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2.4 The derivation of gravity from the electromagnetic force

2.4.1 The plasma droplet model

In section 2.2.1 it was shown that magnetism can be explained in a very clear and convincing manner by superimposing the individual forces of many moving electrical charges (see figures and This section shows that the same applies to gravity.

Figure How are the two charge quantities to be weighted so that the object appears neutral towards the outside?
For this purpose, an object is postulated with positive and negative point charges within a very small volume of space. Both charge quantities are to be in permanent motion in the form of a gas or plasma. In the following we want to call this plasma droplet a "point mass". The average value of the velocity distribution is initially zero for both charge quantities, i. e. the object should rest as a whole. However, the variances $\nu_p$ and $\nu_n$ shall differ. The question that will be examined in the following is whether both charge quantities have to be equal in size, so that the object appears to be electrically neutral towards the outside. Figure illustrates the model with a sketch, in which the variance of the positive charge quantity is much smaller than that of the negative one.

In order for such an object to appear electrically neutral towards the outside, the resulting force on a distant test charge $q_d$ must disappear. Obviously, the total force of one of these objects on a test charge consists of two components, namely
  1. the force of the positive charge cloud and
  2. the force of the negative charge cloud.
The calculation of the partial forces can be performed by using equation ( To do this, the force of each possible velocity must be weighted and integrated according to the frequency distribution.

It shall be assumed in the following, that the velocities $\vec{u}$ within the charge clouds are Gauss-distributed. Furthermore, it is assumed that none of the space-directions are privileged or special in any respect. The velocity distribution of a charge cloud resting in the temporal average thus has the form
$$p_u(\vec{u},\nu) = g(\vec{u},\nu) = g(u_x,\nu)\cdot g(u_y,\nu)\cdot g(u_z,\nu),$$ (
whereby $g(u,\nu)$ represents the Gaussian function
$$g(u,\nu) = \frac{1}{\sqrt{2\,\pi\,\nu}}\,\exp\left(-\frac{u^2}{2\,\nu}\right)$$ (
and the parameter $\nu$ the variance.

With the help of the velocity distribution (, the formula ( can be generalized in such a way that it is also valid for charge clouds. Basically, for the force $\vec{F}_t(\nu,q_s,q_d)$ of a source charge cloud with the total charge $q_s$ and the speed variance $\nu$ onto a target charge $q_d$ located at the position $\vec{r}$ with the speed $\vec{v}$, the relationship
$$\vec{F}_t(\nu,q_s,q_d) = \iiint\limits_{-\infty}^{+\infty}\,\vec{F}_{R}(\vec{r},\vec{u}-\vec{v})\,p_u(\vec{u},\nu)\,\d{\vec{u}}$$ (
applies. Inserting formulas ( and ( gives the force
$$\vec{F}_t(\vec{r},\vec{v},\nu,q_s,q_d) = \left(1 + \frac{v^2}{c^2} - \frac{3}{2}\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2 + \frac{3}{2}\frac{\nu}{c^2}\right)\,\frac{q_d\,q_s}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}$$ (
after the integration. As we can see, this formula is only the same as force ( for $\nu = 0$, but it is stronger for $\nu \neq 0$. This means that electric point charges enclosed in a small volume of space generate a greater force when they move like gas molecules. This again shows that in Quantino theory the electrical charge is relative.

