11 Tensors and Local Symmetries - University of New...

11

Tensors and Local Symmetries

11.1 Points and Coordinates

A point on a curved surface or in a curved space also is a point in a higher-dimensional flat space called an embedding space. For instance, a point ona sphere also is a point in three-dimensional euclidean space and in four-dimensional space-time. One always can add extra dimensions, but it’ssimpler to use as few as possible, three in the case of a sphere.

On a su�ciently small scale, any reasonably smooth space locally lookslike n-dimensional euclidean space. Such a space is called a manifold. In-cidentally, according to Whitney’s embedding theorem, every n-dimensionalconnected, smooth manifold can be embedded in 2n-dimensional euclideanspace R2n. So the embedding space for such spaces in general relativity hasno more than eight dimensions.

We use coordinates to label points. For example, we can choose a polaraxis and a meridian and label a point on the sphere by its polar and az-imuthal angles (✓,�) with respect to that axis and meridian. If we use a dif-ferent axis and meridian, then the coordinates (✓0,�0) for the same point willchange. Points are physical, coordinates are metaphysical. Whenwe change our system of coordinates, the points don’t change, buttheir coordinates do.

Most points p have unique coordinates xi(p) and x0i(p) in their coordinatesystems. For instance, polar coordinates (✓,�) are unique for all points ona sphere — except the north and south poles which are labeled by ✓ = 0and ✓ = ⇡ and all 0 � < 2⇡. By using more than one coordinate system,one usually can arrange to label every point uniquely in some coordinatesystem. In the flat three-dimensional space in which the sphere is a surface,each point of the sphere has unique coordinates, ~p = (x, y, z).

436 Tensors and Local Symmetries

We will use coordinate systems that represent points on the manifolduniquely and smoothly at least in local patches, so that the maps

x0i = x0i(p) = x0i(p(x)) = x0i(x) (11.1)

and

xi = xi(p) = xi(p(x0)) = xi(x0) (11.2)

are well defined, di↵erentiable, and one to one in the patches. We’ll oftengroup the n coordinates xi together and write them collectively as x withouta superscript. Since the coordinates x(p) label the point p, we sometimeswill call them “the point x.” But p and x are di↵erent. The point p is uniquewith infinitely many coordinates x, x0, x00, . . . in infinitely many coordinatesystems.

11.2 Scalars

A scalar is a quantity B that is the same in all coordinate systems

B0 = B. (11.3)

If it also depends upon the coordinates x of the space-time point p, and

B0(x0) = B(x) (11.4)

then it is a scalar field.

11.3 Contravariant Vectors

The change dx0i due to changes in the unprimed coordinates is

dx0i =Xj

@x0i

@xjdxj . (11.5)

This rule defines contravariant vectors: a quantity Ai is a contravariantvector if it transforms like dxi

A0i =Xj

@x0i

@xjAj . (11.6)

The coordinate di↵erentials dxi form a contravariant vector. A contravariantvector Ai(x) that depends on the coordinates x and transforms as

A0i(x0) =Xj

@x0i

@xjAj(x) (11.7)

is a contravariant vector field.

11.4 Covariant Vectors 437

11.4 Covariant Vectors

The chain rule for partial derivatives

@

@x0i=Xj

@xj

@x0i@

@xj(11.8)

defines covariant vectors: a vector Ci that transforms as

C 0i =

Xj

@xj

@x0iCj (11.9)

is a covariant vector. If it also is a function of x, then it is a covariantvector field and

C 0i(x

0) =Xj

@xj

@x0iCj(x). (11.10)

Example 11.1 (Gradient of a Scalar) The derivatives of a scalar fieldform a covariant vector field. For by using the chain rule to di↵erentiate theequation B0(x0) = B(x) that defines a scalar field, one finds

@B0(x0)

@x0i=@B(x)

@x0i=Xj

@xj

@x0i@B(x)

@xj(11.11)

which shows that the gradient @B(x)/@xj is a covariant vector field.

11.5 Euclidean Space in Euclidean Coordinates

If we use euclidean coordinates to describe points in euclidean space, thencovariant and contravariant vectors are the same.

Euclidean space has a natural inner product (section 1.6), the usual dot-product, which is real and symmetric. In a euclidean space of n dimensions,we may choose any n fixed, orthonormal basis vectors ei

(ei, ej) ⌘ ei · ej =nX

k=1

eki ekj = �ij (11.12)

and use them to represent any point p as the linear combination

p =nX

i=1

ei xi. (11.13)

The coe�cients xi are the euclidean coordinates in the ei basis. Since the


basis vectors ei are orthonormal, each xi is an inner product or dot product

xi = ei · p =nX

j=1

ei · ej xj =nX

j=1

�ij xj . (11.14)

The dual vectors ei are defined as those vectors whose inner products withthe ej are (ei, ej) = �ij . In this section, they are the same as the vectors ei,and so we shall not bother to distinguish ei from ei = ei.If we use di↵erent orthonormal vectors e0i as a basis

p =nX

i=1

e0i x0i (11.15)

then we get new euclidean coordinates x0i = e0i ·p for the same point p. Thesetwo sets of coordinates are related by the equations

x0i = e0i · p =nX

j=1

e0i · ej xj

xj = ej · p =nX

k=1

ej · e0k x0k.(11.16)

Because the basis vectors e and e0 are all independent of x, the co-e�cients @x0i/@xj of the transformation laws for contravariant (11.6) andcovariant (11.9) vectors are

contravariant@x0i

@xj= e0i · ej and

@xj

@x0i= ej · e0i covariant. (11.17)

But the dot-product (1.82) is symmetric, and so these are the same:

@x0i

@xj= e0i · ej = ej · e0i =

@xj

@x0i. (11.18)

Contravariant and covariant vectors transform the same way ineuclidean space with euclidean coordinates.The relations between x0i and xj imply that

x0i =nX

j,k=1

�e0i · ej

� �ej · e0k

�x0k. (11.19)

Since this holds for all coordinates x0i, we have

nXj=1

�e0i · ej

� �ej · e0k

�= �ik. (11.20)

11.6 Summation Conventions 439

The coe�cients e0i · ej form an orthogonal matrix, and the linear operator

nXi=1

eie0iT =

nXi=1

|eiihe0i| (11.21)

is an orthogonal (real, unitary) transformation. The change x ! x0 is arotation plus a possible reflection (exercise 11.2).

Example 11.2 (A Euclidean Space of Two Dimensions) In two-dimensionaleuclidean space, one can describe the same point by euclidean (x, y) andpolar (r, ✓) coordinates. The derivatives

@r

@x=

x

r=@x

@rand

@r

@y=

y

r=@y

@r(11.22)

respect the symmetry (11.18), but (exercise 11.1) these derivatives

@✓

@x= � y

r26= @x

@✓= �y and

@✓

@y=

x

r26= @y

@✓= x (11.23)

do not.

11.6 Summation Conventions

When a given index is repeated in a product, that index usually is beingsummed over. So to avoid distracting summation symbols, one writes

AiBi ⌘nX

i=1

AiBi. (11.24)

The sum is understood to be over the relevant range of indices, usually from0 or 1 to 3 or n. Where the distinction between covariant and contravariantindices matters, an index that appears twice in the same monomial, onceas a subscript and once as a superscript, is a dummy index that is summedover as in

AiBi ⌘

nXi=1

AiBi. (11.25)

These summation conventions make tensor notation almost as compactas matrix notation. They make equations easier to read and write.


Example 11.3 (The Kronecker Delta) The summation convention andthe chain rule imply that

@x0i

@xk@xk

@x0j=@x0i

@x0j= �ij =

⇢1 if i = j0 if i 6= j.

(11.26)

The repeated index k has disappeared in this contraction.

11.7 Minkowski Space

Minkowski space has one time dimension, labeled by k = 0, and n spacedimensions. In special relativity n = 3, and the Minkowski metric ⌘

⌘k` = ⌘k` =

8<:�1 if k = ` = 01 if 1 k = ` 30 if k 6= `

(11.27)

defines an inner product between points p and q with coordinates xkp and x`qas

(p, q) = p · q = pk ⌘k` q` = (q, p). (11.28)

If one time component vanishes, the Minkowski inner product reduces tothe euclidean dot-product (1.82).We can use di↵erent sets {ei} and {e0i} of n + 1 Lorentz-orthonormal

basis vectors

(ei, ej) = ei · ej = eki ⌘k` e`j = ⌘ij = e0i · e0j = (e0i, e

0j) (11.29)

to represent any point p in the space either as a linear combination of thevectors ei with coe�cients xi or as a linear combination of the vectors e0iwith coe�cients x0i

p = ei xi = e0i x

0i. (11.30)

The dual vectors, which carry upper indices, are defined as

ei = ⌘ij ej and e0i = ⌘ij e0j . (11.31)

They are orthonormal to the vectors ei and e0i because

(ei, ej) = ei · ej = ⌘ik ek · ej = ⌘ik ⌘kj = �ij (11.32)

and similarly (e0i, e0j) = e0i · e0j = �ij . Since the square of the matrix ⌘ is the

identity matrix ⌘`i⌘ij = �j` , it follows that

ei = ⌘ij ej and e0i = ⌘ij e

0j . (11.33)

11.7 Minkowski Space 441

The metric ⌘ raises (11.31) and lowers (11.33) the index of a basis vector.The component x0i is related to the components xj by the linear map

x0i = e0i · p = e0i · ej xj . (11.34)

Such a map from a 4-vector x to a 4-vector x0 is a Lorentz transformation

x0i = Lij x

j with matrix Lij = e0i · ej . (11.35)

The inner product (p, q) of two points p = ei xi = e0i x0i and q = ek yk =

e0k y0k is physical and so is invariant under Lorentz transformations

(p, q) = xi yk ei · ek = ⌘ik xi yk = x0i y0k e0i · e0k = ⌘ik x

0i y0k. (11.36)

With x0i = Lir x

r and y0k = Lks y

s, this invariance is

⌘rs xrys = ⌘ik L

ir x

r Lks y

s= ⌘ik x0iy0k (11.37)

or since xr and ys are arbitrary

⌘rs = ⌘ik Lir L

ks = Li

r ⌘ik Lks. (11.38)

In matrix notation, a left index labels a row, and a right index labels acolumn. Transposition interchanges rows and columns Li

r = LT ir , so

⌘rs = LT ir ⌘ik L

ks or ⌘ = LT ⌘L (11.39)

in matrix notation. In such matrix products, the height of an index—whether it is up or down—determines whether it is contravariant or covariantbut does not a↵ect its place in its matrix.

Example 11.4 (A Boost) The matrix

L =

0BBB@�

p�2 � 1 0 0p

�2 � 1 � 0 00 0 1 00 0 0 1

1CCCA (11.40)

where � = 1/p1� v2/c2 represents a Lorentz transformation that is a boost

in the x-direction. Boosts and rotations are Lorentz transformations. Work-ing with 4⇥ 4 matrices can get tedious, so students are advised to think interms of scalars, like p · x = pi⌘ijxj = p · x� Et whenever possible.

If the basis vectors e and e0 are independent of p and of x, then thecoe�cients of the transformation law (11.6) for contravariant vectors are

@x0i

@xj= e0i · ej . (11.41)


Similarly, the component xj is xj = ej · p = ej · e0i x0i, so the coe�cients ofthe transformation law (11.9) for covariant vectors are

@xj

@x0i= ej · e0i. (11.42)

Using ⌘ to raise and lower the indices in the formula (11.41) for the coef-ficients of the transformation law (11.6) for contravariant vectors, we find

@x0i

@xj= e0i · ej = ⌘ik ⌘j` e

0k · e` = ⌘ik ⌘j`

@x`

@x0k(11.43)

which is ± @xj/@x0i. So if we use coordinates associated with fixed basisvectors ei in Minkowski space, then the coe�cients for the two kinds oftransformation laws di↵er only by occasional minus signs.Thus if Ai is a contravariant vector

A0i =

@x0i

@xjAj (11.44)

then the relation (11.43) between the two kinds of coe�cients implies that

⌘siA0i = ⌘si

@x0i

@xjAj = ⌘si ⌘

ik ⌘j`@x`

@x0kAj = �ks

@x`

@x0k⌘j`A

j =@x`

@x0s⌘`j A

j

(11.45)which shows that A` = ⌘`j Aj transforms covariantly

A0s =

@x`

@x0sA`. (11.46)

The metric ⌘ turns a contravariant vector into a covariant one. It alsoswitches a covariant vector A` back to its contravariant form Ak

⌘k`A` = ⌘k`⌘`jAj = �kjA

j = Ak. (11.47)

In Minkowski space, one uses ⌘ to raise and lower indices

Ai = ⌘ij Aj and Ai = ⌘ij Aj . (11.48)

In general relativity, the space-time metric g raises and lowers indices.

11.8 Lorentz Transformations

In section 11.7, Lorentz transformations arose as linear maps of the coor-dinates due to a change of basis. They also are linear maps of the basisvectors ei that preserve the inner products

(ei, ej) = ei · ej = ⌘ij = e0i · e0j = (e0i, e0j). (11.49)

11.9 Special Relativity 443

The vectors ei are four linearly independent four-dimensional vectors, and sothey span four-dimensional Minkowski space and can represent the vectorse0i as

e0i = ⇤k

i ek (11.50)

where the coe�cients ⇤ ki are real numbers. The requirement that the new

basis vectors e0i are Lorentz orthonormal gives

⌘ij = e0i · e0j = ⇤ ki ek · ⇤ `

j e` = ⇤k

i ek · e` ⇤ `j = ⇤ k

i ⌘k` ⇤`

j (11.51)

or in matrix notation

⌘ = ⇤ ⌘⇤T (11.52)

where ⇤T is the transpose (⇤T)` j = ⇤`

j . Evidently ⇤T satisfies the definition(11.39) of a Lorentz transformation. What Lorentz transformation is it? Thepoint p must remain invariant, so by (11.35 & 11.50) one has

p = e0i x0i = ⇤ k

i ek Lijx

j = �kj ek xj = ej x

j (11.53)

whence ⇤ ki Li

j = �kj or ⇤TL = I. So ⇤T = L�1.By multiplying condition (11.52) by the metric ⌘ first from the left and

then from the right and using the fact that ⌘2 = I, we find

1 = ⌘2 = ⌘⇤ ⌘⇤T = ⇤ ⌘⇤T ⌘ (11.54)

which gives us the inverse matrices

⇤�1 = ⌘⇤T ⌘ = LT and (⇤T)�1 = ⌘⇤ ⌘ = L. (11.55)

In special relativity, contravariant vectors transform as

dx0i = Lij dx

j (11.56)

and since xj = L�1ji x

0i, the covariant ones transform as

@

@x0i=@xj

@x0i@

@xj= L�1j

i

@

@xj= ⇤ j

i

@

@xj. (11.57)

By taking the determinant of both sides of (11.52) and using the transpose(1.194) and product (1.207) rules for determinants, we find that det⇤ = ±1.

11.9 Special Relativity

The space-time of special relativity is flat, four-dimensional Minkowski space.The inner product (p� q) · (p� q) of the interval p� q between two points isphysical and independent of the coordinates and therefore invariant. If the


points p and q are close neighbors with coordinates xi + dxi for p and xi forq, then that invariant inner product is

(p� q) · (p� q) = ei dxi · ej dxj = dxi ⌘ij dx

j = dx2 � (dx0)2 (11.58)

with dx0 = c dt. (At some point in what follows, we’ll measure distance inlight-seconds so that c = 1.) If the points p and q are on the trajectory ofa massive particle moving at velocity v, then this invariant quantity is thesquare of the invariant distance

ds2 = dx2 � c2dt2 =�v2 � c2

�dt2 (11.59)

which always is negative since v < c. The time in the rest frame of theparticle is the proper time. The square of its di↵erential element is

d⌧2 = � ds2/c2 =�1� v2/c2

�dt2. (11.60)

A particle of mass zero moves at the speed of light, and so its element d⌧of proper time is zero. But for a particle of mass m > 0 moving at speed v,the element of proper time d⌧ is smaller than the corresponding element oflaboratory time dt by the factor

p1� v2/c2. The proper time is the time in

the rest frame of the particle, d⌧ = dt when v = 0. So if T (0) is the lifetimeof a particle at rest, then the apparent lifetime T (v) when the particle ismoving at speed v is

T (v) = dt =d⌧p

1� v2/c2=

T (0)p1� v2/c2

(11.61)

which is longer — an e↵ect known as time dilation.

Example 11.5 (Time Dilation in Muon Decay ) A muon at rest has amean life of T (0) = 2.2 ⇥ 10�6 seconds. Cosmic rays hitting nitrogen andoxygen nuclei make pions high in the Earth’s atmosphere. The pions rapidlydecay into muons in 2.6⇥10�8 s. A muon moving at the speed of light from10 km takes at least t = 10 km/300, 000 (km/sec) = 3.3 ⇥ 10�5 s to hit theground. Were it not for time dilation, the probability P of such a muonreaching the ground as a muon would be

P = e�t/T (0) = exp(�33/2.2) = e�15 = 2.6⇥ 10�7. (11.62)

The (rest) mass of a muon is 105.66 MeV. So a muon of energy E = 749MeV has by (11.69) a time-dilation factor of

1p1� v2/c2

=E

mc2=

749

105.7= 7.089 =

1p1� (0.99)2

. (11.63)

11.10 Kinematics 445

So a muon moving at a speed of v = 0.99 c has an apparent mean life T (v)given by equation (11.61) as

T (v) =E

mc2T (0) =

T (0)p1� v2/c2

=2.2⇥ 10�6 sp1� (0.99)2

= 1.6⇥ 10�5 s. (11.64)

The probability of survival with time dilation is

P = e�t/T (v) = exp(�33/16) = 0.12 (11.65)

so that 12% survive. Time dilation increases the chance of survival by afactor of 460,000 — no small e↵ect.

11.10 Kinematics

From the scalar d⌧ , and the contravariant vector dxi, we can make the4-vector

ui =dxi

d⌧=

dt

d⌧

✓dx0

dt,dx

dt

◆=

1p1� v2/c2

(c,v) (11.66)

in which u0 = c dt/d⌧ = c/p1� v2/c2 and u = u0 v/c. The product mui is

the energy-momentum 4-vector pi

pi = mui = mdxi

d⌧= m

dt

d⌧

dxi

dt=

mp1� v2/c2

dxi

dt

=mp

1� v2/c2(c,v) =

✓E

c,p

◆. (11.67)

Its invariant inner product is a constant characteristic of the particle andproportional to the square of its mass

c2 pi pi = mcuimcui = �E2 + c2 p 2 = �m2 c4. (11.68)

Note that the time-dilation factor is the ratio of the energy of a particle toits rest energy

1p1� v2/c2

=E

mc2(11.69)

and the velocity of the particle is its momentum divided by its equivalentmass E/c2

v =p

E/c2. (11.70)


The analog of F = ma is

md2xi

d⌧2= m

dui

d⌧=

dpi

d⌧= f i (11.71)

in which p0 = E/c, and f i is a 4-vector force.

Example 11.6 (Time Dilation and Proper Time) In the frame of a lab-oratory, a particle of mass m with 4-momentum pilab = (E/c, p, 0, 0) travelsa distance L in a time t for a 4-vector displacement of xilab = (ct, L, 0, 0).In its own rest frame, the particle’s 4-momentum and 4-displacement arepirest = (mc, 0, 0, 0) and xirest = (c⌧, 0, 0, 0). Since the Minkowski innerproduct of two 4-vectors is Lorentz invariant, we have�

pixi�rest

=�pixi

�lab

or Et� pL = mc2⌧ = mc2tp1� v2/c2 (11.72)

so a massive particle’s phase exp(�ipixi/~) is exp(imc2⌧/~).

