General Relativity (225A) Fall 2013 Assignment 5 Solutions

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Univesity of Califonia at San Diego Depatment of Physics Pof. John McGeevy Geneal Relativity 225A Fall 2013 Assignment 5 Solutions Posted Octobe 23, 2013 Due Monday, Novembe 4, 2013 1. A constant vecto field. Conside the vecto field W in the flat plane with metic ds 2 d 2 + dy 2. W W + W ϕ ϕ and compute thei patial deivatives. Then compute the metic-compatible covaiant deivative in pola coodinates. The point of this poblem is to convince ouselves that when we say a constant vecto field what we mean is a covaiantly constant vecto field, because the latte statement is meaningful independent of ou choice of coodinates. The pola coodinates ae cos ϕ, y sin ϕ. Away fom 0 whee the pola coodinates, ϕ beak down, the jacobian mati J is given by d cos ϕ sin ϕ d d J. dy sin ϕ cos ϕ dϕ dϕ Witing this equation as d i Jad i a, the coodinate vecto fields ae elated by the invese-tanspose mati: J 1 a i i a which is J 1 a cos ϕ i sin ϕ sin ϕ a cos ϕ i J i a J 1 b i δb a notice that i is the ow inde and a is the column inde you can check this by dimensional analysis, since and ϕ have diffeent dimensions so W J 1 + J 1 ϕ ϕ cos ϕ sin ϕ θ. So the patial deivatives of the components in pola coods ae cetainly not constant: ϕ W sin ϕ, W 0, ϕ W ϕ cos ϕ, W ϕ + sin ϕ. 2 The covaiant deivative in the flat! plane in catesian coodinates is just the same as the patial deivative so i W j i W j 0. By constuction, this must also be tue in pola coodinates. Moe eplicitly the metic in pola coodinates is ds 2 d 2 + 2 dϕ 2 1

which means that the nonzeo Chistoffel symbols ae Γ ϕϕ, Γ ϕ ϕ 1 Γϕ ϕ. This means that a W b W + 0 ϕ W + Γ ϕϕw ϕ W ϕ + Γ ϕ ϕw ϕ ϕ W ϕ + Γ ϕ ϕw the lowe inde is the ow inde. b a sin ϕ 2 + 1 0 sin ϕ + sin ϕ sin ϕ cos ϕ + 1 cos ϕ 2. Vecto fields on the 2-sphee. [fom Oogui] A two-dimensional sphee S 2 of unit adius can be embedded in the thee-dimensional euclidean space IR 3 by the equation 2 + y 2 + z 2 1. Fo coodinates on the sphee we can use θ, ϕ defined by sin θ cos ϕ, y sin θ sin ϕ, z cos θ, ecept at the noth and south poles θ 0, π whee the value of ϕ is ambiguous. An infinitesimal otation of IR 3 aound its oigin induces a tangent vecto field on S 2, which is said to geneate the otation 1. Show that thee ae thee linealy-independent vecto fields 2 of this type and compute thei commutatos [σ i, σ j ]. A otation about the oigin is a linea map R : IR 3 IR 3 which peseves the euclidean metic ds 2 d i d i ; if it also peseves the oientation det R 1, then it is continuously connected to the identity map. Since such an R is linea, it is defined by its action on an othonomal basis î, i 1, 2, 3: Rî R ij ĵ, î ĵ δ ij Rî Rĵ R ik R kj. which gives si conditions on the nine elements R ij. This means that the space of such R the goup SO3 is 3 dimensional. Fom the definition above, it is a closed subset of IR 9 and hence a smooth manifold. A goup which is a smooth manifold is a Lie goup. A Lie goup G is geneated by a Lie algeba g this is just the tangent space to G at the identity element. It has a natual poduct, which is the Lie backet the commutato of tangent vecto fields. The tangent vecto fields on S 2 induced by such infinitesimal otations ae theefoe a ealization of this Lie algeba so3, and hence thee can only be thee linealy independent such vecto fields. The thee independent 1 Moe pecisely, conside the esult of acting with a otation by an infinitesimal angle θ on an abitay smooth function: f fr f + θa f + θa i i f +... the vecto field A i i geneates the otation. 2 A vecto field v on M is linealy dependent on some othes {v α } if thee eist constants a α s.t. v a α v α eveywhee in M. b a 0 2

infinitesimal otations, which I ll denote i, i 1, 2, 3 can be epesented on functions by vecto fields v: fr f + θvf + Oθ 2 as follows, beginning with the otation about the z ais: cos θ sin θ 0 0 θ 0 R z ij sin θ cos θ 0 1 ij + θ 0 0 0 0 1 0 0 0 It maps the coodinate vecto i to + θy R z i σ z ij j y θ + Oθ 2 z ij i ij + Oθ 2 fσ f + θy, y θ, z + Oθ 2 f + θ y y f + Oθ 2 Cyclically pemuting, y, z, we have σ z y y. σ z y y z, σ y z z. Witing these in tems of θ, ϕ, we can make clea that these estict to vecto fields tangent to S 2 IR 3. This amounts to changing to pola coods and showing that these vecto fields have no components along. The Jacobian mati evaluated at 1 is θ ϕ y θ y ϕ y z θ z ϕ z whee ρ 2 + y 2. And so And so z ρ zy ρ y ρ 2 ρ 2 y ρ 0 z cos θ cos ϕ sin θ cos ϕ sin θ cos ϕ cos θ sin ϕ sin θ sin ϕ sin θ sin θ 0 cos θ sin ϕ θ θ + ϕ ϕ + cos θ cos ϕ θ sin ϕ sin θ ϕ + sin θ cos ϕ σ z y y y cos θ sin ϕ θ + cos ϕ sin θ ϕ + sin θ sin ϕ z sin θ θ + cos θ zy θ + z ρ ρ 2 ϕ + y y ρ θ y ρ 2 ϕ y ϕ. σ y z z y y sin θ θ z cos θ sin ϕ ϕ + cos ϕ sin θ ϕ sin 2 θ sin ϕ cos 2 θ sin ϕ θ cot θ cos ϕ ϕ sin ϕ θ cot θ cos ϕ ϕ. 3

