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2.7.4 Quantino pressure of uniformly moving charges

For uniformly moving point charges, the centers of emission spheres $\vec{r}_c(t,\tau)$ are always exactly where the source is located, i.e. with $\vec{v}_s := \dot{\vec{r}}_s(\tau)$ as constant speed of a source it is
$$\vec{r}_c(t,\tau) = \vec{r}_s(t) = \vec{r}_s(\tau) + \vec{v}_s\,(t-\tau).$$ (
If this is used in equations ( and (,
$$\vec{w}(\tau) = \frac{\vec{r}_d(t)-\vec{r}_s(t)}{t-\tau}$$ (
$$\vec{u}(\tau) = \frac{\vec{r}_d(t)-\vec{r}_s(t)}{t-\tau} - (\vec{v}_d - \vec{v}_s)$$ (
are obtained, where $\vec{v}_d := \dot{\vec{r}}_d(t)$ represents the constant velocity of the receiver charge.

Using equations ( and (, the quantino pressure ( can be converted into
$$\vec{\mathcal{P}}_e = \frac{a_c\,\mathcal{m}_{pho}}{8\,\pi} \,\int\limits_{0}^{\infty} \left\Vert\frac{\vec{r}}{\tau_s}-\vec{v}\right\Vert^2\,\frac{\vec{r}}{\tau_s}\,\sgn\left(\frac{r^2}{\tau_s}-\vec{r}\,\vec{v}\right) \,\Gamma\left(\frac{r}{\tau_s}\right)\,\frac{\intfunc_{0}^{c}\left(\left\Vert \frac{\vec{r}}{\tau_s}-\vec{v}\right\Vert\right)}{r^3}\,\d{\tau_s}$$ (
using the substitution $\tau_s := t-\tau$. For reasons of convenience, the abbreviations $\vec{r} := \vec{r}_d(t)-\vec{r}_s(t)$, $\vec{v} := \vec{v}_d-\vec{v}_s$ and $r := \Vert\vec{r}\Vert$ were introduced. The interval function can be omitted by using the facts that the lower condition is always true due to the calculation of the absolute value and the upper condition is fulfilled if $\tau_s$ exceeded a certain value, namely
$$\tau_c = \frac{\sqrt{r^2 c^2 - \Vert\vec{r}\times\vec{v}\Vert^2} - \vec{r}\,\vec{v}}{c^2 - v^2}.$$ (
Note that the expression below the root must remain real. However, if this term becomes imaginary, the condition in the interval function has no solution at all and therefore returns zero in the entire integration area. This is the case, for example, when $\vec{r}$ is perpendicular to $\vec{v}$ and $v$ reaches the speed of light. This shows that the quantino pressure and thus also the electrical force disappear completely when a charge moves transversely past another charge and the differential speed is greater than or equal to the speed of light.

With the help of $\tau_c$ it is possible to simplify expression ( to
$$\vec{\mathcal{P}}_e = \frac{a_c\,\mathcal{m}_{pho}}{8\,\pi}\,\frac{\vec{r}}{r^3}\,\int\limits_{\tau_c}^{\infty} \left\Vert\frac{\vec{r}}{\tau_s}-\vec{v}\right\Vert^2\,\frac{\Gamma\left(\frac{r}{\tau_s}\right)}{\tau_s}\,\sgn\left(\frac{r^2}{\tau_s}-\vec{r}\,\vec{v}\right)\,\d{\tau_s}.$$ (
To solve the integral, the emission velocity distribution $\Gamma(w)$ is expressed by a Taylor series, i.e. the ansatz
$$\Gamma(w) = \sum\limits_{k=1}^{\infty} \Gamma_k\,w^k$$ (
is chosen. With this,
$$\vec{\mathcal{P}}_e = \frac{a_c\,\mathcal{m}_{pho}}{8\,\pi}\,\frac{\vec{r}}{r^3}\,\sum\limits_{k=1}^{\infty} \Gamma_k\,\mathcal{I}_k$$ (
follows with
$$\mathcal{I}_k := \int\limits_{\tau_c}^{\infty} \left\Vert\frac{\vec{r}}{\tau_s}-\vec{v}\right\Vert^2\,\frac{r^{k}}{\tau_s^{k+1}}\,\sgn\left(\frac{r^2}{\tau_s}-\vec{r}\,\vec{v}\right)\,\d{\tau_s}.$$ (

