Trapping of bodies by the gravitational waves endowed with angular - - PowerPoint PPT Presentation

Trapping of bodies by the gravitational waves endowed with angular momentum

Iwo Bialynicki-Birula

Center for Theoretical Physics, Warsaw, Poland Humboldt Kolleg Vilnius Aug 2018

Prologue “What the principal investigator would like to be“trapped”? If these are elementary particles, then gravity is much too weak, as compared to other forces at those scales If they are massive bodies (e.g. stars or compact objects), then there are no physically motivated sufficiently strong gravitational waves to achieve this”

Electromagnetism Gravitation

Electromagnetism Gravitation

Angular momentum of a binary system

• f two equal masses

L = √ Gm3r = Gm2 c √ 2r rS Example: Black holes with 30 solar masses spiraling at the distance of 10 Schwarzschild radii have the angular momentum hundred thousand times the angular momentum

• f the Earth on its orbit around the Sun

Angular momentum luminosity for a binary system of two masses dL dt = 32 √ 2 5 G7/2 c5 m9/2 r7/2 = 4mc2 5(r/rS)7/2 Example: Black holes with 30 solar masses spiraling at the distance of 10 Schwarzschild radii emit per second six million times the angular momentum

• f the Earth on its orbit around the Sun

Bessel beams carry angular momentum Bessel beams in electromagnetism and gravity can be obtained as derivatives of the superpotential χλMqzq⊥(ρ, φ, z, t) = e−iλ(ωqt−qzz−Mφ)JM(q⊥ρ) Quantum numbers: λ - helicity ωq - energy M - z-component of total angular momentum qz - z-component of momentum q⊥ - transverse momentum

Geodesic motion

minert d2ξµ dτ 2 + mgravΓµ

αβ

dξα dτ dξβ dτ = 0 τ is the proper time Galileo-Einstein minert = mgrav Mass disappears Motion is universal d2ξµ dτ 2 + Γµ

αβ

dξα dτ dξβ dτ = 0

The simplest geodesic trajectory

Experts in numerical gravity claim that the dominant contribution to gravitational radiation emitted by inspiraling binaries comes from the M = 2 component of radiation and I will consider only this case Owing to the axial symmetry of Bessel beams there is one natural candidate for a geodesic: the z-axis

Uniform motion along the z-axis

4-velocity has the form: dξα dτ = {u0, 0, 0, u3} Relevant components of the Christoffel symbol vanish Γµ

αβ = 0 (α = 0, 3 β = 0, 3) ⇛ Γµ αβ

dξα dτ dξβ dτ = 0 Therefore, the motion is uniform d2ξµ dτ 2 = 0

Geodesic deviation

Is the geodesic trajectory along the z-axis stable? The answer is obtained from the equation of geodesic deviation d2ηµ dτ 2 = Rµ

αβν

dξα dτ dξβ dτ ην Rµ

αβν

Riemann curvature tensor

What is geodesic deviation?

ηi λ ξk λ

Reference geodesic is drawn in red

Riemann curvature tensor

R1

001 = −2k4 +BM−2 − 4k2 ⊥k2 +BM − 2k4 ⊥BM+2,

R1

002 = 2k4 +BM−2 − 2k4 ⊥BM+2,

R1

003 = −4k⊥k3 +BM−1 − 4k3 ⊥k+BM+1,

R2

002 = 2k4 +BM−2 − 4k2 ⊥k2 +BM − 2k4 ⊥BM+2,

· · · where BM = AJM(k⊥ρ) cos(k∥ + Mφ − ωt) A is the amplitude that determines the strength Looks very complicated, however. . .

Motion near the reference trajectory

Reference trajectory ξi(τ) along the z-axis d2η1 dt2 = −γω2(η1 cos(ωt) + η2 sin(ωt)), d2η2 dt2 = −γω2(η1 sin(ωt) − η2 cos(ωt)), d2η3 dt2 = 0.

In the frame rotating with ω/2 Coriolis force overcomes repulsion

d2η1 dt2 = (1/4 − γ)ω2η1 + ωdη2 dt d2η2 dt2 = (1/4 + γ)ω2η2 − ωdη1 dt Eigenfrequencies in the rotating frame: Ω± = ω √ 1/4 ± γ

In the inertial frame

The motion in the inertial frame is characterized by 4 frequencies: Ω±

± = ω(1/2 ±

√ 1/4 ± γ)

Conclusion Gravitational waves may carry with them all kinds of cosmic debris The trapping of these bodies near the beam axis is due to the Coriolis force