Part E · curved space · spine

Metric, connection, geodesics — free fall is a straight line

A metric measures. A connection differentiates. Together they decide what straight means — and the straight lines turn out to be the worldlines of bodies nobody is pushing.

needs 31, 11~3 h slides + 6 h problemsproblems: ./mr problems 32

Unit 31 built a space with no coordinates in it, put tangent vectors on it, and found exactly one derivative available for free: the Lie derivative, which differentiates along a flow. That is not enough. To say that a vector field is constant, or that a curve is straight, you need to compare a vector at one point with a vector at another, and nothing built so far does that. This unit buys the missing structure, at the price of one new object — the connection — and then spends the rest of the course computing with the single formula that object reduces to.

What the physics is waiting for

Unit 30 proved two things and left them unreconciled. First: a freely falling frame kills the first derivatives of the metric at a point but cannot kill the second, so something survives every change of coordinates. Second: gravity acts identically on all bodies, so a trajectory depends on nothing but the starting event and the starting four-velocity. The second statement is the definition of a connection, read backwards; the first is the statement that the connection is not flat. This unit does the first half, and unit 33 the second.

Conventions, fixed here and never changed again

Signature $(-,+,+,+)$ throughout, as in units 26, 27 and 30.
Latin $i,j,k,l$ index a chart on an $n$-manifold; Greek $\mu,\nu,\rho,\sigma$ index a chart on a 4-dimensional spacetime. Repeated upper–lower indices are summed (unit 16).
$\nabla_iV^j = \partial_iV^j + \Gamma^j{}_{ik}V^k$; the lower pair of $\Gamma$ is the pair that is symmetric for the connection this course uses.
$A^a{}_i = \partial\tilde x^a/\partial x^i$ and $S^i{}_a = \partial x^i/\partial\tilde x^a$ for a change of chart, as in unit 31.

How to read this deck

One surface runs through all six sections: the round sphere of radius $R$ in the ordinary spherical chart. Its metric, its volume element, its two nonzero Christoffel symbols, one parallel transport around one circle of latitude, its geodesics derived twice, and normal coordinates at its pole — all of it is on the same surface, so there is always a number to compare against. The test latitude is $\theta_0 = \pi/3$, chosen because the transport answer comes out at exactly $180^\circ$.

Proofs are on the slides where they fit and in these notes under Full proof where they do not. A result imported from elsewhere is marked cited, with where it is proved; none of the unit's four checkpoint items rests on one.

32 · overview

One sphere, six sections, one formula

The aim, on the round sphere: carry a vector once round the latitude $\theta_0 = \pi/3$ without turning it, and it comes back reversed.

The latitude $\theta_0 = \pi/3$ and the cap above it. The vector that "does not turn" (section 3 makes it precise) returns reversed; the cap's area is $\pi R^2$ — the same $\pi$.

the route, in four steps

1. A metric, $g = R^2(\mathrm d\theta^2 + \sin^2\theta\,\mathrm d\varphi^2)$, measures lengths, angles, areas. Section 1.

2. A metric fixes exactly one way to differentiate a vector field, the Levi-Civita connection: one formula in the first derivatives of $g$, the formula of the title. Section 2.

3. On the sphere that formula leaves two nonzero coefficients, $-\sin\theta\cos\theta$ and $\cot\theta$, and "not turning" along a curve is a linear ODE. Section 3.

4. In the unit frame (south, east) the ODE is a rotation at rate $\cos\theta_0$ per radian of longitude. Once round: $2\pi\cos\theta_0 = \pi$.

The plan. 1 the metric · 2 the affine connection, and the one formula · 3 parallel transport · 4 geodesics, two ways · 5 normal coordinates · 6 free fall: the geodesic equation replaces $\vec F = m\vec a$, and Newton returns with a factor of two.

The route, slide by slide. Step 1 is Worked example 1 — the round sphere, measured, which also measures the latitude: length $2\pi R\sin\theta_0$. Step 2 is The fundamental theorem, in six lines (exactly one connection is compatible with the metric and torsion-free) and The one formula the rest of the course computes with. Step 3 is Worked example 2 for the two coefficients and Differentiating along a curve for the ODE. Step 4 is Worked example 3; the slide after it watches the vector turn station by station, and Why half a turn: unroll the tangent cone shows the same half-turn with no calculus at all. The equality with the cap's area, $2\pi R^2(1-\cos\theta_0) = \pi R^2$, is checked there and explained in unit 33.

The one formula of the title is $\Gamma^k{}_{ij} = \tfrac12 g^{kl}(\partial_ig_{jl} + \partial_jg_{il} - \partial_lg_{ij})$. Units 33 to 40 compute with it and with nothing else at this level; everything after it is built by differentiating it. It is worth noticing, when it arrives, that it is derived here rather than posited: the fundamental theorem of Riemannian geometry says that one connection and only one is both torsion-free and metric, and the formula is what that connection's components have to be.

The order of the sections is forced except for one choice. Metric before connection, connection before transport, transport before geodesics, geodesics before normal coordinates (the coordinates are built out of geodesics). The free choice was where to put the physics; it is all in section 6, at the end, rather than sprinkled through, because every physical statement here — free fall is geodesic, a gyroscope parallel-transports, $g_{00}$ carries the Newtonian potential — is a one-line reading of a piece of mathematics that has to exist first.

section 1

What a metric measures

Lengths, angles, volumes — and the isomorphism that makes an index's position a choice.

32 · metric

The definition, with the positivity dropped

definition

A pseudo-Riemannian metric on a smooth $n$-manifold $M$ is a $(0,2)$-tensor field $g$ that at every point is symmetric and nondegenerate, with constant signature. Riemannian means positive definite; Lorentzian means signature $(-,+,\dots,+)$.

Nondegenerate: $g_p(X,Y) = 0$ for all $Y$ forces $X = 0$. Equivalently the matrix $g_{ij} = g(\partial_i,\partial_j)$ is invertible in every chart — a chart-independent statement, since $\tilde g = S^{\mathsf T}gS$ with $S$ invertible.

Unit 31 gave the Riemannian and Lorentzian cases. Both are this; only the signature differs, and only the signature is physics.

what it is for

A metric is the one structure that turns a manifold into a geometry. Without it there is no length, no angle, no volume, no orthogonality, no "unit vector" — a smooth manifold is a shapeless thing on which only counting and smoothness make sense.

signature is not a convenience

In Lorentzian signature $g(X,X)$ can be negative, zero or positive, and the three cases are timelike, null and spacelike (unit 26). "Nondegenerate" is what survives of "positive definite", and it is all the algebra below needs.

Why constant signature. By Sylvester's law of inertia (unit 16, and unit 26 for the Lorentzian case) each $g_p$ has a well-defined signature. It is an integer-valued function of $p$, and on a connected manifold a continuous integer-valued function is constant — so on a connected $M$ constancy is automatic once $g$ is smooth and nondegenerate everywhere. The clause is in the definition only so the disconnected case is not a special case.

Existence. Every smooth manifold that is second countable and Hausdorff carries a Riemannian metric: take any atlas, put the Euclidean metric on each chart, and glue with a partition of unity — sums of positive definite forms are positive definite. Cited: Lee, Introduction to Smooth Manifolds (2nd ed.), Prop. 13.3. The second countability of unit 31's definition is exactly what pays for this. A Lorentzian metric is a different matter: sums of Lorentzian forms need not be Lorentzian, and a compact manifold admits one only if its Euler characteristic vanishes. Cited: O'Neill, Semi-Riemannian Geometry, ch. 5. The sphere $S^2$, Euler characteristic 2, admits no Lorentzian metric at all — which is one reason no one models a spacetime on it.

The physics angle. Unit 30 wrote down two metrics that were not constant — the Rindler metric on flat spacetime and the rotating-disc metric — and used them to compute clock rates without ever saying what a metric on a manifold is. This slide is the payment for that debt. Nothing computed in unit 30 changes; what changes is that "$\mathrm ds^2 = \ldots$" is now a tensor field rather than a formula somebody wrote down.

32 · metric

Raising and lowering is an isomorphism, not a notation

theorem

$g$ is nondegenerate at $p$ if and only if the musical isomorphism $$\flat: T_pM\to T_p^*M,\qquad X\mapsto X^\flat := g_p(X,\cdot)$$ is a linear isomorphism.

proof

$\flat$ is linear because $g$ is bilinear. Its kernel is $\{X : g(X,Y)=0\ \forall Y\}$, which is $\{0\}$ exactly when $g$ is nondegenerate. Since $\dim T_pM = \dim T_p^*M = n$, injective and isomorphism are the same thing. $\square$

Write $\sharp := \flat^{-1}$. In components $(X^\flat)_i = g_{ij}X^j$, and $(\omega^\sharp)^i = g^{ij}\omega_j$ where $g^{ij}$ is the inverse metric, the matrix inverse of $g_{ij}$.

two checks, one line each

$g^{ij}$ are the components of a $(2,0)$-tensor: $(\tilde g^{-1}) = (S^{\mathsf T}gS)^{-1} = A\,g^{-1}A^{\mathsf T}$, which in indices is $\tilde g^{ab} = A^a{}_iA^b{}_jg^{ij}$ — the $(2,0)$ law.

Raising then lowering is the identity, because $g^{ij}g_{jk} = \delta^i{}_k$ by definition of inverse.

the thing to internalise

$V^i$ and $V_i$ are two descriptions of one object only once a metric is chosen. Unit 16's inertia tensor and unit 04's gradient were the same point made without the word. Without $g$ — on a bare manifold — there is no way to turn a vector into a covector, and the Lie derivative of unit 31 knows this.

Full proof that $g^{ij}$ is a tensor, in index form. By unit 31, $\tilde g_{ab} = S^i{}_aS^j{}_bg_{ij}$. Define $h^{ab}$ by $h^{ab}\tilde g_{bc} = \delta^a{}_c$ and set $k^{ab} := A^a{}_iA^b{}_jg^{ij}$. Then $k^{ab}\tilde g_{bc} = A^a{}_iA^b{}_jg^{ij}S^k{}_bS^l{}_cg_{kl} = A^a{}_i\delta^k{}_jg^{ij}S^l{}_cg_{kl} = A^a{}_ig^{ij}g_{jl}S^l{}_c = A^a{}_i\delta^i{}_lS^l{}_c = A^a{}_iS^i{}_c = \delta^a{}_c$, using $A^b{}_jS^k{}_b = \delta^k{}_j$. Inverses are unique, so $k = h$: the matrix inverse of the components is the components of a $(2,0)$-tensor. This is the kind of statement that looks like bookkeeping and is not — it is why $g^{\mu\nu}\partial_\mu\phi\,\partial_\nu\phi$ is a scalar.

Why "musical". $\flat$ lowers a pitch and $\sharp$ raises it, and the index goes down and up respectively. The notation is worth keeping because it names the map rather than its effect on symbols: $(\mathrm df)^\sharp$, for instance, is the gradient of unit 04, and writing it that way says exactly which structure the gradient needed.

A warning about $g^{ij}$. It is the inverse of the matrix, not the components of $g$ with indices raised in some other sense — though, pleasingly, raising both indices of $g_{ij}$ with $g$ gives $g^{ik}g^{jl}g_{kl} = g^{ij}$, so the two readings agree. Raising one index of $g$ gives $g^{i}{}_{j} = \delta^i{}_j$, which is why the Kronecker delta is the one tensor whose components are the same in every chart on any manifold.

32 · metric

Length, proper time, and the volume element

definitions

The arc length of a curve $\gamma:[a,b]\to M$ with $g(\dot\gamma,\dot\gamma)\gt0$: $$L[\gamma] = \int_a^b\sqrt{g_{ij}\,\dot x^i\dot x^j}\ \mathrm d\lambda .$$ For a timelike curve in Lorentzian signature the same integral with $-g$ inside and a $1/c$ outside is the proper time of unit 27: $\tau = \tfrac1c\int\sqrt{-g_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm d\lambda$.

Both are reparametrisation invariant: substituting $\lambda = \lambda(\sigma)$ multiplies the integrand by $|\mathrm d\lambda/\mathrm d\sigma|^{-1}$ and the measure by $|\mathrm d\lambda/\mathrm d\sigma|$. That innocent fact causes the whole of the affine-parameter discussion later.

the volume element

$\sqrt{|\det g_{ij}|}\ \mathrm d^nx$ is the same on every chart overlap, so integration of scalars over $M$ is well defined.

proof

$\tilde g = S^{\mathsf T}gS$ gives $\det\tilde g = (\det S)^2\det g$, so $\sqrt{|\det\tilde g|} = |\det S|\sqrt{|\det g|}$. Unit 05's change of variables gives $\mathrm d^n\tilde x = |\det A|\,\mathrm d^nx = |\det S|^{-1}\mathrm d^nx$. Multiplying, the Jacobians cancel. $\square$

Check. Flat plane, polar chart: $g = \mathrm{diag}(1,r^2)$, $\sqrt{\det g} = r$, so the element is $r\,\mathrm dr\,\mathrm d\theta$ — unit 05's Jacobian, now derived from a tensor rather than from a picture of a small wedge.

The proof, written out. Let $\mathbf x$ and $\tilde{\mathbf x}$ be overlapping charts, $A^a{}_i = \partial\tilde x^a/\partial x^i$, $S = A^{-1}$. Unit 31's transformation law for a $(0,2)$-tensor is $\tilde g_{ab} = S^i{}_aS^j{}_bg_{ij}$, i.e. $\tilde g = S^{\mathsf T}gS$ as matrices, so $\det\tilde g = (\det S)^2\det g$ and $\sqrt{|\det\tilde g|} = |\det S|\,\sqrt{|\det g|}$. Unit 05's change of variables says that for a function $f$, $\int f\,\mathrm d^n\tilde x = \int (f\circ\tilde{\mathbf x}\circ\mathbf x^{-1})\,|\det A|\,\mathrm d^nx$. Since $|\det A| = |\det S|^{-1}$, the product $\sqrt{|\det \tilde g|}\,\mathrm d^n\tilde x$ carries $|\det S|\cdot|\det S|^{-1} = 1$ relative to $\sqrt{|\det g|}\,\mathrm d^nx$. Both charts give the same number. $\square$

Why the absolute value. In Lorentzian signature $\det g \lt 0$ — for $\eta = \mathrm{diag}(-1,1,1,1)$ it is $-1$ — so the bare determinant has no square root. The standard abbreviation is $g := \det g_{\mu\nu}$ and the element is written $\sqrt{-g}\,\mathrm d^4x$. This course writes $\sqrt{|g|}$ where the signature is not fixed and $\sqrt{-g}$ in spacetime.

What this is not, yet. A volume element is not a volume form: the factor $|\det S|$ has an absolute value, so the construction needs no orientation, whereas a form changes sign under an orientation-reversing chart change. Unit 34 builds the form $\sqrt{|g|}\,\mathrm dx^1\wedge\dots\wedge\mathrm dx^n$ on an oriented manifold and shows that the two agree there. Until then, integration of scalars is all this course needs — and it is enough for the Einstein–Hilbert action of unit 35.

History. Gauss's Disquisitiones generales circa superficies curvas (1827) computed exactly $\sqrt{EG-F^2}$ for a surface in $\mathbb R^3$ — the same determinant, in the notation $E,F,G$ for $g_{11},g_{12},g_{22}$, which survives in older differential geometry. Riemann's 1854 habilitation lecture is what generalised it to $n$ dimensions and to intrinsic data, and Riemann never wrote a Christoffel symbol; those are Christoffel's, 1869.

32 · metric · worked example

Worked example 1 — the round sphere, measured

the metric of $S^2_R$, and three quantities read off it

1 · the metric, from the embedding. $\mathbf X(\theta,\varphi) = R(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta)$. Then $$\mathbf X_\theta = R(\cos\theta\cos\varphi,\cos\theta\sin\varphi,-\sin\theta),$$ $$\mathbf X_\varphi = R(-\sin\theta\sin\varphi,\sin\theta\cos\varphi,0).$$ Euclidean dot products: $\mathbf X_\theta\cdot\mathbf X_\theta = R^2$, $\mathbf X_\theta\cdot\mathbf X_\varphi = 0$, $\mathbf X_\varphi\cdot\mathbf X_\varphi = R^2\sin^2\theta$. So $$g_{\theta\theta} = R^2,\quad g_{\theta\varphi}=0,\quad g_{\varphi\varphi} = R^2\sin^2\theta.$$

2 · a length. The circle $\theta = \theta_0$, $\varphi\in[0,2\pi]$ has $\dot\theta = 0$, $\dot\varphi = 1$, so $L = \int_0^{2\pi}\!\sqrt{R^2\sin^2\theta_0}\ \mathrm d\varphi = 2\pi R\sin\theta_0$. At $\theta_0 = \pi/3$: $L = \pi\sqrt3\,R = 5.4414\,R$.

The two coordinate vectors at a point of the latitude, drawn at half scale. They are orthogonal, and $\mathbf X_\varphi$ is shorter by $\sin\theta_0$ because the latitude is a circle of radius $R\sin\theta_0$ about the axis — hence its length.

