GRNotesDecSchwarzschildGeodesicsPost

# GRNotesDecSchwarzschildGeodesicsPost - Radial geodesics in...

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Unformatted text preview: Radial geodesics in Schwarzschild spacetime Spherically symmetric solutions to the Einstein equation take the form ds 2 =- 1 + a r dt 2 + dr 2 1 + a r + r 2 dθ 2 + r 2 sin 2 θdϕ 2 where a is constant. We also have the connection components, which now take the form (using e ν = e- λ = 1+ a r , and therefore ν = ln ( 1 + a r ) and ν , 1 =- a r 2 e- ν Γ 00 = 0 Γ 01 = Γ 10 =- a 2 r 2 ( 1 + a r ) Γ 1 00 =- 1 2 1 + a r a r 2 Γ 1 11 = a 2 r 2 ( 1 + a r ) Γ 1 01 = Γ 1 10 = 0 Γ 11 = 0 Γ 2 12 = Γ 2 21 = 1 r Γ 1 22 =- r 1 + a r Γ 3 13 = Γ 3 31 = 1 r Γ 1 33 =- 1 + a r r sin 2 θ Γ 3 23 = Γ 3 32 = cos θ sin θ Γ 2 33 =- sin θ cos θ Now consider the geodesic equation for a particle which starts from rest at time τ = t = 0 . 0 = du a dτ + Γ a bc u b u c It might seem that the initial velocity 4-vector is u a = (1 , , , 0) , but this is not allowed. Using the line element, ds 2 =- dτ 2 , we must have- 1 =- 1 + a r dt dτ 2 + 1 c 2 ( 1 + a r ) dr dτ 2 + r 2 c 2 dθ dτ 2 + r 2 c 2 sin 2 θ dϕ dτ 2 where c is the speed of light. If there is no initial motion in the spatial directions we must have u = dt dτ = 1 q 1 + a r where r is the initial radial position. The geodesic equation for each component is then (at the initial time), 0 = 1 c du dτ + 2Γ 01 u u 1 1 0 = 1 c du 1 dτ + Γ 1 00 u u + Γ 1 11 u 1 u 1 + Γ 1 22 u 2 u 2 + Γ 1 33 u 3 u 3 0 = 1 c du 2 dτ + 2Γ 2 21 u 2 u 1 + Γ 2 33 u 3 u 3 0 = 1 c du 3 dτ + 2Γ 3 13 u 1 u 3 + Γ 3 23 u 2 u 3 Notice that if u 2 = 0 or u 3 = 0 then the corresponding accelerations also vanish, so they remain zero. However, u 1 cannot remain zero. For radial motion, substituting for the connection coefficients, we therefore have 0 = 1 c du dτ- a r 2 ( 1 + a r ) u u 1 0 = 1 c du 1 dτ- 1 + a r a 2 r 2 u u + a 2 r 2 ( 1 + a r ) u 1 u 1 We also have the relation given by the line element,- 1 =- 1 + a r dt dτ 2 + 1 1 + a r 1 c dr dτ 2 First equation Integrate the first equation, 0 = du u- adr r 2 ( 1 + a r ) = ln u- ˆ adr r ( r + a ) = ln u- ˆ 1 r- 1 r + a dr = ln u- ln r + ln( r + a )- ln b for some constant b, and therefore 1 + a r u = b Evaluating at the initial condition, we see that b = q 1 + a r . Radial component of velocity from the line element Now substitute into the line element equation to find u 1 ,- 1 =- 1 + a r ( u ) 2 + 1 1 + a r ( u 1 ) 2- 1 =- 1 + a r b 1 + a r 2 + 1 1 + a r ( u 1 ) 2 ( u 1 ) 2 = b 2- 1 + a r u 1 = r b 2- 1 + a r = r a r- a r 2 Now integrate to find r ( τ ) , cτ = ˆ dr q a r- a r = 1 √ a ˆ √ rdr q r r- 1 Let y = √ r . Then dy = 1 2 dr √ r , so cτ = 2 √ a ˆ y 2 dy q y 2 r- 1 Now let y = √ r cosh ξ...
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