(X,X)$ may be positive, negative or zero. So Chapter 2.3's three-way classification applies at each point separately: $X$ is timelike if $g(X,X)>0$, spacelike if $g(X,X)<0$, and null if $g(X,X)=0$ with $X\neq0$.

Since the signature is the same everywhere, so is the classification. That is why the light cone at each point is a genuine structure rather than an artefact of a chart.

2 · Flat space in polar coordinates — a warning delivered early

This section exists to destroy a belief before it forms. The belief is: if the metric components depend on position, the space is curved.

It is false. It is also the single most common misunderstanding at this stage, and it will wreck the next two chapters for anyone holding it. So we work the counterexample now, at length, before curvature has even been defined.

2.1 · The change of chart, one step at a time

Take the ordinary Euclidean plane, with Cartesian coordinates $(x,y)$ and the metric everyone agrees on,

Now change chart to polar coordinates by $x=r\cos\theta$, $y=r\sin\theta$. This is a legitimate chart on the plane minus the origin: smooth, with a smooth inverse, exactly as Chapter 3.2 §2 requires.

We want the line element rewritten in the new coordinates, and the only way in is to express $\dd x$ and $\dd y$ in terms of $\dd r$ and $\dd\theta$. So differentiate each coordinate, using the product rule on $r\cos\theta$ and treating $r$ and $\theta$ as the two variables:

What the line element needs is $\dd x^{2}$ and $\dd y^{2}$, so square each of those two expressions in turn. Here is the first one, expanded one term at a time:

The second one squares the same way. Watch the middle term, because its sign comes out opposite to the one above, and that is the whole reason the sum is going to be tidy:

Now add the two of them, which is what $\dd s^{2}$ asks for. Three things happen. The cross terms are equal and opposite, so they cancel outright. The two $\dd r^{2}$ coefficients combine by $\cos^{2}\theta+\sin^{2}\theta=1$. The two $\dd\theta^{2}$ coefficients combine by that same identity, once $r^{2}$ has been factored out. The result is

2.2 · Nothing happened to the plane

Let's look at what changed and what did not. The component $g_{\theta\theta}=r^{2}$ is a function of position. It is $1$ at $r=1$, it is $100$ at $r=10$, and it grows without bound. By the eye test the metric now looks dramatically more complicated than (3.3.6).

And yet the plane is the plane. The change of chart was a smooth bijection. No point was added, removed, stretched or bent. Every geometric statement one could make survives verbatim. Three of them are worth checking explicitly, because they are the ones people reach for when asked what "flat" means.

Circumference. Take the curve $r=R$ constant, $\theta$ running from $0$ to $2\pi$. Along it $\dd r=0$, so (3.3.10) gives $\dd s = R\,\dd\theta$, and the length is

which is Euclid's answer exactly, with no correction of any order.

Straight lines. The curve $\theta=\theta_{0}$ constant has $\dd s=\dd r$, so its length between $r=a$ and $r=b$ is $b-a$. Radial rays are still straight lines and still have the length they always had.

Angles. The off-diagonal component vanishes, so the coordinate directions $\partial_{r}$ and $\partial_{\theta}$ are perpendicular at every point, as radial and circular directions ought to be.

Two more things will be shown about (3.3.10) before this chapter ends. It is worth stating them now so that you know what is coming. In §7 we compute the connection coefficients of the flat plane in polar coordinates and find them non-zero: $\Gamma^{r}{}_{\theta\theta}=-r$ and $\Gamma^{\theta}{}_{r\theta}=1/r$. In Chapter 3.4 we compute the curvature of the same metric and find it identically zero, component by component.

Non-zero connection, zero curvature, in a space that is flat by construction. That pair of facts is the warning in its final form, and it also explains where Chapter 1.1's centrifugal and Coriolis terms came from.

Those two coefficients are quoted here, five sections early, because the figure below needs them. The comparison the box above asks for cannot be made on the page. It can only be run. One routine transports one vector round one coordinate rectangle, and the only thing that differs between the two panels is which $\Gamma$ it is handed.

plane Δr

1.60

plane Δθ

1.10 rad

sphere Δθ

0.80

sphere Δφ

1.20 rad

marker

32% round

shrink both loops

show the connection at the marker

reset

plane |V| start 0.61846584384265 end 0.61846584384265 drift 2.2e-16

sphere |V| start 0.61638163516818 end 0.61638163516818 drift 3.3e-16

plane marker r=2.200 θ=0.658 V^r=+0.617238 V^θ=-0.017705 Cartesian (+0.512189, +0.346645)

sphere marker θ=1.400 φ=0.686 V^θ=+0.607079 V^φ=+0.108259

closing angle: plane 4.16e-15° sphere 45.059816966° enclosed area/a² 45.059816966° difference 3.9e-12°

