The book / chapter 04
CHAPTER 04

Manifolds, maps, and metrics

Coordinates locate events. The metric tells us how to measure the separations between them.

3 worked examples in this chapter
Before you begin
THE QUESTION

Why can a position-dependent metric describe a perfectly flat plane?

BRING WITH YOU

By the end: Read a metric, invert it, and calculate the size of a coordinate cell.

4.1 A manifold is a place where local coordinates work#

Imagine finding a place on Earth using latitude and longitude. Two numbers locate a point on the surface, even though we draw the globe in three-dimensional space. The point needs two independent coordinates because we can move along the surface in two independent directions.

The labels have limits. At the north pole, all longitudes meet. A small map centered on that pole can instead use two directions across the map. The place is ordinary; our first set of labels was unsuitable there.

A manifold is a space that can be described this way: each point has a surrounding region with an ordinary coordinate list. The number of entries is its dimension. A sphere’s surface is two-dimensional; spacetime is four-dimensional, with one time coordinate and three space coordinates. We use a collection of overlapping maps when one set of coordinates cannot cover everything.

A coordinate chart is the rule assigning a coordinate list to each point in one such region. It must work both ways: each point has one list, and each list in the chart’s range identifies one point. Nearby points must have nearby lists, and vice versa. Symbolically,

x:UMx(U)Rn.x:U\subset M\longrightarrow x(U)\subset\mathbb R^n.

Here MM is the manifold, UU is the region covered by this chart, and x(U)x(U) is its set of coordinate lists. The notation Rn\mathbb R^n means lists of nn real numbers. In this formula xx names the whole chart, rather than just one horizontal coordinate.

An atlas is a collection of charts covering the space. Where two charts overlap, we can translate between their labels. Starting with one list, use the first chart’s inverse to find the point, then the second chart to label it:

yx1:x(UV)y(UV).y\circ x^{-1}:x(U\cap V)\longrightarrow y(U\cap V).

The symbol \cap means the common region of the two patches; \circ means composition, applying the right-hand map first. This translation is a transition map. On a smooth manifold, the transition maps and their inverses can be differentiated repeatedly, with continuous derivatives of every order. That is what lets us use calculus consistently on overlapping maps.

A complete two-chart example. Describe the unit sphere temporarily by X2+Y2+Z2=1X^2+Y^2+Z^2=1 in ordinary three-dimensional space. Capital letters are only a convenient construction aid. The surface itself has two independent coordinates. Let N=(0,0,1)N=(0,0,1) and S=(0,0,1)S=(0,0,-1) be its poles.

On the sphere with NN removed, use the chart

(u,v)=(X1Z,Y1Z).(u,v)=\left(\frac{X}{1-Z},\frac{Y}{1-Z}\right).

Every finite pair (u,v)(u,v) corresponds to exactly one point of that patch. To see this without trusting a picture, put s=u2+v2s=u^2+v^2 and write the inverse:

(X,Y,Z)=(2u1+s,2v1+s,s11+s).(X,Y,Z)=\left(\frac{2u}{1+s},\frac{2v}{1+s},\frac{s-1}{1+s}\right).

The squared components sum to one, and substituting them back into the chart returns uu and vv. The denominator 1+s1+s never vanishes. The north pole is approached only as the coordinate radius becomes unbounded; the south pole is the perfectly ordinary coordinate pair (0,0)(0,0).

The second chart removes SS instead:

(p,q)=(X1+Z,Y1+Z).(p,q)=\left(\frac{X}{1+Z},\frac{Y}{1+Z}\right).

It covers the missing north pole, where (p,q)=(0,0)(p,q)=(0,0). On the overlap, substitute the first chart’s inverse into the second chart. Since 1+Z=2s/(1+s)1+Z=2s/(1+s), the transition is

(p,q)=(uu2+v2,vu2+v2),(u,v)(0,0).(p,q)=\left(\frac{u}{u^2+v^2},\frac{v}{u^2+v^2}\right), \qquad (u,v)\ne(0,0).

The excluded origin represents SS, which is outside the second chart. The transition applies only where both charts cover the point. Applying the same formula to (p,q)(p,q) returns (u,v)(u,v), so the transition has a smooth inverse everywhere on the overlap. Its Jacobian has determinant

det(p,q)(u,v)=1(u2+v2)20.\det\frac{\partial(p,q)}{\partial(u,v)} =-\frac{1}{(u^2+v^2)^2}\ne0.

The minus sign reverses the orientation of these coordinate lists; the nonzero value says no infinitesimal direction has been collapsed. Two overlapping charts therefore cover the whole sphere, even though neither chart does so alone. These coordinate lists locate points. To calculate distances between them, we still need a measuring rule.

As a numerical check, the first coordinates (u,v)=(2,1)(u,v)=(2,1) locate (X,Y,Z)=(2/3,1/3,2/3)(X,Y,Z)=(2/3,1/3,2/3) and give second coordinates (p,q)=(2/5,1/5)(p,q)=(2/5,1/5). Those are two addresses for one point. Chapter 2’s Jacobian rule tells us how a tangent’s components change between them; §4.3 will make the tangent itself precise.

A small globe and a globe twice its size can use the same latitude and longitude labels. The distance between two given labels doubles. This separates the two ingredients we need: the manifold and its charts locate points; a metric supplies the measuring rule.

We used an outside picture to construct the sphere’s charts, but the final coordinate translations work without it. In the same way, describing four-dimensional spacetime does not require an additional physical space surrounding it.