Why the embedding is allowed here. Nothing in this unit depends on $S^2$ sitting in $\mathbb R^3$ — the metric, once written, is intrinsic data on an abstract 2-manifold, and every later computation uses only the array $g_{ij}$. The embedding is used once, on this slide, as a quick way to obtain the array, exactly as unit 31 obtained the round metric in stereographic coordinates. The pullback being used is unit 31's: for $F: M\to N$ and a metric $h$ on $N$, $(F^*h)(X,Y) = h(\mathrm dF X,\mathrm dF Y)$, and here $\mathrm dF\partial_\theta = \mathbf X_\theta$.

The chart's honest domain. $(\theta,\varphi)$ is a chart only on $0\lt\theta\lt\pi$, $0\lt\varphi\lt2\pi$ — it misses a closed half-meridian, and at $\theta = 0,\pi$ the metric degenerates ($g_{\varphi\varphi}\to0$), which is a failure of the chart, not of the sphere. Unit 31 proved no single chart covers $S^2$, and unit 05 flagged the same coordinate singularity in spherical polars. The computation of the area above integrates over $\theta\in[0,\pi]$ anyway; that is legitimate because the missing set has measure zero, and a careful version splits the sphere into two chart domains and adds.

Reading the figure. $\mathbf X_\theta$ points due south along the meridian and has length $R$ everywhere; $\mathbf X_\varphi$ points due east along the latitude and has length $R\sin\theta$, the distance from the axis (dashed). That is the whole content of $g = \mathrm{diag}(R^2, R^2\sin^2\theta)$: the diagonal entries are the squared lengths of the two coordinate vectors and the zero off-diagonal entry says they are perpendicular.

32 · metric · worked example

Worked example 1, continued — an angle and an area

worked example 1, continued

3 · an angle. At the point $\theta_0 = \pi/3$ take $X = \partial_\theta + \partial_\varphi$ and $Y = \partial_\theta$. Then $g(X,Y) = g_{\theta\theta} = R^2$, $\ \lVert Y\rVert^2 = R^2$, and $\lVert X\rVert^2 = R^2 + R^2\sin^2\tfrac\pi3 = \tfrac74R^2$. So $$\cos\alpha = \frac{R^2}{R\cdot R\sqrt{7}/2} = \frac{2}{\sqrt7} = 0.75593,$$ $\alpha = 40.893^\circ$. Note the answer has no $R$ in it: angles are scale-free, as they must be.

4 · the area. $\det g = R^4\sin^2\theta$, so $\sqrt{\det g} = R^2\sin\theta$ (positive for $0\lt\theta\lt\pi$) and $$\mathrm{Area} = \int_0^{2\pi}\!\!\int_0^{\pi}\!R^2\sin\theta\,\mathrm d\theta\,\mathrm d\varphi = 4\pi R^2 .$$

Drawn in the chart, $X = \partial_\theta + \partial_\varphi$ sits at $45^\circ$ to $Y = \partial_\theta$. The metric says $\partial_\varphi$ is only $0.866$ times as long as $\partial_\theta$ at this latitude, so the true angle is $\arctan(\sin\theta_0) = 40.9^\circ$.

Checks. Dimensions: $g_{ij}$ carries $[\text{length}]^2$ because $\theta,\varphi$ are dimensionless, so $L$ is a length and the area element is an area ✓. Limit $\theta_0\to0$: the circle shrinks to a point and $L\to0$ ✓. Special case $\theta_0 = \pi/2$: $L = 2\pi R$, the equator ✓. The area agrees with Archimedes ✓.

The angle, from the figure. In the orthonormal directions (south, east) the vector $X$ has components $R$ and $R\sin\theta_0$, so $\tan\alpha = \sin\theta_0 = \sqrt3/2$ and $\alpha = 40.893^\circ$ — the same as $\arccos(2/\sqrt7)$ on the slide, since $\cos(\arctan t) = 1/\sqrt{1+t^2} = 1/\sqrt{7/4} = 2/\sqrt7$. The chart's picture is not wrong; it is a picture of the chart, which knows nothing about lengths.

Comparing with unit 31. The same sphere in the stereographic chart had $g = 4(\mathrm du^2+\mathrm dv^2)/(1+u^2+v^2)^2$ with $R=1$, in which $g_{11}=g_{22}$ and $g_{12}=0$: the chart is conformal, angles are the Euclidean ones, and that is what stereographic projection is famous for. The spherical chart is not conformal ($g_{\theta\theta}\ne g_{\varphi\varphi}$), which is why the $40.893^\circ$ above is not the naive $45^\circ$. Both charts describe one metric.

The number to keep. $\theta_0 = \pi/3$ is the test latitude for the whole unit. $\sin\theta_0 = \sqrt3/2$, $\cos\theta_0 = 1/2$, $\cot\theta_0 = 1/\sqrt3$. The $\cos\theta_0 = 1/2$ is what makes the parallel-transport answer come out at exactly half a turn.

section 2

A derivative that is a tensor

The partial derivative of a vector field is not one. Repairing that costs exactly one new object — and then a theorem says which one to use.

32 · connection

The problem, in one computation

$\partial_iV^j$ is not a tensor

With $\tilde V^a = A^a{}_jV^j$ and $\tilde\partial_b = S^i{}_b\partial_i$, $$\tilde\partial_b\tilde V^a \;=\; \underbrace{S^i{}_bA^a{}_j\,\partial_iV^j}_{\text{the }(1,1)\text{ law}} \;+\; \underbrace{S^i{}_b\big(\partial_iA^a{}_j\big)V^j}_{\text{the offender}} .$$

proof

Product rule on $A^a{}_jV^j$, then contract with $S^i{}_b$. That is the whole computation. The offending term contains $\partial_iA^a{}_j = \partial^2\tilde x^a/\partial x^i\partial x^j$, a second derivative of the chart change. $\square$

It vanishes for every $V$ exactly when the transition map is affine. Unit 31's point was that a manifold has no preferred charts at all, so on a manifold this term never goes away.

what fails, precisely

Not the differentiation — the comparison. $V_{p+\epsilon}$ and $V_p$ live in different vector spaces, $T_{p+\epsilon}M$ and $T_pM$, and unit 31 gave no map between them. Subtracting them is meaningless, and a derivative is a limit of a difference.

The next slide shows the offending term at work on the flattest example there is.

Why the Lie derivative does not solve it. Unit 31 did build a tensorial derivative, $\mathcal L_XT$. But $\mathcal L_XY$ at $p$ depends on $X$ in a whole neighbourhood, not just on $X_p$ — indeed $\mathcal L_XY = [X,Y]$, which is antisymmetric in the two fields. There is no such thing as "the Lie derivative in the direction $X_p$", so it cannot define a derivative along a curve, and cannot say what a straight line is. Everything in this unit is about getting a derivative that is tensorial in the direction slot.

Historical note. The offending second-derivative term is the reason Ricci and Levi-Civita called their subject the "absolute differential calculus" (1900): the point was to build derivatives whose results were absolute, that is, independent of the coordinate system. Einstein spent 1912 to 1915 learning it with Marcel Grossmann's help, and wrote to Sommerfeld in October 1912 that he had acquired a great respect for mathematics, whose subtler parts he had until then regarded as pure luxury.

32 · connection

A constant field with moving components

One arrow, $V = \partial_x$, the same everywhere. At $p$ and at $q$ the polar basis $(\partial_r, \partial_\theta)$ points in different directions and $\partial_\theta$ has length $r$, so the same arrow has different components: they vary although $V$ does not.

seen on the plane

$V = \partial_x$, the most constant field there is. In polar coordinates $$V = \cos\theta\ \partial_r - \frac{\sin\theta}{r}\ \partial_\theta,$$ whose components depend on position. $\partial_r V^\theta \ne 0$ although $V$ is constant.

So $\partial_r V^\theta$ and $\partial_\theta V^r$ measure the turning of the basis, not any change in $V$. The second term of the previous slide is exactly that turning: here the chart change from $(x,y)$ to $(r,\theta)$ is not affine.

Full derivation of the polar example. $r = \sqrt{x^2+y^2}$, $\theta = \operatorname{atan2}(y,x)$. A tangent vector's components transform as $V^a = (\partial\tilde x^a/\partial x^j)V^j$ (unit 31, definition C), and $V = \partial_x$ has $V^x = 1$, $V^y = 0$. So $V^r = \partial r/\partial x = x/r = \cos\theta$ and $V^\theta = \partial\theta/\partial x = -y/r^2 = -\sin\theta/r$. Both depend on $r$ and $\theta$. This is the entire difficulty of the unit in one line: a field that is constant by any reasonable standard has non-constant components, so "constant components" cannot be the definition of constant, and "the derivative of the components" cannot be the definition of the derivative.

What to look for in the figure. At $p$, $\theta = 30^\circ$ and $r = 2$: $V^r = \cos30^\circ = 0.87$, $V^\theta = -\sin30^\circ/2 = -0.25$. At $q$, $\theta = 122^\circ$ and $r = 1.55$: $V^r = -0.53$, $V^\theta = -0.55$. The arrow $V$ (heavy) is the same at both points; what differs is the basis, and $\partial_\theta$ in particular grows with $r$. A derivative that is to see only the change in $V$ must subtract the change in the basis — which is what the connection coefficients will do.

32 · connection

What a connection is

definition

An affine connection on $M$ is a map $$\nabla: \mathfrak X(M)\times\mathfrak X(M)\to\mathfrak X(M),\quad (X,Y)\mapsto\nabla_XY ,$$ where $\mathfrak X(M)$ is the set of smooth vector fields on $M$, that is (i) $C^\infty(M)$-linear in $X$, (ii) $\mathbb R$-linear in $Y$, and (iii) Leibniz: $$\nabla_X(fY) = (Xf)\,Y + f\,\nabla_XY .$$

(i) is the crucial asymmetry: $\nabla_{fX}Y = f\nabla_XY$, so $(\nabla_XY)_p$ depends on $X$ only through $X_p$. The direction slot is tensorial; the differentiated slot is not, and cannot be.

$V_p\in T_pM$ and $V_q\in T_qM$ live in different planes, and nothing so far relates them. A connection is the extra structure that will, given a curve $\gamma$ from $p$ to $q$, carry one into the other (section 3).

Full proof that (i) makes $\nabla_XY$ pointwise in $X$. Suppose $X_p = X'_p$. Write $Z = X - X'$, so $Z_p = 0$. In a chart, $Z = Z^i\partial_i$ with $Z^i(p) = 0$, hence $\nabla_ZY = Z^i\nabla_{\partial_i}Y$ by $C^\infty$-linearity, and evaluating at $p$ gives $Z^i(p)(\nabla_{\partial_i}Y)_p = 0$. So $(\nabla_XY)_p = (\nabla_{X'}Y)_p$. The same argument run on the second slot fails precisely because of the $(Xf)Y$ term in (iii) — and that failure is the derivative doing its job. The technical name for an operator that is $C^\infty$-linear in a slot is tensorial in that slot; the general statement, that a $C^\infty$-multilinear map of fields is induced by a tensor field, is the tensor characterisation lemma. Cited: Lee, Introduction to Smooth Manifolds (2nd ed.), Lemma 12.24 — the proof is a partition-of-unity bump argument of exactly the kind unit 11 used for the fundamental lemma of the calculus of variations.

Why "connection". The structure connects the tangent spaces at different points: given a curve from $p$ to $q$ it will produce, in the next section, a definite linear map $T_pM\to T_qM$. That is the missing comparison. The word predates the modern definition and comes from Weyl (1918), Raum, Zeit, Materie, who called it a "Zusammenhang" — literally a hanging-together.

32 · connection

Christoffel symbols: the connection in a chart

its components

In a chart, $\nabla_{\partial_i}\partial_j$ is a vector field, so it has components: $$\nabla_{\partial_i}\partial_j =: \Gamma^k{}_{ij}\,\partial_k .$$ The $n^3$ functions $\Gamma^k{}_{ij}$ are the Christoffel symbols (or connection coefficients) of that chart.

the covariant derivative in components

$$\nabla_iV^j := (\nabla_{\partial_i}V)^j = \partial_iV^j + \Gamma^j{}_{ik}V^k .$$ Proof: $\nabla_{\partial_i}(V^k\partial_k) = (\partial_iV^k)\partial_k + V^k\Gamma^j{}_{ik}\partial_j$ by (iii); relabel. $\square$

the convention, fixed once

The first lower index of $\Gamma^k{}_{ij}$ is the direction of differentiation. For the connection this course uses the two lower indices are symmetric, so nothing depends on this — but not every book's connection is symmetric, and not every book puts the direction first.

Why components are enough. By (i) and (iii), $\nabla_XY$ for any two fields is fixed once the $n^2$ vector fields $\nabla_{\partial_i}\partial_j$ are known: expand $X = X^i\partial_i$ and $Y = Y^j\partial_j$, pull $X^i$ out of the direction slot by (i), and differentiate $Y^j\partial_j$ by (iii). That is the proof on the slide, and it is why a connection in a chart is its $n^3$ functions $\Gamma^k{}_{ij}$.

How much freedom is there? $n^3$ functions per chart, $= 64$ in four dimensions, with no constraint at all beyond smoothness. A connection is genuinely extra structure: a manifold admits many, and they are not related to each other by anything. That is what makes the fundamental theorem of this section important — the metric picks one of the $64$-function families out uniquely.

The physics angle, in advance. Section 6 reads $\Gamma^\mu{}_{\nu\rho}$ as the gravitational field. A field with $n^3$ components, no tensorial character, and a value that can be set to zero at any chosen point by a choice of chart is exactly what the equivalence principle demands of gravity, and exactly what no unit before 30 could have accommodated.

32 · connection

$\Gamma$ is not a tensor, and that is the point

transformation law

$$\begin{aligned}\tilde\Gamma^c{}_{ab} = {}&A^c{}_k\,S^i{}_a S^j{}_b\,\Gamma^k{}_{ij}\\ &+ A^c{}_k\,\frac{\partial^2x^k}{\partial\tilde x^a\partial\tilde x^b}.\end{aligned}$$ The first term is the $(1,2)$-tensor law; the second is inhomogeneous, so $\Gamma$ is not a tensor.

proof

$\tilde\Gamma^c{}_{ab}\tilde\partial_c = \nabla_{\tilde\partial_a}\tilde\partial_b = \nabla_{S^i{}_a\partial_i}(S^j{}_b\partial_j)$. By (i), pull out $S^i{}_a$; by (iii), $= S^i{}_a\big[(\partial_iS^j{}_b)\partial_j + S^j{}_b\Gamma^k{}_{ij}\partial_k\big]$. Now $S^i{}_a\partial_i = \partial/\partial\tilde x^a$, so $S^i{}_a\partial_iS^j{}_b = \partial^2x^j/\partial\tilde x^a\partial\tilde x^b$. Convert $\partial_k$ to $\tilde\partial_c$ with $A^c{}_k$. $\square$

two corollaries, free

Differences are tensors. Two connections have the same inhomogeneous term, so $\Gamma^k{}_{ij} - \Gamma'^k{}_{ij}$ transforms as a $(1,2)$-tensor. The set of connections is an affine space over the $(1,2)$-tensor fields.

$\nabla$ works. The offending term of The problem, in one computation was $S^i{}_b(\partial_iA^a{}_j)V^j$. Differentiating $A^a{}_kS^k{}_j = \delta^a{}_j$ shows it is minus the inhomogeneous term above, so it cancels in $\partial_iV^j+\Gamma^j{}_{ik}V^k$.

A connection is precisely an object whose non-tensorial transformation law is designed to absorb the non-tensorial part of $\partial$.

Full proof of the cancellation, which the corollary states without doing. Write $\tilde\nabla_b\tilde V^a = \tilde\partial_b\tilde V^a + \tilde\Gamma^a{}_{bc}\tilde V^c$ and insert the expansion from The problem, in one computation of $\tilde\partial_b\tilde V^a$ and this slide's law for $\tilde\Gamma$, with $\tilde V^c = A^c{}_lV^l$: $$\tilde\nabla_b\tilde V^a = S^i{}_bA^a{}_j\partial_iV^j + S^i{}_b(\partial_iA^a{}_j)V^j + A^a{}_kS^i{}_bS^j{}_c\Gamma^k{}_{ij}A^c{}_lV^l + A^a{}_k\frac{\partial^2x^k}{\partial\tilde x^b\partial\tilde x^c}A^c{}_lV^l .$$ The third term is $A^a{}_kS^i{}_b\Gamma^k{}_{il}V^l$, since $S^j{}_cA^c{}_l = \delta^j{}_l$. For the fourth, differentiate the identity $S^k{}_cA^c{}_l = \delta^k{}_l$ along $\tilde x^b$, writing $\tilde\partial_b = S^i{}_b\partial_i$ and noting $\tilde\partial_bS^k{}_c = \partial^2x^k/\partial\tilde x^b\partial\tilde x^c$: $$\frac{\partial^2x^k}{\partial\tilde x^b\partial\tilde x^c}\,A^c{}_l = -S^k{}_c\,S^i{}_b\,\partial_iA^c{}_l .$$ Contracting with $A^a{}_k$ turns $S^k{}_c$ into $\delta^a{}_c$, so the fourth term is $-S^i{}_b(\partial_iA^a{}_l)V^l$ — exactly minus the second. What remains is $S^i{}_bA^a{}_j\big(\partial_iV^j + \Gamma^j{}_{il}V^l\big)$, the $(1,1)$ law. $\square$

A trap worth naming. "$\Gamma$ is not a tensor" is sometimes said as "$\Gamma$ is not a physical object". That does not follow and is misleading: $\Gamma$ is exactly as physical as an electromagnetic potential, which is also not gauge-invariant. What is true is that the value of $\Gamma$ at a point carries no chart-independent information, because the inhomogeneous term can be given any symmetric value at any one point by choosing the second derivatives of the chart change. Section 5 turns that observation into the equivalence principle.

