General Relativity: Homework 4 Solutions - Department of Physics ...

More documents

Recommendations

Info

and Einstein’s equation now reads ( R µν − 1 2 Rg µν + β − 1 ∇ 2[ λ ∇ µ (F λρ F νσ g ρσ ) − ∇ λ ∇ γ (F λρ F γσ g ρσ )g µν + ∇ λ ∇ µ (F νρ F λσ g ρσ ) −∇ λ ∇ λ (F µρ F νσ g ρσ ) ] + F µρ F νσ R ρσ − 1 2 Rλγ g ρσ F λρ F γσ g µν ) = T µν where T µν is the energy momentum tensor associated with the electromagnetic field derived in part a. 2. The point of this exercise is to compute the first correction to the Newtonian solution caused by a source that rotates. In Newtonian gravity, the angular momentum of the source does not affect the field: two sources with identical densities, ρ(x i ), but different angular momenta have the same gravitational field. This is not so in General Relativity, since all components of T µν generate the field. This is an example of a post-Newtonian effect. a) Suppose a spherical body of uniform density ρ and radius R rotates rigidly about the x 3 axis with constant angular velocity ω. Write down the components T 0ν in a Lorentz frame at rest with respect to the center of mass of the body, assuming ρ, ω, and R are independent of time. For each component, work to the lowest non-vanishing order in ωR. b) The general solution to the equation ∇ 2 f = g, that vanishes at infinity, is the generalization of Φ(r) = −Gm/r, f(x) = − 1 ∫ 4π g(y) |x − y| d3 y Use this to solve Einstein’s Equation, keeping only terms linear in the perturbation, for the components of the trace-reversed perturbation h 00 and h 0i , where h µν = h µν − 1 2 ηµν h, for the source described in part (a). Obtain solutions only outside the body and only to the lowest non-vanishing order in r −1 , where r is the distance from the center of the body. Express the result for h 0i in terms of ω. Find the metric tensor within this approximation and transform it to spherical coordinates. c) Since the metric is independent of t and the azimuthal angle, φ, particles orbiting this body will have constant p 0 and p φ along their trajectories. Consider a particle of non-zero rest mass in a circular orbit of radius r in the equitorial plane. To lowest order in ω, calculate the difference between its orbital period when it’s orbiting with the direction of the body’s rotation (prograde) and when it’s orbiting against the direction of the body’s rotation (retrograde). (Define the period to be the coordinate time taken for one orbit of ∆φ = 2π). Solution: a) Given that we are doing a small perturbation to the Newtonian solution, we should assume that the rotation of the body is non-relativistic (ωR
1 from this we can determine the factor γ = √ , where we will use the unperturbed metric η dx µ dx ν µν to gµν dt dt calculate. 1 ⇒ γ = 1 − ω 2 r 2 ≈ 1 + ω2 r 2 Now we can write out the 4-velocity U µ = dxµ dτ = γ dxµ dt = γ(1, −ωr sin ωt sin θ, ωr cos ωt sin θ, 0) which to first order in ωR is just dxµ dt (i.e. γ = 1). Now we can calculate the components T 0ν for the energy-momentum tensor (for r = √ x 2 + y 2 + z 2 ≤ R): T 00 = ρU 0 U 0 = ρ T 01 = ρU 0 U 1 = −ρωy T 02 = ρU 0 U 2 = ρωx T 03 = ρU 0 U 3 = 0 b) We will now use Einstein’s equation (with upper indices) to solve for the perturbation to the metric. R µν − 1 2 Rgµν = 8πGT µν where g µν = η µν − h µν We will be dealing with the case where |h µν |
Page 1 and 2: General Relativity: Homework 4 Solu
Page 3: Now we must consider variations of
Page 7 and 8: where we have used the notation < y
Page 9 and 10: 3. In the presence of a cosmologica

General Relativity: Homework 4 Solutions - Department of Physics ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?