examgr - General Relativity Exam Michael Good Problem 3...

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General Relativity Exam Michael Good April 25, 2006 Problem 3 Taking Robertson-Walker metric ds 2 = - dt 2 + R ( t ) 2 ± dr 2 1 - kr 2 + r 2 d Ω 2 ² with k = +1 , 0 , - 1 and R ( t ) the cosmological scale. Use G μν = 8 πT μν and T μν ; ν = 0 to obtain the Friedmann equation ˙ R 2 + k = 8 π 3 ρR 2 and the equation of energy conservation d dR ( ρR 3 ) = - 3 pR 2 assuming a perfect fluid stress enrgy tensor where ρ ( t ) is the matter total energy density and p is the isotropic pressure. Specialize to the case of a flat universe k = 0 and compute the resulting time dependence of the scale factor when the universe is filled with dust p = 0 and in the case where the universe is filled with radiation p = ρ/ 3. Solution: Writing Einstein’s equation in the form R μν = 8 π ( T μν - 1 2 g μν T ) We see that we need the Ricci tensor and the stress-energy tensor for a perfect fluid. The Ricci tensor involves calculating the Christoffel symbols from the metric which is a tedious but easy calculation. The stress-energy tensor can be found more quickly. As our fluid is perfect, the fluid may be at rest in comoving coordinates with a 4-velocity of U μ = (1 , 0 , 0 , 0) and the stress-energy tensor T μν = ( ρ + p ) U μ U ν + pg μν is, with an index raised: 1
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T μ ν = diag ( - ρ,p,p,p ) with a trace of T = - ρ + 3 p We will remember this for later. For now, lets focus on the Ricci tensor. To find the Christoffel symbols, simply use: Γ σ μν = 1 2 g σρ ( μ g νρ + ν g ρμ - ρ g μν ) Lets explicitly calculate a few. Try Γ 3 03 . Γ 3 03 = 1 2 g 3 ρ ( 0 g 3 ρ + 3 g ρ 0 - ρ g 03 ) This is just the first term and only with ρ = 3 , as the others vanish Γ 3 03 = 1 2 g 33 0 g 33 = 1 2 R 2 ( t R 2 ) = ˙ R R Lets try another: Γ 0 22 = 1 2 g 00 ( - 0 g 22 ) = 1 2 t ( R 2 r 2 ) = r 2 2 ( t R 2 ) = r 2 R ˙ R The rest follow in exactly the same way. Here are all of them: Γ 0 11 = R ˙ R/ (1 - kr 2 ) Γ 0 22 = R ˙ Rr 2 Γ 0 33 = R ˙ Rr 2 sin 2 θ Γ 1 01 = ˙ R/R Γ 1 11 = kr/ (1 - kr 2 ) Γ 1 22 = - r (1 - kr 2 ) Γ 1 33 = - r (1 - kr 2 ) sin 2 θ Γ 2 12 , Γ 3 13 = 1 /r Γ 3 23 = cot θ Γ 2 02 = ˙ R/R Γ 3 03 = ˙ R/R Γ 2 33 = - sin θ cos θ With our Christoffel symbols in hand, we are in a position to calculate the Ricci tensor from: R σν = λ Γ λ νσ - ν Γ λ λσ + Γ λ λμ Γ μ νσ - Γ λ νμ Γ μ λσ An important shortcut is that because our space is isotropic, we only care about two equations from the Ricci tensor, that is the tt and one of the space- space ones, I’ll go with just 22. I’ll explicitly go through the tt one.
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This note was uploaded on 03/17/2012 for the course PHYS 202 taught by Professor Atkin during the Spring '12 term at Amity University.

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examgr - General Relativity Exam Michael Good Problem 3...

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