A.1Using the Einstein summation convention, we can write

B Solutions Chapter 1 •A.1Using the Einstein summation convention, we can write ∂f ∂x i ∂x i ∂x k ⌘ 3 Â i=1 ∂f ∂x i ∂x i ∂x k = ∂f ∂x 1 ∂x 1 ∂x k + ∂f ∂x 2 ∂x 2 ∂x k + ∂f ∂x 3 ∂x 3 ∂x k = ∂f ∂x ∂x ∂x k + ∂f ∂y ∂y ∂x k + ∂f ∂z ∂z ∂x k ; where we summed over the repeated Latin indexi(running from 1 to 3). Also, in the last equality we render explicit the choice of a Cartesian coordinate systemx j ⌘ {x,y,z}. Note that the only free index on the very left and on the vary right of the previous equation isk, that can assume any value from 1 to 3, In analogous way, we can write ∂ 2 f ∂x i ∂x j ∂x i ∂x m ∂x j ∂x n = 3 Â i=1 3 Â j=1 ∂ 2 f ∂x i ∂x j ∂x i ∂x m ∂x j ∂x n = 3 Â j=1 ∂ 2 f ∂x∂x j ∂x ∂x m ∂x j ∂x n + ∂ 2 f ∂y∂x j ∂y ∂x m ∂x j ∂x n + ∂ 2 f ∂z∂x j ∂z ∂x m ∂x j ∂x n = ∂ 2 f ∂x 2 ∂x ∂x m ∂x ∂x n + ∂ 2 f ∂x∂y ∂x ∂x m ∂y ∂x n + ∂ 2 f ∂x∂z ∂x ∂x m ∂z ∂x n + ∂ 2 f ∂y∂x ∂y ∂x m ∂x ∂x n + ∂ 2 f ∂y 2 ∂y ∂x m ∂y ∂x n + ∂ 2 f ∂y∂z ∂y ∂x m ∂z ∂x n + ∂ 2 f ∂z∂x ∂z ∂x m ∂x ∂x n + ∂ 2 f ∂z∂y ∂z ∂x m ∂y ∂x n + ∂ 2 f ∂z 2 ∂z ∂x m ∂z ∂x n where we summed again over the repeated Latin indexesiandj(each running from

• to 3). We rendered also here explicit the choice of a Cartesian coordinate system

x i ⌘{x,y,z}. Note that now the only free indexes aremandn, that can assume any value from 1 to 3.

•A.2

In cartesian coordinates, the displacement vector is given byd~`=dxˆx+dyˆy+dzˆz.To pass to cylindrical coordinates, we have to consider the following transformation: x=rcosq;y=rsinq;z=z. Then, we can write d~`= ✓ ∂x ∂r dr+ ∂x ∂q dq ◆ ˆx+ ✓ ∂y ∂r dr+ ∂y ∂q dq ◆ ˆy+dzˆz In cylindrical coordinates, we would like to writed~`=drˆe r+dqˆe q+dzˆe z. So, we can identify the new basis vectors by considering in the previous equation the terms 251 Solutions Manual for An Overview of General Relativity and Space-Time (Series in Astronomy and Astrophysics), 1e by Nicola Vittorio (All Chapters, 100% Original Verified, A+ Grade) 1 / 3

252The physics of the space-time proportional todr,dqanddz, respectively ˆe r= ∂x ∂r ˆx+ ∂y ∂r ˆy=cosqˆx+sinqˆy; ˆe q= ∂x ∂q ˆx+ ∂y ∂q ˆy="rsinqˆx+rcosqˆy; ˆe z=ˆz After defining ˆe i={ˆer,ˆeq,ˆez}, we can finally write g mn⌘ˆem·ˆen=

@ 100 0r 2

001 1 A By using Eq.(1.4), we finally getd` 2 =dr 2 +r 2 dq 2 +dz 2

. QED.

