Cylindrical gravitational waves in an expanding universe

Cylindrical gravitational waves in an expanding universe Surprise: Rogue Waves! Robert H. Gowdy Virginia Commonwealth University June 21, 2007 R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 1 / 31

Credits Einstein for his gravitational eld equations. 1915 Einstein and Rosen for the original cylindrical wave solutions. 1937 Doug Edmonds for helping with the paper. Phys. Rev. D15, 75, 084011 (2007) R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 2 / 31

The Plan Describing Spacetime Cylindrical coordinates and frames Cylindrically symmetric spacetimes Solving the Field Equations Specialized to this case The cylinder area function The wave and red-shift functions Properties of the new solutions Extendability Weber-Wheeler-Bonnor-style Pulses The Assumed Amplitudes The Easy Part of the Integration Elliptic Function Adventures Rogue Waves! R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 3 / 31

Describing Spacetime: A curved spacetime Here is a cylindrically symmetric, curved spacetime. d t = e a dt d x = e a dr dȳ = p Re ψ dφ d z = p Re ψ dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 4 / 31

Describing Spacetime: A curved spacetime Here is a cylindrically symmetric, curved spacetime. d t = e a dt d x = e a dr dȳ = p Re ψ dφ d z = p Re ψ dz R determines the area element: dȳd z = Rdφdz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 4 / 31

Describing Spacetime: A curved spacetime Here is a cylindrically symmetric, curved spacetime. d t = e a dt d x = e a dr dȳ = p Re ψ dφ d z = p Re ψ dz R determines the area element: dȳd z = Rdφdz a determines the gravitational red-shift. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 4 / 31

Describing Spacetime: A curved spacetime Here is a cylindrically symmetric, curved spacetime. d t = e a dt d x = e a dr dȳ = p Re ψ dφ d z = p Re ψ dz R determines the area element: dȳd z = Rdφdz a determines the gravitational red-shift. ψ describes the gravitational wave. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 4 / 31

Solving the Field Equations: Specialized to this case d t = e a dt, d x = e a dr, dȳ = p Re ψ dφ, d z = p Re ψ dz Note that the advanced and retarded time coordinates u = t + r v = t r are always lightlike in these spacetimes. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 5 / 31

Solving the Field Equations: Specialized to this case d t = e a dt, d x = e a dr, dȳ = p Re ψ dφ, d z = p Re ψ dz Note that the advanced and retarded time coordinates u = t + r v = t r Use derivatives with respect to these variables to simplify the eld equations. R + = R u, R = R v are always lightlike in these spacetimes. R + = 2 R u v = 2 R t 2 2 R r 2 R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 5 / 31

Solving the Field Equations: Specialized to this case d t = e a dt, d x = e a dr, dȳ = p Re ψ dφ, d z = p Re ψ dz Einstein s Field Equations yield these di erential equations for the functions a, R, ψ Two of the functions satisfy wave equations R + = 2 R u v = 2 R t 2 Rψ + + Rψ + = 0 2 R r 2 = 0 while the remaining function, a is found by integrating the equations R + a + = Rψ 2 + + 1 2 R ++ and 1 R+ 2 4 R R a = Rψ 2 + 1 2 R 1 R 2 4 R R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 6 / 31

Solving the Field Equations: The Area Function The area element dȳd z = Rdφdz determines what these cylinders actually look like. Choose a solution of the vibrating string equation: 2 R 2 R = 0 t 2 r 2 R = r! cylindrical waves. R = t! expanding universe. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 7 / 31

Solving the Field Equations: The Area Function The area element dȳd z = Rdφdz determines what these cylinders actually look like. Choose a solution of the vibrating string equation: 2 R 2 R = 0 t 2 r 2 R = r! cylindrical waves. R = t! expanding universe. R = sin r sin t! closed S 3 and S 1 S 2 universes. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 7 / 31

Solving the Field Equations: The Area Function The area element dȳd z = Rdφdz determines what these cylinders actually look like. Choose a solution of the vibrating string equation: 2 R 2 R = 0 t 2 r 2 R = r! cylindrical waves. R = t! expanding universe. R = sin r sin t! closed S 3 and S 1 S 2 universes. New choice: R = rt! cylindrical waves in an expanding universe. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 7 / 31