The precise topological conditions

Charts cover open regions: around each point there is a smaller patch still inside the region. Openness on a surface is measured within the surface; a patch need not contain a three-dimensional ball.

A chart and its inverse must be continuous. The general definition says that the preimage of every open set is open. A preimage consists of all starting points mapped into the chosen set. This defines continuity before any particular distance formula is chosen. Continuity alone does not imply differentiability: the function f(x)=xf(x)=|x| is continuous but has a corner at zero.

The usual manifold definition includes two further conditions. Hausdorff separation means that two distinct points have disjoint surrounding neighborhoods. On a sphere, sufficiently small separate disks around any two distinct points illustrate this condition. Second countability means that a countable collection of basic open regions suffices to build all open regions by unions. In the plane, rectangles with rational-number corners form such a collection. These conditions exclude spaces with pathological global behavior even when their small patches resemble ordinary coordinate space.

The sphere and spacetime models used here satisfy these conditions. The calculations below use their charts and smooth transitions; no further topology theorem is needed to follow them.

Explore a manifold

A curved world. A flat address.

Move one point between a surface and its map. Then change the map while the point stays put.

Drag to rotate
The flat map(u,v)(u,v)

Drag inside the solid circle to move the point. The dashed circle marks the other patch.

Read the model and its limits

A surface is the curved object. A chart assigns two numbers to points in one open patch. Here the map FAF_A takes a pair of numbers back to the surface:

FA(u,v)=(u0.55, v, h(u0.55,v))F_A(u,v)=\bigl(u-0.55,\ v,\ h(u-0.55,v)\bigr)

The inverse chart simply reads (u,v)=(x+0.55,y)(u,v)=(x+0.55,y). The height is determined by those two values; it is not a third independent coordinate. Our smooth example is

h(x,y)=0.38sin(1.3x)cos(1.15y)+0.16(x2y2)h(x,y)=0.38\sin(1.3x)\cos(1.15y)+0.16(x^2-y^2)

Chart B uses a shifted origin and axes rotated through π/5\pi/5. On the overlap, the change of coordinates is an ordinary invertible rotation and translation:

(uv)=(cosαsinαsinαcosα)[(uv)(1.10.15)],α=π5\begin{pmatrix}u'\\v'\end{pmatrix}=\begin{pmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{pmatrix}\left[\begin{pmatrix}u\\v\end{pmatrix}-\begin{pmatrix}1.1\\0.15\end{pmatrix}\right],\quad\alpha=\frac\pi5

Each colored disk represents an open coordinate domain u2+v2<1.352u^2+v^2<1.35^2; the rim marks its limiting boundary. The two derivatives of FAF_A are independent because their first two components are (1,0)(1,0) and (0,1)(0,1). That is why two coordinates work smoothly even where the surface bends.

This particular graph also admits a global projection chart. We deliberately restrict two charts to explain overlap; we are not claiming every manifold has a global chart. The faded outer rim is a rendering window, not an edge of the mathematical surface. Coordinates and heights are dimensionless. This is a two-dimensional example embedded in Euclidean three-space, not a depiction of spacetime.

Background: Gualtieri’s differential-geometry notes, coordinate charts and smooth maps.

WORKED EXAMPLE

Give one sphere two reliable addresses

What should a map do when the place you want to visit is missing from it?

See the idea

A chart is an invertible assignment of coordinates on an open patch. It need not cover the whole surface. On a unit sphere, two stereographic charts complement one another: the (u,v)(u,v) chart omits the north pole and the (p,q)(p,q) chart omits the south pole. Move the point and watch both addresses. A coordinate becoming large is a warning about that map, not evidence that the sphere is breaking.

Work it out
  1. Build an address by projecting from a pole

    Write the sphere as X2+Y2+Z2=1X^2+Y^2+Z^2=1. A line from N=(0,0,1)N=(0,0,1) through a surface point meets the plane Z=0Z=0 at (u,v,0)(u,v,0). All quantities here are dimensionless.

    u=X1Z,v=Y1Z.u=\frac{X}{1-Z},\qquad v=\frac{Y}{1-Z}.

    Why this step works Setting the height of the projection line to zero fixes the scale factor between the sphere coordinates and the plane coordinates.

  2. Recover the surface point before changing its address

    Let s=u2+v2s=u^2+v^2. The inverse below turns the sample address (2,0)(2,0) into (4/5,0,3/5)(4/5,0,3/5). It also shows why every finite address gives a point away from the north pole.

    (X,Y,Z)=(2u1+s,2v1+s,s11+s).(X,Y,Z)=\left(\frac{2u}{1+s},\frac{2v}{1+s},\frac{s-1}{1+s}\right).

    Why this step works Substitute the projection formulas into the sphere equation and select the solution away from the projection pole.

  3. Read the other chart at that same point

    Project from the south pole instead: p=X/(1+Z)p=X/(1+Z) and q=Y/(1+Z)q=Y/(1+Z). Substituting the inverse map gives the transition. The sample point has (p,q)=(1/2,0)(p,q)=(1/2,0).

    (p,q)=(u,v)u2+v2,(u,v)(0,0).(p,q)=\frac{(u,v)}{u^2+v^2},\qquad (u,v)\ne(0,0).

    Why this step works The transition first recovers the sphere point and then reads the second address. Its excluded origin represents the south pole, outside the second chart.

