1 Full derivation of the Schwarzschild solution

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1 EPGY Summe Institute SRGR Gay Oas 1 Full deivation of the Schwazschild solution The goal of this document is to povide a full, thooughly detailed deivation of the Schwazschild solution. Much of the diffeential geometic foundations can be found elsewhee (and may be added at a late date). Fo details on how to get the fom of the Reimann cuvatue tenso and the stess-enegy tenso, see the othe notes. Hee we begin with a statement of the question. Question: What is the metic solution fo spacetime exteio to a spheically symmetic, static body of adius R and mass M? To begin answeing this poblem we will examine the metic of spacetime in spheical coodinates and in a geneal fom. We will wok with a west coast signatue ( ) and set c = 1 fo claity. We will einset c at the end. The flat metic in spheical coodinates is expessed as follows ds = µν g µν dx µ dx ν = dt d dθ sin θdφ Whee we can expess the non-zeo components of the metic tenso as g 00 = 1, g 11 = 1, g =, g 33 = sin θ. As we poceed we will use Geek indices to feely ange ove 0 though 3 (0 = t, 1 =, = θ, 3 = φ) we will late use Latin indices (i, j, k) to efe to only spatial indices (, θ, φ). We now genealize this to a moe geneal fom, ds = µν g µν dx µ dx ν = Udt V d W dθ X sin θdφ The functions U, V, W, X ae to be detemined but can be limited by the popeties of ou solution: 1. Spheical symmety. The functions should not depend on θ of φ. Also, the angula pat of the metic should have W = X. In fact, we can limit this to W = X = 1 without loss of geneality.. Static. The functions ae not functions of time. Also, any deivatives with espect to time of any metic component vanishes. Thus U = U(), V = V (). 3. Vacuum solution. Outside of the object that is the souce of cuvatue thee is no matte o enegy. Thus the fom of Einstein s solution will simplify a lot in the next section. Thus ae limited geneal metic is, ds = µν g µν dx µ dx ν = U()dt V ()d dθ sin θdφ (1) The metic components ae explicitly, g 00 = U, g 11 = V, g =, g 33 = sin θ The metic tenso is symmetic and has the following popety, g µ µ = g µν g µν g µν g µν =. () µ ν Whee in the last expession we intoduced the Einstein summation convention. Wheneve two indices ae epeated (one up, one down) in an expession they ae to be summed ove all dimensions. In geneal, F αβµ λβg γµ 3 3 F αβµ λβg γµ = (FG) α λγ β=0 µ=0 Thus we see we can find the metic tenso with aised indices to be just the invese of (), g 00 = 1 U, g11 = 1 V, g = 1, 1 g33 = sin θ 1