Solving the Field Equations: The Waves With the new choice for R d t = e a dt d x = e a dr dȳ = p rte ψ dφ d z = p rte ψ dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 8 / 31

Solving the Field Equations: The Waves With the new choice for R d t = e a dt d x = e a dr dȳ = p rte ψ dφ d z = p rte ψ dz Compare to at spacetime in the same coordinates. d t = dt d x = dr dȳ = rd φ d z = dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 8 / 31

Solving the Field Equations: The Waves With the new choice for R d t = e a dt d x = e a dr dȳ = p rte ψ dφ d z = p rte ψ dz The function ψ obeys the wave equation 1 t t t ψ t 1 r r r ψ r = 0 Compare to at spacetime in the same coordinates. d t = dt d x = dr dȳ = rd φ d z = dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 8 / 31

Solving the Field Equations: The Waves With the new choice for R d t = e a dt d x = e a dr dȳ = p rte ψ dφ d z = p rte ψ dz Compare to at spacetime in the same coordinates. The function ψ obeys the wave equation 1 t t t ψ t Pull out a singular solution ψ = W 1 r r r ψ r = 0 1 ln (r/t) 2 d t = dt d x = dr dȳ = rd φ d z = dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 8 / 31

Solving the Field Equations: The Waves With the new choice for R d t = e a dt d x = e a dr dȳ = p rte ψ dφ d z = p rte ψ dz Compare to at spacetime in the same coordinates. The function ψ obeys the wave equation 1 t t t ψ t Pull out a singular solution ψ = W 1 r r r ψ r = 0 1 ln (r/t) 2 d t = dt d x = dr dȳ = rd φ d z = dz W obeys the same equation. 1 t t t W t 1 r r r W r = 0 R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 8 / 31

Solving the Field Equations: The Waves In terms of W d t = e a dt d x = e a dr dȳ = re W dφ d z = te W dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 9 / 31

Solving the Field Equations: The Waves In terms of W d t = e a dt d x = e a dr dȳ = re W dφ d z = te W dz Compare to at spacetime in the same coordinates. d t = dt d x = dr dȳ = rd φ d z = dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 9 / 31

Solving the Field Equations: The Waves In terms of W d t = e a dt d x = e a dr dȳ = re W dφ d z = te W dz Compare to at spacetime in the same coordinates. The constant-z surfaces are regular distortions of at 3D Minkowski spacetime. d t = dt d x = dr dȳ = rd φ d z = dz R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 9 / 31

Solving the Field Equations: The Waves In terms of W d t = e a dt d x = e a dr dȳ = re W dφ d z = te W dz Compare to at spacetime in the same coordinates. d t = dt d x = dr dȳ = rd φ d z = dz The constant-z surfaces are regular distortions of at 3D Minkowski spacetime. There is expansion along the z direction from an initial Big Bang at t = 0. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 9 / 31

Solving the Field Equations: The Waves The detailed solutions look like this: Z W = dk [A (k) J 0 (kt) + C (k) N 0 (kt)] J o (kr) where J 0 and N 0 are Bessel functions of the rst and second kind. The amplitude functions need to obey the constraint Z dk C (k) = 0 or π k so that the remaining equations will have a consistent solution for the red-shift function a. where a (x, t) = W (x, t) + Z = Wr 2 + Wt 2 Z x 0 rdr t 2 r 2 t2 Z 2 r t W r W t 2 r t 2 W r + 2 t W t R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 10 / 31

Properties of the Solutions: Extendability The background geometry d t = dt d x = dr dȳ = rd φ d z = tdz corresponds to Minkowski coordinates T = t cosh z X = r cos ϕ Y = r sin ϕ Z = t sinh z R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 11 / 31

Properties of the Solutions: Extendability The background geometry d t = dt d x = dr dȳ = rd φ d z = tdz Our solutions correspond to the shaded area. corresponds to Minkowski coordinates T = t cosh z X = r cos ϕ Y = r sin ϕ Z = t sinh z R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 11 / 31

Properties of the Solutions: Extendability The background geometry d t = dt d x = dr dȳ = rd φ d z = tdz Our solutions correspond to the shaded area. corresponds to Minkowski coordinates T = t cosh z X = r cos ϕ Y = r sin ϕ Z = t sinh z Translation along the z-axis corresponds to a Lorentz boost in the underlying Minkowski space. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 11 / 31