  4. Check that a direction survives the change

    Differentiate the transition at (u,v)=(2,0)(u,v)=(2,0). A tangent rate (1,2)(1,2) becomes (1/4,1/2)(-1/4,1/2). Negative entries can reverse coordinate directions; they do not tear or rotate the sphere.

    (p,q)(u,v)(2,0)=(1/4001/4).\left.\frac{\partial(p,q)}{\partial(u,v)}\right|_{(2,0)}=\begin{pmatrix}-1/4&0\\0&1/4\end{pmatrix}.

    Why this step works The Jacobian has nonzero determinant, so it does not collapse any infinitesimal direction.

Go deeper

The first chart maps S2{N}S^2\setminus\{N\} onto R2\mathbb R^2 and the second maps S2{S}S^2\setminus\{S\} onto R2\mathbb R^2. Their overlap corresponds to R2{(0,0)}\mathbb R^2\setminus\{(0,0)\}. The transition is its own inverse and has determinant 1/(u2+v2)2-1/(u^2+v^2)^2, which never vanishes on that overlap. This proves smooth compatibility of these two charts and constructs an atlas covering the sphere. We have not yet assigned distances to coordinate differences. This two-dimensional sphere embedded in three-dimensional space is a convenient construction here; the definition of a manifold does not require an embedding.

A STATE YOU CAN CHECK

On the unit sphere, the point (0.8, 0, 0.6) has two regular addresses. Neither chart changes the point.

North-pole projection
(u,v)=(2,0)(u,v)=(2,0)
South-pole projection
(p,q)=(0.5,0)(p,q)=(0.5,0)

The worked steps explain these measurements. Interactive controls appear when available.

Test the idea

FIRST, PREDICT

A point approaches the north pole while its (u,v)(u,v) coordinates grow without bound. What happens at the pole itself?

Compare the reasoning

The north pole is an infinitely distant physical point.

We have discussed coordinate labels, not a distance measurement. The ordinary sphere is smooth at its north pole.

Both charts fail because any pole is a surface singularity.

The omitted poles are different. Each chart covers the point omitted by the other.

The (u,v)(u,v) chart has no address there; the (p,q)(p,q) chart assigns (0,0)(0,0).

Exactly. The second chart supplies a regular neighborhood of the point that the first chart excludes.

A hint

Insert X=Y=0X=Y=0, Z=1Z=1 directly into the second chart.

NOW CHANGE THE EXAMPLE

At (u,v)=(0,2)(u,v)=(0,2), a path has tangent rates (du/dλ,dv/dλ)=(3,4)(du/d\lambda,dv/d\lambda)=(3,4). What is dp/dλdp/d\lambda for p=u/(u2+v2)p=u/(u^2+v^2)?

A hint

Differentiate the quotient along the path. Terms multiplied by uu vanish at this point.

Work through the solution

At (0,2)(0,2), p/u=1/4\partial p/\partial u=1/4 and p/v=0\partial p/\partial v=0. Therefore dp/dλ=(1/4)(3)+0(4)=3/4dp/d\lambda=(1/4)(3)+0(4)=3/4. Transform the tangent locally; do not put its component list into the nonlinear point map.

An atlas covers a surface with compatible local maps, and a map can fail where the surface remains perfectly regular.

4.2 Where coordinates fail#

On an ordinary flat plane, polar coordinates describe a point by its distance rr from the origin and the angle θ\theta from the positive xx axis. We measure the angle in radians: arc length divided by radius, so a full turn is 2π2\pi. Before calculating, try changing the distance and angle separately. A change in distance takes you along a spoke; a change in angle takes you around a circle.

GEOMETRY LAB / 01

One point. Two addresses.

Walk outward from O, then turn. Your distance and angle locate the same point that the horizontal and vertical coordinates do.

x [m]y [m] -2-2-1-11122 rθO P Square grid: x and y · circles and spokes: r and θ
Horizontal position x1.43 m
Vertical position y1.09 m

Move P, or use the sliders. The square grid and the polar grid give two addresses for the same point. A full turn is 2π radians; it brings P back to the same place.

From the picture to a distance

An angle in radians is defined by arc length divided by radius. Therefore a turn through Δθ\Delta\theta sweeps an arc of length rΔθr\,\Delta\theta. The teal arc is part of a circle; the blue chord joins its endpoints directly. For a finite turn, the chord is shorter.

Cut the isosceles triangle OPQ in half. Each half has hypotenuse rr and angle Δθ/2\Delta\theta/2, so the whole chord is 2rsin(Δθ/2)2r\sin(\Delta\theta/2). For a very small angle, sine approaches its argument, and both lengths approach rΔθr\,\Delta\theta. This is the angular length that will appear in the polar metric.

All lengths here are measured in metres. Angles run counterclockwise from the positive horizontal axis. Both grids describe an ordinary flat plane; curved coordinate lines do not make the plane curved.

The horizontal and vertical sides of the triangle from the origin to P are rcosθr\cos\theta and rsinθr\sin\theta. Thus the same point has Cartesian coordinates

x=rcosθ,y=rsinθ.x=r\cos\theta,\qquad y=r\sin\theta.

At r=0r=0, every value of θ\theta labels the same point. Polar coordinates fail there. A Cartesian map still describes the origin normally.

Likewise, longitude fails to distinguish directions at a sphere’s poles. One can use another chart near a pole. The original labels were inadequate; the geometry is regular.

