13 Tensors and general relativity

13.1 Points and their coordinates

We use coordinates to label the physical points of a spacetime and the mathematical points of an abstract object. For example, we may label a point on a sphere by its latitude and longitude with respect to a polar axis and meridian. If we use a di↵erent axis and meridian, our coordinates for the point will change, but the point remains as it was. Physical and mathematical points exist independently of the coordinates we use to talk about them. When we change our system of coordinates, we change our labels for the points, but the points remain as they were. At each point p, we can set up various coordinate systems that assign i i unique coordinates x (p) and x0 (p)top and to points near it. For instance, polar coordinates (✓,) are unique for all points on a sphere—except the north and south poles which are labeled by ✓ = 0 and ✓ = ⇡ and all 0 <2⇡. By using a second coordinate system with ✓ = 0 and ✓ = ⇡  0 0 on the equator in the (✓,) system, we can assign unique coordinates to the north and south poles in that system. Embedding simplifies labeling. In a 3-dimensional euclidian space and in the 4-dimensional Minkowski space- time in which the sphere is a surface, each point of the sphere has unique coordinates, (x, y, z) and (t, x, y, z). We will use coordinate systems that represent the points of a space or spacetime uniquely and smoothly at least in local patches, so that the maps 492 Tensors and general relativity

i i i i x = x (p)=x (p(x0)) = x (x0) i i i i (13.1) x0 = x0 (p)=x0 (p(x)) = x0 (x) are well defined, di↵erentiable, and one to one in the patches. We’ll often group the n coordinates xi together and write them collectively as x without superscripts. Since the coordinates x(p) label the point p, we sometimes will call them “the point x.” But p and x are di↵erent. The point p is unique with infinitely many coordinates x, x0, x00, . . . in infinitely many coordinate systems. We begin this chapter by noticing carefully how things change as we change our coordinates. Our goal is to write physical theories so their equa- tions look the same in all systems of coordinates as Einstein taught us.

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

B0 = B. (13.2)

If it also depends upon the coordinates of the spacetime point p(x)=p(x0), then it is a scalar field, and

B0(x0)=B(x). (13.3)

13.3 Contravariant vectors i By the chain rule, the change in dx0 due to changes in the unprimed coor- dinates is i i @x0 k dx0 = dx . (13.4) @xk Xk This transformation defines contravariant vectors: a quantity Ai is a com- ponent of a contravariant vector if it transforms like dxi i i @x0 k A0 = A . (13.5) @xk Xk The coordinate di↵erentials dxi form a contravariant vector. A contravariant vector Ai(x) that depends on the coordinates is a contravariant vector 13.4 Covariant vectors 493 field and transforms as i i @x0 k A0 (x0)= A (x). (13.6) @xk Xk

13.4 Covariant vectors The chain rule for partial derivatives

@ @xk @ i = i k (13.7) @x0 @x0 @x Xk defines covariant vectors: a quantity Ci that transforms as @xk Ci0 = i Ck (13.8) @x0 Xk is a covariant vector. A covariant vector Ci(x) that depends on the coor- dinates and transforms as @xk Ci0(x0)= i Ck(x) (13.9) @x0 Xk is a covariant vector field.

Example 13.1 (Gradient of a scalar) The derivatives of a scalar field B0(x0)=B(x) form a covariant vector field because

k @B0(x0) @B(x) @x @B(x) i = i = i k , (13.10) @x0 @x0 @x0 @x Xk which shows that the gradient @B(x)/@xk fits the definition (13.9) of a covariant vector field.

13.5 Tensors Tensors are structures that transform like products of vectors. A rank-zero tensor is a scalar. A rank-one tensor is a covariant or contravariant vector. Second-rank tensors are distinguished by how they transform under changes 494 Tensors and general relativity of coordinates: @xk @xl covariant Fij0 = i j Fkl @x0 @x0 i j ij @x0 @x0 kl contravariant M 0 = M (13.11) @xk @xl i l i @x0 @x k mixed Nj0 = k j Nl . @x @x0 We can define tensors of higher rank by extending these definitions to quan- tities with more indices. The rank of a tensor also is called its order and its degree. If S(x) is a scalar field, then its derivatives with respect to the coordinates are covariant vectors (13.10) and tensors @S @2S @3S V = ,T= , and U = . (13.12) i @xi ik @xi@xk ik` @xi@xk@x`

Example 13.2 (Rank-2 tensors) If Ak and B` are covariant vectors, and m n C and D are contravariant vectors, then the product Ak B` is a second- rank covariant tensor; Cm Dn is a second-rank contravariant tensor; and m n m n Ak C , Ak D , Bk C , and Bk D are second-rank mixed tensors. Since the transformation laws that define tensors are linear, any linear combination (with constant coecients) of tensors of a given rank and kind is a tensor of that rank and kind. Thus if Fij and Gij are both second-rank covariant tensors, so is their sum Hij = Fij + Gij.

13.6 Summation convention and contractions An index that appears in the same monomial once as a covariant subscript and once as a contravariant superscript, is a dummy index that is summed over n A Bi A Bi (13.13) i ⌘ i Xi=1 usually from 0 to 3. Such a sum in which an index is repeated once covari- antly and once contravariantly is a contraction.Therank of a tensor is the number of its uncontracted indices. ` Although the product Ak C is a mixed second-rank tensor, the contrac- k tion Ak C is a scalar because ` k ` k @x @x0 m @x m ` m ` Ak0 C0 = k m A` C = m A` C = mA` C = A` C . (13.14) @x0 @x @x 13.7 Symmetric and antisymmetric tensors 495

ik Similarly, the doubly contracted product F Fik is a scalar.

Example 13.3 (Kronecker delta) The summation convention and the chain rule imply that

@x i @xk @x i 1ifi = ` 0 = 0 = i = (13.15) @xk @x ` @x ` ` 0ifi = `. 0 0 ⇢ 6 The repeated index k has disappeared in this contraction. The Kronecker i delta j is a mixed second-rank tensor; it transforms as

i l i k i i @x0 @x k @x0 @x @x0 i j0 = k j l = k j = j = j (13.16) @x @x0 @x @x0 @x0 and is invariant under changes of coordinates.

13.7 Symmetric and antisymmetric tensors A covariant tensor is symmetric if it is independent of the order of its indices. That is, if Sik = Ski,thenS is symmetric. Similarly a contravari- ant tensor Sk`m is symmetric if permutations of its indices k, `, m leave it unchanged. The metric of spacetime gik(x)=gki(x) is a symmetric rank-2 covariant tensor because it is an inner product of two tangent basis vectors. A covariant or contravariant tensor is antisymmetric if it changes sign when any two of its indices are interchanged. The Maxwell field strength Fk`(x)= F`k(x) is an antisymmetric rank-2 covariant tensor. ik 12 21 If T ✏ik =0where✏12 = ✏21 = 1 is antisymmetric, then T T = 0. ik ik Thus T ✏ik = 0 means that the tensor T is symmetric.

13.8 Quotient theorem Suppose that the product BAof a quantity B with unknown transformation properties with all tensors A a given rank and kind is a tensor. Then B must be a tensor. i The simplest example is when BiA is a scalar for all contravariant vectors Ai i j Bi0A0 = BjA . (13.17) 496 Tensors and general relativity Then since Ai is a contravariant vector i i @x0 j j B0A0 = B0 A = B A (13.18) i i @xj j or i @x0 j B0 B A =0. (13.19) i @xj j ✓ ◆ Since this equation holds for all vectors A, we may promote it to the level of a vector equation i @x0 B0 B =0. (13.20) i @xj j j k Multiplying both sides by @x /@x0 and summing over j, we get i j j @x0 @x @x Bi0 j k = Bj k (13.21) @x @x0 @x0 which shows that the unknown quantity Bi transforms as a covariant vector

@xj Bk0 = k Bj. (13.22) @x0 The quotient rule works for tensors A and B of arbitrary rank and kind. The proof in each case is similar to the one given here.

13.9 Tensor equations Maxwell’s homogeneous equations (12.45) relate the derivatives of the field- strength tensor to each other as

0=@iFjk + @kFij + @jFki. (13.23) They are generally covariant tensor equations (sections 13.19 & 13.20). They follow from the Bianchi identity (12.71) dF = ddA =0. (13.24) Maxwell’s inhomegneous equations (12.46) 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 13.12) @(pgFik) = µ pgji. (13.25) @xk 0 They are generally covariant tensor equations. We’ll write them as the diver- gence of a contravariant vector in section 13.29, derive them from an action principle in section 13.31, and write them as invariant forms in section 14.7. 13.10 Comma notation for derivatives 497 If we can write a physical law in one coordinate system as a tensor equation Gj`(x) = 0, then in any other coordinate system the corresponding tensor ik equation G0 (x0) = 0 is valid because

i k ik @x0 @x0 j` G0 (x0)= G (x)=0. (13.26) @xj @x` Physical laws also remain the same if expressed in terms of invariant forms. A theory written in terms of tensors or forms has equations that are true in all coordinate systems if they are true in any coordinate system. Only such generally covariant theories have a chance at being right because we can’t be sure that our particular coordinate system is the chosen one. One can make a theory the same in all coordinate systems by applying the principle of stationary action (section 13.31) to an action that is invariant under all coordinate transformations.

13.10 Comma notation for derivatives Commas are used to denote derivatives. If f(✓,) is a function of ✓ and , we can write its derivatives with respect to these coordinates as @f @f f,✓ = @✓f = and f, = @f = . (13.27) @✓ @ And we can write its double derivatives as @2f @2f @2f f,✓✓ = ,f,✓ = , and f, = . (13.28) @✓2 @✓@ @2

If we use indices i, k, . . . to label the coordinates xi,xk, then we can write the derivatives of a scalar f as

@f @2f f,i = @if = and f,ik = @k@if = (13.29) @xi @xk@xi

ik and those of tensors T and Fik as

2 ik 2 ik @ T @ Fik T = and Fik,j` = (13.30) ,j` @xj@x` @xj@x` and so forth. Semicolons are used to denote covariant derivatives (section 13.15). 498 Tensors and general relativity 13.11 Basis vectors and tangent vectors A point p(x) in a space or spacetime with coordinates x is a scalar (13.3) because it is the same point p0(x0)=p(x0)=p(x) in any other system of coordinates x0. Thus its derivatives with respect to the coordinates @p(x) = e (x) (13.31) @xi i form a covariant vector ei(x)

k k @p0(x0) @p(x) @x @p(x) @x ei0 (x0)= i = i = i k = i ek(x). (13.32) @x0 @x0 @x0 @x @x0 Small changes dxi in the coordinates (in any fixed system of coordinates) lead to small changes in the point p(x)

i dp(x)=ei(x) dx . (13.33)

The covariant vectors ei(x) therefore form a basis (1.49) for the space or spacetime at the point p(x). These basis vectors ei(x) are tangent to the curved space or spacetime at the point x and so are called tangent vectors. Although complex and fermionic manifolds may be of interest, the manifolds, points, and vectors of this chapter are assumed to be real.

13.12 Metric tensor A Riemann manifold of dimension d is a space that locally looks like d- dimensional euclidian space Ed and that is smooth enough for the derivatives (13.31) that define tangent vectors to exist. The surface of the Earth, for example, looks flat at distances less than a kilometer. Just as the surface of a sphere can be embedded in flat 3-dimensional space, so too every Riemann manifold can be embedded without change of shape (isometrically) in a euclidian space En of suitably high dimen- sion (Nash, 1956). In particular, every Riemann manifold of dimension d =3 (or 4) can be isometrically embedded in a euclidian space of at most n = 14 (or 19) dimensions, E14 or E19 (G¨unther, 1989). The euclidian dot products (example 1.15) of the tangent vectors (13.31) define the metric of the manifold n g (x)=e (x) e (x)= e↵(x) e↵(x)=e (x) e (x)=g (x) (13.34) ik i · k i k k · i ki ↵=1 X which is symmetric, g (x)=g (x). Here 1 i, k d and 1 ↵ n. ik ki     13.12 Metric tensor 499 The dot product of this equation is the dot product of the n-dimensional euclidian embedding space En. Because the tangent vectors ei(x) are covariant vectors, the metric tensor transforms as a covariant tensor if we change coordinates from x to x0 @xj @x` gik0 (x0)= i k gj`(x). (13.35) @x0 @x0 The squared distance ds2 between two nearby points is the dot product of the small change dp(x) (13.33) with itself

2 i i ds = dp(x) dp(x)=(ei(x) dx ) (ei(x) dx ) · · (13.36) = e (x) e (x) dxidxk = g (x) dxidxk. i · i ik So by measuring the distances ds between nearby points, one can determine the metric gik(x) of a Riemann space.

Example 13.4 (The sphere S2 in E3) In polar coordinates, a point p on the 2-dimensional surface of a sphere of radius R has coordinates p = R(sin ✓ cos , sin ✓ sin , cos ✓) in an embedding space E3. The tangent space E2 at p is spanned by the tangent vectors @p e = p = = R (cos ✓ cos , cos ✓ sin , sin ✓) ✓ ,✓ @✓ (13.37) @p e = p = = R ( sin ✓ sin , sin ✓ cos , 0). , @ The dot products of these tangent vectors are easy to compute in the em- bedding space E3. They form the metric tensor of the sphere g g e e e e R2 0 g = ✓✓ ✓ = ✓ · ✓ ✓ · = . (13.38) ik g g e e e e 0 R2 sin2 ✓ ✓ ✓ ◆ ✓ · ✓ · ◆ ✓ ◆ Its determinant is det(g )=R4 sin2 ✓.Sincee e = 0, the squared in- ik ✓ · finitesimal distance (13.36) is

ds2 = e e d✓2 + e e d2 = R2d✓2 + R2 sin2 ✓d2. (13.39) ✓ · ✓ · We change coordinates from the angle ✓ to a radius r = R sin ✓/a in which a is a dimensionless scale factor. Then R2d✓2 = a2dr2/ cos2 ✓, and cos2 ✓ =1 sin2 ✓ =1 a2r2/R2 =1 kr2 where k =(a/R)2.Inthese coordinates, the squared distance (13.39) is

a2 ds2 = dr2 + a2r2 d2 (13.40) 1 kr2 500 Tensors and general relativity and the r, metric of the sphere and its inverse are

2 1 2 2 (1 kr ) 0 ik 2 1 kr 0 g = a and g = a . (13.41) ik 0 r2 0 r 2 ✓ ◆ ✓ ◆ The sphere is a maximally symmetric space (section 13.24).

Example 13.5 (Graph paper) Imagine a piece of slightly crumpled graph paper with horizontal and vertical lines. The lines give us a two-dimensional coordinate system (x1,x2) that labels each point p(x) on the paper. The vec- tors e1(x)=@1p(x) and e2(x)=@2p(x) define how a point moves dp(x)= i 1 2 ei(x) dx when we change its coordinates by dx and dx . The vectors e1(x) and e2(x) span a di↵erent tangent space at the intersection of every horizon- tal line with every vertical line. Each tangent space is like the tiny square of the graph paper at that intersection. We can think of the two vectors ei(x) as three-component vectors in the three-dimensional embedding space we live in. The squared distance between any two nearby points separated by dp(x)isds2 dp2(x)=e2(x)(dx1)2 +2e (x) e (x) dx1dx2 + e2(x)(dx2)2 ⌘ 1 1 · 2 2 in which the inner products g = e (x) e (x) are defined by the euclidian ij i · j metric of the embedding euclidian space R3.

(p,d p) But our universe has time. A semi-euclidian spacetime E of di- mension d is a flat spacetime with a dot product that has p minus signs and q = d p plus signs. A semi-riemannian manifold of dimension d (p,d p) is a spacetime that locally looks like a semi-euclidian spacetime E and that is smooth enough for the derivatives (13.31) that define its tangent vectors to exist. Every semi-riemannian manifold can be embedded without change of (u,n u) shape (isometrically) in a semi-euclidian spacetime E for suciently large u and n (Greene, 1970; Clarke, 1970). Every physically reasonable (globally hyperbolic) semi-riemannian manifold with 1 dimension of time and 3 dimensions of space can be embedded without change of shape (iso- metrically) in a flat semi-euclidian spacetime of 1 temporal and at most 19 spatial dimensions E(1,19) (M¨uller and S´anchez, 2011; Ak´eet al., 2018). The semi-euclidian dot products of the tangent vectors of a semi-riemannian manifold of d dimensions define its metric as u n g (x)=e (x) e (x)= e↵(x) e↵(x)+ e↵(x) e↵(x) (13.42) ik i · k i k i k ↵=1 ↵=u+1 X X for 0 i, k d 1. The metric (13.42) is symmetric g (x)=g .In   ik ki 13.12 Metric tensor 501 an extended summation convention, the dot product (13.42) is gik(x)= ↵ ei↵(x) ek (x). The squared pseudo-distance or line element ds2 between two nearby points is the inner product of the small change dp(x) (13.33) with itself

2 i i ds = dp(x) dp(x)=(ei(x) dx ) (ei(x) dx ) · · (13.43) = e (x) e (x) dxidxk = g (x) dxidxk. i · i ik 2 Thus measurements of line elements ds determine the metric gik(x) of the spacetime. Some Riemann spaces have natural embeddings in semi-euclidian spaces. One example is the hyperboloid H2.