The solution of the integrals of type $\mathcal{I}_k$ is a purely mathematical problem. Because of $\Vert\vec{x}\Vert^2 = \vec{x}\cdot\vec{x}$,
$$\mathcal{I}_k = \int\limits_{\tau_c}^{\infty} \left(\frac{r^2}{\tau_s^2} - \frac{2\,\vec{r}\,\vec{v}}{\tau_s} + v^2\right)\,\frac{r^{k}}{\tau_s^{k+1}}\,\sgn\left(\frac{r^2}{\tau_s}-\vec{r}\,\vec{v}\right)\,\d{\tau_s}$$ (
applies. By substitution of $r^2/\tau_s-\vec{r}\,\vec{v}$ with $z$ follows
$$\mathcal{I}_k = \frac{1}{r^{k+2}} \int\limits_{-\vec{r}\,\vec{v}}^{r^2/\tau_c-\vec{r}\,\vec{v}} (\vec{r}\,\vec{v} + z)^{k-1}\,\left(z^2 + r^2\,v^2 -(\vec{r}\,\vec{v})^2\right)\,\sgn\left(z\right)\,\d{z}.$$ (
This integral can be solved with help of formula
$$\int\limits_{a}^{b}\,f(x)\,\sgn(x)\,\d{x} = \sgn(a)\int\limits_{a}^{0}f(x)\,\d{x} + \sgn(b)\int\limits_{0}^{b}f(x)\,\d{x}$$ (
and after some rearranging of the terms we get
$$\begin{eqnarray}\mathcal{I}_k & = & \left(\sgn(\vec{r}\,\vec{v}) + \sgn\left(\frac{r^2}{\tau_c}-\vec{r}\,\vec{v}\right)\right) \left(\frac{\vec{r}\,\vec{v}}{r}\right)^k \left(\frac{(k+3)\,(\vec{r}\,\vec{v})^{2}}{(k+1)(k+2)\,r^{2}} - \frac{v^2}{k}\right) + \\ & & \sgn\left(\frac{r^2}{\tau_c}-\vec{r}\,\vec{v}\right) \left(\frac{r}{\tau_c}\right)^k\left(\frac{r^2}{(k+2)\,\tau_c^2} - \frac{2\,\vec{r}\,\vec{v}}{(k+1)\,\tau_c} + \frac{v^2}{k}\right).\end{eqnarray}$$ (
This can be simplified by analyzing the term $r^2/\tau_c-\vec{r}\,\vec{v}$ more closely. Without loss of generality it is always possible to choose the orientation of the coordinate system so that $\vec{r} = (r,0,0)$ and $\vec{v}=(v_{\parallel},v_{\perp},0)$ applies. If we insert this into the equation (, we get
$$\tau_c = \frac{r^2}{r\,v_{\parallel} + c\,r\,\sqrt{1-\frac{v_{\perp}^2}{c^2}}}.$$ (
And this in turn results in the equation
$$\sgn\left(\frac{r^2}{\tau_c}-\vec{r}\,\vec{v}\right) = \sgn\left(c\,r\,\sqrt{1-\frac{v_{\perp}^2}{c^2}}\right) = 1.$$ (
This simplifies formula ( further to
$$\begin{eqnarray}\mathcal{I}_k & = & \left(1 + \sgn(\vec{r}\,\vec{v})\right) \left(\frac{\vec{r}\,\vec{v}}{r}\right)^k \left(\frac{(k+3)\,(\vec{r}\,\vec{v})^{2}}{(k+1)(k+2)\,r^{2}} - \frac{v^2}{k}\right) + \\ & & \left(\frac{r}{\tau_c}\right)^k\left(\frac{r^2}{(k+2)\,\tau_c^2} - \frac{2\,\vec{r}\,\vec{v}}{(k+1)\,\tau_c} + \frac{v^2}{k}\right).\end{eqnarray}$$ ( Pure transversal motion