We will meet the same distinction at a black-hole horizon in Chapter 17. A coordinate expression can become infinite while a different chart describes the same location regularly. To establish a physical problem, we need more than the behavior of one set of labels.

A coordinate list is also different from a vector. Consider the simple relabeling x=x+10x'=x+10. A point formerly labeled 2 is now labeled 12. But a displacement of 3 coordinate units is still a displacement of 3 units: the added 10 cancels when we subtract the endpoints. The point labels and displacement components follow different rules.

For nonlinear changes, even subtracting two widely separated coordinate lists does not generally produce a vector. Tangent vectors use the local Jacobian rule from Chapter 2. We now make their local meaning precise.

4.3 Tangent vectors live at events#

Take a curve through an event pp, written xμ(λ)x^\mu(\lambda). Its tangent components are

Vμ=dxμdλp.V^\mu=\left.\frac{dx^\mu}{d\lambda}\right|_p.

We can identify this tangent by the rate of change it produces in a scalar field ff along the curve:

V[f]=ddλf(x(λ))p=Vμμfp.V[f]=\left.\frac{d}{d\lambda}f(x(\lambda))\right|_p =V^\mu\partial_\mu f\big|_p.

For example, take f(x,y)=x+2yf(x,y)=x+2y and a tangent with components (3,4)(3,4). Then V[f]=3+2(4)=11V[f]=3+2(4)=11, the same vector–covector pairing we calculated in Chapter 2. Here we are viewing the vector as an instruction to differentiate a function in a specified direction. The brackets in V[f]V[f] mean applying that instruction to ff.

The operator satisfies linearity and the product rule,

V[af+bg]=aV[f]+bV[g],V[fg]=V[f]g(p)+f(p)V[g].V[af+bg]=aV[f]+bV[g], \qquad V[fg]=V[f]g(p)+f(p)V[g].

All tangent vectors at pp form the tangent space TpMT_pM. A coordinate basis is

eμ=μp,V=Vμeμ.e_\mu=\partial_\mu\big|_p, \qquad V=V^\mu e_\mu.

Its dual basis is dxμpdx^\mu|_p, with dxμ(eν)=δμνdx^\mu(e_\nu)=\delta^\mu{}_{\nu}.

There is an additional question when we compare two different points. A vector at pp belongs to TpMT_pM; a vector at a neighboring event qq belongs to TqMT_qM. They are elements of different vector spaces. Subtracting them requires a rule for comparing those spaces.

On a flat Cartesian grid, an obvious translation rule is already operating in the background. On a general manifold there is no preferred rule supplied by the manifold alone. A connection will provide the missing comparison procedure. We will construct and use that rule in Chapters 6 and 7.

4.4 The metric is a local measuring operation#

In Chapter 3, the Minkowski metric turned a small displacement into an interval. Now allow the measuring rule to vary from place to place. We also give it two vector inputs, because this lets us describe their relation as well as each vector separately.

The ordinary dot product is a familiar example: VVV\cdot V gives squared length and VW=VWcosθV\cdot W=|V||W|\cos\theta relates two lengths to the angle between the arrows. If WW has unit length, the latter measures the component of VV along WW. Expanding the square of V+WV+W gives

2VW=V+W2V2W2.2V\cdot W=|V+W|^2-|V|^2-|W|^2.

So the rule for squared lengths already determines a rule for pairs. The same identity works for a symmetric bilinear form even when its squared values can be negative. At every event, the metric is such a form on the tangent vectors:

g(V,W)=gμνVμWν.g(V,W)=g_{\mu\nu}V^\mu W^\nu.

It is symmetric,

gμν=gνμ,g_{\mu\nu}=g_{\nu\mu},

and nondegenerate: no nonzero vector pairs to zero with every vector. Here orthogonal means g(V,W)=0g(V,W)=0, extending the zero-dot-product condition. In GR it has Lorentzian signature (,+,+,+)(-,+,+,+). At a point, there is a basis in which its components are ημν\eta_{\mu\nu}, with one negative and three positive eigenvalue directions. The number of positive and negative directions cannot be changed by a nonsingular real basis transformation.

Nondegenerate does not mean g(V,V)g(V,V) is nonzero for every nonzero VV. Null vectors have g(V,V)=0g(V,V)=0, but a nonzero null vector still has nonzero pairing with some other vectors. For example, with the two-dimensional matrix diag(1,1)\operatorname{diag}(-1,1), V=(1,1)V=(1,1) pairs to zero with itself but gives g(V,W)=2g(V,W)=-2 for W=(1,1)W=(1,-1). The matrix is invertible despite having nonzero vectors with zero self-pairing.

The line element

ds2=gμν(x)dxμdxνds^2=g_{\mu\nu}(x)dx^\mu dx^\nu

encodes the local geometry. For a timelike worldline, dτ=ds2/cd\tau=\sqrt{-ds^2}/c. For a spacelike curve, its length is obtained from ds2\int\sqrt{ds^2}. Defining a finite spatial distance between distant observers requires a specified spacelike slice or measurement protocol; the metric does not hand everyone the same universal “distance right now.”

The metric also defines the light cone by g(V,V)=0g(V,V)=0. The metric therefore tells us about clock readings and possible light signals as well as spatial lengths.

In four dimensions, a symmetric 4×44\times4 metric has 4(4+1)/2=104(4+1)/2=10 independent components. Those are ten functions in a coordinate description, not ten independent propagating gravitational modes. Coordinate freedom and the structure of the field equations will substantially change that counting.

4.5 A complete metric calculation in flat polar coordinates#

Return to the angular-step view in the polar-coordinate experiment. Moving outward by a small amount changes your position by that same length. Turning through a small angle changes it by a length proportional to rr: the little arc is longer on a larger circle. At the point where you stand, these radial and tangential directions are perpendicular.

This gives us a prediction. The two perpendicular legs of a very small displacement have lengths drdr and rdθr\,d\theta, so Pythagoras should give d2=dr2+r2dθ2d\ell^2=dr^2+r^2d\theta^2. The differentials describe the local limit; a finite straight chord need not equal a finite arc. We can now check the prediction by differentiating the coordinate formulas:

dx=cosθdrrsinθdθ,dy=sinθdr+rcosθdθ.dx=\cos\theta\,dr-r\sin\theta\,d\theta, \qquad dy=\sin\theta\,dr+r\cos\theta\,d\theta.

Square and add. The mixed terms cancel, and sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1 gives

d2=dx2+dy2=dr2+r2dθ2.d\ell^2=dx^2+dy^2=dr^2+r^2d\theta^2.

Therefore the spatial metric is

(gij)=(100r2).(g_{ij})= \begin{pmatrix}1&0\\0&r^2\end{pmatrix}.

The coefficient r2r^2 is necessary because one radian of angular change spans a longer arc farther from the origin. It is not evidence of curved space. We obtained the metric from the flat Euclidean plane by changing labels.

Units deserve attention. If rr is measured in meters and θ\theta is dimensionless, grrg_{rr} is dimensionless but gθθg_{\theta\theta} has units of square meters. It is the full line element that must have consistent units. The statement “the metric is dimensionless” is only true in suitable coordinate conventions.

Off-diagonal terms also have a simple meaning. On the same plane, choose oblique coordinates with position

r(q1,q2)=q1x^+q2(x^+y^).\mathbf r(q^1,q^2)=q^1\hat{\mathbf x} +q^2(\hat{\mathbf x}+\hat{\mathbf y}).

Then

d2=(dq1)2+2dq1dq2+2(dq2)2,(gij)=(1112).d\ell^2=(dq^1)^2+2dq^1dq^2+2(dq^2)^2, \qquad (g_{ij})=\begin{pmatrix}1&1\\1&2\end{pmatrix}.

The cross term tells you the coordinate basis directions are not orthogonal. Its coefficient is 2g122g_{12} because the Einstein sum includes both g12dq1dq2g_{12}dq^1dq^2 and g21dq2dq1g_{21}dq^2dq^1.

WORKED EXAMPLE

Teach a coordinate grid how to measure

Why does the same angular change mean a longer trip when you are farther from the origin?

See the idea

Take two nearby spokes on a flat plane. Walking across their separation at radius rr covers a circular arc of length rΔθr\,\Delta\theta. The coordinate difference Δθ\Delta\theta alone is an angle, not a length. The metric supplies the missing conversion. In the figure, shrink the coordinate cell to distinguish an exact local measuring rule from an approximation for a finite step.

Work it out
  1. Differentiate the map, not the picture

    For x=rcosθx=r\cos\theta and y=rsinθy=r\sin\theta, a tangent change splits into a radial part and an angular part. The angle is measured in radians and rr in meters.

    dx=cosθdrrsinθdθ,dy=sinθdr+rcosθdθ.\begin{aligned}dx&=\cos\theta\,dr-r\sin\theta\,d\theta,\\dy&=\sin\theta\,dr+r\cos\theta\,d\theta.\end{aligned}

    Why this step works The product rule and the derivatives of sine and cosine give the Cartesian components of the same local displacement.

  2. Apply the familiar measuring rule

    Square the two expressions and add them. Their mixed terms cancel because the radial and angular directions are perpendicular.

    ds2=dx2+dy2=dr2+r2dθ2.ds^2=dx^2+dy^2=dr^2+r^2d\theta^2.

    Why this step works The squared sine and cosine terms add to one, leaving one radial term and one angular term.

  3. Read the metric as two calibrated rulers

    In the coordinate order (r,θ)(r,\theta), the metric matrix is diagonal. At r=3mr=3\,\mathrm m, the local changes dr=0.03mdr=0.03\,\mathrm m and dθ=0.01d\theta=0.01 contribute equal physical lengths.

    gij=(100r2),ds=(0.03m)2+(0.03m)20.04243m.g_{ij}=\begin{pmatrix}1&0\\0&r^2\end{pmatrix},\qquad ds=\sqrt{(0.03\,\mathrm m)^2+(0.03\,\mathrm m)^2}\approx0.04243\,\mathrm m.

    Why this step works Multiply the angular increment by the radius before combining perpendicular physical lengths.

  4. Separate local vectors from finite journeys

    For finite coordinate differences the preceding number estimates the straight distance between nearby endpoints. A path that first walks radially and then follows an arc is longer and has its own exact length. For a continuously specified path, the metric gives an exact instantaneous speed.

    dsdt=(drdt)2+r(t)2(dθdt)2.\frac{ds}{dt}=\sqrt{\left(\frac{dr}{dt}\right)^2+r(t)^2\left(\frac{d\theta}{dt}\right)^2}.

    Why this step works The line element measures tangent vectors exactly. Integrating their lengths measures a chosen finite path.