Counting the freedom in the inhomogeneous term. At one point, $\partial^2x^k/\partial\tilde x^a\partial\tilde x^b$ is free: $n$ values of $k$ times $n(n+1)/2$ symmetric index pairs, so $n^2(n+1)/2$ numbers — $40$ in four dimensions. And a symmetric connection has exactly $n^2(n+1)/2 = 40$ independent components at a point. Equal counts: that is the arithmetic behind "$\Gamma$ can always be set to zero at a point", and it is done properly in section 5.

32 · connection

The covariant derivative of anything

Two demands fix $\nabla$ on all tensors once it is known on vectors: it agrees with $\partial$ on scalars, and it obeys Leibniz for tensor products and commutes with contraction (unit 16).

the covector rule, derived

$\omega_jV^j$ is a scalar, so $$\partial_i(\omega_jV^j) = (\nabla_i\omega_j)V^j + \omega_j\nabla_iV^j .$$ Expand both sides: $(\partial_i\omega_j)V^j + \omega_j\partial_iV^j = (\nabla_i\omega_j)V^j + \omega_j\partial_iV^j + \omega_j\Gamma^j{}_{ik}V^k$. Cancel, relabel $j\leftrightarrow k$, and use that $V$ is arbitrary: $$\nabla_i\omega_j = \partial_i\omega_j - \Gamma^k{}_{ij}\omega_k .\ \square$$

the general rule

$$\begin{aligned}\nabla_kT^{i_1\dots}{}_{j_1\dots} = {}&\partial_kT^{i_1\dots}{}_{j_1\dots} + \textstyle\sum_{r}\Gamma^{i_r}{}_{kl}\,T^{\dots l\dots}{}_{j_1\dots}\\ &- \textstyle\sum_{s}\Gamma^{l}{}_{kj_s}\,T^{i_1\dots}{}_{\dots l\dots}\end{aligned}$$ One $+\Gamma$ per upper index, one $-\Gamma$ per lower index, the direction index always first on $\Gamma$.

the case used most

$$\nabla_kT_{ij} = \partial_kT_{ij} - \Gamma^l{}_{ki}T_{lj} - \Gamma^l{}_{kj}T_{il}.$$ Mnemonic that is also a proof: each index is corrected as if it were alone, and the corrections add because the rule is Leibniz on a tensor product.

Check. $\nabla_i\delta^j{}_k = \partial_i\delta^j{}_k + \Gamma^j{}_{il}\delta^l{}_k - \Gamma^l{}_{ik}\delta^j{}_l = 0 + \Gamma^j{}_{ik} - \Gamma^j{}_{ik} = 0$ ✓ — as it must be, since $\delta$ is the identity map of $T_pM$ and contraction commutes with $\nabla$.

Full proof of the general rule. Every $(p,q)$-tensor is, pointwise, a sum of tensor products of vectors and covectors (unit 16: the products $e_{i_1}\otimes\dots\otimes e^{j_q}$ are a basis). Leibniz for $\otimes$ says $\nabla_k(S\otimes T) = (\nabla_kS)\otimes T + S\otimes\nabla_kT$, so it is enough to know the rule on one factor of each type, which the two boxes give. Writing it out on $T = V\otimes\omega$ with components $T^i{}_j = V^i\omega_j$: $\nabla_kT^i{}_j = (\partial_kV^i + \Gamma^i{}_{kl}V^l)\omega_j + V^i(\partial_k\omega_j - \Gamma^l{}_{kj}\omega_l) = \partial_kT^i{}_j + \Gamma^i{}_{kl}T^l{}_j - \Gamma^l{}_{kj}T^i{}_l$. Linearity extends it from products to sums, and smoothness of the coefficients extends it from a point to a field. $\square$

The alternative derivation, for the sceptical. One can instead define $(\nabla_XT)(\omega^1,\dots,Y_1,\dots) := X\big(T(\omega^1,\dots,Y_1,\dots)\big) - \sum T(\dots,\nabla_X\omega^r,\dots) - \sum T(\dots,\nabla_XY_s,\dots)$ and check that the result is $C^\infty$-multilinear, hence a tensor. That is the invariant statement of the same rule, and it is worth doing once by hand on a $(0,2)$-tensor, because it is the form in which metric compatibility is stated on the next slide.

Notation warning. Many texts write $T^{i}{}_{j;k}$ for $\nabla_kT^i{}_j$ and $T^i{}_{j,k}$ for $\partial_kT^i{}_j$ — comma for partial, semicolon for covariant. That is where the name "comma goes to semicolon" in section 6 comes from. This course writes $\nabla$, because the semicolon notation puts the direction index last while $\Gamma$ carries it first, and mixing the two has cost many people an afternoon.

32 · connection

Torsion, and metric compatibility

definition

The torsion of $\nabla$ is $$T(X,Y) := \nabla_XY - \nabla_YX - [X,Y].$$ $\nabla$ is torsion-free (or symmetric) if $T\equiv0$.

$T$ is a $(1,2)$-tensor

Proof. Antisymmetry is clear. $C^\infty$-linearity in $X$: replace $X$ by $fX$. The three terms give $f\nabla_XY$, $-(Yf)X - f\nabla_YX$ and $-[fX,Y] = -f[X,Y] + (Yf)X$ (unit 31's bracket identity). The $(Yf)X$ terms cancel. $\square$

Components: $T^k{}_{ij} = \Gamma^k{}_{ij} - \Gamma^k{}_{ji}$, since $[\partial_i,\partial_j]=0$. So torsion-free $\iff\Gamma^k{}_{ij}=\Gamma^k{}_{ji}$.

definition

$\nabla$ is metric compatible (or metric) if $\nabla g = 0$, that is $$\partial_kg_{ij} = \Gamma^l{}_{ki}g_{lj} + \Gamma^l{}_{kj}g_{il}$$ in every chart. Equivalently, for all fields, $X\,g(Y,Z) = g(\nabla_XY,Z) + g(Y,\nabla_XZ)$.

what each demand buys

Metric: transport preserves lengths and angles, and $\nabla$ commutes with lowering: $\nabla_kV_i = g_{ij}\nabla_kV^j$.

Torsion-free: second covariant derivatives of a scalar commute, $\nabla_i\nabla_jf = \nabla_j\nabla_if$ — unit 03's Clairaut survives.

Full proof that $\nabla g = 0$ lets $\nabla$ pass through the metric. $\nabla_kV_i = \nabla_k(g_{ij}V^j) = (\nabla_kg_{ij})V^j + g_{ij}\nabla_kV^j = g_{ij}\nabla_kV^j$. So "lower the index then differentiate" and "differentiate then lower" agree, and the notation $\nabla_kV_i$ is unambiguous. Without metric compatibility it is not, and every index-gymnastics manipulation in units 33 to 40 would need a rule about which order was meant.

Full proof that $\nabla_i\nabla_jf = \nabla_j\nabla_if$ for scalars. $\nabla_jf = \partial_jf$, a covector; so $\nabla_i\nabla_jf = \partial_i\partial_jf - \Gamma^k{}_{ij}\partial_kf$. The first term is symmetric in $i,j$ by Clairaut (unit 03), so the commutator is $-(\Gamma^k{}_{ij}-\Gamma^k{}_{ji})\partial_kf = -T^k{}_{ij}\partial_kf$. Torsion is exactly the obstruction. For a vector field the commutator does not vanish even with zero torsion, and what survives is the Riemann tensor — unit 33's definition.

Does anything in physics have torsion? Einstein–Cartan theory lets torsion be sourced by intrinsic spin; the torsion it predicts is non-propagating — it vanishes outside matter — and inside ordinary matter its effects are far below anything measured. General relativity takes torsion to be zero, and this course does too. The honest statement is that the choice is an assumption, not a theorem — what is a theorem is the next slide: once torsion is set to zero, metric compatibility leaves no freedom at all.

The name. "Torsion" is Cartan's (1922). For a connection on a surface it measures how a frame twists as it is transported, and the visual image is that the two ways of building a small parallelogram — go along $X$ then $Y$, or $Y$ then $X$ — fail to close by $T(X,Y)$ even after the bracket, unit 31's non-closure, is accounted for.

32 · connection

The fundamental theorem, in six lines

fundamental theorem of (pseudo-)Riemannian geometry

On a pseudo-Riemannian manifold $(M,g)$ there is exactly one torsion-free metric-compatible connection, the Levi-Civita connection. It satisfies the Koszul formula $$2g(\nabla_XY,Z) = Xg(Y,Z) + Yg(Z,X) - Zg(X,Y) + g([X,Y],Z) - g([X,Z],Y) - g([Y,Z],X).$$

uniqueness — and the formula falls out of it

Write metric compatibility three times, on the three cyclic orderings, and form (1)+(2)−(3):

$\ \ (1)\ \ Xg(Y,Z) = g(\nabla_XY,Z) + g(Y,\nabla_XZ)$
$\ \ (2)\ \ Yg(Z,X) = g(\nabla_YZ,X) + g(Z,\nabla_YX)$
$\ \ (3)\ \ Zg(X,Y) = g(\nabla_ZX,Y) + g(X,\nabla_ZY)$

Torsion-freeness three times — $\nabla_XZ-\nabla_ZX=[X,Z]$, $\nabla_YZ-\nabla_ZY=[Y,Z]$, $\nabla_YX = \nabla_XY-[X,Y]$ — pairs off the six terms: two pairs give brackets, the third a second $g(\nabla_XY,Z)$. The result is Koszul.

The right-hand side involves only $g$ and brackets. So $g(\nabla_XY,Z)$ is determined for every $Z$; nondegeneracy then determines $\nabla_XY$. Uniqueness. $\square$

Existence: define $\nabla_XY$ by the formula and check the five properties — in the notes.

Full proof of the pairing step. Expanding (1)+(2)−(3) gives $$g(\nabla_XY,Z) + g(Y,\nabla_XZ) + g(\nabla_YZ,X) + g(Z,\nabla_YX) - g(\nabla_ZX,Y) - g(X,\nabla_ZY).$$ Pair terms 2 and 5: $g(Y,\nabla_XZ) - g(\nabla_ZX,Y) = g([X,Z],Y)$. Pair terms 3 and 6: $g(\nabla_YZ,X) - g(X,\nabla_ZY) = g([Y,Z],X)$. Term 4: $g(Z,\nabla_YX) = g(Z,\nabla_XY) - g(Z,[X,Y])$. So the sum is $2g(\nabla_XY,Z) + g([X,Z],Y) + g([Y,Z],X) - g([X,Y],Z)$, and moving the last three across gives the Koszul formula exactly as boxed. $\square$

Full proof of existence. Let $K(X,Y,Z)$ denote the right-hand side of Koszul. Five checks, each by replacing one argument by $f$ times it and collecting the terms in which $f$ is differentiated; unit 31's $[fX,Y] = f[X,Y] - (Yf)X$ and $[X,fY] = f[X,Y] + (Xf)Y$ are all that is used.

$C^\infty$-linear in $Z$. The derivative terms from $Z\to fZ$ are $(Xf)g(Y,Z)$ from the first term, $(Yf)g(Z,X)$ from the second, $-(Xf)g(Z,Y)$ from $-g([X,fZ],Y)$ and $-(Yf)g(Z,X)$ from $-g([Y,fZ],X)$. They cancel in pairs. So for fixed $X,Y$, $Z\mapsto\tfrac12K(X,Y,Z)$ is a covector field, and by nondegeneracy there is a unique vector field — call it $\nabla_XY$ — with $g(\nabla_XY,Z) = \tfrac12K(X,Y,Z)$ for all $Z$.
$C^\infty$-linear in $X$. From $X\to fX$: $(Yf)g(Z,X)$, $-(Zf)g(X,Y)$, $-(Yf)g(X,Z)$ from $g([fX,Y],Z)$, and $+(Zf)g(X,Y)$ from $-g([fX,Z],Y)$. They cancel in pairs.
Leibniz in $Y$. From $Y\to fY$: $(Xf)g(Y,Z)$, $-(Zf)g(X,Y)$, $+(Xf)g(Y,Z)$ from $g([X,fY],Z)$, and $+(Zf)g(Y,X)$ from $-g([fY,Z],X)$. The $Z f$ terms cancel and $K(X,fY,Z) = fK(X,Y,Z) + 2(Xf)g(Y,Z)$, which is $\nabla_X(fY) = f\nabla_XY + (Xf)Y$. $\mathbb R$-linearity in $Y$ is plain.
Torsion-free. In $K(X,Y,Z) - K(Y,X,Z)$ the three derivative terms cancel by symmetry of $g$, the two brackets containing $Z$ cancel, and $g([X,Y],Z) - g([Y,X],Z) = 2g([X,Y],Z)$ survives. So $g(\nabla_XY - \nabla_YX,Z) = g([X,Y],Z)$ for all $Z$.
Metric. In $K(X,Y,Z) + K(X,Z,Y)$ the terms $\pm Yg(Z,X)$, $\pm Zg(X,Y)$, $\pm g([X,Y],Z)$, $\pm g([X,Z],Y)$ cancel, the two terms $-g([Y,Z],X)$ and $-g([Z,Y],X)$ cancel by antisymmetry of the bracket, and $2Xg(Y,Z)$ survives. Halving: $g(\nabla_XY,Z) + g(Y,\nabla_XZ) = Xg(Y,Z)$. $\square$

Problem P5 asks for all five without looking.

Who. Levi-Civita's 1917 paper introduced parallel transport on a surface and showed that the Christoffel symbols, which Christoffel had written in 1869 as bookkeeping for the transformation of quadratic forms, were the components of a geometrically meaningful object. Hermann Weyl gave the axiomatic definition of a connection a year later. Koszul's formula is from the 1950s and is later than the theorem, but it is the cleanest proof, which is why it is the one used here: it produces the answer while proving there is only one.

32 · connection

The one formula the rest of the course computes with

Christoffel symbols of the Levi-Civita connection

$$\boxed{\ \Gamma^k{}_{ij} = \tfrac12\,g^{kl}\big(\partial_ig_{jl} + \partial_jg_{il} - \partial_lg_{ij}\big)\ }$$

from Koszul, in one step

Put $X = \partial_i$, $Y=\partial_j$, $Z=\partial_l$. All three brackets vanish, so $$2g(\nabla_{\partial_i}\partial_j,\partial_l) = \partial_ig_{jl}+\partial_jg_{il}-\partial_lg_{ij}.$$ The left side is $2\Gamma^k{}_{ij}g_{kl}$. Contract with $g^{lm}$ and relabel. $\square$

three checks

Symmetry. $i\leftrightarrow j$ swaps the first two terms and fixes the third ✓ — torsion-free, as promised.

Flat. $g_{ij}$ constant $\Rightarrow\Gamma=0$, so $\nabla=\partial$ in Cartesian coordinates on $\mathbb R^n$, and in an inertial chart on Minkowski space ✓.

Dimensions. $\Gamma^k{}_{ij}$ has the dimension of $1/x$ when all coordinates share a dimension: $[g]/([x][g]) = 1/[x]$ ✓.

Count: $n^2(n+1)/2$ independent symbols — $6$ on a surface, $\mathbf{40}$ in four dimensions.

Read the formula properly. Three derivatives of the metric, in a fixed pattern: two with the "outer" indices differentiated and the summation index down, minus one with the summation index differentiating. A useful way to remember it is that the antisymmetric part in $(i,j)$ must vanish, which the first two terms guarantee by their symmetry and the third by being symmetric in $i,j$ already.

What it costs to use. Naively: $n^2(n+1)/2$ symbols, each an $n$-term sum, each term needing up to three partial derivatives. In four dimensions that is $40\times4\times3 = 480$ derivative evaluations, most of them zero. Nobody does it this way twice — the geodesic Lagrangian trick in section 4 replaces it with $n$ Euler–Lagrange equations, and for Schwarzschild that is four.

A lemma worth having now, proved in section 6. Contracting the formula on $k$ and $i$ gives $\Gamma^i{}_{ij} = \tfrac12g^{il}\partial_jg_{il} = \partial_j\ln\sqrt{|g|}$, using Jacobi's formula for the derivative of a determinant. That single identity turns the covariant divergence into $\nabla_iV^i = |g|^{-1/2}\partial_i(|g|^{1/2}V^i)$, which is how divergences are actually computed for the rest of the course — and it is where unit 05's $\nabla\cdot\vec F$ in spherical coordinates, with its $r^{-2}\partial_r(r^2F_r)$, comes from.

Why "boxed formula" is not a lapse into recipe. Everything about it has been derived: that a connection exists, that exactly one is torsion-free and metric, and that its components must be these. The formula is the end of an argument, not the start of one. That is the difference between this unit and a book that opens by defining $\Gamma$ by this expression and then verifies it transforms correctly — the verification is possible but it answers no question.

