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2.3 Proofs of the conservation laws

2.3.1 Conservation of momentum

In quantino theory, the electric force of a point charge $q_s$ at the coordinate origin onto a second point charge $q_d$ at the place $\vec{r}$ with the velocity $\vec{w}$ is given by equation (

The force of a point charge $q_i$ at the location $\vec{r}_i$ with the speed $\dot{\vec{r}}_i$ onto another point charge $q_j$ at the location $\vec{r}_j$ with the speed $\dot{\vec{r}}_j$ is thus
$$\vec{F}_{ij} = \vec{F}_R(\vec{r}_j-\vec{r}_i,\dot{\vec{r}}_j-\dot{\vec{r}}_i).$$ (
Likewise, the force of a charge $q_j$ onto a charge $q_i$ is represented by
$$\vec{F}_{ji} = \vec{F}_R(\vec{r}_i-\vec{r}_j,\dot{\vec{r}}_i-\dot{\vec{r}}_j).$$ (
Furthermore, due to the structure of force formula (, relationship
$$\vec{F}_{ji} = -\vec{F}_R(\vec{r}_j-\vec{r}_i,\dot{\vec{r}}_j-\dot{\vec{r}}_i)$$ (
applies, i. e.
$$\vec{F}_{ij} = -\vec{F}_{ji}.$$ (
For an isolated system of $n$ point charges in total, the equation
$$\dot{\vec{p}}_i = \sum\limits_{j=1}^{n} \vec{F}_{ji}$$ (
applies to the temporal change of the momentum $\vec{p}_i$ of the $i$-th point charge, provided that it is defined that $\vec{F}_{ii} := 0$. The change of the total momentum of the system is then the sum of all individual momentum changes and
$$\sum_{i=1}^{n} \dot{\vec{p}}_i = \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} \vec{F}_{ji}.$$ (
Because of property (,
$$\sum_{i=1}^{n} \dot{\vec{p}}_i = -\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} \vec{F}_{ij} = -\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{n} \vec{F}_{ij} = -\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} \vec{F}_{ji}$$ (
also holds true. From this it follows that for the change of the total momentum it must be
$$\sum_{i=1}^{n} \dot{\vec{p}}_i = 0.$$ (
Consequently, the total momentum of the system does not change over time and is therefore a conserved quantity.

2.3.2 Conservation of angular momentum

The angular momentum $\vec{L}$ is defined as a cross product of the vector $\vec{r}$ and the momentum $\vec{p}$. For a point charge $q_d$ at the place $\vec{r}_d$ therefore equation
$$\vec{L} := \vec{r}_d \times \vec{p}_d$$ (
applies. Derivation according to time results in
$$\dot{\vec{L}} = \dot{\vec{r}}_d \times \vec{p}_d + \vec{r}_d \times \dot{\vec{p}}_d.$$ (
Because the momentum $\vec{p}_d$ is always parallel to velocity $\dot{\vec{r}}_d$, relation $\dot{\vec{r}}_d \times \vec{p}_d = 0$ follows. This turns equation ( into
$$\dot{\vec{L}} = \vec{r}_d \times \dot{\vec{p}}_d.$$ (
The temporal change of the momentum $\dot{\vec{p}}_d$ is equal to the force $\vec{F}_R$ acting on the point charge $q_d$. For this reason,
$$\dot{\vec{L}} = \vec{r}_d \times \vec{F}_R.$$ (
Equation (, however, describes a central force, i. e. $\vec{F}_R$ is always parallel to $\vec{r}_d$. Hence, $\vec{r}_d \times \vec{F}_R = 0$ and we get
$$\dot{\vec{L}} = 0.$$ (
This means that the angular momentum does not change with time and represents a conserved quantity. Furthermore, it is clear that with this the second of Kepler's laws is fulfilled.

2.3.3 Conservation of energy

Note: This proof is wrong. The correction can be found here.
The force given by formula ( depends not only on the distance vector of the point charges $\vec{r}$ to each other, but also on the relative speed $\dot{\vec{r}}$. Furthermore, the force is a central force, but not radially symmetric and therefore not vortex-free. For these reasons, Equation ( does not represent a conservative force in the original sense. Furthermore, it is not possible to determine the force by means of gradient calculation.

Nonetheless, it is possible to define a formula for the potential energy $V_R$ of a point charge $q_d$ with the inertial mass $m_d$ in the field of a point charge $q_s$. The equation is
$$V_R(\vec{r},\dot{\vec{r}}) := \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\,\frac{1}{r}\left(1 - \frac{\dot{r}^2}{c^2} + \frac{(\dot{\vec{r}}\,\vec{r})^2}{2\,c^2\,r^2} + \frac{q_s\,q_d}{8\,\pi\,\varepsilon_0\,m_d\,c^2\,r}\right).$$ (
In contrast to formulas of potential energy for conservative force fields, potential energy in this case is not only a function of the distance $r$, but also a function of the relative speed $\dot{\vec{r}}$.

