The line element for the hyperbolic plane was given in problem 12, of chapter 8 in Hartle

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1 Physics 4445 Solution for homework Fall 20 Cornell University (46 points) I. HARTLE CHAPTER 2, PROBLEM (8 POINTS) The line element for the hyperbolic plane was given in problem 2, of chapter 8 in Hartle ds 2 = y 2 dx 2 + y 2 dy 2, (y 0). (.) We want to calculate the Ricci curvature scalar R, which is the fully contracted Riemann tensor From the general form of the Riemann tensor and from g αβ = y 2 δ βδ, we have Instead of computing the Γ α βγ homework 6] R = g βδ R γ βγδ (.2) R = y 2 δ βδ ( Γ γ βδ,γ Γγ βγ,δ + Γγ γɛγ ɛ βδ Γ γ δɛ Γɛ βγ ). (.3) directly from their definition, we will first compute [Eq. (.2) on the solutions to g γλ ẍ λ = ( ) 2 g αβ,γ g γα,β ẋ α ẋ β, (.4) where the overdot represents a derivative with respect to the curve parametrization. Setting γ to x and y gives ẍ = 2 y ẋẏ, ÿ = y ẋ2 + y ẏ2. (.5) Comparing these to the geodesic equation ẍ α = Γ α βγẋβ ẋ γ gives Γ x xy = Γ x yx = Γ y yy = Γ y xx = y, Γx xx = Γ x yy = Γ y xy = Γ y yx = 0. (.6) Substituting these into the expression for the Ricci curvature scalar (.3) gives R = 2. (.7) This confirms that the hyperbolic plane (.) has constant negative curvature. II. INTERPRETATION OF THE RIEMANN TENSOR (0 POINTS) A. As we are doing the derivation, anything that is symmetric in µ and ν, like µ ν w λ or Γ λ µν, is automatically zero. µ ν w λ ν µ w λ = µ ( ν w λ ) Γ ρ µν( ρ w λ ) + Γ λ µρ( ν w ρ ) {µ ν} = µ ν w λ + µ (Γ λ νρw ρ ) 0 + Γ λ µρ( ν w ρ + Γ ρ νσw σ ) {µ ν} = 0 + ( µ Γ λ νρ)w ρ + Γ λ νρ( µ w ρ ) + Γ λ µρ( ν w ρ ) + Γ λ µργ ρ νσw σ {µ ν} = ( µ Γ λ νρ)w ρ Γ λ µργ ρ νσw σ {µ ν} = ( µ Γ λ νσ ν Γ λ µσ + Γ λ µργ ρ νσ Γ λ νργ ρ µσ)w σ = R λ µνσ w σ

2 2 B. We simply expand u µ µ (v ν ν w λ ) v ν ν (u µ µ w λ ) (2.) = u µ v ν ( µ ν w λ ) + (u µ µ v ν )( ν w λ ) v ν u µ ( ν µ w λ ) (v ν ν u µ ) µ w λ ) (2.2) = u µ v ν ( µ ν w λ ν µ w λ ) + (u σ σ v ρ v σ σ u ρ ) ρ w λ (2.3) = s λ + t ρ ρ w λ (2.4) where t = u v v u. C. We are free to choose coordinates so that the point P is at the origin and so that the metric is flat at that point, so that the connection coefficients are zero at P. Our path is to start at a point A, go along u to B, along v to C, along u to D, and along v back to E. The point A is P, and E = A to order ɛ, but the value of w will be different after going around the loop so it is clearer to give the point two different names. Each path is ɛ units of affine parameter, and we will write x AB for the path from A to B and so on. The path to first order in λ is x AB = 0 + λu, x BC = ɛu + λv, x CD = ɛu + ɛv λu, x DE = ɛv λv (2.5) and it is obvious that these satisfy the integral curves dxα dλ = uα (x), etc. We will define B γ δαβ = δγ γ αβ, and then the Taylor expansion of the connection coefficients is Γγ αβ (0 + y) = Bγ δαβ yδ. The geodesic equation for the first part of the path will be dλ = Γ αβ(x AB )u α w β = (B γ δαβ λuδ )u α w β A (2.6) where u and w don t need to be Taylor expanded because they would contribute at too high an order in λ. The solution for w along this path is w AB = w A 2 λ2 B γ δαβ uδ u α w β A, w B = w A 2 ɛ2 B γ δαβ uδ u α w β A (2.7) We have the value of w at point B, so now we evolve it to point C We know that λ and ɛ are of the same order, so we can drop higher-order terms. The geodesic equation and solution for the next part is dλ = Γ αβ(x BC )v α w β = (B γ δαβ (ɛuδ + λv δ )v α w β B (2.8) w BC = w B ɛλb γ δαβ uδ v α w β B 2 λ2 B γ δαβ vδ v α w β B (2.9) w C = w A 2 ɛ2 B γ δαβ (uδ u α + 2u δ v α + v δ v α )w β A (2.0) where any time ɛ is multiplied by w B we get w A because we only keep terms to order ɛ 2. The next part is dλ = Γ αβ(x CD )( u α )w β = (B γ δαβ (ɛuδ + ɛv δ λu δ )v α w β C (2.) w CD = w C + ɛλb γ δαβ uδ u α w β C + ɛλbγ δαβ uδ v α w β C 2 λ2 B γ δαβ uδ u α w β C (2.2) w D = w A 2 ɛ2 B γ δαβ (2uδ v α + v δ v α 2v δ u α )w β A (2.3) The final result is dλ = Γ αβ(x DE )( v α )w β = (B γ δαβ (ɛvδ λv δ )v α w β D (2.4) w DE = w D + ɛλb γ δαβ vδ v α w β D 2 λ2 B γ δαβ vδ v α w β D (2.5) w E = w A ɛ 2 B γ δαβ (uδ v α v δ u α )w β A (2.6) The change in w α is ɛ 2 ( ν γ α µσ µ γ α νσ)u µ v ν w σ which is the Riemann tensor in Fermi normal coordinates, so δw α = R α µνσu µ v ν w σ.

