Spin as a Manifestation of a Nonlinear Constitutive Tensor and a Non-Riemannian Geometry Fredrick W. Cotton

http://sites.google.com/site/fwcotton/em-25.pdf

[email protected]

Abstract

The electromagnetic constitutive tensor can be used to introduce a classical form of spin. For charged particles at rest, with no external forces, the spin does not aect the energy density of the particle. The particle solutions satisfy a non-Riemannian form of the Einstein-Maxwell equations.

© 2013 Fredrick W. Cotton, rev. 21 October

PACS numbers: 03.50.De, 04.40.Nr, 04.50.Kd

1. Introduction

The 4-dimensional electromagnetic constitutive tensor can be combined with a generalized stress-energy tensor to describe electric monopole particles with a classical form of spin. We will begin by looking at the problem in a 3-dimensional notation in order to develop a physical understanding of the role of the constitutive tensor. In particular, there are terms which do not aect the energy density. We will construct particular solutions which have some of the properties required for the elementary particles. We will then proceed to a 4-dimensional notation and will show that the solutions satisfy a generalized form of the Einstein-Maxwell equations. The spin portion of the electromagnetic stress-energy tensor will be equal to non-Riemannian curvature terms arising from a general symmetric connection. There are also electromagnetic wave solutions that couple only to the non-Riemannian part of the curvature tensor. 2. Maxwell's Equations in 3-Dimensions

Maxwell's equations can be written in 3-dimensions, using SI units, as: Ei = −ϕ;i − ∂t Ai Di = ϵij E − γji B j

B i = εijk Ak;j j

j

Hi = αij B + γij E

j

j i = εijk Hk;j − ∂t Di

i

ρ = D ;i

(2.1a) (2.1b) (2.1c)

where εijk is the Levi-Civita tensor and αij is the inverse permeability. In free space, with metric gij , ϵij = ϵ0 gij

αij = µ−1 0 gij

c2 ϵ0 µ0 = 1

(2.2)

The following vector-dyadic notation will also be useful: D =ϵ·E−B·γ

H =α·B+γ·E

(2.3)

The γij have not traditionally been written explicitly in classical electromagnetic theory. However, they arise from the fact that, in the 4-dimensional formulation, the constitutive relations are described by a fourth rank tensor. 1

We will generalize the traditional denitions of the energy density, the symmetric stress tensor and the Poynting vector. En = 12 (αij B i B j + ϵij E i E j ) − Q T

ij i

=

N =

− 12 (E i Dj + E j Di + H i B j 1 ijk (Ej Hk + c2 Dj Bk ) 2ε

j

i

+H B )+

m n 1 ij 2 g (αmn B B

m

n

ij

+ ϵmn E E ) + g Q

(2.4a) (2.4b) (2.4c)

The reason for these denitions is to make the 4-dimensional stress-energy tensor symmetric, thus ensuring that angular momentum is conserved and that gravitation can be included via a symmetric gravitational stress-energy tensor. The function Q will be chosen so that the particle solutions are force-free. The determination of Q has to be done for each particle separately. T ij is dened with the opposite sign from what is usually used in 3-dimensions. It is useful because it lets T ij be the spatial part of T µν , which is dened so that T 44 = −En . Since the energy density does not include any contribution from the γij terms, they can represent internal degrees of freedom. We will dene the force density and the power loss density. Fi = −Ti j;j − c−2 ∂t Ni

Pwr = −N i;i − ∂t En

(2.5)

In this paper, we will show that time-independent solutions for which B = 0 and γ ̸= 0 can be used to represent particles with spin. 3. Electric Monopole Solutions With Spin

In spherical coordinates (r, θ, φ) , let E = fe (r)er = −er ϕ′ (r) A=0 α = c2 ϵ = c2 ϵ0 fϵ (r)(er er + eθ eθ + eφ eφ ) γ = h(r)[(2er er − eθ eθ − eφ eφ ) cos(θ) + (er eθ + eθ er ) sin(θ))]

(3.1a) (3.1b) (3.1c) (3.1d)