2 1.1 Einstein s equation The goal is to find a solution of Einstein s equation fo ou metic (1), R µν 1 g µν = 8πG c T µν (3) FIst some teminology: R µν Ricci tenso, R Ricci scala, and T µν stess-enegy tenso (the last tem will vanish fo the Schwazschild solution). Befoe we can poceed we need to intoduce some quantities and unavel the expession above. The Ricci tenso and scala in tems of the Reimann cuvatue tenso The Ricci tenso and scala ae obtained fom the Reimann cuvatue tenso, R β νρσ that is intoduced in the othe set of notes. The Ricci tenso is a contaction of the full cuvatue tenso, The Ricci scala is a contaction of the Ricci tenso, R µν R α µνα R R µ µ = g µα R µα The full Reimann cuvatue tenso is most compactly descibed in tems of the Chistoffel symbols (of the second kind), Γ β nσ, which will be defined in tems of the metic aftewads. R β νρσ = = Γ β νσ,ρ Γ β νρ,σ + Γ α νσγ β αρ Γ α νργ β ασ () In this expession a shot hand notation fo patial deivatives is employed. Any index offset by a comma is meant to signify a deivative with espect to that coodinate. I.e. Γ β νσ,ρ x Γ β ρ νσ = ρ Γ β νσ. The Chistoffel symbols ae expessed in tems of the metic tenso, Γ µ νσ = 1 gµλ {g λν,σ + g λσ,ν g νσ,λ } (5) We now see what needs to be done. We have the components of the metic tenso in tems of ou functions to be detemined, U, V the next step is to find all of the Chistoffel symbols. The last step is to find the Ricci tenso and scala. 1. The Chistoffel symbols To simplify the numeous calculations to come we establish some shot cuts utilizing the popeties of ou solution. 1. Any deivatives with espect to time vanish. That is, any tem like g µν,0 = 0. Thus we can quickly dismiss many tems.. All off diagonal tems vanish. That is, any g µν = 0 if µ ν. 3. The Chistoffel symbols ae symmetic in the lowe two indices (veify fo youself). I.e. Γ α µν = Γ α νµ thus cutting down on the numbe of tems significantly.. The fist tem in the geneal expession of (5) must be on the diagonal. Notice in (5) that the second uppe index is to be summed ove. Howeve we can set it at the outset because the two uppe indices must match, othewise the whole expession vanishes. 5. Also note that only deivatives of U, V with espect to (o 1) ae non-vanishing. We will poceed tem by tem, in ode of the uppe index and ecall Geek indices un ove all fou values while Latin only un ove 1,,3.

3 Γ µ νσ = 1 gµλ [g λν,σ + g λσ,ν g νσ,λ ] and ou non-vanishing metic components ae, g 00 = U, g 11 = V, g =, g 33 = sin θ g 00 = 1 U, g11 = 1 V, g = 1, 1 g33 = sin θ Γ 0 µν Γ 0 00 = 1 g00 [g 00,0 + g 00,0 g 00,0 ] = 0 Γ 0 0i = 1 g00 [g 00,i + g 0i,0 g 0i,0 ] = 1 g00 1 g 00 = 1 1 U U Γ 0 ij = 1 g00 [g 0i,j + g 0j,i g ij,0 ] = 0 Γ 1 µν Γ 1 00 = 1 g11 [g 10,0 + g 10,0 g 00,1 ] = 1 g11 1 g 00 = 1 1 V U Γ 1 0i = 1 g11 [g 10,i + g 1i,0 g 0i,1 ] = 0 Γ 1 i,j i = 1 g11 [g 1i,j + g 1j,i g ij,1 ] = 0 because V is a function of only Γ 1 11 = 1 g11 [g 11,1 + g 11,1 g 11,1 ] = 1 g11 1 g 11 = 1 1 V V Γ 1 = 1 g11 [g 1, + g 1, g,1 ] = 1 g11 1 g = 1 1 V = V Γ 1 33 = 1 g11 [g 13,3 + g 13,3 g 33,1 ] = 1 g11 1 g 33 = 1 1 V sin θ = V sin θ Γ µν Γ 00 = 1 g [g 0,0 + g 0,0 g 00, ] = 0 Γ 0i = 1 g [g 0,i + g i,0 g 0i, ] = 0 Γ ii = 1 g [g i,i + g i,i g ii, ] = 1 g g 33 = 1 θ sin θ = cosθ sin θ = Γ 33 Γ 1 = 1 g [g 1, + g,1 g 1, ] = 1 g 1 g = 1 = 1 Γ 13 = 1 g [g 1,3 + g 3,1 g 13, ] = 0 Γ 3 = 1 g [g,3 + g 3, g 3, ] = 0 Γ 3 µν Γ 3 00 = 1 g33 [g 30,0 + g 30,0 g 00,3 ] = 0 Γ 3 0i = 1 g33 [g 30,i + g 3i,0 g 0i,3 ] = 0 Γ 3 ii = 1 g33 [g 3i,i + g 3i,i g ii,3 ] = 0 Γ 3 i,j i = 1 g33 [g 3i,j + g 3j,i g ij,3 ] = 1 g33 [g 3i,j + g 3j,i ] Γ 3 13 = = 1 g33 [g 31,3 + g 33,1 ] = 1 1 g33 1 g 33 = sin θ sin θ = 1 Γ 3 3 = = 1 g33 [g 3,3 + g 33, ] = 1 1 g33 g 33 = sin θ θ( sin θ) = cosθ sinθ 3