32 · connection · worked example

Worked example 2 — every Christoffel symbol of the sphere

from the boxed formula, all six, on $S^2_R$

Setup. $g_{\theta\theta}=R^2$, $g_{\varphi\varphi}=R^2\sin^2\theta$, $g_{\theta\varphi}=0$; hence $g^{\theta\theta}=R^{-2}$, $g^{\varphi\varphi}=R^{-2}\sin^{-2}\theta$. The only nonzero derivative of the metric is $$\partial_\theta g_{\varphi\varphi} = 2R^2\sin\theta\cos\theta .$$ Because $g$ is diagonal, $g^{kl}$ is diagonal and only $l=k$ contributes.

$\Gamma^\theta{}_{\theta\theta}$ $=\tfrac12g^{\theta\theta}(\partial_\theta g_{\theta\theta}+\partial_\theta g_{\theta\theta}-\partial_\theta g_{\theta\theta}) = \tfrac12R^{-2}\partial_\theta(R^2) = 0$.

$\Gamma^\theta{}_{\theta\varphi}$ $=\tfrac12g^{\theta\theta}(\partial_\theta g_{\varphi\theta}+\partial_\varphi g_{\theta\theta}-\partial_\theta g_{\theta\varphi}) = 0$: all three vanish.

$\Gamma^\theta{}_{\varphi\varphi}$ $=\tfrac12g^{\theta\theta}(2\partial_\varphi g_{\varphi\theta}-\partial_\theta g_{\varphi\varphi}) = -\tfrac12R^{-2}\cdot2R^2\sin\theta\cos\theta$ $= -\sin\theta\cos\theta$.

$\Gamma^\varphi{}_{\theta\theta}$ $=\tfrac12g^{\varphi\varphi}(2\partial_\theta g_{\theta\varphi}-\partial_\varphi g_{\theta\theta}) = 0$.

$\Gamma^\varphi{}_{\theta\varphi}$ $=\tfrac12g^{\varphi\varphi}(\partial_\theta g_{\varphi\varphi}+\partial_\varphi g_{\theta\varphi}-\partial_\varphi g_{\theta\varphi})$ $=\tfrac12\cdot\dfrac{2R^2\sin\theta\cos\theta}{R^2\sin^2\theta} = \cot\theta$.

$\Gamma^\varphi{}_{\varphi\varphi}$ $=\tfrac12g^{\varphi\varphi}\partial_\varphi g_{\varphi\varphi} = 0$, nothing depending on $\varphi$.

the answer

$\Gamma^\theta{}_{\varphi\varphi} = -\sin\theta\cos\theta$, $\ \Gamma^\varphi{}_{\theta\varphi}=\Gamma^\varphi{}_{\varphi\theta}=\cot\theta$, all others zero. At $\theta_0=\pi/3$: $-\sqrt3/4 = -0.43301$ and $1/\sqrt3 = 0.57735$.

Checks. No $R$ survives — correct, since rescaling a metric by a constant leaves $\Gamma$ unchanged (the formula is homogeneous of degree zero in $g$) ✓. Dimensions: both coordinates are angles, so both symbols are pure numbers ✓. Limit $\theta\to\pi/2$: $\Gamma^\theta{}_{\varphi\varphi}\to0$ and $\cot\theta\to0$, so the equator has no correction terms at all — a hint that it is a geodesic ✓. Limit $\theta\to0$: $\cot\theta$ blows up, the chart's own singularity, not the sphere's ✓.

Six, not eight. $\Gamma^k{}_{ij}$ on a surface has $2\times3 = 6$ independent components once the lower pair is symmetric, and all six are listed above. Two are nonzero.

The arithmetic saved by noticing things. Every term of the boxed formula contains a derivative of $g$, and here only one derivative of $g$ is nonzero. Scanning for that first reduces six computations to two. The same tactic on Schwarzschild in section 4 reduces $40$ to $9$ — with the Lagrangian trick doing the scanning automatically.

Sanity test against unit 31. The same sphere in stereographic coordinates has $g = 4\delta_{ab}/(1+u^2+v^2)^2$, a conformally flat metric, whose Christoffel symbols are $\Gamma^a{}_{bc} = \delta^a{}_b\partial_c\Omega + \delta^a{}_c\partial_b\Omega - \delta_{bc}\partial^a\Omega$ with $\Omega = \ln\big(2/(1+u^2+v^2)\big)$ — a completely different-looking array of numbers for the same connection. That is the transformation law of the earlier slide doing its work. Components are not the object.

Where $\cot\theta$ comes from physically. It is the reason a compass course of "due east" is not a straight line on the Earth: holding $\theta$ constant requires a continual acceleration toward the pole of magnitude $\Gamma^\theta{}_{\varphi\varphi}\dot\varphi^2$ in the $\theta$-direction. Section 4 computes that failure explicitly for $\theta_0 = \pi/3$.

section 3

Parallel transport, and what a gyroscope does

The connection's reason for existing: a definite map from one tangent space to another, along a chosen path.

32 · transport

Differentiating along a curve

definition

For a smooth curve $\gamma$ and a vector field $V$ along it, the covariant derivative along a curve is the unique operator $D/\mathrm d\lambda$ that is $\mathbb R$-linear, Leibniz in $fV$, and agrees with $\nabla_{\dot\gamma}$ on restrictions of fields. In components: $$\frac{DV^k}{\mathrm d\lambda} = \frac{\mathrm dV^k}{\mathrm d\lambda} + \Gamma^k{}_{ij}\,\dot x^i V^j .$$ $V$ is parallel along $\gamma$ if $DV/\mathrm d\lambda = 0$.

This is a linear first-order ODE system in the $n$ unknowns $V^k(\lambda)$, with coefficients $-\Gamma^k{}_{ij}\dot x^i$ that are smooth along the curve.

parallel transport

For each $\lambda_1$ the map $P_\gamma^{\lambda_0\to\lambda_1}: T_{\gamma(\lambda_0)}M\to T_{\gamma(\lambda_1)}M$ sending $V(\lambda_0)$ to the parallel $V(\lambda_1)$ is a well-defined linear isomorphism.

proof

Unit 01's Picard–Lindelöf gives a unique solution for given initial data; for a linear system the solution extends to the whole interval, because the Gronwall bound forbids blow-up in finite time. Linearity of the equation makes $V(\lambda_0)\mapsto V(\lambda_1)$ linear; running the ODE backwards inverts it. $\square$

and it is an isometry

If $\nabla g=0$ and $V,W$ are parallel then $\frac{\mathrm d}{\mathrm d\lambda}g(V,W) = g\big(\tfrac{DV}{\mathrm d\lambda},W\big)+g\big(V,\tfrac{DW}{\mathrm d\lambda}\big) = 0$. $\square$

Full proof that $D/\mathrm d\lambda$ exists and is unique. Uniqueness: write $V = V^k\partial_k|_{\gamma}$ along the curve. Leibniz and the third property force $DV/\mathrm d\lambda = (\mathrm dV^k/\mathrm d\lambda)\partial_k + V^k\nabla_{\dot\gamma}\partial_k$, and $\nabla_{\dot\gamma}\partial_k = \dot x^i\Gamma^j{}_{ik}\partial_j$ since the direction slot is tensorial. That is the boxed formula, so the operator is determined. Existence: define it by that formula in a chart, and check by the transformation law that two overlapping charts agree — which they do, precisely because the inhomogeneous term in $\tilde\Gamma$ cancels the $\mathrm d/\mathrm d\lambda$ of $A^a{}_k$. $\square$ A subtlety worth naming: a field along $\gamma$ need not extend to a field on $M$ (the curve may cross itself), which is why the operator has to be constructed rather than inherited.

Why "for a linear system the solution is global". Unit 01 proved Picard–Lindelöf locally and Gronwall's inequality separately. For $\dot V = A(\lambda)V$ with $A$ continuous on a compact interval, $\lVert V(\lambda)\rVert \le \lVert V(\lambda_0)\rVert\exp\int\lVert A\rVert$, so the solution cannot escape to infinity, and the standard continuation argument extends it to the whole interval. This matters: transport is defined along the entire curve, however long, with no smallness assumption.

The image to carry. Transporting a vector is not "keeping its components fixed" — that is chart-dependent nonsense — and it is not "keeping it pointing the same way in the ambient space", because there is no ambient space. It is: at each instant, remove exactly the change that the connection says is due to the chart. On a sphere, transporting along a meridian keeps a vector at a constant angle to that meridian, which is what the next example makes quantitative.

32 · transport · physics

What a gyroscope does, and why the path matters

the physical reading

An ideal gyroscope carried along a worldline $\gamma$ — no torque, its spin $S^\mu$ orthogonal to the four-velocity — parallel-transports $S$: $\ DS^\mu/\mathrm d\tau = 0$ whenever $\gamma$ is a geodesic.

No new postulate beyond section 6's rule: the flat-space law for a free, torque-free gyroscope, $\mathrm dS^\mu/\mathrm d\tau = 0$, with $\mathrm d$ promoted to $D$. Off a geodesic it gains a term — Fermi–Walker transport, in the notes — that vanishes with the acceleration.

The figure transports one vector from $A$ to $B$ along two routes. The ODE of the previous slide does each leg, and the answers differ by a quarter-turn.

$V$ starts at $A$ pointing north. Along the equator it arrives at $B$ still pointing north; up the meridian to $N$ and down the next it arrives pointing east. Same start, same end, different answers.

The honest version of the gyroscope claim. Cited: Misner, Thorne and Wheeler, Gravitation, §6.5, for the derivation. The correct transport law for a spin carried by an observer with four-acceleration $a^\mu$ is Fermi–Walker: $DS^\mu/\mathrm d\tau = c^{-2}\big(u^\mu a_\nu - a^\mu u_\nu\big)S^\nu$, which preserves $S\cdot u = 0$ and $S\cdot S$. On a geodesic $a = 0$ and it reduces to parallel transport. So "a gyroscope parallel-transports its spin" is exactly right for a freely falling gyroscope, and needs the correction for one held in place by a rocket or a tower. Gravity Probe B (2004–2011) measured the resulting rotation of a gyroscope in Earth orbit: $-6\,601.8\pm18.3$ milliarcseconds per year of geodetic precession, against a prediction of $-6\,606.1$. That number is computed in unit 37, not here.

Each leg, from the ODE. Use the unit frame $a = RV^\theta$ (south), $b = R\sin\theta\,V^\varphi$ (east) and the sphere's two symbols. Along the equator ($\theta = \pi/2$, $\dot\varphi = 1$): both $\Gamma^\theta{}_{\varphi\varphi} = -\sin\theta\cos\theta$ and $\Gamma^\varphi{}_{\theta\varphi} = \cot\theta$ vanish, so the components are constant and north stays north. Along a meridian ($\dot\theta = \pm1$, $\dot\varphi = 0$): the $\theta$-equation has no $\Gamma$ term, so $V^\theta$ is constant; the $\varphi$-equation is $\dot V^\varphi = \mp\cot\theta\,V^\varphi$, solved by $V^\varphi\propto1/\sin\theta$, so $b = R\sin\theta\,V^\varphi$ is constant too. A vector along the meridian stays along it. At the pole the chart fails, but the vector does not: it arrives pointing along the meridian $\varphi = 0$, continued, which is at right angles to the meridian $\varphi = \pi/2$ — so it leaves down that meridian pointing east, and keeps $(a,b) = (0,1)$ to $B$. The quarter-turn is the angle between the two meridians at $N$.

32 · transport

Holonomy: transport round a closed loop

holonomy

For a loop $\gamma$ based at $p$, $P_\gamma\in GL(T_pM)$ — in the metric case $O(T_pM,g_p)$. The set of all such maps is a group, the holonomy group of $(M,g)$ at $p$.

On the flat plane every loop gives the identity, whatever the chart. Compute it on one circle of latitude and the answer is half a turn.

The figure's loop $A\to N\to B\to A$ gives a quarter-turn: the two routes of the last slide, one run backwards. The triangle it bounds is an eighth of the sphere, area $\pi R^2/2$ — and $\pi R^2/2$ over $R^2$ is a quarter-turn.

transport is path-dependent

$P_\gamma$ for a loop is generally not the identity, whatever the chart or parametrisation: the first chart-independent quantity built from $\Gamma$. Its infinitesimal version is unit 33's Riemann tensor.

Holonomy as a group. Composition of loops composes transports, and reversing a loop inverts it, so the holonomy maps form a subgroup of $O(n)$ (or $O(1,n-1)$). The Ambrose–Singer theorem says its Lie algebra is spanned by the curvature operators — cited: Kobayashi and Nomizu, vol. I, ch. II. Berger's classification of the possible holonomy groups of an irreducible Riemannian manifold is one of the striking results of the subject; nothing in this course needs it, but it is where the subject goes.

Why path-dependence is the whole point. If transport were path-independent, then fixing a basis at one point and transporting it everywhere would produce a global parallel frame, in which $\Gamma$ vanishes identically — a global inertial frame. Unit 30 proved there is none in a gravitational field. So in a gravitational field transport must be path-dependent, and the amount by which it is, is the gravitational field's irreducible content.

32 · transport · worked example

Worked example 3 — half a turn around a circle of latitude

transport around $\theta = \theta_0 = \pi/3$, from the ODE

1 · the curve and the equations. $\theta(\lambda)=\theta_0$, $\varphi(\lambda)=\lambda$, so $\dot\theta = 0$, $\dot\varphi = 1$. With the two symbols: $$\frac{\mathrm dV^\theta}{\mathrm d\lambda} + \Gamma^\theta{}_{\varphi\varphi}V^\varphi = 0,\qquad \frac{\mathrm dV^\varphi}{\mathrm d\lambda} + \Gamma^\varphi{}_{\varphi\theta}V^\theta = 0,$$ that is $\dot V^\theta = \sin\theta_0\cos\theta_0\,V^\varphi$ and $\dot V^\varphi = -\cot\theta_0\,V^\theta$.

2 · orthonormal frame. $a = RV^\theta$, $b = R\sin\theta_0\,V^\varphi$ are the components in $\hat e_\theta = R^{-1}\partial_\theta$, $\hat e_\varphi = (R\sin\theta_0)^{-1}\partial_\varphi$. Then $$\dot a = \cos\theta_0\,b,\qquad \dot b = -\cos\theta_0\,a .$$

3 · solve. $\dot{\overline{(a+\mathrm ib)}} = -\mathrm i\cos\theta_0\,(a+\mathrm ib)$, so $$a+\mathrm ib = e^{-\mathrm i\lambda\cos\theta_0}\,(a_0+\mathrm ib_0).$$ Unit 06's complex exponential, used as the rotation it is. After one circuit $\lambda = 2\pi$ the vector has rotated by $$\Delta = 2\pi\cos\theta_0 .$$

4 · the number. $\cos\theta_0 = \tfrac12$, so $\Delta = \pi$: the transported vector comes back reversed. Starting from $\hat e_\theta$, it returns as $-\hat e_\theta$.

and the area

The cap $\theta\lt\theta_0$ has area $\int_0^{\theta_0}\!\!\int_0^{2\pi}R^2\sin\theta = 2\pi R^2(1-\cos\theta_0) = \pi R^2$. Divided by $R^2$: $\pi$ — the same angle, mod $2\pi$.

checks

$a,b$ are lengths and $R$ has cancelled from the angle ✓. Equator $\theta_0=\pi/2$: $\Delta = 0$, as for a geodesic ✓. Tiny loop $\theta_0\to0$: $\Delta\to2\pi\equiv0$, area $\to0$ ✓. $a^2+b^2$ is constant: an isometry ✓.

Where the mod $2\pi$ comes from, honestly. The ODE delivers the rotation angle $-2\pi\cos\theta_0$ in the frame $(\hat e_\theta,\hat e_\varphi)$ — negative, that is, clockwise in that frame, for a loop run anticlockwise seen from outside. The Gauss–Bonnet statement delivers $+2\pi(1-\cos\theta_0)$, the integral of the Gaussian curvature $1/R^2$ over the enclosed cap. The two differ by exactly $2\pi$, which is the identity rotation, so they describe the same element of $SO(2)$. The full turn is not an error: it is the rotation of the frame itself as it goes once round the pole, and separating "the vector rotated" from "the frame rotated" is exactly the ambiguity that Gauss–Bonnet resolves by counting the geodesic curvature of the boundary as well. The Gauss–Bonnet theorem is cited here — do Carmo, Differential Geometry of Curves and Surfaces, §4-5 — and unit 33 checks it on a geodesic triangle once curvature has a definition. Here the arithmetic is checked at three values of $\theta_0$, which is all this unit claims.

Full check at a third value. $\theta_0 = \pi/3$ gives $\Delta = \pi$ and area $\pi R^2$, angle $\pi$ ✓. $\theta_0 = \arccos(3/4)$ gives $\Delta = 3\pi/2$ and cap area $2\pi R^2/4 = \pi R^2/2$, angle $\pi/2$ — and $3\pi/2 + \pi/2 = 2\pi$ ✓, the same relation. $\theta_0 = \pi/2$ gives $0$ and $2\pi$ ✓.

