The Heim gravity differential equation

The Heim differential equation for gravity is:

, where

, and

The solution of this equation is

, where

Where does this differential equation come from? It is clear that the field mass of gravity is involved. According to general relativity the total field mass density is where is the pure mass density of the field and is the pressure component of the field. For gravitational fields and

, which gives: . ( is gravity potential) The minus sign indicates that the field mass of gravity is negative. This negative field mass density will decrease the total mass of a body as you move outward from the centre of the body. We have

(1)

According to Heim: where M(r) is the varying mass.(Point mass +field mass).

This gives if you solve for M(r):

Substitute this in the differential equation (1)

Compare this to the Heim equation when

This is exactly the equation above apart from the coefficient 32/3!

Why do we have first order differential equation and not a second order equation?

In General Relativity we have the weak field equation:

(2)

The solution is .

Let us look at the assumption according to Heim. (M(r) = point mass+ field mass).

It looks quite plausible to identify the total mass in this way. However there is another

way of looking at it. If you instead identify the total mass with the relation

, which is Newton law of gravitation with a variable mass M(r). (Mass inside r)

Then you have

Substitute this in equation (1)

There we are! This is exactly the General Relativity equation (2)!

It looks like that one of the differences between Heim gravity and Einstein gravity

Is where to put the mass :

Heim :

General Relativity:

Which is the correct way? I think the Heim way is wrong!

As Anton Mueller points out in his “Response to Borje Mansson’s analysis”

Heim assumes that you do not have to consider contributions to potential field from the field mass outside the radius r. The potential field which is a scalar field do not add up in the same way as the gravitational field (a vector field.) If the field mass density = we have:

The second integral is the contributions from outside r. This contribution is not zero if !

Differentiate:

, where is the total mass (Point mass + field mass) inside r.

This means that it is the strength of gravity field (and not the strength of potential field) that only depends on the mass inside r.

But this is exactly the assumption leading to the General relativity equation (2)!

(With second order derivatives) So if the field mass is treated correctly the Heim solution when just becomes the general relativity solution. (As it should)

The incorrect treatment of field mass means we can not expect the original Heim gravity solution to be correct in strong fields.

Is it correct in weak fields?

General Relativity is a very reliable theory which predicts the outcome of different experiments with a high degree of accuracy.

Can Heim gravity do the same?

In the weak field limit we have with The Heim solution:

, where I have neglected terms of higher order than 2 .

We have (I have used the fact that <<)

With we have

The first term is the Newton potential and the second a first order correction term.

The exact solution of general relativity which gives the correct static field measured locally (Even in strong fields)

, where is the weak field solution

If we compare Heim and General relativity solution:

Heim:

General relativity:

As we know that the Einstein solution will match experimental tests very well (The advancement of perihelion of mercury, the bending of light rays for example)

We conclude that Heim original solution will not be able to predict the correct values

of these tests!

However, if the factor is changed to The Heim solution will have the same first order correction term as general relativity solution.

Then the Heim solution is:

Does all this mean we should dismiss Heim gravity altogether? Maybe not.

I think it is possible to adapt the core of Heim gravity to a general relativity consistent theory. If we solve the original Heim equation without field energy term we should get

a solution that corresponds to Newton gravity corrected for the fact that

, and not: Without field energy we have: (Proper Heim solution)

If we use the Puthoff polarizable vacuum representation of gravity, which is known to predict most general relativity tests correctly, the line element is:

where and is the gravity potential field.

If we substitute the above Heim solution for we have

, and the acceleration field

Here is a plot of the fields: The red line is gravity potential field and the green the acceleration field.

What could generate such a field?

The matter distribution kg/m³ that generate K can be calculated with

. If you substitute

and calculate K you get :

This corresponds to a negative effective mass distribution .This looks strange, but this is a effective mass distribution and not the real mass distribution.

There are hidden negative pressure components added to the real mass distribution.

This is not obvious in the Puthoff representation but is transparent in general relativity

where . The negative sign in indicates presence of dark energy. Dark energy has the peculiar feature that its inertial mass is positive, but its effective gravitational mass is negative (Because of negative pressure), and will anti gravitate.

