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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 (2.2.1.4).

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
$$\vec{F}_{ij} = \vec{F}_R(\vec{r}_j-\vec{r}_i,\dot{\vec{r}}_j-\dot{\vec{r}}_i).$$ (2.3.1.1)
Likewise, the force of a charge $q_j$ onto a charge $q_i$ is represented by
$$\vec{F}_{ji} = \vec{F}_R(\vec{r}_i-\vec{r}_j,\dot{\vec{r}}_i-\dot{\vec{r}}_j).$$ (2.3.1.2)
Furthermore, due to the structure of force formula (2.2.1.4), relationship
$$\vec{F}_{ji} = -\vec{F}_R(\vec{r}_j-\vec{r}_i,\dot{\vec{r}}_j-\dot{\vec{r}}_i)$$ (2.3.1.3)
applies, i. e.
$$\vec{F}_{ij} = -\vec{F}_{ji}.$$ (2.3.1.4)
For an isolated system of $n$ point charges in total, the equation
$$\dot{\vec{p}}_i = \sum\limits_{j=1}^{n} \vec{F}_{ji}$$ (2.3.1.5)
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
$$\sum_{i=1}^{n} \dot{\vec{p}}_i = \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} \vec{F}_{ji}.$$ (2.3.1.6)
Because of property (2.3.1.4),
$$\sum_{i=1}^{n} \dot{\vec{p}}_i = -\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} \vec{F}_{ij} = -\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{n} \vec{F}_{ij} = -\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} \vec{F}_{ji}$$ (2.3.1.7)
also holds true. From this it follows that for the change of the total momentum it must be
$$\sum_{i=1}^{n} \dot{\vec{p}}_i = 0.$$ (2.3.1.8)
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
$$\vec{L} := \vec{r}_d \times \vec{p}_d$$ (2.3.2.1)
applies. Derivation according to time results in
$$\dot{\vec{L}} = \dot{\vec{r}}_d \times \vec{p}_d + \vec{r}_d \times \dot{\vec{p}}_d.$$ (2.3.2.2)
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 (2.3.2.2) into
$$\dot{\vec{L}} = \vec{r}_d \times \dot{\vec{p}}_d.$$ (2.3.2.3)
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,
$$\dot{\vec{L}} = \vec{r}_d \times \vec{F}_R.$$ (2.3.2.4)
Equation (2.2.1.4), 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
$$\dot{\vec{L}} = 0.$$ (2.3.2.5)
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

The force given by formula (2.2.1.4) 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 (2.2.1.4) 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
$$V_R(\vec{r},\dot{\vec{r}}) := \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\,\frac{1}{r}\left(1 - \frac{(\dot{\vec{r}}\,\vec{r})^2}{2\,c^2\,r^2} - \frac{q_s\,q_d}{8\,\pi\,\varepsilon_0\,m_d\,c^2\,r}\right).$$ (2.3.3.1)
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 (2.3.3.1) 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 ensures 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
$$\vec{r} = \left(\begin{matrix}\rho\,\cos\left(\varphi\right) \\ \rho\,\sin\left(\varphi\right)\end{matrix}\right).$$ (2.3.3.2)
With this we get
$$r = \rho\quad\text{and}\quad\dot{\vec{r}}\,\vec{r}=\dot{\rho}\rho.$$ (2.3.3.3)
Insertion of both in equation (2.3.3.1) results in
$$V_R = \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\,\frac{1}{\rho}\left(1 - \frac{\dot{\rho}^2}{2\,c^2} - \frac{q_s\,q_d}{8\,\pi\,\varepsilon_0\,c^2\,m_d\,\rho}\right).$$ (2.3.3.4)
We derive the generalized potential $V_R$ after the time $t$ and get
$$\dot{V}_R = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left(\frac{\dot{\rho}}{\rho^2} -\frac{\dot{\rho}^3}{2\,c^2\,\rho^2}+\frac{\dot{\rho}\,\ddot{\rho}}{c^2\,\rho} - \frac{q_d\,q_s\,\dot{\rho}}{4\,\pi\,c^2\,\varepsilon_0\,m_d\,\rho^3} \right).$$ (2.3.3.5)
For the next step we need the equation of motion $\vec{F}_R = m\,\ddot{\vec{r}}$. Multiplying both sides by $\vec{r}$ results in
$$\vec{F}_R\,\vec{r} = \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left(\frac{1}{\rho}-\frac{\dot{\rho}^2}{2\,c^2\,\rho}+\frac{\rho\,\dot{\varphi}^2}{c^2}\right)$$ (2.3.3.6)
and
$$m\,\ddot{\vec{r}}\,\vec{r} = m\,\rho\,\ddot{\rho} - m\,\rho^2\,\dot{\varphi}^2.$$ (2.3.3.7)
From this follows the formula
$$\ddot{\rho} = \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0\,m}\left(\frac{1}{\rho^2}-\frac{\dot{\rho}^2}{2\,c^2\,\rho^2}+\frac{\dot{\varphi}^2}{c^2}\right) + \rho\,\dot{\varphi}^2.$$ (2.3.3.8)
Insertion into equation (2.3.3.5) yields
$$\dot{V}_R = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left(\frac{\dot{\rho}}{\rho^2} -\frac{\dot{\rho}^3}{2\,c^2\,\rho^2} +\frac{\dot{\rho}\,\dot{\varphi}^2}{c^2} + \frac{1}{c^4}\,\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0\,m}\,\left(\frac{\dot{\rho}\,\dot{\varphi}^2}{\rho}-\frac{\dot{\rho}^3}{2\,\rho^3}\right)\right).$$ (2.3.3.9)
But this corresponds exactly to
$$-\vec{F}_R\,\dot{\vec{r}} = -\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\left(\frac{\dot{\rho}}{\rho^2}-\frac{\dot{\rho}^3}{2\,c^2\,\rho^2} + \frac{\dot{\rho}\,\dot{\varphi}^2}{c^2}\right)$$ (2.3.3.10)
if we neglect the term multiplied by $1/c^4$, what is possible for not too high speeds. And this was the basic assumption for the validity of equation (2.2.1.4).

