Energy Conservation in a Charged Retarded Field Engine

Abstract

Energy conservation, rooted in the time invariance of physical laws and formalized by Noether’s theorem, requires that systems with space-time translational symmetry conserve momentum and energy. This work examines how this principle applies to a charged retarded field engine, where the rate of change of total energy—mechanical plus field energy—is balanced by the energy flux through the system’s boundary. Using electric and magnetic field expressions from a Taylor expansion to incorporate retardation effects, we analyze the energy equation order by order for two arbitrary charged bodies. Our results show that total energy is conserved up to the fourth order, with mechanical and field energy changes exactly offset by boundary energy flux. Consequently, the work done by the internal electromagnetic field precisely equals the engine’s gained mechanical kinetic energy, addressing the central focus of this study.

1. Introduction

Charged retarded systems represent a ground-breaking approach to space propulsion, aimed at addressing the major limitations of traditional rocket fuel. Robertson et al. [1] reviewed the principal barriers slowing advancements in space propulsion technologies and highlighted several theoretical ideas that, within the last twenty years, have evolved from speculative fiction into scientific theory. Thus, it is essential to explore various aspects of these systems, such as linear momentum and energy conservation, as all engines operate within the constraints of the laws of mechanics and the first law of thermodynamics: energy cannot be created or destroyed—only transformed.

To articulate clearly the research gap, we must show explicitly what is the source of energy of the charged retarded field engine and how it converts stored electromagnetic energy into kinetic energy. The present study specifically advances prior work, in which we calculated engine characteristics such as force and gained linear momentum, but no attention was given to the source of the engine’s kinetic energy. Thus, we aim to explicit connect the problem statement, which is the source of energy, to a methodology of analytical calculations, which will allow us to pinpoint this source. This illuminating calculation is the main result of the current paper.

This work discusses the retardation effects caused by the relativistic finite speed of signal propagation in a charged engine. As signals must propagate at a finite speed, it follows that an action and its corresponding reaction cannot occur simultaneously. Although retardation already exists in Maxwell’s equations, relativity went even further to postulate that no object, message, signal (of any kind, including non-electromagnetic), or field can travel faster than the speed of light in a vacuum [2,3,4]. In their paper, Griffiths and Heald [5] highlighted in that Coulomb’s law and the Biot–Savart law are strictly valid only for determining electric and magnetic fields from static sources. To extend their applicability beyond the static domain, Jefimenko’s time-dependent generalizations [6] of these laws were formulated, taking into account the retardation effect. In this case, momentum and energy are exchanged between the field and the material, enabling the relativistic engine effect. This result had already been observed and highlighted by several authors [7,8,9,10,11]. This should not be confused with photonic sails [12], in which external photons need to transfer their momentum to the device. Basically, photons can only be used to gain linear momentum (p, which is the concept of photonic sales) at amounts of $p={\displaystyle \frac{E}{c}}$, in which E is the photons’ energy and c is the speed of light in vacuum; thus, it seems much better to store the energy (E) and use it as a source for a relativistic kinetic engine.

This process differs significantly from the transfer of momentum and energy between two material bodies, which is the basis of standard transportation (a rocket ejects material to propagate and a car moves forward by pushing against the road). Relativistic effects indicate a novel type of motor, consisting not of two material components but of a combination of matter and an electromagnetic field. In [13], we provided a detailed analysis of the retardation effect and presented the conceptual framework and design principles underlying a charged relativistic motor. In particular, we have shown that if retardation is not neglected in a two-body charged system, then the system will be affected by a total force (despite the lack of an external force) and will gain linear momentum. The system’s center of mass will move with a velocity equal to the total linear momentum divided by the system’s mass. Mathematical expressions for total force, total linear momentum, and system velocity were derived and shown to be of order ${\displaystyle \frac{1}{c^2}}$. Recently, we also investigated the interaction between charged bodies and magnetic currents generated by permanent magnetic materials, analyzing the potential implications for a charged relativistic engine utilizing a permanent magnet [14]. Energy conservation has already been discussed in the case of an uncharged relativistic engine [15], in which the energy transfers between the mechanical component of the relativistic engine and the electromagnetic field for a non charged engine. In this case, the total utilized electromagnetic energy is six times the kinetic energy gained by the relativistic motor. Furthermore, previous research [16] has provided a mathematical proof that demonstrates that linear momentum is conserved in a retarded electric motor that is uncharged. More explicitly, we calculated the electromagnetic linear momentum in a retarded electric motor and found it to be equal in quantity and opposite in sign to the gained mechanical linear momentum in the same retarded electric motor, as calculated in [17].

In the current work, we investigate a relativistic charged engine consisting of two subsystems of arbitrary geometry as in [13], with a clear intention to show that the kinetic energy acquired by the system is deducted from its stored field energy. We begin by analyzing the energy exchange between mechanical and electromagnetic components in the relativistic regime. Subsequently, we carry out a detailed examination of the engine’s energy dynamics by expanding the relevant field quantities as a power series in $1/c$, where c denotes the speed of light in vacuum. This expansion systematically incorporates the effects of retardation. The total energy, comprising mechanical energy, electromagnetic field energy, and energy flux via the Poynting vector, is analyzed to verify the energy balance equation at each order. Specifically, we evaluate the contributions to the energy equation for $n=0,1,2,3,$ and 4, demonstrating how retardation influences energy exchange and confirming that the energy balance is satisfied at each order within the accuracy of the expansion. Finally, we demonstrate that the kinetic energy acquired by the engine exactly equals the amount of electromagnetic energy lost.

2. Energy Conservation

The conservation of energy in the general electromagnetic system is expressed by the following equation:

$$\begin{array}{c}\hfill {\displaystyle \frac{d{E}_{mech}}{dt}}+{\displaystyle \frac{d{E}_{field}}{dt}}=-{\oint }_S{\overrightarrow{S}}_P·\widehat{n}da.\end{array}$$

(1)

This equation reflects the fact that any decrease (or increase) in the total energy within the system is accounted for by the energy carried away (or supplied) by the electromagnetic field through the system’s boundary. Here, S denotes a closed surface enclosing the volume of interest, $\widehat{n}$ is a unit vector normal to the surface, and $da$ represents a surface element.