3 3 III. HARTLE, CHAPTER 22, PROBLEM 6 (8 POINTS) A. We have the energy density, momentum density, and stress tensor, so we use Hartle Equation Although we are using units where c 2 = ɛ 0µ 0 =, we will explicitly write the factor of c in the stress tensor because it will help 2 in the second part of the problem. The stress tensor should be roughly a pressure, not an energy density. T tt = ɛ = ( ɛ 0E 2 + ) B 2 (3.) 2 µ 0 T it = T ti = ( E µ B) i (3.2) 0 T ij = [ ( c 2 ɛ 0 E i E j + ) ( 2 δij E 2 + µ0 B i B j + )] 2 δij B 2 (3.3) (3.4) We need to check that α T αβ = 0, and since β is a free index there are four equations. The first is B. t T tt + i T it = ɛ 0 2 t E 2 + 2µ 0 t B 2 + µ 0 ( E B) (3.5) = ɛ 0 E t E + µ 0 B t B + µ 0 ( B ( E) ( E ( B)) = ɛ 0 E ( t E ɛ 0 µ 0 B) + µ 0 B ( t B + E) = 0. Both quantities in parentheses are Maxwell equations and so are zero. t T tj + i T ij = t ( E µ B) j + ɛ 0 0 c 2 ( Ei E j + 2 δij E 2 ) + µ 0 c 2 ( Bi B j + 2 δij B 2 ) (3.6) = [ ] ( te) B + E ( tb) + ɛ 0 [ ( µ E)E j + 2 ] j E 2 ( E )E j 0 + µ 0 c 2 c 2 [ ( B)B j + 2 j B 2 ( B )B j ] The two divergences are zero because of the Maxwell equations. We use the vector identity 2 ( V 2 ) ( V ) V = V ( V ) and replace the /c 2 with ɛ 0 µ 0 to get [ µ ] B ( te) + E ( tb) + [ ] E ( E) + [ ] B ( B) 0 µ 0 ɛ 0 µ 2 (3.7) 0 = [ ] E tb + E + [ ] B t E + B µ 0 µ 0 ɛ 0 µ 0 which is zero by the other two Maxwell equations.. IV. THE SCHWARZSCHILD SOLUTION (20 POINTS) We want to show that the Schwarzschild line element, which can be written as ds 2 = e ν(r) dt 2 + e λ(r) dr 2 + r 2 dθ 2 + r 2 sin 2 θ dϕ 2, (4.) where ν = λ = ln( 2M/r), satisfies the vacuum Einstein equation G αβ = 0. (4.2)

4 4 The Einstein tensor G αβ is defined in terms of the Ricci curvature tensor R αβ = Γ γ αβ,γ Γγ αγ,β + Γγ γɛγ ɛ αβ Γ γ βɛ Γɛ αγ, (4.3) and its trace, the curvature scalar R = Rα, α by the relation G αβ = R αβ 2 g αβr. (4.4) We calculate the connection coefficients Γ α βγ, by using the relation from the solutions to homework 6 ( ) g γλ ẍ λ = 2 g αβ,γ g γα,β ẋ α ẋ β, (4.5) where an overdot represents a derivative with respect to proper time. Setting γ to each of t, r, θ, and ϕ gives ẗ = ν ṫṙ, (4.6) r = ν 2 e2ν ṫ 2 + ν 2 ṙ2 + e ν r θ 2 + e ν r sin 2 θ ϕ 2, (4.7) θ = sin θ cos θ ϕ 2 2 r ṙ θ, (4.8) ϕ = 2 r ϕṙ 2 cot θ ϕ θ, (4.9) where a prime represents a derivative with respect to r. By comparing to the geodesic equation ẍ α = Γ α βγẋβ ẋ γ, we find that the only non-vanishing connection coefficients are Γ t rt = Γ t tr = Γ r rr = ν /2, (4.0) Γ r tt = ν e 2ν /2, (4.) Γ r θθ = re ν, (4.2) Γ r ϕϕ = re ν sin 2 θ, (4.3) Γ θ ϕϕ = sin θ cos θ, (4.4) Γ θ rθ = Γ θ θr = Γ ϕ rϕ = Γ ϕ ϕr = /r, (4.5) Γ ϕ ϕθ = Γϕ θϕ = cot θ. (4.6) Next we calculate the components of the Ricci tensor. The Ricci tensor is symmetric: R αβ = R γ αγβ = gγλ R λαγβ = g γλ R γβλα = R λ βλα = R βα, (4.7) where we have used R λαγβ = R γβλα, which is shown in Sec. I of the solutions to homework 0. Since the metric is diagonal, the off-diagonal components of the Ricci tensor are equal to the off-diagonal components of the Einstein tensor [see Eq. (4.4)]. For each of the components replace α and β appropriately in the general expression for the Ricci tensor Eq. (4.3). Then, with frequent references back to the list of non-vanishing connection coefficients (4.0)-(4.6), This can easily be rederived by computing Γ α βγẋβ ẋ γ from the general expression for Γ α βγ.