Example 11.7 (p + ⇡ ! ⌃ + K) What is the minimum energy that abeam of pions must have to produce a sigma hyperon and a kaon by strikinga proton at rest? Conservation of the energy-momentum 4-vector givespp + p⇡ = p

⌃

+ pK . We set c = 1 and use this equality in the invariant form(pp+p⇡)2 = (p

⌃

+pK)2. We compute (pp+p⇡)2 in the the pp = (mp,0) frameand set it equal to (p

⌃

+ pK)2 in the frame in which the spatial momenta ofthe ⌃ and the K cancel:

(pp + p⇡)2 = p2p + p2⇡ + 2pp · p⇡ = �m2

p �m2

⇡ � 2mpE⇡

= (p⌃

+ pK)2 = � (m⌃

+mK)2 . (11.73)

Thus, since the relevant masses (in MeV) are m⌃

+ = 1189.4, mK+ = 493.7,mp = 938.3, and m⇡+ = 139.6, the minimum total energy of the pion is

E⇡ =(m

⌃

+mK)2 �m2

p �m2

⇡

2mp⇡ 1030 MeV (11.74)

of which 890 MeV is kinetic.

11.11 Electrodynamics 447

11.11 Electrodynamics

In electrodynamics and in mksa (si) units, the three-dimensional vectorpotential A and the scalar potential � form a covariant 4-vector potential

Ai =

✓��c

,A

◆. (11.75)

The contravariant 4-vector potential is Ai = (�/c,A). The magnetic in-duction is

B = r⇥A or Bi = ✏ijk@jAk (11.76)

in which @j = @/@xj , the sum over the repeated indices j and k runs from1 to 3, and ✏ijk is totally antisymmetric with ✏

123

= 1. The electric field is

Ei = c

✓@A

0

@xi� @Ai

@x0

◆= � @�

@xi� @Ai

@t(11.77)

where x0 = ct. In 3-vector notation, E is given by the gradient of � and thetime-derivative of A

E = �r�� A. (11.78)

In terms of the second-rank, antisymmetric Faraday field-strength tensor

Fij =@Aj

@xi� @Ai

@xj= �Fji (11.79)

the electric field is Ei = c Fi0 and the magnetic field Bi is

Bi =1

2✏ijk Fjk =

1

2✏ijk

✓@Ak

@xj� @Aj

@xk

◆= (r⇥A)i (11.80)

where the sum over repeated indices runs from 1 to 3. The inverse equationFjk = ✏jkiBi for spatial j and k follows from the Levi-Civita identity (1.449)

✏jkiBi =1

2✏jki✏inm Fnm =

1

2✏ijk✏inm Fnm

=1

2(�jn �km � �jm �kn) Fnm =

1

2(Fjk � Fkj) = Fjk. (11.81)

In 3-vector notation and mksa = si units, Maxwell’s equations are a banon magnetic monopoles and Faraday’s law, both homogeneous,

r ·B = 0 and r⇥E + B = 0 (11.82)

and Gauss’s law and the Maxwell-Ampere law, both inhomogeneous,

r ·D = ⇢f and r⇥H = jf + D. (11.83)


Here ⇢f is the density of free charge and jf is the free current density. Byfree, we understand charges and currents that do not arise from polarizationand are not restrained by chemical bonds. The divergence ofr⇥H vanishes(like that of any curl), and so the Maxwell-Ampere law and Gauss’s lawimply that free charge is conserved

0 = r · (r⇥H) = r · jf +r · D = r · jf + ⇢f . (11.84)

If we use this continuity equation to replace r ·jf with � ⇢f in its middle

form 0 = r ·jf +r ·D, then we see that the Maxwell-Ampere law preservesthe Gauss-law constraint in time

0 = r · jf +r · D =@

@t(�⇢f +r ·D) . (11.85)

Similarly, Faraday’s law preserves the constraint r ·B = 0

0 = �r · (r⇥E) =@

@tr ·B = 0. (11.86)

In a linear, isotropic medium, the electric displacement D is relatedto the electric field E by the permittivity ✏, D = ✏E, and the mag-netic or magnetizing field H di↵ers from the magnetic induction B bythe permeability µ, H = B/µ.On a sub-nanometer scale, the microscopic form of Maxwell’s equations

applies. On this scale, the homogeneous equations (11.82) are unchanged,but the inhomogeneous ones are

r ·E =⇢

✏0

and r⇥B = µ0

j + ✏0

µ0

E = µ0

j +E

c2(11.87)

in which ⇢ and j are the total charge and current densities, and ✏0

=8.854⇥10�12 F/m and µ

0

= 4⇡⇥10�7 N/A2 are the electric and magneticconstants, whose product is the inverse of the square of the speed of light,✏0

µ0

= 1/c2. Gauss’s law and the Maxwell-Ampere law (11.87) imply (ex-ercise 11.6) that the microscopic (total) current 4-vector j = (c⇢, j) obeysthe continuity equation ⇢+r · j = 0. Electric charge is conserved.In vacuum, ⇢ = j = 0, D = ✏

0

E, and H = B/µ0

, and Maxwell’sequations become

r ·B = 0 and r⇥E + B = 0

r ·E = 0 and r⇥B =1

c2E.

(11.88)

Two of these equations r · B = 0 and r · E = 0 are constraints. Taking

11.11 Electrodynamics 449

the curl of the other two equations, we find

r⇥ (r⇥E) = � 1

c2E and r⇥ (r⇥B) = � 1

c2B. (11.89)

One may use the Levi-Civita identity (1.449) to show (exercise 11.8) that

r⇥ (r⇥E) = r (r · E)�4E and r⇥ (r⇥B) = r (r · B)�4B

(11.90)in which 4 ⌘ r2. Since in vacuum the divergence of E vanishes, andsince that of B always vanishes, these identities and the curl-curl equations(11.89) tell us that waves of E and B move at the speed of light

1

c2E �4E = 0 and

1

c2B �4B = 0. (11.91)

We may write the two homogeneous Maxwell equations (11.82) as

@iFjk + @kFij + @jFki = @i (@jAk � @kAj) + @k (@iAj � @jAi)

+ @j (@kAi � @iAk) = 0 (11.92)

(exercise 11.9). This relation, known as the Bianchi identity, actually isa generally covariant tensor equation

✏`ijk@iFjk = 0 (11.93)

in which ✏`ijk is totally antisymmetric, as explained in Sec. 11.32. Thereare four versions of this identity (corresponding to the four ways of choosingthree di↵erent indices i, j, k from among four and leaving out one, `). The` = 0 case gives the scalar equation r · B = 0, and the three that have` 6= 0 give the vector equation r⇥E + B = 0.In tensor notation, the microscopic form of the two inhomogeneous equa-

tions (11.87)—the laws of Gauss and Ampere—are

@iFki = µ

0

jk (11.94)

in which jk is the current 4-vector

jk = (c⇢, j) . (11.95)

The Lorentz force law for a particle of charge q is

md2xi

d⌧2= m

dui

d⌧=

dpi

d⌧= f i = q F ij dxj

d⌧= q F ij uj . (11.96)

We may cancel a factor of dt/d⌧ from both sides and find for i = 1, 2, 3

dpi

dt= q

��F i0 + ✏ijkBkvj

�or

dp

dt= q (E + v ⇥B) (11.97)


and for i = 0dE

dt= qE · v (11.98)

which shows that only the electric field does work. The only special-relativisticcorrection needed in Maxwell’s electrodynamics is a factor of 1/

p1� v2/c2

in these equations. That is, we use p = mu = mv/p1� v2/c2 not p = mv

in (11.97), and we use the total energy E not the kinetic energy in (11.98).The reason why so little of classical electrodynamics was changed by specialrelativity is that electric and magnetic e↵ects were accessible to measure-ment during the 1800’s. Classical electrodynamics was almost perfect.Keeping track of factors of the speed of light is a lot of trouble and a

distraction; in what follows, we’ll often use units with c = 1.

11.12 Tensors

Tensors are structures that transform like products of vectors. A first-ranktensor is a covariant or a contravariant vector. Second-rank tensors also aredistiguished by how they transform under changes of coordinates:

contravariant M 0ij =@x0i

@xk@x0j

@xlMkl

mixed N 0ij =

@x0i

@xk@xl

@x0jNk

l (11.99)

covariant F 0ij =

@xk

@x0i@xl

@x0jFkl.

We can define tensors of higher rank by extending the these definitions toquantities with more indices.

Example 11.8 (Some Second-Rank Tensors) If Ak and B` are covariantvectors, and Cm and Dn are contravariant vectors, then the product CmDn

is a second-rank contravariant tensor, and all four products Ak Cm, Ak Dn,Bk Cm, and Bk Dn are second-rank mixed tensors, while CmDn as well asCmCn and DmDn are second-rank contravariant tensors.

Since the transformation laws that define tensors are linear, any linearcombination of tensors of a given rank and kind is a tensor of that rank andkind. Thus if Fij and Gij are both second-rank covariant tensors, then so istheir sum

Hij = Fij +Gij . (11.100)

11.13 Di↵erential Forms 451

A covariant tensor is symmetric if it is independent of the order of itsindices. That is, if Sik = Ski, then S is symmetric. Similarly, a contravarianttensor is symmetric if permutations of its indices leave it unchanged. ThusA is symmetric if Aik = Aki.

A covariant or contravariant tensor is antisymmetric if it changes signwhen any two of its indices are interchanged. So Aik, Bik, and Cijk areantisymmetric if

Aik = �Aki and Bik = �Bki and

Cijk =Cjki = Ckij = � Cjik = � Cikj = � Ckji.(11.101)

Example 11.9 (Three Important Tensors) The Maxwell field strengthFkl(x) is a second-rank covariant tensor; so is the metric of spacetime gij(x).The Kronecker delta �ij is a mixed second-rank tensor; it transforms as

�0ij =@x0i

@xk@xl

@x0j�kl =

@x0i

@xk@xk

@x0j=@x0i

@xj= �ij . (11.102)

So it is invariant under changes of coordinates.

Example 11.10 (Contractions) Although the product Ak C` is a mixedsecond-rank tensor, the product Ak Ck transforms as a scalar because

A0k C

0k =@x`

@x0k@x0k

@xmA`C

m =@x`

@xmA`C

m = �`mA`Cm = A`C

`. (11.103)

A sum in which an index is repeated once covariantly and once contravari-antly is a contraction as in the Kronecker-delta equation (11.26). In gen-eral, the rank of a tensor is the number of uncontracted indices.

11.13 Di↵erential Forms

By (11.10 & 11.5), a covariant vector field contracted with contravariant co-ordinate di↵erentials is invariant under arbitrary coordinate transformations

A0 = A0i dx

0i =@xj

@x0iAj

@x0i

@xkdxk = �jk Aj dx

k = Ak dxk = A. (11.104)

This invariant quantity A = Ak dxk is a called a 1-form in the language ofdi↵erential forms introduced about a century ago by Elie Cartan (1869–1951, son of a blacksmith).


The wedge product dx^dy of two coordinate di↵erentials is the directedarea spanned by the two di↵erentials and is defined to be antisymmetric

dx ^ dy = � dy ^ dx and dx ^ dx = dy ^ dy = 0 (11.105)

so as to transform correctly under a change of coordinates. In terms of thecoordinates u = u(x, y) and v = v(x, y), the new element of area is

du ^ dv =

✓@u

@xdx+

@u

@ydy

◆^✓@v

@xdx+

@v

@ydy

◆. (11.106)

Labeling partial derivatives by subscripts (6.20) and using the antisymmetry(11.105), we see that the new element of area du^ dv is the old area dx^ dymultiplied by the Jacobian J(u, v;x, y) of the transformation x, y ! u, v

du ^ dv = (uxdx+ uydy) ^ (vxdx+ vydy)

= ux vx dx ^ dx+ ux vy dx ^ dy + uy vx dy ^ dx+ uy vy dy ^ dy

= (ux vy � uyvx) dx ^ dy

=

�� ux uyvx vy

�� dx ^ dy = J(u, v;x, y) dx ^ dy. (11.107)

A contraction H = 1

2

Hik dxi ^ dxk of a second-rank covariant tensor witha wedge product of two di↵erentials is a 2-form. A p -form is a rank-pcovariant tensor contracted with a wedge product of p di↵erentials

K =1

p!Ki1...ip dx

i1 ^ . . . dxip . (11.108)

The exterior derivative d di↵erentiates and adds a di↵erential; it turnsa p-form into a (p + 1)-form. It converts a function or a 0-form f into a1-form

df =@f

@xidxi (11.109)

and a 1-form A = Aj dxj into a 2-form dA = d(Aj dxj)= (@iAj) dxi ^ dxj .

Example 11.11 (The Curl) The exterior derivative of the 1-form

A = Ax dx+Ay dy +Az dz (11.110)

11.13 Di↵erential Forms 453

is the 2-form

dA = @yAx dy ^ dx+ @zAx dz ^ dx

+ @xAy dx ^ dy + @zAy dz ^ dy

+ @xAz dx ^ dz + @yAz dy ^ dz

= (@yAz � @zAy) dy ^ dz (11.111)

+ (@zAx � @xAz) dz ^ dx

+ (@xAy � @yAx) dx ^ dy

= (r⇥A)x dy ^ dz + (r⇥A)y dz ^ dx+ (r⇥A)z dx ^ dy

in which we recognize the curl (6.39) of A.

The exterior derivative of the 1-form A = Aj dxj is the 2-form

dA = dAj ^ dxj = @iAj dxi ^ dxj = 1

2

Fij dxi ^ dxj = F (11.112)

in which @i = @/@xi. So d turns the electromagnetic 1-form A—the 4-vectorpotential or gauge field Aj—into the Faraday 2-form—the tensor Fij . Itssquare dd vanishes: dd applied to any p-form Q is zero

ddQi...dxi^· · · = d(@rQi...)^dxr^dxi^· · · = (@s@rQi...)dx

s^dxr^dxi^· · · = 0(11.113)

because @s@rQ is symmetric in r and s while dxs ^ dxr is anti-symmetric.Some writers drop the wedges and write dxi^dxj as dxidxj while keeping

the rules of antisymmetry dxidxj = �dxjdxi and�dxi�2

= 0. But this econ-omy prevents one from using invariant quantities like S = 1

2

Sik dxidxk inwhich Sik is a second-rank covariant symmetric tensor. If Mik is a covariantsecond-rank tensor with no particular symmetry, then (exercise 11.7) onlyits antisymmetric part contributes to the 2-form Mik dxi ^ dxk and only itssymmetric part contributes to the quantity Mik dxidxk.The exterior derivative d applied to the Faraday 2-form F = dA gives

dF = ddA = 0 (11.114)

which is the Bianchi identity (11.93). A p-form H is closed if dH = 0. By(11.114), the Faraday 2-form is closed, dF = 0.A p-form H is exact if there is a (p � 1)-form K whose di↵erential is

H = dK. The identity (11.113) or dd = 0 implies that every exact formis closed. The lemma of Poincare shows that every closed form is locallyexact.If the Ai in the 1-form A = Aidxi commute with each other, then the

2-form A ^ A = 0. But if the Ai don’t commute because they are matricesor operators or Grassmann variables, then A ^A need not vanish.


Example 11.12 (A Static Electric Field Is Closed and Locally Exact) IfB = 0, then by Faraday’s law (11.82) the curl of the electric field vanishes,r ⇥ E = 0. Writing the electrostatic field as the 1-form E = Ei dxi fori = 1, 2, 3, we may express the vanishing of its curl as

dE = @jEi dxj ^ dxi =

1

2(@jEi � @iEj) dx

j ^ dxi = 0 (11.115)

which says that E is closed. We can define a quantity VP (x) as a line integralof the 1-form E along a path P to x from some starting point x

0

VP (x) = �Z

x

P,x0

Ei dxi = �

ZPE (11.116)

and so VP (x) will depend on the path P as well as on x0

and x. But ifr ⇥ E = 0 in some ball (or neighborhood) around x and x

0

, then withinthat ball the dependence on the path P drops out because the di↵erenceVP 0(x) � VP (x) is the line integral of E around a closed loop in the ballwhich by Stokes’s theorem (6.44) is an integral of the vanishing curl r ⇥ E

over any surface S in the ball whose boundary @S is the closed curve P 0�P

VP 0(x)� VP (x) =

IP 0�P

Ei dxi =

ZS(r ⇥ E) · da = 0 (11.117)

or

VP 0(x)� VP (x) =

Z@S

E =

ZSdE = 0 (11.118)

in the language of forms (George Stokes, 1819–1903). Thus the potentialVP (x) = V (x) is independent of the path, E = �rV (x), and the 1-formE = Ei dxi = �@iV dxi = �dV is locally exact.

The general form of Stokes’s theorem is that the integral of any p-formH over the boundary @R of any (p + 1)-dimensional, simply connected,orientable region R is equal to the integral of the (p+ 1)-form dH over RZ

@RH =

ZRdH (11.119)

which for p = 1 gives (6.44).

Example 11.13 (Stokes’s Theorem for 0-forms) Here p = 0, the regionR = [a, b] is 1-dimensional, H is a 0-form, and Stokes’s theorem is

H(b)�H(a) =

Z@R

H =

ZRdH =

Z b

adH(x) =

Z b

aH 0(x) dx (11.120)

familiar from elementary calculus.

11.14 Tensor Equations 455

Example 11.14 (Exterior Derivatives Anticommute with Di↵erentials) Theexterior derivative acting on two one-forms A = Aidxi and B = Bjdxj is

d(A^B) = d(Aidxi ^Bjdx

j) = @k(AiBj) dxk ^ dxi ^ dxj (11.121)

= (@kAi)Bj dxk ^ dxi ^ dxj +Ai(@kBj) dx

k ^ dxi ^ dxj

= (@kAi)Bj dxk ^ dxi ^ dxj �Ai(@kBj) dx

i ^ dxk ^ dxj

= (@kAi)dxk ^ dxi ^Bjdx

j �Aidxi ^ (@kBj) dx

k ^ dxj

= dA ^B �A ^ dB.

If A is a p-form, then d(A^B) = dA^B+(�1)pA^dB (exercise 11.10).

11.14 Tensor Equations

Maxwell’s homogeneous equations (11.93) relate the derivatives of the field-strength tensor to each other as

0 = @iFjk + @kFij + @jFki. (11.122)

They are generally covariant tensor equations (sections 11.31 & 11.32).In terms of invariant forms, they are the Bianchi identity (11.114)

dF = ddA = 0. (11.123)

Maxwell’s inhomegneous equations (11.94) relate the derivatives of the field-strength tensor to the current density ji and to the square-root of the mod-ulus g of the determinant of the metric tensor gij (section 11.16)

@(pg F ik)

@xk= µ

0

pg ji. (11.124)

We’ll write them as invariant forms in section 11.26 and derive them froman action principle in section 11.38.

If we can write a physical law in one coordinate system as a tensor equation

Kkl = 0 (11.125)

then in any other coordinate system, the corresponding tensor equation

K 0ij = 0 (11.126)

also is valid since

K 0ij =@x0i

@xk@x0j

@xlKkl = 0. (11.127)


Similarly, physical laws remain the same when expressed in terms of invari-ant forms. Thus by writing a theory in terms of tensors or forms,one gets a theory that is true in all coordinate systems if it istrue in any. Only such generally covariant theories have a chance at beingright in our coordinate system, which is not special. One way to make agenerally covariant theory is to start with an action that is invariant underall coordinate transformations.

11.15 The Quotient Theorem

Suppose that the product BA of a quantity B (with unknown transforma-tion properties) with an arbitrary tensor A (of a given rank and kind) is atensor. Then B is itself a tensor. The simplest example is when BiAi is ascalar for all contravariant vectors Ai

B0iA

0i = BjAj . (11.128)

Then since Ai is a contravariant vector

B0iA

0i = B0i

@x0i

@xjAj = BjA

j (11.129)

or ✓B0

i

@x0i

@xj�Bj

◆Aj = 0. (11.130)

Since this equation holds for all vectors A, we may promote it to the levelof a vector equation

B0i

@x0i

@xj�Bj = 0. (11.131)

Multiplying both sides by @xj/@x0k and summing over j

B0i

@x0i

@xj@xj

@x0k= Bj

@xj

@x0k(11.132)

we see that the unknown quantity Bi does transform as a covariant vector

B0k =

@xj

@x0kBj . (11.133)

The quotient rule works for unknowns B and tensors A of arbitrary rankand kind. The proof in each case is very similar to the one given here.