Poynting’s vector is defined as follows:

$$\begin{array}{c}\hfill {\overrightarrow{S}}_P={\displaystyle \frac{1}{{\mu }_0}}\overrightarrow{E}\times \overrightarrow{B}={\displaystyle \frac{{ϵ}_0}{{\mu }_0{ϵ}_0}}\overrightarrow{E}\times \overrightarrow{B}={ϵ}_0{c}^2\overrightarrow{E}\times \overrightarrow{B},\end{array}$$

(2)

where ${ϵ}_0=8.85\times {10}^{-12}$ ${\mathrm{Fm}}^{-1}$ is the vacuum permittivity, ${\mu }_0=4\pi \times {10}^{-7}$ (H/m), and $c\simeq 3\times {10}^8$ m/s is the velocity of light in vacuum. The term ${\oint }_S{\overrightarrow{S}}_P·\widehat{n}da$ represents the total electromagnetic power flux through the closed surface (S). The field energy density is defined as follows:

$$\begin{array}{c}\hfill E_{field}={\displaystyle \frac{{ϵ}_0}{2}}\int {\displaystyle \left({\overrightarrow{E}}^2+c^2{\overrightarrow{B}}^2\right)}{d}^{3}x.\end{array}$$

(3)

The derivative of the mechanical energy is the mechanical power given by the following expression:

$$\begin{array}{c}\hfill Power\equiv {\displaystyle \frac{d{E}_{mech}}{dt}}=\int d^{3}x\overrightarrow{J}·\overrightarrow{E},\end{array}$$

(4)

where $\overrightarrow{J}$ is the current density. Hence,

$$\begin{array}{c}\hfill \int d^{3}x\overrightarrow{J}·\overrightarrow{E}+{\displaystyle \frac{d{E}_{field}}{dt}}=-{\oint }_S{\overrightarrow{S}}_P·\widehat{n}da\end{array}$$

(5)

or

$$\begin{array}{c}\hfill \int d^{3}x\overrightarrow{J}·\overrightarrow{E}=-{\displaystyle \frac{d{E}_{field}}{dt}}-{\oint }_S{\overrightarrow{S}}_P·\widehat{n}da.\end{array}$$

(6)

In what follows, we interpret Equation (6) in terms of its role in governing energy conservation for a charged relativistic engine system.

Kinetic Energy

According to [13] (see also Appendix K), the mechanical momentum associated with the time-dependent charged retarded field engine is expressed as follows:

$$\begin{array}{c}\hfill \overrightarrow{P}\left(t\right)={\displaystyle \frac{{\mu }_0}{4\pi }}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[{\displaystyle \frac{1}{2}}{\displaystyle \left({\rho }_2{\partial }_t{\rho }_1-{\rho }_1{\partial }_t{\rho }_2\right)}\widehat{R}-{\displaystyle \left({\rho }_1\overrightarrow{J_2}+{\rho }_2\overrightarrow{J_1}\right)}{R}^{-1}\right]},\end{array}$$

(7)

or taking into account the fact that $c={\displaystyle \frac{1}{\sqrt{{\mu }_0{ϵ}_0}}}$ and, as is customary, defining $k={\displaystyle \frac{1}{4\pi {ϵ}_0}}$,

$$\begin{array}{c}\hfill \overrightarrow{P}\left(t\right)={\displaystyle \frac{k}{c^2}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[{\displaystyle \frac{1}{2}}{\displaystyle \left({\rho }_2{\partial }_t{\rho }_1-{\rho }_1{\partial }_t{\rho }_2\right)}\widehat{R}-{\displaystyle \left({\rho }_1\overrightarrow{J_2}+{\rho }_2\overrightarrow{J_1}\right)}{R}^{-1}\right]}.\end{array}$$

(8)

It is clear from the expression that the mechanical momentum is of the order of ${\displaystyle \frac{1}{c^2}}$. The kinetic mechanical energy associated with this momentum is given by the following expression:

$$\begin{array}{c}\hfill E_k={\displaystyle \frac{{\overrightarrow{P}}^2}{2M}}={\displaystyle \frac{1}{2}}\overrightarrow{P}·{\overrightarrow{v}}_s,\end{array}$$

(9)

where M is the mass of the relativistic engine and the engine velocity (${\overrightarrow{v}}_s$) is defined as follows:

$$\begin{array}{c}\hfill {\overrightarrow{v}}_s\equiv {\displaystyle \frac{\overrightarrow{P}}{M}}\propto {\displaystyle \frac{1}{c^2}}\end{array}$$

(10)

We notice that the expression for kinetic energy is of order ${\displaystyle \frac{1}{c^4}}$; lower order corrections do not exist, and higher order corrections are neglected.

3. Field Energy for Two Charged Sub-Systems

Consider two arbitrary charged sub-systems denoted system 1 and system 2 (see Figure 1).

The charged subsystems are far apart such that their interaction is negligible. In this case, Equation (6) is correct for each sub-system separately, that is,

$$\begin{array}{c}\hfill \int d^{3}x\overrightarrow{J_1}·\overrightarrow{E_1}=-{\displaystyle \frac{d{E}_{field1}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P1}·\widehat{n}da.\end{array}$$

(11)

$$\begin{array}{c}\hfill \int d^{3}x\overrightarrow{J_2}·\overrightarrow{E_2}=-{\displaystyle \frac{d{E}_{field2}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P2}·\widehat{n}da.\end{array}$$

(12)

Next, we put them closer together such that they may interact but without modifying the charge or the current densities of each of the subsystems. The total fields of the combined system are expressed as follows:

$$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2,\overrightarrow{B}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2.$$

(13)

Since both the field energy (Equation (3)) and Poynting’s vector Equation (2) are quadratic in the fields, the following result is obtained:

$$\begin{array}{c}\hfill E_{field}\equiv {\displaystyle \frac{{ϵ}_0}{2}}\int {\displaystyle \left({\overrightarrow{E}}^2+c^2{\overrightarrow{B}}^2\right)}{d}^{3}x=E_{field1}+E_{field2}+E_{field12}.\end{array}$$