5 5 see that each goes to zero. G tr = R tr = Γ γ tr,γ Γ γ tγ,r + Γ γ γɛγ ɛ tr Γ γ rɛγ ɛ tγ = Γ t tr,t + Γ γ γtγ t tr Γ γ rɛγ ɛ tγ = Γ γ rɛγ ɛ tγ = 0, (4.8) G tθ = R tθ = Γ γ tθ,γ Γγ tγ,θ + Γγ γɛγ ɛ tθ Γ γ θɛ Γɛ tγ = Γ γ θɛ Γɛ tγ = Γ r θɛγ ɛ tr = 0, (4.9) G tϕ = R tϕ = Γ γ tϕ,γ Γ γ tγ,ϕ + Γ γ γɛγ ɛ tϕ Γ γ ϕɛγ ɛ tγ = Γ γ ϕɛγ ɛ tγ = Γ r ϕɛγ ɛ tr = 0, (4.20) G θr = R θr = Γ γ θr,γ Γγ θγ,r + Γγ γɛγ ɛ θr Γ γ rɛγ ɛ θγ = Γ γ γɛγ ɛ θr Γ γ rɛγ ɛ θγ = Γ ϕ ϕθ Γθ θr Γ ϕ rϕγ ϕ θϕ = 0, (4.2) G rϕ = R rϕ = Γ γ rϕ,γ Γ γ rγ,ϕ + Γ γ γɛγ ɛ rϕ Γ γ ϕɛγ ɛ rγ = Γ γ γɛγ ɛ rϕ Γ γ ϕɛγ ɛ rγ = Γ γ ϕɛγ ɛ rγ = 0, (4.22) G θϕ = R θϕ = Γ γ θϕ,γ Γγ θγ,ϕ + Γγ γɛγ ɛ θϕ Γ γ ϕɛγ ɛ θγ = Γ γ γɛγ ɛ θϕ Γ γ ϕɛγ ɛ θγ = Γ γ ϕɛγ ɛ θγ = 0. (4.23) We must now deal with the diagonal elements. The calculation for the tt component goes as follows: R tt = Γ γ tt,γ Γ γ tγ,t + Γ γ γɛγ ɛ tt Γ γ tɛγ ɛ tγ = Γ r tt,r + Γ γ γrγ r tt ( Γ r tɛγ ɛ tr + Γ t tɛγ ɛ ) tt = Γ r tt,r + Γ r tt ( Γ t tr + Γ r rr + Γ θ θr + Γ ϕ ϕr) 2Γ r tt Γ t tr = Γ r tt,r + 2Γ r ttγ θ θr 2Γ r ttγ t tr. (4.24) Unlike the off-diagonal elements, we now insert the values for the connection coefficients to find R tt = r ( ν 2 e2ν Then by using the definition ν = ln( 2M/r), along with ν = 2Me ν /r 2, we find R tt = ( ) M r r 2 eν + 2Mr 3 eν 2M 2 r 4 ) + ν r e2ν (ν ) 2 2 e2ν. (4.25) = ν M r 2 eν 2M 2 r 4 = 0. (4.26) The calculation for the other components is very similar, and the results are identical R rr = R θθ = R ϕϕ = 0. (4.27)

6 Since all of the R αβ = 0, the trace R = R α α = g αβ R αβ, certainly vanishes as well. So from the definition of the Einstein tensor (4.4), we have the expected result that the Schwarzschild line element (4.) satisfies the vacuum Einstein equation As a side note, remember that 2 G αβ = 0. (4.28) R αβ = 0 G αβ = 0. (4.29) So we should not be surprised that we found R αβ = 0, in a problem where we were asked to show that G αβ = One way to see this is to first show that G γ γ = R γ γ. Then if G αβ = 0, we have that G γ γ = R γ γ = 0, which by Eq. (4.4) means that R αβ = 0. The proof in the other direction R αβ = 0 G αβ = 0, is trivial.