With Two Gravitational Fields
by Louis Nielsen
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.
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:
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.
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.
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.
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.
From here we get the distance |O'P'|:
where s is defined by: s = (x² + (1-ß²) ·
(y² + z²))1/2
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.
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:
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.
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
operating with the rotation vector
on both sides of the equation sign in (36).
The last term in (37) we can rewrite as follows:
which gives the change of
per time unit,
if we follow the field, which is then constant.
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 .
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.
An argument for equation (48) might be formulated by the force law,
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.
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
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.
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.
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
of the order 10-14.
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.
which also gives an acceleration towards Mg.
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.
As K0 is also negative, the energy density given by (76) will also be equivalent to a negative mass.