Why $\theta_0=\pi/3$ was chosen for this unit. It is the only latitude at which the two answers are numerically equal rather than merely equal mod $2\pi$: $\cos\theta_0 = 1/2$ makes $2\pi\cos\theta_0 = 2\pi(1-\cos\theta_0) = \pi$. A reader checking the arithmetic for the first time should not have to think about the $2\pi$ at the same time.

The physical version. The Foucault pendulum. Over one sidereal day the Earth carries the pendulum's support once round its circle of latitude, and the plane of swing, which no force turns about the local vertical, is parallel-transported along that circle — on a surface in $\mathbb R^3$, "no rotation about the normal" is exactly the Levi-Civita transport. So relative to the local north–east frame the plane has turned by $2\pi\cos\theta_0$ per day, a rate $\omega\cos\theta_0$, which at geographical latitude $\ell$ is the textbook $\omega\sin\ell$ — the complementary angle appearing because latitude is measured from the equator and $\theta$ from the pole.

32 · transport · worked example

Worked example 3, continued — watching the vector turn

The latitude cut open at $\varphi = 0$, with the unit frame $\hat e_\theta$ (south), $\hat e_\varphi$ (east) at nine stations. The transported vector $a\,\hat e_\theta + b\,\hat e_\varphi = \cos\tfrac\varphi2\,\hat e_\theta - \sin\tfrac\varphi2\,\hat e_\varphi$ turns steadily against the frame, half as fast as the longitude advances: south, west, north.

32 · transport

Why half a turn: unroll the tangent cone

Left: the cone that touches the sphere along the latitude. Right: the same cone cut along one line from the apex and laid flat — a half-disc. On the flat sheet "not turning" is just the same arrow at every point; at the far edge it points inward, opposite to $\hat e_\theta$.

The cone and the sphere have the same tangent plane at every point of the latitude, and for surfaces in $\mathbb R^3$ that is all transport along the curve depends on (cited; notes). A cone is flat: it unrolls with no stretching, so on it transport is the ordinary parallel shift of the plane.

the same answer as the ODE

Slant $R\tan\theta_0 = \sqrt3R$, rim $2\pi R\sin\theta_0$, so the flat sector's angle is $2\pi\sin\theta_0/\tan\theta_0 = 2\pi\cos\theta_0 = \pi$. After longitude $\varphi$ the frame has turned by $\varphi\cos\theta_0$ on the sheet while the arrow has not: $a = \cos\tfrac\varphi2$, $b = -\sin\tfrac\varphi2$ ✓.

The geometry of the cone. The tangent line to the meridian at the point of colatitude $\theta_0$ meets the axis at distance $R/\cos\theta_0$ from the centre: that is the apex, at $2R$ for $\theta_0 = \pi/3$. The distance from apex to latitude along the cone (the slant) is $R\tan\theta_0$, and the rim is the latitude itself, of length $2\pi R\sin\theta_0$. Unrolled, the cone is a sector of a disc of radius $R\tan\theta_0$ whose arc has that length, so its angle is $2\pi R\sin\theta_0/(R\tan\theta_0) = 2\pi\cos\theta_0$. At $\theta_0 = \pi/3$ that is $\pi$: a half-disc, with rim length $\pi\sqrt3R = 5.4414R$, matching worked example 1.

What is cited, and what is checked. For a surface in $\mathbb R^3$ the Levi-Civita transport along a curve is "differentiate in $\mathbb R^3$, then keep only the part tangent to the surface", which depends only on the tangent planes along the curve; two surfaces tangent along a curve therefore have the same transport along it. Cited: do Carmo, Differential Geometry of Curves and Surfaces, §4-4, which draws exactly this cone. The slide does not lean on the citation for the number: the box checks that the flat picture reproduces step 3's solution at every longitude, so the picture is a picture of the proof already given.

Limits. As $\theta_0\to\pi/2$ the apex runs off to infinity and the cone becomes the cylinder touching the equator; it unrolls into a strip, the sector angle $2\pi\cos\theta_0\to0$, and nothing turns — the equator is a geodesic. As $\theta_0\to0$ the cone flattens into the tangent plane at the pole, the sector becomes a whole disc, and the turn is $2\pi$, which is no turn at all.

section 4

Geodesics, two ways

Straightest, and shortest. Two definitions, one set of curves — and a trick that computes every Christoffel symbol for free.

32 · geodesics

Straightest: the curve that transports its own tangent

definition

A geodesic is a curve whose tangent is parallel along itself: $\nabla_{\dot\gamma}\dot\gamma = 0$. In a chart, the geodesic equation $$\ddot x^k + \Gamma^k{}_{ij}\,\dot x^i\dot x^j = 0 .$$ Such a curve is auto-parallel.

existence and uniqueness

Through each $p$ and each $v\in T_pM$ there is exactly one maximal geodesic with $\gamma(0)=p$, $\dot\gamma(0)=v$. Proof: the equation is a second-order ODE system with smooth (indeed quadratic) right-hand side; unit 01's Picard–Lindelöf applies to the first-order system on $TM$. $\square$

constant speed, and the causal character

$\dfrac{\mathrm d}{\mathrm d\lambda}g(\dot\gamma,\dot\gamma) = 2g\big(\tfrac{D\dot\gamma}{\mathrm d\lambda},\dot\gamma\big) = 0$. So $g(\dot\gamma,\dot\gamma)$ is constant: a geodesic that starts timelike stays timelike, and one that starts null stays null. $\square$

affine parameter

If $\lambda$ satisfies the geodesic equation, so does $s = a\lambda+b$ ($a\ne0$), and no other reparametrisation does. Proof: under $\lambda\mapsto s(\lambda)$ the equation acquires $-\,(s''/s')\,\mathrm dx^k/\mathrm ds$, which vanishes for all geodesics iff $s''=0$. $\square$

Timelike: proper time is affine. Null: there is no arc length, only the affine parameter.

Full proof of the affine-parameter claim. Let $\gamma$ satisfy $\ddot x^k + \Gamma^k{}_{ij}\dot x^i\dot x^j = 0$ in $\lambda$, and reparametrise by $\lambda\mapsto s$. With $x^k(s)$ and $' = \mathrm d/\mathrm ds$: $\dot x^k = x'^k s'$ and $\ddot x^k = x''^k (s')^2 + x'^k s''$. Substituting, $(s')^2\big(x''^k + \Gamma^k{}_{ij}x'^ix'^j\big) + s''x'^k = 0$, so $$x''^k + \Gamma^k{}_{ij}x'^ix'^j = -\frac{s''}{(s')^2}\,x'^k .$$ This is the non-affinely parametrised geodesic equation: the right-hand side is proportional to the tangent, so the curve is unchanged, only its parametrisation. It vanishes for a non-constant curve exactly when $s''=0$, i.e. $s$ affine in $\lambda$. $\square$ The form $\nabla_{\dot\gamma}\dot\gamma = f\dot\gamma$ is how a geodesic is recognised when it arrives with an inconvenient parameter — for instance in unit 36, where a null geodesic is first found in terms of $r$.

Why the equation needs $TM$. Unit 01's theorem is about first-order systems. Write $y = (x^k, v^k)$ on the tangent bundle (unit 31 made $TM$ a $2n$-manifold) and the geodesic equation becomes $\dot x^k = v^k$, $\dot v^k = -\Gamma^k{}_{ij}v^iv^j$: a first-order autonomous system with smooth right-hand side. Its flow is the geodesic flow, one of the most studied dynamical systems there is: on a compact negatively curved surface nearby geodesics separate exponentially fast, and it is the standard example of chaotic dynamics.

Maximal, and not always complete. "Maximal" means defined on the largest interval it can be; that interval need not be all of $\mathbb R$. On $\mathbb R^2$ with the origin removed, the geodesic heading for the origin simply stops. A manifold whose geodesics all extend forever is geodesically complete; unit 36 will find that Schwarzschild in its original chart is not, and that the failure is a chart's fault in one case and a genuine singularity in the other.

32 · geodesics

Shortest: the same curves, from unit 11

the energy functional gives the geodesic equation

Let $E[\gamma] = \tfrac12\int_a^b g_{ij}(x)\,\dot x^i\dot x^j\,\mathrm d\lambda$. Unit 11's Euler–Lagrange equation for the Lagrangian $\mathcal L = \tfrac12g_{ij}\dot x^i\dot x^j$ is $$\frac{\mathrm d}{\mathrm d\lambda}\big(g_{kj}\dot x^j\big) - \tfrac12\,\partial_kg_{ij}\,\dot x^i\dot x^j = 0 .$$

and it is the geodesic equation

Expand the first term: $g_{kj}\ddot x^j + \partial_ig_{kj}\dot x^i\dot x^j$. Split the second factor into its symmetric part, $\tfrac12(\partial_ig_{kj}+\partial_jg_{ki})\dot x^i\dot x^j$ — legitimate because $\dot x^i\dot x^j$ is symmetric. So $$g_{kj}\ddot x^j + \tfrac12\big(\partial_ig_{kj}+\partial_jg_{ki}-\partial_kg_{ij}\big)\dot x^i\dot x^j = 0 .$$ Contract with $g^{lk}$: that bracket is exactly $2\Gamma^l{}_{ij}g_{lk}$, and what remains is $\ddot x^l + \Gamma^l{}_{ij}\dot x^i\dot x^j = 0$. $\square$

read it twice

Forwards it says: stationary energy $\Rightarrow$ geodesic. Backwards it says something more useful — the Christoffel combination is what the Euler–Lagrange operator produces from $g_{ij}\dot x^i\dot x^j$, whatever $g$ is.

That is the trick, at the end of this section.

why $E$ and not $L$

$L$ is reparametrisation invariant, so its Euler–Lagrange system is degenerate: every stationary curve comes in every parametrisation. $E$ is not invariant, and picks out the affine ones (notes).

Full statement and proof of the $L$-versus-$E$ relation. For $\mathcal L_{\!L} = \sqrt{F}$ with $F = g_{ij}\dot x^i\dot x^j$, the Euler–Lagrange equation is $\frac{\mathrm d}{\mathrm d\lambda}\big(\frac{g_{kj}\dot x^j}{\sqrt F}\big) - \frac{\partial_kg_{ij}\dot x^i\dot x^j}{2\sqrt F} = 0$. Multiplying by $\sqrt F$ and using $\frac{\mathrm d}{\mathrm d\lambda}\frac{1}{\sqrt F} = -\frac{\dot F}{2F^{3/2}}$ gives $$\ddot x^k + \Gamma^k{}_{ij}\dot x^i\dot x^j = \frac{\dot F}{2F}\,\dot x^k ,$$ which is exactly the non-affine form derived in the previous slide's notes with $f = \dot F/2F = \mathrm d(\ln\sqrt F)/\mathrm d\lambda$. So: stationary points of $L$ are geodesics up to reparametrisation, and they are affinely parametrised precisely when $F$ is constant, i.e. when the curve is traversed at constant speed. $\square$

The inequality behind the preference. By Cauchy–Schwarz on $[a,b]$, $L[\gamma]^2 = \big(\int\sqrt F\big)^2 \le (b-a)\int F = 2(b-a)E[\gamma]$, with equality iff $F$ is constant. So minimising $E$ over curves with fixed endpoints and fixed parameter interval both minimises $L$ and forces constant speed. This is the standard reason every Riemannian geometry text minimises energy: it is a smooth functional with a non-degenerate Euler–Lagrange system, and the square root is a genuine analytic nuisance where $F$ vanishes — which in Lorentzian signature it does, on every null curve.

Lorentzian signature reverses the extremum. Unit 30 stated it and unit 27's twin calculation showed it: among timelike curves between two timelike-separated events, the geodesic maximises proper time. A neighbouring wiggle always ages less. The mechanism is the minus sign: $-g_{\mu\nu}\dot x^\mu\dot x^\nu$ is what is under the root, and deviating in a spatial direction adds a positive quantity that is subtracted. Problem P8 works the sign out.

Where this came from. Unit 12's Lagrangian for a free system in generalised coordinates is $T = \tfrac12m_{ij}(q)\dot q^i\dot q^j$ — the energy functional of the kinetic-energy metric on configuration space. So its Euler–Lagrange equations were already this slide's geodesic equation: free motion under workless constraints has always been geodesic motion, and the course now has the vocabulary to say so.

32 · geodesics · worked example

Worked example 4 — great circles, twice, and one curve that is not

the geodesics of $S^2_R$

Route A — auto-parallel. With the two symbols, $$\ddot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0,\qquad \ddot\varphi + 2\cot\theta\,\dot\theta\dot\varphi = 0 .$$ Try $\theta\equiv\pi/2$: the first is $0 = 0$ since $\cos\tfrac\pi2=0$; the second is $\ddot\varphi = 0$. So the equator at constant rate is a geodesic.

Every great circle is the equator for some pole, and rotations are isometries (unit 31), which map geodesics to geodesics (notes). So all great circles are geodesics, and by uniqueness given $(p,v)$ there are no others.

Route B — stationary length. Write the curve as $\theta(\varphi)$. Its length is $$L = R\!\int\!\sqrt{\theta'^2+\sin^2\theta}\ \mathrm d\varphi .$$ The integrand has no explicit $\varphi$, so unit 11's Beltrami identity gives $F - \theta'\,\partial F/\partial\theta' = \text{const}$: $$\frac{\sin^2\theta}{\sqrt{\theta'^2+\sin^2\theta}} = C \ \Rightarrow\ \theta'^2 = \frac{\sin^4\theta}{C^2} - \sin^2\theta .$$ The constant is Clairaut's relation. With $w = \cot\theta$, $w' = -\theta'/\sin^2\theta$, so $w'^2 = C^{-2} - 1 - w^2$: solutions $w = A\cos\varphi + B\sin\varphi$, bar spurious constants (notes). Multiplying by $R\sin\theta$ gives $z = Ax + By$ — a plane through the origin. $\square$

Checks. $L$ is $R$ times an angle, $C$ a pure number in $(0,1]$ ✓. $C = 1$ forces $\theta=\pi/2$, the equator of route A ✓. Meridians are not graphs $\theta(\varphi)$; they are the $A,B\to\infty$ limit ✓. Both routes give the same two-parameter family ✓.

Full working of route B. With $F = \sqrt{\theta'^2+\sin^2\theta}$, $\partial F/\partial\theta' = \theta'/F$, so the Beltrami combination is $F - \theta'^2/F = \sin^2\theta/F$, constant $= C$. Squaring, $\theta'^2+\sin^2\theta = \sin^4\theta/C^2$. Note $C = \sin^2\theta/F \le \sin\theta\le1$. Now $w = \cot\theta$ has $w' = -\csc^2\theta\,\theta'$, so $w'^2 = \csc^4\theta\,\theta'^2 = C^{-2} - \csc^2\theta = C^{-2} - 1 - w^2$. Put $D^2 = C^{-2}-1\ge0$; then $w'^2+w^2 = D^2$. Differentiating, $2w'(w''+w) = 0$.

Why the two routes must agree. The slide "Shortest: the same curves" proved that stationary points of length are geodesics up to parametrisation. Route B found the stationary curves of $L$ as point sets; route A found the geodesics with their affine parametrisation. They are the same set of curves, and the example shows it by computation rather than by appeal.

Why route A needed the symmetry argument. Solving the coupled nonlinear system directly is unpleasant; noticing that one solution plus a transitive group of isometries gives them all is not a dodge but the standard and better argument. Formally: if $\Phi$ is an isometry then $\Phi$ maps the Levi-Civita connection to itself (uniqueness in the fundamental theorem does the work — an isometry carries a torsion-free metric connection to a torsion-free metric connection, and there is only one), so it carries geodesics to geodesics. $SO(3)$ acts transitively on unit tangent vectors of $S^2$, and uniqueness given $(p,v)$ finishes it.

Clairaut's relation, named. The Beltrami constant $C$ is in disguise a conserved momentum: with the energy Lagrangian $\mathcal L = \tfrac12R^2(\dot\theta^2+\sin^2\theta\,\dot\varphi^2)$, the coordinate $\varphi$ is absent, so $p_\varphi = R^2\sin^2\theta\,\dot\varphi$ is constant along every geodesic, and $C = p_\varphi/(R\,\lVert\dot\gamma\rVert)$. Geometrically $r\sin\psi$ is constant, with $r = R\sin\theta$ the distance from the axis and $\psi$ the angle to the meridian: Clairaut's theorem, true on any surface of revolution. It is unit 13's Noether theorem applied to the rotational Killing field, and unit 36 will use precisely this mechanism — a Killing field gives $g(\dot\gamma,K)$ constant along a geodesic — to get Schwarzschild's conserved energy and angular momentum without solving anything.

32 · geodesics · worked example

Great circles in the chart, and a latitude that mimics one

Route B's answer, $\cot\theta = D\cos(\varphi-\varphi_0)$, drawn in the chart for three values of $D$. Each oscillates about the equator; the larger $D$, the closer the curve comes to the poles.

The first integral $w'^2 + w^2 = D^2$ has one more solution, $w\equiv D$: the latitude $\cot\theta = D$, which touches the great circle at its crest, where $w' = 0$, and then leaves it.

It satisfies the squared equation, not the Euler–Lagrange equation — the trap of unit 11's Beltrami identity. For $D = 1/\sqrt3$ it is the test latitude $\theta_0 = \pi/3$.