In order to show that equation ( is actually the potential energy of the charge $q_d$ in the field of the charge $q_s$, we take advantage of the fact that the conservation of angular momentum has already been proven. This makes it clear that the movement of the point charge $q_d$ always takes place in the same plane. For this reason, the task is inherently two-dimensional, which makes it possible to express $\vec{r}$ in cylinder coordinates
$$\vec{r} = \left(\begin{matrix}\rho\,\cos\left(\varphi\right) \\ \rho\,\sin\left(\varphi\right)\end{matrix}\right).$$ (
This results in equations
$$r = \rho,\quad\dot{r}^2 = \rho^2\,\dot{\varphi}^2 + \dot{\rho}^2\quad\text{and}\quad\dot{\vec{r}}\,\vec{r}=\dot{\rho}\rho.$$ (
Their insertion in equation ( yields
$$V_R = \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\,\frac{1}{\rho}\left(1 - \frac{\rho^2\,\dot{\varphi}^2}{c^2} - \frac{\dot{\rho}^2}{2\,c^2} - \frac{q_s\,q_d}{8\,\pi\,\varepsilon_0\,c^2\,m_d\,\rho}\right).$$ (
By deriving after the time $t$, the power
$$\dot{V}_R = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left( \frac{\dot{\varphi}^2\,\dot{\rho}}{c^2}+\frac{2\,\rho\,\dot{\varphi}\,\ddot{\varphi}}{c^2}-\frac{\dot{\rho}^3}{2\,c^2\,\rho^2}+\frac{\dot{\rho}\,\ddot{\rho}}{c^2\,\rho}+\frac{\dot{\rho}}{\rho^2} - \frac{q_d\,q_s\,\dot{\rho}}{4\,\pi\,c^2\,\varepsilon_0\,m_d\,\rho^3} \right)$$ (
follows. Since force ( is a central force acting only in radial direction, acceleration $\ddot{\varphi}$ can only be zero. For $\ddot{\rho}$, however, the equation
$$\ddot{\rho} = \frac{\Vert\vec{F}_R\Vert}{m_d} = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0\,m_d}\,\left(\frac{1}{\rho^2} + \frac{\dot{\varphi}^2}{c^2}-\frac{\dot{\rho}^2}{2\,c^2\rho^2}\right)$$ (
applies because of the formulae ( and ( If we use both, ( becomes
$$\dot{V}_R = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left(\frac{\dot{\rho}}{\rho^2} + \frac{\dot{\varphi}^2\,\dot{\rho}}{c^2}-\frac{\dot{\rho}^3}{2\,c^2\,\rho^2}+\frac{1}{c^4}\,\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0\,m_d}\,\left(\frac{\dot{\varphi}^2\,\dot{\rho}}{\rho}-\frac{\dot{\rho}^3}{2\,\rho^3}\right) \right).$$ (
The term multiplied by $1/c^4$ can be completely neglected for not too high velocities, which was the basic assumption for the validity of equation ( It follows
$$\dot{V}_R = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left(\frac{1}{\rho^2} + \frac{\dot{\varphi}^2}{c^2}-\frac{\dot{\rho}^2}{2\,c^2\,\rho^2}\right)\,\dot{\rho}.$$ (
However, this corresponds exactly to the dot product of force ( and velocity $\dot{\vec{r}}$ in cylinder coordinates ( with a negative sign. In other words,
$$\dot{V}_R = -\vec{F}_R\,\dot{\vec{r}}$$ (
applies. Because of the relationship
$$\vec{F}_R = m_d\,\ddot{\vec{r}},$$ (
then follows
$$\dot{V}_R = -m_d\,\ddot{\vec{r}}\,\dot{\vec{r}} = -\dot{T}$$ (
with the kinetic energy
$$T(\dot{\vec{r}}) = \frac{1}{2}\,m_d\,\dot{\vec{r}}\,\dot{\vec{r}} = \frac{1}{2}\,m_d\,\dot{r}^2.$$ (
From equation ( we get finally
$$\dot{V}_R + \dot{T} = 0.$$ (
As the total energy $\mathcal{E}$ is defined as the sum of potential energy $V_R$ and kinetic energy $T$,
$$\dot{\mathcal{E}} = \dot{V}_R + \dot{T} = 0$$ (
applies. This means that the change in total energy over time is zero. It is therefore a conserved quantity for the electromagnetic force (