Then D = ϵ0 fϵ (r)fe (r)er ρ(r) = ϵ0 r

−2

{∂r [r fe (r)fϵ (r)]} 2

H = fe (r)h(r)[2 cos(θ)er + sin(θ)eθ ] j = r−1 {2fe (r)h(r) + ∂r [rfe (r)h(r)]} sin(θ)eφ ∫ ∞ Q(r) = 12 ϵ0 fe2 (r)fϵ (r) − 2ϵ0 dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ )

(3.2a) (3.2b) (3.2c) (3.2d) (3.2e)

r

T = 12 ϵ0 fe2 (r)fϵ (r)[−er er + eθ eθ + eφ eφ ] + Q(r)[er er + eθ eθ + eφ eφ ] ∫ ∞ En(r) = 12 ϵ0 fe2 (r)fϵ (r) − Q(r) = 2ϵ0 dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ )

(3.2f) (3.2g)

r

(3.2h)

N = 12 h(r)fe2 (r) sin(θ)eφ

For continuously dierentiable functions, these solutions are force free and radiationless. At r = 0 , we must have

fe (0) = 0 . We must also have

lim fe (r) = q(4πϵ0 r2 )−1

(3.3a)

lim fϵ (r) = 1

(3.3b)

r→∞

r→∞

lim H = γ(4πr3 )−1 [2 cos(θ)er + sin(θ)eθ ]

r→∞

The rest mass

∫ π ∫ 2π dθ sin(θ) dφ En(r) dr r2 0 0 ∫ ∞ 0 ∫ ∞ dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ ) dr r2 = 8πϵ0 c−2

m0 = c−2

∫

∞

0

© 2013 Fredrick W. Cotton

(3.3c)

2

(3.4)

r

http://sites.google.com/site/fwcotton/em-25.pdf

Since er = sin(θ) cos(φ)ex + sin(θ) sin(φ)ey + cos(θ)ez eθ = cos(θ) cos(φ)ex + cos(θ) sin(φ)ey − sin(θ)ez

(3.5)

eφ = − sin(φ)ex + cos(φ)ey

the total angular momentum J T = c−2

∫

∫

π

dr r2

∫

0

−2 4 3 πc

=

∫

∞

2π

dφ r × N

dθ sin(θ) 0

∞

dr r

3

0

(3.6)

h(r)fe2 (r)ez

0

The total current and the total angular moment of the current are dened by ∫

jT =

∫

∞

dr r

∫

0

0

dr r ∫ ∞

∫

π

2

8 3 πez

=

dφ j

0 ∞

2π

dθ sin(θ)

0

=0 ∫ MT =

∫

π

2

2π

dφ r × j

dθ sin(θ) 0

(3.7a)

0

(3.7b)

dr r {∂r [rfe (r)h(r)] + 2fe (r)h(r)} 2

0

If there are external elds with potentials ϕext = −(E0x x + E0y y + E0z z) Aext = 12 [(B0y z − B0z y)ex + (B0z x − B0x z)ey + (B0x y − B0y x)ez ]

(3.8)

then we have the constant elds E 0 = E0x ex + E0y ey + E0z ez

B 0 = B0x ex + B0y ey + B0z ez

(3.9)

If we assume that the external elds do not, to a rst approximation, modify fϵ (r) , h(r) and Q(r) , then the total force and the total torque are ∫

FT =

∫

∞

dr r 0

∫

π

2

2π

dθ sin(θ) 0

dφ F 0

∞

= 4πϵ0 r2 fϵ (r)fe (r) |r=0 E 0 = qE 0 ∫ ∞ ∫ π ∫ 2π 2 WT = dr r dθ sin(θ) dφ r × F 0 0 0 ∞ = 2πr3 h(r)fe (r) r=0 ez × B 0 =

1 2 γez

(3.10a)

(3.10b)

× B0

The factor of 12 in W T distinguishes this result from the normal magnetic dipole, W T = µm ez × B . Thus the numerical values for γ are related to the numerical values reported for µm by (3.11)

|γ| = 2 |µm |

We can dene an eective rest mass energy for the particle by subtracting the unperturbed energy density of the external eld. In this case, referring back to the general denition in (2.4a), ∫

me c2 =

∞

∫

dr r2

∫

π

2

= m0 c +

dφ [En − 12 ϵ0 (B02 c2 + E02 )]

