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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern