Interacting KerrInteracting Kerr--Newman fieldsNewman fields

Kjell Rosquist

MG12, Paris 2009

(with Lars Samuelsson and Mikael von Strauss)

Main points of talk

� Microscopic gravity dominated by non-Newtonian spin effects

� Gravitational field induces electromagnetic effects at Compton scale

(even though the gravitational force is small)

� Experimental data + GR

� magnetic force is possible source of the nuclear force

Kerr-Newman metric: ds2 = −h(r)(M 0)2 + h(r)−1(M1)2 + (M 2)2 + (M 3)2

h(r) =1−2GMr −GQ2

r2 + a2

where ds2 = −(M0)2 + (M1)2 + (M 2)2 + (M 3)2

is the Minkowski metric in

Boyer-Lindquist coordinates

M 0 =ρ0

ρ(dt − asin2θdφ)

M1 =ρρ0

dr

M 2 = ρdθ

M 0 =sinθ

ρ(−adt + ρ0

2dφ)

ρ2 = r2 + a2 cos2θ , ρ0

2 = r2 + a2

Kerr-Newman geometry

Kerr-Newman objects

(1) mass (M)

(2) angular momentum per unit mass (a = S /M)

(3) charge (Q)

(4) magnetic dipole moment per unit charge (Θ = µ /Q = a)

Characterized by:

Black hole uniqueness theorem ⇒

Only three independent parameters for stationary black holes: M, a, Q, a2 +Q2 ≤ M 2

Only possible black hole in Einstein- Maxwell theory is the Kerr - Newman solution

Then g = 2 and all higher moments (infinitely many) are fixed

Note : the g - factor is defined by g = 2Θa

a2 +Q2 < M 2 : underextreme black hole

a2 +Q2 = M 2 : extreme black hole

a2 +Q2 > M 2 : overextreme case, no horizon, no ergo region (⇒ not a black hole)

"weak" ring singularity, two asymptotically flat regions

Angular momentum vs. mass

K. Rosquist -- May 11, 2009 6

electron (1044)

proton (1038)

CD disk (1019)

Earth (700)

Sun/neutron star (0.2)

solar system (40)

galaxy (1?)

a/M

M (cm)

black hole region: a<M

Assorted spin values

� a/M typically larger for smaller size systems

Kerr-Newman electromagnetic field

K. Rosquist -- May 11, 2009 7

EM field tensor in ON frame expressed

in Boyer-Lindquist coordinates

Characteristic distance a = S/M (spin

radius)

The Einstein-Maxwell equations

Rµν = κ FµαFνβ gαβ − 1

4gµν g

αβgγδFαγFβδ( )

Try Coulomb for RHS and

Schwarzschild for LHS

⇒ Rµν = 0 but RHS ≠ 0

Reissner-Nordström is solution but has no magnetic moment

Instead, try preferred Einstein-Maxwell solution: Kerr-Newman

Question:

What are the external gravitational and EM

fields of the electron?

Moments (known) of the electron

• Charge (e)

• Magnetic dipole (µ)• Mass (m)

• Spin (gravitomagnetic dipole) (S=ma)

K. Rosquist -- May 11, 2009 10

g - factor : g = 2 +ε (QED correction : ε ~ 10−3)

a >> e >> m → overextreme (not BH!)

S = h /2 → a = h /2m = 12λC (half reduced Compton wave length)

Main reasons for Kerr-Newman as candidate

Einstein-Maxwell field for electron

� Unique Einstein-Maxwell field with additional conserved

quantity (Carter’s constant). Necessary for QM treatment

of hydrogen atom e.g. (cf. Gair 2002)

� Finite Lagrangian in G � 0 limit (cf. Coulomb divergence)

(to be discussed later)

� Can model all four known moments (Q,µ,m,S) with correct classical g-factor (g=2)

K. Rosquist -- May 11, 2009 11

Nuclear force characteristics

o Proton size: rp = 1 fm (= 10-13 cm)

o Range of nuclear force ~ 2 fm

o Deuteron binding energy: Eb = 2.2 MeV

K. Rosquist -- May 11, 2009 15

Electromagnetic vs.nuclear force

Combined Coulomb and magnetic dipole-dipole interaction

Q = e, µ = µproton (proton-proton)

Range of nuclear force: ~ 2 fm

Proton radius: 1 fm

Deuteron binding energy: 2.2 MeV

� elementary, but little known fact:

Range and strength of magnetic force ≈ range and strength of nuclear force

The Kerr-Newman EM field (G=0)

=Q

x2 + y2 + (z − ib)2

Spheroidal coordinates Cartesian coordinates

The KN EM field can be obtained from the Coulomb field by a

complex displacement of the origin [Newman]

The Kerr-Newman EM Lagrangian

L = 18π (B

2 − E 2)dV∫

E = F10 =Q(r 2 − a 2 cos2θ)

r 2 + a 2 cos2θ( )2B = F23 =

2Qar cosθ

r 2 + a 2 cos2θ( )2

Note: L = 0 if a≠ 0 (if a = 0 then L = ∞)

The spin regularizes the field! No spin → “infinite field”

Interacting Kerr-Newman “particles”

Two interacting fields/particles with aligned spins in G � 0 limit

Fµ (x) = fµ (x) + ′ f µ ( ′ x ) = fµ (x) + Lµλ (v) fλ( ′ x )

( ′ x , ′ y , ′ z ) = (x,y,z) + (sx,sy,sz )

′ x µ = xµ + sµ

sµ = (0,sx,sy ,sz)

Lµλ is a boost with velocity v =

ds

dt

EM field Fµ = Eµ + iBµ

� To include non-aligned spins requires also a rotation involving two

angles describing the direction of the spin

� Non-standard kinetic energy

� Must go to next higher order in G (linearized gravity) for dynamics

The potential of two interacting Kerr-

Newman particles

Setting v = 0 in the Lagrangian gives the potential:

Fµ (x) = fµ (x) + ′ f µ ( ′ x ) = fµ (x) + fµ ( ′ x )

′ x µ = xµ + sµ

sµ = (0,s x ,s y ,sz )

V = Lv=0

= 18π Re(FµF

µ )∫ d 3x = 18π Re ( fµ + ′ f µ )( f µ + ′ f µ )[ ]∫ d 3x

= 18π Re( fµ f

µ )∫ d 3x

=0 for KN field

1 2 4 4 4 3 4 4 4 + 1

8π Re( ′ f µ ′ f µ )∫ d 3x

=0 for KN field

1 2 4 4 4 3 4 4 4 + 1

4π Re( fµ ′ f µ )∫ d 3x

interaction potential, V (s)

1 2 4 4 4 3 4 4 4

The interaction potential for aligned spins

Two interacting Kerr-Newman fields, (Q, a) and (Q’, a’),

aligned along a common axis of symmetry

V =Q ′ Q z

z2 + (a + ′ a )2

z = relative distance between the two fields,along spin axis in this case

Interacting identical

Kerr-Newman “particles”

Interacting

Kerr-Newman – anti-Kerr-Newman

“Proton” - “charged Neutron” potential

(“deuteron”)

Deuteron binding energy

Aligned vs. anti-aligned interaction

Anti-aligned configuration more attractive

Same behavior as nuclear force in this respect

Concluding remarks

• Kerr-Newman geometry � finite self-interaction

• Microscopic gravity dominated by spin

• Gravitational field relevant at Compton scale

• Experimental data + GR

� magnetic force is possible source of nuclear force

K. Rosquist -- May 11, 2009 27