A Maxwell Analog Gravitation Theory
With Two Gravitational Fields

by Louis Nielsen
Senior Physics Master, M.Sc.
E-mail: louis44nielsen@gmail.com


I shall in this article show that the gravitational rotation field, which I introduced in an earlier article (Louis Nielsen, Gamma No. 9, 1972, Niels Bohr Institute, Copenhagen), is a consequence of the special theory of relativity.
The calculations are used to show that the magnetic induction field is a relativistic correction to electrostatics. It can namely be shown that Maxwell's equations are a consequence of:

  1. Coulomb's electrostatic forces,
  2. the special theory of relativity, and
  3. the invariance of electric charge.
Likewise it has been shown that the origin of the magnetic induction field is found in the relativistic Thomas rotation. An analog argumentation leads to the existence of a gravitational rotation field.

Deduction of the gravitational equations

I shall now deduct the field equations, which must be fulfilled by the gravitostatic field and the gravitational rotation field . I shall use the following starting points and assumptions:

    a) Newton's gravitostatic force law,
    b) the transformation equations for positions, times, velocities, and forces of the special theory of relativity, and
    c) the assumption that the gravitational mass is Lorentz invariant.
Ad a):
Newton's force law for resting gravitational charges (equal to gravitational masses) I shall write as follows:


where m1 and m2 are the gravitational forces between two particles, r the distance between them, and a negative parameter, connected to Newton's gravitational constant G by the formula:


We shall everywhere use the SI unit system. As we do not have a minus sign in (1), must necessarily be negative for the gravitational masses, as we know them in our part of the Universe. Our experience tells us that gravitational masses with the same sign attract each other.
We see that equation (1) is mathematically identical to Coulomb's force law for resting electrical charges:


where now q1 and q2 are the electrical charges, and the dielectricity constant in vacuum. The physics behind equations (1) and (3) is not identical. This f.i. corresponds to the fact that the physics behind water waves and electromagnetic waves is not identical, but the mathematical description is identical. Another example among many, where the mathematical structure is identical, but the physics different, can be found in the phenomena spin and isospin. This mathematical isomorphism (same form) is an experience which should be remembered.

Ad b):
We shall assume that we can use the transformation equations of the special theory of relativity. We shall use the Lorentz transformation:



and the transformation equations for the components of force:




The unmarked quantities are measured in the inertial frame I, the marked in the inertial frame I', which moves with the velocity v along the X-axis of I.   u' is the velocity of a particle in I'. The quantity c0 is a nature constant, corresponding to the velocity of a photon in field free vacuum. By field free vacuum I shall understand a domain of space, absolutely free from both fields and matter.
Besides equations (4) to (8) we shall use the transformation equations for velocities.

Ad c):
The assumption that the gravitational mass is Lorentz invariant collides with the equivalence principle. – The equivalence principle can be formulated as the principle that the inertial and the gravitational masses are identical.
As the inertial mass increases by velocity, the gravitational mass will also increase in the same way. This seems intuitively to be very strange.
Let us consider two initial systems, I and I'.

In the inertial frame I the gravitational mass m2 is moving with velocity v. The inertial frame I' is moving with velocity v in relation to frame I and along this frame's X-axis. The mass m2 will then be at rest in frame I', and we shall presume that it is situated at the origo O', the starting point for I'. We shall calculate the force on a particle with the mass m1, with at the time t=0 is situated in the point P, measured in relation to the frame I. The particle m2 is then just in the point O in the frame I. At the time t=0 the mass m1 has the velocity = (ux, uy, uz) in frame I and similarly ' = (u'x, u'y, u'z) in frame I'. In the inertial frame I' we have at time t=0 in frame I:





From here we get the distance |O'P'|:


where s is defined by: s = (x² + (1-ß²) · (y² + z²))1/2
The numerical value of the force F' on the mass m1 in point P' will at the time t' = - v/c0²·x·(1-ß²)-1/2, corresponding to time t=0 in system I be given by:


or in components along the x'-, y'-, z'-axes:




Inserting (15) in the transformation equation (6) we get:


This can be written as:




where g is defined by:


For the other components we get by means of the transformation equations:



Equations (20), (22) and (23) give the force components as measured in the inertial frame I. We see that the force between the particles depends on their velocities in frame I.
Equations (20), (22) and (23) now allow us to define a new gravitational field in such a way that it can be expressed by a gravitational force law, as follows:


or written in components:




Writing equation (20) as:


and comparing with (25), we conclude that the -field is defined by:


Likewise we get by comparing the equation:


by (26):


Finally we compare:


with (27), and get:


We have thus shown that in the inertial field I the -field is determined by:


where is the position vector for the point P. We see that the field depends on the velocity v, whereby the mass m2 is moving in relation to system I.
Collecting the components for the -field we get:


and, we see that it has the nature of a rotation field. The equations in (35) mean that the following vector equation is valid:


We can now set up a rotation equation for the -field, operating with the rotation vector on both sides of the equation sign in (36).
We get:


The last term in (37) we can rewrite as follows:
As follows from this, that the total derivative of is given by:


which gives the change of per time unit, if we follow the field, which is then constant.
The total derivative is thus zero, and we get:


which inserted in (37) gives:


By introduction of "the gravitational permeability of vacuum" and the vector for mass current density we finally get the equation:


In equation (37) we have used that:


which is an equivalent, but still more general equation than Newton's gravitostatic force law (1). Equation (42) can be understood as follows: Let the mass m2 be placed in the center of a sphere with radius r. On the surface the -field is radial and given by:


Integrating over the sphere and using Gauß's integral theorem we get:


At the limit to a small volume around m2 we get (42) as:


We now only need equations for and .
As an assumption, but definitely not sure, we shall consider the -field as free of sources, viz. as a pure whirl field. We thus assume that the divergence of the -field is negligible, which means:


It is quite possible that -field monopoles exist, and (46) must then be modified as:


where is a constant, and is the density of the -field monopoles.
We shall now deduct an equation for . We could now go back to the original transformation equations and deduct expressed by and . We shall not do that, but simply give the result:


An argument for equation (48) might be formulated by the force law, equation (24).
Let there be given an -field, and let us consider a definite point within it. Let a particle arrive to this point and here have the velocity . This requires that the -field in the point shall change in time in a way so that the force on the particle is zero. If we assume that a change of the -field in time is introducing a -field, we see from (24) that , which gives (48). By operation with the rotation operator we get from (48):


as and .
I have hereby shown that the special theory of relativity requires the existence of two gravitational vector fields characterized by the -field and the -field.
I have thus demonstrated that the following equations must be valid (pending the absence of -field monopoles):





together with the force law defining the - and -fields:


These equations are in their primary structure mathematically identical to Maxwell's electrodynamic field equations for the electric -field and the magnetic induction field , as these fields obey the equations:





together with the force law, defining the - and -fields:


Here is the dielectric constant of vacuum and the magnetic permeability of vacuum, is the gravitational mass density, and the electrical charge density.

Cross references:
  1. A two-vector field gravitation theory with variable coupling »constants«
  2. Centrifugal and Coriolis forces are identical to the N¯ field

Non-linear field equations. Introduction of negative gravitational masses

In spite of the primary mathematical resemblance between the two theories, expressed by the two sets of equations (50) to (54) and (55) to (59), the physical difference, however, is great. Maxwell's electrodynamical theory is linear while the present theory is non-linear. This non-linearity is a consequence of Einstein's mass-energy relation, coupled with the equivalence principle. Einstein's mass-energy relation tells that any energy E has an inertia corresponding to an inertial mass mi given by:


This is a result of the special theory of relativity and is thus only relating to the inertial mass. The equivalence principle, which is basical for Einstein's general theory of relativity, can be formulated as follows:
If a particle has an inertial mass mi then there will to this particle also be attached an equivalent gravitational mass mg of the same size as mi, viz. the following is valid:


This result means that to every energy E there is also attached an equivalent gravitational mass mg given by:


As there is, in the gravitational field, an energy density this corresponds to an equivalent mass density as given by (62). This mass density causes a new gravitational field etc.
This special interchange between field and mass causes that the gravitational field equations become non-linear. We shall investigate this below.   We shall, for simplicity, consider a resting gravitational mass mg. This mass is creating a gravitostatic field according to the equation:


It can be shown that this field (similar to the field around a resting electrical charge) contains an energy density Ug given by:


As is negative, we see that the energy density in the field is negative. According to (62) this corresponds to a negative gravitational mass density:


If (62) is exact and valid we must conclude that negative gravitational masses exist. This result is highly interesting and simultaneously it fulfils our sense for symmetry. If the interpretation is true can, however, be discussed. The equivalence principle, as expressed by equation (61) causes that a negative gravitational mass must also have a negative inertial mass, which seems to be rather remarkable. If (61) is generally valid, negative inertial masses must be the consequence of negative gravitational masses.
The mass density (65) gives raise to a -field which couples with the original, so that the total -field must be determined by the equation:


Equation (66) is a non-linear differential equation, which causes that it is rather complicated to solve. If the gravitational fields are relatively weak, the equations can be linearized.  F.i. the last term in (66) will be vanishing in comparison with , if we consider the Earth.   will be of the order 10-5 and of the order 10-14.
It could now be interesting to calculate how much negative mass the gravitostatic -field around the Earth is equivalent to. The energy density Ug (J/m³) is, in the distance r from the center of the Earth, given by:




By integration from the surface of the Earth to infinity we get for the total energy Eg in the field:


Here R is particle radius (radius of the Earth) and M particle mass (mass of the Earth). As we note is still negative, and the total energy in the field is negative. Expressed by Newton's gravitational constant G we can write:


This expression formally corresponds to the potential energy of a particle with the mass M, being in the distance 2·R from another particle, also with the mass M. The energy (70) is equivalent to a negative gravitational mass:


Using the values M = 1024 kg, R = 6·106 m, G = 6.6·10-11 Nm²/kg² we get a negative mass of the order mg = -1014 kg.

Interaction between positive and negative masses

In this section we shall investigate how negative and positive masses interact. Premarily we shall assume that the equivalence principle is valid. This causes, as earlier mentioned, that a negative gravitational mass must be allocated a negative inertial mass. Let us consider a positive gravitational mass mg being in a field from a fixed positive gravitational mass Mg.

The equation of motion for mg is given by:


which gives an acceleration towards Mg.
Now we consider a negative gravitational mass -mg which will have the equation of motion:


which also gives an acceleration towards Mg.
Now let us consider a negative gravitational mass -Mg which is fixed (f.i. by means of supporting forces). Let in the field first a positive, then a negative mass move.
The positive mass has the equation of motion:


and thus an acceleration way from the mass M. The negative mass has the equation of motion:


which also means an acceleration away from Mg.
The results of (72) to (75) are based on the assumption that the equivalence principle is valid so that se must operate with negative inertial mass. If we limit the equivalence principle to be valid only for positive gravitational masses and thus furnish a negative gravitational mass with a positive inertial mass, then we get the following interaction: gravitational masses with the same sign attract each other, whereas masses with different sign repulse.
In the previous section we considered a resting gravitational mass. If a gravitational mass moves in relation to an observer, there will besides the -field also be an -field. The energy density Ug, which is in the field, is then given by:


As K0 is also negative, the energy density given by (76) will also be equivalent to a negative mass.

Louis Nielsen
E-mail: louis44nielsen@gmail.com

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