Problem for the gravitational field

Advances in Physics Theories and Applications
ISSN 2224-719X (Paper) ISSN 2225-0638 (Online)
Vol.23, 2013

www.iiste.org

4/3 Problem for the Gravitational Field
Sergey G. Fedosin
PO box 614088, Sviazeva str. 22-79, Perm, Russia
E-mail: intelli@list.ru
Abstract
The gravitational field potentials outside and inside a uniform massive ball were determined using the
superposition principle, the method of retarded potentials and Lorentz transformations. The gravitational field
strength, the torsion field, the energy and the momentum of the field, as well as the effective masses associated
with the field energy and its momentum were calculated. It was shown that 4/3 problem existed for the
gravitational field as well as in the case of the electromagnetic field.
Keywords: energy, momentum, theory of relativity, gravitation, field potentials
1. Introduction
In field theory, there are a number of unsolved problems, which need deeper analysis and logical understanding.
An example is the problem of choosing a universal form of the stress-energy tensor of the body, which would
include the rest energy of the substance as well as the field energy and at the same time would provide an
univocal connection with thermodynamic variables of the substance in the language of four-vectors and tensors.
Another interesting problem is 4/3 problem, according to which the effective mass of the body field, which is
calculated through the field momentum, and the effective mass of the field, found through the field energy, for
some reason do not coincide with each other, with the ratio of the masses approximately equal to 4/3.
The problem of 4/3 is known for a long time for the mass of electromagnetic field of a moving charge. Joseph
John Thomson, George Francis FitzGerald, Oliver Heaviside, George Frederick Charles Searle and many others
write about it (Heaviside 1888/1894), (Searle 1897), (Hajra 1991). We also discuss this question with respect to
the gravitational field of a moving ball (Fedosin 2008). Now we present a more accurate description of the
problem, not limited to the approximation of small velocities.
2. Methods
In the calculation of the energy and the momentum of gravitational field of a uniform massive ball, we will use
the superposition principle by means of summing up the field energies and momenta from all point particles
forming the moving ball. This approach is reasonable in the case of a weak field, when the general theory of
relativity changes to gravitomagnetism and the covariant theory of gravitation – to the Lorentz-invariant theory
of gravitation (Fedosin 2009a). The field equations then become linear, allowing the use of the superposition
principle. We will note that the gravitational field can be considered weak if the spacetime metric differs
insignificantly from the Minkowski spacetime metric (the spacetime metric of the special theory of relativity). If
the effects of gravitational time dilation and sizes contraction are significantly less than the similar effects due to
the motion velocity of the reference frame under consideration, then this gravitational field can be considered
weak.
3. Results and Discussions
3.1 The Gravitational Field Outside a Uniform Massive Ball
We will first define the gravitational field potentials for a ball moving at a constant velocity V along the axis
OХ of the reference frame K . We will proceed from the so-called Liénard-Wiechert potentials (Liénard 1898;
Wiechert 1900) for any point particles that make up the ball. Popular presentation of the problem (for the
electromagnetic field) can be found in Feynman’s book (Feynman at all. 1964). Similarly to this, the differential
scalar Liénard-Wiechert potential for the gravitational field from a point particle with mass dM has the
following form:

dψ = −

γ dM
r ′ − V ⋅ r ′ / cg

,

(1)

r ′ is the vector
connecting the early position of the point particle at time t ′ and the position r = ( x, y , z ) at which the
potential is determined at time t . In this case, the equation must hold:
Where

γ

is the gravitational constant, cg is the velocity of gravitation propagation, vector

19

Vol.23, 2013

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t′ = t −

r′
.
cg

(2)

The meaning of equation (2) is that during the time period t − t ′ the gravitational effect of the mass dM must
cover the distance r ′ at velocity cg up to the position r = ( x, y , z ) so that at this position the potential dψ
would appear.
Suppose there is continuous distribution of point particles and at t = 0 these particles are described by the
coordinates ( x0 , y0 , z0 ) and the center of distribution of point particles coincides with the origin of the
reference frame. Then at time t the distribution center of the point particles would move along the axis OX to
the position x = Vt , and the radius vector of an arbitrary particle of distribution would equal
r2 = ( x0 + Vt , y0 , z0 ) . At the early time t ′ the position of this point particle is specified by the vector

r1 = ( x0 + Vt ′, y0 , z0 ) . Since r ′ = r − r1 and r ′ = cg (t − t ′ ) according to (2), then for the square r ′2 we
can write down:
2
r ′2 = ( x − x0 − Vt ′) 2 + ( y − y0 ) 2 + ( z − z0 ) 2 = cg (t − t ′ ) 2 .

