Physics/Astronomy 226, Problem set 5, Due 2/17 Solutions

Transcript

1 Physics/Astronomy 6, Problem set 5, Due /17 Solutions Reding: Ch. 3, strt Ch Prove tht R λναβ + R λαβν + R λβνα = 0. Solution: We cn evlute this in locl inertil frme, in which the Christoffel symbols but not their derivtives) vnish, so tht: R λναβ = g λρ α Γ ρ βν βγ ρ αν) 1) = 1 α ν g λβ α λ g βν β ν g λα + β λ g αν ). ) Under the permuttion of indicies ν α β ν, R λαβν = 1 β α g λν β λ g να ν α g λβ + ν λ g βα ). 3) Under nother permuttion of ν α β ν, Tking the sum of the terms, R λβνα = 1 ν β g λα ν λ g αβ α β g λν + α λ g νβ ). 4) R λ[ναβ] = 1 [ α ν g λβ ν α g λβ ) + β α g λν α β g λν ) + ν β g λα β ν g λα )+ α λ g νβ g βν ) + β λ g αν g να ) + ν λ g βα g αβ )]. 5) The first three terms vnish becuse the prtil derivtives commute, nd the lst three terms vnish becuse the metric g µν is symmetric. Therefore, R λναβ + R λαβν + R λβνα = R λ[ναβ] = 0. 6) Becuse this is tensoril eqution, nd is true in our loclly inertil coordintes, it is lwys true.. Picture donut in 3d Eucliden spce. It my be either chocolte or glzed, s long s when viewed from the top it looks like two concentric circles of rdius r 1 nd r > r 1. Let b = r 1 + r )/ nd = r r 1 )/. 1

2 ) Set up coordintes θ, φ on the donut surfce for consistency, let θ lbel the ngle bout the center of the donut s mesured from bove, nd φ mesure the ngle round circulr cross-section of the donut.) b) Write down the metric g ij this surfce inherits from the Eucliden spce it is embedded in. c) Compute ll non-vnishing connection coefficients Γ u αβ. d) Compute ll nonzero components of R µναβ, R µν, nd R. Solution: Given torus with outer rdius R nd inner rdius r, we mke the ssignments = R r)/ nd b = R + r)/. We embed it in R 3 with cylindricl coordintes r, θ, z, such tht: z = sinφ) r = b + cosφ) θ = θ. 7) The metric is then: ds = dz + dr + r dθ = + b + cosφ)) dθ 8) or, g µν = 0 0 b + cosφ)) ) 9) nd g µν = 0 0 b + cosφ)) ). 10) We strt by finding the non-zero connection coefficients. For upper index θ, we must evlute Γ θ θθ, Γθ φθ, Γθ θφ nd Γθ φφ. Given tht we hve: Γ σ µν = 1 gσσ µ g νσ + ν g σµ σ g µν ), 11) Γ θ θθ = 0 Γ θ φφ = 0 Γ θ φθ = 1 gθθ φ g θθ sinφ) = b+ cosφ)) Γ θ θφ = Γθ φθ. 1) For upper index φ, we evlute the terms Γ φ φφ, Γφ θφ, Γφ φθ nd Γφ θθ.

3 Γ φ φφ = 0 Γ φ θφ = 0 Γ φ θθ = 1 gφφ φ g θθ Γ φ φθ = sinφ) b + cosφ)) 13) We now wnt to find the non-zero components of the curvture tensor. Before evluting component t rndom, let us consider these properties: If n = nd there re 1 1 n n 1) independent components of the tensor, we only expect one independent component overll. R ρσµν = R σρµν so non-zero component must hve its first two indicies distinct i.e. one must be θ nd the other φ). R ρσµν = R µνρσ, so the lst two indicies must lso be distinct for the component to be non-zero. Therefore, we conclude tht R θφθφ must be non-zero. It s esiest to compute it with one rised index: R ρ σµν = µ Γ ρ νσ ν Γ ρ µσ + Γ ρ µλ Γλ νσ Γ ρ νλ Γλ µσ R θ φθφ = φγ θ θφ Γθ φθ ) sinφ) sinφ) = φ ) b+ cosφ)) b+ cosφ)) ) = cosφ) ). b+ cosφ) 14) Lowering the index by contrction: R θφθφ = g θλ R λ φθφ = g θθ R θ φθφ = cosφ)b + cosφ)). 15) All the other non-zero components cn be found by symmetry. R φθφθ = R φθθφ = R θφφθ = R θφθφ 16) We find the Ricci tensor by contrction, R µν = g αβ R αµβν, so tht: R θθ R φφ = g φφ R φθφθ = cosφ) b + cosφ)) = g θθ R θφθφ = cosφ). b+ cosφ) 17) The Ricci sclr R is lso found by contrction, where R = R µ µ = g µν R µν. Therefore, R = g θθ R θθ + g φφ R φφ = cosφ) b+ cosφ)). 18) 3

