[ home ] Schwarzschild Metric General Relativity. Note: « 9 »
Unnamed Web page Schwarzschild metric is a static spherically symmetric solution of the vacuum Einstein's equations, Rab=0:
ds2 = (1 − rg/r ) dt2 − (1 − rg/r )−1 dr2 − r2(dθ2 +sin2θ dφ2),
where rg=2Gmc−2 is the gravitational (Schwarzschild) radius of the central body with mass m.

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

In the Schwarzschild metric the gravitational radius rg=2m is a singular point. Inside the Schwarzschild radius time and radial coordinates interchange. A transformation to the new coordinates τ, ρ

dτ = dt + √[ rg/r ] 1/(1− rg/r ) dr ,   dρ = dt + √[ r/rg ] 1/(1− rg/r ) dr
leads to the Lemaitre metric
ds2 = dτ2 − rg/r dρ2 − r2(dθ2 +sin2θ dφ2) ,    r=[ 3/2 (ρ−τ)]2/3rg1/3
which is related to free particles radially falling towards the center.

For a free falling body, dρ=0, in the region r~=rg

dt~= − rg/r−rg dr  ,   ⇒   r−rg = (r0−rg)\exp(− t−t0/rg ),
it takes a free falling body infinitely long time to reach the Schwarzschild radius (for the outer observer).

In τ, ρ coordinates the free falling particle reaches the Schwarzschild radius and the origin within finite time

∫r0rg dτ = 2/3 ( r03/2−rg3/2/rg1/2 )
Along the tragectory of a (radial) light ray
dr=√[ r/rg ](±√[ r/rg ]−1)dτ ,
therefore no signal can escape from inside the Schwarzschild radius where always dr<0 and the light rays emitted radially inwards and outwards both end up at the origin.

Excercises

  1. Show that Bianchi identities (DaRdebc+DcRdeab+DbRdeca) imply that the Einstein tensor, Gab=Rab− 1/2 gab R, has vanishing divergence, DaGab=0.

    Hint: multiply Bianchi identities by two Kronneker delta's with carefully chosen indexes.

  2. Calculate the Schwarzschild radius for the Sun (M⊗~= 2.0⋅1030 kg), the Earth (M⊕~= 6.0⋅1024 kg) and the proton (Mp~= 938 MeV).

  3. 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 Dur=0. Answer: like in Newtonian theory, ω2 = M/r3.

Copyleft © 2005 D.V.Fedorov (fedorov @ phys au dk)