∫

0

0

2π

dθ sin(θ) 0

2πϵ0 (B02 c2

+

∞

E02 )

(3.12)

dr r [fϵ (r) − 1] 2

0

© 2013 Fredrick W. Cotton

3

http://sites.google.com/site/fwcotton/em-25.pdf

Even though the particle is at rest, we can dene an eective total eld momentum and an eective total angular momentum by subtracting the unperturbed Poynting vector at innity. In this case, (3.13a)

N 0 = c2 ϵ0 [q(4πϵ0 r2 )−1 er + E 0 ] × B 0 ∫ ∞ ∫ π ∫ 2π P e = c−2 dr r2 dθ sin(θ) dφ (N − N 0 ) 0 0 0 ∫ ∞ = 4πϵ0 E 0 × B 0 dr r2 [fϵ (r) − 1] J e = c−2

∫

∞

∫

0

∫

π

(3.13b)

2π

dφ r × (N − N 0 ) ∫ ∞ ∫ ∞ dr r3 h(r)fe2 (r)ez + 32 B 0 dr r[q − 4πϵ0 r2 fϵ (r)fe (r)] = 34 πc−2 dr r2

dθ sin(θ)

0

0

0

0

(3.13c)

0

Thus we can dene the angular momentum about another point P as (3.14)

J P = J e + r P × P e

where rP is the radius vector from the point P to the center of the particle. In standard quantum theory, this is valid for orbital angular momenta, but not for spin. The total current and the total angular moment of the current are jT = 0

∫

M T = 83 πez

∞ 0

∫ dr r2 {∂r [rfe (r)h(r)] + 2fe (r)h(r)} − 38 πϵ0 c2 B 0

∞

(3.15a) dr r3 fϵ′ (r)

(3.15b)

0

In classical theory, every static magnetic eld is associated with a current of moving charges. In quantum theory, static magnetic elds are associated either with a current of moving charges or with a xed array of particles that have spin. In this theory, B = 0 for spin elds, but not for moving charges. This would seem to be one reason why it has been dicult to explore the mathematical transition between quantum theory and classical theory. For a particle moving with a uniform velocity in an inertial frame, a transformation of the elds to the rest frame of the particle will show that the Lorentz force law will be modied if the source of the external magnetic eld is due to spins rather than to moving charges. 4. Particular Solutions

For any given total charge q , it is obviously possible to construct particles with many dierent rest masses. As an example of solutions for which fe (r) approaches q(4πϵ0 r2 )−1 exponentially, we can look at fe (r) = q(4πϵ0 )−1 {1 − exp[−(r/r0 )3 ]}r−2 3

fϵ (r) = 1 + λ exp[−(r/r0 ) ] h(r) = γϵ0 q −1 {1 − exp[−(r/r0 )3 ]}r−1

(4.1a) (4.1b) (4.1c)

Then m0 c2 = q 2 (6πϵ0 r0 )−1 [2 − 21/3 + (−1 + 24/3 − 31/3 )λ]Γ(2/3) 2

J T = γq[4c πϵ0 r0 ] 2

me c = P e =

−1

(1 − 2

1/3

2

m0 c + 23 πϵ0 λr03 (B02 c2 3 4 3 πϵ0 λr0 E 0 × B 0

+3

+

−2/3

)Γ(2/3)ez

E02 )

J e = J T + 19 [2 + (−2 + 21/3 )λ]qr02 Γ(2/3)B 0 M T = 23 γez + 38 πϵ0 c2 λr03 B 0

© 2013 Fredrick W. Cotton

4

(4.2a) (4.2b) (4.2c) (4.2d) (4.2e) (4.2f)

http://sites.google.com/site/fwcotton/em-25.pdf

If we set the z -component of J T to 12 ~ and use (3.11), (4.2a) and (4.2b) with λ = −1 , then we obtain the standard magneton result, 2µm = γ = q~m−1 0 . Adding additional terms to the example above makes it possible to satisfy the conditions me = m0 , P e = 0 and J e = J T . fe (r) = q(4πϵ0 r2 )−1 {1 − exp[−(r/r0 )3 ]} 3