(14)

$$\begin{array}{c}\hfill E_{field1}\equiv E_{Efield1}+E_{Mfield1}\equiv {\displaystyle \frac{{ϵ}_0}{2}}\int {\displaystyle \left(\overrightarrow{E_1^2}+c^2\overrightarrow{B_1^2}\right)}{d}^{3}x,\end{array}$$

(15)

$$\begin{array}{c}\hfill E_{field2}\equiv E_{Efield2}+E_{Mfield2}\equiv {\displaystyle \frac{{ϵ}_0}{2}}\int {\displaystyle \left(\overrightarrow{E_2^2}+c^2\overrightarrow{B_2^2}\right)}{d}^{3}x,\end{array}$$

(16)

$$\begin{array}{c}\hfill E_{field12}\equiv E_{Efield12}+E_{Mfield12}\equiv {ϵ}_0\int {\displaystyle \left(\overrightarrow{E_1}·\overrightarrow{E_2}+c^2\overrightarrow{B_1}·\overrightarrow{B_2}\right)}{d}^{3}x.\end{array}$$

(17)

$$\begin{array}{c}\hfill {\overrightarrow{S}}_P\equiv {ϵ}_0{c}^2\overrightarrow{E}\times \overrightarrow{B}={\overrightarrow{S}}_{P1}+{\overrightarrow{S}}_{P2}+{\overrightarrow{S}}_{P12}.\end{array}$$

(18)

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P1}\equiv {ϵ}_0{c}^2\overrightarrow{E_1}\times \overrightarrow{B_1},\end{array}$$

(19)

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P2}\equiv {ϵ}_0{c}^2\overrightarrow{E_2}\times \overrightarrow{B_2},\end{array}$$

(20)

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}\equiv {ϵ}_0{c}^2{\displaystyle \left(\overrightarrow{E_1}\times \overrightarrow{B_2}+\overrightarrow{E_2}\times \overrightarrow{B_1}\right)}.\end{array}$$

(21)

The power invested in the combined system is bilinear for the current density and electric field according to Equation (4). This will lead to the following expression:

$$\begin{array}{c}\hfill Power=\int d^{3}x\overrightarrow{J}·\overrightarrow{E}=Powe{r}_1+Powe{r}_2+Powe{r}_{12}.\end{array}$$

(22)

$$\begin{array}{c}\hfill Powe{r}_1=\int d^{3}x\overrightarrow{J_1}·\overrightarrow{E_1},\end{array}$$

(23)

$$\begin{array}{c}\hfill Powe{r}_2=\int d^{3}x\overrightarrow{J_2}·\overrightarrow{E_2},\end{array}$$

(24)

$$\begin{array}{c}\hfill Powe{r}_{12}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·\overrightarrow{E_2}+\overrightarrow{J_2}·\overrightarrow{E_1}\right)}.\end{array}$$

(25)

Subtracting the expressions given in Equations (11) and (12) from Equation (6) yields the following:

$$\begin{array}{c}\hfill Power-Powe{r}_1-Powe{r}_2=-{\displaystyle \frac{d{\displaystyle \left(E_{field}-E_{field1}-E_{field2}\right)}}{dt}}-{\oint }_S{\displaystyle \left({\overrightarrow{S}}_P-{\overrightarrow{S}}_{P1}-{\overrightarrow{S}}_{P2}\right)}·\widehat{n}da.\end{array}$$

(26)

Taking into account Equations (14)–(25), we arrive at the following:

$$\begin{array}{c}\hfill Powe{r}_{12}=-{\displaystyle \frac{d{E}_{field12}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}·\widehat{n}da.\end{array}$$

(27)

The significance of the above equation is best illustrated by the flow chart presented in Figure 2, which illustrates the possible exchange between stored electromagnetic energy, mechanical work and radiated energy.

4. Power Expansion in ${\displaystyle \frac{\mathbf{1}}{\mathbf{c}}}$

Let us now expand Equation (27) in power of ${\displaystyle \frac{1}{c}}$; to this end, we introduce the notation of $G^{\left[n\right]}$ to designate a term of order ${\displaystyle \frac{1}{c}}$ and quantity G. Thus, any quantity (G) may be written as the following sum:

$$G=\sum _{n=0}^{\infty }{G}^{\left[n\right]}.$$

(28)

Specifically,

$$Powe{r}_{12}^{\left[n\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[n\right]}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}^{\left[n\right]}·\widehat{n}da.$$

(29)

It follows from Equation (25) that

$$Powe{r}_{12}^{\left[n\right]}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·{\overrightarrow{E}}_2^{\left[n\right]}+\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[n\right]}\right)}.$$

(30)

It also follows from Equation (17) that

$$E_{field12}^{\left[n\right]}={ϵ}_0\int {\displaystyle \left(\sum _{k=0}^n{\overrightarrow{E}}_1^{[n-k]}·{\overrightarrow{E}}_2^{\left[k\right]}+c^2\sum _{k=0}^{n+2}{\overrightarrow{B}}_1^{[n+2-k]}·{\overrightarrow{B}}_2^{\left[k\right]}\right)}{d}^{3}x.$$

(31)

Finally, from Equation (21), it follows that

$${\overrightarrow{S}}_{P12}^{\left[n\right]}={ϵ}_0{c}^2\sum _{k=0}^{n+2}{\displaystyle \left({\overrightarrow{E}}_1^{[n+2-k]}\times {\overrightarrow{B}}_2^{\left[k\right]}+{\overrightarrow{E}}_2^{[n+2-k]}\times {\overrightarrow{B}}_1^{\left[k\right]}\right)}.$$

(32)

In the subsections below, we discuss Equation (29) for different values of n—from 0 up to 4. An analysis of the field contributions can now be carried out, considering each order in turn.

