Trapping of bodies by thegravitational waves endowedwith 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

of 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

of 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

c5m9/2

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

of 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

minertd2ξµ

dτ 2+mgravΓ

µαβ

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

Γµαβ = 0 (α = 0, 3 β = 0, 3) V 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

αβνdξα

dτ

dξβ

dτην

Rµαβν Riemann curvature tensor

What is geodesic deviation?

ηi λ

ξk λ

Reference geodesic is drawn in red

Riemann curvature tensor

R1001 = −2k4

+BM−2 − 4k2⊥k

2+BM − 2k4

⊥BM+2,

R1002 = 2k4

+BM−2 − 2k4⊥BM+2,

R1003 = −4k⊥k

3+BM−1 − 4k3

⊥k+BM+1,

R2002 = 2k4

+BM−2 − 4k2⊥k

2+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/2Coriolis force overcomes repulsion

d2η1

dt2= (1/4− γ)ω2η1 + ω

dη2

dtd2η2

dt2= (1/4 + γ)ω2η2 − ω

dη1

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