Example 13.6 (The hyperboloid H2) If we embed a hyperboloid H2 of radius R in a semi-euclidian spacetime E(1,2), then a point p =(x, y, z) on the 2-dimensional surface of H2 obeys the equation R2 = x2 y2 z2 and has polar coordinates p = R(cosh ✓, sinh ✓ cos , sinh ✓ sin ). The tangent vectors are @p e = p = = R (sinh ✓, cosh ✓ cos , cosh ✓ sin ) ✓ ,✓ @✓ (13.44) @p e = p = = R (0, sinh ✓ sin , sinh ✓ cos ). , @ The line element dp2 = ds2 between nearby points is

ds2 = e e d✓2 + e e d2. (13.45) ✓ · ✓ · The metric and line element (13.45) are

10 R2 and ds2 = R2 d✓2 + R2 sinh2 ✓d2. (13.46) 0sinh2 ✓ ✓ ◆ We change coordinates from the angle ✓ to a radius r = R sinh ✓/a in which a is a dimensionless scale factor. Then in terms of the parameter k =(a/R)2, the metric and line element (13.46) are (exercise 13.7)

(1 + r2) 1 0 dr2 a2 and ds2 = a2 + r2 d2 (13.47) 0 r2 1+kr2 ✓ ◆ ✓ ◆ which describe one of only three maximally symmetric (section 13.24) two- dimensional spaces. The other two are the sphere S2 (13.40) and the plane. 502 Tensors and general relativity 13.13 Inverse of metric tensor

The metric gik is a nonsingular matrix (exercise 13.4), and so it has an inverse gik that satisfies

ik i ik g gk` = ` = g0 gk0 ` (13.48) in all coordinate systems. The inverse metric gik is a rank-2 contravariant tensor (13.11) because the metric gk` is a rank-2 covariant tensor (13.35). To show this, we combine the transformation law (13.35) with the definition (13.48) of the inverse of the metric tensor

r s i ik ik @x @x ` = g0 gk0 ` = g0 k ` grs (13.49) @x0 @x0 and multiply both sides by

@x ` @x v gtu 0 0 . (13.50) @xt @xu Use of the Kronecker-delta chain rule (13.15) now leads (exercise 13.5) to

i v iv @x0 @x0 tu g0 (x0)= g (x) (13.51) @xt @xu which shows that the inverse metric gik transforms as a rank-2 contravariant tensor. i The contravariant vector A associated with any covariant vector Ak is i ik i defined as A = g Ak which ensures that A transforms contravariantly (exercise 13.6). This is called raising an index. It follows that the covari- i i ant vector corresponding to the contravariant vector A is Ak = gki A = i` ` gki g A` = k A` = Ak which is called lowering an index.Thesedefini- ik` ij km `n tions apply to all tensors, so T = g g g Tjmn, and so forth.

Example 13.7 (Making scalars) Fully contracted products of vectors and tensors are scalars. Two contravariant vectors Ai and Bk contracted with the i k k ik k metric tensor form the scalar gikA B = AkB . Similarly, g AiBk = A Bk. Derivatives of scalar fields with respect to the coordinates are covariant vectors S,i (example 13.1) and covariant tensors S,ik (section 13.5). If S is ik a scalar, then S,i is a covariant vector, g S,k is a contravariant vector, and ik the contraction g S,i S,k is a scalar.

In what follows, I will often use space to mean either space or spacetime. 13.14 Dual vectors, cotangent vectors 503 13.14 Dual vectors, cotangent vectors Since the inverse metric gik is a rank-2 contravariant tensor, dual vectors

i ik e = g ek (13.52) are contravariant vectors. They are orthonormal to the tangent vectors e` because ei e = gike e = gik g = i. (13.53) · ` k · ` k` ` Here and throughout these sections, the dot product is that (13.34) of eu- d n (p,d p) clidian space E (or E ) or that (13.42) of semi-euclidian space E (or (u,n u) i E . The dual vectors e are called cotangent vectors or tangent i covectors. The tangent vector ek is the sum ek = gki e because

i i` ` ek = gki e = gki g e` = k e` = ek. (13.54) The definition (13.52) of the dual vectors and their orthonormality (13.53) to the tangent vectors imply that their inner products are the matrix elements of the inverse of the metric tensor

ei e` = gik e e` = gik ` = gi`. (13.55) · k · k k The outer product of a tangent vector with its cotangent vector P = eke (summed over the dimensions of the space) is both a projection matrix P from the embedding space onto the tangent space and an identity matrix T k for the tangent space because Pei = ei. Its transpose P = e ek is both a projection matrix P from the embedding space onto the cotangent space and an identity matrix for the cotangent space because P Tei = ei.So

k T k P = ek e = It and P = e ek = Ict. (13.56)

Details and examples are in the file tensors.pdf in Tensors and general relativity at github.com/kevinecahill.

13.15 Covariant derivatives of contravariant vectors k k The covariant derivative D`V of a contravariant vector V is a derivative of V k that transforms like a mixed rank-2 tensor. An easy way to make such i a derivative is to note that the invariant description V (x)=V (x) ei(x) of a i contravariant vector field V (x) in terms of tangent vectors ei(x) is a scalar. 504 Tensors and general relativity Its derivative @V @V i @e = e + V i i (13.57) @x` @x` i @x` is therefore a covariant vector. And the inner product of that covariant vector k V,` with a contravariant tangent vector e is a mixed rank-2 tensor k k k i i k i k i D`V = e V,` = e V,`ei + ei,`V = i V,` + e ei,`V · · · (13.58) k k i = V + e ei,` V . ,` · k The inner product e ei,` is usually written as · k k @ei k e ei,` = e (13.59) · · @x` ⌘ i` and is variously called an ane connection (it relates tangent spaces lack- ing a common origin), a Christo↵el connection, and a Christo↵el sym- bol of the second kind. The covariant derivative itself often is written with a semicolon, thus k k k k i k k i D`V = V = V + e ei,` V = V + V . (13.60) ;` ,` · ,` i`

Example 13.8 (Covariant derivatives of cotangent vectors) Using the identity k k k k 0= =(e ei),` = e ei + e ei,` (13.61) i,` · ,` · · and the projection matrix (13.56), we find that k k k i k k i k k D`e = e + e ei,` e = e e ei e = e e = 0 (13.62) ,` · ,` ,` · ,` ,` the covariant derivatives of cotangent vectors vanish. k Under general coordinate transformations, D`V transforms as a rank-2 mixed tensor k m k k @x0 @x p k m p D V 0(x0)= V 0(x0)= V (x)=x0 x V (x). (13.63) ` ;` p ` ;m ,p ,`0 ;m @x @x0 Tangent basis vectors ei are derivatives (13.31) of the spacetime point p i with respect to the coordinates x , and so ei,` = e`,i because partial deriva- tives commute 2 2 @ei @ p @ p ei,` = = = = e`,i. (13.64) @x` @x`@xi @xi@x` Thus the ane connection (13.59) is symmetric in its lower indices k k k k = e ei,` = e e`,i = . (13.65) i` · · `i 13.16 Covariant derivatives of covariant vectors 505

i Although the covariant derivative V;` (13.60) is a rank-2 mixed tensor, the k ane connection i` transforms inhomogeneously (exercise 13.8) k m n k 2 p k k @ei0 @x0 @x @x p @x0 @ x 0 i` = e0 ` = p ` i nm + p ` i · @x0 @x @x0 @x0 @x @x0 @x0 (13.66) k m n p k p = x0 x x + x0 x ,p ,`0 ,i0 nm ,p ,`0i0 and so is not a tensor. Its variation k =k k is a tensor, however, i` 0 i` i` because the inhomogeneous terms in the di↵erence cancel. k Since the Levi-Civita connection i` is symmetric in i and `, in four- dimensional spacetime, there are 10 ’s for each k, or 40 in all. The 10 correspond to 3 rotations, 3 boosts, and 4 translations.

13.16 Covariant derivatives of covariant vectors k The derivative of the scalar V = Vk e is the covariant vector k k k V,` =(Vk e ),` = Vk,` e + Vk e,`. (13.67)

Its inner product with the covariant vector ei transforms as a rank-2 covari- ant tensor. Thus using again the identity (13.61), we see that the covariant derivative of a covariant vector is k k k k D`Vi = Vi;` = ei V,` = ei Vk,` e + Vk e,` = i Vk,` + ei e,` Vk · · · (13.68) k k = Vi,` ei,` e Vk = Vi,` Vk. · i` D`Vi transforms as a rank-2 covariant tensor because it is the inner product of a covariant tangent vector ei with the derivative V,` of a scalar. Note that k k i` appears with a minus sign in Vi;` and a plus sign in V;` . Example 13.9 (Covariant derivatives of tangent vectors) Using again the projection matrix (13.56), we find that k D`ei = ei` = ei,` ei,` e ek = ei,` ei,` = 0 (13.69) · that covariant derivatives of tangent vectors vanish.

13.17 Covariant derivatives of tensors Tensors transform like products of vectors. So we can make the derivative of a tensor transform covariantly by using Leibniz’s rule (5.49) to di↵erentiate products of vectors and by turning the derivatives of the vectors into their covariant derivatives (13.60) and/or (13.68). 506 Tensors and general relativity Example 13.10 (Covariant derivative of a rank-2 contravariant tensor) An arbitrary rank-2 contravariant tensor T ik transforms like the product of i k ik two contravariant vectors A B . So its derivative @`T transforms like the derivative of the product of the vectors Ai Bk

i k i k i k @`(A B )=(@`A ) B + A @`B . (13.70) By using twice the formula (13.60) for the covariant derivative of a con- i travariant vector, we can convert these two ordinary derivatives @`A and k @`B into tensors i k i k i i j k i k k j D`(A B )=(A B );` =(A,` +j` A )B + A (B,` + j` B ) (13.71) i k i j k k i j =(A B ),` +j` A B + j` A B . Thus the covariant derivative of a rank-2 contravariant tensor is

ik ik ik i jk k ij D`T = T ;` = T ,` +j` T + j` T . (13.72) It transforms as a rank-3 tensor with one covariant index.

Example 13.11 (Covariant derivative of a rank-2 mixed tensor) A rank- i i 2 mixed tensor T k transforms like the product A Bk of a contravariant i i vector A and a covariant vector Bk. Its derivative @`T k transforms like the i derivative of the product of the vectors A Bk

i i i @`(A Bk)=(@`A ) Bk + A @`Bk. (13.73) We can make these derivatives transform like tensors by using the formulas (13.60) and (13.68)

i i i i j i j D`(A Bk)=(A Bk);` =(A,` +j` A )Bk + A (Bk,` Bj) k` (13.74) i i j j i =(A Bk),` + A Bk A Bj. j` k` Thus the covariant derivative of a mixed rank-2 tensor is

D T i = T i = T i +i T j j T i . (13.75) ` k k;` k,` j` k k` j It transforms as a rank-3 tensor with two covariant indices.

Example 13.12 (Covariant derivative of a rank-2 covariant tensor) A rank-2 covariant tensor Tik transforms like the product Ai Bk of two covari- ant vectors Ai and Bk. Its derivative @`Tik transforms like the derivative of the product of the vectors Ai Bk

@`(Ai Bk)=(@`Ai) Bk + Ai @`Bk. (13.76) 13.17 Covariant derivatives of tensors 507 We can make these derivatives transform like tensors by twice using the formula (13.68)

D`(Ai Bk)=(Ai Bk);` = Ai;` Bk + Ai Bk` j j =(Ai,` Aj)Bk + Ai(Bk,` Bj) (13.77) i` k` j j =(Ai Bk),` Aj Bk Ai Bj. i` k`

Thus the covariant derivative of a rank-2 covariant tensor Tik is

j j D` Tik = Tik;` = Tik,` Tjk Tij. (13.78) i` k` It transforms as a rank-3 covariant tensor. Another way to derive the same result is to note that the scalar form of a rank-2 covariant tensor T is T = ei ek T . So its derivative is a covariant ik ⌦ ik vector

i k i k i k T,` = e e Tik,` + e e Tik + e e Tik. (13.79) ⌦ ,` ⌦ ⌦ ,` j i i Using the projector Pt = e ej (13.56), the duality e en = n of tangent k · k k and cotangent vectors (13.53), and the relation ej e = e ej,` = · ,` · j` (13.59 & 13.61), we can project this derivative onto the tangent space and find after shu✏ing some indices

n j i k n k i i j k (e en e ej) T,` = e e Tik,` + e e (en e )Tik + e e (ej e ) Tik ⌦ ⌦ ⌦ · ,` ⌦ · ,` i k n k i i j k = e e Tik,` e e Tik e e Tik ⌦ ⌦ n` ⌦ j` i k j j =(e e ) Tik,` Tjk Tij ⌦ i` k` ⇣ ⌘ which again gives us the formula (13.78).

As in these examples, covariant derivatives are derivations:

Dk(AB)=(AB);k = A;k B + AB;k =(DkA) B + ADkB. (13.80)

The rule for a general tensor is to treat every contravariant index as in (13.60) and every covariant index as in (13.68). The covariant derivative of a mixed rank-4 tensor, for instance, is

T ab = T ab + T jba + T amb T abj T ab m . (13.81) xy;k xy,k xy jk xy mk jy xk xm yk 508 Tensors and general relativity 13.18 The covariant derivative of the metric tensor vanishes The metric tensor is the inner product (13.42) of tangent basis vectors ↵ gik = ei ⌘↵ ek (13.82) in which ↵ and are summed over the dimensions of the embedding space. Thus by the product rule (13.77), the covariant derivative of the metric ↵ ↵ ↵ D` gik = gik;` = D` (ei ⌘↵ ek )=(D` ei ) ⌘↵ ek + ei ⌘↵ D` ek = 0 (13.83) vanishes because covariant derivatives of tangent vectors vanish (13.69), ↵ ↵ D` ei = ei;` = 0 and D` ek = ek;` =0.

13.19 Covariant curls k Because the connection i` is symmetric (13.65) in its lower indices, the covariant curl of a covariant vector Vi is simply its ordinary curl k k V`;i Vi;` = V`,i Vk Vi,` + Vk = V`,i Vi,`. (13.84) `i i` Thus the Faraday field-strength tensor Fi` = A`,i Ai,` being the curl of the covariant vector field Ai is a generally covariant second-rank tensor.

13.20 Covariant derivatives and antisymmetry m m The covariant derivative (13.78) Ai`;k is Ai`;k = Ai`,k Am` Aim . ik `k If the tensor A is antisymmetric A = A , then by adding together the i` `i three cyclic permutations of the indices i`k, we find that the antisymmetry m m of the tensor and the symmetry (13.65) of the ane connection ik = ki conspire to cancel the terms with s m m Ai`;k + Aki;` + A`k;i = Ai`,k Am` ik Aim `k m m + Aki,` Ami k` Akm i` m m + A`k,i Amk A`m `i ki = Ai`,k + Aki,` + A`k,i (13.85) an identity named after Luigi Bianchi (1856–1928). The Maxwell field-strength tensor Fi` is antisymmetric by construction (Fi` = A`,i Ai,`), and so Maxwell’s homogeneous equations 1 ijk` ✏ Fjk,` = Fjk,` + Fk`,j + F`j,k 2 (13.86) = Ak,j` Aj,k` + A`,kj Ak,`j + Aj,`k A`,jk =0 13.21 What is the ane connection? 509 are tensor equations valid in all coordinate systems.

13.21 What is the ane connection? j We insert the identity matrix (13.56) of the tangent space in the form e ej k k into the formula (13.59) for the ane connection i` = e ei,`.Inthe k k j k j · resulting combination = e e ej e`,i we recognize e e as the inverse i` · · · (13.55) of the metric tensor ek ej = gkj. Repeated use of the relation · ei,k = ek,i (13.64) then leads to a formula for the ane connection

k k k j k j 1 kj = e ei,` = e e ej ei,` = e e ej e`,i = g ej ei,` + ej e`,i i` · · · · · 2 · · 1 kj = g (ej ei),` ej,` ei +(ej e`),i ej,i e` 2 · · · · 1 kj = g gji,` + gj`,i ej,` ei ej,i e` (13.87) 2 · · 1 kj = g gji,` + gj`,i e`,j ei ei,j e` 2 · · 1 kj 1 kj = g gji,` + gj`,i (ei e`),j = g gji,` + gj`,i gi`,j 2 · 2 in terms of the inverse of the metric tensor and a combination of its deriva- k tives. The metric gik determines the ane connection i`. The ane connection with all lower indices is

k 1 ni` = gnk = gni,` + gn`,i gi`,n . (13.88) i` 2

13.22 Parallel transport The movement of a vector along a curve on a manifold so that its length and direction in successive tangent spaces do not change is called parallel k k transport. In parallel transport, a vector V = V ek = Vk e may change ` i i dV = V,` dx , but the projection of the change PdV = e ei dV = eie dV into the tangent space must vanish, PdV = 0. In terms of its contravari- k ant components V = V ek, this condition for parallel transport is just the vanishing of its covariant derivative (13.60)

i i ` i k ` i k k ` 0=e dV = e V,`dx = e (V ek),`dx = e V,` ek + V ek,` dx (13.89) i k i k ` i i ⇣ k ` ⌘ = V + e ek,` V dx = V + V dx . k ,` · ,` k` ⇣ ⌘ ⇣ ⌘ 510 Tensors and general relativity

k In terms of its covariant components V = Vke , the condition of parallel transport is just the vanishing of its covariant derivative (13.68)

` k ` k k ` 0=eidV = eiV,`dx = ei(Vke ),`dx = ei Vk,`e + Vke,` dx (13.90) k k k ` k⇣ ` ⌘ = Vk,` + ei e V dx = Vi,` Vk dx . i · ,` i` ⇣ ⌘ ⇣ ⌘ If the curve is x`(u), then these conditions (13.89 & 13.90) for parallel trans- port are i ` ` ` ` dV i dx i k dx dVi dx k dx = V = V and = Vi,` = Vk . du ,` du k` du du du i` du (13.91) Example 13.13 (Parallel transport on a sphere) We parallel-transport the vector v = e =(0, 1, 0) up from the equator along the line of longitude = 0. Along this path, the vector v =(0, 1, 0) = e is constant, so @✓v =0 ✓ k k and so both e e,✓ = 0 and e e,✓ = 0. Thus D✓v = v =0between · · ;✓ the equator and the north pole. As ✓ 0 along the meridian = 0, the ! vector v =(0, 1, 0) approaches the vector e✓. We then parallel-transport v = e✓ down from the north pole along the line of longitude = ⇡/2tothe equator. Along this path, the vector v = e /r =(0, cos ✓, sin ✓) obeys the ✓ parallel-transport condition (13.90) because its ✓-derivative is 1 v,✓ = r e,✓ =(0, cos ✓, sin ✓) = (0, sin ✓, cos ✓)= rˆ . ,✓ |=⇡/2 (13.92) So v,✓ is perpendicular to the tangent vectors e✓ and e along the curve k = ⇡/2. Thus e v,✓ = 0 for k = ✓ and k = and so v;✓ = 0, along · the meridian = ⇡/2. When e reaches the equator, it is e =(0, 0, 1). ✓ ✓ Finally, we parallel-transport v along the equator back to the starting point = 0. Along this path, the vector v =(0, 0, 1) = e✓ is constant, so v, =0 and v = 0. The change from v =(0, 1, 0) to v =(0, 0, 1) is due to the ; curvature of the sphere.