4 Summaizing the non-vanishing tems (whee pime denotes a deivative with espect to ), Γ 0 01 = Γ 0 10 = Γ 1 00 = Γ 1 = Γ 1 = 1 Γ 3 13 = Γ3 31 = 1 V Γ 1 11 = V Γ 33 Γ 1 = V = cosθ sin θ Γ 3 3 = Γ3 3 = cotθ Γ 1 33 = V sin θ 1.3 The Ricci tenso Now that we have the Chistoffel symbols we can stat to assemble the Ricci tenso and scala. Fom equation () we can wite the Ricci tenso in tems of the Chistoffel symbols, R µν = R β µνβ = Γβ µβ,ν Γβ µν,β + Γα µβγ β αν Γ α µνγ β αβ = Γ 0 µ0,ν Γ0 µν,0 + Γα µ0 Γ0 αν Γα µν Γ0 α0 +Γ 1 µ1,ν Γ 1 µν,1 + Γ α µ1γ 1 αν Γ α µνγ 1 α1 +Γ µ,ν Γ µν, + Γ α µγ αν Γ α µνγ α +Γ 3 µ3,ν Γ3 µν,3 + Γα µ3 Γ3 αν Γα µν Γ3 α3 Fist we establish the off diagonal tems of the metic tenso. R µν, µ ν R 0i = Γ 0 00,i Γ0 0i,0 + Γα 00 Γ0 αi Γα 0i Γ0 α Γ α 00 Γ0 αi Γα 0i Γ0 α0 +Γ 1 01,i Γ1 0i,1 + Γα 01 Γ1 αi Γα 0i Γ1 α Γ 0,i Γ 0i, + Γ α 0Γ αi Γ α 0iΓ α Γ 3 03,i Γ 3 0i,3 + Γ α 03Γ 3 αi Γ α 0iΓ 3 α All tems vanish since fo Γ α 00 Γ0 αi eithe has α = 0, which has the fist tem vanishing o α = j and i which makes Γ0 µν vanish. Thus R 0i = 0. R i,j i = Γ 0 i0,j Γ0 ij,0 + Γα i0 Γ0 αj Γα ij Γ0 α Γ α i0 Γ0 αj Γα ij Γ0 α0 +Γ 1 i1,j Γ 1 ij,1 + Γ α i1γ 1 αj Γ α ijγ 1 α Γ α i1γ 1 αj 0 +Γ i,j Γ ij, + Γ α iγ αj Γ α ijγ α Γ α iγ αj 0 +Γ 3 i3,j Γ3 ij,3 + Γα i3 Γ3 αj Γα ij Γ3 α3 Γ 3 i3,j 0 + Γα i3 Γ3 αj Γα ij Γ3 α3 The tems that ae obviously 0 (fo one of the 5 easons stated ealie) ae witten to the ight. Let s examine the emaining tems closely, line by line. Γ α i0 Γ0 αj. Hee if α o j =, 3 the second symbol vanishes. If α = 0 then fo this tem to be non-vanishing i must equal 1, then j must equal o 3 and the whole tem vanishes. If α = 1 then i must equal 0 which is not one of the options (i = 1,, o 3). Γ α ij Γ0 α0 Hee we see the only option fo α is 1 o else the second tem vanishes. Then the fist tem must vanish since Γ 1 ij if i j. Γ α i1 Γ1 αj Hee we see that α = j fo the second tem to be nonzeo. Thus we ae left with: Γ1 i1 Γ Γ i1 Γ1 + Γ 3 i1 Γ1 33 The fist tem vanishes because i j. The second vanishes because i = 1 o 3 and the fist symbol vanishes. The thid tem vanishes because i = 1 o fo which the fist symbol vanishes. Γ α i Γ αj Fist note that α 0 o else the fist symbol vanishes. If α = 1 then i = fo the fist tem to suvive but j must be fo the second tem to suvive. This is not possible. If α = then i = 1 and j must be 1 but this is not possible (i j). Lastly, if α = 3 then i = 3 fo the fist tem to suvive but then j = 3 fo the second tem to suvive, which it can t. Thus the whole tem is 0.