6

9

3

fϵ (r) = 1 + [λ0 + λ1 (r/r0 ) + λ2 (r/r0 ) + λ3 (r/r0 ) ] exp[−(r/r0 ) ] h(r) = γϵ0 q −1 {1 − [1 + λh (r/r0 )3 ] exp[−(r/r0 )3 ]}r−1 λ2 = [324 + 4(−41 + 38 · 21/3 )λ0 + (−56 + 44 · 21/3 )λ1 ][5(8 − 5 · 21/3 )]−1 λ3 = − 16 (λ0 + λ1 + 2λ2 )

(4.3a) (4.3b) (4.3c) (4.3d) (4.3e)

m0 c2 = q 2 Γ(2/3)(8748πϵ0 r0 )−1 × [1458(2 − 21/3 ) + 1458(−1 + 2 · 21/3 − 31/3 )λ0 + 162(3 − 3 · 21/3 + 31 /3 )λ1 + 18(18 − 9 · 21/3 + 2 · 31/3 )λ2 + 5(108 − 27 · 21/3 + 4 · 31/3 )λ3 ] −1

2

J T = γqΓ(2/3)(108c πϵ0 r0 ) MT =

(3 − 3 · 2

1/3

+3

1/3

)(9 − λh )ez

2 3 γez

(4.3f) (4.3g) (4.3h)

An example of a neutral particle for which F T = 0 and W T = 0 and for which me = m0 , P e = 0 and

J e = J T is

fe (r) = βr exp(−r/r0 ) 2

3

fϵ (r) = 1 + [λ0 + λ1 (r/r0 ) + λ2 (r/r0 ) + λ3 (r/r0 ) ] exp(−r/r0 ) h(r) = ηr exp(−r/r0 ) λ3 = λ2 = m0 c2 =

1 90 (256 + 3λ0 + 5λ1 ) 1 − 36 (512 + 9λ0 + 19λ1 ) 2 2β πϵ0 r05 (−45853 + 1104λ0 320πβ 2 ηr07 (729c2 )−1 ez

+ 880λ1 )(19683)−1

JT = MT = 0

(4.4a) (4.4b) (4.4c) (4.4d) (4.4e) (4.4f) (4.4g) (4.4h)

There are, as yet, no quantization conditions to specify the allowable solutions. 5. Maxwell's Equations in 4-Dimensions

The electromagnetic elds and the current density are dened by (5.1a) (5.1b) (5.1c)

fµν = Aν ,µ − Aµ,ν = Aν ;µ − Aµ;ν µν

p = jµ =

1 µνρσ fρσ 2χ pµν ;ν

where fµν and pµν are antisymmetric and the constitutive tensor, χµνρσ , has the symmetries χµνρσ = −χνµρσ

χµνρσ = −χµνσρ

χµνρσ = χρσµν

(5.2)

In terms of the 3-dimensional potentials, Aµ = c(A, − ϕ) . We will dene the stress-energy tensor and the force density. (5.3a) (5.3b)

T µν = 12 (f µτ pντ + f ντ pµτ ) − g µν ( 14 fκτ pκτ − Q) fµ = −Tµ ;ν ν

6. Einstein-Maxwell Equations

Eisenhart [1] shows that the most general symmetric connection can be written in the form ˜ µ = aµ + Γµ Γ αβ αβ αβ

© 2013 Fredrick W. Cotton

aµαβ = aµβα

5

Γµαβ = Γµβα

(6.1)