(3)

The right side of (3) is a quadratic equation for the time t ′ . After we find t ′ from (3), we can then find r ′ from
(2). If we consider that in (1) the product of vectors is V ⋅ r ′ = V ( x − x0 − Vt ′ ) , then substituting r ′ also in
(1), we obtain the following expression (Fedosin 2009b):

γ dM

dψ = −

( x − x0 − V t ) 2
+ ( y − y0 ) 2 + ( z − z0 ) 2
2
1 − V 2 cg

2
1 − V 2 cg

.

(4)

dψ of the point mass dM at the time t during its
motion along the axis OX depends on the initial position ( x0 , y0 , z0 ) of this mass at t = 0 . If we use the
According to (4), the differential gravitational potential

extended Lorentz transformations for the spatial coordinates in (4):

x∗ =

x − x0 − V t

1−V c
2

2
g

,

y ∗ = y − y0 ,

z ∗ = z − z0 ,

and then let the velocity V tend to zero, we obtain the formula for the potential in the reference frame
origin of which coincides with the point mass dM :

dψ ∗ = −

γ dM
∗2

x + y ∗2 + z ∗2

.

(5)

K ∗ the
(6)

K ∗ the vector r ∗ = ( x ∗ , y ∗ , z ∗ ) at the proper time t ∗ specifies the same point in
space as the vector r = ( x, y , z ) in the reference frame K at the time t . If we introduce the gravitational four-

In (6) in the reference frame

ψ

Dµ =  , − D  , including the scalar potential ψ and the vector potential D (Fedosin 1999), then
 сg



∗
the relation between the scalar potential (6) in the reference frame K and the scalar potential (4) in the
reference frame K can be considered as the consequence of extended Lorentz transformations in four-

potential

dimensional formalism, which are applied to the differential four-potential of a single point particle. These
transformations are carried out by multiplying the corresponding transformation matrix by the four-potential,
which gives the four-potential in a different reference frame with its own coordinates and time.

K ∗ the point mass is at rest, its vector potential is dD ∗ = 0 , and the four-potential
 dψ ∗ 
∗
has the form: dDµ = 
, 0  . In order to move to the reference frame K , in which the reference frame
 сg



K ∗ is moving at the constant velocity V along the axis OX , we must use the matrix of inverse partial Lorentz

Since in the reference frame

transformation (Fedosin 2009a):

20

Vol.23, 2013

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1


2
1 − V 2 / cg


V
µ
Lk =  −
2
 cg 1 − V 2 / cg

0


0


−


0 0


0 0 ,


1 0

0 1

V
cg 1 − V / c
2

2
g

1
2
1 − V 2 / cg

0
0


  dψ

dψ ∗
V dψ ∗
∗
dDk = Lkµ dDµ = 
,−
, 0, 0  = 
, − dDx , 0, 0  .
2
2
2

 сg 1 − V 2 / cg
  с
cg 1 − V 2 / cg


  g

(7)

From (7) taking into account (6) and (5) we obtain the following relations:

dψ =

dψ ∗

1−V / c
2

2
g

γ dM

=−

1−V c
2

( x − x0 − V t )2
+ ( y − y0 ) 2 + ( z − z0 ) 2
2
2
1 − V cg

2
g

dDx =

dψ V
,
2
cg

dDx = 0 ,

,

dDz = 0 .

(8)

The first equation in (8) coincides with (4) and the differential vector potential of the point mass is directed along
its motion velocity.
After integration of (8) over all point masses inside the ball on the basis of the principle of superposition, the
standard formulas are obtained for the potentials of gravitational field around the moving ball, with retardation of
the gravitational interaction taken into account:

ψ =−

γM

( x − V t ) + (1 − V 2 cg2 )( y 2 + z 2 )
2

D=

,

ψV
2
cg

,

(9)

Where ψ – the scalar potential of the moving ball,

M – the mass of the ball, ( x, y, z ) – the coordinates of the
point at which the potential is determined at the time t (on the condition that the center of the ball at t = 0 was
in the origin of coordinate system), D – the vector potential of the ball.
ψV
In (9) it is assumed that the ball is moving along the axis OX at a constant speed V , so that Dx = 2 ,
cg
Dy = 0 , Dz = 0 . With the help of the field potentials we can calculate the field strengths around the ball by
the formulas (Fedosin 1999):