Properties of the Solutions: Extendability In some cases (certainly for a = W = 0) there is a smooth extension through the apparent singularity at t = 0. Those cases correspond to the boost-rotation symmetric solutions analyzed by Biµcak and Schmidt, Phys. Rev. D 40, 1827 (1989). R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 12 / 31

Weber-Wheeler-style Pulses: The Assumed Amplitudes The formal solution W (r, t) = Z 0 dk [A (k) J 0 (kt) + C (k) N 0 (kt)] J o (kr) is only a starting point since we need to choose amplitude functions A, C and then do the integral. The function C is subject to a constraint, which we can solve trivially by taking it to be zero. For the other amplitude, try the suggestion (due to Bonnor?) A (k) = A 0 αe αk that leads to the Weber-Wheeler pulse solution when it is applied to Einstein-Rosen waves. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 13 / 31

Weber-Wheeler-style Pulses: The Assumed Amplitudes The formal solution W (r, t) = Z 0 dk [A (k) J 0 (kt) + C (k) N 0 (kt)] J o (kr) is only a starting point since we need to choose amplitude functions A, C and then do the integral. The function C is subject to a constraint, which we can solve trivially by taking it to be zero. For the other amplitude, try the suggestion (due to Bonnor?) A (k) = A 0 αe αk that leads to the Weber-Wheeler pulse solution when it is applied to Einstein-Rosen waves. The task is to do the integral W 0 (α, r, t) = α Z 0 e αk J 0 (kt) J o (kr) dk R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 13 / 31

Weber-Wheeler style Pulses: The Easy Part of the Integration Use integral representations of the Bessel functions: J 0 (z) = 1 Z π e iz sin φ dφ 2π π to get W 0 into the form W 0 = α Z π Z π Z 4π 2 dζ dξ π π 0 k( α+it sin ζ+ir sin ξ) dke The integral over k is trivial and the subsequent integral over dξ forces us to think about the behavior of the arctangent function in the complex plane but eventually comes down to with W 0 = 1 2πr w = α r Z π π α p 1 + w 2 dζ i t r sin ζ R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 14 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The nal integral can be gotten into the general form of an elliptic integral: W 0 = α π p r 2 + α 2 Z 1 1 dy p (1 y) (1 + y) (1 ρ y) (1 + ρy) where ρ = r t iα The Legendre Normal Form of the Elliptic function looks like this: K k 2 = 1 2 Z 1 1 dx p (1 x) (1 + x) (1 kx) (1 + kx) The general form of an elliptic integral is preserved by a bilinear transformation. We need to nd one that brings the integral into the Legendre Normal Form, with all of the branch points on the real axis. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 15 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The needed bilinear transform is y = x + ih ihx + 1 which preserves the branch points at x = 1. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 16 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The needed bilinear transform is y = x + ih ihx + 1 which preserves the branch points at x = 1. To see where the modi ed integration path is going, note that Im (y) = h h 2 x 2 + 1 1 x 2 For h < 0 it arcs below the real axis for the whole integration interval. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 16 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The needed bilinear transform is y = x + ih ihx + 1 which preserves the branch points at x = 1. To see where the modi ed integration path is going, note that Im (y) = h h 2 x 2 + 1 1 x 2 For h < 0 it arcs below the real axis for the whole integration interval. We would really like h (x) < 0 for x between 1 and +1. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 16 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The function that brings the branch points onto the axis is h = 1 r 2 t 2 + α 2 q(r 2tα 2 t 2 + α 2 ) 2 + 4t 2 α 2 R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 17 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The function that brings the branch points onto the axis is h = 1 r 2 t 2 + α 2 q(r 2tα 2 t 2 + α 2 ) 2 + 4t 2 α 2 To keep the integration path below the axis (and away from the branch points), choose h = 1 r 2 t 2 + α 2 q (r 2tα 2 t 2 + α 2 ) 2 + 4t 2 α 2 The elliptic modulus, k then becomes m = k 2 r 2 + α 2 = r 2 + (α ht) 2 h + tα! 2 r r 2 + α 2 + 2 t 2 (r 2 + α 2 ) 2 which is real, positive, and less than or equal to 1, exactly as it should be. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 17 / 31