If the two point charges move directly sidewards past each other, the speed is purely transversal and orthogonal to the distance vector. In this special case is $v_{\parallel} = 0$. With this, $\vec{r}\,\vec{v} = 0$ and formula
$$\tau_{c,\perp} = \frac{r}{c\,\sqrt{1-\frac{v^2}{c^2}}}$$ (
follows from equation ( If we use this in formula (, we get
$$\begin{eqnarray}\mathcal{I}_k & = & \left(\frac{r}{\tau_{c,\perp}}\right)^k\left(\frac{r^2}{(k+2)\,\tau_{c,\perp}^2} + \frac{v^2}{k}\right) \\ & = & \left(c\,\sqrt{1-\frac{v^2}{c^2}}\right)^k\left(\frac{c^2 -v^2}{k+2} + \frac{v^2}{k}\right).\end{eqnarray}$$ (
As we can see immediately, $\mathcal{I}_k$ goes towards zero for $v \to c$. This means that the quantino pressure and thus also the electrical force disappears when the charges move laterally past each other at the speed of light or more. The two point charges are thus "invisible" to each other for very high speeds. Pure longitudinal motion

If the two point charges move in a direct line towards each other, the equations $v_{\perp} = 0$ and $\vec{r}\,\vec{v} = -r\,v$ apply. For equation ( then follows
$$\tau_{c,\parallel} = \frac{r}{c-v}.$$ (
If this is used in the formula (, we get
$$\begin{eqnarray}\mathcal{I}_k & = & \left(c-v\right)^k\left(\frac{(c-v)^2}{(k+2)} + \frac{2\,v\,(c-v)}{(k+1)} + \frac{v^2}{k}\right).\end{eqnarray}$$ (
Here, too, it is obvious that the electrical force becomes zero when the speed at which the two objects approach each other reaches the speed of light. If $v$ exceeds the speed of light, the force is also zero, since in integral ( the interval function for $v > c$ is zero.

In the case that the two point charges move away from each other on a direct line, $\vec{r}\,\vec{v} = r\,v$. This time the term $1 + \sgn(\vec{r}\,\vec{v})$ in expression ( does not disappear, but gives the value $2$ and we get
$$\mathcal{I}_k = 2\, v^k \left(\frac{(k+3)\,v^2}{(k+1)(k+2)} - \frac{v^2}{k}\right) + \left(c+v\right)^k\left(\frac{(c+v)^2}{(k+2)} - \frac{2\,v\,(c+v)}{(k+1)} + \frac{v^2}{k}\right).$$ (
If we set $k=1$, we get
$$\mathcal{I}_1 = \frac{1}{3}(c^3-v^3).$$ (
However, since $\mathcal{I}_1$ dominates the emission velocity distribution (, it follows that even in this case the electrical force becomes small for high speeds.

2.7.5 Quantino pressure for slowly uniformly moving charges

Since the difference speed $\vec{v}$ between source and receiver in everyday physics is always much smaller than $c$, it is worth considering this special case separately by approximating the integrals ( in the quantino pressure ( by truncated Taylor series.