But back to the plasma droplet in Figure We want to calculate the ratio of positive charge $q_p$ and negative charge $q_n$, which is necessary to ensure that the force of the entire plasma droplet onto a stationary sample charge $q_d$ is zero. The total force
$$\vec{F} = \vec{F}_t(\vec{r},0,\nu_p,q_p,q_d) + \vec{F}_t(\vec{r},0,\nu_n,q_n,q_d)$$ (
on a resting test charge $q_d$ at any location $\vec{r}$ should therefore be zero. By inserting the previously calculated force formula (,
$$\vec{F} = \frac{q_d\,q_p}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}\,\left(1 + \frac{3}{2}\,\frac{\nu_p}{c^2}\right) + \frac{q_d\,q_n}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}\,\left(1 + \frac{3}{2}\,\frac{\nu_n}{c^2}\right)$$ (
follows. And this force can only be zero for $\nu_n \neq \nu_p$ if
$$q_p = \frac{2\,c^2 + 3\,\nu_n}{2\,c^2 + 3\,\nu_p}\,(-q_n)$$ (
is valid, as can be verified by inserting. If the total charge quantity $Q$ is defined by
$$Q := \vert q_n\vert + \vert q_p\vert = -q_n + q_p,$$ (
the formulas
$$q_n = -\frac{2\,c^2 + 3\,\nu_p}{4\,c^2 + 3\,(\nu_n + \nu_p)}\,Q$$ (
$$q_p = +\frac{2\,c^2 + 3\,\nu_n}{4\,c^2 + 3\,(\nu_n + \nu_p)}\,Q$$ (
follow from equation ( With these values the "point mass" is electrically neutral, or in other words, a static electrical charge in the environment is neither attracted nor repelled.

2.4.2 The cause of gravity

Figure Is there a force between these objects even though they are electrically neutral?
In this section, two plasma droplets are examined. One of it shall be at the origin of the coordinates and the other at the location $\vec{r}$. Furthermore, there shall be a differential velocity $\vec{v}$. We are interested on the force between the two objects.

At first, we consider that the total force of one object onto the other consists of four parts, namely
  1. the force $\vec{F}_{pp}$ of the positive charge cloud onto the other positive charge cloud,
  2. the force $\vec{F}_{nn}$ of the negative charge cloud onto the other negative charge cloud,
  3. the force $\vec{F}_{pn}$ of the positive charge cloud onto the negative charge cloud and
  4. the force $\vec{F}_{np}$ of the negative charge cloud onto the positive charge cloud.
With the aid of formula (, each of these partial forces can be stated:
  1. $\vec{F}_{pp} = \vec{F}_t(\vec{r},\vec{v},\nu_{p1}+\nu_{p2},q_{p1},q_{p2})$
  2. $\vec{F}_{nn} = \vec{F}_t(\vec{r},\vec{v},\nu_{n1}+\nu_{n2},q_{n1},q_{n2})$
  3. $\vec{F}_{pn} = \vec{F}_t(\vec{r},\vec{v},\nu_{p1}+\nu_{n2},q_{p1},q_{n2})$
  4. $\vec{F}_{np} = \vec{F}_t(\vec{r},\vec{v},\nu_{n1}+\nu_{p2},q_{n1},q_{p2})$
The total force $\vec{F}_G$ is the sum of all individual forces $\vec{F}_{pp} + \vec{F}_{nn} + \vec{F}_{pn} + \vec{F}_{np}$. By inserting formula (,
$$\begin{eqnarray}\vec{F}_G & = & \frac{\vec{r}}{4\,\pi\,\varepsilon_0\,r^3}\,\left(1 + \frac{v^2}{c^2} - \frac{3}{2}\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2 + \frac{3}{2}\frac{\nu_{p1} + \nu_{p2}}{c^2}\right)\,q_{p1}\,q_{p2} + \\ & & \frac{\vec{r}}{4\,\pi\,\varepsilon_0\,r^3}\,\left(1 + \frac{v^2}{c^2} - \frac{3}{2}\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2 + \frac{3}{2}\frac{\nu_{n1} + \nu_{n2}}{c^2}\right)\,q_{n1}\,q_{n2} + \\ & & \frac{\vec{r}}{4\,\pi\,\varepsilon_0\,r^3}\,\left(1 + \frac{v^2}{c^2} - \frac{3}{2}\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2 + \frac{3}{2}\frac{\nu_{p1} + \nu_{n2}}{c^2}\right)\,q_{p1}\,q_{n2} + \\ & & \frac{\vec{r}}{4\,\pi\,\varepsilon_0\,r^3}\,\left(1 + \frac{v^2}{c^2} - \frac{3}{2}\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2 + \frac{3}{2}\frac{\nu_{n1} + \nu_{p2}}{c^2}\right)\,q_{n1}\,q_{p2} \end{eqnarray}$$ (
follows. By using equations ( and (, this can be simplified to
$$\vec{F}_G = -\frac{3\,(\nu_{p1} - \nu_{n1})}{4\,c^2 + 3\,(\nu_{p1} + \nu_{n1})}\,\frac{3\,(\nu_{p2} - \nu_{n2})}{4\,c^2 + 3\,(\nu_{p2} + \nu_{n2})}\,\left(1 - \frac{v^2}{c^2} + \frac{3}{2}\left(\frac{\vec{r}}{r}\frac{\vec{v}}{c}\right)^2\right)\,\frac{Q_1\,Q_2}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}.$$ (
Now we define
$$M_i := \frac{3\,(\nu_{pi} - \nu_{ni})}{4\,c^2 + 3\,(\nu_{pi} + \nu_{ni})}\,\frac{Q_i}{\sqrt{4\,\pi\,\varepsilon_0\,G}},$$ (
where $G$ stands for the gravitational constant. By inserting in formula (,
$$\vec{F}_G(\vec{r},\vec{v}) = -G\,\left(1 - \frac{v^2}{c^2} + \frac{3}{2}\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2\right)\,M_1\,M_2\,\frac{\vec{r}}{r^3}$$ (
follows. It is remarkable that for $\vec{v}=0$ or $c \to \infty$ this corresponds to the Newtonian gravitational law
$$\vec{F}_G = -G\,M_1\,M_2\,\frac{\vec{r}}{r^3},$$ (
assuming that the $M_i$s are the gravitational masses of both objects. This means that it is possible to interpret the entire mass of an object as an electrical multi-particle effect, provided the hypothesis is accepted that gravitational mass (gravity charge) is in the end only moving electrical charge in imbalance.