Go deeper

For r>0r>0 and an angular interval avoiding a seam, gijg_{ij} is the matrix of inner products of the coordinate basis vectors. The radial basis has unit length; the angular basis has length rr. Thus gθθg_{\theta\theta} has units of length squared, while grrg_{rr} is dimensionless. These different units compensate for the units of the coordinates. At r=0r=0 the polar chart ceases to distinguish angular directions; the Euclidean metric itself remains regular in Cartesian coordinates. This calculation proves the polar form of the flat metric. Position-dependent metric components alone do not prove curvature.

A STATE YOU CAN CHECK

At radius 3 m, compare a radial step of 0.03 m with an angular step of 0.01 rad. The tangent prediction is a local approximation.

Radial length
Δr=0.03m\Delta r=0.03\,\mathrm m
Angular length
rΔθ=0.03mr\Delta\theta=0.03\,\mathrm m
Tangent prediction
tan=0.04243m\ell_{\rm tan}=0.04243\,\mathrm m

The worked steps explain these measurements. Interactive controls appear when available.

Open the reference diagram
Coordinates are not rulersA polar grid has an annular sector highlighted; its radial and angular edges have different measuring factors. For an infinitesimal cell, the angular edge is r dθ and the radial edge is dr. The finite cell is enlarged for visibility; the plane is intrinsically flat.07 / COORDINATES ARE NOT RULERSA flat plane, with polar labelsA grid can bend without the surface curving.Two physical edge lengths
07 /
Coordinates are not rulers. For an infinitesimal cell, the angular edge is rdθr\,d\theta and the radial edge is drdr. The finite cell is enlarged for visibility; the plane is intrinsically flat.
Test the idea

FIRST, PREDICT

Double rr while keeping a purely angular increment dθd\theta fixed. How do its length and squared length change?

Compare the reasoning

Neither changes because the coordinate increment is unchanged.

The angular coordinate counts radians. The metric tells us how many meters correspond to that increment at the chosen radius.

The length doubles; its contribution to ds2ds^2 quadruples.

Yes: ds=rdθds=r|d\theta| for this tangent, so squaring produces the factor of four.

Both double.

The metric computes a squared length. The square of a doubled length is four times as large.

A hint

Set dr=0dr=0 in the line element, then take its positive square root.

NOW CHANGE THE EXAMPLE

A particle has r=4mr=4\,\mathrm m, dr/dt=3ms1dr/dt=3\,\mathrm{m\,s^{-1}}, and dθ/dt=1s1d\theta/dt=1\,\mathrm{s^{-1}}. What is its instantaneous speed?

A hint

The angular part of the physical velocity has magnitude rdθ/dtr\,d\theta/dt. Combine it with the radial part.

Work through the solution

The perpendicular velocity components have magnitudes 3ms13\,\mathrm{m\,s^{-1}} and 4ms14\,\mathrm{m\,s^{-1}}, so ds/dt=32+42=5ms1ds/dt=\sqrt{3^2+4^2}=5\,\mathrm{m\,s^{-1}}. This is exact at the specified instant; no small-time approximation is needed.

A metric turns coordinate rates into physical measurements while keeping track of scale, direction, and units.

WORKED EXAMPLE

Units belong to coordinates, too

When we replace seconds by the distance light travels in those seconds, how does the interval formula keep giving the same answer?

See the idea

A coordinate is a label with a chosen unit. You can label a journey by elapsed seconds tt or by the distance X=ctX=ct that light travels in that time. These are two labels for the same events. Write the units before interpreting a coefficient: a dimensionless angle is different from a distance even when both are called coordinates.

Work it out
  1. Require each term to measure the same thing

    Take the spacetime interval ds2=c2dt2+dx2ds^2=-c^2dt^2+dx^2, where xx is in metres and tt in seconds. Each term has units of square metres. The minus sign distinguishes time from space; it does not change the dimensional check.

    [gtt]=m2/s2,[gxx]=1.[g_{tt}]=\mathrm{m^2/s^2},\qquad [g_{xx}]=1.

    Why this step works Terms can be added only when their dimensions agree.

  2. Relabel time without changing the interval

    Set X=ctX=ct, so dX=cdtdX=c\,dt. Substitution gives ds2=dX2+dx2ds^2=-dX^2+dx^2. Now gXX=1g_{XX}=-1. This coefficient changed because its coordinate unit changed. The measured interval did not.

    gXX=gtt(dtdX)2=c21c2=1.g_{XX}=g_{tt}\left(\frac{dt}{dX}\right)^2=-c^2\frac1{c^2}=-1.

    Why this step works The differential and the metric coefficient transform together.

  3. Check a curving coordinate grid

    On an ordinary Euclidean plane, a tiny angular change dθd\theta at radius rr sweeps length rdθr\,d\theta. Pythagoras gives dl2=dr2+r2dθ2dl^2=dr^2+r^2d\theta^2. Thus gθθ=r2g_{\theta\theta}=r^2 carries length squared, even though grr=1g_{rr}=1.

    [gab]=length2[xa][xb].[g_{ab}]=\frac{\mathrm{length}^2}{[x^a][x^b]}.

    Why this step works Metric components inherit the units of the chosen labels.