Because $\vec{v}$ is a vector, the best way to do this is to use the vector $\kappa\,\vec{v}$ instead of $\vec{v}$. The actual series expansion is then performed at $\kappa = 0$. If all terms of order greater than $\mathcal{O}\{\kappa^2\}$ are neglected, we get
$$\mathcal{I}_k \approx c^k \left(\frac{c^2}{k+2} + \frac{(k-1)\,c}{k + 1} \left(\frac{\vec{r}}{r}\cdot\vec{v}\right) + \frac{k-2}{2}\left(\frac{\vec{r}}{r}\cdot\vec{v}\right)^2 - \frac{k - 2}{2\,k} v^2\right).$$ (
This means that the quantino pressure ( for speeds $v \ll c$ is
$$\begin{eqnarray}\vec{\mathcal{P}}_e(\vec{r},\vec{v}) & \approx & \frac{a_c\,\mathcal{m}_{pho}}{8\,\pi}\,\frac{\vec{r}}{r^3}\,\sum\limits_{k=1}^{\infty} \,\Gamma_k\,c^k \left(\frac{c^2}{k+2} + \frac{(k-1)\,c}{k + 1} \left(\frac{\vec{r}}{r}\cdot\vec{v}\right) \right. \\ & + & \left. \frac{k-2}{2}\left(\frac{\vec{r}}{r}\cdot\vec{v}\right)^2 - \frac{k - 2}{2\,k} v^2\right).\end{eqnarray}$$ (
As already mentioned, the quantino pressure is directly proportional to the electric force $\vec{F}$. In order to get a force from this "pressure" with the unit force per surface, it is obvious to multiply it by the cross section of the receiver elementary charge $\sigma_e$, which an elementary charge must have according to the messenger particle model of quantino theory. If we want to generalize the resulting formula to any amount of charge, we also have to multiply it by the relative charge of the source $q_s/e$ and the relative charge of the receiver $q_d/e$. It follows
$$\vec{F}(\vec{r},\vec{v}) = \frac{\sigma_e\,q_d\,q_s}{e^2}\,\vec{\mathcal{P}}_e(\vec{r},\vec{v}).$$ (
The physical constants $a_c$, $\mathcal{m}_{pho}$ and $\sigma_e$ are all unknown in their numerical value. However, one of the constants must be chosen in such a way that equation
$$\mu_0 = \frac{a_c\,\mathcal{m}_{pho}\,\sigma_e}{e^2}$$ (
is fulfilled. If we insert this and the equation ( into the formula (, the equation
$$\begin{eqnarray}\vec{F}(\vec{r},\vec{v}) & = & \frac{\mu_0\,q_d\,q_s}{4\,\pi}\,\frac{\vec{r}}{r^3}\,\sum\limits_{k=1}^{\infty} \,\Gamma_k\,\frac{c^k}{2} \left(\frac{c^2}{k+2} + \frac{(k-1)\,c}{k + 1} \left(\frac{\vec{r}}{r}\cdot\vec{v}\right) \right. \\ & + & \left. \frac{k-2}{2}\left(\frac{\vec{r}}{r}\cdot\vec{v}\right)^2 - \frac{k - 2}{2\,k} v^2\right)\end{eqnarray}$$ (
follows after some rearrangement of the terms.

This equation is similar to the force equation (, which, as already shown, explains both magnetism and gravity for slow, uniformly moving charges. The parameters $\Gamma_k$ are however unknown. Furthermore, they cannot be uniquely determined, since there are only four equations, whose fulfilment leads to a complete agreement with the formula ( These equations are:
$$\sum\limits_{k=1}^{\infty}\Gamma_k\,c^k\,\frac{1}{2(k+2)} \stackrel{!}{=} 1,\quad\sum\limits_{k=1}^{\infty}\Gamma_k\,c^k\,\frac{(k-1)\,c}{2(k+1)} \stackrel{!}{=} 0,$$ (
$$\sum\limits_{k=1}^{\infty}\Gamma_k\,c^k\,\frac{k-2}{4} \stackrel{!}{=} -\frac{3}{2}\quad\text{and}\quad\sum\limits_{k=1}^{\infty}\Gamma_k\,c^k\,\frac{k-2}{4\,k} \stackrel{!}{=} -1.$$ (
Possible parameters $\Gamma_k$ which satisfy these equations are
$$\Gamma_2 = \frac{4397}{265\,c^2},\,\Gamma_4 = -\frac{1053}{53\,c^4},\,\Gamma_6=\frac{546}{53\,c^6},\,\Gamma_8=-\frac{66}{53\,c^8}\quad\,\left(\Gamma_k=0\,\,\text{otherwise}\right).$$ (
If we insert these parameters ( into the formula (, we obtain
$$\vec{F}(\vec{r},\vec{v}) = \frac{\mu_0\,q_d\,q_s}{4\,\pi}\,\frac{\vec{r}}{r^3}\,\left(1 - \frac{3}{2}\,\left(\frac{\vec{r}}{r}\cdot\vec{v}\right)^2 + v^2\right).$$ (
With the known relation $\mu_0 = 1/(\varepsilon_0\,c^2)$ the equation
$$\vec{F}(\vec{r},\vec{v}) = \frac{q_d\,q_s}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}\,\left(1 - \frac{3}{2}\,\left(\frac{\vec{r}}{r}\cdot\frac{\vec{v}}{c}\right)^2 + \frac{v^2}{c^2}\right),$$ (
follows, which corresponds exactly to formula ( This proves that the quantino mechanism is capable of reproducing the initially postulated force law for slow, approximately nonaccelerated electrical charges.