It is also remarkable that the gravitational mass in definition ( can be positive ($\nu_p > \nu_n$), negative ($\nu_p < \nu_n$) or zero ($\nu_p = \nu_n$). Assuming that the variances of the negative charge quantities $\nu_{n1}$ and $\nu_{n2}$ are zero, positive values follow for the two masses $M_1$ and $M_2$, since the total charge quantities $Q_1$ and $Q_2$ cannot be negative due to their definition ( The force $\vec{F}_G$ is attractive.

If, however, the variances of the positive charge quantities are zero, as shown in Figure, so the masses are negative. This seems pointless in the first moment, but after a short reflection it is clear that this must be antimatter. The force $\vec{F}_G$ is also attractive between two antimasses. It is interesting, however, that a mass and antimass repel each other! In this aspect, electrical force and gravitation differ, because the electric force has a repulsive effect on similar objects and attracts opposite ones. In gravity, this is exactly the opposite. This is consistent with the fact that only matter is present in our immediate surroundings and explains why there is no antimatter in the vicinity, since it has either been annihilated long ago or pushed away due to antigravity.

The hypothesis that in normal matter the positive charge quantity possesses the greater velocity variance is also supported by the fact that the mass of a hydrogen atom is slightly smaller than the sum of the two individual masses of electron and proton. The system consists of a fast, negatively charged electron and a slow, positively charged core. This together forms some antimass, whereby the total mass of the atom appears to be reduced in comparison with the individual masses of proton and electron.

2.4.3 The laws of conservation for gravity

Since the plasma droplet model of gravity is based solely on electrical force ( and the laws of conservation have already been proven for this force, it follows immediately that the laws of conservation must also apply to gravity. However, it is possible to carry out the proofs without the electric force only on the basis of formula (

The fulfillment of momentum conservation is easy to show, because the symmetry property $\vec{F}_G(\vec{r},\vec{v}) = -\vec{F}_G(-\vec{r},-\vec{v})$ is also valid for the formula of gravity (, which means that the proof from section 2.3.1 is directly transferable to gravity. The conservation of angular momentum is also immediately apparent, as ( is a central force. This had been in section 2.3.2 the only premise for the proof of angular momentum conservation.