3 · Proper time, and the action of a free particle

Chapter 2.3 §5 defined proper time along a worldline by $c^{2}\dd\tau^{2}=\dd s^{2}$, the interval being positive along timelike curves in our signature. That definition referred to $\eta_{\mu\nu}$ only through $\dd s^{2}$, so replacing the flat metric by the general one carries it across untouched:

Parametrise a worldline by anything at all, say $\lambda$, write $\dot x^{\mu}=\dd x^{\mu}/\dd\lambda$, and divide (3.3.12) by $\dd\lambda^{2}$ to get the elapsed proper time as an integral:

Two immediate readings of (3.3.12), both of which are collected debts.

A clock standing still. An observer who stays at fixed spatial coordinates has $\dd x^{i}=0$, so only the $00$ term survives. With $x^{0}=ct$ as always,

Chapter 3.1 §6.5 wrote exactly this line in advance and used it, together with the redshift derived from the accelerating cabin, to read off $g_{00}=1+2\Phi/c^{2}$ to first order. That argument used no general relativity, and it is now expressed in the notation it was waiting for.

It is also the reason the metric cannot be $\eta_{\mu\nu}$ near a mass. Two static clocks at different heights tick at different rates, and only a position-dependent $g_{00}$ can say so.

The action. Chapter 2.5 §5 obtained the action of a free relativistic particle by demanding a Lorentz-invariant number attached to each history and finding that only one was available: $S=-mc^{2}\!\int\dd\tau$. Nothing in that argument used flatness. So

Check the units before going on, since the constants are being kept explicit: $g_{\mu\nu}$ is dimensionless, $x^{\mu}$ is a length, so the square root has the dimensions of a length per unit $\lambda$, and $mc\times$ length is $\mathrm{kg\,m^{2}\,s^{-1}}$, which is an action. And setting $g_{\mu\nu}=\eta_{\mu\nu}$ returns Chapter 2.5's expression character for character.

4 · The coordinate derivative is not a tensor

Here is the obstruction, in full. Chapter 3.2 §5.4 asserted it and deferred the computation to this section. Chapter 3.2 §7 listed it as the single entry of Chapter 2.4's table that goes wrong on a manifold. Everything from §5 onward exists to remove it, so it is shown completely rather than summarised.

Here is the destination, announced before we set out. We differentiate the vector transformation law, find two terms where a tensor would have one, and identify the extra term exactly. It contains a second derivative of the coordinate change, and it vanishes if and only if that change is affine.

The calculation is three lines long, and each line is labelled below.

Line 1 · what we are differentiating. Chapter 3.2 (3.2.16) established

To differentiate that, we need to know what the primed derivative operator even means in terms of the unprimed one, so that is the next thing to get.

Line 2 · the derivative operator in the new chart. By the chain rule (Chapter 0.6 §5), differentiating with respect to a primed coordinate means going through the unprimed ones:

Line 3 · apply it. Act with (3.3.17) on (3.3.16). The object being differentiated is a product of two position-dependent things, the Jacobian factor and the components, so the product rule of Chapter 0.1 §4 applies and produces two terms:

Read the two terms carefully, because the whole chapter turns on the difference between them.

The first term is exactly what Chapter 3.2 §7's definition demands of a type $(1,1)$ tensor: one factor of $\partial x'/\partial x$ for the upper index, one factor of $\partial x/\partial x'$ for the lower one, contracted against the components in the old chart. Had the calculation stopped there, $\partial_{\alpha}V^{\beta}$ would be a tensor and this chapter would not exist.

The second term is the obstruction, and note precisely what it is made of. It carries a second derivative of the new coordinates with respect to the old ones, and it is proportional to the components $V^{\beta}$ themselves rather than to their derivatives. Nothing about it can be absorbed into a redefinition of $\partial_{\alpha}V^{\beta}$, because it does not contain that object at all.

4.1 · When the obstruction vanishes, and why Part II never met it

The extra term dies exactly when $\partial^{2}x'^{\nu}/\partial x^{\alpha}\partial x^{\beta}=0$ for all indices. That is a differential equation for the coordinate change, and integrating it twice means

which is an affine change of coordinates. Lorentz transformations and translations are precisely of this form. That is why Chapter 2.4 could list $\partial_{\nu}V^{\mu}$ in its table of tensors without qualification and never be caught out. The moment arbitrary smooth relabellings are admitted, as Chapter 3.2 §7 said they must be, the second derivative is generically non-zero and the entry is false.