•A.3

Consider the vector~Z=~V+~U. In terms of contravariant components we can write Z i =V i +U i = ∂x i ∂¯x m ¯V m + ∂x i ∂¯x m ¯U m = ∂x i ∂¯x m (¯V m +¯U m )= ∂x i ∂¯x m ¯Z m where we exploit the transformation law given in Eq.(1.16), first to transform~Vand ~Uin the barred reference frame and, then, to show that~Ztransforms according to the same law. Thus,~Zis a vector. QED.•A.4. The procedure of lowering a contravariant index does not depend on the cho- sen coordinate system. So, in theK

reference frame we havedx

i =g

ij dx 0j =[c.f.Eq.(1.17)]. Moreover,g

ij =gmnL m i L n j [c.f.Eq.(1.12)]. It follows that dx

i =g

ij dx 0j =gmnL m i L n j dx 0j =gmnL m i dx n =L m i dxm This is exactly the transformation law for the covariant components of a vector [c.f.Eq.(1.16)], QED.

•A.5

. According to definitions given in Eq.(1.13) and Eq.(1.16) we can write A k Bk=L k m A 0m M j k B

j =A 0j B

j asL k m M j k =d k j

•A.6

. Consider a 2D Euclidean space with Cartesian ' x l ⌘{x,y} ( or polar ' x i ⌘ {r,f} ( coordinates. Let’s pass from the Cartesian to the polar coordinate system by writing the Jacobian matrix and its inverse ∂x l ∂x j ⌘ ✓ cosf"rsinf sinfrcosf ◆ ; ∂x i ∂x l ⌘ ✓ cosf sinf "sinf/rcos(f)/r

◆ 2 / 3

Solutions253 beingx=rcosfandy=rsinf. It is then straightforward to write ∂ 2 x 1∂x j ∂x k ⌘ ✓

• "sinf

"sinf"rcosf ◆ ; ∂ 2 x 2 ∂x j ∂x k ⌘ ✓

• cosf

cosf"rsinf ◆ By exploiting the definition given in Eq(B1.2.f), we can write G i jk = ∂x i ∂x l ∂x l ∂x j ∂x k = ∂x i ∂x 1 ∂x 1 ∂x j ∂x k + ∂x i ∂x 2 ∂x 2 ∂x j ∂x k = cosf " sinf r !✓

• "sinf

"sinf"rcosf ◆ + sinf cosf r !✓

• cosf

cosf"rsinf ◆ Thus, G 1 jk =cosf ✓

• "sinf

"sinf"rcosf ◆ +sinf ✓

• cosf

cosf"rsinf ◆ = ✓ 00 0"r ◆ and G 2 jk =" sinf r ✓

• "sinf

"sinf"rcosf ◆ + cosf r ✓

• cosf

cosf"rsinf ◆ =

B @

1 r 1 r

1 C A consistently with the finding of Eq.(B1.2.e). QED.Chapter 2 •Exercise A.7: Let’s use the definitions given in Eq.(B2.1.a), Eq.(B2.1.e) and Eq.(B2.1.f). Then, we can write the inner product of two vectors,AandBas fol- lows

A·B=A

aB a =M µ a A

µ L a n B 0n =A

µ B 0n SinceM=L "1 andM µ a L a n =d µ n . We can then conclude thatAaB a =A

µ B 0n , QED.

•A.8: A rotation in thex

2 "x 3 pla is described by the following rotation matrixes L (1)a b ⌘

B B @

10 0 0

01 0 0

• 0 cosfsinf

00" sinfcosf 1 C C A ;M (1)µ a ⌘

B B @

10 0 0

01 0 0

• 0 cosf"sinf
• 0 sinfcosf

1 C C A where the superscript(1)indicates that the rotation is around thex 1 axis. It is easy to show that in this case the transformations given in Eq.(2.7) provide x 0a ⌘

B B @ x

x 1 x 2 cosf"x 3 sinf x 3 cosf+x 2 sinf 1 C C A

• / 3