4 FIG. 1: Torus figure stolen from Mthworld website). It is worth noticing two things bout this problem. Firstly, the torus is not mximlly symmetric s defined by Crroll, i.e. there does not exist constnt α such tht R ρσµν = α g ρµ g σν g ρν g σµ ). We might hve suspected this, since the torus doesn t hve s mny symmetries s the spce in which it is embedded. Secondly, if one ignores the pesky word embedded in the problem set, one cn simply slice open the torus nd put it in lovely flt R where ll connection coefficients nd components of the curvture tensor re zero! 3. The donut hole left over from the bove donut hs roughly sphericl surfce. A sphere with coordintes θ, φ) hs metric ds = dθ + sin θ. ) Show tht lines of constnt φ longitude) re geodesics, nd tht the only line of constnt θ ltitude) tht is geodesic is the θ = π/ the equtor). b) Tke vector with components V µ = V θ, V φ ) = 1, 0) nd prllel-trnsport it once round circle of constnt ltitude, θ = θ 0. Wht re the components of the resulting vector, s function of θ 0? Solution: Consider -sphere with the metric: ds = dθ + sin θ ) We must first compute the connection coefficients, given by: Γ λ µν = 1 g λλ µ g νλ + ν g λµ λ g µν ) The only nonzero terms will come from θ g φφ. The non-zero connection coefficients re therefore: Γ φ θφ = 1 θ g φφ = 1 g φφ sin θ θsin θ) = cot θ Γ φ φθ = 1 g φφ θ g φφ = cot θ Γ θ φφ = 1 g θθ θ g φφ = 1 θsin θ) = sin θ cos θ We my now look t the geodesic eqution: For µ = φ, we hve: d x µ + dx ρ dx σ Γµ ρσ = 0 d x φ + dx θ dx φ Γφ θφ + dx φ dx θ Γφ φθ = 0 4

5 Substituting for the connection coefficients: For µ = θ, we hve: d x θ d φ + cot θ + dx φ dx φ Γθ φφ = 0 = d θ dθ = 0 19) sin θ cos θ ) 0) It cn be seen tht if φ were constnt, then geodesic eqution 7) will be stisfied trivilly, nd Eq. 8) will be stisfied if τ is linerly relted to θ. If θ were constnt, then Eq. 7) is likewise stisfied for τ linerly relted to φ. But geodesic eqution 8) is: ) sin θ c cos θ c = 0 This will be stisfied while vrying φ only if the constnt θ c is 0, π, π, 3π, π...the choices θ c = 0, π, π re uninteresting becuse they represent motion on the poles, so vrying φ does nothing. The choices θ c = π, 3π correspond to motion long the equtor. b) Consider vector with components V µ = V θ, V φ ) = 1, 0). If the vector is prllel propgted round circle of constnt θ = θ c, it must stisfy the eqution of prllel trnsport long its pth: dv µ dλ + dx σ Γµ σρ dλ V ρ = 0 For µ = θ nd µ = φ, this becomes: Re-writing dv dλ dv s dv θ dλ + dx φ Γθ φφ dλ V φ = dv θ dλ sin θ o cos θ o dλ V φ = 0 dv φ dλ + dx φ Γφ φθ dλ V φ = dv φ dλ + cot θ o dλ V θ = 0, we hve coupled 1st order equtions to integrte: dλ dv θ sin θ o cos θ o V φ = 0 dv φ + cot θ ov θ = 0 These decouple into two nd order equtions: d V φ + cos θ o V φ = 0 d V θ + cos θ o V θ = 0 5

6 The generl solution to these is: V φ = A cos φ cos θ o ) + B sin φ cos θ o ) V θ = C cos φ cos θ o ) + D sin φ cos θ o ) We cn solve for the constnts using the initil conditions on V nd dv evluting the coupled 1st order equtions bove t φ = 0): V θ 0), V φ 0)) = 1, 0) A = 0 nd C = 1 dv θ, dv ) φ = 0, cot θ o ) φ=0 φ=0 B = 1 sin θ o nd D = 0 Therefore, V φ = π) is given by: V θ, V φ ) = cos [π cos θ o ], sin [π cos θ ) o] sin θ o found by If θ o = π prllel trnsport long the equitoril geodesic), then we get bck our originl vector t φ = π s expected. Also, the norm of V is equl to one s it should be: g µν V µ V ν = g θθ V θ V θ + g φφ θ o )V φ V φ = cos [φ cos θ o ] + sin θ o sin [φ cos θ o ] sin θ o = 1 6