5 Γ 3 i3,j This only has i = 1 and as non vanishing tems. Fo the fist j = o 3 but the symbol is only a function of, thus those tems vanish. If i = then j must be 1 o 3 but the symbol then is only a function of (θ) and it vanishes entiely. Γ α i3 Γ3 αj Hee we have α 0. Fo α = 1 both i and j must be 3 which is not possible. Fo α =, then i = 3 and j = 3 which can t be and the tem vanishes. Lastly, fo α = 3 we have that i is eithe 1 o and j must be o 1 espectively. Those this tem does not vanish and we ae left with Γ 3 13 Γ3 3 + Γ3 3 Γ3 31 Γ α ij Γ3 α3 Fo this tem we can only have α = 1 o. If α = 1 then, fo the fist symbol not to vanish, we must have i = j which is not the case. Fo α = we ae left with, Γ 1Γ 3 3 Γ 1Γ 3 3 We ae left with non vanishing tems. Let s wite them out explicitly. Γ 3 13Γ Γ 3 3Γ 3 31 Γ 1Γ 3 3 Γ 1Γ 3 3 = 1 cotθ + cotθ1 1 cotθ 1 cotθ = 0 Afte all that wok we have found out that R µν = 0 fo µ ν. R µµ R 00 = Γ 0 00,0 Γ 0 00,0 + Γ α 00Γ 0 α0 Γ α 00Γ 0 α0 cancelpaiwise +Γ 1 01,0 Γ 1 00,1 + Γ α 01Γ 1 α0 Γ α 00Γ 1 α1 0 Γ 1 00,1 + Γ 0 01Γ 1 00 Γ 1 00Γ Γ 0,0 Γ 00, + Γα 0 Γ α0 Γα 00 Γ α Γ 1 00 Γ 1 +Γ 3 03,0 Γ3 00,3 + Γα 03 Γ3 α0 Γα 00 Γ3 α Γ 1 00 Γ3 13 Inseting the values we find, R 00 = Γ 1 00,1 + Γ 0 01Γ 1 00 Γ 1 00Γ 1 11 Γ 1 00Γ 1 Γ 1 00Γ 3 13 = V + V V V 1 V 1 V = V 1 V + ( ) UV V 1 V = V + V + ( ) UV V 1 V = V + V + ( ) UV 1 V (6) R 11 = Γ 0 10,1 Γ0 11,0 + Γα 10 Γ0 α1 Γα 11 Γ0 α0 Γ 0 10,1 0 + Γ0 10 Γ0 01 Γ1 11 Γ0 10 +Γ 1 11,1 Γ 1 11,1 + Γ α 11Γ 1 α1 Γ α 11Γ 1 α1 cancel paiwise +Γ 1,1 Γ 11, + Γ α 1Γ α1 Γ α 11Γ α Γ 1,1 0 + Γ 1Γ 1 Γ 1 11Γ 1 +Γ 3 13,1 Γ3 11,3 + Γα 13 Γ3 α1 Γα 11 Γ3 α3 Γ 3 13,1 0 + Γ3 13 Γ3 31 Γ1 11 Γ3 13 Now collect tems and inset values, R 11 = Γ 0 10,1 + Γ 0 10Γ 0 01 Γ 1 11Γ Γ 1,1 + Γ 1Γ 1 Γ 1 11Γ 1 + Γ 3 13,1 + Γ 3 13Γ 3 31 Γ 1 11Γ 3 13 = + U = + U = U V V V V V V V R = Γ 0 0, Γ 0,0 + Γ α 0Γ 0 α Γ α Γ 0 α Γ 1 Γ Γ 1 1, Γ 1,1 + Γ α 1Γ 1 α Γ α Γ 1 α1 0 Γ 1,1 + Γ 1Γ 1 Γ 1 Γ Γ, Γ, + Γα Γ α Γα Γ α cancel paiwise +Γ 3 3, Γ3,3 + Γα 3 Γ3 α Γα Γ3 α3 Γ 3 3, 0 + Γ3 3 Γ3 3 Γ1 Γ3 13 5

6 Collecting tems and inseting values we get, R = Γ 1 Γ 0 10 Γ 1,1 + Γ 1Γ 1 Γ 1 Γ Γ 3 3, + Γ 3 3Γ 3 3 Γ 1 Γ 3 13 = V + ( 1 V V ) 1 V + V + θ cotθ + cot θ + 1 V = V + 1 V V + ( 1 cot θ) + cot θ = V + 1 V V 1 R 33 = Γ 0 30,3 Γ0 33,0 + Γα 30 Γ0 α3 Γα 33 Γ0 α Γ 1 33 Γ0 10 +Γ 1 31,3 Γ1 33,1 + Γα 31 Γ1 α3 Γα 33 Γ1 α1 0 Γ 1 33,1 + Γ3 31 Γ1 33 Γ1 33 Γ1 11 +Γ 3,3 Γ 33, + Γ α 3Γ α3 Γ α 33Γ α 0 Γ 33, + Γ 3 3Γ 33 Γ 1 33Γ 1 +Γ 3 33,3 Γ 3 33,3 + Γ α 33Γ 3 α3 Γ α 33Γ 3 α3 cancel paiwise Collecting tems and inseting values we get, R 33 = Γ 1 33Γ 0 10 Γ 1 33,1 + Γ 3 31Γ 1 33 Γ 1 33Γ 1 11 Γ 33, + Γ 3 3Γ 33 Γ 1 33Γ 1 The Ricci scala = V sin θ + ( sin θ V sin θ V ) sin θ + V V sin θ + θ (cosθ sin θ) cosθ sin θ cotθ + sin θ V = V sin θ + sin θ V V sin θ + ( sin θ + cos θ) cos θ = V sin θ + sin θ V V sin θ sin θ ( U = V + 1 V ) V 1 sin θ = sin θr The Ricci scala is obtained fom the Ricci tenso, R = R µ µ = gµν R µν = g 00 R 00 + g 11 R 11 + g R + g 33 R 33 = g 00 R 00 + g 11 R 11 + g R + g 33 sin θr = 1 U R 00 1 V R 11 1 R 1 ( sin θ )sin θr = 1 U R 00 1 V R 11 R = V + UV + ( ) V + U V + UV V + = UV + UV U V 1 V + V V + UV + ( ) V UV + V + (1 1 V ) The Einstein equation fo Schwazschild spacetime Since the Schwazschild solution concens with the exteio spacetime of a spheically symmetic body, Einstein s equation (3) takes a simple fom R µν 1 g µνr = 0 (7) 6

7 This leads to fou equations that must be satisfied. R 00 1 g 00R = 0 = R 00 U R 0 = V + 0 = 1 V + ( ) UV 1 V + V V ( ) UV + 1 V + 1 U V + U (1 1 V ) V + 1 (1 1 V ) (8) R 11 1 g 11R = 0 = R 11 + V R 0 = U UV V + 0 = 1 U + V (1 1 V ) UV + ( ) U 1 U + 1 V + V (1 1 V ) 0 = UV + 1 (1 1 V ) (9) R 1 g R = 0 = R + R 0 = V + 1 V V 1 V + 0 = V + V V + 0 = U + U V U + UV + ( ) U V UV + UV + ( ) U V UV + ( ) Note that the last equation is not independent of the pevious, V + (1 1 V ) R 33 1 g 33R = 0 = sin θr + sin θ R = R + R (10).1 Solving the Einstein equation We ae nealy thee. Notice that the fist equation fo G 00, (8), is only a function of V. Thus we can solve this diffeential equation to find V and then use that in the othe equations to find U. Re-expessing that equation we get, Integating this and using the fomula d 0 = V + 1 (V 1) 0 = = dx ax+bx = 1 a+bx a ln x V (V 1) + 1 dv V (V 1) we get, (11) This is half of ou solution. ln + C = ln 1 + C = ln C (V 1) V = V 1 V V 1 = C V V (1 C ) = 1 V = 1 1 C 7

8 To find U we inset this solution into (9) to get, 0 = U (1 C ) 1 (1 (1 C )) 0 = U (1 C ) C U = C (1 C ) = C C du U = Cd C Now we integate both sides of this equation (using the same fomula we used befoe) to get, du U = lnu = C d C = C C ln C = ln(1 C ) Exponentiating both sides we obtain ae othe answe. Thus ou Schwazschild metic is, U = (1 C ) ds = (1 C )c dt d (1 C ) dθ sin θdφ (1) Fist we