4.2 · A concrete demonstration, in the plane

Abstract failures are easy to nod at and hard to believe. So here is the failure with numbers in it, staged in the flat plane of §2. There is no curvature anywhere in what follows, only a change of chart.

Take the constant vector field $V=\partial_{x}$: the same arrow at every point of the plane, pointing along the $x$-axis, with Cartesian components $(V^{x},V^{y})=(1,0)$. Every derivative of every component vanishes in the Cartesian chart. Now find its components in the polar chart. By (3.3.16), with primed meaning polar,

using $r=\sqrt{x^{2}+y^{2}}$ so that $\partial r/\partial x = x/r = \cos\theta$, and $\theta=\arctan(y/x)$ so that $\partial\theta/\partial x = -y/r^{2} = -\sin\theta/r$.

Those polar components depend on position. The question we care about is whether their coordinate derivatives come out zero, as they did in the Cartesian chart, so differentiate them and look:

So the array $\partial_{\mu}V^{\nu}$ is zero in one chart and non-zero in another, for a field that is manifestly the same field. No tensor behaves like that. The transformation law (3.3.3) is linear and homogeneous in the components, so a tensor vanishing in one chart vanishes in all of them (Chapter 2.4 §6). The array $\partial_{\mu}V^{\nu}$ fails that test, and (3.3.18) says why: the polar-to-Cartesian map is not affine.

5 · The connection, defined as the repair

We do not now guess an object and check that it works. We demand that the disease be cured, and read off what the cure must be.

Here is the destination. Write down the most general first-order correction to the coordinate derivative, require the corrected object to be a tensor, and derive the transformation law that the correction is thereby forced to obey. The correction does not itself come out a tensor, which is the surprise, and that is exactly why it can cancel something which is not a tensor either.

5.1 · The definition

The correction has been taken linear in $V$ and free of derivatives of $V$. That is the least it can be if it is to cancel the obstruction term of (3.3.18), since that term is itself linear in $V$ and free of derivatives of $V$. Nothing else about $\Gamma$ is assumed.

5.2 · What the connection must do, derived

Everything now follows from one demand, so let's write the demand down first. We want $\nabla_{\mu}V^{\nu}$ to transform as a type $(1,1)$ tensor, which says exactly this:

The plan is to write each side of that demand out in full and see what is left over. Start with the left side. Use the definition of $\nabla$, then substitute (3.3.18) for the coordinate-derivative piece:

Now do the same to the other side, so that the two can be compared term against term. The right side of (3.3.22) carries $\nabla_{\alpha}V^{\beta}$, which is a coordinate derivative plus a connection term, so it opens up into two pieces as well:

The first terms of (3.3.23) and (3.3.24) are identical and cancel between the two sides. What is left is a requirement on $\Gamma'$:

We want a statement about $\Gamma$ alone, with no $V$ in it. To strip the vector off, every term has to be expressed through the same $V'^{\lambda}$. So use (3.3.16) in reverse, $V^{\gamma}=(\partial x^{\gamma}/\partial x'^{\lambda})V'^{\lambda}$, on both right-hand terms. In the second term the summed index is called $\beta$, so relabel it $\gamma$ first and the two terms then carry the same letter. Every term is now proportional to $V'^{\lambda}$, and $V'^{\lambda}$ is arbitrary, so the coefficients must agree:

5.3 · Three consequences, and the surprising one first

(a) The connection is not a tensor. Compare (3.3.26) with the tensor law of Chapter 3.2 §7. The first term is exactly right, with one Jacobian per index, in the correct places, for a type $(1,2)$ tensor. The second term has no business being there. A tensor law is homogeneous in the components. This one is inhomogeneous, carrying an additive piece that does not involve $\Gamma$ at all.

The consequence is startling on first meeting, and it is worth stating baldly: the connection can be zero in one chart and non-zero in another. Set $\Gamma^{\beta}{}_{\alpha\gamma}=0$ in (3.3.26) and the first term dies, but the second does not, so $\Gamma'\neq0$ in general.

That is not a defect. It is precisely the property required, because $\Gamma$'s job is to cancel a term that is itself not tensorial, and only a non-tensor can do that. Section 7 will exhibit the phenomenon concretely. The flat plane has $\Gamma=0$ in Cartesian coordinates and $\Gamma\neq0$ in polar coordinates, and (3.3.26) is what relates the two. Chapter 3.1 §2.1 already met the same phenomenon in physical dress, when a uniform gravitational field was deleted by changing coordinates.