Great circles in the $(\varphi,\theta)$ chart, north at the top. The dashed latitude $\theta = \pi/3$ meets the great circle with $D = 1/\sqrt3$ at one point, tangentially, and is not a solution.

The trap in that last line. It allows $w' = 0$ as well as $w''+w = 0$, and $w\equiv\pm D$ — a latitude $\theta\equiv\theta_1$ — does satisfy $w'^2 + w^2 = D^2$. It is not a stationary curve. Unit 11 flagged exactly this: the Beltrami combination has derivative $\theta'$ times the Euler–Lagrange expression $\partial F/\partial\theta - (\partial F/\partial\theta')'$, so Beltrami follows from Euler–Lagrange but gives it back only where $\theta'\ne0$. On a stretch with $\theta'\equiv0$ that expression is $\sin\theta\cos\theta/F = \cos\theta_1$, and setting it to zero forces $\theta_1 = \pi/2$: the equator, which is $w\equiv0$ with $D = 0$ and already in the family. So on a stationary curve with $D\gt0$ the set where $w' = 0$ contains no interval; $w''+w = 0$ holds on its dense complement, hence everywhere by continuity ($\theta$ is $C^2$), and $w = D\cos(\varphi-\varphi_0)$ — unit 06's equation. The latitude $\theta_0 = \pi/3$ on the slide is this spurious solution caught by the other route. Now $z = R\cos\theta$, $x = R\sin\theta\cos\varphi$, $y = R\sin\theta\sin\varphi$; multiplying $\cos\theta/\sin\theta = D\cos\varphi_0\cos\varphi + D\sin\varphi_0\sin\varphi$ by $R\sin\theta$ gives $z = (D\cos\varphi_0)x + (D\sin\varphi_0)y$: a plane through the origin, cut with the sphere, which is a great circle. $\square$

32 · geodesics · worked example

The curve that is not: a latitude against the polar route

$P$ and $Q$ on the latitude $\theta_0 = \pi/3$, $180^\circ$ of longitude apart. Along the latitude: $2.7207R$. Over the pole, on a great circle: $2.0944R$, $23.0\%$ shorter. Staying on the latitude needs a push towards the pole.

The curve that is not. $\theta\equiv\pi/3$, $\varphi=\lambda$: the first equation reads $0 - \sin\tfrac\pi3\cos\tfrac\pi3 = -\sqrt3/4 = -0.43301 \ne 0$. A constant-latitude flight is not the shortest route.

The sign says which way: the geodesic equation wants $\ddot\theta = +\tfrac{\sqrt3}4\dot\varphi^2 \gt 0$, a drift towards the equator. A body held on the latitude must be pushed towards the pole, as the arrow shows.

the numbers

Half the latitude: $\tfrac12\cdot2\pi R\sin\theta_0 = \pi\sqrt3R/2 = 2.7207R$. Over the pole: the angle $2\theta_0$ at the centre, so $2\pi R/3 = 2.0944R$. Ratio $0.770$.

Numbers to keep. Two points on the latitude $\theta_0 = \pi/3$, $180^\circ$ of longitude apart. Along the latitude the distance is half of $\pi\sqrt3R$, that is $2.7207R$. Over the pole, along the meridian great circle, it is $2\theta_0R = 2\pi R/3 = 2.0944R$. The geodesic is $23.0\%$ shorter, and that is why polar routes exist.

Why the push points poleward. With $\theta(\lambda)\equiv\theta_0$ the geodesic $\theta$-equation $\ddot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0$ fails by $-\sin\theta_0\cos\theta_0\,\dot\varphi^2$. Adding a force term $f^\theta$ on the right to make the latitude a solution requires $f^\theta = -\sin\theta_0\cos\theta_0\,\dot\varphi^2 \lt 0$: towards smaller $\theta$, the pole. The physical acceleration is $R$ times that, $R\sin\theta_0\cos\theta_0\,\dot\varphi^2$ — the component along the sphere of the centripetal acceleration $R\sin\theta_0\,\dot\varphi^2$ towards the axis, which is what a body on a latitude circle must have.

32 · geodesics

The trick: read the Christoffel symbols off a Lagrangian

the method, four steps

Write $\mathcal L = g_{ij}(x)\,\dot x^i\dot x^j$ — the metric with $\mathrm dx^i\mathrm dx^j$ replaced by $\dot x^i\dot x^j$. (Any constant multiple works.)
Write the $n$ Euler–Lagrange equations $\frac{\mathrm d}{\mathrm d\lambda}\frac{\partial\mathcal L}{\partial\dot x^k} = \frac{\partial\mathcal L}{\partial x^k}$.
Divide each by the coefficient of $\ddot x^k$ so it reads $\ddot x^k + (\cdots) = 0$.
Read off $\Gamma^k{}_{ij}$ as the coefficient of $\dot x^i\dot x^j$, symmetrised: a term $2A\,\dot x^i\dot x^j$ with $i\ne j$ means $\Gamma^k{}_{ij}=\Gamma^k{}_{ji}=A$.

the one place it goes wrong

Forgetting step 4's factor of two. $\ddot r + \frac2r\dot r\dot\theta$ means $\Gamma^r{}_{r\theta} = \Gamma^r{}_{\theta r} = 1/r$, not $2/r$: the sum $\Gamma^k{}_{ij}\dot x^i\dot x^j$ counts the off-diagonal pair twice.

why it is legitimate

Not a mnemonic: the slide "Shortest" proved the Euler–Lagrange operator turns $g_{ij}\dot x^i\dot x^j$ into $\ddot x^k + \Gamma^k{}_{ij}\dot x^i\dot x^j$.

The saving, in four dimensions: $40$ symbols $\times\,4$ terms $\times\,3$ derivatives $= 480$ by the formula; $4$ equations $\times\,2$ differentiations $=$ eight by the trick.

The step that is easy to skip. Step 3 matters when $g$ is not diagonal or when $g_{kk}$ is not 1: the Euler–Lagrange equation for $x^k$ starts with $\frac{\mathrm d}{\mathrm d\lambda}(2g_{kj}\dot x^j)$, which is $2g_{kk}\ddot x^k$ plus other second derivatives when $g$ has off-diagonal entries. For a diagonal metric, dividing by $2g_{kk}$ is the whole of step 3. For a non-diagonal one, the honest route is to assemble the $n$ equations as $2g_{kj}\ddot x^j + (\cdots)_k = 0$ and contract with $g^{lk}$ — which is what the proof did. Every metric in this course from here to unit 40 is diagonal or block-diagonal, so the easy case is the case.

A second use of the same Lagrangian. Because $\mathcal L$ does not depend on $\lambda$, unit 14's conserved quantity for an autonomous Lagrangian says $\dot x^k\partial\mathcal L/\partial\dot x^k - \mathcal L = \mathcal L$ is constant along solutions, which is the constant-speed statement again, now for free. And any coordinate absent from $\mathcal L$ gives a conserved momentum by unit 13 — which is exactly how unit 36 will get Schwarzschild's $E$ and $L$ out of $t$ and $\varphi$ being cyclic. The geodesic Lagrangian is therefore not only a calculational trick: it is the bridge from all of Part B's machinery into general relativity.

Numerically checking $\Gamma$, when it matters. A cheap independent test of a computed $\Gamma$ is metric compatibility, $\partial_kg_{ij} = \Gamma^l{}_{ki}g_{lj} + \Gamma^l{}_{kj}g_{il}$, checked on one or two index triples. It is a different equation from the one used to find $\Gamma$ and catches sign errors immediately. Problem P3 asks for exactly this check.

32 · geodesics · worked example

Worked example 5 — the Schwarzschild form, by the trick

$\mathrm ds^2 = -f(r)c^2\mathrm dt^2 + f(r)^{-1}\mathrm dr^2 + r^2(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\varphi^2)$

$\mathcal L = -fc^2\dot t^2 + f^{-1}\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\varphi^2$, with $f = f(r)$ and ${}' = \mathrm d/\mathrm dr$.

The $t$ equation. $t$ is absent from $\mathcal L$, so $\partial\mathcal L/\partial\dot t = -2fc^2\dot t$ is conserved. Differentiating: $f\ddot t + f'\dot r\dot t = 0$, i.e. $$\ddot t + \frac{f'}{f}\,\dot r\dot t = 0\ \Rightarrow\ \Gamma^t{}_{tr} = \Gamma^t{}_{rt} = \frac{f'}{2f}.$$

The $r$ equation. $\frac{\mathrm d}{\mathrm d\lambda}(2f^{-1}\dot r) = 2f^{-1}\ddot r - 2f^{-2}f'\dot r^2$, and $\partial_r\mathcal L = -f'c^2\dot t^2 - f^{-2}f'\dot r^2 + 2r\dot\theta^2 + 2r\sin^2\theta\dot\varphi^2$.

Equating and multiplying by $f/2$: $$\ddot r + \tfrac12 ff'c^2\,\dot t^2 - \tfrac{f'}{2f}\,\dot r^2 - rf\,\dot\theta^2 - rf\sin^2\theta\,\dot\varphi^2 = 0 .$$ Four symbols at once: $\Gamma^r{}_{tt} = \tfrac12ff'c^2$, $\ \Gamma^r{}_{rr} = -\tfrac{f'}{2f}$, $\ \Gamma^r{}_{\theta\theta} = -rf$, $\ \Gamma^r{}_{\varphi\varphi} = -rf\sin^2\theta$.

The $\theta$, $\varphi$ equations are the sphere's with $r$ variable: $\ddot\theta + \tfrac2r\dot r\dot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0$ and $\ddot\varphi + \tfrac2r\dot r\dot\varphi + 2\cot\theta\,\dot\theta\dot\varphi = 0$, giving $\Gamma^\theta{}_{r\theta}=\Gamma^\varphi{}_{r\varphi}=1/r$, $\ \Gamma^\theta{}_{\varphi\varphi}=-\sin\theta\cos\theta$, $\ \Gamma^\varphi{}_{\theta\varphi}=\cot\theta$.

count

Nine nonzero symbols of $40$, from four equations.

Checks. $[\Gamma^r{}_{tt}] = [c^2/\text{length}]$ and $[\dot t^2] = \mathrm s^2/[\lambda]^2$, so $\Gamma^r{}_{tt}\dot t^2$ is a length per $[\lambda]^2$, like $\ddot r$ ✓. $f\equiv1$: flat spacetime in spherical coordinates ✓. $f = 1 - 2GM/(c^2r)$: $\Gamma^r{}_{tt} = (GM/r^2)f$, Newton to one part in $10^9$ at the Earth's surface ✓.

What this metric is, and what it is not, yet. It is a form: the most general static, spherically symmetric metric with the areal radius chosen as the coordinate $r$. Nothing here says what $f$ is. Unit 36 solves the vacuum field equations with this ansatz and finds $f = 1 - r_s/r$ with $r_s = 2GM/c^2$; unit 35 fixes the constant by the Newtonian limit computed in section 6 of this unit. Practising the trick on it now means unit 36 can spend its time on the physics rather than on arithmetic.

Full check against the coordinate formula, for one symbol. $\Gamma^r{}_{tt} = \tfrac12g^{rr}(2\partial_tg_{rt} - \partial_rg_{tt}) = \tfrac12f\cdot(-\partial_r(-fc^2)) = \tfrac12ff'c^2$ ✓. And $\Gamma^r{}_{rr} = \tfrac12g^{rr}\partial_rg_{rr} = \tfrac12f\cdot\partial_r(f^{-1}) = -\tfrac{f'}{2f}$ ✓. Two checks out of nine is the right amount of paranoia; problem P3 does all of them on a different metric.

The number that will matter. With $f = 1-r_s/r$ one has $f' = r_s/r^2$, so $\Gamma^r{}_{tt} = \tfrac12c^2(r_s/r^2)f$. Since $c^2r_s/2 = GM$, that is $GM f/r^2$ — Newton's $GM/r^2$ times a correction $f$ that at the Earth's surface differs from 1 by $1.39\times10^{-9}$. Section 6 gets the same thing from the other direction, by demanding that the geodesic equation reduce to Newton's law.

A convention note. Writing $c$ explicitly, as here, keeps dimensions checkable at the cost of clutter. Most of the literature sets $c=1$ and restores it at the end by dimensions; the conversion is always "insert the unique powers of $c$ that make the dimensions work". This unit keeps $c$ throughout.

section 5

Normal coordinates, and the equivalence principle stated exactly

What can be made to vanish at a point, what cannot, and why the difference is the whole of unit 30 in one sentence.

32 · normal coordinates

The exponential map

definition

For $v\in T_pM$ let $\gamma_v$ be the geodesic with $\gamma_v(0)=p$, $\dot\gamma_v(0)=v$. The exponential map is $$\exp_p(v) := \gamma_v(1),$$ defined on the open set of $v$ for which $\gamma_v$ reaches parameter 1.

it is a local diffeomorphism at $0$

Proof. Homogeneity: $\gamma_{av}(s) = \gamma_v(as)$, since both sides solve the geodesic equation with the same initial data (the equation is quadratic in $\dot x$). Hence $\exp_p(tv) = \gamma_v(t)$ and $$\mathrm d(\exp_p)_0(v) = \tfrac{\mathrm d}{\mathrm dt}\big|_0\gamma_v(t) = v,$$ so the differential at $0$ is the identity. Unit 04's inverse function theorem finishes it. $\square$

On the sphere, a cross-section through $N$. A tangent vector $v$ is wrapped onto the sphere along the geodesic it starts: $\exp_N(v)$ is at distance $|v|$ in the direction of $v$. Near $0$ this is one-to-one; at $|v| = \pi R$ every direction lands on $S$.

Full proof of homogeneity. Fix $v$ and $a\ne0$, and set $\sigma(s) := \gamma_v(as)$. Then $\dot\sigma(s) = a\dot\gamma_v(as)$ and $\ddot\sigma(s) = a^2\ddot\gamma_v(as)$, so $\ddot\sigma^k + \Gamma^k{}_{ij}\dot\sigma^i\dot\sigma^j = a^2\big(\ddot\gamma^k_v + \Gamma^k{}_{ij}\dot\gamma^i_v\dot\gamma^j_v\big)(as) = 0$: $\sigma$ is a geodesic. Its initial data are $\sigma(0)=p$, $\dot\sigma(0) = av$, so by uniqueness $\sigma = \gamma_{av}$. Hence $\gamma_{av}(s) = \gamma_v(as)$, and putting $s=1$, $\exp_p(av) = \gamma_v(a)$. $\square$ The scaling is possible only because the geodesic equation is homogeneous of degree two in $\dot x$ — which in turn is because $\Gamma$ has two lower indices and no more.

Why $\exp_p$ is smooth, which the inverse function theorem needs. $\exp_p(v)$ is the $x$-part of the geodesic flow on $TM$ at time 1 started from $(p,v)$, and the flow of a smooth first-order system depends smoothly on its initial point. Unit 01 proved the continuous version with Gronwall; the smooth version is cited: Hartman, Ordinary Differential Equations, ch. V. It is the same fact unit 31 used to differentiate flows for the Lie derivative, and it also makes the set of $v$ whose geodesic survives to parameter 1 open.

How large a neighbourhood? The inverse function theorem gives some ball; the largest radius for which $\exp_p$ is a diffeomorphism is the injectivity radius, and it is finite whenever geodesics refocus. On the unit sphere it is $\pi$: geodesics from the north pole all meet again at the south pole, and beyond that they cross. Unit 11's conjugate points are the same phenomenon, met there for the brachistochrone; unit 33 relates them to curvature by the Jacobi equation. The physical reading is the one unit 30 made with a number: a freely falling frame is valid only over a region small compared with the scale on which tidal effects act.

Why the name. $\exp$ because on a Lie group with a bi-invariant metric the geodesics through the identity are the one-parameter subgroups $t\mapsto e^{tX}$, and this map is exactly the exponential map of unit 16. The same word, and in the group case the same function.

32 · normal coordinates

Normal coordinates: name a point by the geodesic that reaches it

normal coordinates

Choose a basis $(e_1,\dots,e_n)$ of $T_pM$ with $g_p(e_i,e_j) = \eta_{ij}$ (Gram–Schmidt, or Sylvester in the indefinite case — unit 16, unit 26). Define $$\mathbf y\big(\exp_p(y^ie_i)\big) := (y^1,\dots,y^n)$$ on a neighbourhood of $p$ small enough for $\exp_p$ to be a diffeomorphism.

the defining property

Geodesics through $p$ are the lines $y^i(t) = tv^i$, affinely parametrised. The next slide's proof uses nothing else.

Physically: fire test particles from $p$ in every direction and label events "which particle, how long by its clock" — a freely falling laboratory.

What the chart is, concretely. $\exp_p$ is a diffeomorphism from a small ball in $T_pM$ onto a neighbourhood of $p$; an orthonormal basis identifies $T_pM$ with $\mathbb R^n$; the chart is the inverse of the composite. On the sphere at $N$, $y^1 = R\theta\cos\varphi$, $y^2 = R\theta\sin\varphi$: polar coordinates in the tangent plane, with the radius measured along the sphere. That is the example two slides on.