G = −∇ψ −

∂D
,
∂t

Ω = ∇× D ,

(10)

Where G is the gravitational field strength, Ω – the gravitational torsion in Lorentz-invariant theory of
gravitation (gravitomagnetic field in gravitomagnetism).
In view of (9) and (10) we find:

Gx = −

2
γ M ( x − V t ) (1 − V 2 cg )

2
[( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

Gz = −

Gy = −

,

2
γ M y (1 − V 2 cg )
2
[( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

2
γ M z (1 − V 2 cg )
2
[( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

21

,

Ωx = 0 ,

,

(11)

Vol.23, 2013

Ωy =

www.iiste.org

2
γ M z V (1 − V 2 cg )

,

2
2
cg [( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

Ωz = −

2
γ M y V (1 − V 2 cg )
2
2
cg [( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

.

The energy density of the gravitational field is determined by the formula (Fedosin 1999):

u=−

1
8π γ

(G

2

+c Ω
2
g

2

)=−

2
2
γ M 2 (1 − V 2 cg )[( x − V t ) + (1 + V 2 cg )( y 2 + z 2 )]
2

.

2
8π [( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

(12)

The total energy of the field outside the ball at a constant velocity should not depend on time. So it is possible to
integrate the energy density of the field (12) over the external space volume at t = 0 . For this purpose we shall
introduce new coordinates:

y = r sin θ cos ϕ ,

2
x = 1 − V 2 cg r cos θ ,

The volume element is determined by the formula
matrix:

2

(13)

d ϒ = J dr dθ dϕ , where J is determinant of Jacobian

∂x
∂r
∂ ( x, y, z ) ∂y
J=
=
∂ (r , θ , ϕ ) ∂r
∂z
∂r
It follows that d ϒ = r sin θ

z = r sin θ sin ϕ .

∂x
∂θ
∂y
∂θ
∂z
∂θ

∂x
∂ϕ
∂y
.
∂ϕ
∂z
∂ϕ

2
1 − V 2 cg dr dθ dϕ . The integral over the space of the energy density (12)

will equal:

Ub = ∫ u d ϒ = −

γM2
2
2
8 π cg 1 − V 2 c g

∫

2
[cg + V 2 (sin 2 θ − cos2 θ )] sin θ dr dθ dϕ

r2

We shall take into account that due to the Lorentz contraction during the motion along the axis
must be as Heaviside ellipsoid, the surface equation of which at t = 0 is the following:

x2
+ y2 + z 2 = R2 .
2
1 − V 2 cg

.

(14)

OX the ball
(15)

After substituting (13) in (15), it becomes apparent that the radius r at the integration in (14) must change from
R to ∞ , and the angles θ and ϕ change the same way as in spherical coordinates (from 0 to π for the angle

θ , and from 0 to 2π

for the angle

ϕ ). For the energy of the gravitational field outside the moving ball we find:
2
2
γ M 2 (1 + V 2 3 cg ) U b 0 (1 + V 2 3 cg )
Ub = −

Where U b 0 = −

γM2
2R

2
2 R 1 − V 2 cg

=

,

2
1 − V 2 cg

(16)

is the field energy around the stationary ball.

We can introduce the effective relativistic mass of the field related to the energy of moving ball:

mg b =

2
U b 1 − V 2 cg
2
cg

=

2
U b 0 (1 + V 2 3 cg )
2
cg

.

(17)

We shall now consider the momentum density of the gravitational field:

g=

22

H
,
2
cg

(18)

Vol.23, 2013

Where

H =−

2
cg

4π γ

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[G × Ω ] is the vector of energy flux density of the gravitational field (Heaviside vector)

(Fedosin 1999).
Substituting in (18) the components of the field (11), we find:

gx = −
gy =
gz =

2
γ M 2 (1 − V 2 cg )2 ( y 2 + z 2 ) V
2
2
4π cg [( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

2
γ M 2 (1 − V 2 cg ) 2 ( x − V t ) y V
2

2
2
4π cg [( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2

(19)

,

2
2
4π cg [( x − V t ) + (1 − V 2 cg )( y 2 + z 2 )]3
2
γ M 2 (1 − V 2 cg ) 2 ( x − V t ) z V

,

.