(b) The difference of two connections is a tensor. Suppose $\Gamma$ and $\tilde\Gamma$ both satisfy (3.3.26). Subtract the two statements. The inhomogeneous term is identical in both and cancels exactly, leaving

which is the tensor law exactly. So the set of connections is not a vector space but an affine space: you cannot add two connections, but you can subtract them and get a genuine tensor. This is used twice more in the book, once in §7 and once in Part VI where the same structure reappears with an internal space in place of the tangent space.

(c) The obstruction is now exactly cancelled. By construction, not by luck. That was the demand from which (3.3.26) was derived.

5.4 · Extending $\nabla$ to everything else

We have defined $\nabla$ on vector fields. Two further requirements fix it on everything else, and both are natural rather than arbitrary:

• On a scalar field, $\nabla_{\mu}f\equiv\partial_{\mu}f$. This is forced, because $\partial_{\mu}f$ is already a tensor. It is the covector $\dd f$ of Chapter 3.2 §6, and nothing needs repairing.
• $\nabla$ obeys the Leibniz rule on products of tensors, as any object calling itself a derivative must.

From those two, the rule for covectors is derived rather than posited. Take a covector $\omega_{\nu}$ and a vector $V^{\nu}$. Their contraction $\omega_{\nu}V^{\nu}$ is a scalar, so the first bullet says $\nabla$ acts on it as a plain $\partial$, while the second says it must split across the product:

The unknown in that line is $\nabla_{\mu}\omega_{\nu}$, so the goal is to isolate it. Expand both sides. On the left, the ordinary product rule gives $(\partial_{\mu}\omega_{\nu})V^{\nu}+\omega_{\nu}\partial_{\mu}V^{\nu}$. On the right, substitute the definition of $\nabla_{\mu}V^{\nu}$:

The terms $\omega_{\nu}\partial_{\mu}V^{\nu}$ appear on both sides and cancel. In the surviving $\Gamma$ term the summed indices are $\nu$ and $\lambda$. Relabel them, swapping the two letters, so that the free vector index reads $\nu$ like everywhere else. Both are dummies, so that changes nothing. The term becomes $\omega_{\lambda}\Gamma^{\lambda}{}_{\mu\nu}V^{\nu}$, and now every term carries a factor $V^{\nu}$ with $V$ arbitrary. Matching coefficients:

Note the minus sign, and note that it was not chosen. It was forced by the Leibniz rule and the requirement that the pairing be a scalar. Iterating the same argument on a general tensor gives the pattern: one $+\Gamma$ term per upper index and one $-\Gamma$ term per lower index.

Two particular cases are used later in this chapter and in the next, so let's write them out. First a tensor with two lower indices, which by the pattern picks up two minus terms:

and then one index up and one down, where the pattern gives a plus term for the upper index and a minus term for the lower one:

Equation (3.3.32) is the one the whole of Chapter 3.4 will lean on, so it is worth a second look now. A $(1,1)$ tensor gets three terms: a derivative, and one connection term per index, with the sign set by whether the index is up or down.

6 · Parallel transport

Now that we have a derivative that behaves itself, we can spend it. This section turns it into the one thing Chapter 3.2 said was missing: a way of carrying a vector from one point to another.

The covariant derivative measures how a field changes. Setting it to zero along a curve is therefore the statement that the field does not change along that curve, which is exactly the comparison Chapter 3.2 §5.3 could not supply.

That definition is compact but not yet usable. What we want is an equation we could hand to a computer, so let's unpack it. Expand the covariant derivative first, and deal with the leading term afterwards:

The first term is a directional derivative along the curve, and by the chain rule (Chapter 0.6 §5) that is an ordinary derivative with respect to the parameter:

Put that back into the first equation and set the whole thing to zero, as the definition demands. Parallel transport is then a system of ordinary differential equations for the components:

Three observations, all of which matter later.

It always has a solution, and only one. Equation (3.3.35) is a linear first-order system in the $n$ unknowns $V^{\mu}(\lambda)$, with coefficients that are smooth functions of $\lambda$ along the curve. ⚑ By the standard existence-and-uniqueness theorem for such systems, quoted here and used without proof, a value of $V$ at one point of the curve determines $V$ at every other point of it. So the comparison Chapter 3.2 lacked now exists: given a path, a vector at one end can be carried to the other.

The answer depends on the path. Equation (3.3.35) is an equation along a curve, and different curves joining the same two points give different coefficient functions and hence, in general, different answers. Chapter 3.2 §5.3's figure showed exactly this happening on a sphere, before there was any machinery to describe it. Chapter 3.4 takes the path-dependence, makes it quantitative, and calls it curvature.