Go deeper

In ds2=gabdxadxbds^2=g_{ab}\,dx^a dx^b, the whole expression has dimensions of length squared. Thus [gab]=L2/([xa][xb])[g_{ab}]=L^2/([x^a][x^b]), where square brackets mean dimensions. Mixed terms follow the same rule. For example, keep tt and replace xx by y=xvty=x-vt for constant speed vv. Then dx=dy+vdtdx=dy+v\,dt and ds2=(v2c2)dt2+2vdtdy+dy2ds^2=(v^2-c^2)dt^2+2v\,dt\,dy+dy^2. The mixed coefficient gty=gyt=vg_{ty}=g_{yt}=v has speed units; the factor two comes from the two equal terms in the sum. Different component units are consistent because every complete term has square-length units. Setting c=1c=1 chooses compatible time and length units; it does not make all coordinates, such as angles, lengths.

Test the idea

FIRST, PREDICT

If X=ctX=ct, what happens to a fixed pair of events when you replace tt by XX?

Compare the reasoning

Their time labels change; the interval is unchanged.

The coordinate differential and metric coefficient compensate exactly.

Their physical separation grows by a factor of c.

A unit conversion changes a numerical label, not the physical events.

All metric coefficients become dimensionless in every chart.

Angular coordinates already provide a counterexample: the polar angular coefficient has length squared.

A hint

Substitute dX = c dt into the interval.

NOW CHANGE THE EXAMPLE

In dl2=dr2+r2dθ2dl^2=dr^2+r^2d\theta^2, r=3mr=3\,\mathrm m. What is gθθg_{\theta\theta}, in square metres?

A hint

Read the coefficient multiplying the square of the angular differential.

Work through the solution

gθθ=32=9m2g_{\theta\theta}=3^2=9\,\mathrm{m^2}.

A component means nothing until you know the coordinate units and the measurement it participates in.

4.6 Inverse metrics and raising or lowering indices#

The inverse metric is defined by

gμαgαν=δμν.g^{\mu\alpha}g_{\alpha\nu}=\delta^\mu{}_{\nu}.

It is the matrix inverse, not component-by-component reciprocation except when the metric is diagonal. For our oblique metric,

(gij)=(2111).(g^{ij})=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}.

Multiplying the two matrices verifies the identity directly.

The metric converts a vector into a covector:

Vμ=gμνVν.V_\mu=g_{\mu\nu}V^\nu.

The new covector asks another vector WW the question VμWμ=g(V,W)V_\mu W^\mu=g(V,W). The inverse conversion is

ωμ=gμνων.\omega^\mu=g^{\mu\nu}\omega_\nu.

For the oblique example, Vi=(1,2)V^i=(1,2) gives Vi=(3,5)V_i=(3,5), and

g(V,V)=ViVi=31+52=13.g(V,V)=V_iV^i=3\cdot1+5\cdot2=13.

The corresponding ordinary Cartesian vector is 3x^+2y^3\hat{\mathbf x}+2\hat{\mathbf y}, whose Euclidean norm squared is indeed 9+4=139+4=13. The components changed from (1,2)(1,2) to (3,5)(3,5) because the lowered object is a measuring rule. Applying it to the original vector gives the squared length in either description.

In inertial spacetime coordinates,

Vμ=(V0,V1,V2,V3)Vμ=(V0,V1,V2,V3).V^\mu=(V^0,V^1,V^2,V^3) \quad\Longrightarrow\quad V_\mu=(-V^0,V^1,V^2,V^3).

For example, u0=γcu_0=-\gamma c while u0=γcu^0=\gamma c. Lowering the time index accounts for the minus sign in the energy contraction below.

The gradient distinction from Chapter 2 now resolves:

(df)μ=μf,(gradf)μ=gμννf.(df)_\mu=\partial_\mu f, \qquad (\operatorname{grad}f)^\mu=g^{\mu\nu}\partial_\nu f.

The differential was available before the metric. The gradient vector was not. In Lorentzian geometry it is also unsafe to carry over every Euclidean slogan about a gradient “pointing uphill most steeply,” because the unit-vector set and norm structure are different.

4.7 Measuring volume with a determinant#

On a plane, a narrow polar cell has physical area approximately dr×rdθdr\times r\,d\theta, so

dA=rdrdθ=detgijdrdθ.dA=r\,dr\,d\theta=\sqrt{\det g_{ij}}\,dr\,d\theta.

For a small coordinate cell in more dimensions, the same idea applies. A matrix that converts coordinate components to components in a laboratory basis with metric matrix η\eta changes cell volumes by the absolute value of its determinant. The metric contains this conversion twice, once for each vector input, so its determinant contains the square of the volume factor. Let

gdet(gμν).g\equiv\det(g_{\mu\nu}).

With Lorentzian signature (,+,+,+)(-,+,+,+), g<0g<0 in every regular coordinate chart. Write d4x=dx0dx1dx2dx3d^4x=dx^0dx^1dx^2dx^3 for the product of the four small coordinate widths. This notation is not a fourth power of one coordinate. The invariant spacetime volume measure is

dV4=gd4x.\boxed{dV_4=\sqrt{-g}\,d^4x.}

Why exactly this factor? Under xxx\mapsto x', the covariant metric transforms as a matrix according to

g=KTgK,det(g)=(detK)2det(g),g'=K^{\mathsf T}gK, \qquad \det(g')=(\det K)^2\det(g),

where the first equation uses gg for the metric matrix and the second its determinant. Thus the square root picks up detK|\det K|. Meanwhile d4x=detJd4xd^4x'=|\det J|d^4x, and K=J1K=J^{-1}. The factors cancel:

gd4x=gd4x.\sqrt{-g'}\,d^4x'=\sqrt{-g}\,d^4x.