13.23 Curvature

To find the curvature at a point p(x), we parallel-transport a vector Vi along ` acurvex (u) that runs around a tiny square about the point p(x)=p(x0) as u runs from 0 to 1. We then measure the change in the vector

k ` Vi = i` Vk dx . (13.93) I 13.23 Curvature 511

` k On the curve x (u), we approximate i`(x(u)) and Vk(u) as

k k k n (x)= (x0)+ (x0)(x x0) i` i` i`,n (13.94) V (u)=V (0) + m (x ) V (0) (x x )n. k k kn 0 m 0 So keeping only terms linear in (x x )n,wehave 0

k ` Vi = i` Vk dx (13.95) I = k (x ) V (0) + k (x )m (x ) V (0) (x x )n dx` i`,n 0 k i` 0 kn 0 m 0 h i I = k (x ) V (0) + m (x )k (x ) V (0) (x x )n dx` i`,n 0 k i` 0 mn 0 k 0 h i I after interchanging the dummy indices k and m in the second term within the square brackets. The integral around the square is antisymmetric in n and ` and equal in absolute value to the area a2 of the tiny square

(x x )n dx` = a2 ✏ . (13.96) 0 ± n` I The overall sign depends upon whether the integral is clockwise or counter- clockwise, what n and ` are, and what we mean by positive area. The integral picks out the part of the term between the brackets in the formula (13.95) that is antisymmetric in n and `. We choose minus signs in (13.96) so that the change in the vector is

V = a2 k k +k m k m V . (13.97) i in,` i`,n `m in nm i` k h i The quantity between the brackets is Riemann’s curvature tensor

Rk =k k +k m k m . (13.98) i`n ni,` `i,n `m ni nm `i The sign convention is that of (Zee, 2013; Misner et al., 1973; Carroll, 2003; Schutz, 2009; Hartle, 2003; Cheng, 2010; Padmanabhan, 2010). Wein- berg (Weinberg, 1972) uses the opposite sign. The covariant form Rijk` of k Riemann’s tensor is related to R i`n by

n i in Rijk` = ginR jk` and R jk` = g Rnjk`. (13.99)

The Riemann curvature tensor is the commutator of two covariant deriva- tives. To see why, we first use the formula (13.78) for the covariant derivative 512 Tensors and general relativity

DnD`Vi of the second-rank covariant tensor D`Vi

k DnD`Vi = Dn Vi,` Vk `i ⇣ k ⌘ k = Vi,`n Vk Vk,n (13.100) `i,n `i j m m q Vj,` Vm Vi,m Vq . ni `j `n im ⇣ ⌘ Subtracting D`DnVi, we find the commutator [Dn,D`]Vi to be the contrac- k tion of the curvature tensor R i`n (13.98) with the covariant vector Vk

[D ,D ]V = k k +k j k j V = Rk V . n ` i ni,` `i,n `j ni nj `i k i`n k ⇣ ⌘ (13.101)

Since [Dn,D`]Vi is a rank-3 covariant tensor and Vk is an arbitrary covariant vector, the quotient theorem (section 13.8) implies that the curvature tensor is a rank-4 tensor with one contravariant index. k If we define the matrix ` with row index k and column index i as i` 0 0 0 0 `0 `1 `2 `3 1 1 1 1 `0 `1 `2 `3 ` = 0 2 2 2 2 1 , (13.102) `0 `1 `2 `3 B3 3 3 3 C B `0 `1 `2 `3C @ A then we may write the covariant derivatives appearing in the curvature ten- k sor R i`n as D` = @` +` and Dn = @n +n. In these terms, the curvature tensor is the i, k matrix element of their commutator

k k k R i`n =[@` +`,@n +n] i =[D`,Dn] i . (13.103) The curvature tensor is therefore antisymmetric in its last two indexes

Rk = Rk . (13.104) i`n in` The curvature tensor with all lower indices shares this symmetry

R = g Rk = g Rk = R (13.105) ji`n jk i`n jk in` jin` and has three others. In Riemann normal coordinates the derivatives of the metric vanish at any particular point x . In these coordinates, the ’s all ⇤ vanish, and the curvature tensor in terms of the ’s with all lower indices (13.88) is after a cancellation

1 Rki`n =kni,` k`i,n = gkn,i` gni,k` gk`,in + g`i,kn . (13.106) 2 In these coordinates and therefore in all coordinates, Rki`n is antisymmetric 13.23 Curvature 513 in its first two indexes and symmetric under the interchange of its first and second pairs of indexes R = R and R = R . (13.107) ijk` jik` ijk` k`ij Cartan’s equations of structure (13.328 & 13.330) imply (13.342) that the curvature tensor is antisymmetric in its last three indexes 1 0=Rj = Rj + Rj + Rj Rj Rj Rj (13.108) [ik`] 3! ik` `ik k`i ki` i`k `ki ⇣ ⌘ and obeys the cyclic identity j j j 0=R ik` + R `ik + R k`i. (13.109)

The vanishing (13.108) of Ri[jk`] implies that the completely antisymmetric part of the Riemann tensor also vanishes 1 0=R = (R R R R + R ) . (13.110) [ijk`] 4! ijk` jik` ikj` ij`k jki` ··· The Riemann tensor also satisfies a Bianchi identity

i 0=Rj[k`;m]. (13.111)

These symmetries reduce 256 di↵erent functions Rijk`(x) to 20. The Ricci tensor is the contraction

k Rin = R ikn. (13.112) The curvature scalar is the further contraction

ni R = g Rin. (13.113)

Example 13.14 (Curvature of the sphere S2) While in four-dimensional spacetime indices run from 0 to 3, on the everyday sphere S2 (example 13.4) they are just ✓ and . There are only eight possible ane connections, and i i because of the symmetry (13.65) in their lower indices ✓ =✓, only six are independent. In the euclidian embedding space E3, the point p on a sphere of radius R has cartesian coordinates p = R (sin ✓ cos , sin ✓ sin , cos ✓), so the two tangent 3-vectors are (13.37) ˆ e✓ = p,✓ = R (cos ✓ cos , cos ✓ sin , sin ✓)=R ✓ (13.114) e = p = R sin ✓ ( sin , cos , 0) = R sin ✓ ˆ. , 514 Tensors and general relativity Their dot products form the metric (13.38)

g g e e e e R2 0 g = ✓✓ ✓ = ✓ · ✓ ✓ · = (13.115) ik g g e e e e 0 R2 sin2 ✓ ✓ ✓ ◆ ✓ · ✓ · ◆ ✓ ◆ 2 2 2 which is diagonal with g✓✓ = R and g = R sin ✓. Di↵erentiating the vectors e✓ and e,wefind

e✓,✓ = R (sin ✓ cos , sin ✓ sin , cos ✓)= R rˆ ˆ e✓, =R cos ✓ ( sin , cos , 0) = R cos ✓ (13.116) e,✓ =e✓, e, = R sin ✓ (cos , sin , 0) . ij The metric with upper indices g is the inverse of the metric gij

R 2 0 (gij)= , (13.117) 0 R 2 sin 2 ✓ ✓ ◆ i ik so the dual vectors e = g ek are

✓ 1 1 e = R (cos ✓ cos , cos ✓ sin , sin ✓)=R ✓ˆ 1 1 e = ( sin , cos , 0) = ˆ. (13.118) R sin ✓ R sin ✓ The ane connections are given by (13.59) as

i i i = = e ej,k. (13.119) jk kj · ✓ ✓ Since both e and e are perpendicular tor ˆ, the ane connections ✓✓ ˆ and ✓✓ both vanish. Also, e, is orthogonal to ,so = 0 as well. ˆ ✓ ✓ Similarly, e✓, is perpendicular to ✓,so✓ = ✓ also vanishes. The two nonzero ane connections are

1 1 ˆ ˆ = e e✓, = R sin ✓ R cos ✓ = cot ✓ (13.120) ✓ · · and

✓ ✓ = e e, = sin ✓ (cos ✓ cos , cos ✓ sin , sin ✓) (cos , sin , 0) · · = sin ✓ cos ✓. (13.121) The nonzero connections are = cot ✓ and ✓ = sin ✓ cos ✓.Sothe ✓ 13.23 Curvature 515 matrices ✓ and , the derivative ,✓, and the commutator [✓, ] are 00 0 sin ✓ cos ✓ = and = (13.122) ✓ 0 cot ✓ cot ✓ 0 ✓ ◆ ✓ ◆ 0sin2 ✓ cos2 ✓ 0 cos2 ✓ = and [ , ]= . ,✓ csc2 ✓ 0 ✓ cot2 ✓ 0 ✓ ◆ ✓ ◆ Both [✓, ✓] and [, ] vanish. So the commutator formula (13.103) gives for Riemann’s curvature tensor

✓ ✓ R ✓✓✓ =[@✓ +✓,@✓ +✓] ✓ =0 R ✓✓ =[@ +,@✓ +✓] ✓ = ✓, ✓ +[, ✓] ✓ =1 ✓ ✓ ✓ ✓ 2 R =[@✓ +✓,@ +] = ✓, +[✓, ] =sin ✓ ✓ R =[@ +,@ +] =0. (13.123) n The Ricci tensor (13.112) is the contraction Rmk = Rmnk, and so ✓ R✓✓ = R ✓✓✓ + R ✓✓ =1 (13.124) ✓ 2 R = R ✓ + R =sin ✓. km The curvature scalar (13.113) is the contraction R = g Rmk, and so since ✓✓ 2 2 2 g = R and g = R sin ✓,itis

✓✓ 2 2 2 R = g R + g R = R + R = (13.125) ✓✓ R2 for a 2-sphere of radius R. The scalar curvature is a constant because the sphere is a maximally symmetric space (section 13.24). Gauss invented a formula for the curvature K of a surface; for all two- dimensional surfaces, his K = R/2.

Example 13.15 (Curvature of a cylindrical hyperboloid) The points of a cylindrical hyperboloid in 3-space satisfy R2 = x2 y2 + z2 and may be parameterized as p = r(sinh ✓ cos , sinh ✓ sin , cosh ✓). The (orthogonal) coordinate basis vectors are

e✓ = p,✓ = r(cosh ✓ cos , cosh ✓ sin , sinh ✓) (13.126) e = p, = r( sinh ✓ sin , sinh ✓ cos , 0). The squared distance ds2 between nearby points is ds2 = e e d✓2 + e e d2. (13.127) ✓ · ✓ · 516 Tensors and general relativity If the embedding metric is m = diag(1, 1, 1), then ds2 is ds2 = R2 d✓2 + R2 sinh2 ✓d2 (13.128) and 10 (g )=r2 . (13.129) ij 0sinh2 ✓ ✓ ◆ The Mathematica scripts great.m and cylindrical hyperboloid.nb com- pute the scalar curvature as R = 2/r2. The surface is maximally symmetric with constant negative curvature. This chapter’s programs and scripts are in Tensors and general relativity at github.com/kevinecahill.

Example 13.16 (Curvature of the sphere S3) The three-dimensional sphere S3 may be embedded isometrically in four-dimensional flat euclidian space E4 as the set of points p =(x, y, z, w) that satisfy L2 = x2 + y2 + z2 + w2. If we label its points as p(,✓,)=L(sin sin ✓ cos , sin sin ✓ sin , sin cos ✓, cos ), (13.130) then its coordinate basis vectors are

e = p, = L(cos sin ✓ cos , cos sin ✓ sin , cos cos ✓, sin ) e✓ = p,✓ = L(sin cos ✓ cos , sin cos ✓ sin , sin sin ✓, 0) (13.131) e = p, = L( sin sin ✓ sin , sin sin ✓ cos , 0, 0). The inner product of E4 is the four-dimensional dot-product. The basis vec- tors are orthogonal. In terms of the radial variable r = L sin , the squared distance ds2 between two nearby points is ds2 = e e d2 + e e d✓2 + e e d2 · ✓ · ✓ · = L2 d2 +sin2 d✓2 +sin2 sin2 ✓d2 (13.132) dr2 dr2 = + r2d✓2 + r2 sin2 ✓d2 = + r2d⌦2 1 sin2 1 (r/L)2 where d⌦2 = d✓2 +sin2 ✓d2. In these coordinates, r, ✓,,themetricis 1/(1 (r/L)2)0 0 g = 0 r2 0 . (13.133) ik 0 1 00r2 sin2 ✓ @ A The Mathematica scripts great.m and sphere S3.nb compute the scalar curvature as 6 R = (13.134) L2 13.24 Maximally symmetric spaces 517 which is a constant because S3 is maximally symmetric (section 13.24).

Example 13.17 (Curvature of the hyperboloid H3) The hyperboloid H3 is a three-dimensional surface that can be isometrically embedded in the semi- euclidian spacetime E(1,3) in which distances are ds2 = dx2 +dy2 +dz2 dw2, and w is a time coordinate. The points of H3 satisfy L2 = x2 y2 z2 +w2. If we label them as p(,✓,)=L (sinh sin ✓ cos , sinh sin ✓ sin , sinh cos ✓, cosh) (13.135) then the coordinate basis vectors or tangent vectors of H3 are

e = p, = L(cosh sin ✓ cos , cosh sin ✓ sin , cosh cos ✓, sinh) e✓ = p,✓ = L(sinh cos ✓ cos , sinh cos ✓ sin , sinh sin ✓, 0) (13.136) e = p, = L( sinh sin ✓ sin , sinh sin ✓ cos , 0, 0). The basis vectors are orthogonal. In terms of the radial variable r = L sinh /a, the squared distance ds2 between two nearby points is ds2 = e e d2 + e e d✓2 + e e d2 · ✓ · ✓ · = L2 d2 +sinh2 d✓2 +sinh2 sin2 ✓d2 (13.137) dr2 dr2 = + r2d✓2 + r2 sin2 ✓d2 = + r2d⌦2. 1+sinh2 1+(r/L)2 The Mathematica scripts great.m and hyperboloid H3.nb compute the scalar curvature of H3 as 6 R = . (13.138) L2 Its curvature is a constant because H3 is maximally symmetric (section 13.24). The only maximally symmetric 3-dimensional manifolds are S3, H3, and eu- clidian space E3 whose line element is ds2 = dr2 + r2d⌦2. They are the spatial parts of Friedmann-Lemaˆıtre-Robinson-Walker cos- mologies (section 13.42).

13.24 Maximally symmetric spaces The spheres S2 and S3 (examples 13.4 & 13.16) and the hyperboloids H2 and H3 (examples 13.6 & 13.17) are maximally symmetric spaces. A space 518 Tensors and general relativity described by a metric gik(x) is symmetric under a transformation x x0 i k i k ! if the distances gik(x0)dx0 dx0 and gik(x)dx dx are the same. To see what this symmetry condition means, we consider the infinitesimal transformation ` ` ` ` x0 = x + ✏y (x) under which to lowest order gik(x0)=gik(x)+gik,`✏y and i i i j dx0 = dx + ✏y,jdx . The symmetry condition requires

i k ` i i j k k m gik(x)dx dx =(gik(x)+gik,`✏y )(dx + ✏y,jdx )(dx + ✏y,mdx ) (13.139) or ` m j 0=gik,` y + gim y,k + gjk y,i. (13.140)

i i i i The vector field y (x) must satisfy this condition if x0 = x + ✏y (x)isto be a symmetry of the metric gik(x). By using the vanishing (13.83) of the covariant derivative of the metric tensor, we may write the condition on the symmetry vector y`(x) as (exercise 13.9)

0=yi;k + yk;i. (13.141)

The symmetry vector y` is a Killing vector (Wilhelm Killing, 1847–1923). We may use symmetry condition (13.140) or (13.141) either to find the symmetries of a space with a known metric or to find the metric with given symmetries.

Example 13.18 (Killing vectors of the sphere S2) The first Killing vector ✓ is (y1,y1 )=(0, 1). Since the components of y1 are constants, the symmetry condition (13.140) says gik, = 0 which tells us that the metric is indepen- ✓ dent of . The other two Killing vectors are (y2,y2 )=(sin, cot ✓ cos ) and (y✓,y) = (cos , cot ✓ sin ). The symmetry condition (13.140) for 3 3 i = k = ✓ and Killing vectors y2 and y3 tell us that g✓ = 0 and that g✓✓,✓ = 0. So g✓✓ is a constant, which we set equal to unity. Finally, the symmetry condition (13.140) for i = k = and the Killing vectors y2 and 2 y3 tell us that g,✓ = 2 cot ✓g which we integrate to g =sin✓.The 2-dimensional space with Killing vectors y1,y2,y3 therefore has the metric (13.115) of the sphere S2.