4.1. n = 0

Let us look at Equation (27) and study it for the zeroth order in ${\displaystyle \frac{1}{c}}$:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[0\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[0\right]}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}^{\left[0\right]}·\widehat{n}da.\end{array}$$

(33)

4.1.1. Power

We start by calculating $Powe{r}_{12}^{\left[0\right]}$ according to Equation (30):

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[0\right]}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·{\overrightarrow{E}}_2^{\left[0\right]}+\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[0\right]}\right)}.\end{array}$$

(34)

After some calculations elaborated upon in Appendix A, it follows from Equation (A18) that

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[0\right]}=-{\displaystyle \frac{d}{dt}}{\displaystyle \left[k\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \frac{{\rho }_1{\rho }_2}{R}}\right]}.\end{array}$$

(35)

4.1.2. Field Energy

Next, we calculate field energy according to Equation (31).

$$\begin{array}{c}\hfill E_{field12}^{\left[0\right]}={ϵ}_0\int {\displaystyle \left({\overrightarrow{E}}_1^{\left[0\right]}·{\overrightarrow{E}}_2^{\left[0\right]}+c^2{\displaystyle \left[{\overrightarrow{B}}_1^{\left[2\right]}·{\overrightarrow{B}}_2^{\left[0\right]}+{\overrightarrow{B}}_1^{\left[1\right]}·{\overrightarrow{B}}_2^{\left[1\right]}+{\overrightarrow{B}}_1^{\left[0\right]}·{\overrightarrow{B}}_2^{\left[2\right]}\right]}\right)}{d}^{3}x\end{array}$$

(36)

The magnetic field does not have terms that are lower than second order. We use the null magnetic field expressions from Equation (58) of [13]:

$$\begin{array}{c}\hfill {\overrightarrow{B}}_1^{\left[0\right]}={\overrightarrow{B}}_2^{\left[0\right]}={\overrightarrow{B}}_1^{\left[1\right]}={\overrightarrow{B}}_2^{\left[1\right]}=0.\end{array}$$

(37)

Accordingly, the field energy takes the following form:

$$\begin{array}{c}\hfill E_{field12}^{\left[0\right]}={ϵ}_0\int {\displaystyle \left({\overrightarrow{E}}_1^{\left[0\right]}·{\overrightarrow{E}}_2^{\left[0\right]}\right)}{d}^{3}x.\end{array}$$

(38)

After calculations elaborated upon in Appendix B and taking into account the identity expressed as $k={\displaystyle \frac{1}{4\pi {ϵ}_0}}$, it follows from Equation (A28) that

$$\begin{array}{c}\hfill E_{field12}^{\left[0\right]}=k{\displaystyle \left[\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \frac{{\rho }_1{\rho }_2}{R}}\right]}.\end{array}$$

(39)

Based on Equations (35) and (39), it is now clear that

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[0\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[0\right]}}{dt}}.\end{array}$$

(40)

It follows that the power related to mechanical work equals less thanthe temporal rate of change of the stored field energy.

4.1.3. Poynting Vector

Finally, we study the Poynting vector according to Equation (32):

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}^{\left[0\right]}={ϵ}_0{c}^2{\displaystyle \left({\overrightarrow{E}}_1^{\left[2\right]}\times {\overrightarrow{B}}_2^{\left[0\right]}+{\overrightarrow{E}}_1^{\left[1\right]}\times {\overrightarrow{B}}_2^{\left[1\right]}+{\overrightarrow{E}}_1^{\left[0\right]}\times {\overrightarrow{B}}_2^{\left[2\right]}+{\overrightarrow{E}}_2^{\left[2\right]}\times {\overrightarrow{B}}_1^{\left[0\right]}+{\overrightarrow{E}}_2^{\left[1\right]}\times {\overrightarrow{B}}_1^{\left[1\right]}+{\overrightarrow{E}}_2^{\left[0\right]}\times {\overrightarrow{B}}_1^{\left[2\right]}\right)}.\end{array}$$

(41)

It is clear from Equation (37) that

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}^{\left[0\right]}={ϵ}_0{c}^2{\displaystyle \left({\overrightarrow{E}}_1^{\left[0\right]}\times {\overrightarrow{B}}_2^{\left[2\right]}+{\overrightarrow{E}}_2^{\left[0\right]}\times {\overrightarrow{B}}_1^{\left[2\right]}\right)},\end{array}$$

(42)

As both ${\overrightarrow{E}}^{\left[0\right]}$ and ${\overrightarrow{B}}^{\left[2\right]}$ decrease as ${\displaystyle \frac{1}{r^2}}$ at infinity, it follows that their cross product decreases as ${\displaystyle \frac{1}{r^4}}$; thus, a surface integral over an infinite sphere will also vanish, which is what should be expected from Equation (40), balancing the changes in the electromagnetic energy and mechanical power without any Poynting contribution. For $n=0$, we conclude that the energy equation of order zero is, indeed, balanced, as is expected.

4.1.4. Summary

The key results of the $n=0$ calculation is that in this order, stored field energy, which is basically electric (Coulomb type), can be transferred to mechanical energy and vice-versa, but no energy can be radiated. This is illustrated in Figure 3.

4.2. n = 1

Let us look at Equation (27) and study it for the first order in ${\displaystyle \frac{1}{c}}$:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[1\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[1\right]}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}^{\left[1\right]}·\widehat{n}da.\end{array}$$

(43)

4.2.1. Power

The first-order expression for the power invested in mechanical work is expressed as follows:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[1\right]}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·{\overrightarrow{E}}_2^{\left[1\right]}+\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[1\right]}\right)}.\end{array}$$

(44)

According to Equations (51) and (54) of [13], the first-order term in the electric field expansion is expressed as follows:

$$\begin{array}{c}\hfill {\overrightarrow{E}}_1^{\left[1\right]}={\overrightarrow{E}}_2^{\left[1\right]}=0.\end{array}$$

(45)

Therefore,

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[1\right]}=0.\end{array}$$

(46)

4.2.2. Field Energy

Now, we calculate the field energy according to Equation (31) for $n=1$.

$$\begin{array}{c}\hfill E_{field12}^{\left[1\right]}={ϵ}_0\int {\displaystyle \left({\overrightarrow{E}}_1^{\left[0\right]}·{\overrightarrow{E}}_2^{\left[1\right]}+{\overrightarrow{E}}_1^{\left[1\right]}·{\overrightarrow{E}}_2^{\left[0\right]}+c^2\sum _{k=0}^3{\overrightarrow{B}}_1^{[3-k]}·{\overrightarrow{B}}_2^{\left[k\right]}\right)}{d}^{3}x.\end{array}$$

(47)

From Equations (37) and (45), we obtain the following:

$$\begin{array}{c}\hfill E_{field12}^{\left[1\right]}=0.\end{array}$$

(48)

4.2.3. Poynting Vector

Taking into account Equations (37) and (45), the first-order expression for the Poynting vector is given by the following expression:

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}^{\left[1\right]}={ϵ}_0{c}^2{\displaystyle \left({\overrightarrow{E}}_1^{\left[0\right]}\times {\overrightarrow{B}}_2^{\left[3\right]}+{\overrightarrow{E}}_2^{\left[0\right]}\times {\overrightarrow{B}}_1^{\left[3\right]}\right)},\end{array}$$

(49)

According to Equation (61) of [13], it follows that

$${\overrightarrow{B}}^{\left[3\right]}=0.$$

(50)

Hence,

$${\overrightarrow{S}}_{P12}^{\left[1\right]}=0.$$

(51)

Thus, for $n=1$, we conclude that the energy equation of order one is, indeed, balanced in a trivial way. Equation (27) is satisfied in the sense that 0 = 0.

