zotero/storage/4T5TEBC7/.zotero-ft-cache

Gravitation and electromagnetism
Valery P. Dmitriyev Lomonosov University P.O.Box 160, Moscow 117574, Russia∗ (Dated: 23 July 2002)
Maxwell’s equations comprise both electromagnetic and gravitational fields. The transverse part of the vector potential belongs to magnetism, the longitudinal one is concerned with gravitation. The Coulomb gauge indicates that longitudinal components of the fields propagate instantaneously. The delta-function singularity of the field of the divergence of the vector potential, referred to as the dilatation center, represents an elementary agent of gravitation. Viewing a particle as a source or a scattering center of the point dilatation, the Newton’s gravitation law can be reproduced.

arXiv:physics/0207091v1 [physics.gen-ph] 23 Jul 2002

1. MAXWELL’S EQUATIONS IN THE KELVIN-HELMHOLTZ REPRESENTATION

The general form of Maxwell’s equations is given by

1 c

∂A ∂t

+

E

+

∇ϕ

=

0,

∂E ∂t

−

c

∇

×

∇

×

A

+

4πj

=

0

,

(1.1) (1.2)

∇· E = 4πρ .

(1.3)

The Helmholtz theorem: a vanishing at infinity vector field u can be expanded into the sum of solenoidal ur and potential ug components. We have for the electric field:

E = Er + Eg ,

(1.4)

where

∇ · Er = 0 ,

(1.5)

∇× Eg = 0 .

The respective expansion for the vector potential can be written as

A = Ar +

c cg

Ag ,

where

∇· Ar = 0 ,

(1.6) (1.7) (1.8)

∇ × Ag = 0 ,

and cg is a constant. Substitute (1.4) and (1.7) into (1.1):

1 c

∂Ar ∂t

+

Er

+

1 cg

∂Ag ∂t

+

Eg

+

∇ϕ

=

0.

Taking the curl of (1.10), we get through (1.6) and (1.9)

∇×

1 c

∂Ar ∂t

+ Er

= 0.

(1.9) (1.10) (1.11)

2

On the other hand, by (1.5) and (1.8) we have

∇·

1 c

∂Ar ∂t

+ Er

= 0.

(1.12)

If the divergence and curl of a field equal to zero, then the very field is vanishing. Hence (1.11) and (1.12) imply that

Subtracting (1.13) from (1.10) we get also

1 c

∂Ar ∂t

+

Er

=

0.

(1.13)

1 cg

∂Ag ∂t

+ Eg + ∇ϕ = 0 .

Similarly, expanding as well the density of the current

(1.14)

j = jr + jg ,

(1.15)

∇· jr = 0 ,

(1.16)

∇ × jg = 0 ,

(1.2) can be broken up in two equations

∂Er ∂t

−

c∇

×

∇

×

Ar

+

4πjr

=

0,

Through (1.4) and (1.5) equation (1.3) will be

∂Eg ∂t

+

4πjg

=

0.

∇ · Eg = 4πρ .

(1.17) (1.18) (1.19) (1.20)

2. WAVE EQUATIONS

Let us derive from (1.13), (1.14), (1.18), (1.19) and (1.20) the wave equations for the solenoidal (transverse) and potential (longitudinal) components of the fields. In what follows we will use the general vector relation

∇ (∇ · u) = ∇2u + ∇ × ∇ × u .

(2.1)

The wave equation for Ar can be found thus. Differentiate (1.13) with respect to time:

1 c

∂ 2Ar ∂t2

+

∂Er ∂t

=

0.

Substitute (1.18) into (2.2). With the account of (2.1) we get

(2.2)

∂ 2Ar ∂t2

−

c2∇2Ar

=

4πcjr .

The wave equation for Er can be found as follows. Differentiate (1.18) with respect to time

∂ 2Er ∂t2

−

c∇×

∇×

∂Ar ∂t

+

4π

∂jr ∂t

=

0.

Substitute (1.13) into (2.4). With the account of (2.1) we get

(2.3) (2.4)

∂2Er ∂t2

−

c2∇2Er

=

− 4πc∂tjr

.

(2.5)

3

In order to find the wave equations for the potential fields we need a gauge relation. Let us postulate for the potential part of the vector potential the specific Lorentz gauge

∇· Ag +

1 cg

∂ϕ ∂t

=

0,

(2.6)

where in general cg = c. The solenoidal part of the vector potential meets automatically the Coulomb gauge (1.8). The wave equation for Ag can be found as follows. Differentiate (1.14) with respect to time:

1 cg

∂ 2 Ag ∂t2

+

∂Eg ∂t

+

∂∇ϕ ∂t

= 0.

(2.7)

Take the gradient of (2.6):

∇ (∇· Ag) +

1 cg

∇

∂ϕ ∂t

=

0.

(2.8)

Combine (2.7), (2.8) and (1.19). With the account of (2.1) we get

∂ 2Ag ∂t2

−

c2g ∇2 Ag

=

4πcgjg

.

Next, we will find the wave equation for ϕ. Take the divergence of (1.14):

(2.9)

1 cg

∂∇·Ag ∂t

+ ∇· Eg

+ ∇2ϕ

=

0.

(2.10)

Combine (2.10), (2.6) and (1.20):

∂2ϕ ∂t2

−

c2g ∇2 ϕ

=

4πc2gρ

.

(2.11)

The wave equation for Eg we will find from the wave equations of Ag and ϕ, using (1.14). Differentiate (2.9) with respect to time

Take the gradient of (2.11)

∂2 ∂t2

∂Ag ∂t

−

c2g

∇2

∂Ag ∂t

=

4πcg

∂jg ∂t

.

(2.12)

∂2∇ϕ ∂t2

−

c2g ∇2 ∇ϕ

=

4πc2g∇ρ .

Summing (2.12) and (2.13), we get with the account of (1.14)

(2.13)

∂2Eg ∂t2

−

c2g ∇2 Eg

=

− 4π

c2g ∇ρ

+

∂jg ∂t

.

(2.14)

Thus, Maxwell’s equations (1.1)-(1.3) with the specific Lorentz gauge (2.6) imply that the solenoidal and potential components of the fields propagate with different velocities. Solenoidal components propagate with the speed c of light. Their wave equations are (2.3) and (2.5). Potential components and the electrostatic potential propagate with a speed cg. Their wave equations are (2.9), (2.11) and (2.14).

3. QUASIELASTICITY
Equations (2.3) and (2.9) have the character of the elastic equations. In this connection, the vector potential A can be correspondent1 with a certain displacement field s, and the density j of the current – with the density f of an external force . The gauge relation (2.6) is interpreted as a linearized continuity equation, in which the constant cg has directly the meaning of the speed of an expansion-contraction wave2. We are interested in the interaction of two

4

external forces f1 and f2, which produce elastic fields s1 and s2, respectively. The energy of the elastic interaction is given by the general relation

U12 = − ς f1 · s2d3x = − ς f2 · s1d3x ,

(3.1)

where the sign minus in (3.1) corresponds to conditions of the Clapeyron theorem3, ς is a constant. The energy of the static interaction can be found substituting into (3.1)

s∼

1 c

A,

(3.2)

f ∼ 4πcj ,

(3.3)

We have for the transverse interaction

ς

∼

1 4πc

.

Ur

=

−

1 c

jr · Ard3x .

(3.4) (3.5)

Suppose that

cg >> c . Then, through (1.7) relations (2.6) and (1.8) turn to the Coulomb gauge

(3.6)

∇· A = 0 .

(3.7)

We have according to (1.16) and (1.17)

jr = ∇ × R, jg = ∇G ,

(3.8)

where R and G are vector and scalar fields. Using (1.15), (3.8), (3.7), (1.7) and (1.9) take the following integral by parts:

j · Ad3x = (jr + ∇G) · Ad3x = jr · Ad3x = (∇× R) · (Ar + c Ag/cg)d3x = jr · Ard3x . From (3.9) and (3.5) we get the regular expression for the energy of magnetostatic interaction

(3.9)

Ur

=

−

1 c

j · Ad3x .

(3.10)

Elementary sources of the magnetic field correspond to the two forms of the external force density f (3.3). The point force at x′:

f = 4πcqvδ (x − x′) ,

(3.11)

and the torsion center at x′:

fr = 4πca∇ × [µδ (x − x′)] ,

(3.12)

where qv and aµ are constant vectors, |µ| = 1. They describe a moving electric charge and a point magnetic dipole, respectively1. Substituting (3.11) and (3.12) into the right-hand part of the equation (2.3) we can find the fields A produced by these forces. Then, substituting these fields into (3.10) and (3.5), we arrive at the well-known expressions for the interaction energies of electric currents and point magnetic dipoles.
The elementary source of the longitudinal part Ag of the vector potential is given by the density of the external force of the form

fg = − 4πcgb∇δ (x − x′) ,

(3.13)

5

where b is the strength of the dilatation center (3.13)4. Substitute (3.13) into the right-hand part of the static variant of the equation (2.9):

cg∇2Ag = 4πb∇δ (x − x′) .

(3.14)

With the account of (2.1) and (1.9) we get from (3.14)

cg∇· Ag = 4πbδ (x − x′) .

(3.15)

Following (3.1)-(3.4) we have for the energy of longitudinal interaction:

Ug

=

−

1 cg

jg · Agd3x .

(3.16)

Substitute (3.13) with the account of (3.3) into (3.16):

U12

=

b1 cg

∇δ

(x

−

x1)

·

A2d3x

=

−

b1 cg

δ (x − x1) ∇· A2d3x .

(3.17)

Substituting (3.15) into (3.17), we get

U12

=

−

4πb1b2 c2g

δ (x − x1) δ (x − x2)d3x

(3.18)

=−

4πb1b2 c2g

δ (x1 − x2)

.

(3.19)

Expression (3.19) implies, that two dilatation centers (3.15) interact with each other only if they are in a direct contact. The sign of (3.18), or (3.19), indicates that this is the attraction.
Take notice that solenoidal and potential fields are orthogonal to each other in the sense of (3.1). Indeed, using (3.8), (1.8) and (1.9), we find that

jg · Ard3x = ∇G · Ard3x = − G∇· Ard3x = 0 ,

(3.20)

jr · Agd3x = ∇× R · Agd3x = R · ∇ × Agd3x = 0 .

(3.21)

4. GRAVITATION

We consider dilatation centers distributed with the volume density bp (x). Then equation (3.15) becomes

cg∇ · Ag = 4πbp (x) .

(4.1)

The interaction energy of the two clusters, or clouds, of dilatation centers can be found substituting delta-functions in (3.18) by the reduced densities p (x) of the distributions. This gives

U12

=

−

4πb1b2 c2g

p1 (x) p2 (x)d3x .

(4.2)

Consider a weak source at x∗, which emits dilatation centers with a sufficiently high linear velocity υg. Such a source will create a quick-formed stationary distribution of the point dilatation with the reduced density

p (x)

=

4πυg

g (x −

x∗) 2

,

(4.3)

where g is a universal constant. Substituting (4.3) into (4.2), we find the interaction energy for two sources of the point dilatation

U12

=

−

g2b1b2 4πc2gυg2

d3x (x − x1)2 (x − x2)2

=

−

π2g2 4c2gυg2

b1b2 |x1 − x2|

.

(4.4)

6

We will assume that each particle is a weak source of the point dilatation (3.15) or a scattering center in a dynamic sea of the point dilatation, the strength b of the source being proportional to the particle’s mass. Then relation (4.4) will be a model of the Newton’s law of gravitation.
Notice that in the model thus constructed we must distinguish the speed υg, with which the gravitational interaction is transmitted, and the speed cg of the longitudinal wave. The latter can be interpreted as the gravitational wave.
Thus, gravitation enters into the general structure of Maxwell’s equations. A gravitating center is formally modeled by a potential component of the current having the form

jg

=

−

gb 4πυg

∇ (x

1 − x∗)2

.

(4.5)

And the gravitational interaction is calculated by means of the general relation (3.16), where the longitudinal component Ag of the vector potential is found substituting (4.5) into the longitudinal part (2.9) of Maxwell’s equations.

5. CONCLUSION
Maxwell’s equations (1.1)-(1.3) describe both electromagnetic and gravitational fields. The transverse part of the vector potential belongs to magnetism, and longitudinal one is concerned with gravitation. Transverse fields propagate with the speed of light. The Coulomb gauge (3.7) indicates that longitudinal waves propagate in effect instantaneously, comparing with transverse waves. Choosing properly expressions for the current density, magnetic and gravitational interactions can be modeled. An elementary agent of the gravitational interaction corresponds to the dilaton, which is a delta-function singularity (3.15) of the field of the divergence of the vector potential . The sources of longitudinal and transverse fields do not interact with each other. This signifies that gravitation can not be detected with the aid of light.
In the end it should be noted that some of the questions considered here and in1 were recently approached in5.

∗ Electronic address: dmitr@cc.nifhi.ac.ru 1 V.P.Dmitriyev, “The elastic model of physical vacuum”, Mech.Solids, 26, No 6, 60–71 (1992). 2 V.P.Dmitriyev, “Mechanical analogies for the Lorentz gauge, particles and antiparticles”, Apeiron, 7, No 3/4, 173-183 (2000);
http://xxx.arXiv.org/abs/physics/9904049. 3 W.Nowacki, The theory of elasticity (Warsaw, 1970), §4.16. 4 J.D.Eshelby, “The continuum theory of lattice defects”, Solid State Physics, 3, 79-144 (New York 1956). 5 David M.Drury, “Irrotational and solenoidal components of Maxwell’s equations”, Galilean Electrodynamics, 13, No 4 , 72-75
(2002).