11.16 The Metric Tensor 457

11.16 The Metric Tensor

So far we have been considering coordinate systems with constant basisvectors ei that do not vary with the physical point p. Now we shall assumeonly that we can write the change in the point p(x) due to an infinitesimalchange dxi(p) in its coordinates xi(p) as

dp(x) = ei(x) dxi. (11.134)

In a di↵erent system of coordinates x0, this displacement is dp = e0i(x0) dx0i.

The basis vectors ei and e0i are partial derivatives of the point p

ei(x) =@p

@xiand e0i(x

0) =@p

@x0i. (11.135)

They are linearly related to each other, transforming as covariant vectors

e0i(x0) =

@p

@x0i=@xj

@x0i@p

@xj=@xj

@x0iej(x). (11.136)

They also are vectors in the n-dimensional embedding space with innerproduct

ei(x) · ej(x) =nX

a=1

nXb=1

eai (x) ⌘ab ebj(x) (11.137)

which will be positive-definite (1.75) if all the eigenvalues of the real symmet-ric matrix ⌘ are positive. For instance, the eigenvalues are positive in euclid-ian 3-space with cylindrical or spherical coordinates but not in Minkowski4-space where ⌘ is a diagonal matrix with main diagonal (�1, 1, 1, 1).

The basis vectors ei(x) constitute a moving frame, a concept introducedby Elie Cartan. In general, they are not normalized or orthogonal. Theirinner products define the metric of the manifold or of space-time

gij(x) = ei(x) · ej(x). (11.138)

An inner product by definition (1.73) satisfies (f, g) = (g, f)⇤ and so a realinner product is symmetric. For real coordinates on a real manifold thebasis vectors are real, so the metric tensor is real and symmetric

gij = gji. (11.139)

The basis vectors e0j(x0) of a di↵erent coordinate system define the metric

in that coordinate system g0ij(x0) = e0i(x

0) · e0j(x0). Since the basis vectors eiare covariant vectors, the metric gij is a second-rank covariant tensor

g0ij(x0) = e0i(x

0) ·e0j(x0) =@xk

@x0iek(x) ·

@x`

@x0je`(x) =

@xk

@x0i@x`

@x0jgk`(x). (11.140)


Example 11.15 (The Sphere) Let the point p be a euclidean 3-vectorrepresenting a point on the two-dimensional surface of a sphere of radius r.The spherical coordinates (✓,�) label the point p, and the basis vectors are

e✓ =@p

@✓= r ✓ and e� =

@p

@�= r sin ✓ �. (11.141)

Their inner products are the components (11.138) of the sphere’s metrictensor which is the matrix✓

g✓✓ g✓�g�✓ g��

◆=

✓e✓ · e✓ e✓ · e�e� · e✓ e� · e�

◆=

✓r2 00 r2 sin2 ✓

◆(11.142)

with determinant r4 sin2 �.

11.17 A Basic Axiom

Points are physical, coordinate systems metaphysical. So p, q, p � q, and(p� q) · (p� q) are all invariant quantities. When p and q = p+ dp both lieon the (space-time) manifold and are infinitesimally close to each other, thevector dp = ei dxi is the sum of the basis vectors multiplied by the changesin the coordinates xi. Both dp and the inner product dp · dp are physicaland so are independent of the coordinates. The (squared) distance dp2 isthe same in one coordinate system

dp2 ⌘ dp · dp = (ei dxi) · (ej dxj) = gij dx

idxj (11.143)

as in another

dp2 ⌘ dp · dp = (e0i dx0i) · (e0j dx0j) = g0ij dx

0idx0j . (11.144)

This invariance and the quotient rule provide a second reason why gij is asecond-rank covariant tensor.We want dp to be infinitesimal so that it is tangent to the manifold.

11.18 The Contravariant Metric Tensor

The inverse gik of the covariant metric tensor gkj satisfies

g0ikg0kj = �ij = gikgkj (11.145)

in all coordinate systems. To see how it transforms, we use the transforma-tion law (11.140) of gkj

�ij = g0ikg0kj = g0ik@xt

@x0kgtu

@xu

@x0j. (11.146)

11.19 Raising and Lowering Indices 459

Thus in matrix notation, we have I = g0�1H gH which implies g0�1 =H�1 g�1H�1 or in tensor notation

g0i` =@x0i

@xv@x0`

@xwgvw. (11.147)

Thus the inverse gik of the covariant metric tensor is a second-rank con-travariant tensor called the contravariant metric tensor.

11.19 Raising and Lowering Indices

The contraction of a contravariant vectorAi with any rank-2 covariant tensorgives a covariant vector. We reserve the symbol Ai for the covariant vectorthat is the contraction of Aj with the metric tensor

Ai = gijAj . (11.148)

This operation is called lowering the index on Aj .Similarly the contraction of a covariant vector Bj with any rank-2 con-

travariant tensor is a contravariant vector. But we reserve the symbol Bi

for contravariant vector that is the contraction

Bi = gijBj (11.149)

of Bj with the inverse of the metric tensor. This is called raising the indexon Bj .

The vectors ei, for instance, are given by

ei = gijej . (11.150)

They are therefore orthonormal or dual to the basis vectors ei

ei · ej = ei · gjkek = gjkei · ek = gjkgik = gjkgki = �ji . (11.151)

11.20 Orthogonal Coordinates in Euclidian n-Space

In flat n-dimensional euclidian space, it is convenient to use orthogonalbasis vectors and orthogonal coordinates. A change dxi in the coordi-nates moves the point p by (11.134)

dp = ei dxi. (11.152)

The metric gij is the inner product (11.138)

gij = ei · ej . (11.153)


Since the vectors ei are orthogonal, the metric is diagonal

gij = ei · ej = h2i �ij . (11.154)

The inverse metric

gij = h�2

i �ij (11.155)

raises indices. For instance, the dual vectors

e i = gij ej = h�2

i ei satisfy ei · ek = �ik. (11.156)

The invariant squared distance dp2 between nearby points (11.143) is

dp2 = dp · dp = gij dxi dxj = h2i (dx

i)2 (11.157)

and the invariant volume element is

dV = dnp = h1

. . . hn dx1 ^ . . . ^ dxn = g dx1 ^ . . . ^ dxn = g dnx (11.158)

in which g =p

det gij is the square-root of the positive determinant of gij .The important special case in which all the scale factors hi are unity is

cartesian coordinates in euclidean space (section 11.5).We also can use basis vectors ei that are orthonormal. By (11.154 &

11.156), these vectors

ei = ei/hi = hi ei satisfy ei · ej = �ij . (11.159)

In terms of them, a physical and invariant vector V takes the form

V = ei Vi = hi ei V

i = e i Vi = h�1

i ei Vi = ei V i (11.160)

where

V i ⌘ hi Vi = h�1

i Vi (no sum). (11.161)

The dot-product is then

V · U = gij Vi U j = V i U i. (11.162)

In euclidian n-space, we even can choose coordinates xi so that the vectorsei defined by dp = ei dxi are orthonormal. The metric tensor is then then ⇥ n identity matrix gik = ei · ek = Iik = �ik. But since this is euclidiann-space, we also can expand the n fixed orthonormal cartesian unit vectors ˆ

in terms of the ei(x) which vary with the coordinates as ˆ= ei(x)(ei(x) · ˆ).

11.21 Polar Coordinates 461

11.21 Polar Coordinates

In polar coordinates in flat 2-space, the change dp in a point p due to achange in its coordinates is dp = r dr + ✓ r d✓ so dp = er dr + e✓ d✓ wither = er = r and e✓ = r e✓ = r ✓. The metric tensor for polar coordinates is

(gij) = (ei · ej) =✓1 00 r2

◆. (11.163)

The contravariant basis vectors are e r = r and e ✓ = e✓/r. A physical vectorV is V = V i ei = Vi e i = V r r + V ✓ ✓.

11.22 Cylindrical Coordinates

For cylindrical coordinates in flat 3-space, the change dp in a point p dueto a change in its coordinates is

dp = ⇢ d⇢+ � ⇢ d�+ z dz = e⇢ d⇢+ e� d�+ ez dz (11.164)

with e⇢ = e⇢ = ⇢, e� = ⇢ e� = ⇢ �, and ez = ez = z. The metric tensor forcylindrical coordinates is

(gij) = (ei

· ej

) =

0@1 0 00 ⇢2 00 0 1

1A (11.165)

with determinant det gij ⌘ g = ⇢2. The invariant volume element is

dV = ⇢ dx1 ^ dx2 ^ dx3 =pg d⇢ d� dz = ⇢ d⇢ d� dz. (11.166)

The contravariant basis vectors are e ⇢ = ⇢, e� = e�/⇢, and e z = z. Aphysical vector V is

V = V i ei = Vi ei = V ⇢ ⇢+ V � �+ V z z. (11.167)

Incidentally, since

p = (⇢ cos�, ⇢ sin�, z) , (11.168)

the formulas for the basis vectors of cylindrical coordinates in terms of thoseof rectangular coordinates are (exercise 11.13)

⇢ = cos� x+ sin� y

� = � sin� x+ cos� y

z = z. (11.169)


11.23 Spherical Coordinates

For spherical coordinates in flat 3-space, the change dp in a point p due toa change in its coordinates is

dp = r dr + ✓ r d✓ + � r sin ✓ d� = er dr + e✓ d✓ + e� d� (11.170)

so er = r, e✓ = r ✓, and e� = r sin ✓ �. The metric tensor for sphericalcoordinates is

(gij) = (ei · ej) =

0@1 0 00 r2 00 0 r2 sin2 ✓

1A (11.171)

with determinant det gij ⌘ g = r4 sin2 ✓. The invariant volume element is

dV = r2 sin2 ✓ dx1 ^ dx2 ^ dx3 =pg dr d✓ d� = r2 sin ✓ dr d✓ d�. (11.172)

The orthonormal basis vectors are er = r, e✓ = ✓, and e� = �. Thecontravariant basis vectors are e r = r, e ✓ = ✓/r, e� = �/r sin ✓. A physicalvector V is

V = V i ei = Vi ei = V r r + V ✓ ✓ + V � �. (11.173)

Incidentally, since

p = (r sin ✓ cos�, r sin ✓ sin�, r cos ✓) , (11.174)

the formulas for the basis vectors of spherical coordinates in terms of thoseof rectangular coordinates are (exercise 11.14)

r = sin ✓ cos� x+ sin ✓ sin� y + cos ✓ z

✓ = cos ✓ cos� x+ cos ✓ sin� y � sin ✓ z

� = � sin� x+ cos� y. (11.175)

11.24 The Gradient of a Scalar Field

If f(x) is a scalar field, then the di↵erence between it and f(x+ dx) definesthe gradient rf as (6.26)

df(x) = f(x+ dx)� f(x) =@f(x)

@xidxi = rf(x) · dp. (11.176)

Since dp = ej dxj , the invariant form

rf = ei@f

@xi=

eihi

@f

@xi(11.177)

11.25 Levi-Civita’s Tensor 463

satisfies this definition (11.176) of the gradient

rf · dp =@f

@xiei · ejdxj =

@f

@xi�ij dx

j =@f

@xidxi = df. (11.178)

In two polar coordinates, the gradient is

rf = ei@f

@xi=

eihi

@f

@xi= r

@f

@r+✓

r

@f

@✓. (11.179)

In three cylindrical coordinates, it is (6.27)

rf = ei@f

@xi=

eihi

@f

@xi=@f

@⇢⇢+

1

⇢

@f

@��+

@f

@zz (11.180)

and in three spherical coordinates it is (6.28)

rf =@f

@xiei =

eihi

@f

@xi=@f

@rr +

1

r

@f

@✓✓ +

1

r sin ✓

@f

@��. (11.181)

11.25 Levi-Civita’s Tensor

In 3 dimensions, Levi-Civita’s symbol ✏ijk ⌘ ✏ijk is totally antisymmetricwith ✏

123

= 1 in all coordinate systems.We can turn his symbol into something that transforms as a tensor by

multiplying it by the square-root of the determinant of a rank-2 covarianttensor. A natural choice is the metric tensor. Thus the Levi-Civita tensor⌘ijk is the totally antisymmetric rank-3 covariant (pseudo-)tensor

⌘ijk =pg ✏ijk (11.182)

in which g = | det gmn| is the absolute value of the determinant of the metrictensor gmn. The determinant’s definition (1.184) and product rule (1.207)imply that Levi-Civita’s tensor ⌘ijk transforms as

⌘0ijk =pg0 ✏0ijk =

pg0 ✏ijk =

s��det✓ @xt

@x0m@xu

@x0ngtu

◆�� ✏ijk=

s��det✓ @xt

@x0m

◆det

✓@xu

@x0n

◆det (gtu)

�� ✏ijk=

��det✓ @x@x0◆�� pg ✏ijk = � det

✓@x

@x0

◆pg ✏ijk

= �@xt

@x0i@xu

@x0j@xv

@x0kpg ✏tuv = �

@xt

@x0i@xu

@x0j@xv

@x0k⌘tuv (11.183)


in which � is the sign of the Jacobian det(@x/@x0). Levi-Civita’s tensor isa pseudo-tensor because it doesn’t change sign under the parity transfor-mation x0i = � xi.We get ⌘ with upper indices by using the inverse gnm of the metric tensor

⌘ijk = git gju gkv ⌘tuv = git gju gkvpg ✏tuv=

pg ✏ijk det(gmn)

=pg ✏ijk/ det(gmn) = s✏ijk/

pg = s✏ijk/

pg

(11.184)

in which s is the sign of the determinant det gij = sg.Similarly in 4 dimensions, Levi-Civita’s symbol ✏ijk` ⌘ ✏ijk` is totally

antisymmetric with ✏0123

= 1 in all coordinate systems. No meaning attachesto whether the indices of the Levi-Civita symbol are up or down; someauthors even use the notation ✏(ijk`) or ✏[ijk`] to emphasize this fact.In 4 dimensions, the Levi-Civita pseudo-tensor is

⌘ijk` =pg ✏ijk`. (11.185)

It transforms as

⌘0ijk` =pg0 ✏ijk` =

��det✓ @x@x0◆��pg ✏ijk` = � det

✓@x

@x0

◆pg ✏ijk`

= �@xt

@x0i@xu

@x0j@xv

@x0k@xw

@x0`pg ✏tuvw = �

@xt

@x0i@xu

@x0j@xv

@x0k@xw

@x0`⌘tuvw

(11.186)

where � is the sign of the Jacobian det(@x/@x0).Raising the indices on ⌘ with det gij = sg we have

⌘ijk` = git gju gkv g`w ⌘tuvw = git gju gkv g`wpg ✏tuvw =

pg ✏ijk` det(g

mn)

=pg ✏ijk`/ det(gmn) = s ✏ijk`/

pg ⌘ s ✏ijk`/

pg.

(11.187)

In n dimensions, one may define Levi-Civita’s symbol ✏(i1

. . . in) as totallyantisymmetric with ✏(1 . . . n) = 1 and his tensor as ⌘i1...in =

pg ✏(i

1

. . . in).

11.26 The Hodge Star

In 3 cartesian coordinates, the Hodge dual turns 1-forms into 2-forms

⇤ dx = dy ^ dz ⇤ dy = dz ^ dx ⇤ dz = dx ^ dy (11.188)

and 2-forms into 1-forms

⇤ (dx ^ dy) = dz ⇤ (dy ^ dz) = dx ⇤ (dz ^ dx) = dy. (11.189)

11.26 The Hodge Star 465

It also maps the 0-form 1 and the volume 3-form into each other

⇤1 = dx ^ dy ^ dz ⇤ (dx ^ dy ^ dz) = 1 (11.190)

(William Vallance Douglas Hodge, 1903–1975). More generally in 3-space,we define the Hodge dual, also called the Hodge star, as

⇤1 =1

3!⌘`jkdx

` ^ dxj ^ dxk ⇤ (dx` ^ dxj ^ dxk) = g`tgjugkv⌘tuv

⇤ dxi = 1

2gi` ⌘`jk dx

j ^ dxk ⇤ (dxi ^ dxj) = gik gj` ⌘k`m dxm(11.191)

and so if the sign of det gij is s = +1, then ⇤ ⇤ 1 = 1, ⇤ ⇤ dxi = dxi,⇤ ⇤ (dxi ^ dxk) = dxi ^ dxk, and ⇤ ⇤ (dxi ^ dxj ^ dxk) = dxi ^ dxj ^ dxk.

Example 11.16 (Divergence and Laplacian) The dual of the 1-form

df =@f

@xdx+

@f

@ydy +

@f

@zdz (11.192)

is the 2-form

⇤ df =@f

@xdy ^ dz +

@f

@ydz ^ dx+

@f

@zdx ^ dy (11.193)

and its exterior derivative is the laplacian

d ⇤ df =

✓@2f

@x2+@2f

@y2+@2f

@z2

◆dx ^ dy ^ dz (11.194)

multiplied by the volume 3-form.Similarly, the dual of the one form

A = Ax dx+Ay dy +Az dz (11.195)

is the 2-form

⇤A = Ax dy ^ dz +Ay dz ^ dx+Az dx ^ dy (11.196)

and its exterior derivative is the divergence

d ⇤A =

✓@Ax

@x+@Ay

@y+@Az

@z

◆dx ^ dy ^ dz (11.197)

times dx ^ dy ^ dz.

In flat Minkowski 4-space with c = 1, the Hodge dual turns 1-forms into3-forms

⇤ dt = � dx ^ dy ^ dz ⇤ dx = � dy ^ dz ^ dt

⇤ dy = � dz ^ dx ^ dt ⇤ dz = � dx ^ dy ^ dt(11.198)


2-forms into 2-forms

⇤(dx ^ dt) = dy ^ dz ⇤ (dx ^ dy) = � dz ^ dt

⇤(dy ^ dt) = dz ^ dx ⇤ (dy ^ dz) = � dx ^ dt (11.199)

⇤(dz ^ dt) = dx ^ dy ⇤ (dz ^ dx) = � dy ^ dt

3-forms into 1-forms

⇤ (dx ^ dy ^ dz) = � dt ⇤ (dy ^ dz ^ dt) = � dx

⇤ (dz ^ dx ^ dt) = � dy ⇤ (dx ^ dy ^ dt) = � dz(11.200)

and interchanges 0-forms and 4-forms

⇤1 = dt ^ dx ^ dy ^ dz ⇤ (dt ^ dx ^ dy ^ dz) = � 1. (11.201)

More generally in 4 dimensions, we define the Hodge star as

⇤1 =1

4!⌘k`mn dx

k ^ dx` ^ dxm ^ dxn

⇤ dxi = 1

3!gik ⌘k`mn dx

` ^ dxm ^ dxn

⇤ (dxi ^ dxj) =1

2gik gj` ⌘k`mn dx

m ^ dxn (11.202)

⇤ (dxi ^ dxj ^ dxk) = git gju gkv ⌘tuvw dxw

⇤⇣dxi ^ dxj ^ dxk ^ dx`

⌘= git gju gkv g`w⌘tuvw = ⌘ijk`.

Thus (exercise 11.16) if the determinant det gij of the metric is negative,then

⇤ ⇤ dxi = dxi ⇤ ⇤ (dxi ^ dxj) = � dxi ^ dxj

⇤ ⇤ (dxi ^ dxj ^ dxk) = dxi ^ dxj ^ dxk ⇤ ⇤1 = � 1.(11.203)

In n dimensions, the Hodge star turns p-forms into n� p-forms

⇤�dxi1 ^ . . . ^ dxip

�= gi1k1 . . . gipkp

⌘k1...kp`1...`n�p

(n� p)!dx`1 ^ . . . ^ dx`n�p .