Why "uses nothing else" matters. The local flatness theorem on the next slide is proved from the single fact that the radial lines $y(t) = tv$ are affinely parametrised geodesics. Any chart with that property would do; normal coordinates are simply the one in which it holds by construction. Their freedom is the choice of orthonormal basis at $p$ — a rotation, or in spacetime a Lorentz transformation, which is the freedom a freely falling observer has in pointing the axes of the laboratory.

32 · normal coordinates

The local flatness theorem

at any point, and only at that point

For any $p\in M$ there are coordinates with $$g_{ij}(p) = \eta_{ij},\qquad \partial_kg_{ij}(p) = 0,\qquad\text{hence}\qquad \Gamma^k{}_{ij}(p) = 0 .$$ They cannot in general be chosen so that $\partial_k\partial_lg_{ij}(p) = 0$.

proof of the three that can

Take normal coordinates at $p$. Geodesics through the origin are $y^i(t)=tv^i$, so $\ddot y = 0$, and the geodesic equation gives $$\Gamma^k{}_{ij}(tv)\,v^iv^j = 0\quad\text{for all small }t,\ \text{all }v .$$ At $t=0$, $\Gamma^k{}_{ij}(p)v^iv^j = 0$ for all $v$; $\Gamma$ is symmetric (torsion-free), so polarising ($v\to v+w$) gives $\Gamma(p)=0$.

Then metric compatibility at $p$: $\partial_kg_{ij} = \Gamma^l{}_{ki}g_{lj}+\Gamma^l{}_{kj}g_{il} = 0$. And $g(p)=\eta$ by the choice of basis. $\square$

the physical sentence

Read the "hence" one way only: $\Gamma(p)=0$ does not give $\Gamma=0$ near $p$.

"A gravitational field can be transformed away at a point but not in a neighbourhood" is exactly "$\Gamma$ can be set to zero at a point but $\partial\Gamma$ cannot". Unit 30 argued it with elevators and a Rindler chart; the local flatness theorem is the same claim with a proof.

Torsion-free is load-bearing: with torsion, the antisymmetric part $\Gamma^k{}_{[ij]} := \tfrac12(\Gamma^k{}_{ij} - \Gamma^k{}_{ji}) = \tfrac12T^k{}_{ij}$ is a tensor and cannot be removed.

Full proof of the polarisation step. Suppose $S^k{}_{ij} = S^k{}_{ji}$ and $S^k{}_{ij}v^iv^j=0$ for all $v$. Put $v = a+b$: $0 = S^k{}_{ij}(a^i+b^i)(a^j+b^j) = S^k{}_{ij}a^ia^j + S^k{}_{ij}b^ib^j + 2S^k{}_{ij}a^ib^j = 2S^k{}_{ij}a^ib^j$, using symmetry to combine the two cross terms. Taking $a,b$ to run over a basis gives every component zero. $\square$ Without symmetry only the symmetric part is forced to vanish, which is exactly why the theorem needs torsion-freeness.

Full proof that the second derivatives cannot all be killed — the counting. Work at one point and ask what a chart change can do. Expand the new coordinates in the old: $\tilde x^a = M^a{}_ix^i + \tfrac12 N^a{}_{ij}x^ix^j + \tfrac16 P^a{}_{ijk}x^ix^jx^k+\dots$, with $N$ symmetric in its two lower indices and $P$ in its three.

Order 0. $g(p)$ has $n(n+1)/2 = 10$ components in four dimensions; $M$ has $n^2 = 16$ entries. Set $g$ to $\eta$: $16-10 = 6$ parameters left over — and those six are the Lorentz group of unit 26, which is why a freely falling frame is fixed only up to a boost and a rotation.
Order 1. $\partial g(p)$ has $n\cdot n(n+1)/2 = 40$ components; $N$ has $n\cdot n(n+1)/2 = 40$ entries. Exactly enough: all first derivatives can be killed, with nothing spare. That is the theorem above, recounted.
Order 2. $\partial\partial g(p)$ has $\big(n(n+1)/2\big)^2 = 100$ components; $P$ has $n\cdot\binom{n+2}{3} = 4\times20 = 80$. Deficit $100-80 = \mathbf{20}$.

Twenty numbers at each point survive every coordinate change. They are the independent components of the Riemann tensor in four dimensions, and finding them is unit 33. In two dimensions $g$ has 3 components, $\partial\partial g$ has $3\times3 = 9$, and $P$ has $2\times\binom{4}{3} = 8$, so $9-8=1$: the single Gaussian curvature of a surface, which is why the sphere needed exactly one number, $1/R^2$, in the transport example. The counting shows that at least this many numbers survive; that exactly this many do, and that they assemble into a tensor, is unit 33's work. For a surface, problem P11 exhibits the one survivor and proves no chart removes it.

Who. Riemann, 1854, in the habilitation lecture, asked when a metric can be brought to constant form and answered with a count of the same kind — $n(n+1)/2$ component functions against $n$ coordinate functions, leaving $n(n-1)/2$ that describe the geometry. The order-by-order Taylor count above is the standard modern refinement of his argument. Einstein reached the same question from the physics end in 1912 and learnt from Grossmann that the mathematics already existed.

32 · normal coordinates · example

Normal coordinates on the sphere, and what survives

At the north pole of $S^2_R$: geodesics from $N$ are meridians, so $\exp_N$ sends $\mathbf y$ with $|\mathbf y| = \rho$ to the point at geodesic distance $\rho$ in the direction $\mathbf y/\rho$. In polar form $\rho = R\theta$, and the metric is $$g = \mathrm d\rho^2 + R^2\sin^2(\rho/R)\,\mathrm d\phi^2 .$$ Writing $y^1 = \rho\cos\phi$, $y^2 = \rho\sin\phi$ and expanding to second order:

the expansion

$$g_{ab}(\mathbf y) = \delta_{ab} - \frac{1}{3R^2}\big(\delta_{ab}|\mathbf y|^2 - y_ay_b\big) + O(|\mathbf y|^4).$$

everything the theorem promised

$g_{ab}(0) = \delta_{ab}$ ✓, every first derivative vanishes at $0$ ✓ (the correction is quadratic), so $\Gamma(0)=0$ ✓.

And the second derivatives do not vanish: $\partial_2\partial_2g_{11} = -2/(3R^2)$. One number, $1/R^2$ — the curvature the counting argument said a surface keeps.

Full derivation of the expansion. Start from $g = \mathrm d\rho^2 + R^2\sin^2(\rho/R)\mathrm d\phi^2$, which is the round metric with $\rho = R\theta$. In Cartesian normal coordinates, $\mathrm d\rho = (y_a/\rho)\mathrm dy^a$ and $\mathrm d\phi = \epsilon_{ab}y^a\mathrm dy^b/\rho^2$ with $\epsilon_{12}=1=-\epsilon_{21}$, so $$g_{ab} = \frac{y_ay_b}{\rho^2} + \frac{R^2\sin^2(\rho/R)}{\rho^4}\big(\rho^2\delta_{ab} - y_ay_b\big),$$ using $\epsilon_{ac}y^c\epsilon_{bd}y^d = \rho^2\delta_{ab}-y_ay_b$ in two dimensions. Now $R^2\sin^2(\rho/R)/\rho^2 = 1 - \rho^2/(3R^2) + O(\rho^4)$, from $\sin u = u - u^3/6+\dots$ squared. Substituting, $g_{ab} = \delta_{ab} - \frac{\rho^2\delta_{ab}-y_ay_b}{3R^2} + O(\rho^4)$, which is the boxed line. Component by component: $g_{11} = 1 - y_2^2/(3R^2)$, $g_{22} = 1 - y_1^2/(3R^2)$, $g_{12} = y_1y_2/(3R^2)$ — verified symbolically, and the pattern is the general one.

The general form, for later. In normal coordinates at $p$ on any manifold, $g_{ij}(\mathbf y) = \eta_{ij} - \tfrac13R_{ikjl}(p)\,y^ky^l + O(|\mathbf y|^3)$. Cited: Misner, Thorne and Wheeler, Gravitation, §11.6. The sphere's Riemann tensor is $R_{ikjl} = R^{-2}(g_{ij}g_{kl}-g_{il}g_{kj})$, and substituting it at $p$, where $g=\delta$, gives precisely $-\tfrac1{3R^2}(\delta_{ij}|\mathbf y|^2 - y_iy_j)$. So the computation above is a worked instance of the general expansion, one unit before the tensor in it is defined — which is the honest order, since the expansion is what motivates the definition.

32 · normal coordinates · example

What survives, plotted

$g_{11}$ along the line $y^1 = 0$, exact (solid) against the quadratic term of the expansion (dashed), with distance from the pole $u = |\mathbf y|/R$. Flat space would give the horizontal line. The deviation starts quadratically, with coefficient $1/3$.

A number. At $|\mathbf y| = R/10$ — a laboratory a tenth of a radius across — $g_{11}$ differs from 1 by at most $\tfrac13(1/10)^2 = 1/300 = 0.00333$. Unit 30's "how local is local" question, answered by an expansion instead of an estimate.

Where the chart ends. At $u = \pi$ the exact curve reaches $0$: the geodesic circle of radius $\pi R$ is the single point $S$, and the normal chart has run out — the injectivity radius of the exponential map.

How the curve was computed. On the line $y^1 = 0$, $y = (0,\rho)$, the exact formula in the previous slide's notes gives $g_{11} = R^2\sin^2(\rho/R)/\rho^2 = \sin^2u/u^2$: the ratio of the circumference of the geodesic circle of radius $\rho$, squared, to the flat one. At $u = 0.1$ it is $1 - 0.003329$, against $1 - 0.003333$ from the quadratic term; at $u = 1$, $0.708$ against $0.667$. The two agree to the plotting accuracy up to about $u = 0.5$.

The physical reading of the $1/300$. A freely falling laboratory of size $\ell$ on a manifold with curvature radius $\mathcal R$ sees metric deviations of order $\ell^2/\mathcal R^2$. Unit 30 estimated the same thing for a laboratory falling toward the Earth by comparing tidal accelerations; here it comes out of a Taylor series with a definite coefficient $1/3$. Unit 33 makes the link exact by putting the Riemann tensor into the geodesic deviation equation and setting it beside unit 05's Newtonian tidal tensor, term for term.

section 6

Free fall is a straight line

The geodesic equation in place of Newton's second law, the factor of two in $g_{00}$, and the rule for moving every other law onto a curved manifold.

32 · free fall · physics

The geodesic postulate, and why it is the only candidate

the geodesic postulate

A test body subject to no non-gravitational force follows a timelike geodesic, affinely parametrised by its proper time: $$\frac{\mathrm d^2x^\mu}{\mathrm d\tau^2} + \Gamma^\mu{}_{\nu\rho}\frac{\mathrm dx^\nu}{\mathrm d\tau}\frac{\mathrm dx^\rho}{\mathrm d\tau} = 0 .$$ Light follows a null geodesic, with an affine parameter that is not proper time.

This replaces unit 00's $m\ddot{\vec x} = \vec F$. Gravity is not on the right-hand side; it is in $\Gamma$ on the left.

Not new: a frictionless bead on the sphere has unit 12's $L = \tfrac12mR^2(\dot\theta^2+\sin^2\theta\,\dot\varphi^2)$, $m$ times the geodesic Lagrangian, so it runs on great circles.

why a connection, and not something else

Unit 30's weak equivalence principle says: the trajectory of a freely falling body is determined by its initial event and initial four-velocity, with no dependence on mass, composition or internal structure. A law of that shape is a second-order ODE $\ddot x^\mu = F^\mu(x,\dot x)$ with universal $F$.

Assume the admissible parameters are closed under $\lambda\mapsto a\lambda$ (no preferred clock rate). Then $F$ is quadratic in $\dot x$, $F^\mu = -C^\mu{}_{\nu\rho}\dot x^\nu\dot x^\rho$, and chart-independence forces $C$ to transform exactly as a connection — both proved in the notes. The equivalence principle is the statement that gravity is a connection.

Checks. With $x^0 = ct$, $[\Gamma] = 1/\text{length}$ and $[\mathrm dx/\mathrm d\tau] = \text{length}/\text{time}$, so both terms are accelerations ✓. Inertial chart on Minkowski space: $\Gamma = 0$, $\ddot x^\mu = 0$, unit 27's free particle ✓.

Full proof, step 1: quadratic. If $x(\lambda)$ solves $\ddot x = F(x,\dot x)$ and $a\ne0$, then $y(s) := x(as)$ has $\ddot y = a^2\ddot x = a^2F(x,\dot x)$, while the law applied to $y$ demands $\ddot y = F(y,\dot y) = F(x,a\dot x)$. So $F(x,av) = a^2F(x,v)$ for every $a\ne0$, and by continuity for $a = 0$. Differentiate twice in $a$ at $a=0$: $D_v^2F(x,0)(v,v) = 2F(x,v)$. So $F(x,\cdot)$ is the quadratic form $v\mapsto\tfrac12D_v^2F(x,0)(v,v)$, and its symmetric coefficient array is $-C^\mu{}_{\nu\rho}(x)$. $\square$

Full proof, step 2: the transformation law. Let the law hold in chart $x$ and ask what it says in chart $\tilde x$. With $\dot{\tilde x}^a = A^a{}_i\dot x^i$, $$\ddot{\tilde x}^a = A^a{}_i\ddot x^i + \frac{\partial^2\tilde x^a}{\partial x^i\partial x^j}\dot x^i\dot x^j = -\Big(A^a{}_kC^k{}_{ij} - \frac{\partial^2\tilde x^a}{\partial x^i\partial x^j}\Big)S^i{}_bS^j{}_c\,\dot{\tilde x}^b\dot{\tilde x}^c .$$ For this to read $\ddot{\tilde x}^a = -\tilde C^a{}_{bc}\dot{\tilde x}^b\dot{\tilde x}^c$ for every initial velocity, polarisation (the local-flatness slide's notes) forces $\tilde C^a{}_{bc} = A^a{}_kS^i{}_bS^j{}_cC^k{}_{ij} - S^i{}_bS^j{}_c\,\partial^2\tilde x^a/\partial x^i\partial x^j$. Differentiating $A^a{}_kS^k{}_c = \delta^a{}_c$ along $\tilde x^b$ turns the last term into $+A^a{}_k\,\partial^2x^k/\partial\tilde x^b\partial\tilde x^c$, and the result is the transformation law of a symmetric connection, term for term. $\square$

How much of the argument is a proof. The step from "trajectory depends only on $(x,\dot x)$" to "second-order ODE" is a definition unpacked. The step to "quadratic" is where an assumption hides: closure of the preferred parameters under rescaling, which says a freely falling clock has no preferred rate. A second assumption hides in "differentiate twice at $a = 0$": smoothness of $F$ at zero velocity. Keep the rescaling for $a\gt0$ only and drop that smoothness, and the laws allowed are the geodesic sprays of Finsler geometry, homogeneous of degree two without being quadratic. The honest summary is: the equivalence principle plus a mild regularity demand forces a connection, and general relativity then takes that connection to be the Levi-Civita connection of a metric, which is a further assumption — the one unit 35 will cash out.

Why the Levi-Civita connection specifically. Two further physical demands do it. Metric compatibility says that parallel transport preserves the invariant interval — so a rod carried around does not change length relative to local standards, and unit 27's $u\cdot u=-c^2$ is preserved along a worldline. Torsion-freeness says that free-fall trajectories are the only structure gravity has; a torsion field would be an independent field with its own source, which is Einstein–Cartan theory. The fundamental theorem then says there is exactly one such connection. Nothing is left free.

"Test body" is doing work. The body must be light enough not to disturb the metric, and small enough for tidal couplings to its own structure to be negligible. Neither is automatic — the slide "Geodesic motion is a theorem" says what happens when the first is relaxed, and the Mathisson–Papapetrou–Dixon equations say what happens when the second is. A neutron star in a tight binary is not a test body; a satellite in Earth orbit very nearly is.

32 · free fall · physics

Newton recovered, and the factor of two

the Newtonian limit

Take a static, weak field with $$g_{00} = -\Big(1+\frac{2\Phi}{c^2}\Big),\qquad g_{0i}=0,\qquad g_{ij}=\delta_{ij},$$ and slow motion, $|\mathrm dx^i/\mathrm dt|\ll c$. The geodesic equation becomes $$\frac{\mathrm d^2x^i}{\mathrm dt^2} = -\partial_i\Phi ,$$ — unit 05's Newtonian gravity. Checks: $\Phi/c^2$ is dimensionless, so $[\Phi] = \mathrm{m^2s^{-2}}$ and $\partial_i\Phi$ is an acceleration ✓; $\Phi\to0$ gives $\ddot x^i = 0$, free motion ✓.

proof, four steps

1. Slow motion: $\mathrm dx^i/\mathrm d\tau \ll \mathrm dx^0/\mathrm d\tau$, so in $\Gamma^\mu{}_{\nu\rho}\dot x^\nu\dot x^\rho$ only $\nu=\rho=0$ survives.
2. Static and $g_{ij}=\delta_{ij}$: $\ \Gamma^i{}_{00} = \tfrac12g^{ij}(2\partial_0g_{j0}-\partial_jg_{00}) = -\tfrac12\partial_ig_{00}$.
3. Substitute $g_{00}$: $\ -\tfrac12\partial_i\big[-(1+2\Phi/c^2)\big] = +\partial_i\Phi/c^2$.
4. $\mathrm d^2x^i/\mathrm d\tau^2 = -(\partial_i\Phi/c^2)(c\,\mathrm dt/\mathrm d\tau)^2$; divide by $(\mathrm dt/\mathrm d\tau)^2$ (notes). $\square$

where the 2 comes from, and what it costs

Step 2 supplies a $\tfrac12$; step 3 must cancel it. With $g_{00}=-(1+\Phi/c^2)$ bodies fall at $4.9\ \mathrm{m\,s^{-2}}$.

an independent determination of the same 2

A static clock has $\mathrm d\tau = \sqrt{-g_{00}}\,\mathrm dt = \sqrt{1+2\Phi/c^2}\ \mathrm dt \approx (1+\Phi/c^2)\,\mathrm dt$. That is unit 30's gravitational redshift $\Delta\nu/\nu = \Delta\Phi/c^2$, derived there from the equivalence principle alone with no metric at all.