4.2.4. Summary

The key result of the $n=1$ calculation is that in this order, both electric and magnetic-field components vanish. Thus, no mechanical work can be done, no energy can be stored, and no energy can be radiated.

4.3. n = 2

Let us look at Equation (27) and study it for the second order in ${\displaystyle \frac{1}{c}}$:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[2\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[2\right]}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}^{\left[2\right]}·\widehat{n}da.\end{array}$$

(52)

4.3.1. Power

The second-order expression for the power invested in mechanical work is given below:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[2\right]}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·{\overrightarrow{E}}_2^{\left[2\right]}+\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[2\right]}\right)}.\end{array}$$

(53)

After calculations elaborated upon in Appendix C and following Equation (A40), we obtain the following:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[2\right]}=-{\displaystyle \frac{k}{c^2}}{\displaystyle \frac{d}{dt}}{\displaystyle \left[\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{R{\displaystyle \left({\partial }_t{\rho }_1\left({\overrightarrow{x}}_1\right){\partial }_t{\rho }_2\left({\overrightarrow{x}}_2\right)\right)}}{2}}+{\displaystyle \frac{\overrightarrow{J_1}\left({\overrightarrow{x}}_1\right)·\overrightarrow{J_2}\left({\overrightarrow{x}}_2\right)}{R}}\right)}\right]}.\end{array}$$

(54)

Thus, we obtain a power expression accurate to the second order.

4.3.2. Field Energy

We now proceed to calculate the field energy using Equation (31) for $n=2$:

$$E_{field12}^{\left[2\right]}={ϵ}_0\int {\displaystyle \left(\sum _{k=0}^2{\overrightarrow{E}}_1^{[2-k]}·{\overrightarrow{E}}_2^{\left[k\right]}+c^2\sum _{k=0}^4{\overrightarrow{B}}_1^{[4-k]}·{\overrightarrow{B}}_2^{\left[k\right]}\right)}{d}^{3}x$$

(55)

As we already know that

$$\begin{array}{c}\hfill {\overrightarrow{E}}_1^{\left[1\right]}={\overrightarrow{E}}_2^{\left[1\right]}=0,{\overrightarrow{B}}_1^{\left[0\right]}={\overrightarrow{B}}_2^{\left[0\right]}=0,{\overrightarrow{B}}_1^{\left[1\right]}={\overrightarrow{B}}_2^{\left[1\right]}=0,\end{array}$$

(56)

it follows that the field energy takes the following simplified form:

$$\begin{array}{c}\hfill E_{field12}^{\left[2\right]}={ϵ}_0\int {\displaystyle \left({\overrightarrow{E}}_1^{\left[0\right]}·{\overrightarrow{E}}_2^{\left[2\right]}+{\overrightarrow{E}}_1^{\left[2\right]}·{\overrightarrow{E}}_2^{\left[0\right]}+c^2{\overrightarrow{B}}_1^{\left[2\right]}·{\overrightarrow{B}}_2^{\left[2\right]}\right)}{d}^{3}x.\end{array}$$

(57)

After calculations elaborated upon in Appendix D and combining Equations (A73), (A64) and (A65), we obtain the following expression for the second-order field energy:

$$\begin{array}{c}\hfill E_{field12}^{\left[2\right]}={\displaystyle \frac{k}{c^2}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[{\displaystyle \frac{\overrightarrow{J_1}·\overrightarrow{J_2}}{R}}+{\displaystyle \frac{R}{2}}{\partial }_t{\rho }_1{\partial }_t{\rho }_2\right]}.\end{array}$$

(58)

4.3.3. Poynting Vector

Next, we study the Poynting vector according to Equation (32) for $n=2$:

$${\overrightarrow{S}}_{P12}^{\left[2\right]}={ϵ}_0{c}^2\sum _{k=0}^4{\displaystyle \left({\overrightarrow{E}}_1^{[4-k]}\times {\overrightarrow{B}}_2^{\left[k\right]}+{\overrightarrow{E}}_2^{[4-k]}\times {\overrightarrow{B}}_1^{\left[k\right]}\right)},$$

(59)

According to Equation (56), we obtain the following resulting form of the Poynting vector:

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}^{\left[2\right]}={ϵ}_0{c}^2{\displaystyle \left({\overrightarrow{E}}_1^{\left[2\right]}\times {\overrightarrow{B}}_2^{\left[2\right]}+{\overrightarrow{E}}_1^{\left[0\right]}\times {\overrightarrow{B}}_2^{\left[4\right]}+{\overrightarrow{E}}_2^{\left[2\right]}\times {\overrightarrow{B}}_1^{\left[2\right]}+{\overrightarrow{E}}_2^{\left[0\right]}\times {\overrightarrow{B}}_1^{\left[4\right]}\right)}.\end{array}$$

(60)

The Poynting vector term, typically associated with radiation, contributes to the energy balance only through the surface integral over a closed surface enclosing the system. When this surface is taken to be spherical and located at a far distance ($R_{max}$), only the asymptotic forms of the electric and magnetic fields are relevant for evaluating the Poynting vector contribution. Accordingly, the expression for the Poynting flux on a far-away sphere of radius $R_{max}$ is given by the following expression:

$$\begin{array}{c}\hfill P{F}^{\left[2\right]}=\oint {\overrightarrow{S}}_{P12}^{\left[2\right]}·\widehat{n}da=R_{max}^2{\oint }_{R_{max}}{\overrightarrow{S}}_{P12}^{\left[2\right]}·\widehat{r}d\Omega .\end{array}$$