Example 13.19 (Killing vectors of the hyperboloid H2) The metric (13.46) 2 2 2 2 of the hyperboloid H is diagonal with g✓✓ = R and g = R sinh ✓.The ✓ Killing vector (y1,y1 )=(0, 1) satisfies the symmetry condition (13.140). Since g✓✓ is independent of ✓ and ,the✓✓ component of (13.140) im- ✓ 2 2 plies that y,✓ = 0. Since g = R sinh ✓,the component of (13.140) says that y = coth ✓y✓.The✓ and ✓ components of (13.140) give , 13.24 Maximally symmetric spaces 519 y✓ = sinh2 ✓y . The vectors y =(y✓,y)=(sin, coth ✓ sin ) and , ,✓ 2 2 2 y =(y✓,y) = (cos , coth ✓ sin ) satisfy both of these equations. 3 3 3

The Lie derivative y of a scalar field A is defined in terms of a vector ` ` L field y (x) as yA = y A,`. The Lie derivative y of a contravariant vector L L F i is F i = y`F i F `yi = y`F i F `yi (13.142) Ly ,` ,` ;` ;` ` i k ` i k in which the second equality follows from y `kF = F `ky . The Lie derivative of a covariant vector V is Ly i ` ` ` ` yVi = y Vi,` + V`y = y Vi;` + V`y . (13.143) L ,i ;i Similarly, the Lie derivative of a rank-2 covariant tensor T is Ly ik ` ` ` yTik = y Tik,` + T`ky + Ti`y . (13.144) L ,i ,k We see now that the condition (13.140) that a vector field y` be a symmetry of a metric gjm is that its Lie derivative

` m j ygik = gik,` y + gim y + gjk y = 0 (13.145) L ,k ,i must vanish. A maximally symmetric space (or spacetime) in d dimensions has d trans- lation symmetries and d(d 1)/2 rotational symmetries which gives a total of d(d + 1)/2 symmetries associated with d(d + 1)/2 Killing vectors. Thus for d = 2, there is one rotation and two translations. For d = 3, there are three rotations and three translations. For d = 4, there are six rotations and four translations. A maximally symmetric space has a curvature tensor (13.99) that is simply related to its metric tensor

R = c (g g g g ) (13.146) ijk` ik j` i` jk ki k where c is a constant (Zee, 2013, IX.6). Since g gik = gk = d is the number of dimensions of the space(time), the Ricci tensor (13.112) and the curvature scalar (13.113) of a maximally symmetric space are

R = gkiR = c (d 1) g and R = g`jR = cd(d 1). (13.147) j` ijk` j` j` 520 Tensors and general relativity 13.25 Principle of equivalence

Since the metric tensor gij(x) is real and symmetric, it can be diagonalized at any point p(x)bya4 4 orthogonal matrix O(x) ⇥ e0 000 T k ` 0 e1 00 O i gk` O j = 0 1 (13.148) 00e2 0 B 000e C B 3C @ A which arranges the four real eigenvalues ei of the matrix gij(x) in the order e e e e . Thus the coordinate transformation 0  1  2  3 k T k @x O i i = (13.149) @x0 e | i| takes any spacetime metric gk`(x) with onep negative and three positive eigen- values into the Minkowski metric ⌘ij of flat spacetime 1000 @xk @x` 0100 gk`(x) i j = gij0 (x0)=⌘ij = 0 1 (13.150) @x0 @x0 0010 B 0001C B C @ A at the point p(x)=p(x0). The principle of equivalence says that in these free-fall coordinates x0, the physical laws of gravity-free special relativity apply in a suitably small region about the point p(x)=p(x0). It follows from this principle that the metric gij of spacetime accounts for all the e↵ects of gravity. 2 In the x0 coordinates, the invariant squared separation dp is 2 i j i j dp = g0 dx0 dx0 = e0 (x0) e0 (x0) dx0 dx0 ij i · j a b i j a b i j = ei0 (x0)⌘abej0 (x0) dx0 dx0 = i ⌘abj dx0 dx0 (13.151) i j 2 0 2 2 = ⌘ dx0 dx0 =(dx0) (dx0 ) = ds . ij 2 If dx0 = 0, then dt = p ds /c is the proper time elapsed between events 0 p and p+dp.Ifdt0 = 0, then ds is the proper distance between the events. The x0 coordinates are not unique because every Lorentz transformation (section 12.1) leaves the metric ⌘ invariant. Coordinate systems in which gij(x0)=⌘ij are called Lorentz, inertial, or free-fall coordinate systems. The congruency transformation (1.351 & 13.148–13.150) preserves the signs of the eigenvalues e which make up the signature ( 1, 1, 1, 1) of i the metric tensor. 13.26 Tetrads 521

13.26 Tetrads We defined the metric tensor as the dot product (13.34) or (13.42) of tangent vectors, g (x)=e (x) e (x). If instead we invert the equation (13.150) that k` k · ` relates the metric tensor to the flat metric @x a @x b g (x)= 0 ⌘ 0 (13.152) k` @xk ab @x` then we can express the metric in terms of four 4-vectors @x a ca(x)= 0 as g (x)=ca(x) ⌘ cb(x) (13.153) k @xk k` k ab ` in which ⌘ij is the 4 4 metric (13.150) of flat Minkowski space. Cartan’s a ⇥ four 4-vectors ci (x) are called a moving frame,atetrad, and a vierbein. a b Because L c(x) ⌘ab L d(x)=⌘cd, every spacetime-dependent Lorentz trans- c formation L(x) maps one set of tetrads ck(x) to another set of tetrads a a c ck0 (x)=L c(x) ck(x) that represent the same metric a b a c b d c0 (x) ⌘ab c0 (x)=L c(x) c (x) ⌘ab L (x) c (x) k ` k d ` (13.154) c d = ck(x) ⌘cd c` (x)=gk`(x).

Cartan’s tetrad is four 4-vectors ci that give the metric tensor as gik = c c = ~c ~c c0c0. The dual tetrads ci = gik⌘ cb satisfy i · k i · k i k a ab k i a i i b b cack = k and caci = a. (13.155)

The metric gk`(x)issymmetric,gk`(x)=g`k(x), so it has 10 indepen- a dent components at each spacetime point x. The four 4-vectors ck have 16 components, but a Lorentz transformation L(x) has 6 components. So the tetrads have 16 6 = 10 independent components at each spacetime point. The distinction between tetrads and tangent basis vectors is that each a ↵ tetrad c k has 4 components, a =0, 1, 2, 3, while each basis vector e k(x) has as many components ↵ =0, 1, 2, 3,... as there are dimensions in the minimal semi-euclidian embedding space E1,n where n 19 (Ak´eet al.,  2018). Both represent the metric

3 n a b ↵ gk`(x)= c k(x) ⌘ab c `(x)= e k(x) ⌘↵0 e `(x) (13.156) a,bX=0 ↵X,=0 522 Tensors and general relativity in which ⌘0 is like ⌘ but with n diagonal elements that are unity. (Elie´ Cartan, 1869–1951)

13.27 Scalar densities and g = det(g ) | ik | Let g be the absolute value of the determinant of the metric tensor gik g = g(x)= det(g (x)) . (13.157) | ik | Under a coordinate transformation, pg becomes

@xj @x` g = g (x )= det(g (x )) = det g (x) . (13.158) 0 0 0 | ik0 0 | @x i @x k j` s ✓ 0 0 ◆ p p q The definition (1.204) of a determinant and the product rule (1.225) for determinants tell us that @xj @x` g (x )= det det det(g ) = J(x/x0) g(x) (13.159) 0 0 @x i @x k j` | | s ✓ 0 ◆ ✓ 0 ◆ p p where J(x/x0) is the jacobian (section 1.21) of the coordinate transformation

@xj J(x/x0)= det . (13.160) @x i ✓ 0 ◆ A quantity s(x)isascalar density of weight w if it transforms as w s0(x0)=[J(x0/x)] s(x). (13.161)

Thus the transformation rule (13.159) says that the determinant det(gik)is a scalar density of weight minus two 2 2 det(gik0 (x0)) = = [J(x/x0)] g(x)=[J(x0/x)] det(gj`(x)). (13.162)

We saw in section 1.21 that under a coordinate transformation x x0 d ! the d-dimensional element of volume in the new coordinates d x0 is related to that in the old coordinates ddx by a jacobian i d d @x0 d d x0 = J(x0/x) d x =det d x. (13.163) @xj ✓ ◆ d Thus the product pgdx changes at most by the sign of the jacobian J(x0/x) when x x ! 0 d d d g d x0 = J(x/x0) J(x0/x) g(x) d x = g(x) d x. (13.164) 0 | | ± p p p 13.28 Levi-Civita’s symbol and tensor 523

The quantity pgd4x is the invariant scalar pg d4x so that if L(x)isa | | scalar, then the integral over spacetime

L(x) pgd4x (13.165) Z is invariant under general coordinate transformations. The Levi-Civita ten- sor provides a fancier definition.

13.28 Levi-Civita’s symbol and tensor In 3 dimensions, Levi-Civita’s symbol ✏ ✏ijk is totally antisymmetric ijk ⌘ with ✏123 = 1 in all coordinate systems. In 4 space or spacetime dimensions, Levi-Civita’s symbol ✏ ✏ijk` is totally antisymmetric with ✏ =1 ijk` ⌘ 1234 or equivalently with ✏0123 = 1 in all coordinate systems. In n dimensions,

Levi-Civita’s symbol ✏i1i2...in is totally antisymmetric with ✏123...n = 1 or ✏012...n 1 = 1. We can turn his symbol into a pseudotensor by multiplying it by the square root of the absolute value of the determinant of a rank-2 covariant tensor. A natural choice is the metric tensor. In a right-handed coordinate system in which the tangent vector e0 points (orthochronously) toward the future, the Levi-Civita tensor ⌘ijk` is the totally antisymmetric rank-4 covariant tensor

⌘ijk`(x)= g(x) ✏ijk` (13.166) in which g(x)= det g (x) is (13.157)p the absolute value of the determi- | mn | nant of the metric tensor gmn. In a di↵erent system of coordinates x0,the Levi-Civita tensor ⌘ijk`(x0) di↵ers from (13.166) by the sign s of the jaco- bian J(x0/x) of any coordinate transformation to x0 from a right-handed, orthochronous coordinate system x

⌘ijk`(x0)=s(x0) g(x0) ✏ijk`. (13.167) The transformation rule (13.159) and thep definition (1.204) and product rule (1.225) of determinants show that ⌘ijk` transforms as a rank-4 covariant tensor

⌘0 (x0)=s(x0) g (x ) ✏ = s(x0) J(x/x0) g(x) ✏ ijk` 0 0 ijk` | | ijk` p @x p = J(x/x0) g(x) ✏ =det pg✏ (13.168) ijk` @x ijk` ✓ 0 ◆ @xt @xpu @xv @xw @xt @xu @xv @xw = i j k ` pg✏tuvw = i j k ` ⌘tuvw. @x0 @x0 @x0 @x0 @x0 @x0 @x0 @x0 524 Tensors and general relativity

Raising the indices of ⌘,wehaveusing as the sign of det(gik) ijk` it ju kv `w it ju kv `w mn ⌘ = g g g g ⌘tuvw = g g g g pg✏tuvw = pg✏ijk` det(g ) = pg✏ / det(g )=✏ /pg ✏ijk`/pg. (13.169) ijk` mn ijk` ⌘ In terms of the Hodge star (14.151), the invariant volume element is 1 pg d4x = 1= ⌘ dxi dxj dxk dx`. (13.170) | | ⇤ 4! ijk` ^ ^ ^

13.29 Divergence of a contravariant vector The contracted covariant derivative of a contravariant vector is a scalar known as the divergence, V = V i = V i + V k i . (13.171) r· ;i ,i ki Because gik = gki,inthesumoveri of the connection (13.59) i 1 i` = g gi`,k + g`k,i gki,` (13.172) ki 2 the last two terms cancel because they di↵er only by the interchange of the dummy indices i and ` i` `i i` g g`k,i = g gik,` = g gki,`. (13.173) So the contracted connection collapses to

i 1 i` ki = 2 g gi`,k. (13.174) There is a nice formula for this last expression. To derive it, let g g be ⌘ i` the 4 4 matrix whose elements are those of the covariant metric tensor g . ⇥ i` Its determinant, like that of any matrix, is the cofactor sum (1.213) along any row or column, that is, over ` for fixed i or over i for fixed `

det(g)= gi` Ci` (13.175) iXor ` in which the cofactor C is ( 1)i+` times the determinant of the reduced i` matrix consisting of the matrix g with row i and column ` omitted. Thus the partial derivative of det g with respect to the i`th element gi` is @ det(g) = Ci` (13.176) @gi` 13.29 Divergence of a contravariant vector 525 in which we allow gi` and g`i to be independent variables for the purposes of this di↵erentiation. The inverse gi` of the metric tensor g,liketheinverse (1.215) of any matrix, is the transpose of the cofactor matrix divided by its determinant det(g)

C 1 @ det(g) gi` = `i = . (13.177) det(g) det(g) @g`i Using this formula and the chain rule, we may write the derivative of the determinant det(g) as

@ det(g) `i det(g),k = gi`,k =det(g) g gi`,k (13.178) @gi` and so since gi` = g`i, the contracted connection (13.174) is

i 1 i` det(g),k det(g) ,k g,k (pg),k = g gi`,k = = | | = = (13.179) ki 2 2det(g) 2 det(g) 2g pg | | in which g det(g) is the absolute value of the determinant of the metric ⌘ tensor. Thus from (13.171 & 13.179), we arrive at our formula for the covariant divergence of a contravariant vector:

k i i i k k (pg),k k (pgV ),k V = V;i = V,i +kiV = V,k + V = . (13.180) r· pg pg Example 13.20 (Maxwell’s inhomogeneous equations) An important ap- plication of this divergence formula (13.180) is the generally covariant form (14.157) of Maxwell’s inhomogeneous equations

1 k` k pgF = µ0j . (13.181) pg ,` ⇣ ⌘

Example 13.21 (Energy-momentum tensor) Another application is to the divergence of the symmetric energy-momentum tensor T ij = T ji

ij ij i kj j im T;i = T,i +ki T + mi T (pgT kj) (13.182) = k +j T im. pg mi 526 Tensors and general relativity 13.30 Covariant laplacian In flat 3-space, we write the laplacian as = 2 or as . In euclidian 2 2 2 r·r r 4 coordinates, both mean @x + @y + @z . In flat minkowski space, one often 2 turns the triangle into a square and writes the 4-laplacian as 2 = @0 . ,i4ik The gradient f,k of a scalar field f is a covariant vector, and f = g f,k is its contravariant form. The invariant laplacian 2f of a scalar field f ,i is the covariant divergence f;i . We may use our formula (13.180) for the divergence of a contravariant vector to write it in these equivalent ways

,i ik ,i ik (pgf ),i (pgg f,k),i 2f = f =(g f,k);i = = . (13.183) ;i pg pg

13.31 Principle of stationary action in general relativity

The invariant proper time for a particle to move along a path xi(t)

⌧2 1 1 i ` 2 T = d⌧ = gi`dx dx (13.184) ⌧ c Z 1 Z ⇣ ⌘ is extremal and stationary on free-fall paths called geodesics. We can iden- tify a geodesic by computing the variation d⌧

i ` i ` i ` (gi`)dx dx 2gi`dx dx cd⌧ = gi`dx dx = (13.185) 2 g dxidx` i` pgi`,k g gi`,k g = xkuiu`d⌧ i` uipdx` = xkuiu`d⌧ i` uidx` 2c c 2c c in which u` = dx`/d⌧ is the 4-velocity (12.20). The path is extremal if

⌧2 ⌧2 ` 1 1 k i ` i dx 0=cT = c d⌧ = gi`,kx u u + gi`u d⌧ (13.186) c 2 d⌧ Z⌧1 Z⌧1 ✓ ◆ ` ` which we integrate by parts keeping in mind that x (⌧2)=x (⌧1)=0

⌧2 i 1 k i ` d(gi`u ) ` 0= gi`,kx u u x d⌧ 2 d⌧ Z⌧1 ✓ ◆ ⌧2 i 1 k i ` i k ` du ` = gi`,kx u u gi`,ku u x gi` x d⌧. (13.187) 2 d⌧ Z⌧1 ✓ ◆ Now interchanging the dummy indices ` and k on the second and third 13.31 Principle of stationary action in general relativity 527 terms, we have

⌧2 i 1 i ` i ` du k 0= gi`,ku u gik,`u u gik x d⌧ (13.188) 2 d⌧ Z⌧1 ✓ ◆ or since xk is arbitrary i 1 i ` i ` du 0= gi`,ku u gik,`u u gik . (13.189) 2 d⌧ rk i ` If we multiply this equation of motion by g and note that gik,`u u = i ` g`k,iu u ,thenwefind r du 1 rk i ` 0= + g gik,` + g`k,i gi`,k u u . (13.190) d⌧ 2 r So using the symmetry gi` = g`i and the formula (13.87) for i`, we get dur d2xr dxi dx` 0= +r uiu` or 0 = +r (13.191) d⌧ i` d⌧ 2 i` d⌧ d⌧ which is the geodesic equation. In empty space, particles fall along geodesics independently of their masses. One gets the same geodesic equation from the simpler action principle 2 dxi dx` d2xr dxi dx` 0= g (x) d = 0= +r . (13.192) i` d d ) d2 i` d d Z1 The right-hand side of the geodesic equation (13.191) is a contravariant vector because (Weinberg, 1972) under general coordinate transformations, r r i ` the inhomogeneous terms arising fromx ¨ cancel those from i`x˙ x˙ . Here and often in what follows we’ll use dots to mean proper-time derivatives. The action for a particle of mass m and charge q in a gravitational field r i` and an electromagnetic field Ai is

1 ⌧2 i ` 2 q i S = mc gi`dx dx + Ai(x) dx (13.193) c ⌧ Z ⇣ ⌘ Z 1 i in which the interaction q Aidx is invariant under general coordinate trans- formations. By (12.59 & 13.188), the first-order change in S is R ⌧2 i 1 i ` i ` du q i k S = m gi`,ku u gik,`u u gik + Ai,k Ak,i u x d⌧ 2 d⌧ mc Z⌧1  (13.194) and so by combining the Lorentz force law (12.60) and the geodesic equation ri r i (13.191) and by writing F x˙ i as F i x˙ ,wehave d2xr dxi dx` q dxi 0= +r F r (13.195) d⌧ 2 i` d⌧ d⌧ m i d⌧ 528 Tensors and general relativity as the equation of motion of a particle of mass m and charge q.Itisstriking how nearly perfect the electromagnetism of Faraday and Maxwell is. The action of the electromagnetic field interacting with an electric current jk in a gravitational field is

S = 1 F F k` + µ A jk pgd4x (13.196) 4 k` 0 k Z h i in which pgd4x is the invariant volume element. After an integration by parts, the first-order change in the action is @ S = F k` pg + µ jk pg A d4x, (13.197) @x` 0 k Z  ⇣ ⌘ and so the inhomogeneous Maxwell equations in a gravitational field are @ pgFk` = µ pgjk. (13.198) @x` 0 ⇣ ⌘ The action of a scalar field of mass m in a gravitational field is

1 ik 2 2 4 S = ,i g ,k m pgd x. (13.199) 2 Z ⇣ ⌘ After an integration by parts, the first-order change in the action is

ik 2 4 S = pgg ,k m pg d x (13.200) ,i Z ⇣ ⌘ which yields the equation of motion

ik 2 pgg ,k m pg=0. (13.201) ,i ⇣ ⌘ The action of the gravitational field itself is a spacetime integral of the Riemann scalar (13.113) divided by Newton’s constant c3 S = R pgd4x. (13.202) 16⇡G Z Its variation leads to Einstein’s equations (section 13.35).

13.32 Equivalence principle and geodesic equation The principle of equivalence (section 13.25) says that in any gravitational field, one may choose free-fall coordinates in which all physical laws take the same form as in special relativity without acceleration or gravitation—at least over a suitably small volume of spacetime. Within this volume and in 13.32 Equivalence principle and geodesic equation 529 these coordinates, 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 the gravitational field is constant throughout it. Such free-fall coordinate systems are called local Lorentz frames and local inertial frames.