(11.204)

Example 11.17 (The Inhomogeneous Maxwell Equations) Since the ho-mogeneous Maxwell equations are

dF = ddA = 0 (11.205)

we first form the dual ⇤F = ⇤dA

⇤F = 1

2

Fij ⇤�dxi ^ dxj

�= 1

4

Fijgikgj`⌘k`mndx

m^dxn = 1

4

F k`⌘k`mndxm^dxn

11.26 The Hodge Star 467

and then apply the exterior derivative

d ⇤ F = 1

4

d⇣F k`⌘k`mndx

m ^ dxn⌘= 1

4

@p⇣F k`⌘k`mn

⌘dxp ^ dxm ^ dxn.

To get back to a 1-form like j = jk dxk, we apply a second Hodge star

⇤d ⇤ F = 1

4

@p⇣F k` ⌘k`mn

⌘⇤ (dxp ^ dxm ^ dxn)

= 1

4

@p⇣F k` ⌘k`mn

⌘gpsgmtgnu⌘stuv dx

v

= 1

4

@p⇣p

g F k`⌘✏k`mn g

psgmtgnupg ✏stuv dx

v (11.206)

= 1

4

@p⇣p

g F k`⌘✏k`mn g

psgmtgnugwv ✏stuvpg dxw

= 1

4

@p⇣p

g F k`⌘✏k`mn ✏pmnw

pg

det gijdxw

=s

4pg@p⇣p

g F k`⌘✏k`mn ✏pmnw dxw

in which we used the definition (1.184) of the determinant. Levi-Civita’s4-symbol obeys the identity (exercise 11.17)

✏k`mn ✏pwmn = 2

��pk �

w` � �wk �

p`

�. (11.207)

Applying it to ⇤ d ⇤ F , we get

⇤ d ⇤ F =s

2pg@p⇣p

g F k`⌘ ��pk �

w` � �wk �

p`

�dxw = � s

pg@p⇣p

g F kp⌘dxk.

In our space-time s = �1. Setting ⇤ d ⇤ F equal to j = jk dxk = jk dxkmultiplied by the permeability µ

0

of the vacuum, we arrive at expressions forthe microscopic inhomogeneous Maxwell equations in terms of both tensorsand forms

@p⇣p

g F kp⌘= µ

0

pg jk and ⇤ d ⇤ F = µ

0

j. (11.208)

They and the homogeneous Bianchi identity (11.93, 11.114, & 11.247)

✏ijk` @`Fjk = dF = d dA = 0 (11.209)

are invariant under general coordinate transformations.


11.27 Derivatives and A�ne Connections

If F (x) is a vector field, then its invariant description in terms of space-time-dependent basis vectors ei(x) is

F (x) = F i(x) ei(x). (11.210)

Since the basis vectors ei(x) vary with x, the derivative of F (x) containstwo terms

@F

@x`=@F i

@xèi + F i @ei

@x`. (11.211)

In general, the derivative of a vector ei is not a linear combination of thebasis vectors ek. For instance, on the 2-dimensional surface of a sphere in3-dimensions, the derivative

@e✓@✓

= �r (11.212)

points to the sphere’s center and isn’t a linear combination of e✓ and e�.The inner product of a derivative @ei/@x` with a dual basis vector ek is

the Levi-Civita a�ne connection

�kì = ek · @ei@x`

(11.213)

which relates spaces that are tangent to the manifold at infinitesimally sep-arated points. It is called an a�ne connection because the di↵erent tangentspaces lack a common origin.In terms of the a�ne connection (11.213 ), the inner product of the deriva-

tive (11.211) with ek is

ek · @F@x`

= ek · @Fi

@xèi + F i ek · @ei

@x`=@F k

@x`+ �kì F

i (11.214)

a combination that is called a covariant derivative (section 11.30)

D`Fk ⌘r`F

k ⌘ @F k

@x`+ �kì F

i= ek · @F@x`

. (11.215)

It is a second-rank mixed tensor.Some physicists write the a�ne connection �ki` as⇢

ki`

�= �ki` (11.216)

and call it a Christo↵el symbol of the second kind.

11.28 Parallel Transport 469

The vectors ei are the space-time derivatives (11.135) of the point p, andso the a�ne connection (11.213) is a double derivative of p

�kì = ek · @ei@x`

= ek · @2p

@x`@xi= ek · @2p

@xi@x`= ek · @e`

@xi= �ki` (11.217)

and thus is symmetric in its two lower indices

�ki` = �kì. (11.218)

A�ne connections are not tensors. Tensors transform homogeneously;connections transform inhomogeneously. The connection �ki` transforms as

�0ki` = e0k · @e0`

@x0i=@x0k

@xpep · @x

m

@x0i@

@xm

✓@xn

@x0èn

◆=@x0k

@xp@xm

@x0i@xn

@x0èp · @en

@xm+@x0k

@xp@2xp

@x0i@x0`(11.219)

=@x0k

@xp@xm

@x0i@xn

@x0`�pmn +

@x0k

@xp@2xp

@x0i@x0`.

The electromagnetic field Ai(x) and other gauge fields are connections.Since the Levi-Civita connection �ki` is symmetric in i and `, in four-

dimensional space-time, there are 10 of them for each k, or 40 in all. The10 correspond to 3 rotations, 3 boosts, and 4 translations.Einstein-Cartan theories do not assume that the space-time manifold

is embedded in a flat space of higher dimension. So their basis vectorsneed not be partial derivatives of a point in the embedding space, and theira�ne connections �abc need not be symmetric in their lower indices. Theantisymmetric part is the torsion tensor

T abc = �

abc � �acb. (11.220)

11.28 Parallel Transport

The movement of a vector along a curve on a manifold so that its directionin successive tangent spaces does not change is called parallel transport.If the vector is F = F iei, then we want ek · dF to vanish along the curve.But this is just the condition that the covariant derivative (11.215) of Fshould vanish along the curve

ek · @F@x`

=@F k

@x`+ �kì F

i = D`Fk = 0. (11.221)


Example 11.18 (Parallel Transport on a Sphere) The tangent space ona 2-sphere is spanned by the unit basis vectors

✓ = (cos ✓ cos�, cos ✓ sin�, � sin ✓)

� = (� sin�, cos�, 0) .(11.222)

We can parallel-transport the vector � down from the north pole along themeridian � = 0 to the equator; all along this path � = (0, 1, 0). Then wecan parallel-transport it along the equator to � = ⇡/2 where it is (�1, 0, 0).Then we can parallel-transport it along the meridian � = ⇡/2 up to thenorth pole where it is (�1, 0, 0) as it was on the equator. The change from(0, 1, 0) to (�1, 0, 0) is due to the curvature of the sphere.

11.29 Notations for Derivatives

We have various notations for derivatives. We can use the variables x, y,and so forth as subscripts to label derivatives

fx = @xf =@f

@xand fy = @yf =

@f

@y. (11.223)

If we use indices to label variables, then we can use commas

f,i = @if =@f

@xiand f,ik = @k@if =

@2f

@xk@xi(11.224)

and f,k0 = @f/@x0k. For instance, we may write part of (11.217) as ei,` = e`,i.

11.30 Covariant Derivatives

In comma notation, the derivative of a contravariant vector field F = F i eiis

F,` = F i,` ei + F i ei,` (11.225)

which in general lies outside the space spanned by the basis vectors ei. Sowe use the a�ne connections (11.213) to form the inner product

ek · F,` = ek ·�F i,èi + F iei,`

�= F i

,` �ki + F i �kì = F k

,` + �kì F

i. (11.226)

This covariant derivative of a contravariant vector field often is writ-ten with a semicolon

F k;` = ek · F,` = F k

,` + �kì F

i. (11.227)

11.31 The Covariant Curl 471

It transforms as a mixed second-rank tensor. The invariant change dFprojected onto ek is

ek · dF = ek · F,` dx` = F k

;` dx`. (11.228)

In terms of its covariant components, the derivative of a vector V is

V,` = (Vk ek),` = Vk,` e

k + Vk ek,`. (11.229)

To relate the derivatives of the vectors ei to the a�ne connections �ki`, wedi↵erentiate the orthonormality relation

�ki = ek · ei (11.230)

which gives us

0 = ek,` · ei + ek · ei,` or ek,` · ei = �ek · ei,` = ��ki`. (11.231)

Since ei · ek,` = ��ki`, the inner product of ei with the derivative of V is

ei · V,` = ei ·⇣Vk,`e

k + Vk ek,`

⌘= Vi,` � Vk�

ki`. (11.232)

This covariant derivative of a covariant vector field also is often writ-ten with a semicolon

Vi;` = ei · V,` = Vi,` � Vk�ki`. (11.233)

It transforms as a rank-2 covariant tensor. Note the minus sign in Vi;` andthe plus sign in F k

;`. The change ei · dV is

ei · dV = ei · V,` dx` = Vi;` dx

`. (11.234)

Since dV is invariant, ei covariant, and dx` contravariant, the quotient rule(section 11.15) confirms that the covariant derivative Vi;` of a covariantvector Vi is a rank-2 covariant tensor.

11.31 The Covariant Curl

Because the connection �ki` is symmetric (11.218) in its lower indices, thecovariant curl of a covariant vector Vi is simply its ordinary curl

V`;i � Vi;` = V`,i � Vk �k`i � Vi,` + Vk �

ki` = V`,i � Vi,`. (11.235)

Thus the Faraday field-strength tensor Fi` which is defined as the curl ofthe covariant vector field Ai

Fi` = A`,i �Ai,` (11.236)


is a generally covariant second-rank tensor.In orthogonal coordinates, the curl is defined (6.39, 11.111) in terms of

the totally antisymmetric Levi-Civita symbol ✏ijk (with ✏123

= ✏123 = 1), as

r⇥ V =3X

i=1

(r⇥ V )i ei =1

h1

h2

h3

3Xijk=1

ei ✏ijk Vk;j (11.237)

which, in view of (11.235) and the antisymmetry of ✏ijk, is

r⇥ V =3X

i=1

(r⇥ V )i ei =3X

ijk=1

1

hihjhkei ✏

ijk Vk,j (11.238)

or by (11.159 & 11.161)

r⇥ V =3X

ijk=1

1

hihjhkhiei ✏

ijk Vk,j =3X

ijk=1

1

hihjhkhiei ✏

ijk (hkV k),j .

(11.239)Often one writes this as a determinant

r⇥ V =1

h1

h2

h3

��e1

e2

e3

@1

@2

@3

V1

V2

V3

�� = 1

h1

h2

h3

��h1

e1

h2

e2

h3

e3

@1

@2

@3

h1

V1

h2

V2

h3

V3

�� .(11.240)

In cylindrical coordinates, the curl is

r⇥ V =1

⇢

��⇢ ⇢ � z@⇢ @� @zV ⇢ ⇢V � V z

�� . (11.241)

In spherical coordinates, it is

r⇥ V =1

r2 sin ✓

��r r✓ r sin ✓ �@r @✓ @�V r rV ✓ r sin ✓ V �

�� . (11.242)

In more formal language, the curl is

dV = d⇣Vkdx

k⌘= Vk,i dx

i ^ dxk =1

2(Vk,i � Vi,k) dx

i ^ dxk. (11.243)

11.32 Covariant Derivatives and Antisymmetry 473

11.32 Covariant Derivatives and Antisymmetry

By applying our rule (11.233) for the covariant derivative of a covariantvector to a second-rank tensor Ai`, we get

Ai`;k = Ai`,k � �mikAm` � �m`kAim. (11.244)

Suppose now that our tensor is antisymmetric

Ai` = �Aì. (11.245)

Then by adding together the three cyclic permutations of the indices i`k wefind that the antisymmetry of the tensor and the symmetry (11.218) of thea�ne connection conspire to cancel the nonlinear terms

Ai`;k +Aki;` +A`k;i = Ai`,k � �mikAm` � �m`kAim

+ Aki,` � �mkÀmi � �miÀkm

+ A`k,i � �mìAmk � �mkiA`m

= Ai`,k +Aki,` +A`k,i (11.246)

an identity named after Luigi Bianchi (1856–1928).The Maxwell field-strength tensor Fi` is antisymmetric by construction

(Fi` = A`,i �Ai,`), and so the homogeneous Maxwell equations

✏ijk` Fjk,` = Fjk,` + Fk`,j + F`j,k = 0 (11.247)

are tensor equations valid in all coordinate systems. This is another exampleof how amazingly right Maxwell was in the middle of the nineteenth century.

11.33 A�ne Connection and Metric Tensor

To relate the a�ne connection �mì to the derivatives of the metric tensorgk`, we lower the contravariant index m to get

�kì = gkm �mì = gkm �

mi` = �ki` (11.248)

which is symmetric in its last two indices and which some call a Christo↵elsymbol of the first kind, written [ì, k]. One can raise the index k backup by using the inverse of the metric tensor

gmk �kì = gmk gkn �nì = �mn �

nì = �

mì . (11.249)

Although we can raise and lower these indices, the connections �mì and �kìare not tensors.


The definition (11.213) of the a�ne connection tells us that

�kì = gkm �mì = gkm em · e`,i = ek · e`,i = �ki` = ek · ei,`. (11.250)

By di↵erentiating the definition gi` = ei · e` of the metric tensor, we find

gi`,k = ei,k · e` + ei · e`,k = e` · ei,k + ei · e`,k = �ìk + �i`k. (11.251)

Permuting the indices cyclicly, we have

gki,` = �ik` + �ki`

g`k,i = �kì + �`ki.(11.252)

If we now subtract relation (11.251) from the sum of the two formulas(11.252) keeping in mind the symmetry �abc = �acb, then we find that fourof the six terms cancel

gki,` + g`k,i � gi`,k = �ik` + �ki` + �kì + �`ki � �ìk � �i`k = 2�kì (11.253)

leaving a formula for �kì

�kì =1

2

(gki,` + g`k,i � gi`,k) . (11.254)

Thus the connection is three derivatives of the metric tensor

�si` = gsk�kì =1

2

gsk (gki,` + g`k,i � gi`,k) . (11.255)

11.34 Covariant Derivative of the Metric Tensor

Covariant derivatives of second-rank and higher-rank tensors are formed byiterating our formulas for the covariant derivatives of vectors. For instance,the covariant derivative of the metric tensor is

gi`;k ⌘ gi`,k � �mik gm` � �nk` gin. (11.256)

One way to derive this formula is to proceed as in Sec.(11.30) by di↵eren-tiating the invariant metric tensor gi` ei ⌦ e` in which the vector productei ⌦ e` is a kind of direct product

g,k = (gièi ⌦ e`),k = gi`,k e

i ⌦ e` + gi` ei,k ⌦ e` + gi` e

i ⌦ e`,k. (11.257)

We now take the inner product of this derivative with em ⌦ en

(em⌦en, g,k) = gi`,k em ·ei en ·e`+gi` em ·ei,k en ·e`+gi` em ·ei en ·e`,k (11.258)

and use the rules em · ei = �im and em · ei,k = ��imk (11.231) to write

(em ⌦ en, g,k) = gmn;k = gmn,k � gi`�imk�

`n � gi`�

im�

`nk (11.259)

11.35 Divergence of a contravariant vector 475

or

gmn;k = gmn,k � �imkgin � �`nkgm` (11.260)

which is (11.256) inasmuch as both gi` and �ki` are symmetric in their twolower indices.

If we now substitute our formula (11.255) for the connections �lik and �nk`

gi`;k = gi`,k � 1

2

gms (gis,k + gsk,i � gik,s) gm` � 1

2

gns (g`s,k + gsk,` � g`k,s) gin(11.261)

and use the fact (11.145) that the metric tensors gi` and g`k are mutuallyinverse, then we find

gi`;k = gi`,k � 1

2

�s` (gis,k + gsk,i � gik,s)� 1

2

�si (g`s,k + gsk,` � g`k,s)

= gi`,k � 1

2

(gi`,k + g`k,i � gik,`)� 1

2

(g`i,k + gik,` � g`k,i)

= 0. (11.262)

The covariant derivative of the metric tensor vanishes. This result followsfrom our choice of the Levi-Civita connection (11.213); it is not true forsome other connections.

11.35 Divergence of a contravariant vector

The contracted covariant derivative of a contravariant vector is a scalarknown as the divergence,

r · V = V i;i = V i

,i + �iikV

k. (11.263)

Because two indices in the connection

�iik = 1

2

gim (gim,k + gkm,i � gik,m) (11.264)

are contracted, its last two terms cancel because they di↵er only by theinterchange of the dummy indices i and m

gimgkm,i = gmigkm,i = gimgki,m = gimgik,m. (11.265)

So the contracted connection collapses to

�iik = 1

2

gimgim,k. (11.266)

There is a nice formula for this last expression involving the absolutevalue of the determinant det g ⌘ det gmn of the metric tensor considered asa matrix g ⌘ gmn. To derive it, we recall that like any determinant, the


determinant det(g) of the metric tensor is given by the cofactor sum (1.195)

det(g) =X`

gi`Ci` (11.267)

along any row or column, that is, over ` for fixed i or over i for fixed `, whereCi` is the cofactor defined as (�1)i+` times the determinant of the reducedmatrix consisting of the matrix gi` with row i and column ` omitted. Thusthe partial derivative of det g with respect to the i`th element gi` is

@ det(g)

@gi`= Ci`, (11.268)

in which we consider gi` and gì to be independent variables for the purposesof this di↵erentiation. The inverse gi` of the metric tensor g, like the inverse(1.197) of any matrix, is the transpose of the cofactor matrix divided by itsdeterminant det(g)

gi` =Cì

det(g)=

1

det(g)

@ det(g)

@gì. (11.269)

The chain rule gives us the derivative of the determinate det(g) as

det(g),k = gi`,k@ det(g)

@gi`= gi`,k det(g) gì (11.270)

and so since gi` = gì, the contracted connection (11.266) is

�iik = 1

2

gimgim,k =det(g),k2 det(g)

=| det(g)|,k2| det(g)| =

g,k2g

=(pg),kpg

(11.271)

in which g⌘��det(g)�� is the absolute value of the determinant of the metric

tensor.Thus from (11.263), we arrive at our formula for the covariant divergence

of a contravariant vector:

r · V = V i;i = V i

,i + �iikV

k = V k,k +

(pg),kpg

V k =(pg V k),kp

g. (11.272)

More formally, the Hodge dual (11.202) of the 1-form V = Vi dxi is

⇤V = Vi ⇤ dxi = Vi1

3!gik ⌘k`mn dx

` ^ dxm ^ dxn

=1

3!

pg V k ✏k`mn dx

` ^ dxm ^ dxn(11.273)

11.35 Divergence of a contravariant vector 477

in which g is the absolute value of the determinant of the metric tensor gij .The exterior derivative now gives

d ⇤ V =1

3!

⇣pg V k

⌘,p✏k`mn dx

p ^ dx` ^ dxm ^ dxn. (11.274)

So using (11.202) to apply a second Hodge star, we get (exercise 11.19)

⇤ d ⇤ V =1

3!

⇣pg V k

⌘,p✏k`mn ⇤

⇣dxp ^ dx` ^ dxm ^ dxn

⌘=

1

3!

⇣pg V k

⌘,p✏k`mn g

pt g`u gmv gnw⌘tuvw

=1

3!

⇣pg V k

⌘,p✏k`mn g

pt g`u gmv gnw✏tuvwpg

=1

3!

