Homework # 4 Solution Key

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1 PHYSICS 631: Generl Reltivity Homework # 4 Solution Key The metric for the surfce of cylindr of rdius, R (fixed), for coordintes z, φ ( ) 1 0 g µν = 0 R 2 In these coordintes ll derivtives with respect to z nd φ vnish immeditely, so the Christoffel symbols nd Riemnn Curvture does s well. The problem ws ctully bit vgue, but if you insted use the 3-d metric: g µν = 0 R This produces only few non-zero Christoffel symbols: Γ 1 22 = R nd Γ 2 12 = 1 R The only possible non-zero Riemnn term is of the form R212, 1 nd vrious symmetriclly-relted terms. The vlue should be: R = Γ 1 22,1 Γ 1 12,2 + Γ 1 σ1γ σ 22 Γ 1 σ2γ σ 12 = Γ 1 22Γ 2 21 = 0 2. One wy of describing the metric of flt, homogeneous, expnding universe is: g µν = 0 (t) (t) (t) 2 where (t) is function of time only, nd the coordintes re: t x µ = x y z () Compute ll Non-vnishing terms in the Riemnn Tensor. (Express s R α βµν ). First, note tht ll of the non-vnishing Christoffel terms re of the form: Γ 0 ii = ȧ nd Γ i 0i = ȧ

2 For those of you who ve tken cosmology, this should look extremely suggestive. Thus, we only get non-zero derivtives of the form g ii,00, nd thus, we ll only get Riemnn terms of the form R 0i0i nd R ijij : R 0i0i = ä R ijij = 2 ȧ 2 (b) Compute ll Non-vnishing terms in the Ricci Tensor. Clerly, bsed on the symmetry, we re only going to get digonl terms in the Ricci Tensor: R 00 = 3ä R ii = 2ȧ 2 ä (c) Compute the Ricci Sclr. Simply: R = R α α = 6ȧ2 + ä 2 (d) Compute the Einstein Tensor. G µν = R µν 1 2 g µν = 3ȧ ȧ 2 2ä ȧ 2 2ä ȧ 2 2ä This problem essentilly hs you compute the Schwrzschild metric from generl considertions. Becuse the metric is digonl: e 2Φ g µν = 0 e 2Λ r r 2 sin 2 θ It s esy to generte the Christoffel symbols: Γ 0 11 = Φ Γ 1 11 = Λ Γ 2 12 = 1 r Γ 2 33 = sin(θ) cos(θ) Γ 3 13 = 1 r Where primes re derivtives with respect to r, nd where I hven t written out the simple vritions of the indices.

3 From the simple symmetries, only function of single vrible (r), the only non-zero terms re: R 0101, R 0202, R 0303, R 1212, R 1313, R In prticulr, they re: (metrics i,ii, nd iv only) R 0101 = e 2Φ Λ [ Φ + (Φ ) 2 Φ ] R 0202 = re 2Φ 2Λ Φ R 0303 = r sin 2 θe 2Φ 2Λ Φ R 1212 = rλ R 1313 = r sin 2 θλ R 2323 = r 2 [ (cos 2 θ 1)(1 e 2Λ ) ] () Conserved quntities (explicit independnce in the metric): i. Minkowski: E = p 0, p x, p y, p z ii. Schwrzschild: E = p 0, L = p φ iii. Kerr: E = p 0, L = p φ iv. Friedmnn-Robertson-Wlked: L = p φ (b) This is pretty strightforwrd. (c) Solve for p r for source on the equtor. i. Flt, sphericl: ii. Schwrzschild: (p r ) 2 = E 2 m 2 L2 r 2 ( (p r ) 2 = E 2 1 2M r ) ) (m 2 + L2 iii. Kerr I m going to leve this in the form of g tt nd similr, though long the equtoril plne, the metric relly does simplify considerbly. r 2 (p r ) 2 = gtt E 2 m 2 + 2g tφ EL + g φφ L 2 g rr iv. For sphericl symmetry, p θ = p φ = 0, but if k = 0, there is no remining dependnce on r to p r is conserved. 5. Let s do this one just to shore up some issues from the midterm. Consider photon with 4-momentum: p 0 p α = p is moving through universe with nery Minkowski metric: g µν = (1 + 2φ) φ φ φ where φ is function of position nd not time nd is smll. Consider n observer t rest with respect to this coordinte system t position x µ. He observes the photon to hve n energy, E 0.

4 () In terms of E 0, wht re p 0 nd p 1? Recll tht n observer will mesure n energy s: E O = U p In this cse, the since the observer is t rest with respect to the coordintes: (U 0 ) 2 ( 1 2φ) = 1 or So or Since we get: U 0 1 φ E 0 = (1 φ)p 0 (1 + 2φ) = (1 + φ)p 0 p 0 (1 φ)e 0 p p = 0 p 1 = E 0 (1 + φ) (b) Compute p α,0 This looks like job for geodesics: p α,0 = Γ α βγp β dxγ dt where for γ = 0, 1, the velocity on the right is 1. but the only non-zero geodesics re those with with two identicl terms nd those with sptil derivtives. The non-zero terms will be: nd similrly for z. ṗ 0 = Γ 0 01p 1 Γ 0 10p 0 = 2Eφ,x ṗ 1 = Γ 1 00p 0 = Eφ,x ṗ 2 = Γ 2 00p 0 Γ 2 10p 0 Γ 2 01p 1 Γ 2 11p 1 = 2Eφ,y (c) Suppose the photon flls from potentil φ i to potentil φ f. Wht is the energy observed by n observer t the finl potentil? The observed energy is lwys the sme reltion to n observer t rest: but to n observer t rest: so nd p 0 is conserved. So: E f E i E obs = U 0 p 0 (U 0 ) 2 g 00 = 1 U 0 1 φ = 1 φ f 1 φ i 1 φ Just to mke things cler, if the photon drops to lower potentil φ < 0, so the energy increses.

5 (d) Let s do this with rel units. Wht frctionl energy does photon lose if it goes from the surfce of the erth to deep spce? In rel units, our potentil does: φ φ c 2 GM rc 2 Plugging in, t the surfce of the erth: φ So E E =