notice that this metic satisfies ou popeties spelled out at the beginning. It is spheically symmetic. It is asymptotically flat (when it is ou flat spacetime metic). One moe equiement will help us pin down the emaining undetemined paamete C. We expect that in the limit that M 0 we should again obtain the flat metic. We see that this would esult when C = 0 so we hypothesize that it is popotional to mass (as thee ae no othe fee paametes this must be the case). To get the exact fom we need to compae the esults fo geodesics fo this metic fo a vey small test mass when the souce mass, M, is weak. O compaing Einstein s equation in the low mass limit and noting we must egain Newton s law will also wok. The esult is that C = GM c and yields the full Schwazschild metic. ds = (1 GM c )c dt d (1 GM c ) dθ sin θdφ (13) Some loose ends Bikhoff s theoem Recall at the outset that we stated that the coefficients U, V wee not functions of time. Well, you may not believe that. To convince you, we only need to look at one othe tem in the Ricci tenso but no longe assume time-independence. So etun to examine the tem R 0 = R 01. But fist we have to deive the additional Chistoffel symbols that no longe need to vanish due to time independence. The additional tems ae, Γ 0 00 = U, Now etuning to this tem of the Ricci tenso, Γ0 11 = V, Γ1 01 = Γ 1 10 = V V R 01 = Γ 0 00,1 Γ 0 01,0 + Γ α 00Γ 0 α1 Γ α 01Γ 0 α0 Γ Γ α 00Γ 0 α1 Γ α 01Γ 0 α0 +Γ 1 01,1 Γ 1 01,1 + Γ α 01Γ 1 α1 Γ α 0iΓ 1 α1 cancel paiwise +Γ 0,1 Γ 01, + Γα 0 Γ α1 Γα 0i Γ α Γ 3 03,1 Γ3 01,3 + Γα 03 Γ3 α1 Γα 0i Γ3 α

9 We ae left to examine, R 01 = Γ Γ α 00Γ 0 α1 Γ α 01Γ 0 α0 = Γ Γ 0 00Γ 0 01 Γ 0 01Γ Γ 1 00Γ 0 11 Γ 1 01Γ 0 10 = Γ Γ1 00 Γ0 11 Γ1 01 Γ0 10 ( ) U = + V V = 1 ( V V t U + 1 U U t ) U = (1) Note that, fom Einstein s equation, this tem must vanish: R 01 1 g 01R = 0 = R 01, o, 0 = 1 0 = t U + 1 U U t U U t = t U 1 U t U 1 U ( U U ) + t t U (15) 3 Othe Coodinates in Schwazschild spacetime Eddinginton-Finkelstein inwad and outwad coodinates To analytically continue the Schwazschild coodinates though the hoizon and down to the singulaity we define a new adial coodinate (with R s = GM c ), d, define, x = R s, 1 Rs x + Rs = dx = 1 + R s x x dx = x + R s lnx, = R s + ln( R s ) (16) We will ignoe the constant, R s and stick with = + R s ln( R s ). Now define the advanced null coodinate, and etaded null coodinate, v = ct w = ct + 9