Example 13.22 (Elevators) When a modern elevator starts going down from a high floor, it accelerates downward at something less than the lo- cal acceleration of gravity. One feels less pressure on one’s feet; one feels lighter. After accelerating downward for a few seconds, the elevator assumes a constant downward speed, and then one feels the normal pressure of one’s weight on one’s feet. The elevator seems to be slowing down for a stop, but actually it has just stopped accelerating downward. What if the cable snapped, and a frightened passenger dropped his laptop? He could catch it very easily as it would not seem to fall because the elevator, the passenger, and the laptop would all fall at the same rate. The physics in the falling elevator would be the same as if the elevator were at rest in empty space far from any gravitational field. The laptop’s clock would tick as fast as it would at rest in the absence of gravity, but to an observer on the ground it would appear slower. What if a passenger held an electric charge? Observers in the falling el- evator would see a static electric field around the charge, but observers on the ground could detect radiation from the accelerating charge.

Example 13.23 (Proper time) If the events are the ticks of a clock, then the proper time between ticks d⌧/c is the time between the ticks of the clock at rest or at speed zero if the clock is accelerating. The proper lifetime d⌧`/c of an unstable particle is the average time it takes to decay at speed zero. In arbitrary coordinates, this proper lifetime is

c2d⌧ 2 = ds2 = g (x) dxidxk. (13.203) ` ik

Example 13.24 (Clock hypothesis) The apparent lifetime of an unstable particle is independent of the acceleration of the particle even when the particle is subjected to centripetal accelerations of 1019 m/s2(Bailey et al., 1977) and to longitudinal accelerations of 1016m/s2(Roos et al., 1980).

The transformation from arbitrary coordinates xk to free-fall coordinates i y changes the metric gj` to the diagonal metric ⌘ik of flat spacetime ⌘ = 530 Tensors and general relativity diag( 1, 1, 1, 1), which has two indices and is not a Levi-Civita tensor. Al- gebraically, this transformation is a congruence (1.353) @xj @x` ⌘ = g . (13.204) ik @yi j` @yk The geodesic equation (13.191) follows from the principle of equiva- lence (Weinberg, 1972; Hobson et al., 2006). Suppose a particle is moving under 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 = 0 (13.205) d⌧ 2 with d⌧ the proper time d⌧ 2 = ⌘ dyidyk. The chain rule applied to yi(x) ik in (13.205) gives d @yi dxk 0= d⌧ @xk d⌧ ✓ ◆ @yi d2xk @2yi dxk dx` = + . (13.206) @xk d⌧ 2 @xk@x` d⌧ d⌧ We multiply by @xm/@yi and use the identity @xm @yi = m (13.207) @yi @xk k to write the equation of motion (13.205) in the x-coordinates d2xm dxk dx` +m =0. (13.208) d⌧ 2 k` d⌧ d⌧ This is the geodesic equation (13.191) in which the ane connection is @xm @2yi m = . (13.209) k` @yi @xk@x`

13.33 Weak static gravitational fields Newton’s equations describe slow motion in a weak static gravitational field. Because the motion is slow, we neglect ui compared to u0 and simplify the geodesic equation (13.191) to dur 0= +r (u0)2. (13.210) d⌧ 00 13.34 Gravitational time dilation 531

Because the gravitational field is static, we neglect the time derivatives gk0,0 r and g0k,0 in the connection formula (13.87) and find for 00 r 1 rk 1 rk = g g0k,0 + g0k,0 g00,k = g g00,k (13.211) 00 2 2 0 with 00 = 0. Because the field is weak, the metric can di↵er from ⌘ij by only a tiny tensor gij = ⌘ij + hij so that to first order in hij 1we r 1 | |⌧ have = h00,r for r =1, 2, 3. With these simplifications, the geodesic 00 2 equation (13.191) reduces to

2 r 2 r 0 2 d x 1 0 2 d x 1 dx = (u ) h00,r or = h00,r. (13.212) d⌧ 2 2 d⌧ 2 2 d⌧ ✓ ◆ So for slow motion, the ordinary acceleration is described by Newton’s law d2x c2 = h . (13.213) dt2 2 r 00 If is his potential, then for slow motion in weak static fields

g = 1+h = 1 2/c2 and so h = 2/c2. (13.214) 00 00 00 Thus, if the particle is at a distance r from a mass M, then = GM/r and h = 2/c2 =2GM/rc2 and so 00 d2x GM r = = = GM . (13.215) dt2 r r r r3 How weak are the static gravitational fields we know about? The dimen- 2 39 9 sionless ratio /c is 10 on the surface of a proton, 10 on the Earth, 6 4 10 on the surface of the sun, and 10 on the surface of a white dwarf.

13.34 Gravitational time dilation The proper time (example 13.23) interval d⌧ of a clock at rest in the weak, static gravitational field (13.210–13.215) satisfies equation (13.203)

c2d⌧ 2 = ds2 = g (x) dxidxk = g c2dt2 =(1+2/c2)c2dt2. (13.216) ik 00 So if two clocks are at rest at distances r and r + h from the center of the Earth, then the times dtr and dtr+h between ticks of the clock at r and the one at r + h are related by the proper time d⌧ of the ticks of the clock

2 2 2 2 2 2 (c +2(r))dtr = c d⌧ =(c +2(r + h))dtr+h. (13.217) 532 Tensors and general relativity Since (r)= GM/r, the potential at r + h is (r + h) (r)+gh, and ⇡ so the ratio of the tick time dtr of the lower clock at r to the tick time of the upper clock at r + h is dt c2 +2(r + h) c2 +2(r)+2gh gh r = = 1+ . (13.218) 2 2 2 dtr+h p c +2(r) p c +2(r) ⇡ c The lower clockp is slower. p Example 13.25 (Pound and Rebka) Pound and Rebka in 1960 used the M¨ossbauer e↵ect to measure the blue shift of light falling down a 22.6 m shaft. They found (⌫ ⌫ )/⌫ = gh/c2 =2.46 10 15. (Robert Pound ` u ⇥ 1919–2010, Glen Rebka 1931–2015) media.physics.harvard.edu/video/ ?id=LOEB_POUND_092591.flv

Example 13.26 (Redshift of the sun) A photon emitted with frequency ⌫0 at a distance r from a mass M would be observed at spatial infinity to have 2 frequency ⌫ = ⌫0p g00 = ⌫0 1 2MG/c r for a redshift of ⌫ = ⌫0 ⌫. 2 2 6 Since the Sun’s dimensionless potential /c is MG/c r = 2.12 10 p ⇥ at its surface, sunlight is shifted to the red by 2 parts per million.

13.35 Einstein’s equations If we make an action that is a scalar, invariant under general coordinate transformations, and then apply to it the principle of stationary action, we will get tensor field equations that are invariant under general coordinate transformations. If the metric of spacetime is among the fields of the action, then the resulting theory will be a possible theory of gravity. If we make the action as simple as possible, it will be Einstein’s theory. To make the action of the gravitational field, we need a scalar. Apart from the scalar pgd4x = pgcdtd3x,whereg = det(g ) , the simplest scalar | ik | we can form from the metric tensor and its first and second derivatives is the scalar curvature R which gives us the Einstein-Hilbert action c3 c3 S = R pgd4x = gik R pgd4x (13.219) EH 16⇡G 16⇡G ik Z Z 39 2 2 11 3 1 2 in which G =6.7087 10 ~c (GeV/c ) =6.6742 10 m kg s is ⇥ ⇥ Newton’s constant. 13.35 Einstein’s equations 533 If gik(x) is a tiny local change in the inverse metric, then the rule 1 det A =detA Tr(A A) (1.228), valid for any nonsingular, nondefective ik ik ik matrix A) together with the identity 0 = (g gk`)=g gk` + g gk` and the notation g for the metric tensor gj` considered as a matrix imply that

2 ik det g (det g) g gik 1 ik pg = det g = = pggik g . (13.220) 2gpg 2gpg 2 So the first-order change in the action density is

ik ik ik ik g Rik pg = Rik pgg + g Rik pg + g pgRik ⇣ ⌘ 1 (13.221) = R Rg pggik + gik pgR . ik 2 ik ik ✓ ◆ ik The product g Rik is a scalar, so we can evaluate it in any coordinate a system. In a local inertial frame, where bc = 0 and gde is constant, this invariant variation of the Ricci tensor (13.112) is

gikR = gik n n = gik (@ n @ n ) ik in,k ik,n k in n ik = gik @n gin @ k = @ gik n gin k . k in k in k in in ⇣ (13.222)⌘ The transformation law (13.66) for the ane connection shows that the n k variations in and in are tensors although the connections themselves aren’t. Thus, we can evaluate this invariant variation of the Ricci tensor in any coordinate system by replacing the derivatives with covariant ones getting ik ik n in k g Rik = g in g in (13.223) ;k ⇣ ⌘ which we recognize as the covariant divergence (13.180) of a contravariant vector. The last term in the first-order change (13.221) in the action density is therefore a surface term whose variation vanishes for tiny local changes gik of the metric

ik ik n in k pgg Rik = pg g in g in . (13.224) ,k h ⇣ ⌘i Hence the variation of SEH is simply c3 1 S = R g R pggik d4x. (13.225) EH 16⇡G ik 2 ik Z ✓ ◆ The principle of least action SEH = 0 now gives us Einstein’s equations 534 Tensors and general relativity for empty space: 1 R g R =0. (13.226) ik 2 ik The tensor G = R 1 g R is Einstein’s tensor. ik ik 2 ik Taking the trace of Einstein’s equations (13.226), we find that the scalar curvature R and the Ricci tensor Rik are zero in empty space:

R = 0 and Rik =0. (13.227)

The energy-momentum tensor Tik is the source of the gravitational field. It is defined so that the change in the action of the matter fields due to a tiny local change gik(x)inthemetricis 1 1 S = T pggik d4x = T ik pgg d4x (13.228) m 2c ik 2c ik Z Z in which the identity gik = gijg`kg explains the sign change. Now the j` principle of least action S = SEH + Sm = 0 yields Einstein’s equations in the presence of matter and energy 1 8⇡G R g R = T . (13.229) ik 2 ik c4 ik Taking the trace of both sides, we get 8⇡G 8⇡G T R = T and R = T g . (13.230) c4 ik c4 ik 2 ik ✓ ◆

13.36 Energy-momentum tensor

The action Sm of the matter fields is a scalar that is invariant under general coordinate transformations. In particular, a tiny local general coordinate a a a transformation x0 = x + ✏ (x)leavesSm invariant

4 0=Sm = L(i(x)) g(x) d x. (13.231) Z ⇣ p ⌘ The vanishing change Sm = Sm + Smg has a part Sm due to the changes in the fields i(x) and a part Smg due to the change in the metric ik g . The principle of stationary action tells us that the change Sm is zero as long as the fields obey the classical equations of motion. The definition (13.228) of the energy-momentum tensor now tells us that 1 0=S = S = T ik pgg d4x. (13.232) m mg 2c ik Z 13.37 Perfect fluids 535

We take the change in Sm to be

4 4 S = L0(0 (x0)) g (x ) d x0 L( (x)) g(x) d x m i 0 0 i Z Z (13.233) p 4 p 4 = L0(0 (x)) g (x) d x L( (x)) g(x) d x. i 0 i Z p Z p So using the identity gik g = gik g , the definition (13.68) of the k` k` covariant derivative of a covariant vector, and the formula (13.87) for the connection in terms of the metric, we find to lowest order in the change ✏a(x) in xa that the change in the metric is

g = g0 (x) g (x)=g0 (x0) g (x) (g0 (x0) g0 (x)) ik ik ik ik ik ik ik a a b b c =( ✏ )( ✏ )gab ✏ gik,c i ,i k ,k b a c = gib ✏ gak ✏ ✏ gik,c ,k ,i bc ac c = gib(g ✏c),k gak(g ✏c),i ✏ gik,c (13.234) bc ac c = ✏i,k ✏k,i ✏c gib g ✏c gak g ✏ gik,c ,k ,i bc ac c = ✏i,k ✏k,i + ✏c g gib,k + ✏c g gak,i ✏ gik,c ac = ✏i,k ✏k,i + ✏c g gia,k + gak,i gik,a c c = ✏i,k ✏k,i + ✏c + ✏c = ✏i;k ✏k;i. ik ki Combining this result (13.234) with the vanishing (13.232) of the change Smg,wehave

ik 4 0= T pg (✏i;k + ✏k;i) d x. (13.235) Z Since the energy-momentum tensor is symmetric, we may combine the two terms, integrate by parts, divide by pg, and so find that the covariant di- vergence of the energy-momentum tensor is zero

ik ik k ia i ak 1 ik i ak 0=T;k = T,k + ak T +ak T = (pgT ),k +ak T (13.236) pg when the fields obey their equations of motion. In a given inertial frame, only the total energy, momentum, and angular momentum of both the matter and the gravitational field are conserved.

13.37 Perfect fluids In many cosmological models, the energy-momentum tensor is assumed to be that of a perfect fluid, which is isotropic in its rest frame, does not 536 Tensors and general relativity conduct heat, and has zero viscosity. The energy-momentum tensor Tij of a perfect fluid moving with 4-velocity ui (12.20) is p T = pg +( + ⇢) u u (13.237) ij ij c2 i j in which p and ⇢ are the pressure and mass density of the fluid in its rest frame and gij is the spacetime metric. Einstein’s equations (13.229) then are 1 8⇡G 8⇡G p R g R = T = pg +( + ⇢) u u . (13.238) ik 2 ik c4 ik c4 ij c2 i j h i 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 = c2⇢ and the energy-momentum tensor is T = pg = c2⇢g (13.239) ij ij ij 2 in which T00 = c ⇢ is the (presumably constant) value of the energy den- sity of the ground state of the theory. This energy density ⇢ is a plausible candidate for the dark-energy density. It is equivalent to a cosmological constant ⇤=8⇡G⇢.

On small scales, such as that of our solar system, one may neglect mat- ter and dark energy. So in empty space and on small scales, the energy- momentum tensor vanishes Tij = 0 along with its trace and the scalar cur- vature T =0=R, and Einstein’s equations (13.230) are

Rij =0. (13.240)

13.38 Gravitational waves The nonlinear properties of Einstein’s equations (13.229– 13.230) are im- portant on large scales of distance (sections 13.42 & 13.43) and near great masses (sections 13.39 & 13.40). But throughout most of the mature uni- verse, it is helpful to linearize them by writing the metric as the metric ⌘ik of empty, flat spacetime (12.3) plus a tiny deviation hik

gik = ⌘ik + hik. (13.241)

To first order in hik, the ane connection (13.87) is k 1 kj 1 kj = g gji,` + gj`,i gi`,j = ⌘ hji,` + hj`,i hi`,j (13.242) i` 2 2 13.39 Schwarzschild’s solution 537 and the Ricci tensor (13.112) is the contraction

R = Rk =[@ + ,@ + ]k =k k . (13.243) i` ik` k k ` ` i `i,k ki,` k k Since i` = `i and hik = hki, the linearized Ricci tensor is 1 kj 1 kj Ri` = ⌘ hji,` + hj`,i hi`,j ⌘ hji,k + hjk,i hik,j 2 ,k 2 ,` ,k ,k = 1 hk + h h hk . (13.244) 2 `,ik ik,` i`,k k,i` ⇣ ⌘ We can simplify Einstein’s equations (13.230) in empty space Ri` =0by using coordinates in which hik obeys (exercise 13.17) de Donder’s harmonic i 1 j` 1 gauge condition h = (⌘ hj`),k h,k. In this gauge, the linearized k,i 2 ⌘ 2 Einstein equations in empty space are ,k R = 1 h = 0 or c2 2 @2 h =0. (13.245) i` 2 i`,k r 0 i` On 14 September 2015, the LIGO collaboration detected the merger of two black holes of 29 and 36 solar masses which liberated 3M c2 of energy. By 2017, LIGO and Virgo had detected gravitational waves from the mergers of six pairs of black holes and one pair of neutron stars and had set an upper limit of c2m < 2 10 25 eV on the mass of the graviton. g ⇥

13.39 Schwarzschild’s solution In 1916, Schwarzschild solved Einstein’s field equations (13.240) in empty space Rij = 0 outside a static mass M and found as the metric 1 2MG 2MG ds2 = 1 c2dt2 + 1 dr2 + r2 d⌦2 (13.246) c2r c2r ✓ ◆ ✓ ◆ in which d⌦2 = d✓2 +sin2 ✓d2. The Mathematica scripts great.m and Schwarzschild.nb give for the Ricci tensor and the scalar curvature Rik = 0 and R = 0, which show that the metric obeys Einstein’s equations and that the singularity in 1 2MG g = 1 (13.247) rr c2r ✓ ◆ 2 at the Schwarzschild radius rS =2MG/c is an artifact of the coordinates. Schwarzschild.nb also computes the ane connections. 2 The Schwarzschild radius of the Sun rS =2M G/c =2.95 km is far less 538 Tensors and general relativity than the Sun’s radius r =6.955 105 km, beyond which his metric applies. ⇥ (Karl Schwarzschild, 1873–1916)

13.40 Black holes Suppose an uncharged, spherically symmetric star of mass M has collapsed within a sphere of radius rb less than its Schwarzschild radius or horizon 2 rh = rS =2MG/c . Then for r>rb, the Schwarzschild metric (13.246) is correct. The time dt measured on a clock outside the gravitational field is related to the proper time d⌧ on a clock fixed at r 2MG/c2 by (13.216) 2MG p dt = d⌧/ g00 = d⌧/ 1 2 . (13.248) r c r The time dt measured away from the star becomes infinite as r approaches 2 the horizon rh = rS =2MG/c . To outside observers, a clock at the horizon rh seems frozen in time. Due to the gravitational redshift (13.248), light of frequency ⌫p emitted at r 2MG/c2 will have frequency ⌫ 2MG p ⌫ = ⌫p g00 = ⌫p 1 2 (13.249) r c r when observed at great distances. Light coming from the surface at rS = 2MG/c2 is redshifted to zero frequency ⌫ = 0. The star is black. It is a 2 black hole with a horizon at its Schwarzschild radius rh = rS =2MG/c , although there is no singularity there. If the radius of the Sun were less than its Schwarzschild radius of 2.95 km, then the Sun would be a black hole. The radius of the Sun is 6.955 105 km. ⇥ Black holes are not black. Stephen Hawking (1942–2018) showed (Hawk- ing, 1975) that the intense gravitational field of a black hole of mass M radiates at a temperature

~ c3 ~c ~g T = = = (13.250) 8⇡kGM 4⇡krh 2⇡kc 5 1 in which k =8.617 10 eV K is Boltzmann’s constant, ~ is Planck’s ⇥ 34 2 constant h =6.626 10 Js dividedby2⇡, ~ = h/(2⇡), and g = GM/r ⇥ h is the gravitational acceleration at r = rh. More generally, a detector in vacuum subject to a uniform acceleration a (in its instantaneous rest frame) sees a temperature T = ~a/(2⇡kc) (Alsing and Milonni, 2004). In a region of empty space where the pressure p and the chemical poten- 2 tials µj all vanish, the change (7.111) in the internal energy U = c M of a 13.41 Rotating black holes 539 black hole of mass M is c2dM = TdS where S is its entropy. So the change 2 dS in the entropy of a black hole of mass M = c rh/(2G) and temperature T = ~c/(4⇡krh) (13.250) is 2 2 2 3 c 4⇡c krh 4⇡ckrh c ⇡c k dS = dM = dM = drh = 2rhdrh. (13.251) T (~c) ~ 2G G~ Integrating, we get a formula for the entropy of a black hole in terms of its area (Bekenstein, 1973; Hawking, 1975)

3 3 ⇡c k 2 c k A S = rh = (13.252) G~ ~G 4 2 where A =4⇡rh is the area of the horizon of the black hole. A black hole is entirely converted into radiation after a time 5120 ⇡G2 t = M 3 (13.253) ~ c4 proportional to the cube of its mass M.