An oriented volume form additionally tracks the sign of orientation using a wedge product. The positive integration measure above uses absolute Jacobians. These are compatible viewpoints, but changing orientation is where their notation must be handled carefully. A formal treatment is available in Tong’s discussion of the metric volume form.

For a concrete four-dimensional example, keep x0=ctx^0=ct and describe space with spherical coordinates. The radius rr measures distance from the origin, θ\theta measures angle down from the north axis, and ϕ\phi measures angle around that axis:

x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθ.x=r\sin\theta\cos\phi,\qquad y=r\sin\theta\sin\phi,\qquad z=r\cos\theta.

Away from the origin and poles, the three coordinate directions are perpendicular. A radial step has length drdr. A small change in θ\theta has length rdθr\,d\theta. A circle at fixed r,θr,\theta has radius rsinθr\sin\theta, so a small change in ϕ\phi has length rsinθdϕr\sin\theta\,d\phi. Squaring and adding these lengths, and including the time term, gives

ds2=(dx0)2+dr2+r2dθ2+r2sin2θdϕ2,ds^2=-(dx^0)^2+dr^2+r^2d\theta^2+r^2\sin^2\theta\,d\phi^2,

so

g=r4sin2θ,dV4=cdtr2sinθdrdθdϕ.g=-r^4\sin^2\theta, \qquad dV_4=c\,dt\,r^2\sin\theta\,dr\,d\theta\,d\phi.

The factor r2sinθr^2\sin\theta converts the two angular widths and the radial width into physical spatial volume. The factor cdtc\,dt supplies the fourth, time-directed width. The vanishing determinant at r=0r=0 or at a polar axis signals failure of this chart there; it does not mean the regular Minkowski metric becomes physically degenerate.

Why a determinant belongs in the volumeA square transforms into a parallelogram under a linear map; the absolute determinant is its area scaling. This Euclidean two-dimensional example explains the square root of the metric determinant. In four-dimensional Lorentzian geometry the positive measure uses √(−g). The drawn transformation is illustrative.08 / WHY A DETERMINANT BELONGS IN THE VOLUMEThe determinant measures area scalingA unit coordinate square maps to a parallelogram.
08 /
Why a determinant belongs in the volume. This Euclidean two-dimensional example explains the square root of the metric determinant. In four-dimensional Lorentzian geometry the positive measure uses g\sqrt{-g}. The drawn transformation is illustrative.

4.8 Curvature, bending, and local frames#

A sheet can be rolled into a cylinder without stretching it. Distances and angles measured within a sufficiently small patch are unchanged, so its intrinsic curvature is zero. Its embedding in three-dimensional space is visibly bent. A flat sheet cannot smoothly cover a patch of a sphere while preserving all its lengths. The sphere’s intrinsic geometry differs from a plane’s.

Intrinsic curvature concerns measurements available to inhabitants of the geometry. Extrinsic curvature concerns how a chosen surface or other lower-dimensional space sits inside a larger geometry. Such an embedded space is called a submanifold. Both ideas occur in relativity: spacetime has intrinsic curvature, and a spatial slice can have extrinsic curvature within spacetime. They are different objects answering different questions.

Coordinate basis vectors need not be unit length or orthogonal. A physical laboratory instead likes an orthonormal frame ea^e_{\hat a} satisfying

g(ea^,eb^)=ηa^b^.g(e_{\hat a},e_{\hat b})=\eta_{\hat a\hat b}.

Hats here label laboratory directions. For an observer of four-velocity UU, choose e0^=U/ce_{\hat0}=U/c. The other three frame vectors specify their instantaneous spatial axes.

On the polar plane, grr=1g_{rr}=1 and gθθ=r2g_{\theta\theta}=r^2. Thus r\partial_r already has unit length, while θ\partial_\theta has length rr. Divide the latter by rr to obtain an orthonormal frame:

er^=r,eθ^=1rθ.e_{\hat r}=\partial_r, \qquad e_{\hat\theta}=\frac1r\partial_\theta.

Thus Vr^=VrV^{\hat r}=V^r and Vθ^=rVθV^{\hat\theta}=rV^\theta. The physical tangential component includes the conversion from angle to arc length.

There is a subtle distinction between making the metric’s components constant in a frame and finding coordinates in which they are constant throughout a neighborhood. The first is possible locally for a smooth Lorentzian metric. The second would make the neighborhood flat. A varying orthonormal frame generally cannot be the coordinate basis of one coordinate system; its axes can turn relative to one another from point to point. The connection keeps track of that turning.

At any ordinary event of a smooth spacetime, one can choose locally inertial coordinates with

gμν(p)=ημν,ρgμν(p)=0.g_{\mu\nu}(p)=\eta_{\mu\nu}, \qquad \partial_\rho g_{\mu\nu}(p)=0.

This is stronger than merely choosing an orthonormal basis at the point, but it does not generally remove second-order departures from flatness. It is the mathematical doorway to the equivalence principle.

The idea to keep

Changing coordinates changes the metric components. Curvature requires a coordinate-independent test.

In polar coordinates, why is gθθ=r2g_{\theta\theta}=r^2?

A small angular change dθd\theta at radius rr has physical length rdθr\,d\theta. Squaring gives r2dθ2r^2d\theta^2 in the line element.

Figure detail

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