THE SCHWARZSCHILD SOLUTION 1. Birkhoff's Theorem

THE SCHWARZSCHILD SOLUTION

1. Birkhoff’s theorem. This is a uniqueness result for rotationally symmetric solutions of the Vacuum Einstein Equations (VEE). A rotationally symmetric spacetime is a Lorentzian manifold (M, g) diffeomorphic to 3 Ω × I, where I ⊂ R is an interval and Ω ⊂ R is a rotationally symmetric open set.The vector field ∂t corresponding to the t ∈ I coordinate is future- directed timelike, and SO(3) acts on M by isometries, with orbits given by spacelike two-spheres of constant curvature, which foliate each Ωt = Ω×{t}. In addition, the length of ∂t and its projection on spacelike slices (equiva- lently, the lapse function and shift vector field of the natural 3+1 split) are invariant under this isometry group. If the area of the two-sphere orbits is monotone increasing (with respect 3 to enclosure in R ) we may take it as a parameter, defining a diffeomorphism 2 from Ωt to a product S × Jt, Jt ⊂ R, taking the orbit Op of p ∈ Ωt to 2 2 S × {r}, where 4πr = areagt (Op)(r is sometimes called the ‘area radius’.) This introduces global coordinates (t, r, ω)(ω ∈ S2) on M, and the metric may be written in the form:

ds2 = −a2dt2 + 2abdrdt + eλdr2 + r2dω2, with a(t, r) > 0, b(t, r), λ(t, r).

To eliminate the cross-term, complete squares:

ds2 = −(adt − bdr)2 + (b2 + eλ)dr2 + r2dω2 and ﬁnd an integrating factor e−τ/2 for adt + bdr, so that:

e−t/2(adt − bdr) = dσ, eτ dσ2 = (adt − bdr)2, so: ds2 = −eτ dσ2 + (b2 + eλ)dr2 + r2dω2, and changing notation (redeﬁning the time variable):

ds2 = −eτ dt2 + eλdr2 + r2dω2, τ(r, t), λ(r, t).

Remark. Finding the integrating factor amounts to solving the ﬁrst-order PDE: aτr − bτt = 2(ar − bt); this can be done (locally) by standard methods (characteristics.)

1 2 2 This is a skew-product metric P ×r S with ﬁber (S , std), ‘scale function’ 2 r and base P ⊂ R(t,r) (open), with metric on P :

2 τ 2 λ 2 dsP = −e dt + e dr , τ(r, t), λ(r, t).

We use the expressions for Ricci curvature of skew-products with ([ON]) to compute: 2 Ric(∂ , ∂ ) = K h∂ , ∂ i − HessP (r)(∂ , ∂ ), r t P r t P r r t 2 where KP is the Gauss curvature of dsP . Since h∂r, ∂tiP = 0 and 1 1 h∇ ∂ , ∂ i = λ eλ implies (∇ ∂ )r = λ , ∂t r r 2 t ∂t r 2 t we have: 1 1 HessP (r)(∂ , ∂ ) = − λ , Ric(∂ , ∂ ) = λ . t r 2 t t r r t Another formula in [ON] gives:

∆P r 1 Ric(∂ , ∂ ) = (K − − h∇P r, ∇P ri)|∂ |2, θ θ F r r2 θ

1 2 2 where KF = r2 (Gauss curvature of the ﬁber) and |∂θ| = r . Problem 1. Verify that:

λ0 τ 0 HessP (∂ , ∂ ) = − , HessP (∂ , ∂ ) = − eτ−λ, ∇P r = e−λ∂ , h∇P r, ∇P ri = e−λ r r 2 t t 2 r P and therefore: 1 ∆P r = −e−τ HessP (∂ , ∂ ) + e−λHessP (∂ , ∂ ) = e−λ(τ 0 − λ0). t t r r 2 We conclude: r Ric(∂ , ∂ ) = 1 − e−λ( (τ 0 − λ0) + 1). θ θ 2 Now assume g is a solution of the VEE. In particular Ric ≡ 0 and: r λ = 0, 1 + (τ 0 − λ0) = eλ. t 2 So λ is independent of t; hence τ 0 is independent of t, and we may write:

τ(r, t) =τ ¯(r) + c(t).

2 c(t) Making the change of time variable dt¯= e 2 dt we ﬁnd:

eτ(t,r)dt2 = eτ¯(r)(dt¯)2, so (changing notation) we obtain the metric in static form:

ds2 = −eτ(r)dt2 + eλ(r)dr2 + r2dω2.