(61)

After calculations elaborated upon in Appendix E and plugging Equations (A85), (A86), (A90) and (A91) into Equation (A74), we obtain the following:

$$\begin{array}{c}\hfill P{F}^{\left[2\right]}=0,\end{array}$$

(62)

in which the equality sign is only correct forthe current approximation. Hence, there is no Poynting vector contribution to the energy balance. This is expected, as the mechanical work (Equation (54)) is balanced by the field energy loss Equation (58) in the second-order terms of ${\displaystyle \frac{1}{c}}$:

$$Powe{r}_{12}^{\left[2\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[2\right]}}{dt}}.$$

(63)

It is remarkable that no radiation losses need to be considered up to the second-order approximation. This will have implications for the retarded field motor that we discuss in later sections of this paper.

4.3.4. Summary

The key results of the $n=2$ calculation are outlined as follows. First, we notice that this is the first order in which retardation effects are present in a non-trivial way. However, as we have shown, no energy is lost to radiation; thus, the only exchange that is possible is between stored energy and mechanical work and vice-versa. This is very significant for the development of a retarded field engine, as we emphasized above and will elaborate upon later in this paper. This is illustrated in Figure 3.

4.4. n = 3

Let us look at Equation (27) and study it for the third order in ${\displaystyle \frac{1}{c}}$:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[3\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[3\right]}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}^{\left[3\right]}·\widehat{n}da.\end{array}$$

(64)

4.4.1. Power

The third-order expression for the power invested in mechanical work is given below:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[3\right]}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·{\overrightarrow{E}}_2^{\left[3\right]}+\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[3\right]}\right)}\end{array}$$

(65)

After calculations elaborated upon in Appendix F and combining Equations (A104) and (A105), we obtain the following:

$$Powe{r}_{12}^{\left[3\right]}={\displaystyle \left({\displaystyle \frac{2k}{3c^3}}\right)}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left(\overrightarrow{J_1}\left({\overrightarrow{x}}_1\right)·{\partial }_t^2\overrightarrow{J_2}\left({\overrightarrow{x}}_2\right)+\overrightarrow{J_2}\left({\overrightarrow{x}}_2\right)·{\partial }_t^2\overrightarrow{J_1}\left({\overrightarrow{x}}_1\right)\right)}.$$

(66)

The obtained expression represents the third-order mechanical power.

4.4.2. Field Energy

We now proceed to calculate the field energy using Equation (31) for $n=3$:

$$E_{field12}^{\left[3\right]}={ϵ}_0\int {\displaystyle \left(\sum _{k=0}^3{\overrightarrow{E}}_1^{[3-k]}·{\overrightarrow{E}}_2^{\left[k\right]}+c^2\sum _{k=0}^5{\overrightarrow{B}}_1^{[5-k]}·{\overrightarrow{B}}_2^{\left[k\right]}\right)}{d}^{3}x.$$

(67)

It is known from [13] (see also Equation (56)) that

$$\begin{array}{c}\hfill {\overrightarrow{E}}_1^{\left[1\right]}={\overrightarrow{E}}_2^{\left[1\right]}=0,{\overrightarrow{B}}_1^{\left[0\right]}={\overrightarrow{B}}_2^{\left[0\right]}=0,{\overrightarrow{B}}_1^{\left[1\right]}={\overrightarrow{B}}_2^{\left[1\right]}=0,{\overrightarrow{B}}_1^{\left[3\right]}={\overrightarrow{B}}_2^{\left[3\right]}=0.\end{array}$$

(68)

Therefore, the field energy can be written as follows:

$$\begin{array}{c}\hfill E_{field12}^{\left[3\right]}={ϵ}_0\int {\displaystyle \left({\overrightarrow{E}}_1^{\left[3\right]}·{\overrightarrow{E}}_2^{\left[0\right]}+{\overrightarrow{E}}_1^{\left[0\right]}·{\overrightarrow{E}}_2^{\left[3\right]}\right)}{d}^{3}x.\end{array}$$

(69)

After calculations elaborated upon in Appendix G and by inserting the results of Equations (A117) and (A118) into Equation (69), we obtain a null result:

$$\begin{array}{c}\hfill E_{field12}^{\left[3\right]}=0.\end{array}$$

(70)

Remarkably, it follows that no electromagnetic energy can be stored at the third order of ${\displaystyle \frac{1}{c}}$.

4.4.3. Poynting Vector

Next, we study the Poynting vector according to Equation (32) for $n=3$.

$${\overrightarrow{S}}_{P12}^{\left[3\right]}={ϵ}_0{c}^2\sum _{k=0}^5{\displaystyle \left({\overrightarrow{E}}_1^{[5-k]}\times {\overrightarrow{B}}_2^{\left[k\right]}+{\overrightarrow{E}}_2^{[5-k]}\times {\overrightarrow{B}}_1^{\left[k\right]}\right)}$$

(71)

According to Equation (68), the Poynting vector takes the following form:

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}^{\left[3\right]}={ϵ}_0{c}^2{\displaystyle \left({\overrightarrow{E}}_1^{\left[3\right]}\times {\overrightarrow{B}}_2^{\left[2\right]}+{\overrightarrow{E}}_1^{\left[0\right]}\times {\overrightarrow{B}}_2^{\left[5\right]}+{\overrightarrow{E}}_2^{\left[3\right]}\times {\overrightarrow{B}}_1^{\left[2\right]}+{\overrightarrow{E}}_2^{\left[0\right]}\times {\overrightarrow{B}}_1^{\left[5\right]}\right)}.\end{array}$$

(72)

The Poynting flux on a far-away sphere of radius ($R_{max}$) is given by the following expression:

$$\begin{array}{c}\hfill P{F}^{\left[3\right]}=\oint {\overrightarrow{S}}_{P12}^{\left[3\right]}·\widehat{n}da=R_{max}^2{\oint }_{R_{max}}{\overrightarrow{S}}_{P12}^{\left[3\right]}·\widehat{r}d\Omega .\end{array}$$

(73)

After calculations elaborated upon in Appendix H and by inserting Equations (A131), (A132), (A136) and (A137) into Equation (A119), we obtain the total Poynting flux contribution of the third order in ${\displaystyle \frac{1}{c}}$:

$$P{F}^{\left[3\right]}=-{\displaystyle \frac{2k}{3c^3}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left(\overrightarrow{J_1}·{\partial }_t^2\overrightarrow{J_2}+{\partial }_t^2\overrightarrow{J_1}·\overrightarrow{J_2}\right)}.$$

(74)

This term is minus Equation (66), that is, minus the power invested by the electromagnetic field to do mechanical work to the third order in ${\displaystyle \frac{1}{c}}$. It follows that in the third order, no energy can be stored in the electromagnetic field and the energy equation is balanced between radiation and work. That incoming radiation can cause charges to do work, or charges can radiate at the expense of their mechanical energy. However, if the currents in the two subsystems are orthogonal, no radiation is radiated and no work is done.