⇣pg V k

⌘,p✏k`mn

pg

det gij✏p`mn

=spg

⇣pg V k

⌘,p�pk =

spg

⇣pg V k

⌘,k

(11.275)

So in our space-time with det gij = �g

� ⇤ d ⇤ V =1pg

⇣pg V k

⌘,k. (11.276)

In 3-space the Hodge star (11.191) of a 1-form V = Vi dxi is

⇤V = Vi ⇤ dxi = Vi1

2gi` ⌘`jk dx

j ^ dxk =1

2

pg V ` ✏`jk dx

j ^ dxk. (11.277)

Applying the exterior derivative, we get the invariant form

d ⇤ V =1

2

⇣pg V `

⌘,p✏`jk dx

p ^ dxj ^ dxk. (11.278)

We add a star by using the definition (11.191) of the Hodge dual in a 3-spacein which the determinant det gij is positive and the identity (exercise 11.18)

✏`jk ✏pjk = 2 �p` (11.279)


as well as the definition (1.184) of the determinant

⇤ d ⇤ V =1

2

⇣pg V `

⌘,p✏`jk ⇤

⇣dxp ^ dxj ^ dxk

⌘=

1

2

⇣pg V `

⌘,p✏`jk g

ptgjugkv⌘tuv

=1

2

⇣pg V `

⌘,p✏`jk g

ptgjugkv✏tuvpg

=1

2

⇣pg V `

⌘,p✏`jk ✏

pjk

pg

det gij

=1pg

⇣pg V `

⌘,p�p` =

1pg(pg V p),p . (11.280)

Example 11.19 (Divergence in Orthogonal Coordinates) In two orthog-onal coordinates, equations (11.154 & 11.161) imply that

pg = h

1

h2

andV k = V k/hk, and so the divergence (11.272) of a vector V is

r · V =1

h1

h2

2Xk=1

✓h1

h2

hkV k

◆,k

(11.281)

which in polar coordinates (section 11.21) with hr = 1 and h✓ = r, is

r · V =1

r

h�r V r

�,r+�V ✓

�,✓

i=

1

r

h�r V r

�,r+ V ✓,✓

i. (11.282)

In three orthogonal coordinates, equations (11.154 & 11.161) givepg =

h1

h2

h3

and V k = V k/hk, and so the divergence (11.272) of a vector V is(6.29)

r · V =1

h1

h2

h3

3Xk=1

✓h1

h2

h3

hkV k

◆,k

. (11.283)

In cylindrical coordinates (section 11.22), h⇢ = 1, h� = ⇢, and hz = 1; so

r · V =1

⇢

h�⇢V ⇢

�,⇢+�V �

�,�+�⇢V z

�,z

i=

1

⇢

h�⇢V ⇢

�,⇢+ V �,� + ⇢V z,z

i. (11.284)

In spherical coordinates (section 11.23), hr = 1, h✓ = r, h� = r sin ✓, g =| det g| = r4 sin2 ✓ and the inverse gij of the metric tensor is

(gij) =

0@1 0 00 r�2 00 0 r�2 sin�2 ✓

1A . (11.285)

11.36 The Covariant Laplacian 479

So our formula (11.281) gives us

r · V =1

r2 sin ✓

h�r2 sin ✓ V r

�,r+�r sin ✓ V ✓

�,✓+�r V �

�,�

i=

1

r2 sin ✓

hsin ✓

�r2V r

�,r+ r

�sin ✓ V ✓

�,✓+ rV �,�

i(11.286)

as the divergence r · V .

11.36 The Covariant Laplacian

In flat 3-space, we write the laplacian as r ·r = r2 or as 4. In euclidiancoordinates, both mean @2x + @2y + @2z . In flat minkowski space, one oftenturns the triangle into a square and writes the 4-laplacian as 2 = 4� @2

0

.Since the gradient of a scalar field f is a covariant vector, we may use

the inverse metric tensor gij to write the laplacian 2f of a scalar f as thecovariant divergence of the contravariant vector gikf,k

2f = (gikf,k);i. (11.287)

The divergence formula (11.272) now expresses the invariant laplacian as

2f =(pg gikf,k),ip

g=

(pg f ,i),i.p

g(11.288)

To find the laplacian 2f in terms of forms, we apply the exterior derivativeto the Hodge dual (11.202) of the 1-form df = f,idxi

d ⇤ df = d�f,i ⇤ dxi

�= d

✓1

3!f,i g

ik ⌘k`mn dx` ^ dxm ^ dxn

◆=

1

3!

⇣f ,k pg

⌘,p✏k`mn dx

p ^ dx` ^ dxm ^ dxn(11.289)

and then add a star using (11.202)

⇤ d ⇤ df =1

3!

⇣f ,k pg

⌘,p✏k`mn ⇤

⇣dxp ^ dx` ^ dxm ^ dxn

⌘=

1

3!

⇣f ,k pg

⌘,p✏k`mn g

ptg`ugmvgnwpg ✏tuvw.

(11.290)


The definition (1.184) of the determinant now gives (exercise 11.19)

⇤ d ⇤ df =1

3!

⇣f ,k pg

⌘,p✏k`mn ✏

p`mn

pg

det g

=⇣f ,k pg

⌘,p�pk

spg=

spg

⇣f ,k pg

⌘,k.

(11.291)

In our space-time det gij = sg = �g, and so the laplacian is

2f = � ⇤ d ⇤ df =1pg

⇣f ,k pg

⌘,k. (11.292)

Example 11.20 (Invariant Laplacians) In two orthogonal coordinates,equations (11.154 & 11.155) imply that

pg =

p| det(gij)| = h

1

h2

and thatf ,i = gik f,k = h�2

i f,i, and so the laplacian (11.288) of a scalar f is

4f =1

h1

h2

2X

i=1

h1

h2

h2if,i

!,i. (11.293)

In polar coordinates, where h1

= 1, h2

= r, and g = r2, the laplacian is

4f =1

r

h(rf,r),r +

�r�1f,✓

�,✓

i= f,rr + r�1f,r + r�2f,✓✓. (11.294)

In three orthogonal coordinates, equations (11.154 & 11.155) imply thatpg =

p| det(gij)| = h

1

h2

h3

and that f ,i = gik f,k = h�2

i f,i, and so thelaplacian (11.288) of a scalar f is (6.33)

4f =1

h1

h2

h3

3X

i=1

h1

h2

h3

h2if,i

!,i. (11.295)

In cylindrical coordinates (section 11.22), h⇢ = 1, h� = ⇢, hz = 1, g = ⇢2,and the laplacian is

4f =1

⇢

(⇢ f,⇢),⇢ +

1

⇢f,�� + ⇢ f,zz

�= f,⇢⇢+

1

⇢f,⇢+

1

⇢2f,��+f,zz. (11.296)

In spherical coordinates (section 11.23), hr = 1, h✓ = r, h� = r sin ✓, andg = | det g| = r4 sin2 ✓. So (11.295) gives us the laplacian of f as (6.35)

4f =

�r2 sin ✓f,r

�,r+ (sin ✓f,✓),✓ + (f,�/ sin ✓),�

r2 sin ✓

=

�r2f,r

�,r

r2+

(sin ✓f,✓),✓r2 sin ✓

+f,��

r2 sin2 ✓. (11.297)

11.37 The Principle of Stationary Action 481

If the function f is a function only of the radial variable r, then the laplacianis simply

4f(r) =1

r2⇥r2f 0(r)

⇤0=

1

r[rf(r)]00 = f 00(r) +

2

rf 0(r) (11.298)

in which the primes denote r-derivatives.

11.37 The Principle of Stationary Action

It follows from a path-integral formulation of quantum mechanics that theclassical motion of a particle is given by the principle of stationary action�S = 0. In the simplest case of a free non-relativistic particle, the lagrangianis L = mx2/2 and the action is

S =

Z t2

t1

m

2x2 dt. (11.299)

The classical trajectory is the one that when varied slightly by �x (with�x(t

1

) = �x(t2

) = 0) does not change the action to first order in �x. Wefirst note that the change �x in the velocity is the time derivative of thechange in the path

�x = x0 � x =d

dt(x0 � x) =

d

dt�x. (11.300)

So since �x(t1

) = �x(t2

) = 0, the stationary path satisfies

0 = �S =

Z t2

t1

mx · �x dt =

Z t2

t1

mx · d�xdt

dt

=

Z t2

t1

m

d

dt(x · �x)�mx · �x

�dt

= m [x · �x]t2t1 �m

Z t2

t1

x · �x dt = �m

Z t2

t1

x · �x dt. (11.301)

If the first-order change in the action is to vanish for arbitrary small varia-tions �x in the path, then the acceleration must vanish

x = 0 (11.302)

which is the classical equation of motion for a free particle.


If the particle is moving under the influence of a potential V (x), then theaction is

S =

Z t2

t1

⇣m2x2 � V (x)

⌘dt. (11.303)

Since �V (x) = rV (x) · �x, the principle of stationary action requires that

0 = �S =

Z t2

t1

(�mx�rV ) · �x dt (11.304)

or

mx = �rV (11.305)

which is the classical equation of motion for a particle of mass m in a po-tential V .The action for a free particle of mass m in special relativity is

S = �m

Z ⌧2

⌧1

d⌧ = �Z t2

t1

mp1� x2 dt (11.306)

where c = 1 and x = dx/dt. The requirement of stationary action is

0 = �S = � �

Z t2

t1

mp

1� x2 dt = m

Z t2

t1

x · �xp1� x2

dt (11.307)

But 1/p1� x2 = dt/d⌧ and so

0 = �S = m

Z t2

t1

dx

dt· d�xdt

dt

d⌧dt = m

Z ⌧2

⌧1

dx

dt· d�xdt

dt

d⌧

dt

d⌧d⌧

= m

Z ⌧2

⌧1

dx

d⌧· d�xd⌧

d⌧. (11.308)

So integrating by parts, keeping in mind that �x(⌧2

) = �x(⌧1

) = 0, we have

0 = �S = m

Z ⌧2

⌧1

d

d⌧

✓dx

d⌧· �x

◆� d2x

d⌧2· �x

�d⌧ = �m

Z ⌧2

⌧1

d2x

d⌧2· �x d⌧.

(11.309)To have this hold for arbitrary �x, we need

d2x

d⌧2= 0 (11.310)

which is the equation of motion for a free particle in special relativity.

11.38 A Particle in a Gravitational Field 483

What about a charged particle in an electromagnetic field Ai? Its actionis

S = �m

Z ⌧2

⌧1

d⌧ + q

Z x2

x1

Ai(x) dxi =

Z ⌧2

⌧1

✓�m+ qAi(x)

dxi

d⌧

◆d⌧. (11.311)

We now treat the first term in a four-dimensional manner

�d⌧ = �p

�⌘ikdxidxk =�⌘ikdxi�dxkp�⌘ikdxidxk

= �uk�dxk = �ukd�x

k (11.312)

in which uk = dxk/d⌧ is the 4-velocity (11.66) and ⌘ is the Minkowski metric(11.27) of flat space-time. The variation of the other term is

��Ai dx

i�= (�Ai) dx

i +Ai �dxi = Ai,k�x

k dxi +Ai d�xi (11.313)

Putting them together, we get for �S

�S =

Z ⌧2

⌧1

✓muk

d�xk

d⌧+ qAi,k�x

k dxi

d⌧+ qAi

d�xi

d⌧

◆d⌧. (11.314)

After integrating by parts the last term, dropping the boundary terms, andchanging a dummy index, we get

�S =

Z ⌧2

⌧1

✓�m

dukd⌧

�xk + qAi,k�xk dx

i

d⌧� q

dAk

d⌧�xk◆

d⌧

=

Z ⌧2

⌧1

�m

dukd⌧

+ q (Ai,k �Ak,i)dxi

d⌧

��xk d⌧. (11.315)

If this first-order variation of the action is to vanish for arbitrary �xk, thenthe particle must follow the path

0 = �mdukd⌧

+ q (Ai,k �Ak,i)dxi

d⌧or

dpk

d⌧= qF kiui (11.316)

which is the Lorentz force law (11.96).

11.38 A Particle in a Gravitational Field

The invariant action for a particle of mass m moving along a path xi(t) is

S = �m

Z ⌧2

⌧1

d⌧ = �m

Z ⇣� gi`dx

idx`⌘ 1

2. (11.317)


Proceeding as in Eq.(11.312), we compute the variation �d⌧ as

�d⌧ = �p

�gi`dxidx` =��(gi`)dxidx` � 2gi`dxi�dx`

2p�gi`dxidx`

= � 1

2

gi`,k�xkuiu`d⌧ � giù

i�dx`

= � 1

2

gi`,k�xkuiu`d⌧ � giù

id�x` (11.318)

in which u` = dx`/d⌧ is the 4-velocity (11.66). The condition of stationaryaction then is

0 = �S = �m

Z ⌧2

⌧1

�d⌧ = m

Z ⌧2

⌧1

✓1

2

gi`,k�xkuiu` + giù

id�x`

d⌧

◆d⌧ (11.319)

which we integrate by parts keeping in mind that �x`(⌧2

) = �x`(⌧1

) = 0

0 = m

Z ⌧2

⌧1

✓1

2

gi`,k�xkuiu` � d(giùi)

d⌧�x`◆d⌧

= m

Z ⌧2

⌧1

✓1

2

gi`,k�xkuiu` � gi`,ku

iuk�x` � gi`dui

d⌧�x`◆d⌧. (11.320)

Now interchanging the dummy indices ` and k on the second and thirdterms, we have

0 = m

Z ⌧2

⌧1

✓1

2

gi`,kuiu` � gik,ù

iu` � gikdui

d⌧

◆�xkd⌧ (11.321)

or since �xk is arbitrary

0 = 1

2


iu` � gikdui

d⌧. (11.322)

If we multiply this equation of motion by grk and note that gik,ùiu` =g`k,iuiu`, then we find

0 =dur

d⌧+ 1

2

grk (gik,` + g`k,i � gi`,k)uiu`. (11.323)

So using the symmetry gi` = gì and the formula (11.255) for �ri`, we get

0 =dur

d⌧+ �ri` u

iu` or 0 =d2xr

d⌧2+ �ri`

dxi

d⌧

dx`

d⌧(11.324)

which is the geodesic equation. In empty space, particles fall along geodesicsindependently of their masses.The right-hand side of the geodesic equation (11.324) is a contravariant

vector because (Weinberg, 1972) under general coordinate transformations,the inhomogeneous terms arising from xr cancel those from �ri`x

ix`. Hereand often in what follows we’ll use dots to mean proper-time derivatives.

11.39 The Principle of Equivalence 485

The action for a particle of mass m and charge q in a gravitational field�ri` and an electromagnetic field Ai is

S = �m

Z ⇣� gi`dx

idx`⌘ 1

2+ q

Z ⌧2

⌧1

Ai(x) dxi (11.325)

because the interaction qRAidxi is invariant under general coordinate trans-

formations. By (11.315 & 11.321), the first-order change in S is

�S = m

Z ⌧2

⌧1

1

2


iu` � gikdui

d⌧+ q (Ai,k �Ak,i)u

i

��xkd⌧

(11.326)and so by combining the Lorentz force law (11.316) and the geodesic equation(11.324) and by writing F rixi as F r

i xi, we have

0 =d2xr

d⌧2+ �ri`

dxi

d⌧

dx`

d⌧� q

mF r

i

dxi

d⌧(11.327)

as the equation of motion of a particle of mass m and change q. It is strikinghow nearly perfect the electromagnetism of Faraday and Maxwell is.

The action of the electromagnetic field interacting with an electric currentjk in a gravitational field is

S =

Z h�1

4

Fk` Fk` + µ

0

Ak jkip

g d4x (11.328)

in whichpg d4x is the invariant volume element. After an integration by

parts, the first-order change in the action is

�S =

Z � @

@x`

⇣F k`pg

⌘+ µ

0

jkpg

��Ak d

4x, (11.329)

and so the inhomogeneous Maxwell equations in a gravitational field are

@

@x`

⇣pg F k`

⌘= µ

0

pg jk. (11.330)

11.39 The Principle of Equivalence

The principle of equivalence says that in any gravitational field, onemay choose free-fall coordinates in which all physical laws take the sameform as in special relativity without acceleration or gravitation—at leastover a suitably small volume of space-time. Within this volume and in thesecoordinates, things behave as they would at rest deep in empty space far


from any matter or energy. The volume must be small enough so that thegravitational field is constant throughout it.

Example 11.21 (Elevators) When a modern elevator starts going downfrom a high floor, it accelerates downward at something less than the localacceleration of gravity. One feels less pressure on one’s feet; one feels lighter.(This is as close to free fall as I like to get.) After accelerating downward for afew seconds, the elevator assumes a constant downward speed, and then onefeels the normal pressure of one’s weight on one’s feet. The elevator seemsto be slowing down for a stop, but actually it has just stopped acceleratingdownward.If in those first few seconds the elevator really were falling, then the physics

in it would be the same as if it were at rest in empty space far from anygravitational field. A clock in it would tick as fast as it would at rest in theabsence of gravity.

The transformation from arbitrary coordinates xk to free-fall coordinatesyi changes the metric gj` to the diagonal metric ⌘ik of flat space-time⌘ = diag(�1, 1, 1, 1), which has two indices and is not a Levi-Civita ten-sor. Algebraically, this transformation is a congruence (1.308)

⌘ik =@xj

@yigj`

@x`

@yk. (11.331)

The geodesic equation (11.324) follows from the principle of equiva-lence (Weinberg, 1972; Hobson et al., 2006). Suppose a particle is movingunder the influence of gravitation alone. Then one may choose free-fall co-ordinates y(x) so that the particle obeys the force-free equation of motion

d2yi

d⌧2= 0 (11.332)

with d⌧ the proper time d⌧2 = �⌘ik dyidyk. The chain rule applied to yi(x)in (11.332) gives

0 =d

d⌧

✓@yi

@xkdxk

d⌧

◆=@yi

@xkd2xk

d⌧2+

@2yi

@xk@x`dxk

d⌧

dx`

d⌧. (11.333)

We multiply by @xm/@yi and use the identity

@xm

@yi@yi

@xk= �mk (11.334)

11.40 Weak, Static Gravitational Fields 487

to get the equation of motion (11.332) in the x-coordinates

d2xm

d⌧2+ �mk`

dxk

d⌧

dx`

d⌧= 0 (11.335)

in which the a�ne connection is

�mk` =@xm

@yi@2yi

@xk@x`. (11.336)

So the principle of equivalence tells us that a particle in a gravitational fieldobeys the geodesic equation (11.324).

11.40 Weak, Static Gravitational Fields

Slow motion in a weak, static gravitational field is an important example.Because the motion is slow, we neglect ui compared to u0 and simplify thegeodesic equation (11.324) to

0 =dur

d⌧+ �r

00

(u0)2. (11.337)

Because the gravitational field is static, we neglect the time derivatives gk0,0and g

0k,0 in the connection formula (11.255) and find for �r00

�r00

= 1

2

grk (g0k,0 + g

0k,0 � g00,k) = �1

2

grk g00,k (11.338)

with �000

= 0. Because the field is weak, the metric can di↵er from ⌘ij byonly a tiny tensor gij = ⌘ij + hij so that to first order in |hij | ⌧ 1 wehave �r

00

= �1

2

h00,r for r = 1, 2, 3. With these simplifications, the geodesic

equation (11.324) reduces to

d2xr

d⌧2= 1

2

(u0)2 h00,r or

d2xr

d⌧2=

1

2

✓dx0

d⌧

◆2

h00,r. (11.339)

So for slow motion, the ordinary acceleration is described by Newton’s law

d2x

dt2=

c2

2rh

00

. (11.340)

If � is his potential, then for slow motion in weak static fields

g00

= �1 + h00

= �1� 2�/c2 and so h00

= � 2�/c2. (11.341)

Thus, if the particle is at a distance r from a mass M, then � = � GM/rand h

00

= �2�/c2 = 2GM/rc2 and so

d2x

dt2= �r� = rGM

r= �GM

r

r3. (11.342)


How weak are the static gravitational fields we know about? The dimen-sionless ratio �/c2 is 10�39 on the surface of a proton, 10�9 on the Earth,10�6 on the surface of the sun, and 10�4 on the surface of a white dwarf.