Problem 2. Compute the other non-zero components of the Ricci tensor:

τ 00 (τ 0)2 τ 0λ0 τ 0 Ric(∂ , ∂ ) = eτ−λ{ + − + }, t t 2 4 4 r τ 00 (τ 0)2 τ 0λ0 λ0 Ric(∂ , ∂ ) = − − + + . r r 2 4 4 r Thus in the Ricci-ﬂat case (adding the two equations) we obtain the ﬁrst- order system: 2 τ 0 + λ0 = 0, τ 0 − λ0 = (eλ − 1). r Problem 3. Show that the general solution of the system is: A A e−λ = 1 + , eτ = eC (1 + ), r r for arbitrary real constants A, C. By renormalizing the time variable we may take C = 0, and set A = −2M to obtain the Schwarzschild metric: 2M 1 ds2 = −(1 − )dt2 + dr2 + r2dω2. r 2M 1 − r

2 The domain of deiﬁnition of the metric in R(t,r) is the disjoint union PI t PII , where PI = {r > 2M} and PII = {0 < r < 2M}. This gives two skew-product spacetimes:

2 2 N = PI ×r S ,B = PII ×r S , the Schwarzschild exterior and black hole regions, respectivly. We conclude: Theorem. Any rotationally symmetric solution of the VEE is static, and is isometric to an open subset of the Schwarzschild exterior or the Schwarzschild black hole.

3 Remark. Note that no ‘asymptotic conditions’ are required, and neither is completeness. This is a local result.

2. Flamm paraboloid. The Riemannian Schwarzschild metric on 3 3 Rext = R \ B2M : 2 1 2 2 2 ds = 2M dr + r dω 1 − r 4 can be realized as the induced metric on a hypersurface in R :

3 4 3 V = graph(w) ⊂ R = R(r,θ,φ) × Rw, w = w(r).

2 2 2 2 4 The metric induced from the euclidean metric dw + dr + r dω in R is:

2 0 2 2 2 2 dsV = ((w ) + 1)dr + r dω . Comparing this metric with Shwarzschild, we find: 2M 2M 1 + (w0)2 = (1 + )−1, so (w0)2 = , r > 2M. r r − 2M Choosing w0 > 0, and integrating so that w(2M) = 0, we find: 2M √ √ w0 = ( )1/2 → w(r) = 2 2M r − 2M, or w2 = 8M(r−2M), r ≥ 2M. r − 2M Thus the ‘profile’ w is parabolic, the Flamm paraboloid, a geometric way to visualize Riemannian Schwarzschild. Remark. Note that V can be continued into the region {w < 0}, giving the hypersurface:

w2 V˜ ⊂ 4 = {(r, θ, φ, w)|r = 2M + . R 8M WIth the induced metric, this gives a model of the “double Riemannian Schwarzschild manifold”, consisting of two isometric copies of Riemannian Schwarzschild joined at an equatorial S2. Problem 4. (i) Show that, with the induced metric, V˜ is a complete Riemannian manifold. (ii) Show that the “equatorial S2”: {w = 0, r = 2M} ⊂ V˜ is totally geodesic in V˜ (Hint: ﬁxed sets of isometries are totally geodesic (why?) Sectional curvatures. An application of the Flamm paraboloid is the geometric computation of sectional curvatures in Schwarzschild spacetime.

4 2 First, in V , the great circles in Sr (the 2-sphere at ‘height’ w(r)) are principle curves α, with normal curvature given by:

0 1 0 1 w k1 = k2 = hα ,Ni = . r r p1 + (w0)2

The third principal curve through a point of V is the meridian curve itself, (r, w(r), θ0, φ0) for r > 2M, with normal curvature:

w00 k = . 3 [1 + (w0)2]3/2

4 Since R is ﬂat, we have for the sectional curvatures σ of V :

σθφ = k1k2, σrθ = k2k3, σrφ = k1k3, and then the sectional curvature σrt of the Schwarzschild exterior N may be found using the fact the metric is Ricci-ﬂat, since:

Ric(er, er) = σrt + σrθ + σrφ.

Problem 5. Show that the sectional curvatures of N are given by: M 2m σ = σ = σ = σ = − ; σ = σ = . rθ rφ tθ tφ r3 tr θφ r3 Remark. Note that the sectional curvatures are well-behaved as r → 2M from above, suggesting the apparently singular behavior of the metric at the ‘horizon’ r = 2M may be due to the choice of coordinate system. Question. Note that in the “black hole region” B = {0 < r < 2M}, the coordinate vector ﬁeld ∂t is spacelike, while ∂r is timelike. Thus the “Riemannian slices” of B are given by constant r, r = r0 < 2M:

2 2M 2 2 2 Br0 = {(r, t, θ, φ)|r = r0}, dsr0 = ( − 1)dt + r0dω , r0

2 essentially a Riemannian product R × S . But unlike the spacelike hyper- surfaces Nt ∼ V of N, they are not totally geodesic.

Compute the second fundamental form of Br0 in B and use it to answer: what are the sectional curvatures of B, and how do they behave on approach to the horizon ‘from the inside’?

5 3. Schwarzschild geodesics. Let α(s) be a timelike or null geodesic in N t B, parametrized by proper time or an aﬃne parameter (resp.) We may derive the geodesic equations in standard coordinates, α(s) = (t(s), r(s), θ(s), φ(s)), s ∈ I ⊂ R, by considering the ﬁrst variation of the functional: Z L[α] = [−h(r)(t˙)2 + h−1(r)(r ˙)2 + r2(θ˙)2 + r2 sin2 θ(φ˙)2]ds, I under a variation with compact support in I:

α → α + v, v(s) = (v0(s), v1(s), v2(s), v3(s)).