Nothing yet says lengths are preserved. With $\Gamma$ still arbitrary, a transported vector may grow or shrink. Section 7 imposes the condition that stops it, and the figure below measures the result.

path latitude

45°

arc travelled

200°

start bearing

40°

make the path a geodesic

transport the tangent instead

length of the transported vector: start 1.00000000000000 end 1.00000000000001

angle to the direction of travel: start 50.000000° end 268.578636° change -141.421364°

not a great circle: predicted drift −sin(lat)×arc = -141.421356° [tangency 1.6e-15]

7 · The Christoffel formula

This is the chapter's set piece, so we go slowly and announce everything before doing it.

7.1 · Demand 1, and what it actually says

Take two vectors $V$ and $W$, both parallel-transported along a curve with tangent $u$, and ask how their inner product changes along the way. The inner product is a scalar, so its rate of change along the curve is the ordinary derivative, which by (3.3.34) is $u^{\rho}\nabla_{\rho}$ acting on it:

The right-hand side has $\nabla$ acting on a product of three things, so the Leibniz rule of §5.4 applies and gives one term per factor:

The last two terms vanish, because $V$ and $W$ are parallel-transported and that is precisely the statement $u^{\rho}\nabla_{\rho}V^{\mu}=0$. So

Read (3.3.38) both ways. If $\nabla_{\rho}g_{\mu\nu}=0$ then every inner product is constant along every curve, so lengths and angles are preserved by transport. Conversely, if inner products are preserved for all choices of $V$, $W$ and all curves, then the coefficients of $V^{\mu}W^{\nu}$ must vanish for every $u$, giving $\nabla_{\rho}g_{\mu\nu}=0$. The condition and the geometric statement are the same thing.

That is why the figure in §6 reports a length constant to fourteen digits. It is not an accident of the numerical scheme. It is Demand 1, which we are about to impose, showing itself.

7.2 · Demand 2, and why it is a real condition

Apply two covariant derivatives to a scalar $f$. The first gives the covector $\partial_{\nu}f$. The second then differentiates a covector, so (3.3.30) is the rule that applies:

What we want to know is whether the two derivatives commute, so form the difference. Swap $\mu$ and $\nu$ in that line and subtract it from the original. The ordinary second derivatives cancel by Clairaut's theorem (Chapter 0.6 §6.1), leaving

The array $T^{\lambda}{}_{\mu\nu}$ is called the torsion, and two facts about it are worth deriving rather than asserting.

Torsion is a tensor, even though $\Gamma$ is not. Look at the inhomogeneous term in (3.3.26). It carries $\partial^{2}x'^{\nu}/\partial x^{\alpha}\partial x^{\gamma}$, which is symmetric in $\alpha$ and $\gamma$ by Clairaut, and it is contracted with two Jacobians carrying the lower indices $\mu$ and $\lambda$. So the whole inhomogeneous term is symmetric under $\mu\leftrightarrow\lambda$, and it cancels when the antisymmetric part is taken. What survives transforms homogeneously, which is the tensor law exactly. Setting a tensor to zero is a statement independent of chart, so Demand 2 is a legitimate condition rather than a coordinate convention.

Setting it to zero is a genuine choice, with a reason. A connection with torsion is mathematically consistent, and nothing so far forbids it. We set it to zero because (3.3.40) shows that torsion is precisely the failure of second derivatives of a scalar to commute, and Chapter 3.2 built the entire manifold on smooth functions whose mixed partials do commute. ⚑ That general relativity chooses the torsion-free connection is an assumption about nature. Theories with torsion have been constructed and tested, and the observational situation is that no torsion has been detected. We flag the choice and proceed.

7.3 · Counting, before any algebra

It is worth knowing in advance whether a unique answer should be expected. In four dimensions:

• Unknowns. $\Gamma^{\lambda}{}_{\mu\nu}$ has one free upper index (4 values) and a pair of lower indices which Demand 2 makes symmetric, so $4\times5/2=10$ independent pairs. Total $4\times10=40$ functions.
• Equations. $\nabla_{\rho}g_{\mu\nu}=0$ has one free index $\rho$ (4 values) and a symmetric pair $\mu\nu$ (10 values). Total $4\times10=40$ equations.

Forty linear equations for forty unknowns. A unique solution is exactly what one should expect, and the algebra below produces it. Had the counting come out unequal we would have known in advance that either no connection or infinitely many satisfy the demands.