Weber-Wheeler-style Pulses: Elliptic Adventures The gravitational wave amplitude is then W 0 = AK (m) where s A (α, r, t) = 2α 1 + h 2 π (h 2 t 2 2htα + r 2 + α 2 ) m = k 2 r 2 + α 2 = r 2 + (α ht) 2 h + tα! 2 r r 2 + α 2 + 2 t 2 (r 2 + α 2 ) 2 and, as we saw, h = 1 2tα r 2 t 2 + α 2 q (r 2 t 2 + α 2 ) 2 + 4t 2 α 2 R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 18 / 31

Rogue Waves The elliptic function K that appears in the gravitational wave function W is most familiar to physicists as the period of a circular pendulum. Notice that the function diverges suddenly near m = 1. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 19 / 31

Rogue Waves Now check out what the argument of the elliptic function is doing: It is approaching 1 where r = t so K (m) will diverge there. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 20 / 31

Rogue Waves: Near- eld Zone Here is what the gravitational wave amplitude looks like in the near- eld zone just a few pulse-widths out: It is a very broad, well-behaved pulse. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 21 / 31

Rogue Waves: Far- eld Behavior Both the asymptotic forms of the Bessel functions and the argument from the area function, rt suggest that the wave amplitude should fall o as 1/ p rt. Here is the wave amplitude multiplied by p rt. Most of the product is leveling o as it should, but the peak keeps growing and sharpening. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 22 / 31

Rogue Waves: Far- eld Behavior Here is the wave amplitude multiplied by p rt much farther out. The peak is growing into what appears to be a spike. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 23 / 31

Rogue Waves: Asymptotic Behavior The asymptotic behavior of the solution can be found by switching to the variables s = p 1 r t, q = rt r r = 1 sq, t = q s and looking for the dominant terms at the wave peak (q = 1) near s = 0. The gravitational wave amplitude there has the dominant term W 0 1 π s ln s which corresponds to ln r r at the peak of the pulse. For q 6= 1 the amplitude is proportional to s, corresponding to the expected 1 r behavior everywhere else. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 24 / 31

Rogue Waves: Asymptotic Behavior The asymptotic behavior of the solution can be found by switching to the variables s = p 1 r t, q = rt r r = 1 sq, t = q s and looking for the dominant terms at the wave peak (q = 1) near s = 0. The gravitational wave amplitude there has the dominant term W 0 1 π s ln s which corresponds to ln r r at the peak of the pulse. For q 6= 1 the amplitude is proportional to s, corresponding to the expected 1 r behavior everywhere else. We have an isolated logarithmic singularity at in nity. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 24 / 31

Snapshots: The near- eld of a rogue pulse Notice that the pulse is leaving a lot of amplitude behind as it moves outward. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 25 / 31

Snapshots: The shape of the rogue peak The shape does not change. The wave grows but does not become a spike. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 26 / 31

Snapshots: The rogue growth Notice that the rogue pulse has exactly 1/t behavior at the origin and has large amplitude far from its peak. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 27 / 31

Good Waves We can still get solutions that are well-behaved at in nity. If W 0 (r, t) is a solution, then so are the scaled solutions bw 0 (br, bt) and the di erences U b (r, t) = bw 0 (br, bt) W 0 (r, t) In these di erence solutions, the s ln s term cancels out. Thus, we can still build a variety of pulses that are good models for waves from a compact source. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 28 / 31

Snapshots: The near- eld of a good pulse These pulses move out and leave nothing behind. Their length scale (α = 1) is roughly their half-width. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 29 / 31

Snapshots: The far- eld of a good pulse A compact pulse with no precursor, leaves nothing behind, and falls o as 1/r. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 30 / 31

Discussion Rogue waves represent a kind of instability in these expanding, open universes. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 31 / 31

Discussion Rogue waves represent a kind of instability in these expanding, open universes. Do they occur in more realistic cases such as the DeSitter universe? R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 31 / 31

Discussion Rogue waves represent a kind of instability in these expanding, open universes. Do they occur in more realistic cases such as the DeSitter universe? Good pulses are formed by cancelling the scale-invariant part of the solution, U b (r, t) = bw 0 (br, bt) W 0 (r, t), suggesting that the underlying instability is a self-similar gravitational wave. The original motivation of this work was to nd an exact solution model to test how well numerical simulations handle the transition to far- eld behavior. The good pulses will work but one needs to be aware of the rogue-wave instability. R. H. Gowdy (VCU) Still more Einstein-Rosen solutions 06/21 31 / 31