11.41 Gravitational Time Dilation

Suppose we have a system of coordinates xi with a metric gik and a clock atrest in this system. Then the proper time d⌧ between ticks of the clock is

d⌧ = (1/c)q�gij dxi dxj =

p�g

00

dt (11.343)

where dt is the time between ticks in the xi coordinates, which is the labo-ratory frame in the gravitational field g

00

. By the principle of equivalence(section 11.39), the proper time d⌧ between ticks is the same as the timebetween ticks when the same clock is at rest deep in empty space.If the clock is in a weak static gravitational field due to a mass M at a

distance r, then

� g00

= 1 + 2�/c2 = 1� 2GM/c2r (11.344)

is a little less than unity, and the interval of proper time between ticks

d⌧ =p�g

00

dt =p1� 2GM/c2r dt (11.345)

is slightly less than the interval dt between ticks in the coordinate systemof an observer at x in the rest frame of the clock and the mass, and inits gravitational field. Since dt > d⌧ , the laboratory time dt between ticksis greater than the proper or intrinsic time d⌧ between ticks of the clockuna↵ected by any gravitational field. Clocks near big masses run slow.Now suppose we have two identical clocks at di↵erent heights above sea

level. The time T` for the lower clock to make N ticks will be longer thanthe time Tu for the upper clock to make N ticks. The ratio of the clocktimes will be

T`

Tu=

p1� 2GM/c2(r + h)p

1� 2GM/c2r⇡ 1 +

gh

c2. (11.346)

Now imagine that a photon going down passes the upper clock which mea-sures its frequency as ⌫u and then passes the lower clock which measuresits frequency as ⌫`. The slower clock will measure a higher frequency. Theratio of the two frequencies will be the same as the ratio of the clock times

⌫`⌫u

= 1 +gh

c2. (11.347)

As measured by the lower, slower clock, the photon is blue shifted.

11.42 Curvature 489

Example 11.22 (Pound, Rebka, and Mossbauer) Pound and Rebka in1960 used the Mossbauer e↵ect to measure the blue shift of light fallingdown a 22.6 m shaft. They found

⌫` � ⌫u⌫

=gh

c2= 2.46⇥ 10�15 (11.348)

(Robert Pound 1919–2010, Glen Rebka 1931–, Rudolf Mossbauer 1929–2011).

Example 11.23 (Redshift of the Sun) A photon emitted with frequency⌫0

at a distance r from a mass M would be observed at spatial infinity tohave frequency ⌫

⌫ = ⌫0

p�g

00

= ⌫0

p1� 2MG/c2r (11.349)

for a red shift of �⌫ = ⌫0

�⌫. Since the Sun’s dimensionless potential ��/c2

is �MG/c2r = �2.12⇥ 10�6 at its surface, sunlight is shifted to the red by2 parts per million.

11.42 Curvature

The curvature tensor or Riemann tensor is

Rimnk = �imn,k � �imk,n + �ikj �

jnm � �inj �

jkm (11.350)

which we may write as the commutator

Rimnk = (Rnk)

im = [@k + �k, @n + �n]

im

= (�n,k � �k,n + �k �n � �n �k)i m(11.351)

in which the �’s are treated as matrices

�k =

0BB@�0k 0

�0k 1

�0k 2

�0k 3

�1k 0

�1k 1

�1k 2

�1k 3

�2k 0

�2k 1

�2k 2

�2k 3

�3k 0

�3k 1

�3k 2

�3k 3

1CCA (11.352)

with (�k �n)i m = �ikj �jnm and so forth. Just as there are two conventions

for the Faraday tensor Fik which di↵er by a minus sign, so too there are twoconventions for the curvature tensor Ri

mnk. Weinberg (Weinberg, 1972) usesthe definition (11.350); Carroll (Carroll, 2003) uses an extra minus sign.The Ricci tensor is a contraction of the curvature tensor

Rmk = Rnmnk (11.353)


and the curvature scalar is a further contraction

R = gmk Rmk. (11.354)

Example 11.24 (Curvature of a Sphere) While in four-dimensional space-time indices run from 0 to 3, on the sphere they are just ✓ and �. There areonly eight possible a�ne connections, and because of the symmetry (11.218)in their lower indices �i✓� = �i�✓, only six are independent.The point p on a sphere of radius r has cartesian coordinates

p = r (sin ✓ cos�, sin ✓ sin�, cos ✓) (11.355)

so the two 3-vectors are

e✓ =@p

@✓= r (cos ✓ cos�, cos ✓ sin�, � sin ✓) = r ✓

e� =@p

@�= r sin ✓ (� sin�, cos�, 0) = r sin ✓ �

(11.356)

and the metric gij = ei · ej is

(gij) =

✓r2 00 r2 sin2 ✓

◆. (11.357)

Di↵erentiating the vectors e✓ and e�, we find

e✓,✓ = � r (sin ✓ cos�, sin ✓ sin�, cos ✓) = �r r (11.358)

e✓,� = r cos ✓ (� sin�, cos�, 0) = r cos ✓ � (11.359)

e�,✓ = e✓,� (11.360)

e�,� = � r sin ✓ (cos�, sin�, 0) . (11.361)

The metric with upper indices (gij) is the inverse of the metric (gij)

(gij) =

✓r�2 00 r�2 sin�2 ✓

◆(11.362)

so the dual vectors ei are

e ✓ = r�1 (cos ✓ cos�, cos ✓ sin�, � sin ✓) = r�1✓

e� = =1

r sin ✓(� sin�, cos�, 0) =

1

r sin ✓�. (11.363)

The a�ne connections are given by (11.213) as

�ijk = �ikj = e i · ej,k. (11.364)

Since both e✓ and e� are perpendicular to r, the a�ne connections �✓✓✓ and

11.42 Curvature 491

��✓✓ both vanish. Also, e�,� is orthogonal to �, so �� = 0 as well. Similarly,

e✓,� is perpendicular to ✓, so �✓✓� = �✓�✓ also vanishes.The two nonzero a�ne connections are

��✓� = e� · e✓,� = r�1 sin�1 ✓ � · r cos ✓ � = cot ✓ (11.365)

and

�✓�� = e ✓ · e�,�= � sin ✓ (cos ✓ cos�, cos ✓ sin�, � sin ✓) · (cos�, sin�, 0)= � sin ✓ cos ✓. (11.366)

In terms of the two non-zero a�ne connections ��✓� = ��✓ = cot ✓ and

�✓�� = � sin ✓ cos ✓, the two Christo↵el matrices (11.352) are

�✓ =

0 0

0 ��✓�

!=

✓0 00 cot ✓

◆(11.367)

and

�� =

0 �✓��✓ 0

!=

✓0 � sin ✓ cos ✓

cot ✓ 0

◆. (11.368)

Their commutator is

[�✓,��] =

✓0 cos2 ✓

cot2 ✓ 0

◆= �[��,�✓] (11.369)

and both [�✓,�✓] and [��,��] vanish.So the commutator formula (11.351) gives for Riemann’s curvature tensor

R✓✓✓✓ = [@✓ + �✓, @✓ + �✓]

✓✓ = 0

R�✓�✓ = [@✓ + �✓, @� + ��]

�✓ = (��,✓)

�✓ + [�✓,��]

�✓

= (cot ✓),✓ + cot2 ✓ = �1

R✓�✓� = [@� + ��, @✓ + �✓]

✓� = � (��,✓)

✓� + [� �,�✓]

✓�

= cos2 ✓ � sin2 ✓ � cos2 ✓ = � sin2 ✓

R�� = [@� + ��, @� + ��]

�� = 0. (11.370)

The Ricci tensor (11.353) is the contraction Rmk = Rnmnk, and so

R✓✓ = R✓✓✓✓ +R�

✓�✓ = �1

R�� = R✓�✓� +R�

�� = � sin2 ✓.(11.371)


The curvature scalar (11.354) is the contraction R = gkmRmk, and so sinceg✓✓ = r�2 and g�� = r�2 sin�2 ✓, it is

R = g✓✓ R✓✓ + g��R�� = � r�2 � sin2 ✓ r�2 sin�2 ✓ = � 2

r2(11.372)

for a 2-sphere of radius r.Gauss invented a formula for the curvature K of a surface; for all two-

dimensional surfaces, his K = �R/2.

11.43 Einstein’s Equations

The source of the gravitational field is the energy-momentum tensor Tij .In many astrophysical and most cosmological models, the energy-momentumtensor is assumed to be that of a perfect fluid, which is isotropic in its restframe, does not conduct heat, and has zero viscosity. For a perfect fluid ofpressure p and density ⇢ with 4-velocity ui (defined by (11.66)), the energy-momentum or stress-energy tensor Tij is

Tij = p gij + (p+ ⇢)ui uj (11.373)

in which gij is the space-time metric.An important special case is the energy-momentum tensor due to a nonzero

value of the energy density of the vacuum. In this case p = �⇢ and theenergy-momentum tensor is

Tij = �⇢ gij (11.374)

in which ⇢ is the (presumably constant) value of the energy density of theground state of the theory. This energy density ⇢ is a plausible candidatefor the dark-energy density. It is equivalent to a cosmological constant⇤ = 8⇡G⇢.Whatever its nature, the energy-momentum tensor usually is defined so

as to satisfy the conservation law

0 =�T i

j

�;i= @iT

ij + �

iicT

cj � T i

c�cji. (11.375)

Einstein’s equations relate the Ricci tensor (11.353) and the scalar curva-ture (11.354) to the energy-momentum tensor

Rij � 1

2

gij R = �8⇡G

c4Tij (11.376)

11.44 The Action of General Relativity 493

in which G = 6.7087⇥ 10�39 ~c (GeV/c2)�2 = 6.6742⇥ 10�11m3 kg�1 s�2 isNewton’s constant. Taking the trace and using gji gij = �jj = 4, we relate

the scalar curvature to the trace T = T ii of the energy-momentum tensor

R =8⇡G

c4T. (11.377)

So another form of Einstein’s equations (11.376) is

Rij = � 8⇡G

c4

✓Tij �

T

2gij

◆. (11.378)

On small scales, such as that of our solar system, one may neglect darkenergy. So in empty space and on small scales, the energy-momentum tensorvanishes Tij = 0 along with its trace and the scalar curvature T = 0 = R,and Einstein’s equations are

Rij = 0. (11.379)

11.44 The Action of General Relativity

If we make an action that is a scalar, invariant under general coordinatetransformations, and then apply to it the principle of stationary action, wewill get tensor field equations that are invariant under general coordinatetransformations. If the metric of space-time is among the fields of the action,then the resulting theory will be a possible theory of gravity. If we makethe action as simple as possible, it will be Einstein’s theory.

To make the action of the gravitational field, we need a scalar. Apartfrom the volume 4-form ⇤1 =

pg d4x, the simplest scalar we can form from

the metric tensor and its first and second derivatives is the scalar curvatureR which gives us the Einstein-Hilbert action

SEH = � c4

16⇡G

ZRpg d4x = � c4

16⇡G

Zgik Rik

pg d4x. (11.380)

If �gik(x) is a tiny change in the inverse metric, then we may write thefirst-order change in the action SEH as (exercise 11.45)

�SEH = � c4

16⇡G

Z ✓Rik �

1

2gikR

◆pg �gik d4x. (11.381)

Thus the principle of least action �SEH = 0 leads to Einstein’s equations

Gik ⌘Rik �1

2gik R = 0 (11.382)


for empty space in which Gik is Einstein’s tensor.The stress-energy tensor Tik is defined so that the change in the action of

the matter fields due to a tiny change �gik(x) (vanishing at infinity) in themetric is

�Sm = � 1

2

ZTik

pg �gik d4x. (11.383)

So the principle of least action �S = �SEH + �Sm = 0 implies Einstein’sequations (11.376, 11.378) in the presence of matter and energy

Rik �1

2gik R = � 8⇡G

c4Tij or Rij = � 8⇡G

c4

✓Tij �

T

2gij

◆. (11.384)

11.45 Standard Form

Tensor equations are independent of the choice of coordinates, so it’s wiseto choose coordinates that simplify one’s work. For a static and isotropicgravitational field, this choice is the standard form (Weinberg, 1972, ch. 8)

d⌧2 = B(r) dt2 �A(r) dr2 � r2�d✓2 + sin2 ✓ d�2

�(11.385)

in which c = 1, and B(r) and A(r) are functions that one may find by solvingthe field equations (11.376). Since d⌧2 = � ds2 = �gij dxidxj , the nonzerocomponents of the metric tensor are grr = A(r), g✓✓ = r2, g�� = r2 sin2 ✓,and g

00

= �B(r), and those of its inverse are grr = A�1(r), g✓✓ = r�2,g�� = r�2 sin�2 ✓, and g00 = �B�1(r). By di↵erentiating the metric tensorand using (11.255), one gets the components of the connection �ik`, such as�✓�� = � sin ✓ cos ✓, and the components (11.353) of the Ricci tensor Rij ,such as (Weinberg, 1972, ch. 8)

Rrr =B00(r)

2B(r)� 1

4

✓B0(r)

B(r)

◆ ✓A0(r)

A(r)+

B0(r)

B(r)

◆� 1

r

✓A0(r)

A(r)

◆(11.386)

in which the primes mean d/dr.

11.46 Schwarzschild’s Solution

If one ignores the small dark-energy parameter ⇤, one may solve Einstein’sfield equations (11.379) in empty space

Rij = 0 (11.387)

11.47 Black Holes 495

outside a massM for the standard form of the Ricci tensor. One finds (Wein-berg, 1972) that A(r)B(r) = 1 and that r B(r) = r plus a constant, andone determines the constant by invoking the Newtonian limit g

00

= �B !�1 + 2MG/c2r as r ! 1. In 1916, Schwarzschild found the solution

d⌧2 =

✓1� 2MG

c2r

◆c2dt2 �

✓1� 2MG

c2r

◆�1

dr2 � r2�d✓2 + sin2 ✓ d�2

�(11.388)

which one can use to analyze orbits around a star. The singularity in

grr =

✓1� 2MG

c2r

◆�1

(11.389)

at the Schwarzschild radius r = 2MG/c2 is an artifact of the coordinates;the scalar curvature R and other invariant curvatures are not singular atthe Schwarzschild radius. Moreover, for the Sun, the Schwarzschild radiusr = 2MG/c2 is only 2.95 km, far less than the radius of the Sun, which is6.955⇥ 105 km. So the surface at r = 2MG/c2 is far from the empty spacein which Schwarzschild’s solution applies (Karl Schwarzschild 1873–1916) .

11.47 Black Holes

Suppose an uncharged, spherically symmetric star of mass M has collapsedwithin a sphere of radius rb less than its Schwarzschild radius r = 2MG/c2.Then for r > rb, the Schwarzschild metric (11.388) is correct. By Eq.(11.343),the apparent time dt of a process of proper time d⌧ at r � 2MG/c2 is

dt = d⌧/p�g

00

= d⌧/

r1� 2MG

c2r. (11.390)

The apparent time dt becomes infinite as r ! 2MG/c2. To outside ob-servers, the star seems frozen in time.

Due to the gravitational red shift (11.349), light of frequency ⌫p emittedat r � 2MG/c2 will have frequency ⌫

⌫ = ⌫pp�g

00

= ⌫p

r1� 2MG

c2r(11.391)

when observed at great distances. Light coming from the surface at r =2MG/c2 is red-shifted to zero frequency ⌫ = 0. The star is black. It is ablack hole with a surface or horizon at its Schwarzschild radius r = 2MG/c2,although there is no singularity there. If the radius of the Sun were less thanits Schwarzschild radius of 2.95 km, then the Sun would be a black hole. Theradius of the Sun is 6.955⇥ 105 km.


Black holes are not really black. Stephen Hawking (1942–) has shownthat the intense gravitational field of a black hole of mass M radiates attemperature

T =~ c3

8⇡ k GM(11.392)

in which k = 8.617343 ⇥ 10�5 eV K�1 is Boltzmann’s constant, and ~ isPlanck’s constant h = 6.6260693⇥ 10�34 J s divided by 2⇡, ~ = h/(2⇡).The black hole is entirely converted into radiation after a time

t =5120⇡G2

~ c4 M3 (11.393)

proportional to the cube of its mass.

11.48 Cosmology

Astrophysical observations tell us that on the largest observable scales, spaceis flat or very nearly flat; that the visible universe contains at least 1090

particles; and that the cosmic microwave background radiation is isotropicto one part in 105 apart from a Doppler shift due the motion of the Earth.These and other observations suggest that potential energy expanded ouruniverse by exp(60) = 1026 during an era of inflation that could havebeen as brief as 10�35 s. The potential energy that powered inflation becamethe radiation of the Big Bang. During and after inflation, the (negative)gravitational potential energy kept the total energy constant.During the first three minutes, some of that radiation became hydrogen,

helium, neutrinos, and dark matter. But for 50,000 years after the BigBang, most of the energy of the visible universe was radiation. Because themomentum of a particle but not its mass falls with the expansion of theuniverse, this era of radiation gradually gave way to an era of matter.This transition happened when the temperature kT of the universe fell to0.81 eV.The era of matter lasted for 10.3 billion years. After 380,000 years, the

universe had cooled to kT = 0.26 eV, and less than 1% of the atoms wereionized. Photons no longer scattered o↵ a plasma of electrons and ions.The universe became transparent. The photons that last scattered justbefore this initial transparency became the cosmic microwave back-ground radiation or CMBR that now surrounds us, red-shifted to 2.7255±0.0006K. Between 10 and 17 million years after the Big Bang, the temper-ature of the known universe fell from 373 to 273 K. If and where very early,

11.48 Cosmology 497

very heavy stars had produced carbon, nitrogen, and oxygen, biochemistrywould have started.

The era of matter has been followed by the current era of dark energyduring which the energy of the visible universe is mostly a potential energycalled dark energy (something like a cosmological constant). Darkenergy has been accelerating the expansion of the universe for the past 3.5billion years and may continue to do so forever.

It is now 13.817 ± 0.048 billion years after the Big Bang, and the dark-energy density is ⇢de = 5.827⇥ 10�30 c2 g cm�3 or 68.5 percent (± 1.8%) ofthe critical energy density ⇢c = 3H2

0

/8⇡G = 1.87837h2⇥10�29 c2 g cm�3

needed to make the universe flat. Here H0

= 100h km s�1Mpc�1 is theHubble constant, one parsec is 3.262 light-years, the Hubble time is1/H

0

= 9.778h�1 ⇥ 109 years, and h = 0.673± 0.012 is not to be confusedwith Planck’s constant.

Matter makes up 31.5± 1.8% of the critical density, and baryons only4.9± 0.06% of it. Baryons are 15% of the total matter in the visible uni-verse. The other 85% does not interact with light and is called dark mat-ter.

Einstein’s equations (11.376) are second-order, non-linear partial di↵er-ential equations for 10 unknown functions gij(x) in terms of the energy-momentum tensor Tij(x) throughout the universe, which of course we don’tknow. The problem is not quite hopeless, however. The ability to choose ar-bitrary coordinates, the appeal to symmetry, and the choice of a reasonableform for Tij all help.

Hubble showed us that the universe is expanding. The cosmic microwavebackground radiation looks the same in all spatial directions (apart froma Doppler shift due to the motion of the Earth relative to the local super-cluster of galaxies). Observations of clusters of galaxies reveal a universethat is homogeneous on suitably large scales of distance. So it is plausiblethat the universe is homogeneous and isotropic in space, but not in time.One may show (Carroll, 2003) that for a universe of such symmtery, the lineelement in comoving coordinates is

ds2 = �dt2 + a2

dr2

1� k r2+ r2

�d✓2 + sin2 ✓ d�2

��. (11.394)

Whitney’s embedding theorem tells us that any smooth four-dimensionalmanifold can be embedded in a flat space of eight dimensions with a suitablesignature. We need only four or five dimensions to embed the space-timedescribed by the line element (11.394). If the universe is closed, then the


signature is (�1, 1, 1, 1, 1), and our three-dimensional space is the 3-spherewhich is the surface of a four-dimensional sphere in four space dimensions.The points of the universe then are

p = (t, a sin� sin ✓ cos�, a sin� sin ✓ sin�, a sin� cos ✓, a cos�) (11.395)

in which 0 � ⇡, 0 ✓ ⇡, and 0 � 2⇡. If the universe is flat, thenthe embedding space is flat, four-dimensional Minkowski space with points

p = (t, ar sin ✓ cos�, ar sin ✓ sin�, ar cos ✓) (11.396)

in which 0 ✓ ⇡ and 0 � 2⇡. If the universe is open, then the em-bedding space is a flat five-dimensional space with signature (�1, 1, 1, 1,�1),and our three-dimensional space is a hyperboloid in a flat Minkowski spaceof four dimensions. The points of the universe then are

p = (t, a sinh� sin ✓ cos�, a sinh� sin ✓ sin�, a sinh� cos ✓, a cosh�)(11.397)

in which 0 � 1, 0 ✓ ⇡, and 0 � 2⇡.In all three cases, the corresponding Robertson-Walker metric is

gij =

0BB@�1 0 0 00 a2/(1� kr2) 0 00 0 a2 r2 00 0 0 a2 r2 sin2 ✓

1CCA (11.398)

in which the coordinates (t, r, ✓,�) are numbered (0, 1, 2, 3), the speed oflight is c = 1, and k is a constant. One always may choose coordinates(exercise 11.30) such that k is either 0 or ±1. This constant determineswhether the spatial universe is open k = �1, flat k = 0, or closed k = 1.The scale factor a, which in general is a function of time a(t), tells ushow space expands and contracts. These coordinates are called comovingbecause a point at rest (fixed r, ✓,�) sees the same Doppler shift in alldirections.The metric (11.398) is diagonal; its inverse gij also is diagonal; and so we

may use our formula (11.255) to compute the a�ne connections �ki`, such as

�0`` =1

2

g0k (g`k,` + g`k,` � g``,k) =1

2

g00 (g`0,` + g`0,` � g``,0) =1

2

g``,0(11.399)

so that

�011

=aa

1� kr2�022

= aa r2 and �022

= aa r2 sin2 ✓. (11.400)

11.48 Cosmology 499

in which a dot means a time-derivative. The other �0ij ’s vanish. Similarly,for fixed ` = 1, 2, or 3

�`0` =

1

2

g`k (g0k,` + g`k,0 � g

0`,k)

= 1

2

g`` (g0`,` + g``,0 � g

0`,`)

= 1

2

g`` g``,0 =a

a= �``0 no sum over `. (11.401)

The other nonzero �’s are

�122

= �r (1� kr2) �133

= �r (1� kr2) sin2 ✓ (11.402)

�212

= �313

=1

r= �2

21

= �331

(11.403)

�233

= � sin ✓ cos ✓ �323

= cot ✓ = �332

. (11.404)

Our formulas (11.353 & 11.351) for the Ricci and curvature tensors give

R00

= Rn0n0 = [@

0

+ �0

, @n + �n]n0

. (11.405)

Clearly the commutator of �0

with itself vanishes, and one may use theformulas (11.400–11.404) for the other connections to check that

[�0

,�n]n0

= �n0 k �

kn 0

� �nnk �k0 0

= 3

✓a

a

◆2

(11.406)

and that

@0

�nn 0

= 3 @0

✓a

a

◆= 3

a

a� 3

✓a

a

◆2

(11.407)

while @n�n0 0

= 0. So the 00-component of the Ricci tensor is

R00

= 3a

a. (11.408)

Similarly, one may show that the other non-zero components of Ricci’stensor are

R11

= � A

1� kr2R

22

= �r2A and R33

= �r2A sin2 ✓ (11.409)

in which A = aa+ 2a2 + 2k. The scalar curvature (11.354) is

R = gabRba = � 6

a2�aa+ a2 + k

�. (11.410)

In co-moving coordinates such as those of the Robertson-Walker metric


(11.398) ui = (1, 0, 0, 0), and so the energy-momentum tensor (11.373) is

Tij =

0BB@⇢ 0 0 00 p g

11

0 00 0 p g

22

00 0 0 p g

33

1CCA . (11.411)

Its trace is

T = gij Tij = �⇢+ 3p. (11.412)

Thus using our formula (11.398) for g00

= �1, (11.408) for R00

, (11.411)for Tij , and (11.412) for T , we find that the 00 Einstein equation (11.378)becomes the second-order equation

a

a= �4⇡G

3(⇢+ 3p) (11.413)

which is nonlinear because ⇢ and 3p depend upon a. The sum ⇢ + 3p de-termines the acceleration a of the scale factor a(t). When it is negative, itaccelerates the expansion of the universe.Because of the isotropy of the metric, the three nonzero spatial Einstein

equations (11.378) give us only one relation

a

a+ 2

✓a

a

◆2

+ 2k

a2= 4⇡G (⇢� p) . (11.414)

Using the 00-equation (11.413) to eliminate the second derivative a, we have

✓a

a

◆2

=8⇡G

3⇢� k

a2(11.415)

which is a first-order nonlinear equation. It and the second-order equation(11.413) are known as the Friedmann equations.The LHS of the first-order Friedmann equation (11.415) is the square of

the Hubble rate

H =a

a(11.416)

which is an inverse time. Its present value H0

is the Hubble constant. Interms of H, Friedmann’s first-order equation (11.415) is

H2 =8⇡G

3⇢� k

a2. (11.417)

The energy density of a flat universe with k = 0 is the critical energy

11.48 Cosmology 501

density

⇢c =3H2

8⇡G. (11.418)

The ratio of the energy density ⇢ to the critical energy density is called ⌦

⌦ =⇢

⇢c=

8⇡G

3H2

⇢. (11.419)

From (11.417), we see that ⌦ is

⌦ = 1 +k

(aH)2= 1 +

k

a2. (11.420)

Thus ⌦ = 1 both in a flat universe (k = 0) and as aH ! 1. One use ofinflation is to expand a by 1026 so as to force ⌦ almost exactly to unity.

Something like inflation is needed because in a universe in which theenergy density is due to matter and/or radiation, the present value of ⌦

⌦0

= 1.000± 0.036 (11.421)

is unlikely. To see why, we note that conservation of energy ensures that a3

times the matter density ⇢m is constant. Radiation red-shifts by a, so energyconservation implies that a4 times the radiation density ⇢r is constant. Sowith n = 3 for matter and 4 for radiation, ⇢ an ⌘ 3F 2/8⇡G is a constant.In terms of F and n, Friedmann’s first-order equation (11.415) is

a2 =8⇡G

3⇢ a2 � k =

F 2

an�2

� k (11.422)

In the small-a limit of the early Universe, we have

a = F/a(n�2)/2 or a(n�2)/2da = F dt (11.423)

which we integrate to a ⇠ t2/n so that a ⇠ t2/n�1. Now (11.420) says that

|⌦� 1| = 1

a2/ t2�4/n =

⇢t radiationt2/3 matter

. (11.424)

Thus, ⌦ deviated from unity faster than t2/3 during the early Universe. Atthis rate, the inequality |⌦

0

� 1| < 0.036 could last 13.8 billion years onlyif ⌦ at t = 1 second had been unity to within six parts in 1014. The onlyknown explanation for such early flatness is inflation.

Manipulating our relation (11.420) between ⌦ and aH, we see that

(aH)2 =k

⌦� 1. (11.425)

So ⌦ > 1 implies k = 1, and ⌦ < 1 implies k = �1, and as ⌦ ! 1 the


product aH ! 1, which is the essence of flatness since curvature vanishesas the scale factor a ! 1. Imagine blowing up a balloon.Staying for the moment with a universe without inflation and with an

energy density composed of radiation and/or matter, we note that the first-order equation (11.422) in the form a2 = F 2/an�2 � k tells us that for aclosed (k = 1) universe, in the limit a ! 1 we’d have a2 ! �1 which is im-possible. Thus a closed universe eventually collapses, which is incompatiblewith the flatness (11.425) implied by the present value ⌦

0

= 1.000± 0.036.The first-order equation Friedmann (11.415) says that ⇢ a2 � 3k/8⇡G. So

in a closed universe (k = 1), the energy density ⇢ is positive and increaseswithout limit as a ! 0 as in a collapse. In open (k < 0) and flat (k = 0) uni-verses, the same Friedmann equation (11.415) in the form a2 = 8⇡G⇢a2/3�ktells us that if ⇢ is positive, then a2 > 0, which means that a never vanishes.Hubble told us that a > 0 now. So if our universe is open or flat, then italways expands.Due to the expansion of the universe, the wave-length of radiation grows

with the scale factor a(t). A photon emitted at time t and scale factor a(t)with wave-length �(t) will be seen now at time t

0

and scale factor a(t0

) tohave a longer wave-length �(t

0

)

�(t0

)

�(t)=

a(t0

)

a(t)= z + 1 (11.426)

in which the redshift z is the ratio

z =�(t

0

)� �(t)

�(t)=��

�. (11.427)

Now H = a/a = da/(adt) implies dt = da/(aH), and z = a0

/a � 1 impliesdz = �a

0

da/a2, so we find

dt = � dz

(1 + z)H(z)(11.428)

which relates time intervals to redshift intervals. An on-line calculator isavailable for macroscopic intervals (Wright, 2006).

11.49 Model Cosmologies

The 0-component of the energy-momentum conservation law (11.375) is

0 = (T a0

);a = @aT

a0

+ �aacTc0

� T ac�

c0a

= �@0

T00

� �aa0T00

� gccTcc�c0c

= � ⇢� 3a

a⇢� 3p

a

a= � ⇢� 3

a

a(⇢+ p) . (11.429)

11.49 Model Cosmologies 503

ord⇢

da= �3

a(⇢+ p) . (11.430)

The energy density ⇢ is composed of fractions ⇢k each contributing its ownpartial pressure pk according to its own equation of state

pk = wk⇢k (11.431)

in which wk is a constant. In terms of these components, the energy-momentum conservation law (11.430) isX

k

d⇢kda

= �3

a

Xk

(1 + wk) ⇢k (11.432)

with solution

⇢ =Xk

⇢k

✓a

a

◆3(1+w

k

)

=Xk

⇢k

✓a

a

◆3(1+p

k

/⇢k

)

. (11.433)

Simple cosmological models take the energy density and pressure each tohave a single component with p = w⇢, and in this case

⇢ = ⇢

✓a

a

◆3(1+w)

= ⇢

✓a

a

◆3(1+p/⇢)

. (11.434)

Example 11.25 (w = �1/3, No Acceleration) If w = �1/3, then p =w ⇢ = �⇢/3 and ⇢+3p = 0. The second-order Friedmann equation (11.413)then tells us that a = 0. The scale factor does not accelerate.To find its constant speed, we use its equation of state (11.434)

⇢ = ⇢

✓a

a

◆3(1+w)

= ⇢

✓a

a

◆2

. (11.435)

Now all the terms in Friedmann’s first-order equation (11.415) have a com-mon factor of 1/a2 which cancels leaving us with the square of the constantspeed

a2 =8⇡G

3⇢ a2 � k. (11.436)

Incidentally, ⇢ a2 must exceed 3k/8⇡G. The scale factor grows linearly withtime as

a(t) =

✓8⇡G

3⇢ a2 � k

◆1/2

(t� t0

) + a(t0

). (11.437)


Setting t0

= 0 and a(0) = 0, we use the definition of the Hubble parameterH = a/a to write the constant linear growth a as aH and the time as

t =

Z a

0

da0/a0H = (1/aH)

Z a

0

da0 = 1/H. (11.438)

So in a universe without acceleration, the age of the universe is the inverseof the Hubble rate. For our universe, the present Hubble time is 1/H

0

=14.5 billion years, which isn’t far from the actual age of 13.817 ± 0.048billion years. Presumably, a slower Hubble rate during the era of mattercompensates for the higher rate during the era of dark energy.

Example 11.26 (w = �1, Inflation) Inflation occurs when the groundstate of the theory has a positive and constant energy density ⇢ > 0 thatdwarfs the energy densities of the matter and radiation. The internal en-ergy of the universe then is proportional to its volume U = ⇢V , and thepressure p as given by the thermodynamic relation

p = �@U@V

= �⇢ (11.439)

is negative. The equation of state (11.431) tells us that in this case w = �1.The second-order Friedmann equation (11.413) becomes

a

a= �4⇡G

3(⇢+ 3p) =

8⇡G⇢

3⌘ g2 (11.440)

By it and the first-order Friedmann equation (11.415) and by choosing t = 0as the time at which the scale factor a is minimal, one may show (exercise11.37) that in a closed (k = 1) universe

a(t) =cosh g t

g. (11.441)

Similarly in an open (k = �1) universe with a(0) = 0, we have

a(t) =sinh g t

g. (11.442)

Finally in a flat (k = 0) expanding universe, the scale factor is

a(t) = a(0) exp(g t). (11.443)

Studies of the cosmic microwave background radiation suggest that infla-tion did occur in the very early universe—possibly on a time scale as shortas 10�35 s. What is the origin of the vacuum energy density ⇢ that drove


inflation? Current theories attribute it to the assumption by at least onescalar field � of a mean value h�i di↵erent from the one h0|�|0i that min-imizes the energy density of the vacuum. When h�i settled to h0|�|0i, thevacuum energy was released as radiation and matter in a Big Bang.

Example 11.27 (w = 1/3, The Era of Radiation) Until a redshift ofz = 3400 or 50,000 years after inflation, our universe was dominated by ra-diation (Frieman et al., 2008). During The First Three Minutes (Weinberg,1988) of the era of radiation, the quarks and gluons formed hadrons, whichdecayed into protons and neutrons. As the neutrons decayed (⌧ = 885.7s), they and the protons formed the light elements—principally hydrogen,deuterium, and helium in a process called big-bang nucleosynthesis.

We can guess the value of w for radiation by noticing that the energy-momentum tensor of the electromagnetic field (in suitable units)

T ab = F acF

bc � 1

4gabFcdF

cd (11.444)

is traceless

T = T aa = F a

cFc

a � 1

4�aaFcdF

cd = 0. (11.445)

But by (11.412) its trace must be T = 3p� ⇢. So for radiation p = ⇢/3 andw = 1/3. The relation (11.434) between the energy density and the scalefactor then is

⇢ = ⇢

✓a

a

◆4

. (11.446)

The energy drops both with the volume a3 and with the scale factor a dueto a redshift; so it drops as 1/a4. Thus the quantity

f2 ⌘ 8⇡G⇢a4

3(11.447)

is a constant. The Friedmann equations (11.413 & 11.414) now are

a

a= �4⇡G

3(⇢+ 3p) = �8⇡G⇢

3or a = �f2

a3(11.448)

and

a2 + k =f2

a2. (11.449)


With calendars chosen so that a(0) = 0, this last equation (11.449) tellsus that for a flat universe (k = 0)

a(t) = (2f t)1/2 (11.450)

while for a closed universe (k = 1)

a(t) =qf2 � (t� f)2 (11.451)

and for an open universe (k = �1)

a(t) =q(t+ f)2 � f2 (11.452)

as we saw in (6.422). The scale factor (11.451) of a closed universe ofradiation has a maximum a = f at t = f and falls back to zero at t = 2f .

Example 11.28 (w = 0, The Era of Matter) A universe composed onlyof dust or non-relativistic collisionless matter has no pressure. Thusp = w⇢ = 0 with ⇢ 6= 0, and so w = 0. Conservation of energy (11.433), orequivalently (11.434), implies that the energy density falls with the volume

⇢ = ⇢

✓a

a

◆3

. (11.453)

As the scale factor a(t) increases, the matter energy density, which fallsas 1/a3, eventually dominates the radiation energy density, which falls as1/a4. This happened in our universe about 50,000 years after inflation ata temperature of T = 9, 400 K or kT = 0.81 eV. Were baryons mostof the matter, the era of radiation dominance would have lasted for a fewhundred thousand years. But the kind of matter that we know about, whichinteracts with photons, is only about 15% of the total; the rest—an unknownsubstance called dark matter—shortened the era of radiation dominanceby nearly 2 million years.Since ⇢ / 1/a3, the quantity

m2 =4⇡G⇢a3

3(11.454)

is a constant. For a matter-dominated universe, the Friedmann equations(11.413 & 11.414) then are

a

a= �4⇡G

3(⇢+ 3p) = �4⇡G⇢

3or a = �m2

a2(11.455)


and

a2 + k = 2m2/a. (11.456)

For a flat universe, k = 0, we get

a(t) =

3mp2t

�2/3

. (11.457)

For a closed universe, k = 1, we use example 6.47 to integrate

a =p2m2/a� 1 (11.458)

to

t� t0

= �pa(2m2 � a) �m2 arcsin(1� a/m2). (11.459)

With a suitable calendar and choice of t0

, one may parametrize this solutionin terms of the development angle �(t) as

a(t) = m2 [1� cos�(t)]

t = m2 [�(t)� sin�(t)] . (11.460)

For an open universe, k = �1, we use example 6.48 to integrate

a =p2m2/a+ 1 (11.461)

to

t�t0

=⇥a(2m2 + a)

⇤1/2�m2 ln

n2⇥a(2m2 + a)

⇤1/2

+ 2a+ 2m2

o. (11.462)

The conventional parametrization is

a(t) = m2 [cosh�(t)� 1]

t = m2 [sinh�(t)� �(t)] . (11.463)

Transparency: Some 380,000 years after inflation at a redshift of z =1090, the universe had cooled to about T = 3000 K or kT = 0.26 eV—a tem-perature at which less than 1% of the hydrogen is ionized. Ordinary matterbecame a gas of neutral atoms rather than a plasma of ions and electrons,and the universe suddenly became transparent to light. Some scientistscall this moment of last scattering or first transparency recombination.


Example 11.29 (w = �1, The Era of Dark Energy) About 10.3 billionyears after inflation at a redshift of z = 0.30, the matter density falling as1/a3 dropped below the very small but positive value of the energy den-sity ⇢v = (2.23 meV)4 of the vacuum. The present time is 13.817 billionyears after inflation. So for the past 3 billion years, this constant energydensity, called dark energy, has accelerated the expansion of the universeapproximately as (11.442)

a(t) = a(tm) exp⇣(t� tm)

p8⇡G⇢v/3

⌘(11.464)

in which tm = 10.3⇥ 109 years.

Observations and measurements on the largest scales indicate that theuniverse is flat: k = 0. So the evolution of the scale factor a(t) is given bythe k = 0 equations (11.443, 11.450, 11.457, & 11.464) for a flat universe.During the brief era of inflation, the scale factor a(t) grew as (11.443)

a(t) = a(0) exp⇣tp

8⇡G⇢i/3⌘

(11.465)

in which ⇢i is the positive energy density that drove inflation.During the 50,000-year era of radiation, a(t) grew as

pt as in (11.450)

a(t) =⇣2 (t� ti)

p8⇡G⇢(t0r)a

4(t0r)/3⌘1/2

+ a(ti) (11.466)

where ti is the time at the end of inflation, and t0r is any time during theera of radiation. During this era, the energy of highly relativistic particlesdominated the energy density, and ⇢a4 / T 4a4 was approximately con-stant, so that T (t) / 1/a(t) / 1/

pt. When the temperature was in the

range 1012 > T > 1010K or mµc2 > kT > mec2, where mµ is the mass ofthe muon and me that of the electron, the radiation was mostly electrons,positrons, photons, and neutrinos, and the relation between the time t andthe temperature T was (Weinberg, 2010, ch. 3)

t = 0.994 sec⇥1010K

T

�2

+ constant. (11.467)

By 109 K, the positrons had annihilated with electrons, and the neutrinosfallen out of equilibrium. Between 109 K and 106K, when the energy densityof nonrelativistic particles became relevant, the time-temperature relationwas (Weinberg, 2010, ch. 3)

t = 1.78 sec⇥1010K

T

�2

+ constant0. (11.468)

11.50 Yang-Mills Theory 509

During the 10.3 billion years of the matter era, a(t) grew as (11.457)

a(t) =h(t� tr)

p3⇡G⇢(t0m)a(t0m) + a3/2(tr)

i2/3

+ a(tr) (11.469)

where tr is the time at the end of the radiation era, and t0m is any time inthe matter era. By 380,000 years, the temperature had dropped to 3000K,the universe had become transparent, and the CMBR had begun to travelfreely.Over the past 3 billion years of the era of vacuum dominance, a(t) has

been growing exponentially (11.464)

a(t) = a(tm) exp⇣(t� tm)

p8⇡G⇢v/3

⌘(11.470)

in which tm is the time at the end of the matter era, and ⇢v is the densityof dark energy, which while vastly less than the energy density ⇢i that droveinflation, currently amounts to 68.5% of the total energy density.

11.50 Yang-Mills Theory

The gauge transformation of an abelian gauge theory like electrodynam-ics multiplies a single charged field by a space-time-dependent phase factor�0(x) = exp(iq✓(x))�(x). Yang and Mills generalized this gauge transfor-mation to one that multiplies a vector � of matter fields by a space-timedependent unitary matrix U(x)

�0a(x) =nX

b=1

Uab(x)�b(x) or �0(x) = U(x)�(x) (11.471)

and showed how to make the action of the theory invariant under such non-abelian gauge transformations. (The fields � are scalars for simplicity.)Since the matrix U is unitary, inner products like �†(x)�(x) are automat-

ically invariant⇣�†(x)�(x)

⌘0= �†(x)U †(x)U(x)�(x) = �†(x)�(x). (11.472)

But inner products of derivatives @i�† @i� are not invariant because thederivative acts on the matrix U(x) as well as on the field �(x).Yang and Mills made derivatives Di� that transform like the fields �

(Di�)0 = U Di�. (11.473)

To do so, they introduced gauge-field matrices Ai that play the role of


the connections �i in general relativity and set

Di = @i +Ai (11.474)

in which Ai like @i is antihermitian. They required that under the gaugetransformation (11.471), the gauge-field matrix Ai transform to A0

i in sucha way as to make the derivatives transform as in (11.473)

(Di�)0 =

�@i +A0

i

��0 =

�@i +A0

i

�U� = U Di� = U (@i +Ai)�. (11.475)

So they set�@i +A0

i

�U� = U (@i +Ai)� or (@iU)�+A0

i U� = UAi �. (11.476)

and made the gauge-field matrix Ai transform as

A0i = UAiU

�1 � (@iU)U�1. (11.477)

Thus under the gauge transformation (11.471), the derivative Di� trans-forms as in (11.473), like the vector � in (11.471), and the inner product ofcovariant derivativesh�

Di��†

Di�i0

=�Di�

�†U †UDi� =

�Di�

�†Di� (11.478)

remains invariant.To make an invariant action density for the gauge-field matrices Ai, they

used the transformation law (11.475) which implies that D0i U� = UDi � or

D0i = UDi U�1. So they defined their generalized Faraday tensor as

Fik = [Di, Dk] = @iAk � @kAi + [Ai, Ak] (11.479)

which transforms covariantly

F 0ik = UFikU

�1. (11.480)

They then generalized the action density FikF ik of electrodynamics to thetrace Tr

�FikF ik

�of the square of the Faraday matrices which is invariant

under gauge transformations since

Tr⇣UFikU

�1UF ikU�1

⌘= Tr

⇣UFikF

ikU�1

⌘= Tr

⇣FikF

ik⌘. (11.481)

As an action density for fermionic matter fields, they replaced the ordi-nary derivative in Dirac’s formula (�i@i +m) by the covariant derivative(11.474) to get (�iDi + m) (Chen-Ning Yang 1922–, Robert L. Mills1927–1999).In an abelian gauge theory, the square of the 1-form A = Ai dxi vanishes

11.51 Gauge Theory and Vectors 511

A2 = AiAk dxi^dxk = 0, but in a nonabelian gauge theory the gauge fieldsare matrices, and A2 6= 0. The sum dA+A2 is the Faraday 2-form

F = dA+A2 = (@iAk +AiAk) dxi ^ dxk (11.482)

= 1

2

(@iAk � @k Ai + [AiAk]) dxi ^ dxk = 1

2

Fik dxi ^ dxk.

The scalar matter fields � may have self-interactions described by a po-tential V (�) such as V (�) = �(�†��m2/�)2 which is positive unless �†� =m2/�. The kinetic action of these fields is (Di�)†Di�. At low tempera-tures, these scalar fields assume mean values h0|�|0i = �

0

in the vacuumwith �†

0

�0

= m2/� so as to minimize their potential energy density V (�).Their kinetic action (Di�)†Di� = (@i�+ Ai�)†(@i�+ Ai�) then is in e↵ect�†0

AiAi�0. The gauge-field matrix Aiab = i t↵abA

i↵ is a linear combination

of the generators t↵ of the gauge group. So the action of the scalar fieldscontains the term �†

0

AiAi �0 = �M2

↵� Ai↵Ai� in which the mass-squared

matrix for the gauge fields is M2

↵� = �⇤a0

t↵ab t�bc �

c0

. This Higgs mechanismgives masses to those linear combinations b�iA� of the gauge fields for whichM2

↵� b�i = m2

i b↵i 6= 0 .The Higgs mechanism also gives masses to the fermions. The mass termm

in the Yang-Mills-Dirac action is replaced by something like c� in which c is aconstant, di↵erent for each fermion. In the vacuum and at low temperatures,each fermion acquires as its mass c�

0

. On 4 July 2012, physicists at CERN’sLarge Hadron Collider announced the discovery of a Higgs-like particle witha mass near 126 GeV/c2 (Peter Higgs 1929–).

11.51 Gauge Theory and Vectors

This section is optional on a first reading.We can formulate Yang-Mills theory in terms of vectors as we did rela-

tivity. To accomodate noncompact groups, we will generalize the unitarymatrices U(x) of the Yang-Mills gauge group to nonsingular matrices V (x)that act on n matter fields a(x) as

0a(x) =nX

a=1

V ab(x)

b(x). (11.483)

The field

(x) =nX

a=1

ea(x) a(x) (11.484)


will be gauge invariant 0(x) = (x) if the vectors ea(x) transform as

e0a(x) =nX

b=1

eb(x)V�1b

a(x). (11.485)

In what follows, we will sum over repeated indices from 1 to n and often willsuppress explicit mention of the space-time coordinates. In this compressednotation, the field is gauge invariant because

0 = e0a 0a = eb V

�1ba V

ac

c = eb �bc

c = eb b = (11.486)

which is e0T 0 = eTV �1V = eT in matrix notation.The inner product of two basis vectors is an internal “metric tensor”

e⇤a · eb =NX

↵=1

NX�=1

e↵⇤a ⌘↵�e↵b =

NX↵=1

e↵⇤a e↵b = gab (11.487)

in which for simplicity I used the the N -dimensional identity matrix for themetric ⌘. As in relativity, we’ll assume the matrix gab to be nonsingular.We then can use its inverse to construct dual vectors ea = gabeb that satisfyea† · eb = �ab .The free Dirac action density of the invariant field

(�i@i +m) = aea†(�i@i +m)eb

b = a

h�i(�ab@i + ea† · eb,i) +m�ab

i b

(11.488)is the full action of the component fields b

(�i@i +m) = a(�iDa

i b +m �ab) b = a

⇥�i(�ab@i +Aa

i b) +m �ab⇤ b

(11.489)if we identify the gauge-field matrix as Aa

i b = ea† · eb,i in harmony with thedefinition (11.213) of the a�ne connection �ki` = ek · e`,i.Under the gauge transformation e0a = eb V �1b

a, the metric matrix trans-forms as

g0ab = V �1c⇤a gcd V

�1db or as g0 = V �1† g V �1 (11.490)

in matrix notation. Its inverse goes as g0�1 = V g�1 V †.The gauge-field matrix Aa

i b = ea† · eb,i = gace†c · eb,i transforms as

A0ai b = g0ace0†a · e0b,i = V a

cAcidV

�1db + V a

cV�1cb,i (11.491)

or as A0i = V AiV �1 + V @iV �1 = V AiV �1 � (@iV )V �1.

By using the identity ea† · ec,i = � ea†,i · ec, we may write (exercise 11.44)the Faraday tensor as

F aijb = [Di, Dj ]

ab = ea†,i ·eb,j�ea†,i ·ec ec†·eb,j�ea†,j ·eb,i+ea†,j ·ec ec†·eb,i. (11.492)

11.52 Geometry 513

If n = N , thennX

c=1

e↵c e�c⇤ = �↵� and F aijb = 0. (11.493)

The Faraday tensor vanishes when n = N because the dimension of theembedding space is too small to allow the tangent space to have di↵erentorientations at di↵erent points x of space-time. The Faraday tensor, whichrepresents internal curvature, therefore must vanish. One needs at leastthree dimensions in which to bend a sheet of paper. The embedding spacemust have N > 2 dimensions for SU(2), N > 3 for SU(3), and N > 5 forSU(5).The covariant derivative of the internal metric matrix

g;i = g,i � gAi �A†

ig (11.494)

does not vanish and transforms as (g;i)

0 = V �1†g,iV �1. A suitable actiondensity for it is the trace Tr(g

;ig�1g;ig�1). If the metric matrix assumes a(constant, hermitian) mean value g

0

in the vacuum at low temperatures,then its action is

m2Trh(g

0

Ai +A†ig0)g

�1

0

(g0

Ai +Ai†g0

)g�1

0

i(11.495)

which is a mass term for the matrix of gauge bosons

Wi = g1/20

Ai g�1/20

+ g�1/20

A†i g

1/20

. (11.496)

This mass mechanism also gives masses to the fermions. To see how, wewrite the Dirac action density (11.489) as

a

⇥�i(�ab@i +Aa

i b) +m �ab⇤ b =

a ⇥�i(gab@i + gacA

ci b) +mgab

⇤ b.

(11.497)Each fermion now gets a mass mci proportional to an eigenvalue ci of thehermitian matrix g

0

.This mass mechanism does not leave behind scalar bosons. Whether Na-

ture ever uses it is unclear.

11.52 Geometry

This section is optional on a first reading.In gauge theory, what plays the role of space-time? Could it be the group

manifold? Let us consider the gauge group SU(2) whose group manifold isthe 3-sphere in flat euclidian 4-space. A point on the 3-sphere is

p =⇣±p1� r2, r1, r2, r3

⌘(11.498)


as explained in example 10.30. The coordinates ra = ra are not vectors.The three basis vectors are

ea =@p

@ra=

✓⌥ rap

1� r2, �1a, �

2

a, �3

a

◆(11.499)

and so the metric gab = ea · eb is

gab =ra rb1� r2

+ �ab (11.500)

or

k g k= 1

1� r2

0@1� r22

� r23

r1

r2

r1

r3

r2

r1

1� r21

� r23

r2

r3

r3

r1

r3

r2

1� r21

� r22

1A . (11.501)

The inverse matrix is

gbc = �bc � rb rc. (11.502)

The dual vectors

eb = gbcec =⇣⌥rb

p1� r2, �b

1

� rbr1, �b2

� rbr2, �b3

� rbr3⌘

(11.503)

satisfy eb · ea = �ba.There are two kinds of a�ne connections eb · ea,c and eb · ea,i. If we

di↵erentiate ea with respect to an SU(2) coordinate rc, then

Ebc a = eb · ea,c = rb

✓�ac +

ra rc1� r2

◆(11.504)

in which we used E (for Einstein) instead of � for the a�ne connection. Ifwe di↵erentiate ea with respect to a space-time coordinate xi, then

Ebi a = eb · ea,i = eb · ea,c rc,i = rb r

c,i

✓�ac +

ra rc1� r2

◆. (11.505)

But if the group coordinates ra are functions of the space-time coordinatesxi, then there are 4 new basis 4-vectors ei = eara,i. The metric then is a7⇥ 7 matrix k g k with entries ga,b = ea · eb, ga,k = ea · ek, gi,b = ei · eb, andgi,k = ei · ek or

k g k=✓

ga,b ga,b rb,kga,b ra,i ga,b ra,i rb,k.

◆(11.506)

Further Reading

Exercises 515

The classicsGravitation and Cosmology (Weinberg, 1972), Gravitation (Mis-ner et al., 1973), Cosmology (Weinberg, 2010), General Theory of Relativ-ity (Dirac, 1996), and the very accessible Spacetime and Geometry (Carroll,2003) are of special interest.

Exercises

11.1 Compute the derivatives (11.22 & 11.23).11.2 Show that the transformation x ! x0 defined by (11.16) is a rotation

and a reflection.11.3 Show that the matrix (11.40) satisfies the Lorentz condition (11.39).11.4 If ⌘ = L ⌘LT, show that ⇤ = L�1 satisfies the definition (11.39) of a

Lorentz transformation ⌘ = ⇤T⌘⇤.11.5 The LHC is designed to collide 7 TeV protons against 7 TeV protons for

a total collision energy of 14 TeV. Suppose one used a linear acceleratorto fire a beam of protons at a target of protons at rest at one end ofthe accelerator. What energy would you need to see the same physicsas at the LHC?

11.6 Use Gauss’s law and the Maxwell-Ampere law (11.87) to show that themicroscopic (total) current 4-vector j = (c⇢, j) obeys the continuityequation ⇢+r · j = 0.

11.7 Show that if Mik is a covariant second-rank tensor with no particularsymmetry, then only its antisymmetric part contributes to the 2-formMik dxi ^ dxk and only its symmetric part contributes to the quantityMik dxidxk.

11.8 In rectangular coordinates, use the Levi-Civita identity (1.449) to de-rive the curl-curl equations (11.90).

11.9 Derive the Bianchi identity (11.92) from the definition (11.79) of theFaraday field-strength tensor, and show that it implies the two homo-geneous Maxwell equations (11.82).

11.10 Show that if A is a p-form, then d(AB) = dA ^B + (�1)pA ^ dB.11.11 Show that if ! = aijdxi ^ dxj/2 with aij = � aji, then

d! =1

3!(@kaij + @iajk + @jaki) dx

i ^ dxj ^ dxk. (11.507)

11.12 Using tensor notation throughout, derive (11.147) from (11.145 &11.146).

11.13 Use the flat-space formula (11.168) to compute the change dp due tod⇢, d�, and dz, and so derive the expressions (11.169) for the orthonor-mal basis vectors ⇢, �, and z.


11.14 Similarly, derive (11.175) from (11.174).11.15 Use the definition (11.191) to show that in flat 3-space, the dual of the

Hodge dual is the identity: ⇤⇤dxi = dxi and ⇤⇤(dxi^dxk) = dxi^dxk.11.16 Use the definition of the Hodge star (11.202) to derive (a) two of the

four identities (11.203) and (b) the other two.11.17 Show that Levi-Civita’s 4-symbol obeys the identity (11.207).11.18 Show that ✏`mn ✏pmn = 2 �p` .11.19 Show that ✏k`mn ✏p`mn = 3! �pk.11.20 (a) Using the formulas (11.175) for the basis vectors of spherical co-

ordinates in terms of those of rectangular coordinates, compute thederivatives of the unit vectors r, ✓, and � with respect to the variablesr, ✓, and � and express them in terms of the basis vectors r, ✓, and �.(b) Using the formulas of (a) and our expression (6.28) for the gradientin spherical coordinates, derive the formula (11.297) for the laplacianr ·r.

11.21 Consider the torus with coordinates ✓,� labeling the arbitrary point

p = (cos�(R+ r sin ✓), sin�(R+ r sin ✓), r cos ✓) (11.508)

in which R > r. Both ✓ and � run from 0 to 2⇡. (a) Find the basisvectors e✓ and e�. (b) Find the metric tensor and its inverse.

11.22 For the same torus, (a) find the dual vectors e✓ and e� and (b) findthe nonzero connections �ijk where i, j, & k take the values ✓&�.

11.23 For the same torus, (a) find the two Christo↵el matrices �✓ and ��,

(b) find their commutator [�✓,��], and (c) find the elements R✓✓✓✓, R

�✓�✓,

R✓�✓�, and R�

�� of the curvature tensor.11.24 Find the curvature scalar R of the torus with points (11.508). Hint:

In these four problems, you may imitate the corresponding calculationfor the sphere in Sec. 11.42.

11.25 By di↵erentiating the identity gik gk` = �i`, show that �gik = �gisgkt�gst or equivalently that dgik = � gisgktdgst.

11.26 Just to get an idea of the sizes involved in black holes, imagine anisolated sphere of matter of uniform density ⇢ that as an initial con-dition is all at rest within a radius rb. Its radius will be less than itsSchwarzschild radius if

rb <2MG

c2= 2

✓4

3⇡r3b⇢

◆G

c2. (11.509)

If the density ⇢ is that of water under standard conditions (1 gram percc), for what range of radii rb might the sphere be or become a blackhole? Same question if ⇢ is the density of dark energy.

Exercises 517

11.27 For the points (11.395), derive the metric (11.398) with k = 1. Don’tforget to relate d� to dr.

11.28 For the points (11.396), derive the metric (11.398) with k = 0.

11.29 For the points (11.397), derive the metric (11.398) with k = �1. Don’tforget to relate d� to dr.

11.30 Suppose the constant k in the Roberson-Walker metric (11.394 or11.398) is some number other than 0 or ±1. Find a coordinate transfor-mation such that in the new coordinates, the Roberson-Walker metrichas k = k/|k| = ±1. Hint: You also can change the scale factor a.

11.31 Derive the a�ne connections in Eq.(11.402).



11.34 Derive the spatial Einstein equation (11.414) from (11.378, 11.398,11.409, 11.411, & 11.412).

11.35 Assume there had been no inflation, no era of radiation, and no darkenergy. In this case, the magnitude of the di↵erence |⌦�1| would haveincreased as t2/3 over the past 13.8 billion years. Show explicitly howclose to unity ⌦ would have had to have been at t = 1 s so as to satisfythe observational constraint |⌦

0

� 1| < 0.036 on the present value of ⌦.

11.36 Derive the relation (11.434) between the energy density ⇢ and theRobertson-Walker scale factor a(t) from the conservation law (11.430)and the equation of state p = w⇢.

11.37 Use the Friedmann equations (11.413 & 11.415) for constant ⇢ = �pand k = 1 to derive (11.441) subject to the boundary condition thata(t) has its minimum at t = 0.

11.38 Use the Friedmann equations (11.413 & 11.415) with w = �1, ⇢ con-stant, and k = �1 to derive (11.442) subject to the boundary conditionthat a(0) = 0.

11.39 Use the Friedmann equations (11.413 & 11.415) with w = �1, ⇢constant, and k = 0 to derive (11.443). Show why a linear combinationof the two solutions (11.443) does not work.

11.40 Use the conservation equation (11.447) and the Friedmann equations(11.413 & 11.415) with w = 1/3, k = 0, and a(0) = 0 to derive (11.450).

11.41 Show that if the matrix U(x) is nonsingular, then

(@i U)U�1 = � U @i U�1. (11.510)

11.42 The gauge-field matrix is a linear combination Ak = �ig tbAbk of the

generators tb of a representation of the gauge group. The generators


obey the commutation relations

[ta, tb] = ifabctc (11.511)

in which the fabc are the structure constants of the gauge group. Showthat under a gauge transformation (11.477)

A0i = UAiU

�1 � (@iU)U�1 (11.512)

by the unitary matrix U = exp(�ig�ata) in which �a is infinitesimal,the gauge-field matrix Ai transforms as

�igA0ai t

a = �igAai t

a � ig2fabc�aAb

i tc + ig@i�

ata. (11.513)

Show further that the gauge field transforms as

A0ai = Aa

i � @i�a � gfabcA

bi�

c. (11.514)

11.43 Show that if the vectors ea(x) are orthonormal, then ea†·ec,i = �ea†,i ·ec.11.44 Use the identity of exercise 11.43 to derive the formula (11.492) for

the nonabelian Faraday tensor.

11.45 Using the tricks of section 11.35, show that �pg = �1

2

pg gik �gik.

This relation and the definition (11.354) R = gikRnink imply that the

first-order change in the Einstein-Hilbert action is (11.381) apart froman irrelevant surface term (Carroll, 2003, chap 4.3) due to gik

pg �Rn

ink.

11.46 Write Dirac’s action density in the explicitly hermitian form LD =

� 1

2

�i@i � 1

2

⇥ �i@i

⇤†in which the field has the invariant form

= ea a and = i †�0. Use the identity⇥ a�

i b

⇤†= � b�

i a toshow that the gauge-field matrix Ai defined as the coe�cient of a�

i b

as in a�i(@i + iAiab) b is hermitian A⇤

iab = Aiba.

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