(Here dot denotes derivative with respect to s and h(r) = 1 − 2M/r). For example, computing the variation corresponding to v0 only (setting the other vi = 0), we ﬁnd the diﬀerential equation for t(s): Z Z 0 0 0 = δ0L[α] = (−ht˙v˙0)ds = (htv¨ 0 + h r˙tv˙ 0)ds → ht¨+ h r˙t˙ = 0. I I This can be written as a conservation law:

h(r)t˙ ≡ E, a constant of motion.

Problem 6. Compute the variations corresponding to v2 only, then to v3 only (separately) to derive the θ and φ components of the geodesic differential equations: (r2θ˙)· = r2 sin θ cos θ(φ˙)2, (r2 sin2 θφ˙)· = 0, or r2 sin2 θφ˙ ≡ L, a second constant of motion. The θ differential equation shows the initial conditions θ(0) = π/2, θ˙(0) = 0 give rise to the constant solution θ(s) ≡ π/2. These IC can always be achieved by a change of coordinates in S2, so we may assume the motion takes place over a fixed great circle in the ‘sphere of spatial directions’ (or on the ‘surface of the star’). Then the second conservation law takes the form: r2φ˙ ≡ L, reminiscent of Kepler’s law for classical motion under a central force.

6 Problem 6, continued. Under the assumption θ(s) ≡ π/2, the r-component of the geodesic equation reads: 1 (h−1r˙)· = (−h0(t˙)2 + (h−1)0(r ˙)2 + 2r(θ˙)2 + 2r(φ˙)2). 2

The r equation can also be put in the form of a ﬁrst-prder conservation law. Using the fact that the parametr is proper time, we have, for timelike geodesics: E 1 L2 −1 = hα0mα0i = (−h) + (r ˙)2 + r2 , h2 h r4 or: L2 E2 = (r ˙)2 + h(1 + (timelike), r2 while for null geodesics with aﬃne parameter s, the same reasoning gives: L2 E2 = (r ˙)2 + h . r2 The conservation law (‘Energy equation’) for timelike geodesics can be written in ‘kinetic+potential’ form: L2 2M L2 2ML2 E2 = (r ˙)2 + V (r),V (r) = h(1 + ) = 1 − + − . r2 r r2 r3 From this conservation law, one easily derives (assuming r is not constant on any interval: dV 2¨r = − . dr Problem 7. Show this is equivalent to the second-order geodesic ODE for r(s) derived earlier.

Low-energy radial geodesics. Consider a low-energy radial timelike geodesic: physically, a free-falling massive particle (a ‘pebble’, [ON]) starting from rest dr at some point in the exterior region: r(0) = R > 2M, ds (0) = 0, where s is proper time along the particle’s worldline. In addition, θ(s) ≡ π/2, φ(s) ≡ φ0, so L = 0. The energy equation becomes: 2M 2M 1 1 (r ˙)2 − = E2 − 1 = h(R) − 1 = − , or (r ˙)2 = 2M( − ), r R r R so r < R, andr ˙(s) < 0. Thus the ﬁrst-order ODE for r(s) is: √ 1 1 r˙ = 2M( − )1/2, r(0) = R > 2M. r R

7 Note that this ODE is regular (locally Lipschitz) vector field for r > 0, and there is nothing special about the ‘horizon’ r = 2M. It is easy to check that the solution is given parametrically by: r R R R r(η) = (1 + cos η), s(η) = (η + sin η), η ∈ (0, π). 2 2 2M Problem 8. Assume (for simplicity) r(0) = R = 4M. (i) Show that a radially free-falling massive particle (timelike geodesic) with these initial conditions arrives at the horizon (r = 2M) in finite proper time s0. (ii) Integrate the differential equation (conservation law) for t(s) to show that, in contrast, the Schwarzschild time t(s) → ∞ as s → s0 (from below). (iii) Now consider a massive particle in free-falling in the black hole region B. Say r(0) = R = M < 2M,r ˙(0) = 0. Show the particle crashes into the singularity r = 0 in finite proper time. Similar considerations apply to radial null geodesics (radially travelling photons.) Null radial geodesics. The differential equations for a radial null geodesic (θ ≡ π/2, φ ≡ φ0) are: h(r)t˙(s) ≡ E,E2 = (r ˙)2, E > 0.

Eitherr ˙ ≡ E (outgoing), and then r = r0 + Es (deﬁned for all s > 0), or r˙ ≡ −E, so r = r0 − Es (ingoing.) For the time coordinate in the exterior region N = {r > 2M}: 2M rE 2M t˙ = (1 − )−1E = = E(1 + ), r r − 2M r0 ± Es − 2M (+ for outgoing, - for ingoing). The solutions with t(0) = 0 are: r + 0 + Es − 2M t = Es + 2M ln( ), s > 0 (outgoing photon) . r0 − 2M r − 2M − Es r − 2M t = Es − 2M ln( 0 ), with t(s) → ∞ as s → 0 . r0 − 2M E

Problem 9. Analyze the solutions for radial null geodesics in the black hole region B. Show that the area radius is decreasing along any future- directed (radial) null geodesic starting in B, and tends to r = 0 (the singularity) in ﬁnite aﬃne parameter.