13.41 Rotating black holes A half-century after Einstein invented general relativity, Roy Kerr invented the metric for a mass M rotating with angular momentum J = GMa/c. Two years later, Newman and others generalized the Kerr metric to one of charge Q. In Boyer-Lindquist coordinates, its line element is

2 ds2 = dt a sin2 ✓d ⇢2 2 2 (13.254) sin ✓ 2 ⇢ + (r2 + a2)d adt + dr2 + ⇢2d✓2 ⇢2 where ⇢2 = r2 + a2 cos2 ✓ and = r2 + a2 2GMr/c2 + Q2. The Math- ematica script Kerr black hole.nb shows that the Kerr-Newman metric for the uncharged case, Q = 0, has Rik = 0 and R = 0 and so is a solution of Einstein’s equations in empty space (13.230) with zero scalar curvature. A rotating mass drags nearby masses along with it. The daily rotation of the Earth moves satellites to the East by tens of meters per year. The 2 frame dragging of extremal black holes with J . GM /c approaches the speed of light (Ghosh et al., 2018). (Roy Kerr, 1934–) 540 Tensors and general relativity 13.42 Spatially symmetric spacetimes Einstein’s equations (13.230) are second-order, nonlinear partial di↵erential equations for 10 unknown functions gik(x) in terms of the energy-momentum tensor Tik(x) throughout the universe, which of course we don’t know. The problem is not quite hopeless, however. The ability to choose arbitrary co- ordinates, the appeal to symmetry, and the choice of a reasonable form for Tik all help. Astrophysical observations tell us that the universe extends at least 46 billion light years in all directions; that it is flat or very nearly flat; and that the cosmic microwave background (CMB) radiation is isotropic to one part in 105 apart from a Doppler shift due the motion of the Sun at 370 km/s towards the constellation Leo. These microwave photons have been moving freely since the universe became cool enough for hydrogen atoms to be stable. Observations of clusters of galaxies reveal an expanding universe that is homogeneous on suitably large scales of distance. Thus as far as we know, the universe is homogeneous and isotropic in space, but not in time. There are only three maximally symmetric 3-dimensional spaces: eu- clidian space E3,thesphereS3 (example 13.16), and the hyperboloid H3 (example 13.17). Their line elements may be written in terms of a distance L as dr2 ds2 = + r2d⌦2 (13.255) 1 kr2/L2 in which k = 1 for the sphere, k = 0 for euclidian space, and k = 1 for the hyperboloid. The Friedmann-Lemaˆıtre-Robinson-Walker (FLRW) cosmologies add to these spatially symmetric line elements a dimensionless scale factor a(t) that describes the expansion (or contraction) of space dr2 ds2 = c2dt2 + a2(t) + r2 d✓2 + r2 sin2 ✓d2 . (13.256) 1 kr2/L2 ✓ ◆ The FLRW metric is c2 000 0 a2/(1 kr2/L2)0 0 gik(t, r, ✓,)= 0 1 . (13.257) 00a2 r2 0 B 000a2 r2 sin2 ✓C B C @ A The constant k determines whether the spatial universe is open k = 1, flat k = 0, or closed k = 1. The coordinates x0,x1,x2,x3 t, r, ✓, are ⌘ comoving in that a detector at rest at r, ✓, records the CMB as isotropic with no Doppler shift. 13.42 Spatially symmetric spacetimes 541 The metric (13.257) is diagonal; its inverse gij also is diagonal

2 c 000 2 2 2 ik 0(1kr /L )/a 00 g = 0 2 1 . (13.258) 00(ar) 0 B 00 0(ar sin ✓) 2C B C @ A One may use the formula (13.87) to compute the ane connection in k 1 kj terms of the metric and its inverse as = g (gji,` + gj`,i g`i,j). It i` 2 usually is easier, however, to use the action principle (13.192) to derive the i geodesic equation directly and then to read its expressions for the jk’s. So we require that the integral

2 2 2 2 a r0 2 2 2 2 2 2 2 0= c t0 + + a r ✓0 + a r sin ✓0 d, (13.259) 1 kr2/L2 Z ✓ ◆ in which a prime means derivative with respect to , be stationary with respect to the tiny variations t(),r(),✓(), and (). By varying t(), we get the equation

2 aa˙ r0 2 2 2 2 2 t i ` 0=t00 + + r ✓0 + r sin ✓0 = t00 + x0 x0 (13.260) c2 1 kr2/L2 i` ✓ ◆ t which tells us that the nonzero jk’s are

aa˙ aar˙ 2 aar˙ 2 sin2 ✓ t = , t = , and t = . rr c2(1 kr2/L2) ✓✓ c2 c2 (13.261) By varying r() we get (with more e↵ort)

2 2 2 rr0 k/L at˙ 0r0 kr 2 2 2 0=r00 + +2 r (1 )(✓0 +sin ✓0 ). (13.262) (1 kr2/L2) a L2 r So we find that tr =˙a/a,

kr kr3 r = , r = r + , and r =sin2 ✓ r . (13.263) rr L2 kr2 ✓✓ L2 ✓✓ Varying ✓() gives

a˙ 2 2 0=✓00 +2 t0✓0 + ✓0r0 sin ✓ cos ✓0 and a r (13.264) a˙ 1 ✓ = , ✓ = , and ✓ = sin ✓ cos ✓. t✓ a r✓ r 542 Tensors and general relativity Finally, varying () gives

a˙ r00 0=00 +2 t00 +2 + 2 cot ✓✓00 and a r (13.265) a˙ 1 = , = , and = cot ✓. t a r r ✓ k k Other ’s are either zero or related by the symmetry i` = `i. Our formulas for the Ricci (13.112) and curvature (13.103) tensors give

k k k R00 = R 0k0 =[Dk,D0] 0 =[@k +k,@0 +0] 0. (13.266) 1 2 Because [D0,D0] = 0, we need only compute [D1,D0] 0,[D2,D0] 0, and 3 [D3,D0] 0. Using the formulas (13.261–13.265) for the ’s and keeping in mind (13.102) that the element of row r and column c of the `th gamma r matrix is `c,wefind 1 1 1 1 j 1 j 2 [D1,D0] = + = (˙a/a),0 (˙a/a) 0 00,1 10,0 1j 00 0j 10 2 2 2 2 j 2 j 2 [D2,D0] = + = (˙a/a),0 (˙a/a) 0 00,2 20,0 2j 00 0j 20 3 3 3 3 j 3 j 2 [D3,D0] = + = (˙a/a),0 (˙a/a) 0 00,3 30,0 3j 00 0j 30 k 3 Rtt = R00 =[Dk,D0] = 3(˙a/a),0 3(˙a/a) = 3¨a/a. (13.267) 0 k k k Thus for Rrr = R11 = R 1k1 =[Dk,D1] 1 =[@k +k,@1 +1] 1, we get aa¨ +2˙a2 +2kc2/L2 R =[D ,D ]k = (13.268) rr k 1 1 c2(1 kr2/L2) (exercise 13.21), and for R22 = R✓✓ and R33 = R we find

2 2 2 2 2 2 R✓✓ =[(aa¨ +2˙a +2kc /L )r ]/c and R =sin ✓R✓✓ (13.269)

ab (exercises 13.22 & 13.23). And so the scalar curvature R = g Rba is 2 2 ab R00 (1 kr /L )R11 R22 R33 R = g Rba = + + + c2 a2 a2r2 a2r2 sin2 ✓ aa¨ +˙a2 + kc2/L2 =6 . (13.270) c2a2 It is, of course, quicker to use the Mathematica script FLRW.nb.

13.43 Friedmann-Lemaˆıtre-Robinson-Walker cosmologies The energy-momentum tensor (13.237) of a perfect fluid moving at 4-velocity 2 ui is Tik = pgik +(p/c + ⇢)uiuk where p and ⇢ are the pressure and mass density of the fluid in its rest frame. In the comoving coordinates of the 13.43 Friedmann-Lemaˆıtre-Robinson-Walker cosmologies 543 FLRW metric (13.257), the 4-velocity (12.20) is ui =(1, 0, 0, 0), and the energy-momentum tensor (13.237) is c2⇢g 000 00 0 pg11 00 Tij = 0 1 . (13.271) 00pg22 0 B 000pg C B 33C @ A Its trace is T = gij T = c2⇢ +3p. (13.272) ij Thus successively using our formulas (13.257) for g = c2, (13.267) for 00 R = 3¨a/a, (13.271) for T , and (13.272) for T , we can write the 00 00 ij Einstein equation (13.230) as the second-order equation a¨ 4⇡G 3p = ⇢ + . (13.273) a 3 c2 ✓ ◆ It is nonlinear because ⇢ and 3p depend upon a.Thesumc2⇢+3p determines the accelerationa ¨ of the scale factor a(t); when it is negative, it accelerates the expansion. If we combine Einstein’s formula for the scalar curvature R = 8⇡G T/c4 (13.230) with the FLRW formulas for R (13.270) and for the trace T (13.272) of the energy-momentum tensor, we get

a¨ a˙ 2 c2k 4⇡G 3p + + = ⇢ . (13.274) a a L2a2 3 c2 ✓ ◆ ✓ ◆ Using the 00-equation (13.273) to eliminate the second derivativea ¨,wefind

a˙ 2 8⇡G c2k = ⇢ (13.275) a 3 L2a2 ✓ ◆ which is a first-order nonlinear equation. It and the second-order equation (13.273) are known as the Friedmann equations. The left-hand side of the first-order Friedmann equation (13.275) is the square of the Hubble rate a˙ H = (13.276) a which is an inverse time or a frequency. Its present value H0 is the Hubble constant. In terms of H, the first-order Friedmann equation (13.275) is 8⇡G c2k H2 = ⇢ . (13.277) 3 L2a2 544 Tensors and general relativity An absolutely flat universe has k = 0, and therefore its density must be 3H2 ⇢ = (13.278) c 8⇡G which is the critical mass density.

13.44 Density and pressure

The 0-th energy-momentum conservation law (13.236) is

0a 0a a 0c 0 ca 0=T;a = @aT + caT + caT . (13.279) For a perfect fluid of 4-velocity ua, the energy-momentum tensor (13.271) is T ik =(⇢ + p/c2)uiuk + pgik in which ⇢ and p are the mass density and pressure of the fluid in its rest frame. The comoving frame of the Friedmann- Lemaˆıtre-Robinson-Walker metric (13.257) is the rest frame of the fluid. In these coordinates, the 4-velocity ua is (1, 0, 0, 0), and the energy-momentum tensor is diagonal with T 00 = ⇢ and T jj = pgjj for j =1, 2, 3. Our connection 0 0 2 formulas (13.261) tell us that 00 = 0, that jj =˙agjj/(c a), and that j 0j =3˙a/a. Thus the conservation law (13.279) becomes for spatial j

00 j 00 0 jj 0=@0T + 0jT + jjT 3 a˙ ag˙ a˙ p (13.280) =˙⇢ +3 ⇢ + jj pgjj =˙⇢ +3 ⇢ + . a c2a a c2 Xj=1 ⇣ ⌘ Thus 3˙a p d⇢ 3 p ⇢˙ = (⇢ + ), and so = ⇢ + . (13.281) a c2 da a c2 ⇣ ⌘ The energy density ⇢ is composed of fractions ⇢i each contributing its own partial pressure pi according to its own equation of state 2 pi = c wi⇢i (13.282) in which wi is a constant. The rate of change (13.282) of the density ⇢i is then d⇢ 3 i = (1 + w ) ⇢ . (13.283) da a i i

In terms of the present density ⇢i0 and scale factor a0, the solution is

a 3(1+wi) ⇢ = ⇢ 0 . (13.284) i i0 a ⇣ ⌘ 13.45 How the scale factor evolves with time 545

There are three important kinds of density. The dark-energy density ⇢⇤ is assumed to be like a cosmological constant ⇤or like the energy density of the vacuum, so it is independent of the scale factor a and has w = 1. ⇤ A universe composed only of dust or non-relativistic collisionless matter has no pressure. Thus p = w⇢ =0with⇢ = 0, and so w = 0. 6 So the matter density falls inversely with the volume

a 3 ⇢ = ⇢ 0 . (13.285) m m0 a ⇣ ⌘ The density of radiation ⇢r has wr =1/3 because wavelengths scale with the scale factor, and so there’s an extra factor of a

a 4 ⇢ = ⇢ 0 . (13.286) r r0 a ⇣ ⌘ The total density ⇢ varies with a as

a 3 a 4 ⇢ = ⇢ + ⇢ 0 + ⇢ 0 . (13.287) ⇤ m0 a r0 a ⇣ ⌘ ⇣ ⌘ This mass density ⇢, the Friedmann equations (13.273 & 13.275), and the physics of the standard model have caused our universe to evolve as in Fig. 13.1 over the past 14 billion years.

13.45 How the scale factor evolves with time

The first-order Friedmann equation (13.275) expresses the square of the instantaneous Hubble rate H =˙a/a in terms of the density ⇢ and the scale factor a(t)

a˙ 2 8⇡G c2k H2 = = ⇢ (13.288) a 3 L2a2 ✓ ◆ in which k = 1 or 0. The critical density ⇢ =3H2/(8⇡G) (13.278) is the ± c one that satisfies this equation for a flat (k = 0) universe. Its present value is 2 27 3 ⇢c0 =3H0 /(8⇡G)=8.599 10 kg m . Dividing Friedmann’s equation ⇥ 2 by the square of the present Hubble rate H0 , we get

H2 1 a˙ 2 1 8⇡G c2k ⇢ c2k = = ⇢ = (13.289) H2 H2 a H2 3 a2L2 ⇢ a2H2L2 0 0 ✓ ◆ 0 ✓ ◆ 0c 0 546 Tensors and general relativity History of the universe

Figure 13.1 NASA/WMAP Science Team’s timeline of the known universe. in which ⇢ is the total density (13.287)

2 2 H ⇢⇤ ⇢m ⇢r c k 2 = + + 2 2 2 H0 ⇢c0 ⇢c0 ⇢c0 a H0 L 3 4 2 2 (13.290) ⇢⇤ ⇢m0 a0 ⇢r0 a0 c k a0 = + 3 + 4 2 2 2 2 . ⇢c0 ⇢c0 a ⇢c0 a a0H0 L a

The Planck collaboration use a model in which the energy density of the universe is due to radiation, matter, and a cosmological constant ⇤. Only about 18.79% of the matter in their model is composed of baryons, ⌦ =0.05845 0.0003. Most of the matter is transparent and is called dark b ± matter. They assume the dark matter is composed of particles that have masses in excess of a keV so that they are heavy enough to have been nonrela- tivistic or “cold” when the universe was about a year old (Peebles, 1982). The energy density of the cosmological constant ⇤is known as dark energy.The Planck collaboration use this ⇤-cold-dark-matter (⇤CDM) model and their CMB data to estimate the Hubble constant as H0 = 67.66km/(s Mpc) = 2.1927 10 18 s 1 and the density ratios ⌦ = ⇢ /⇢ ,⌦ = ⇢ /⇢ , and ⇥ ⇤ ⇤ c0 m m0 c0 ⌦ c2k/(a H L)2 as listed in the table (13.1) (Aghanim et al., 2018). k ⌘ 0 0 The Riess group use the Gaia observatory to calibrate Cepheid stars and type Ia supernovas as standard candles for measuring distances to remote galaxies. The distances and redshifts of these galaxies give the Hubble con- 13.45 How the scale factor evolves with time 547 stant as H = 73.48 1.66 (Riess et al., 2018). As this book goes to press, 0 ± the 9% discrepancy between the Planck and Riess H0’s is unexplained.