The Puthoff representation gives a clue how to proceed with the real thing:

How do we find the general relativity solution that corresponds to the matter distribution Heim solution indicates?

If we assume a spherical symmetric static line element of the type

where and is the time coordinate, the relativistic Poisson equation for a special matter distribution is:

where is inertial mass density , pressure in radial direction and pressure in direction perpendicular to the radius r, and the cosmological constant.

If we assume to be composed of dark energy with we have according to Irina Dymnicova ( http://www.arxiv.org/abs/gr-qc/0010016)

If we substitute these equations in the Poisson equation we have:

If we now assume the dark energy distribution we have:

, the same as the Puthoff representation gives.

, where can be interpreted as a constant dark energy distribution

, which gives the cosmological constant.

Let us solve the differential equation:

(3)

, where is a constant of integration.

, where is another constant of integration.

If we identify with (Schwarzschild term), we have

Here we can identify constant =1.

So we come up with the solution:

The line element becomes:

(4)

The Heim proper solution contains no cosmological constant. If we put we have the G. R solution that corresponds to Heim proper solution.

, where

Potential field and acceleration field:

The static potential field =

And acceleration field: . (Gravity as seen by local observer for an object at rest).The function M(r) where M is mass is defined by or

This means that the effective mass is negative outside , a result of the dark energy component in Heim gravity.

Plot of the fields shows a close resemblance to the original Heim solution.

However, now we have genuine event horizons when .The lower horizon corresponds (almost) exactly to the usual Schwarzschild horizon and the outer

to the border of visible universe.

However, it is not very plausible that one point mass could determine where the border of visible universe is.

Now, what about the in the line element (4)

We can interpret the term in (3) as a dark energy contribution generated by the mass M. This means that should be the total contribution of all other masses in the universe. If we integrate over the part of universe that can influence our spot we should get.

Here =maximal radius, is the proper volume element, the total mass distribution density of the other type of matter in universe made up of particles (Dark matter and ordinary matter) distribution. (Assumed to be a constant)

Now we can not use the line element (4) as this just use one point mass.

If you solve the G. R equations with a constant ordinary mass density and a constant dark energy density the global line element becomes:

, where .

We observe that the maximum radius then is so we have:

In the integration we assume a constant mean value of.

So we have

I we solve for:

The radius of visible universe:

According to astronomy

We can now solve for and m expressed in known experimental values of and :

, , ,

.

With and =, we can calculate:

,, , (46% of ) and (29% of proton mass). The m-value is a mean value of the particle masses that build up ordinary matter ( about 4% of all matter) and dark matter (about 23% of all matter)

If we assume we have the line element

where is the value of generated by M, and is a mean value generated by all the masses in visible universe. Acceleration field:

-

In the plot I have assumed.

The main difference with is:

1) The acceleration field becomes negative for a value less then because the influences from all the other masses in universe.

2) The radius of visible universe is determined by all the masses in visible universe.

I conclude:

The original Heim gravity differential equation is wrong, because of wrong interpretation of field mass. At least the coefficients must be wrong, as they will not lead to the correct first order correction term to Newton gravity. However, if the particle connection ρ=h²/Gm³ is correct, the Heim proper solution (The solution without field mass), can be made a bases for a solution of ordinary General Relativity field equations. The connection can be interpreted as a dark energy generation from ordinary masses. The cosmological constant could then be interpreted as the result of addition of all mass contributions to dark energy in visible universe. A relation between ρ, cosmological constant and mean density of matter can be established.

I think the dark energy generation could be seen as a repulsion of space itself. All masses make the space around them to expand with a constant acceleration

-GM/ρ². Mass attracts masses, but repels space. This means that the repulsion acceleration is greater than normal attraction for r> ρ. It is not an ordinary repulsion, because it is space itself that “moves”. However for an observer it looks like a “real” repulsion. Of course all these conclusions are useless if Heims derivation of the particle connection also is wrong.

/ Börje Månsson