Curvature singularity

Transcription

1 Cuvatue singulaity We wish to show that thee is a cuvatue singulaity at 0 of the Schwazschild solution. We cannot use eithe of the invaiantsr o R ab R ab since both the Ricci tenso and the Ricci scala vanish. The next siplest invaiant is R abcd R abcd The cuvatues ae given by whee R ν,11e ν λ 1 4 ν,1ν,1 e ν λ 1 4 ν,1λ,1 e ν λ R λ,1e λ R λ,1e λ sin 2 θ R ν,1e ν λ R e λ sin 2 θ R ν,1e ν λ Theefoe, e ν e λ ν,1 1 λ,1 R ν,11 ν,1 ν,1 e ν λ R R sin2 θ R R sin2 θ R Check R ν,11e ν λ 1 4 ν,1ν,1 e ν λ 1 4 ν,1λ,1 e ν λ 1 2 ν,11 ν,1 ν,1 e ν λ 1

2 Loweing the uppe index, R R 1212 R sin 2 θ R R 2323 sin 2 θ R 3030 sin 2 θ and aising all fou indices, R R R sin 2 1 θ R R sin 2 θ R sin 2 θ To copute the invaiant, we will have sus including all eaangeents of the indices of the nonvanishing tes, fo exaple, R 1212 R 1212 R 2112 R 2112 R 1221 R 1221 R 2121 R 2121 So each te ust be counted fou ties. Theefoe R abcd R abcd sin 2 θ 5 sin 2 θ 2

3 sin 2 θ 7 sin 2 θ 4 1 sin 2 θ 5 sin 2 θ so siplifying, R abcd R abcd which diveges stongly at 0, but is egula at. Regulaity at It tuns out that the Schwazschild etic has only a coodinate singulaity at. A change of coodinates eoves the singula facto,. Null coodinates, u and v, such that holding eithe u o v constant gives a null geodesic, tun out to not only eove the singulaity, but also give a cleae pictue of the causal elationships nea the sta. Fist, we find the null geodesics in the t-plane. We have aleady shown that and the line eleent then tells us that u 0 k 0 k2 u 0 u 1 u 1 2 so that Taking the quotient, and integating, u 1 ±k d dt u1 u 0 ± ˆ d ±t ˆ d ˆ ˆ d d ±t c 3

4 Theefoe, defining u t v t we see that u constant gives an ingoing null geodesic and v constant gives an outgoing null geodesic. To wite the line eleent in tes of u and v, copute thei diffeentials, du dt d 1 dt d dv dt d Theefoe, This is still singula at, but now define Then Substituting into the line eleent, we have dudv dt 2 d ds 2 1 dt 2 d2 1 2 dω 2 1 dudv 2 dω 2 U e u 4 V e v 4 dudv 1 u v dudve dudve 16 2 e dudv ds e dudv 2 dω 2 whee U, V. Thee is clealy no poble at, which is called the event hoizon. Radial infall We now conside a paticle falling into the black hole. Since the neighbohood of is now established to be egula, thee can be nothing to keep a paticle fo falling acoss the hoizon. We theefoe conside a paticle falling fo just inside the hoizon towad the singulaity. In this egion, becoes the tielike coodinate. The geodesic equation fo u 0 ay be witten as 1 u 0 k 4

5 with initial value at 0 given by 1 u 0 0 k 0 Since is now the tielike coodinate, we ay choose u 0 0 0, and theefoe, k 0, and u 0 0 along the entie geodesic. Then u 1 is given by Integating, u 1 τ u 1 Letting y, and theefoe, y 2, ˆ 1 dτ 0 d 0 d 1 Now let y sin θ so that τ 2 τ cos 2 θdθ 1 cos 2θ dθ 0 θ 12 0 sin 2θ ˆ y2 dy θ sin θ cos θ 0 0 acsin π 2 acsin 1 0 We ay take the initial position to be 0, which gives τ π which is, in paticula, finite. The infall doesn t take long fo stella sized black holes. Fo a black hole with 10 ties the ass of the sun, we have τ GMπ 4c π sec 5

6 o just ove an hou. Fo a black hole at the cente of a galaxy, which ay have a ass a illion ties as geat, the infall will take just shot of two yeas. The escape of light Finally, conside light which stats nea the event hoizon at t 0. How long does it take to escape fo the egion of the black hole? We conside an outgoing null geodesic, which has v constant, Suppose light leaves 0 δ at tie t 0. Then and the light eaches adius at tie c t c δ δ ct δ δ Copae this tie to the tie it would take in fee space fo light to tavel fo δ to, given by We find t 0 δ c t δ δ t 0 δ δ 1 δ 1 δ δ Conside a black hole with 10 ties the ass of the sun, so that 2GM c The tie to each the obit of Eath fo nea the hoizon would noally be about 500 seconds. This is inceased by t 500 δ δ δ δ Fo a distance of δ 1c above the hoizon, the tie delay is t sec so a collapsing sta will appea to settle to its Schwazschild adius exteely quickly. 6