Table 13.1 Cosmological parameters of the Planck collaboration

H0 (km/(s Mpc) ⌦⇤ ⌦m ⌦k 67.66 0.42 0.6889 0.0056 0.3111 0.0056 0.0007 0.0037 ± ± ± ±

To estimate the ratio ⌦r = ⇢r0/⇢c0 of radiation densities, one may use the present temperature T =2.7255 0.0006 K (Fixsen, 2009) of the CMB 0 ± radiation and the formula (5.110) for the energy density of photons

5 4 8⇡ (kBT0) 31 3 ⇢ = =4.6451 10 kg m . (13.291) 15h3c5 ⇥ 1/3 Adding in three kinds of neutrinos and antineutrinos at T0⌫ =(4/11) T0, we get for the present density of massless and nearly massless particles (Wein- berg, 2010, section 2.1)

4/3 7 4 31 3 ⇢ = 1+3 ⇢ =7.8099 10 kg m . (13.292) r0 8 11 ⇥ " ✓ ◆✓ ◆ #

The fraction ⌦r the present energy density that is due to radiation is then

⇢r0 5 ⌦r = =9.0824 10 . (13.293) ⇢c0 ⇥

In terms of ⌦r and of the ⌦’s in the table (13.1), the formula (13.290) for 2 2 H /H0 is 2 2 3 4 H a0 a0 a0 2 =⌦⇤ +⌦k 2 +⌦m 3 +⌦r 4 . (13.294) H0 a a a 1 Since H =˙a/a, one has dt = da/(aH)=H0 (da/a)(H0/H), and so with x = a/a0, the time interval dt is 1 dx 1 dt = . (13.295) 2 3 4 H0 x ⌦⇤ +⌦k x +⌦m x +⌦r x Integrating and settingp the origin of time t(0) = 0 and the scale factor at the present time equal to unity a0 = 1, we find that the time t(a) that a(t) took to grow from 0 to a(t)is 1 a dx t(a)= . (13.296) H 2 1 2 0 Z0 ⌦⇤ x +⌦k +⌦m x +⌦r x p 548 Tensors and general relativity This integral gives the age of the universe as t(1) = 13.789 Gyr; the Planck- collaboration value is 13.787 0.020 Gyr (Aghanim et al., 2018). Figure 13.2 ± plots the scale factor a(t) and the redshift z(t)=1/a 1 as functions of the time t (13.296) for the first 14 billion years after the time t = 0 of infinite redshift. A photon emitted with wavelength at time t(a) now has wavelength 0 = /a(t). The change in its wavelength is = z(t)= (1/a 1) = . 0 Evolution of scalefactor over 14 Gyr 1 10

0.9 9

0.8 8

0.7 7

0.6 6

0.5 5

0.4 4

0.3 3

0.2 2

0.1 1

0 0 0 2 4 6 8 10 12 14

Figure 13.2 The scale factor a(t) (solid, left axis) and redshift z(t) (dotdash, right axis) are plotted against the time (13.296) in Gyr. This chapter’s Fortran, Matlab, and Mathematica scripts are in Ten- sors and general relativity at github.com/kevinecahill.

13.46 The first hundred thousand years Figure 13.3 plots the scale factor a(t) as given by the integral (13.296) and the densities of radiation ⇢r(t) and matter ⇢m(t) for the first 100,000 years after the time of infinite redshift. Because wavelengths grow with the scale factor, the radiation density (13.286) is proportional to the inverse fourth 4 power of the scale factor ⇢r(t)=⇢r0/a (t). The density of radiation therefore was dominant at early times when the scale factor was small. Keeping only 13.46 The first hundred thousand years 549

⌦r =0.6889 in the integral (13.296), we get 2 a 1/4 t = and a(t)=⌦r 2 H0 t. (13.297) 2H0p⌦r p 4 Since the radiation density ⇢r(t)=⇢r0/a (t) also is proportional to the fourth power of the temperature ⇢ (t) T 4, the temperature varied as the r ⇠ inverse of the scale factor T 1/a(t) t 1/2 during the era of radiation. ⇠ ⇠ Evolution of scalefactor and densities over first 100 kyr -16 10-4 10 4.5 5

4 4.5

4 3.5

3.5 3

3 2.5 2.5 2 2

1.5 1.5

1 1

0.5 0.5

0 0 0 10 20 30 40 50 60 70 80 90 100

Figure 13.3 The Planck-collaboration scale factor a (solid), radiation den- sity ⇢r (dotdash), and matter density ⇢m (dashed) are plotted as functions of the time (13.296) in kyr. The era of radiation ends at t = 50, 506 years 16 3 4 when the two densities are equal to 1.0751 10 kg/m , a =2.919 10 , and z = 3425. ⇥ ⇥

In cold-dark-matter models, when the temperature was in the range 1012 > 10 2 2 T>10 K or mµc >kT>mec ,wheremµ is the mass of the muon and me that of the electron, the radiation was mostly electrons, positrons, photons, and neutrinos, and the relation between the time t and the tem- perature T was t 0.994 sec (1010 K/T )2 (Weinberg, 2010, ch. 3). By ⇠ ⇥ 109 K, the positrons had annihilated with electrons, and the neutrinos fallen out of equilibrium. Between 109 K and 106 K, when the energy density of nonrelativistic particles became relevant, the time-temperature relation was t 1.78 sec (1010 K/T )2 (Weinberg, 2010, ch. 3). During the first three ⇠ ⇥ minutes (Weinberg, 1988) of the era of radiation, quarks and gluons formed 550 Tensors and general relativity hadrons, which decayed into protons and neutrons. As the neutrons decayed (⌧ = 877.7 s), they and the protons formed the light elements—principally hydrogen, deuterium, and helium in a process called big-bang nucleosyn- thesis. The density of nonrelativistic matter (13.285) falls as the third power 3 of the scale factor ⇢m(t)=⇢m0/a (t). The more rapidly falling density of radiation ⇢r(t) crosses it 50,506 years after the Big Bang as indicated by the vertical line in the figure (13.3). This time t = 50, 506 yr and redshift z = 3425 mark the end of the era of radiation.

13.47 The next ten billion years The era of matter began about 50,506 years after the time of infinite redshift when the matter density ⇢m first exceeded the radiation density ⇢r. Some 330,000 years later at t 380, 000 yr, the universe had cooled to about ⇠ T = 3000 K or kT =0.26 eV—a temperature at which less than 1% of the hydrogen was ionized. At this redshift of z = 1090, the plasma of ions and electrons became a transparent gas of neutral hydrogen and helium with trace amounts of deuterium, helium-3, and lithium-7. The photons emitted or scattered at that time as 0.26 eV or 3000 K photons have redshifted down to become the 2.7255 K photons of the cosmic microwave background (CMB) radiation. This time of last scattering and first transparency often and inexplicably is called recombination.

If we approximate time periods t t during the era of matter by keeping m only ⌦m in the integral (13.296), then we get

2/3 2/3 2 a 3H0p⌦m (t tm) t tm = and a(t)= (13.298) 3H p⌦ 2 0 m ✓ ◆ in which tm is a time well inside the era of matter. Between 10 and 17 million years after the Big Bang, the temperature of the known universe fell from 373 to 273 K. If by then the supernovas of very early, very heavy stars had produced carbon, nitrogen, and oxygen, biochemistry may have started during this period of 7 million years. Stars did form at least as early as 180 million years after the Big Bang (Bowman et al., 2018). The era of matter lasted until the energy density of matter ⇢m(t), falling 3 as ⇢m(t)=⇢m0/a (t) had dropped to the energy density of dark energy ⇢ =5.9238 10 27kg/m3. This happened at t = 10.228 Gyr as indicated ⇤ ⇥ by the first vertical line in the figure (13.4). 13.48 Era of dark energy 551 Evolution of scalefactor and densities over 50 Gyr 10-26 9 1

8 0.9

0.8 7

0.7 6

0.6 5 0.5 4 0.4

3 0.3

2 0.2

1 0.1

0 0 0 5 10 15 20 25 30 35 40 45 50

Figure 13.4 The scale factor a (solid), the vacuum density ⇢⇤ (dotdash), and the matter density ⇢m (dashed) are plotted as functions of the time (13.296) in Gyr. The era of matter ends at t = 10.228 Gyr (first vertical line) 27 3 when the two densities are equal to 5.9238 10 kg m and a =0.7672. ⇥ The present time t0 is 13.787 Gyr (second vertical line) at which a(t) = 1.

13.48 Era of dark energy The era of dark energy began 3.6 billion years ago at a redshift of z =0.3034 when the energy density of the universe ⇢m + ⇢⇤ was twice that of empty space, ⇢ =2⇢ =1.185 10 26 kg/m3. The energy density of matter now ⇤ ⇥ is only 31.11% of the energy density of the universe, and it is falling as the 3 cube of the scale factor ⇢m(t)=⇢m0/a (t). In another 20 billion years, the energy density of the universe will have declined almost all the way to the 27 3 4 3 5 dark-energy density ⇢⇤ =5.9238 10 kg/m or (1.5864 meV) /(~ c ). ⇥ At that time t⇤ and in the indefinite future, the only significant term in the integral (13.296) will be the vacuum energy. Neglecting the others and replacing a0 =1witha⇤ = a(t⇤), we find

log(a/a⇤) H0p⌦⇤(t t⇤) t(a/a⇤) t⇤ = or a(t)=e a(t⇤) (13.299) H0p⌦⇤ 552 Tensors and general relativity in which t⇤ & 35 Gyr.

13.49 Before the Big Bang The ⇤CDM model is remarkably successful (Aghanim et al., 2018). But it does not explain why the CMB is so isotropic, apart from a Doppler shift, and why the universe is so flat (Guth, 1981). A brief period of rapid exponential growth like that of the era of dark energy may explain the isotropy and the flatness. Inflation occurs when the potential energy ⇢ dwarfs the energy of matter and radiation. The internal energy of the universe then is proportional to its volume U = c2⇢V, and its pressure p as given by the thermodynamic relation @U p = = c2⇢ (13.300) @V is negative. The second-order Friedmann equation (13.273) a¨ 4⇡G 3p 8⇡G⇢ = ⇢ + = (13.301) a 3 c2 3 ✓ ◆ then implies exponential growth like that of the era of dark energy (13.299)

a(t)=ep8⇡G⇢/3 t a(0). (13.302) The origin of the potential-energy density ⇢ is unknown. In chaotic inflation (Linde, 1983), a scalar field fluctuates to a mean value that makes its potential-energy density ⇢ huge. The field remains h ii at or close to the value , and the scale factor inflates rapidly and expo- h ii nentially (13.302) until time t at which the potential energy of the universe is E = c2⇢ep24⇡G⇢t V (0) (13.303) where V (0) is the spatial volume in which the field held the value . h ii After time t, the field returns to its mean value 0 0 in the ground state 0 h | | i | i of the theory, and the huge energy E is released as radiation in a Big Bang. The energy Eg of the gravitational field caused by inflation is negative, E = E, and energy is conserved. Chaotic inflation plausibly explains g why there is a universe: a quantum fluctuation made it. If the universe did arise from a quantum fluctuation, other quantum fluctuations would occur elsewhere inflating new universes and making a multiverse. 13.50 Yang-Mills theory 553 If a quantum fluctuation gives a field a spatially constant mean value in an initial volume V (0), then the equations for the scale factor h ii ⌘ (13.301) and for the scalar field (13.201) simplify to 8⇡G⇢ 1/2 m2c4 H = and ¨ = 3 H ˙ (13.304) 3 2 ✓ ◆ ~ in which ⇢ is the mass density of the potential energy of the field .The term 3H˙ is a kind of gravitational friction. It may explain why a field sticks at the value long enough to resolve the isotropy and flatness h ii puzzles. The anti-de Sitter (k = 1) spacetime a(t)=c/(L)sin(t) of exam- ple 7.63 and bouncing cosmologies (Steinhardt et al., 2002; Ijjas and Stein- hardt, 2018) explain the flatness and isotropy of the universe as due to repeated collapses and rebirths. Experiments will tell us whether inflation or bouncing or something else actually occurred (Akrami et al., 2018).

13.50 Yang-Mills theory The gauge transformation of an abelian gauge theory like electrodynam- ics multiplies a single charged field by a spacetime-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 spacetime dependent unitary matrix U(x) n

a0 (x)= Uab(x) b(x) or 0(x)=U(x) (x) (13.305) Xb=1 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

0 †(x) (x) = †(x)U †(x)U(x)(x)=†(x)(x). (13.306)

⇣ ⌘ i But inner products of derivatives @ † @i are not invariant because the derivative 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 = UDi. (13.307) 554 Tensors and general relativity

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 (13.308) in which Ai like @i is antihermitian. They required that under the gauge transformation (13.305), the gauge-field matrix Ai transform to Ai0 in such a way as to make the derivatives transform as in (13.307)

(Di)0 = @i + Ai0 0 = @i + Ai0 U = UDi = U (@i + Ai) . (13.309) So they set

@i + Ai0 U = U (@i + Ai) or (@iU) + Ai0 U = UAi (13.310) and made the gauge-field matrix Ai transform as 1 1 A0 = UA U (@ U) U . (13.311) i i i Thus under the gauge transformation (13.305), the derivative Di trans- forms as in (13.307), like the vector in (13.305), and the inner product of covariant derivatives

i 0 i i D † Di = D † U †UDi = D † Di (13.312) remains invariant.h i To make an invariant action density for the gauge-field matrices Ai,they used the transformation law (13.309) which implies that Di0 U = UDi or 1 Di0 = UDi U . So they defined their generalized Faraday tensor as F =[D ,D ]=@ A @ A +[A ,A ] (13.313) ik i k i k k i i k so that it transforms covariantly

1 Fik0 = UFikU . (13.314) ik They then generalized the action density FikF of electrodynamics to the ik trace Tr FikF of the square of the Faraday matrices which is invariant under gauge transformations since 1 ik 1 ik 1 ik Tr UFikU UF U =Tr UFikF U =Tr FikF . (13.315) ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ As an action density for fermionic matter fields, they replaced the ordi- i nary derivative in Dirac’s formula ( @i + m) by the covariant derivative i (13.308) to get ( Di + m) (Chen-Ning Yang 1922–, Robert L. Mills 1927–1999). i In an abelian gauge theory, the square of the 1-form A = Ai dx vanishes 13.51 Cartan’s spin connection and structure equations 555 A2 = A A dxi dxk = 0, but in a nonabelian gauge theory the gauge fields i k ^ are matrices, and A2 = 0. The sum dA + A2 is the Faraday 2-form 6 F = dA + A2 =(@ A + A A ) dxi dxk (13.316) i k i k ^ = 1 (@ A @ A +[A ,A ]) dxi dxk = 1 F dxi dxk. 2 i k k i i k ^ 2 ik ^ The scalar matter fields may have self-interactions described by a po- 2 2 tential V () such as V ()=(† m /) which is positive unless † = 2 i m /. The kinetic action of these fields is (D )†Di. At low temperatures, these scalar fields assume mean values 0 0 = 0 in the vacuum with 2 h | | i 0†0 = m / so as to minimize their potential energy density V () and i i i their kinetic action (D )†Di =(@ + A )†(@i + Ai) is approximately i i ↵ i 0† A Ai 0. The gauge-field matrix Aab = itabA↵ is a linear combination of the generators t↵ of the gauge group. So the action of the scalar fields i 2 i contains the term † A A = M A A in which the mass-squared 0 i 0 ↵ ↵ i 2 a ↵ c matrix for the gauge fields is M↵ = 0⇤ tab tbc 0.ThisHiggs mechanism gives masses to those linear combinations bi A of the gauge fields for which M 2 b = m2b =0. ↵ i i ↵i 6 The Higgs mechanism also gives masses to the fermions. The mass term m in the Yang-Mills-Dirac action is replaced by something like cin which c is a constant, di↵erent for each fermion. In the vacuum and at low temperatures, each fermion acquires as its mass c0 . On 4 July 2012, physicists at CERN’s Large Hadron Collider announced the discovery of a Higgs-like particle with a mass near 12 5 GeV/c2 (Peter Higgs 1929 –).

13.51 Cartan’s spin connection and structure equations a Cartan’s tetrads (13.153) c k(x) are the rows and columns of the orthogo- nal matrix that turns the flat-space metric ⌘ab into the curved-space metric g = ca ⌘ cb . Early-alphabet letters a, b, c, d, =0, 1, 2, 3 are Lorentz in- ik i ab k ··· dexes, and middle-to-late letters i, j, k, `, =0, 1, 2, 3 are spacetime in- ··· dexes. Under a combined local Lorentz (13.154) and general coordinate transformation the tetrads transform as ` a a @x b c0 k(x0)=L b(x0) k c `(x). (13.317) @x0

The covariant derivative of a tetrad D`c must transform as i j a a @x @x b (D` c k)0(x0)=L b(x0) ` k Di c j(x). (13.318) @x0 @x0 556 Tensors and general relativity j We can use the ane connection k` and the formula (13.68) for the co- variant derivative of a covariant vector to cope with the index j. And we can treat the Lorentz index b like an index of a nonabelian group as in a section 13.50 by introducing a gauge field ! `b D ca = ca j ca + !a cb . (13.319) ` k ,` k` j `b k The ane connection is defined so as to make the covariant derivative of the tangent basis vectors vanish

D e↵ = e↵ j e↵ = 0 (13.320) ` k k,` k` j in which the Greek letter ↵ labels the coordinates 0, 1, 2,...,n of the em- bedding space. We may verify this relation by taking the inner product in i the embedding space with the dual tangent vector e↵ i j ↵ i j i i ↵ i e e = = = e e = e ek,` (13.321) ↵ k` j j k` k` ↵ k,` · i i which is the definition (13.59) of the ane connection, k` = e ek,`.Sotoo a · the spin connection ! b` is defined so as to make the covariant derivative of the tetrad vanish

D ca = ca = ca j ca + !a cd =0. (13.322) ` k k;` k,` k` j d` k k The dual tetrads cb are doubly orthonormal: k b k k b b cb ci = i and ca ck = a. (13.323) a d k a d a Thus using their orthonormality, we have ! d`ck cb = ! d`b = ! b`, and so the spin connection is

!a = ck ca j ca = ca ck j ca ck. (13.324) b` b k,` k` j j b k` k,` b In terms of the di↵erential forms (section 12.6)

a a k a a ` c = ck dx and ! b = ! b`dx (13.325) we may use the exterior derivative to express the vanishing (13.322) of the a covariant derivative c k;` as

dca = ca dx` dxk = j ca !a cb dx` dxk. (13.326) k,` ^ k` j b` k ^ j ⇣ ⌘ But the ane connection k` is symmetric in k and ` while the wedge product dx` dxk is antisymmetric in k and `.Thuswehave ^ dca = ca dx` dxk = !a dx` cb dxk (13.327) k,` ^ b` ^ k 13.51 Cartan’s spin connection and structure equations 557 or with c cadxk and ! !a dx` ⌘ k ⌘ b` dc = ! c (13.328) ^ which is Cartan’s first equation of structure. Cartan’s curvature 2-form is a 1 a i j k ` R b = 2 cj cb R ik` dx dx ^ (13.329) = 1 ca ci j j +j n j n dxk dx`. 2 j b `i,k ki,` kn `i `n ki ^ h a i His second equation of structure expresses R b as Ra = d!a + !a !c (13.330) b b c ^ b or more simply as R = d! + ! !. (13.331) ^ A more compact notation, similar to that of Yang-Mills theory, uses Cartan’s covariant exterior derivative