σ y z z z cos θ cos ϕ θ sin ϕ sin θ ϕ sin θ θ cos ϕ θ sin ϕ cot θ ϕ. To check diectly that these ae independent, suppose that thee eist constants a, b, c so that aσ z + bσ y + cσ z 0, θ, ϕ and then evaluate at vaious points. At θ π/2, ϕ 0 we lean 0 a ϕ + b θ a b 0. At θ π/2, ϕ π/2 we lean that 0 c θ, so c 0 as well. It is easiest to compute the commutatos in the catesian coodinates, e.g. [σ, σ y ] [y z z y, z z ] y z z + 1 3 z z y σ z Cyclically pemuting, we see that [σ i, σ j ] ɛ ijk σ k which is the so3 algeba. If we ae feeling like doing some penance, we can also compute the commutatos diectly on the sphee. Fo eample [σ z, σ ] ϕ sin ϕ θ cot θ cos ϕ ϕ sin ϕ θ cot θ cos ϕ ϕ ϕ cos ϕ θ + cot θ sin ϕ ϕ σ y which agees with [σ, σ z ] [σ, σ z ] σ y. 3. E&M in cuved space. Conside EM fields A µ in a cuved spacetime with a geneal metic g µν d µ d ν. a Wite an action functional S[A µ, g µν ] which is geneal-coodinate invaiant and gauge invaiant and which educes to the Mawell action if we evaluate it in Minkowski spacetime S[A µ, η µν ]. S[A, g] d 4 1 g 4 F µνf ρσ g µρ g νσ + A µ j µ + θ 16π 2 F F. b Vay this action with espect to A µ to find the equations of motion govening electodynamics in cuved space. The F F tem does not contibute to the equations of motion because it is a total deivative. The Mawell tem and the souce tem give δ 0 S[A, g] d 4 4 g δa λ y 4 µδν λ δ yf ρσ g µρ g νσ + j λ µ gfρσ g µρ g λσ + j λ. 1 The Bianchi identity is unmodified, since it didn t depend on the metic in the fist place. Note that the above action is gauge invaiant if the cuent j is covaiantly conseved, µ j µ 0: the vaiation unde a gauge tansfomation is δs d 4 g µ λj µ IBP d 4 λ µ gj µ d 4 gλ µ j µ. 4

Hee I have used ou epession fo the covaiant divegence which does not involve Chistoffel symbols eplicitly. Note that if you chose to define the cuent as a tenso density I don t know why you would do this othe than if you wee told to do so by someone on Wikipedia which means that you have absobed the g into the definition of j µ, then you epessions would look accodingly diffeent and... wose. Note also that I have chosen to nomalize my Mawell field so that I didn t have to wite factos of 4πc. 4. The badness cancels. Using the coodinate tansfomation popety of the Chistoffel connection Γ ρ µν, veify that µ ω ν µ ω ν Γ ρ µνω ρ tansfoms as a ank-2 covaiant tenso if ω is a one-fom. Recall that unde, Γ Γ whee Γ ρ µν µ σ κ ρ ν δ Γ κ σδ }{{} κ δ ρ µ κ ν δ }{{} tenso tansf eta badness. 2 This guaantees that µ V ν is a tenso fo any vecto V. The fist tem of µ ω ν tansfoms accoding to µ ω ν µ ω ν µ κ κ ν ρ ω ρ µ κ κ ρ ν ω ρ + µ κ κ ν ρ ω ρ So the whole object becomes µ ω ν µ ω ν Γ ρ µν ω ρ µ κ κ ρ ν ω ρ + µ κ κ ν ρ ω ρ + µ σ κ ρ ν δ Γ κ σδ ρ ξ ω ξ + κ δ ρ µ κ ν δ ρ ξ ω ξ µ κ κ ρ ν ω ρ + µ κ κ ν ρ ω ρ + µ σ ξ ρ Γ ξ σδ ω ξ + κ δ ρ µ κ ν δ ρ ξ ω ξ 3 in the second step we just used the fact that the ρ inde is contacted between tensos to ease a JJ 1, i.e. κ ρ ρ ξ δ ξ κ. The badness will cancel if But on any function f, κ ν ρ ω ρ µ δ δ ρ κ ν δ ω δ 4 [ µ, δ ]f [ µ, δ ρ ρ ]f µ β ρ ρ f and theefoe, the RHS of 4 is µ δ δ ρ κ ν δ ω δ µ δ ρ ρ ξ ν δ ω ξ [ µ, δ ] ξ ν δ ω ξ µ δ ξ β µ ξ ν δ ω ξ 5

Pobably thee is a moe elegant way to show this. µ δ ξ δ δ µ ξ ν δ ω ξ }{{} 0 ν δ δ µ ξ ω ξ ν µ ξ ω ξ µ ν ξ ω ξ. 5 6