http://sites.google.com/site/fwcotton/em-25.pdf

where aµαβ is a tensor and Γµαβ is the metric connection. The curvature tensor for Γ˜ µαβ can be written as [1, eq. 5.15], µ α µ B µνρσ = Rµνρσ + aµνσ;ρ − aµνρ;σ + aα (6.2) νσ aαρ − aνρ aασ where Rµνρσ is the Riemann curvature tensor for the metric gµν . A semicolon denotes covariant dierentiation with respect to the metric connection, Γµαβ ; a colon will denote covariant dierentiation with respect to the general symmetric connection, Γ˜ µαβ ; and a comma will denote partial dierentiation with respect to the coordinates. (This notation is somewhat dierent from that used by Eisenhart. He uses the Christoel symbols for the metric connection and Γµαβ for the general symmetric connection. More importantly, he usually uses a comma to denote covariant dierentiation with respect to the general symmetric connection.) It is important to note that gµν :ρ = α −gαν aα µρ − gµα aνρ ̸= 0 . For that reason, the equations are expressed in terms of covariant dierentiation with respect to the metric connection, Γµαβ . Dene µ α µ Bνσ = B µνµσ = Rµνµσ + aµνσ;µ − aµνµ;σ + aα νσ aαµ − aνµ aασ

(6.3)

Dene symmetric and antisymmetric parts in the following way: µ µ α µ α µ Sνσ = 12 (Bνσ + Bσν ) − Rνσ = aµνσ;µ − 12 (aνµ ;σ + aσµ;ν ) + aνσ aαµ − aνµ aασ

Aνσ =

1 2 (Bνσ

− Bσν ) =

− 12 (aµνµ;σ

−

aµσµ;ν )

(6.4a) (6.4b)

If we consider aµνµ = 0

α β Sµν = aα µν ;α − aβµ aαν

Aνσ = 0

(6.5)

then we can write a generalized form of the Einstein-Maxwell equations Gµν + Sµν = 8πGc−4 Tµν

(6.6)

Gµν = Rµν − 21 gµν R

(6.7)

where

and G is Newton's gravitational constant. For a particle at rest, we will see that the spin is described by the non-Riemannian part of the symmetric connection. In the rest frame of the particle, let the metric be given by ds 2 = gµν dx µ dx ν = fg−1 (r)dr 2 + r2 dθ2 + r2 sin2 (θ)dφ2 − c2 fg (r)dt 2

(6.8)

and let the only non-zero components of aµνσ be a344 = ζ1 (r, θ)

a433 = ζ2 (r, θ)

(6.9)

The only non-zero components of Sµν are (6.10)

S34 = S43 = −a344 a433 = −ζ1 (r, θ)ζ2 (r, θ)

If

∫

Aµ = (0,0,0, − cϕ(r))

ϕ(r) = −

dr fe (r)

(6.11)

and if the metric and non-metric components of the constitutive tensor are specied by χµνρσ = ϵ0 fϵ (r)(gµρ gνσ − gνρ gµσ ) χ3241 = −2r2 h(r) sin(θ) cos(θ) χ3242 = −r2 fg2 (r)χ3141 = −r3 h(r)fg (r) sin2 (θ)

(6.12)

χ2143 = −χ3142 = r2 h(r) sin(θ) cos(θ)

© 2013 Fredrick W. Cotton

6

http://sites.google.com/site/fwcotton/em-25.pdf

then the non-zero components of Tµν and Gµν are ∫

∞

dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ ) ∫ ∞ = ϵ0 r2 [fe2 (r)fϵ (r) − 2 dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ )] sin2 (θ)

T44 = −c2 fg2 (r)T11 = 2c2 ϵ0 fg (r)

(6.13a)

r

T33 = sin2 (θ)T22

(6.13b)

r

(6.13c) (6.13d) (6.13e)

T34 = − 21 rh(r)fe2 (r) sin2 (θ) G44 = G33 =

−c2 fg2 (r)G11 sin2 (θ)G22 =

2 −2

= −c r

fg (r)[−1 + fg (r) +

rfg′ (r)]

[rfg′ (r) + 12 r2 fg′′ (r)] sin2 (θ)

Equations (6.6) reduce to −1 + fg (r) + rfg′ (r) = −16πGc−4 ϵ0 r2

∫

∞

dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ ) ∫ ∞ ′ −4 2 2 1 2 ′′ dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ )] rfg (r) + 2 r fg (r) = 8πGc ϵ0 r [fe (r)fϵ (r) − 2

(6.14a)

r

(6.14b)

r

(6.14c)

ζ1 (r, θ)ζ2 (r, θ) = 4πGc−4 rh(r)fe2 (r) sin2 (θ)

Integrating (6.14b) and substituting into (6.14a) gives fg (r) = 1 − 16πGϵ0 c−4 r−1

∫

r

dr ′ (r′ )2

∫

∞ r′

0

dr ′′ (r′′ )−1 fe2 (r′′ )fϵ (r′′ )

(6.15)

Comparison with the Schwarzschild metric, for which fg (r) = 1 − 2Gm0 c−2 r−1 , shows that −2

∫

∫

∞

m0 = 8πϵ0 c

dr r

∞

2

0

dr ′ (r′ )−1 fe2 (r′ )fϵ (r′ )

(6.16)

r

which agrees with (3.4). The entire rest mass is electromagnetic. 7. Electromagnetic Waves

In a cylindrically symmetric space with the metric given by (7.1)

ds 2 = dr 2 + r2 dφ2 + dz 2 − c2 dt 2

let us look at electromagnetic waves of the form (7.2a) (7.2b) (7.2c) (7.2d)

Aµ = cf (z − ct)fem (r)(1, 0, 0, 0) χµνρσ = ϵ0 (gµρ gνσ − gνρ gµσ ) Q=0 c2 a133 = −ca143 = a144 = [f ′ (z − ct)]2 fa (r)

Then Gµν = 0 and the non-zero components of Tµν and Sµν are 2 T44 = −cT34 = c2 T33 = c4 ϵ0 [f ′ (z − ct)]2 fem (r) ′

S44 = −cS34 = c S33 = [f (z − ct)] 2

2

[fa′ (r)

+r

−1

fa (r)]

(7.3a) (7.3b)

Thus T µν;ν = 0 and S µν;ν = 0 and equations (6.6) reduce to fa (r) = 8πGϵ0 r−1

∫

r

2 dr ′ r′ fem (r′ )

(7.4)

0

© 2013 Fredrick W. Cotton

7

http://sites.google.com/site/fwcotton/em-25.pdf

For this type of wave, jµ ̸= 0 , but jµ j µ = 0 . However, ∫ jT =

∫

∞

µ

dr r 0

∫

2π

∞

dφ 0 ∞

−∞

dz j µ ∞

= 2πϵ0 [rfem (r)]|r=0 f (z − ct)|z=−∞ (0, 0, c, 1)

(7.5)

=0

As an example, consider fem (r) = r exp(−β 2 r2 ) 4 −1

fa (r) = πGϵ0 (rβ )

[1 − (1 + 2β r ) exp(−2β r )] 2 2

2 2

(7.6a) (7.6b)

In free space, electromagnetic waves are usually assumed to have zero current, jµ = 0 . However, if we admit the possibility of non-zero eld currents such that jµ j µ = 0 , then we have a class of force-free wave solutions that have a spatial variation in the plane perpendicular to the direction of propagation. These null-vector eld currents are intrinsic to the structure of the wave; they are not an external source. Furthermore, this class of solutions introduces terms only in the non-Riemannian part of the curvature tensor. 8. Conclusions

We have made use of the properties of the constitutive tensor together with a generalized form of the stress-energy tensor to introduce a classical form of spin for charged particles. The solutions are force-free and radiationless. We have constructed particular solutions that have some of the properties required for the elementary particles. We have also shown that the spin terms can be incorporated into the Einstein-Maxwell equations by adding curvature terms arising from a general symmetric connection. These additional curvature terms can also be used to construct a new class of electromagnetic waves. Acknowledgments

Many of the calculations were done using Mathematica ® 8.01 [2] with the MathTensor — 2.2.1 [3] Application Package. Stephen C. Young double-checked the derivations in an earlier version and suggested several clarications. References

1. L.P. Eisenhart, Non-Riemannian VIII, 1927).

Geometry

(American Mathematical Society, Colloquium Publications, Vol.

2. Wolfram Research, Mathematica ® 8.01 (http://www.wolfram.com/). 3. L. Parker and S.M. Christensen, MathTensor — 2.2.1 (http://smc.vnet.net/MathTensor.html).

© 2013 Fredrick W. Cotton

8

http://sites.google.com/site/fwcotton/em-25.pdf