Two routes, one coefficient: the root eats one 2, the Christoffel $\tfrac12$ the other.

Full check of step 2. $\Gamma^i{}_{00} = \tfrac12g^{i\lambda}(\partial_0g_{\lambda0} + \partial_0g_{\lambda0} - \partial_\lambda g_{00})$. Static kills the two $\partial_0$ terms. $g^{i\lambda}$ is diagonal with $g^{ij}=\delta^{ij}$ and $g^{i0}=0$, so only $\lambda = j$ spatial contributes: $-\tfrac12\delta^{ij}\partial_jg_{00}$. ✓ For the time component, $\Gamma^0{}_{00} = \tfrac12g^{00}\partial_0g_{00} = 0$ by staticity, and $\Gamma^0{}_{0i} = \tfrac12g^{00}\partial_ig_{00} = \partial_i\Phi/c^2 + O(\Phi^2/c^4)$. So $\mathrm d^2x^0/\mathrm d\tau^2 = -2\Gamma^0{}_{0i}\,\dot x^0\dot x^i$, which is smaller than the spatial equation's right-hand side by a factor of order $v/c$. Converting $\mathrm d^2x^i/\mathrm d\tau^2$ to $\mathrm d^2x^i/\mathrm dt^2$ exactly gives $\mathrm d^2x^i/\mathrm dt^2 = (\mathrm d\tau/\mathrm dt)^2\,\mathrm d^2x^i/\mathrm d\tau^2 - (\mathrm dx^i/\mathrm dt)(\mathrm d\tau/\mathrm dt)^2\,\mathrm d^2t/\mathrm d\tau^2$, and the second term is $2(\vec v\cdot\nabla\Phi)v^i/c^2$: of relative size $v^2/c^2$, dropped at this order. That is what lets step 4 divide by $(\mathrm dt/\mathrm d\tau)^2$. Problem P12 puts a number on it.

Numbers. At the Earth's surface $\Phi/c^2 = -GM_\oplus/(R_\oplus c^2) = -6.9535\times10^{-10}$, so $2\Phi/c^2 = -1.3907\times10^{-9}$ and $g_{00} = -(1 - 1.3907\times10^{-9})$: the metric differs from Minkowski's in the ninth decimal place, and that ninth decimal place is what holds a person to the floor. The gradient does the work, not the size: $\partial_r\Phi = GM_\oplus/R_\oplus^2 = 9.7983\ \mathrm{m\,s^{-2}}$ — the measured $9.780$ at the equator differs by the centrifugal $\omega^2R = 0.0339\ \mathrm{m\,s^{-2}}$ of unit 03 and by the Earth's oblateness. At the Sun's surface $\Phi/c^2 = -2.1225\times10^{-6}$, which shifts solar spectral lines by the equivalent of $636.3\ \mathrm{m\,s^{-1}}$ of Doppler — measured, and one of the three classical tests.

Consistency with unit 30's GPS budget. $\mathrm d\tau/\mathrm dt = 1+\Phi/c^2$ at radius $r$; the difference between the ground ($r = R_\oplus$) and a satellite at $a = 26\,561$ km is $\Delta\Phi/c^2 = 5.2837\times10^{-10}$, which over a day is $45.65\ \mu$s — exactly the potential term unit 30 computed from the equivalence principle without a field equation. Nothing has changed; the metric has merely been written down.

What $g_{ij}=\delta_{ij}$ is not good enough for. Slow bodies never probe the spatial metric, because $\dot x^i\dot x^j$ terms are suppressed by $(v/c)^2$. Light does: it moves at $|\mathrm dx^i/\mathrm dt| = c$, and the spatial part contributes at the same order as the temporal. The correct weak-field metric is $g_{ij} = (1-2\Phi/c^2)\delta_{ij}$, and its spatial part supplies exactly the missing half of light bending — which is why unit 30's equivalence-principle argument got half the right answer and said so. Unit 37 computes the whole thing.

32 · free fall

Minimal coupling, and the ambiguity nobody should hide

minimal coupling

To move a special-relativistic law onto a curved manifold: replace $\eta_{\mu\nu}\to g_{\mu\nu}$, $\ \partial_\mu\to\nabla_\mu$, and $\mathrm d^4x\to\sqrt{-g}\,\mathrm d^4x$. The comma-goes-to-semicolon rule.

it is a guess, and it is ambiguous

Ordering. $\partial_\mu\partial_\nu V^\rho = \partial_\nu\partial_\mu V^\rho$, but $\nabla_\mu\nabla_\nu V^\rho \ne \nabla_\nu\nabla_\mu V^\rho$: one flat law becomes two curved ones, differing by curvature (unit 33).

Invisible terms. A term like $\xi R\phi^2$, with $R$ unit 33's scalar curvature, vanishes in flat space whatever $\xi$ is, so the rule cannot see it. Conformal invariance in four dimensions wants $\xi = 1/6$ (cited: Wald, General Relativity, App. D), and no flat-space experiment says so.

one law that needs no rule at all

For any antisymmetric $F_{\mu\nu}$ and any torsion-free connection, $$\nabla_{[\mu}F_{\nu\rho]} = \partial_{[\mu}F_{\nu\rho]} ,$$ where $[\mu\nu\rho]$ antisymmetrises over all three indices; for antisymmetric $F$ it is a third of the cyclic sum, $\partial_{[\mu}F_{\nu\rho]} = \tfrac13(\partial_\mu F_{\nu\rho} + \partial_\nu F_{\rho\mu} + \partial_\rho F_{\mu\nu})$.

Proof. In the cyclic sum the six $\Gamma$ terms pair off, e.g. $-\Gamma^\lambda{}_{\mu\nu}F_{\lambda\rho}-\Gamma^\lambda{}_{\nu\mu}F_{\rho\lambda} = 0$ by symmetry of $\Gamma$ and antisymmetry of $F$. $\square$

An aside, since unit 28 is not among this unit's needs: Maxwell's homogeneous pair is exactly $\partial_{[\mu}F_{\nu\rho]}=0$ for unit 28's field tensor, so it transfers unaltered. Unit 34 writes it $\mathrm dF = 0$: the exterior derivative never needed a connection.

Full proof of the cancellation, written out. Using $\nabla_\mu F_{\nu\rho} = \partial_\mu F_{\nu\rho} - \Gamma^\lambda{}_{\mu\nu}F_{\lambda\rho} - \Gamma^\lambda{}_{\mu\rho}F_{\nu\lambda}$ and cycling $\mu\to\nu\to\rho\to\mu$, the connection terms are $$-\Gamma^\lambda{}_{\mu\nu}F_{\lambda\rho} - \Gamma^\lambda{}_{\mu\rho}F_{\nu\lambda} -\Gamma^\lambda{}_{\nu\rho}F_{\lambda\mu} - \Gamma^\lambda{}_{\nu\mu}F_{\rho\lambda} -\Gamma^\lambda{}_{\rho\mu}F_{\lambda\nu} - \Gamma^\lambda{}_{\rho\nu}F_{\mu\lambda}.$$ Pair the 1st with the 4th: $\Gamma^\lambda{}_{\mu\nu} = \Gamma^\lambda{}_{\nu\mu}$ (torsion-free) and $F_{\lambda\rho} = -F_{\rho\lambda}$ (antisymmetry), so they cancel. Pair the 2nd with the 5th, and the 3rd with the 6th, identically. All six vanish. $\square$ Note what the proof used: symmetry of $\Gamma$ and antisymmetry of $F$, and nothing else — not metric compatibility, not even a metric. That is why the statement belongs to the exterior derivative rather than to the connection.

The divergence identity, which the rule makes constant use of. $\nabla_\mu V^\mu = \partial_\mu V^\mu + \Gamma^\mu{}_{\mu\nu}V^\nu$, and $\Gamma^\mu{}_{\mu\nu} = \tfrac12g^{\mu\lambda}\partial_\nu g_{\mu\lambda}$ (the other two terms of the boxed formula cancel on contraction). Jacobi's formula for a nonsingular matrix $A(\nu)$ says $\partial_\nu\det A = \det A\ \mathrm{tr}(A^{-1}\partial_\nu A)$ — proof: expand $\det$ by cofactors, $\partial\det A/\partial A_{ij} = \mathrm{cof}_{ij} = \det A\,(A^{-1})_{ji}$, then chain rule. Applied to $g$: $\partial_\nu g = g\,g^{\mu\lambda}\partial_\nu g_{\mu\lambda}$, so $\Gamma^\mu{}_{\mu\nu} = \partial_\nu\ln\sqrt{|g|}$ and $$\nabla_\mu V^\mu = \frac{1}{\sqrt{|g|}}\,\partial_\mu\big(\sqrt{|g|}\,V^\mu\big).$$ Check: flat plane in polar coordinates, $\sqrt{|g|} = r$, giving $r^{-1}\partial_r(rV^r) + \partial_\theta V^\theta$ — unit 05's divergence, with the $1/r$ that used to have to be remembered ✓.

The honest status of the rule. Minimal coupling is a choice, defensible on grounds of simplicity and consistent with every experiment so far, not a theorem. The strong equivalence principle is the assertion that it is right; the ordering ambiguity is real, and the two orderings differ by terms of order $R\times(\text{field})$, i.e. by curvature couplings that no local experiment in a freely falling frame can detect. Which is exactly the statement that the equivalence principle cannot fix them.

32 · free fall

Geodesic motion is a theorem, not a postulate

for dust, in three lines

Let $T^{\mu\nu} = \rho\,u^\mu u^\nu$ with $u\cdot u = -c^2$, and suppose $\nabla_\mu T^{\mu\nu} = 0$. Then the integral curves of $u$ are geodesics.

proof

$0 = \nabla_\mu(\rho u^\mu u^\nu) = \big[\nabla_\mu(\rho u^\mu)\big]u^\nu + \rho\,u^\mu\nabla_\mu u^\nu$.

Contract with $u_\nu$. Since $u_\nu u^\nu = -c^2$ is constant, $u_\nu\nabla_\mu u^\nu = \tfrac12\nabla_\mu(u\cdot u) = 0$, so the second term dies and $$-c^2\,\nabla_\mu(\rho u^\mu) = 0 .$$ That is the continuity equation. Feeding it back leaves $\rho\,u^\mu\nabla_\mu u^\nu = 0$, and where $\rho\ne0$ that is the geodesic equation. $\square$

why this matters structurally

In Newton's theory the field equation and the law of motion are separate postulates. Here unit 35's field equations force $\nabla_\mu T^{\mu\nu}=0$ (by unit 33's Bianchi identity), which already contains the motion.

what the general statement needs

For a small body with structure the result holds but the proof is hard: cited, Geroch and Jang (1975), assuming the dominant energy condition. Spinning or tidally sensitive bodies deviate (Mathisson–Papapetrou–Dixon, cited).

What the dust proof assumes. That the matter is pressureless — no internal forces at all — and that $\rho\ne0$ along the curve in question. Pressure changes the conclusion in exactly the way Newtonian intuition expects: for a perfect fluid $T^{\mu\nu} = (\rho + p/c^2)u^\mu u^\nu + p\,g^{\mu\nu}$, the same contraction gives $(\rho+p/c^2)u^\mu\nabla_\mu u^\nu = -\big(g^{\mu\nu}+u^\mu u^\nu/c^2\big)\partial_\mu p$: a fluid element is pushed off its geodesic by the pressure gradient projected orthogonally to its own motion. A person standing on the floor is exactly that: not freely falling, because the floor's pressure gradient is holding them off a geodesic. Unit 39 uses this equation for cosmology.

The history is worth a paragraph. Einstein's 1915 paper stated the geodesic law as a separate postulate alongside the field equations. In 1927 Einstein and Grommer gave a first argument that it was redundant, and the 1938 Einstein–Infeld–Hoffmann paper (Annals of Mathematics 39, 65) derived the motion of slowly moving bodies from the vacuum equations alone, by surrounding each body with a surface and using the field equations' integrability. The result is regarded as one of the structural peculiarities of the theory: no other classical field theory determines the motion of its sources. Maxwell's equations do not tell a charge how to move; the Lorentz force is an extra law.

The citations. R. Geroch and P. S. Jang, "Motion of a body in general relativity", J. Math. Phys. 16 (1975) 65: a conserved stress tensor satisfying the dominant energy condition, supported in arbitrarily thin world-tubes about a curve, forces the curve to be a timelike geodesic. The Mathisson–Papapetrou–Dixon equations for extended and spinning bodies are set out in W. G. Dixon, Proc. R. Soc. A 314 (1970) 499.

Why the Bianchi identity is the real reason. Unit 33 proves the contracted Bianchi identity $\nabla_\mu G^{\mu\nu} = 0$ for the Einstein tensor, as an identity — true of any metric whatsoever. So if the field equations read $G^{\mu\nu} = \kappa T^{\mu\nu}$, then $\nabla_\mu T^{\mu\nu}=0$ follows whether one wants it or not, and by the theorem above the sources move on geodesics whether one postulates it or not. The postulate on the geodesic slide is therefore best read as a provisional statement, made three units before the machinery that makes it unnecessary.

32 · closing

The sphere, closed

what	on $S^2_R$
metric	$R^2(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\varphi^2)$
volume element	$R^2\sin\theta$, total $4\pi R^2$
an angle at $\theta_0$	$\arccos(2/\sqrt7) = 40.893^\circ$
Christoffels	$-\sin\theta\cos\theta$ and $\cot\theta$; two of six
transport round $\theta_0$	rotation by $2\pi\cos\theta_0 = \pi$
enclosed area $/R^2$	$2\pi(1-\cos\theta_0) = \pi$ — the same angle
geodesics	great circles, derived twice
$\theta_0$ is not one	fails by $-\sqrt3/4$
normal coordinates	$\delta_{ab} - \frac{1}{3R^2}(\delta_{ab}\|\mathbf y\|^2 - y_ay_b)$

what it was a rehearsal for

Every line of the table has a Schwarzschild counterpart in unit 36, computed the same way with the same tools. The sphere is the smallest space in which none of the answers is trivial and all of them can be checked by hand.

the one number that did not go away

$1/R^2$. It is what survived the normal-coordinate expansion, what the transport angle measured, and what the counting argument said a surface must keep. Unit 33 gives it a name, extends it to $20$ numbers in four dimensions, and sets it beside unit 05's tidal tensor.

32 · checkpoint

Checkpoint

you can now do all four

For a given metric, compute every Christoffel symbol by the geodesic-Lagrangian trick and verify the answer against the coordinate formula.
Prove that the Levi-Civita connection is the unique torsion-free metric-compatible connection.
Construct normal coordinates at a point and state exactly what can and cannot be made to vanish there.
Recover Newton's law from the geodesic equation, with the correct factor in $g_{00}$.

the unit in one line

A metric fixes one connection; that connection says what straight means; and the straight lines are the worldlines of bodies nobody is pushing.

Unit 33 finds the part of $\Gamma$ that no chart can remove — the twenty numbers the counting argument left over — and sets the result beside unit 05's tidal tensor, term for term.

Thirteen problems, about six hours: ./mr problems 32. When the four items above are true: ./mr done 32.

The falsifiable form. Take the metric $\mathrm ds^2 = (\mathrm dx^2+\mathrm dy^2)/y^2$ on $y\gt0$ (problem P7) and $\mathrm ds^2 = \mathrm dr^2/(1-kr^2) + r^2(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\varphi^2)$ (problem P3). For each: write the geodesic Lagrangian, take its Euler–Lagrange equations, read off every $\Gamma^k{}_{ij}$, verify two of them against $\tfrac12g^{kl}(\partial_ig_{jl}+\partial_jg_{il}-\partial_lg_{ij})$ and one index triple of metric compatibility. Then, with the notes closed, write out the Koszul derivation of uniqueness; build normal coordinates at the origin of $\mathrm d\rho^2 + f(\rho)^2\mathrm d\phi^2$ (problem P11) and say which of $g$, $\partial g$, $\partial\partial g$ you have made trivial and why the last cannot be. If any of that will not come, the checkpoint is false and unit 33 will be a list of symbols rather than a set of statements.

A second test, for item 4: write down the weak-field $g_{00}$, derive $\mathrm d^2x^i/\mathrm dt^2 = -\partial_i\Phi$, and say in one sentence which step would halve the answer if the 2 were dropped, and which independent measurement — Pound and Rebka's, unit 30 — fixes the 2 without reference to the geodesic equation at all.