The conservation of energy can also be proven. If we compare the formula of the gravitational force ( with the formula of the electric force (, so we notice that both have the same structure. It is therefore possible to immediately deduce the form of the potential energy for gravity. For a point mass with the gravitational mass $M_d$ in the field of another point mass $M_s$,
$$V_G(\vec{r},\vec{v}) := -G\,M_s\,M_d\,\frac{1}{r}\left(1 + \frac{(\vec{v}\,\vec{r})^2}{2\,c^2\,r^2} - \frac{G\,M_d\,M_s}{2\,m_d\,c^2\,r}\right)$$ (
must apply. $m_d$ is here the inertial mass of the point mass $M_d$ and identical to the gravitational mass $M_d$ because of the numerical value of $G$. The proof that this is indeed the potential energy is fully equivalent to that shown in Section 2.3.3.

For the classical law of gravity, i. e. for formula ( with $c\to\infty$, besides these three conserved quantities there is a fourth one, which is called Runge-Lenz-Vector. This conserved quantity does not exist for formula (, which has the consequence that, for example, Kepler's orbits of planets slowly change their orientation. More information on this topic can be found in section 3.3.

2.4.4 Forces between moving masses and electric charges

As previously shown, gravity is probably an electrical effect. This means that it must be assumed that a force will occur between a mass and an electrical charge if there is a differential velocity between the both. Nothing like this has ever been observed. Does this mean that the above considerations are wrong? The answer is no, because this interaction is so small, even for higher speeds and masses, that there is no chance to measure this effect directly.

To show this, we use formula ( and calculate the force of a point mass onto a charge $q$ with the relative velocity $\vec{v}$ at the position $\vec{r}$. This force is
$$\vec{F}_{QM} = \vec{F}_t(\vec{r},\vec{v},\nu_p,q_p,q) + \vec{F}_t(\vec{r},\vec{v},\nu_n,q_n,q),$$ (
which for $\vec{v} \neq 0$ corresponds to equation ( Inserting the formulas (, ( and ( yields
$$\vec{F}_{QM} = \frac{3\,(\nu_n - \nu_p)}{4\,c^2 + 3\,(\nu_n + \nu_p)}\,\left(\frac{v^2}{c^2} - \frac{3}{2}\,\left(\frac{\vec{r}}{r}\frac{\vec{v}}{c}\right)^2\right)\frac{q\,Q}{4\,\pi\,\varepsilon_0}\frac{\vec{r}}{r^3}.$$ (
Using the definition (, this becomes
$$\vec{F}_{QM} = \sqrt{\frac{G}{4\,\pi\,\varepsilon_0}}\,\left(\frac{v^2}{c^2} - \frac{3}{2}\,\left(\frac{\vec{r}}{r}\frac{\vec{v}}{c}\right)^2\right)\,M\,q\frac{\vec{r}}{r^3}.$$ (
If we now define the virtual dynamic charge $\tilde{q}$ of a mass by
$$\tilde{q} := \sqrt{G\,4\,\pi\,\varepsilon_0}\,\left(\frac{v^2}{c^2} - \frac{3}{2}\,\left(\frac{\vec{r}}{r}\frac{\vec{v}}{c}\right)^2\right)\,M,$$ (
so the equation
$$\vec{F}_{QM} = \frac{\tilde{q}\,q}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}$$ (
follows for the force, what formally corresponds to Coulomb's law. This formula is only valid for velocities $v \ll c$.

Inserting the numerical values for the constants $G$ and $\varepsilon_0$ shows that a mass of $170\,kg$ with a transverse speed of $1000\frac{m}{s}$ has only a virtual electric charge which corresponds approximately to an elementary charge.