7.4 · Step 1 — write the condition out

The metric is a type $(0,2)$ tensor, so (3.3.31) gives its covariant derivative. Setting that to zero:

7.5 · Step 2 — two cyclic copies

Equation (3.3.41) holds for every assignment of the three free indices, so we may write it again with the letters moved around. Cycle them once, $\rho\to\mu$, $\mu\to\nu$, $\nu\to\rho$, meaning: wherever (A) has $\rho$ write $\mu$, wherever it has $\mu$ write $\nu$, wherever it has $\nu$ write $\rho$. The summed index $\lambda$ is untouched, being private to each term:

Cycle once more, by the same rule applied to (B), so that overall $\rho\to\nu$, $\mu\to\rho$, $\nu\to\mu$ relative to (A):

A third cycle would return (A), which is the check that the permutation was done correctly.

7.6 · Step 3 — the combination (B) + (C) − (A)

Now form that combination. We take it slowly, doing the derivative terms first and then the six connection terms one pair at a time. The derivative terms collect:

The six connection terms are, written out with the signs they carry after (A) has been subtracted:

First cancellation. Compare the second term of the (B) row with the first term of the $-$(A) row:

Two separate symmetries were used and both should be named. The connections match because $\Gamma^{\lambda}{}_{\mu\rho}=\Gamma^{\lambda}{}_{\rho\mu}$, which is Demand 2. The metrics match because $g_{\nu\lambda}=g_{\lambda\nu}$, which is the symmetry of the metric from §1. With both factors equal and the signs opposite, the pair cancels exactly.

Second cancellation. Compare the first term of the (C) row with the second term of the $-$(A) row:

by exactly the same two symmetries, applied to the index pair $\nu\rho$ this time.

What survives. The two remaining terms are the first of the (B) row and the second of the (C) row:

where again both symmetries were used: Demand 2 turns $\Gamma^{\lambda}{}_{\nu\mu}$ into $\Gamma^{\lambda}{}_{\mu\nu}$, and the symmetry of the metric turns $g_{\rho\lambda}$ into $g_{\lambda\rho}$, so the two terms are identical and add.

Putting (3.3.44) and (3.3.48) together, the combination (B) + (C) − (A) reads

7.7 · Step 4 — free the connection

Only one thing remains: the surviving $\Gamma$ carries a metric factor, and we want it bare. Multiply (3.3.49) through by $\half g^{\rho\sigma}$ and sum over $\rho$. On the connection term, $g^{\rho\sigma}g_{\lambda\rho}=\delta^{\sigma}{}_{\lambda}$ by (3.3.5), and the Kronecker delta then does its only job, which is to rename $\lambda$ into $\sigma$ and collapse the sum. Rearranging so that the connection is alone on the left, and finally relabelling the free index $\sigma\to\lambda$ and the summed index $\rho\to\sigma$ so the result reads in the conventional letters:

These are the Christoffel symbols, and (3.3.50) is the Levi-Civita connection of the metric $g$. Two consistency checks, both immediate:

It is symmetric in $\mu\nu$, as Demand 2 requires. Swap the two letters on the right: the first two terms exchange with each other, and the third is already symmetric because $g_{\mu\nu}=g_{\nu\mu}$. Nothing changes. Had the formula come out unsymmetric we would have derived a contradiction.

It vanishes when the components are constant. Suppose $\partial_{\sigma}g_{\mu\nu}=0$ everywhere, as it is in Cartesian coordinates on the plane, or in the Minkowski chart of Part II. Then every term on the right of (3.3.50) is zero, so $\Gamma=0$ and $\nabla_{\mu}=\partial_{\mu}$. All of Part II's calculus is recovered exactly, which it had better be.

7.8 · The flat plane in polar coordinates, computed

Section 2 promised this calculation. The metric is (3.3.10), so $g_{rr}=1$, $g_{\theta\theta}=r^{2}$, $g_{r\theta}=0$, and the inverse is $g^{rr}=1$, $g^{\theta\theta}=1/r^{2}$, $g^{r\theta}=0$. Exactly one derivative of the metric is non-zero:

Feed that into (3.3.50), one component at a time. Since the metric is diagonal, the sum over $\sigma$ has a single surviving term in each case. Take $\Gamma^{r}{}_{\theta\theta}$ first, where $\lambda=r$ picks out $g^{rr}$ and both lower indices are $\theta$:

Now take $\Gamma^{\theta}{}_{r\theta}$, where the upper index picks out $g^{\theta\theta}=1/r^{2}$ and the one surviving metric derivative sits in the first slot rather than the third, so this time it arrives with a plus sign:

By the symmetry just established, $\Gamma^{\theta}{}_{\theta r}=\Gamma^{\theta}{}_{r\theta}=1/r$. The remaining four components all vanish: $\Gamma^{r}{}_{rr}$, $\Gamma^{r}{}_{r\theta}$, $\Gamma^{\theta}{}_{rr}$ and $\Gamma^{\theta}{}_{\theta\theta}$ each involve only derivatives that are zero by (3.3.51), or in the last case $\partial_{\theta}g_{\theta\theta}=\partial_{\theta}(r^{2})=0$. So

8 · Geodesics, twice

A geodesic is the curved-space replacement for a straight line, and there are two entirely different ways to say what a straight line is. It is the curve that never turns, and it is the shortest route between two points. In Euclidean geometry these agree, and the agreement is so familiar it is invisible. Here they are two separate definitions with two separate derivations, and we do both and compare.

8.1 · First route — the curve that transports its own tangent

"Never turns" means: the direction you are travelling in, carried forward by the transport rule of §6, is still the direction you are travelling in. In symbols, the tangent is parallel-transported along its own curve:

Expand it exactly as in §6. The first term becomes a derivative with respect to the parameter by (3.3.34), and the second is the connection term with $V$ replaced by $u$:

Writing $u^{\mu}=\dd x^{\mu}/\dd\lambda$ in both places gives the geodesic equation:

The parameter is not free. Equation (3.3.57) is not invariant under an arbitrary reparametrisation. It holds for affine parameters, meaning those related to each other by $\lambda\to a\lambda+b$ with constants $a,b$. Proper time is one of them, and here is the derivation of that fact.

What we want is a quantity that stays fixed along the curve, since that is what will pin the parameter down. Compute the rate of change of $g_{\mu\nu}u^{\mu}u^{\nu}$ along the curve using (3.3.38) with $V=W=u$. Metric compatibility kills the first factor, and (3.3.55) kills what is left:

So the "speed" $g_{\mu\nu}u^{\mu}u^{\nu}$ is constant along a geodesic. For a timelike geodesic we may therefore normalise it to $c^{2}$ once and for all, and by (3.3.12) that choice is exactly $\lambda=\tau$.

8.2 · Second route — extremise the proper time

Now forget parallel transport entirely. Take the action (3.3.15), which came from Chapter 2.5 and knows nothing about connections, and extremise it by Euler–Lagrange. Chapter 1.2 Worked example 2 did precisely this for the plane and the sphere and promised that the general case would be this chapter's business.

Drop the constant $-mc$ and work with

A remark on the parameter, before varying. $F$ is homogeneous of degree one in $\dot x$: scale $\dot x$ by $k$ and $F$ scales by $k$. So under a change of parameter $\lambda\to\tilde\lambda$ the integrand transforms as $F(x,\dd x/\dd\tilde\lambda)\,\dd\tilde\lambda = F(x,\dd x/\dd\lambda)\,(\dd\lambda/\dd\tilde\lambda)\,\dd\tilde\lambda = F(x,\dd x/\dd\lambda)\,\dd\lambda$. So the integral is the same whichever parameter is used, and therefore so is its Euler–Lagrange equation. That is Chapter 1.2 §7's form-invariance in a particularly clean special case. We may therefore vary with an arbitrary $\lambda$ and choose the parameter afterwards, which is what we do. Choosing it beforehand would restrict the competing paths and is not allowed.

The two derivatives. Euler–Lagrange needs $F$ differentiated with respect to velocity and with respect to position, so take those in turn. The argument of the square root is $g_{\alpha\beta}\dot x^{\alpha}\dot x^{\beta}$. Differentiating it with respect to $\dot x^{\mu}$ hits each of the two velocity factors in turn and, since $g_{\alpha\beta}$ is symmetric, the two contributions are equal, giving $2g_{\mu\beta}\dot x^{\beta}$. The chain rule on the square root supplies $1/(2F)$:

Now the position derivative, which is the easier of the two. Inside $F$ the only thing that depends on $x$ at all is $g_{\alpha\beta}(x)$, so the same $1/(2F)$ from the square root is all that accompanies it:

Assemble Euler–Lagrange. The equation is Chapter 1.2 (1.2.22), in the form $\dv{}{\lambda}\big(\partial F/\partial\dot x^{\mu}\big)-\partial F/\partial x^{\mu}=0$, so drop the two derivatives just computed into their slots:

Now choose the parameter. Those factors of $F$ are what make the line awkward, and a good choice of parameter removes them. Take $\lambda=\tau$. Then (3.3.12) gives $F=\sqrt{g_{\alpha\beta}\dot x^{\alpha}\dot x^{\beta}}=c$, a constant, which therefore passes straight through the derivative. Multiply (3.3.62) through by $c$:

Expand the total derivative. The factor $g_{\mu\beta}$ depends on $\tau$ only through the position, so the chain rule gives $\dd g_{\mu\beta}/\dd\tau=\partial_{\alpha}g_{\mu\beta}\dot x^{\alpha}$:

Symmetrise the middle term. This is the one move in the derivation that is easy to miss, so here it is in full. The middle term is contracted with $\dot x^{\alpha}\dot x^{\beta}$, which is symmetric under exchanging $\alpha$ and $\beta$. Therefore only the symmetric part of $\partial_{\alpha}g_{\mu\beta}$ contributes, and we may replace it by its own average with the $\alpha\leftrightarrow\beta$ relabelled copy:

Substituting (3.3.65) into (3.3.64) collects the three metric-derivative terms under a single factor of $\half$:

Raise the index. Multiply through by $g^{\nu\mu}$ and sum over $\mu$. On the first term $g^{\nu\mu}g_{\mu\beta}=\delta^{\nu}{}_{\beta}$, which renames $\beta$ into $\nu$ and collapses the sum:

Recognise the bracket. Compare the coefficient in (3.3.67) with the Christoffel formula (3.3.50). Take (3.3.50) and rename its indices $\lambda\to\nu$, $\mu\to\alpha$, $\nu\to\beta$ and $\sigma\to\mu$, purely to get the two expressions into the same letters. It becomes

which is character for character the coefficient in (3.3.67), using $g_{\beta\mu}=g_{\mu\beta}$ in the first term. Hence

which is (3.3.57) exactly.

8.3 · The Newtonian limit, and gravity as geometry

The claim of Part III is that a falling body is following a geodesic. That claim earns nothing until it reproduces Newton, so here is the check, with each approximation named as it is made.

Assumption 1 · slow motion. The body moves slowly compared with light, so $\abs{\dd x^{i}/\dd\tau}\ll c\,\dd t/\dd\tau$. Then in the sum $\Gamma^{\mu}{}_{\nu\rho}\dot x^{\nu}\dot x^{\rho}$ the dominant term is by far the one with $\nu=\rho=0$, and we keep only it.

Assumption 2 · a static field. All time derivatives of the metric vanish, $\partial_{0}g_{\mu\nu}=0$.

Assumption 3 · a weak field. $g_{00}=1+2\Phi/c^{2}$ with $\abs{\Phi}/c^{2}\ll1$, which is Chapter 3.1 §6.5's derived result, obtained there from the accelerating cabin and the Doppler effect with no general relativity used. The spatial part of the metric is $g_{ij}=-\delta_{ij}+O(\Phi/c^{2})$. We shall not need to know that correction, and it is worth watching why.

Compute $\Gamma^{i}{}_{00}$ from (3.3.50). Two of the three terms carry $\partial_{0}$ and die by Assumption 2, leaving

where the sum over $\sigma$ collapsed to spatial values because $\partial_{0}g_{00}=0$ removes the $\sigma=0$ term. Now insert $g^{ij}=-\delta^{ij}+O(\Phi/c^{2})$ and $\partial_{j}g_{00}=2\partial_{j}\Phi/c^{2}$. The correction to $g^{ij}$ multiplies a quantity that is already first order in $\Phi$, so it contributes at second order and may be dropped. That is why the spatial metric never had to be specified:

Put that into the geodesic equation (3.3.57) for a spatial component, keeping only the $00$ term by Assumption 1:

For slow motion $\dd t/\dd\tau=1+O(v^{2}/c^{2})$, so to leading order $\tau$ and $t$ may be interchanged and the two factors of $c$ cancel the $c^{2}$ in the denominator:

Newton's law of motion under gravity, recovered with no force anywhere in the derivation. The right-hand side of (3.3.73) came out of $\Gamma$, which came out of the metric, which is the geometry. Chapter 3.1 §2 argued that gravity was eligible to be geometry. Equation (3.3.73) is the first place where it actually is.

One thing is still missing, and it is worth naming so that no more is claimed than has been shown. Nothing here determines $\Phi$, or equivalently $g_{\mu\nu}$. We have shown how matter moves in a given geometry. What geometry a given lump of matter produces requires field equations, which require curvature, which is Chapter 3.4 and then Chapter 3.6.

9 · Worked examples

Two examples, each of which runs the whole chapter's machinery from one end to the other. The first takes a genuinely curved surface, the sphere, and goes from its metric to its straight lines. The second takes the flat plane in polar coordinates and settles the debt from Chapter 1.1.

10 · Your turn