4.4.4. Summary

The key results of the $n=3$ calculation are outlined as follows. First, we notice that in this order, no energy is stored. This is a surprising result and can only be justified based on calculations. However, energy can be radiated by doing work or vice-versa—work can be done by absorbing radiation. From the point of view of the development of a retarded field engine, one should take special care such that radiation losses are avoided. This can be done by avoiding currents in either subsystem or by designing the engine such that the currents are perpendicular. The energy exchange is illustrated in Figure 4.

4.5. n = 4

Let us look at Equation (27) and study it for the fourth order in ${\displaystyle \frac{1}{c}}$:

$$\begin{array}{c}\hfill Powe{r}_{12}^{\left[4\right]}=-{\displaystyle \frac{d{E}_{field12}^{\left[4\right]}}{dt}}-{\oint }_S{\overrightarrow{S}}_{P12}^{\left[4\right]}·\widehat{n}da.\end{array}$$

(75)

4.5.1. Power

The fourth-order expression for the power invested in mechanical work is given below:

$$Powe{r}_{12}^{\left[4\right]}=\int d^{3}x{\displaystyle \left(\overrightarrow{J_1}·{\overrightarrow{E}}_2^{\left[4\right]}+\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[4\right]}\right)}.$$

(76)

According to Equations (51) and (54) of [13], the fourth-order term in the electric field expansion is expressed as follows:

$$\begin{array}{c}\hfill {\overrightarrow{E}}_1^{\left[4\right]}\left(\overrightarrow{x}\right)=-{\displaystyle \frac{k}{2c^4}}\int d^3{x}_1{\displaystyle \left[{\displaystyle \frac{1}{4}}{\partial }_t^4{\rho }_1{R}_1^2{\widehat{R}}_1+{\partial }_t^3\overrightarrow{J_1}{R}_1\right]}.\end{array}$$

(77)

Let us calculate both integrals of Equation (76) one by one:

$$I_1=\int d^{3}x\overrightarrow{J_2}·{\overrightarrow{E}}_1^{\left[4\right]}=-{\displaystyle \frac{k}{2c^4}}\int d^{3}x\overrightarrow{J_2}\left(\overrightarrow{x}\right)·\int d^3{x}_1{\displaystyle \left[{\displaystyle \frac{1}{4}}{\partial }_t^4{\rho }_1\left(\overrightarrow{x_1}\right)R_1^2{\widehat{R}}_1+{\partial }_t^3\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)R_1\right]}.$$

(78)

Changing the integration symbol from $\overrightarrow{x}$ to $\overrightarrow{x_2}$ results in the following form:

$$I_1={\displaystyle \frac{k}{2c^4}}\int d^3{x}_1{d}^3{x}_{2}R{\displaystyle \left[{\displaystyle \frac{1}{4}}{\partial }_t^4{\rho }_1\left(\overrightarrow{x_1}\right){\displaystyle \left(\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)·\overrightarrow{R}\right)}-{\partial }_t^3\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)\right]}.$$

(79)

Exchanging indices 1 and 2, we obtain the second integral of Equation (76):

$$I_2={\displaystyle \frac{k}{2c^4}}\int d^3{x}_1{d}^3{x}_{2}R{\displaystyle \left[-{\displaystyle \frac{1}{4}}{\partial }_t^4{\rho }_2\left(\overrightarrow{x_2}\right){\displaystyle \left(\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·\overrightarrow{R}\right)}-{\partial }_t^3\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)·\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right]}.$$

(80)

Adding the contributions, we finally obtain the following fourth-order power expression:

$$\begin{array}{ccc}\hfill Powe{r}_{12}^{\left[4\right]}& =& {\displaystyle \frac{k}{2c^4}}\int d^3{x}_1{d}^3{x}_{2}R{\displaystyle \left[{\displaystyle \frac{1}{4}}{\displaystyle \left({\partial }_t^4{\rho }_1\left(\overrightarrow{x_1}\right)\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)-{\partial }_t^4{\rho }_2\left(\overrightarrow{x_2}\right)\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right)}·\overrightarrow{R}\right.}\hfill \\ & -& {\displaystyle \left.{\displaystyle \left({\partial }_t^3\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)+{\partial }_t^3\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)·\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right)}.\right]}\hfill \end{array}$$

(81)

4.5.2. Poynting Vector

Next, we analyze the Poynting vector for $n=4$ using Equation (32):

$${\overrightarrow{S}}_{P12}^{\left[4\right]}={ϵ}_0{c}^2\sum _{k=0}^6{\displaystyle \left({\overrightarrow{E}}_1^{[6-k]}\times {\overrightarrow{B}}_2^{\left[k\right]}+{\overrightarrow{E}}_2^{[6-k]}\times {\overrightarrow{B}}_1^{\left[k\right]}\right)}$$

(82)

According to Equation (68), the Poynting vector takes the following fourth-order form:

$$\begin{array}{c}\hfill {\overrightarrow{S}}_{P12}^{\left[4\right]}={ϵ}_0{c}^2{\displaystyle \left({\overrightarrow{E}}_1^{\left[4\right]}\times {\overrightarrow{B}}_2^{\left[2\right]}+{\overrightarrow{E}}_2^{\left[4\right]}\times {\overrightarrow{B}}_1^{\left[2\right]}+{\overrightarrow{E}}_1^{\left[2\right]}\times {\overrightarrow{B}}_2^{\left[4\right]}+{\overrightarrow{E}}_2^{\left[2\right]}\times {\overrightarrow{B}}_1^{\left[4\right]}+{\overrightarrow{E}}_1^{\left[0\right]}\times {\overrightarrow{B}}_2^{\left[6\right]}+{\overrightarrow{E}}_2^{\left[0\right]}\times {\overrightarrow{B}}_1^{\left[6\right]}\right)}.\end{array}$$