D d + ! (13.332) ⌘ ^ to express his two structure equations as

Dc= 0 and R = D!. (13.333)

To derive Cartan’s second structure equation (13.330), we let the exterior a derivative act on the 1-form ! b d!a = d(!a dx`)=!a dxk dx` (13.334) b b` b`,k ^ and add the 2-form !a !c c ^ b !a !c = !a !c dxk dx` (13.335) c ^ b ck b` ^ to get

Sa =(!a + !a !c ) dxk dx` (13.336) b b`,k ck b` ^ a which we want to show is Cartan’s curvature 2-form R b (13.329). First we replace !a with its equivalent (13.324) ca ci j ca ci `b j b i` i,` b a a i j a i a i j a i c p n c p S = (c c c c ),k +(c c c c )(c c c c ) b j b i` i,` b j c ik i,k c n b p` p,` b h dxk dx`. (13.337)i ⇥ ^ j The terms proportional to i`,k are equal to those in the definition (13.329) 558 Tensors and general relativity

a of Cartan’s curvature 2-form. Among the remaining terms in S b, those independent of are after explicit antisymmetrization S = ca ci ca ci + ca ci cc cp ca ci cc cp (13.338) 0 i,k b,` i,` b,k i,k c p,` b i,` c p,k b i i c p a which vanishes (exercise 13.36) because cb,k = cc cp,k cb .ThetermsinS b a that are linear in ’s also vanish (exercise 13.37). Finally, the terms in S b that are quadratic in ’s are ca ci cc cp j n dxk dx` = ca i cp j n dxk dx` j c n b ik p` ^ j n b ik p` ^ = ca cp j n dxk dx` (13.339) j b nk p` ^ a and these match those of Cartan’s curvature 2-form R b (13.329). Since a a S b = R b, Cartan’s second equation of structure (13.330) follows. Example 13.27 (Cyclic identity for the curvature tensor) We can use Cartan’s structure equations to derive the cyclic identity (13.109) of the curvature tensor. We apply the exterior derivative (whose square dd = 0) to Cartan’s first equation of structure (13.328) and then use it and his second equation of structure (13.330) to write the result as 0=d(dc + ! c)=(d!) c ! dc =(d! + ! !) c = R c. (13.340) ^ ^ ^ ^ ^ ^ The definition (13.329) of Cartan’s curvature 2-form R and of his 1-form (13.325) now give 1 a i j k ` b m 0=R c = cj cb R ik` dx dx cmdx ^ 2 ^ ^ (13.341) = 1 ca Rj dxk dx` dxi 2 j ik` ^ ^ which implies that 1 0=Rj = Rj + Rj + Rj Rj Rj Rj . (13.342) [ik`] 3! ik` `ik k`i ki` i`k `ki ⇣ ⌘ But since Riemann’s tensor is antisymmetric in its last two indices (13.104), we can write this result more simply as the cyclic identity (13.109) for the curvature tensor j j j 0=R ik` + R `ik + R k`i. (13.343)

The vanishing of the covariant derivative of the flat-space metric c c 0=⌘ab;k = ⌘ab,k ! ak⌘cb ! bk⌘ac = !bak !abk (13.344) shows that the spin connection is antisymmetric in its Lorentz indexes ! = ! and !ab = !ba . (13.345) abk bak k k 13.52 Spin-one-half fields in general relativity 559 Under a general coordinate transformation and a local Lorentz transfor- mation, the spin connection (13.324) transforms as i a @x a d a 1e !0 b` = ` L d ! ei (@iL e) L b. (13.346) @x0 h i

13.52 Spin-one-half fields in general relativity a The action density (11.329) of a free Dirac field is L = ¯ ( @a + m) 0 in which a =0, 1, 2, 3; is a 4-component Dirac field; ¯ = † = i † ; a and m is a mass. Tetrads ck(x) turn the flat-space indices a into curved- a a ` space indices i, so one first replaces @a by ca@`.Thenextstepistouse the spin connection (13.324) to correct for the e↵ect of the derivative @` on the field . The fully covariant derivative is D = @ 1 !ab [ , ]where ` ` 8 ` a b !ab = !a ⌘bc, and the action density is L = ¯(ac` D + m) . ` c` a `

13.53 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 relativ- ity. To accomodate noncompact groups, we generalize the unitary matrices U(x) of the Yang-Mills gauge group to nonsingular matrices V (x) that act on a a a b a n matter fields (x) as 0 (x)=V b(x) (x). The field (x)=ea(x) (x) will be gauge invariant 0(x)= (x) if the vectors ea(x) transform as 1b ea0 (x)=eb(x) V a(x). We are summing over repeated indices from 1 to n and often will suppress explicit mention of the spacetime coordinates. In this compressed notation, the field is gauge invariant because a 1b a c b c b 0 = ea0 0 = eb V a V c = eb c = eb = (13.347) T T 1 T which is e0 0 = e V V = e in matrix notation. The inner product of two basis vectors is an internal “metric tensor”

N N N ↵ ↵ ↵ ↵ e⇤ e = e ⇤ ⌘ e = e ⇤ e = g (13.348) a · b a ↵ b a b ab ↵=1 ↵=1 X X=1 X in which for simplicity I used the the N-dimensional identity matrix for the metric ⌘. As in relativity, we’ll assume the matrix gab to be nonsingular. We a ab then can use its inverse to construct dual vectors e = g eb that satisfy ea e = a. † · b b 560 Tensors and general relativity The free Dirac action density of the invariant field

i a i b i a a a b ( @i + m) = e †( @i + m)eb = ( @i + e † eb,i)+m a a b · b h (13.349)i is the full action of the component fields b

i i a a b i a a a b ( @i + m) = a( Dib + m b) = a ( b@i + Aib)+m b (13.350) a ⇥ a ⇤ if we identify the gauge-field matrix as Aib = e † eb,i in harmony with the k k· definition (13.59) of the ane connection i` = e e`,i. 1b · Under the gauge transformation ea0 = eb V a, the metric matrix trans- forms as

1c 1d 1 1 gab0 = V ⇤a gcd V b or as g0 = V † gV (13.351)

1 1 in matrix notation. Its inverse goes as g0 = Vg V †. a a ac The gauge-field matrix A = e † eb,i = g ec† eb,i transforms as ib · · a ac a c 1d a 1c A0 = g0 e0† e0 = V A V + V V (13.352) ib a · b,i c id b c b,i 1 1 1 1 or as Ai0 = VAiV + V@iV = VAiV (@iV ) V . a a By using the identity e † ec,i = e † ec, we may write (exercise 13.39) · ,i · the Faraday tensor as

a a a a c a a c F =[Di,Dj] = e † eb,j e † ec e † eb,j e † eb,i+e † ec e † eb,i. (13.353) ijb b ,i · ,i · · ,j · ,j · · If n = N,then n ↵ c ↵ a ec e ⇤ = and Fijb =0. (13.354) c=1 X The Faraday tensor vanishes when n = N because the dimension of the embedding space is too small to allow the tangent space to have di↵erent orientations at di↵erent points x of spacetime. The Faraday tensor, which represents internal curvature, therefore must vanish. One needs at least three dimensions in which to bend a sheet of paper. The embedding space must have N>2 dimensions for SU(2), N>3 for SU(3), and N>5 for SU(5). The covariant derivative of the internal metric matrix

g;i = g,i gAi A†g (13.355) i 1 1 does not vanish and transforms as (g;i)0 = V †g,iV . A suitable action 1 ;i 1 density for it is the trace Tr(g;ig g g ). If the metric matrix assumes a Exercises 561

(constant, hermitian) mean value g0 in the vacuum at low temperatures, then its action is

2 1 i i 1 m Tr (g0Ai + Ai†g0)g0 (g0A + A †g0)g0 (13.356) h i which is a mass term for the matrix of gauge bosons 1/2 1/2 1/2 1/2 Wi = g0 Ai g0 + g0 Ai† g0 . (13.357) This mass mechanism also gives masses to the fermions. To see how, we write the Dirac action density (13.350) as

i a a a b a i c b a ( b@i + Aib)+m b = (gab@i + gacAib)+mgab . (13.358) ⇥ ⇤ ⇥ ⇤ Each fermion now gets a mass mci proportional to an eigenvalue ci of the hermitian matrix g0. This mass mechanism does not leave behind scalar bosons. Whether na- ture ever uses it is unclear. Further reading Einstein Gravity in a Nutshell (Zee, 2013), Gravitation (Misner et al., 1973), Gravitation and Cosmology (Weinberg, 1972), Cosmology (Weinberg, 2010), General Theory of Relativity (Dirac, 1996), Spacetime and Geometry (Car- roll, 2003), Exact Space-Times in Einstein’s General Relativity (Griths and Podolsky, 2009), Gravitation: Foundations and Frontiers (Padmanab- han, 2010), Modern Cosmology (Dodelson, 2003), The primordial density perturbation: Cosmology, inflation and the origin of structure (Lyth and Liddle, 2009), A First Course in General Relativity (Schutz, 2009), Gravity: An Introduction to Einstein’s General Relativity (Hartle, 2003), and Rela- tivity, Gravitation and Cosmology: A Basic Introduction (Cheng, 2010).

Exercises 13.1 Use the flat-space formula dp = xˆ dx + yˆdy + zˆdz to compute the change dp due to d⇢, d, and dz, and so derive expressions for the orthonormal basis vectors ⇢ˆ, ˆ, and zˆ in terms of xˆ, yˆ, and zˆ. 13.2 Similarly, compute the change dp due to dr, d✓, and d, and so derive expressions for the orthonormal basis vectors rˆ, ✓ˆ, and ˆ in terms of xˆ, yˆ, and zˆ. 13.3 (a) Using the formulas you found in exercise 13.2 for the basis vectors of spherical coordinates, compute the derivatives of the unit vectors rˆ, ✓ˆ, and ˆ with respect to the variables r, ✓, and and express them 562 Tensors and general relativity in terms of the basis vectors rˆ, ✓ˆ, and ˆ. (b) Using the formulas of (a) and our expression (2.16) for the gradient in spherical coordinates, derive the formula (2.33) for the laplacian . r · r 13.4 Show that for any set of basis vectors v1,...,vn and an inner product that is either positive definite (1.78–1.81) or indefinite (1.78–1.79 & 1.81 & 1.84), the inner products gik =(vi,vk) define a matrix gik that is nonsingular and that therefore has an inverse. Hint: Show that the matrix gik cannot have a zero eigenvalue without violating either the condition (1.80) that it be positive definite or the condition (1.84) that it be nondegenerate. 13.5 Show that the inverse metric (13.48) transforms as a rank-2 contravari- ant tensor. ik 13.6 Show that if Ak is a covariant vector, then g Ak is a contravariant vector. 13.7 Show that in terms of the parameter k =(a/R)2, the metric and line element (13.46) are given by (13.47). k 13.8 Show that the connection i` transforms as (13.66) and so is not a tensor. 13.9 Use the vanishing (13.83) of the covariant derivative of the metric ten- sor, to write the condition (13.140) in terms of the covariant derivatives of the symmetry vector (13.141). 13.10 Embed the points p = R(cosh ✓, sinh ✓ cos , sinh ✓ sin ) with tangent vectors (13.44) and line element (13.45) in the euclidian space E3.Show that the line element of this embedding is ds2 = R2 cosh2 ✓ +sinh2 ✓ d✓2 + R2 sinh2 ✓d2 (1 + 2kr2)dr2 (13.359) = a2 +r2 d2 1+kr2 ✓ ◆ which describes a hyperboloid that is not maximally symmetric. 13.11 If you have Mathematica, imitate example 13.15 and find the scalar curvature R (13.113) of the line element (13.359) of the cylindrical hyperboloid embedded in euclidian 3-space E3. 13.12 Consider the torus with coordinates ✓, labeling the arbitrary point

p = (cos (R + r sin ✓), sin (R + r sin ✓),rcos ✓) (13.360)

in which R>r. Both ✓ and run from 0 to 2⇡. (a) Find the basis vectors e✓ and e. (b) Find the metric tensor and its inverse. 13.13 For the same torus, (a) find the dual vectors e✓ and e and (b) find i the nonzero connections jk where i, j, k take the values ✓ and . Exercises 563

13.14 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. 13.15 Find the curvature scalar R of the torus with points (13.360). Hint: In these four problems, you may imitate the corresponding calculation for the sphere in Sec. 13.23. ik is kt ik is kt 13.16 Show that g = g g gst or equivalently that dg = g g dgst ik i by di↵erentiating the identity g gk` = `. n n n 13.17 Let gik = ⌘ik + hik and x0 = x + ✏ . To lowest order in ✏ and h, (a) show that in the x0 coordinates hik0 = hik ✏i,k ✏k,i and (b) find an i 1 j` equation for ✏ that puts h0 in de Donder’s gauge h0k,i = 2 (⌘ hj0 `),k. 13.18 Just to get an idea of the sizes involved in black holes, imagine an isolated 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 its Schwarzschild radius if 2MG 4 G r < =2 ⇡r3⇢ . (13.361) b c2 3 b c2 ✓ ◆ If the density ⇢ is that of water under standard conditions (1 gram per cc), for what range of radii rb might the sphere be or become a black hole? Same question if ⇢ is the density of dark energy. 13.19 Embed the points

p =(ct, aL sin sin ✓ cos , aL sin sin ✓ sin , aL sin cos ✓, aL cos ) (13.362) in the flat semi-euclidian space E(1,4) with metric ( 1, 1, 1, 1, 1) and derive the metric (13.257) with k = 1. 13.20 For the points p =(ct, a sin ✓ cos , a sin ✓ sin , a cos ✓), derive the metric (13.257) with k = 0. 13.21 Show that the 11 component of Ricci’s tensor R11 is aa¨ +2˙a2 +2kc2/L2 R =[D ,D ]k = . (13.363) 11 k 1 1 c2(1 kr2/L2) 13.22 Show that the 22 component of Ricci’s tensor R22 is (aa¨ +2˙a2 +2kc2/L2)r2 R =[D ,D ]k = . (13.364) 22 k 2 2 c2

13.23 Show that the 33 component of Ricci’s tensor R33 is (aa¨ +2˙a2 +2kc2/L2)r2 sin2 ✓ R =[D ,D ]k = . (13.365) 33 k 3 3 c2 564 Tensors and general relativity 13.24 Embed the points p =(ct, aL sinh sin ✓ cos , aL sinh sin ✓ sin , aL sinh cos ✓, aL cosh ) (13.366) in the flat semi-euclidian space E(2,3) with metric ( 1, 1, 1, 1, 1) and derive the line element (13.256) and the metric (13.257) with k = 1. 13.25 Derive the second-order FLRW equation (13.273) from the formulas (13.257) for g = c2, (13.267) for R = 3¨a/a, (13.271) for T , 00 00 ij and (13.272) for T . 13.26 Derive the second-order FLRW equation (13.274) from Einstein’s for- mula for the scalar curvature R = 8⇡G T/c4 (13.230), the FLRW formulas for R (13.270) and for the trace T (13.272) of the energy- momentum tensor. 13.27 Assume there had been no inflation, no era of radiation, and no dark energy. In this case, the magnitude of the di↵erence ⌦ 1 would have | | increased as t2/3 over the past 13.8 billion years. Show explicitly how close to unity ⌦would have had to have been at t = 1 s so as to satisfy the observational constraint ⌦ 1 < 0.036 on the present value of ⌦. | 0 | 13.28 Derive the relation (13.284) between the energy density ⇢ and the scale factor a(t) from the conservation law (13.281) and the equation of state pi = wi⇢i. 13.29 For constant ⇢ = p/c2 and k = 1, set g2 =8⇡G⇢/3 and use the Friedmann equations (13.273 & 13.275) and the boundary condition that the minimum of a(t) > 0 is at t = 0 to derive the formula a(t)= c cosh(gt)/(Lg). 13.30 Use the Friedmann equations (13.273 & 13.288) with w = 1, ⇢ con- stant, k = 1, and the boundary conditions a(0) = 0 anda ˙(0) > 0to derive the formula a(t)=c sinh(gt)/(Lg) where again g2 =8⇡G⇢/3. 13.31 Use the Friedmann equations (13.273 & 13.288) with w = 1, ⇢ con- gt stant, and k = 0 to derive the formula a(t)=a(0) e± . 13.32 Use the constancy of 8⇡G⇢a4/3=f 2 for radiation (w =1/3 ) and the Friedmann equations (13.273 & 13.288) to show that if k = 0, a(0) = 0, and a(t) > 0, then a(t)=p2ft where f>0. 13.33 Show that if the matrix U(x) is nonsingular, then 1 1 (@ U) U = U@ U . (13.367) i i 13.34 The gauge-field matrix is a linear combination A = ig tb Ab of the k k generators tb of a representation of the gauge group. The generators obey the commutation relations a b c [t ,t ]=ifabct (13.368) Exercises 565

in which the fabc are the structure constants of the gauge group. Show that under a gauge transformation (13.311) 1 1 A0 = UA U (@ U) U (13.369) i i i by the unitary matrix U =exp( igata)inwhicha is infinitesimal, the gauge-field matrix Ai transforms as a a a a 2 a b c a a igA0 t = igA t ig f A t + ig@ t . (13.370) i i abc i i Show further that the gauge field transforms as a a a b c A0 = A @ gf A . (13.371) i i i abc i a a 13.35 Show that if the vectors ea(x) are orthonormal, then e † ec,i = e † · ,i · ec. a a i i i c p 13.36 Use the equation 0 = =(c c ),k to show that c = c c c . b,k i b b,k c p,k b Then use this result to show that the -free terms S0 (13.338) vanish. a 13.37 Show that terms in S b (13.337) linear in the ’s vanish. 13.38 Derive the formula (13.346) for how the spin connection (13.324) changes under a Lorentz transformation and a general change of co- ordinates. 13.39 Use the identity of exercise 13.35 to derive the formula (13.353) for the nonabelian Faraday tensor. i ik b 13.40 Show that the dual tetrads ca = g ⌘abck are dual (13.155). 13.41 Write Dirac’s action density in the explicitly hermitian form LD = 1 i@ 1 i@ † in which the field has the invariant form 2 i 2 i 0 i i = ea a and ⇥ = i †⇤ . Use the identity a b † = b a to i show that the gauge-field matrix Ai defined as the coecient of a b i ⇥ ⇤ as in a (@i + iAiab) b is hermitian Aiab⇤ = Aiba.