Introduction to General Relativity. Note 8

Introduction to General Relativity. Note « 8 »

Schwarzschild solution

Schwarzschild metric is a static spherically symmetric solution of the vacuum Einstein's equations R_ab=0. In this case the interval ds² can be suitably written as

ds² = A(r) dt² - B(r) dr² - r²(dθ² + sin²θ dφ²),

where A(r) and B(r) are unknown functions to be found from the Einstein equations. From the given metric it is relatively easy to calculate Christoffel symbols and then the Ricci tensor. The resulting vacuum Einstein's equations can be integrated to give the famous Schwarzschild solution

ds² = (1 - ^r_g/_r) dt² - (1 - ^r_g/_r)^-1 dr² - r²(dθ² + sin²θ dφ²),

where r_g=2Gmc^-2 is the gravitational (Schwarzschild) radius of the massive body.

Radiall fall in the Schwarzshild field. Lemaitre coordinates. Event horizons. Black holes.

In the Schwarzschild metric the gravitational radius r_g=2m is a singular point. It takes infinitely long time for a free falling body to reach the Schwarzschild radius. Inside the Schwarzschild radius time and radial coordinates interchange.

A transformation to the new coordinates τ, ρ

dτ = dt + √(2m/r)[¹/_{(1-^2m/_r)}] dr

dρ = dt + √(r/2m)[¹/_{(1-^2m/_r)}] dr

leads to the Lemaitre metric

ds² = dτ² - [³/_4m(ρ - τ)]^-3/2 dρ² - [³/₂(ρ-τ)]^4/3(2m)^2/3(dθ² + sin²θ dφ²)

which is related to free particles radially falling towards the center. In this metric the free falling particle reaches the Schwarzschild radius and the origin within finite time. No signal can escape from inside the Schwarzschild radius where the light rays emitted radially inwards and outwards both end up at the origin.

Excercises

Show that Bianchi identities (D_aR^d_ebc+D_cR^d_eab+D_bR^d_eca) imply that the Einstein tensor, G_ab ≡ R_ab - ¹/₂ g_ab R, has vanishing divergence, D_aG^a_b = 0.
Hint: multiply Bianchi identities by two Kronneker delta's with carefully chosen indexes.
Calculate the Schwarzschild radius for the Sun (M_⊗ ≈ 2.0×10³⁰ kg), the Earth (M_⊕ ≈ 6.0×10²⁴ kg) and the proton (M ≈ 938 MeV).
What is the Kepler's law for a circular orbit (that is the relation between the orbit's period and radius) in the Schwarzschild metric?
Hints: Period=2π/ω, where ω=dφ/dt is the angular frequency which can be found from the geodesic equation Du^r=0.
Answer: like in Newtonian theory, ω² = M/r³.