We can see that the components of the momentum density of gravitational field (19) look the same as if a liquid
flowed around the ball from the axis OX , carrying similar density of the momentum – liquid spreads out to the
sides when meeting with the ball and merges once again on the opposite side of the ball. Integrating the
components of the momentum density of the gravitational field (19) by volume outside the moving ball at t = 0
as in (14), we obtain:

Px = ∫ g x d ϒ = −

γ M 2V
2
2
4 π cg 1 − V 2 c g

sin 3 θ dr dθ dϕ
2γ M 2 V
.
=−
2
2
r2
3 R c g 1 − V 2 cg

∫

Py = ∫ g y d ϒ = 0 ,

(20)

Pz = ∫ g z d ϒ = 0 .

In (20) the total momentum of the field has only the component along the axis OX . By analogy with the
formula for relativistic momentum the coefficient before the velocity V in (20) can be interpreted as the
effective mass of the external gravitational field moving with the ball:

m pb =
Where

Ub0 = −

γM2
2R

2
Px 1 − V 2 cg

V

=−

2γ M 2 4U b 0
=
,
2
2
3 R cg
3 cg

(21)

is the energy of the external static field of the ball at rest.

Comparing (21) and (17) gives:

mg b =

2
3(1 + V 2 3 cg ) m pb

4

.

(22)

The discrepancy between the masses m g b and m p b in (22) shows the existence of the problem of 4/3 for
gravitational field in the Lorentz-invariant theory of gravitation.
3.2 The Gravitational Field Inside a Moving Ball
For a homogeneous ball with the density of substance

ρ0

(measured in the comoving frame), which is moving

along the axis OX , the potentials inside the ball (denoted by subscript i ) depend on time and are as follows
(Fedosin 2009b):

ψi = −

 2 1  ( x − V t )2

+ y 2 + z 2  ,
R − 
2
2

3  1 − V 2 cg
1 − V 2 cg 




2π γ ρ 0

In view of (10) we can calculate the internal field strength and torsion field:

23

Di =

ψi V
2
cg

.

(23)

Vol.23, 2013

Gxi = −

4π γ ρ 0 ( x − V t )
3 1−V

2

c

2
g

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G yi = −

,

Ω yi =

Ω xi = 0 ,

4π γ ρ0 y
3 1−V c
2

4π γ ρ0 zV
2
2
3 cg 1 − V 2 c g

Gzi = −

,

2
g

4π γ ρ0 z
2
3 1 − V 2 cg

Ω zi = −

,

,

4π γ ρ0 yV
2
2
3 cg 1 − V 2 cg

.

(24)

Similarly to (12) for the energy density of the field we find:

ui = −

1
8π γ

(G

2
i

+c Ω
2
g

2
i

)=−

2
2
2π γ ρ0 [( x − V t ) + (1 + V 2 cg )( y 2 + z 2 )]
2

2
9(1 − V 2 cg )

.

(25)

According to (25) the minimum energy density inside a moving ball is achieved on its surface, and in the center
at t = 0 it is zero.
The integral of (25) by volume of the ball at t = 0 in coordinates (13) with the volume element
2
d ϒ = r 2 sin θ 1 − V 2 cg dr dθ dϕ equals:

U i = ∫ ui d ϒ = −

2
2π γ ρ 0

9c

2
g

1−V

2

c

2
g

∫ [c

2
g

+ V 2 (sin 2 θ − cos 2 θ )] r 4 sin θ dr dθ dϕ .

(26)

According to the theory of relativity the moving ball looks like Heaviside ellipsoid with equation of the surface
(15) at t = 0 , and in the coordinates (13) the radius in the integration in (26) varies from 0 to R . With this in
mind for the energy of the gravitational field inside the moving ball, we have:

Ui = −
Where

Ui0 = −

γM2
10 R

2
γ M 2 (1 + V 2 3 cg )
2
10 R 1 − V 2 cg

=

2
U i 0 (1 + V 2 3 cg )

is the field energy inside a stationary ball with radius

2
1 − V 2 cg

,

(27)

R.

The effective mass of the field associated with energy (27) is:

mg i =

2
U i 1 − V 2 cg
2
cg

=

2
U i 0 (1 + V 2 3 cg )
2
cg

.