This shows that
$$\dot{V}_R = -\vec{F}_R\,\dot{\vec{r}}$$ (2.3.3.11)
holds true. Because of
$$\vec{F}_R = m_d\,\ddot{\vec{r}}$$ (2.3.3.12)
then follows
$$\dot{V}_R = -m_d\,\ddot{\vec{r}}\,\dot{\vec{r}} = -\dot{T}$$ (2.3.3.13)
with the kinetic energy
$$T(\dot{\vec{r}}) = \frac{1}{2}\,m_d\,\dot{\vec{r}}\,\dot{\vec{r}} = \frac{1}{2}\,m_d\,\dot{r}^2.$$ (2.3.3.14)
Equation (2.3.3.13) means that
$$\dot{V}_R + \dot{T} = 0.$$ (2.3.3.15)
Since the total energy $\mathcal{E}$ is defined as the sum of potential energy $V_R$ and kinetic energy $T$ we have
$$\dot{\mathcal{E}} = \dot{V}_R + \dot{T} = 0$$ (2.3.3.16)
This implies that the total energy does not change over time. Thus, the electromagnetic force (2.2.1.4) is a conserved quantity.

2.3.4 Conservation of energy in an n-particle system

For reasons of completeness, a proof of energy conservation for an n-particle system is provided here, which applies to conservative forces, but especially also to force (2.2.1.4) and to any other force formula that fulfils the third Newtonian axiom.

Given are $n$ particles interacting with each other. External forces are not present. The $n$ equations of motion for $k=1,\ldots,n$ are
$$\sum\limits_{i=1}^{n}\vec{F}_{ik} = m_k\,\ddot{\vec{r}}_k.$$ (2.3.4.1)
The force of the i-th particle on itself $\vec{F}_{ii}$ is defined to zero here to avoid any exceptions in the summation. Multiplication by $\dot{\vec{r}}_k$ results in
$$\sum\limits_{i=1}^{n}\vec{F}_{ik}\,\dot{\vec{r}}_k = m_k\,\ddot{\vec{r}}_k\,\dot{\vec{r}}_k.$$ (2.3.4.2)
The addition of all $n$ equations of motion yields
$$\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\vec{F}_{ik}\,\dot{\vec{r}}_k = \sum\limits_{k=1}^{n}\,m_k\,\ddot{\vec{r}}_k\,\dot{\vec{r}}_k.$$ (2.3.4.3)
Formal manipulation of the summation indexes results in
$$\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\vec{F}_{ik}\,\dot{\vec{r}}_k = \frac{1}{2}\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\vec{F}_{ik}\,\dot{\vec{r}}_k + \frac{1}{2} \sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\vec{F}_{ki}\,\dot{\vec{r}}_i.$$ (2.3.4.4)
Because of the third Newtonian axiom it is $\vec{F}_{ki} = -\vec{F}_{ik}$ and we get
$$\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\vec{F}_{ik}\,\dot{\vec{r}}_k = \frac{1}{2}\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n} \vec{F}_{ik}\cdot(\dot{\vec{r}}_k - \dot{\vec{r}}_i).$$ (2.3.4.5)
Let $V$ be the antiderivative of $-\vec{F}\cdot\dot{\vec{r}}$, i.e. the following shall apply
$$\dot{V} = -\vec{F}\cdot\dot{\vec{r}}.$$ (2.3.4.6)
(Note: It is not necessary that the function $V$ is elementary. It just has to exist. See: fundamental theorem of calculus). This allows to rearrange equation (2.3.4.5) into
$$\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\vec{F}_{ik}\,\dot{\vec{r}}_k = -\frac{1}{2}\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\dot{V}_{ik},$$ (2.3.4.7)
because $\dot{V}_{ik} = -\vec{F}_{ik}\cdot(\dot{\vec{r}}_k - \dot{\vec{r}}_i)$. We use this in equation (2.3.4.3) and get
$$-\frac{1}{2}\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\dot{V}_{ik} = \sum\limits_{k=1}^{n}\,m_k\,\ddot{\vec{r}}_k\,\dot{\vec{r}}_k.$$ (2.3.4.8)
The term $m_k\,\ddot{\vec{r}}_k\,\dot{\vec{r}}_k$ corresponds to the time derivative of kinetic energy $T_k = \frac{1}{2} m_k \dot{r}_k^2$ of the k-th particle. With this, equation (2.3.4.8) becomes to
$$-\frac{1}{2}\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}\dot{V}_{ik} = \sum\limits_{k=1}^{n}\dot{T}_k.$$ (2.3.4.9)
This shows that the time derivative of the total energy ([Brandt2005], pages 79 and 80)
$$\mathcal{E} := \sum\limits_{k=1}^{n}\frac{1}{2} m_k \dot{r}_k^2 + \frac{1}{2}\sum\limits_{k=1}^{n}\,\sum\limits_{i=1}^{n}V_{ik}$$ (2.3.4.10)
in an n-particle system must be zero. This means, the total energy is constant in time and therefore a conserved quantity.