(28)

Substituting in (18) the components of the field strengths (24), we find the components of the vector of
momentum density of gravitational field:
2
4π γ ρ0 ( y 2 + z 2 ) V
g xi = −
,
2
2
9cg (1 − V 2 cg )

2
4π γ ρ0 ( x − V t ) yV
g yi =
,
2
2
9cg (1 − V 2 cg )

g zi =

2
4π γ ρ0 ( x − V t ) zV
.
2
2
9cg (1 − V 2 cg )

(29)

The vector connecting the origin of coordinate system and center of the ball depends on the time and has the
components (Vt , 0, 0) . From this in the point, coinciding with the center of the ball, the components of the
vector of the momentum density of the gravitational field are always zero. At t = 0 the center of the ball passes
through the origin of the coordinate system, and at the time from (29) it follows that the maximum density of the
2
4π γ ρ0 R 2 V
γ M 2V
=−
field momentum g max = − 2
is achieved on the surface of the ball
2
2
2
9cg (1 − V 2 cg )
4π R 4cg (1 − V 2 cg )
on the circle of radius R in the plane YOZ , which is perpendicular to the line OX of the ball’s motion. The

same follows from (19).
We can integrate the components of the momentum density of gravitational field (29) over the volume inside the
moving ball at t = 0 in the coordinates (13) similar to (20):

24

Vol.23, 2013

Pxi = ∫ g xi d ϒ = −

www.iiste.org

4π γ ρ 02 V
2
2
9c g 1 − V 2 c g

4
3
∫ r sin θ dr dθ dϕ = −

Py i = ∫ g y i d ϒ = 0 ,

2γ M 2 V
2
2
15 R cg 1 − V 2 cg

.

(30)

Pz i = ∫ g z i d ϒ = 0 .

As in (20), the total momentum of the field (30) has only the component along the axis OX . By analogy with
(21) the coefficient before the velocity V in (30) is interpreted as the effective mass of the gravitational field
inside the ball:

mpi =
Where U i 0 = −

γM2
10 R

2
Pxi 1 − V 2 cg

V

=−

2γ M 2 4U i 0
,
=
2
2
15 R cg
3 cg

(31)

is the field energy inside a stationary ball.

Comparing (28) and (31) gives:

mg i =

2
3(1 + V 2 3 cg ) m pi

4

.

(32)

Connection (32) between the masses of the field inside the ball is the same as in (22) for the masses of the
external field, so the problem of 4/3 exists inside the ball too.
4. Conclusion
A characteristic feature of the fundamental fields, which include the gravitational and electromagnetic fields, is
the similarity of their equations for the potentials and the field strengths. As it was shown above, the external
potentials (9) of the gravitational (and similarly, the electromagnetic) field of the moving ball are similar by their
form to the potentials of the point mass (point charge) (8), and can be obtained both using the superposition
principle of potentials of the point masses inside the ball, and using the Lorentz transformation. We also
presented the exact field potentials (23) inside the moving ball, for which both the superposition principle and
the Lorentz transformation are satisfied.
From the stated above we saw that the 4/3 problem was common for both the electromagnetic and the
gravitational field. It also followed from this that considering the contribution of the energy and the momentum
of both fields into the mass of the moving body were to be done in the same way, taking into account the
negative values of the energy and the momentum of gravitational field and the positive values of the energy and
the momentum of electromagnetic field.
References
Fedosin, S.G. (1999), “Fizika i filosofiia podobiia: ot preonov do metagalaktik”, Perm: Style-MG.
Fedosin, S.G. (2008), “Mass, Momentum and Energy of Gravitational Field”, Journal of Vectorial Relativity 3
(3), 30-35.
Fedosin, S.G. (2009a), “Fizicheskie teorii i beskonechnaia vlozhennost’ materii”, Perm.
Fedosin, S.G. (2009b), Comments to the book: “Fizicheskie teorii i beskonechnaia vlozhennost’ materii”, Perm.
Feynman, R.P., Leighton, R., & Sands M. (1964), “The Feynman lectures on physics”. Massachussets: AddisonWesley. (Vol. 2).
Hajra, S. (1991), “Classical Electrodynamics – Reexamined”, Indian Journal of Theoretical Physics 42 (2), 164.
Heaviside, O. (1888/1894), “Electromagnetic waves, the propagation of potential, and the electromagnetic
effects of a moving charge”, Electrical papers 2, 490-499.
Liénard, A.M. (1898), “Champ électrique et Magnétique”, L’éclairage électrique 16 (27-29). 5-14, 53-59, 106112.
Searle, G.F.C. (1897), “On the steady motion of an electrified ellipsoid”, The Philosophical Magazine Series 5,
44 (269), 329-341.
Wiechert, E. (1900), “Elektrodynamische Elementargesetze”, Archives Néerlandaises 5, 549-573.

25

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