(83)

The Poynting flux on a far-away sphere of radius $R_{max}$ is given by the following expression:

$$P{F}^{\left[4\right]}=\oint {\overrightarrow{S}}_{P12}^{\left[4\right]}·\widehat{n}da=R_{max}^2{\oint }_{R_{max}}{\overrightarrow{S}}_{P12}^{\left[4\right]}·\widehat{r}d\Omega .$$

(84)

After calculations elaborated upon in Appendix I and by inserting Equations (A145), (A146), (A150), (A151) and (A153) into Equation (A138), we obtain the total Poynting flux contribution of the fourth order in ${\displaystyle \frac{1}{c}}$:

$$P{F}^{\left[4\right]}={\displaystyle \frac{k{R}_{max}}{2c^4}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{1}{2}}{\displaystyle \left({\partial }_t^3\overrightarrow{J_1}·\overrightarrow{J_2}+{\partial }_t^3\overrightarrow{J_2}·\overrightarrow{J_1}\right)}-{\displaystyle \frac{1}{3}}{\displaystyle \left({\partial }_t\overrightarrow{J_1}·{\partial }_t^2\overrightarrow{J_2}+{\partial }_t\overrightarrow{J_2}·{\partial }_t^2\overrightarrow{J_1}\right)}\right)}.$$

(85)

4.5.3. Field Energy

Now, the contribution of the field energy to the energy balance at the fourth order is evaluated as follows:

$$E_{field12}^{\left[4\right]}={ϵ}_0\int {\displaystyle \left(\sum _{k=0}^4{\overrightarrow{E}}_1^{[4-k]}·{\overrightarrow{E}}_2^{\left[k\right]}+c^2\sum _{k=0}^6{\overrightarrow{B}}_1^{[6-k]}·{\overrightarrow{B}}_2^{\left[k\right]}\right)}{d}^{3}x.$$

(86)

Taking into account Equation (68), the field energy simplifies as follows:

$$\begin{array}{c}\hfill E_{field12}^{\left[4\right]}={ϵ}_0\int {\displaystyle \left[{\overrightarrow{E}}_1^{\left[4\right]}·{\overrightarrow{E}}_2^{\left[0\right]}+{\overrightarrow{E}}_1^{\left[0\right]}·{\overrightarrow{E}}_2^{\left[4\right]}+{\overrightarrow{E}}_1^{\left[2\right]}·{\overrightarrow{E}}_2^{\left[2\right]}+c^2{\displaystyle \left({\overrightarrow{B}}_1^{\left[4\right]}·{\overrightarrow{B}}_2^{\left[2\right]}+{\overrightarrow{B}}_1^{\left[2\right]}·{\overrightarrow{B}}_2^{\left[4\right]}\right)}\right]}{d}^{3}x.\end{array}$$

(87)

After calculations elaborated upon in Appendix J and by adding the fourth-order electric field energy given in Equation (A186) with the fourth-order magnetic field energy contribution given in Equation (A199), we arrive at the total fourth-order field energy:

$$\begin{array}{ccc}\hfill & & E_{field12}^{\left[4\right]}=E_{field12electric}^{\left[4\right]}+E_{field12magnetic}^{\left[4\right]}=\hfill \\ \hfill & & {\displaystyle \frac{k}{8c^4}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[-{\displaystyle \frac{1}{3}}{\partial }_t^2{\rho }_1{\partial }_t^2{\rho }_2{\displaystyle \left(R^3+2R^2{R}_{max}\right)}+{\displaystyle \frac{1}{3}}{R}^3({\partial }_t^3{\rho }_2{\partial }_t{\rho }_1+{\partial }_t^3{\rho }_1{\partial }_t{\rho }_2)+\right.}\hfill \\ & & {\displaystyle \left.4{\partial }_t\overrightarrow{J_1}·{\partial }_t\overrightarrow{J_2}{\displaystyle \left(2R_{max}-R\right)}+4(R-R_{max})({\partial }_t^2\overrightarrow{J_2}·\overrightarrow{J_1}+{\partial }_t^2\overrightarrow{J_1}·\overrightarrow{J_2})\right]}.\hfill \end{array}$$

(88)

This expression can be partitioned into a volume contribution and surface contribution. The latter type of contribution involves $R_{max}$ designating the maximal distance to which our expansion is applicable and, thus, also the radius of the sphere in which our energy calculations are performed. Thus,

$$\begin{array}{ccc}& & E_{field12}^{\left[4\right]}=E_{field12volume}^{\left[4\right]}+E_{field12surface}^{\left[4\right]}\hfill \\ & & E_{field12volume}^{\left[4\right]}\equiv {\displaystyle \frac{k}{8c^4}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[{\displaystyle \frac{1}{3}}{R}^3{\displaystyle \left({\partial }_t^3{\rho }_2{\partial }_t{\rho }_1+{\partial }_t^3{\rho }_1{\partial }_t{\rho }_2-{\partial }_t^2{\rho }_1{\partial }_t^2{\rho }_2\right)}+\right.}\hfill \\ & & {\displaystyle \left.4R{\displaystyle \left({\partial }_t^2\overrightarrow{J_2}·\overrightarrow{J_1}+{\partial }_t^2\overrightarrow{J_1}·\overrightarrow{J_2}-{\partial }_t\overrightarrow{J_1}·{\partial }_t\overrightarrow{J_2}\right)}\right]}.\hfill \\ & & E_{field12surface}^{\left[4\right]}\equiv {\displaystyle \frac{k{R}_{max}}{4c^4}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[-{\displaystyle \frac{1}{3}}{R}^2{\partial }_t^2{\rho }_1{\partial }_t^2{\rho }_2-2{\displaystyle \left({\partial }_t^2\overrightarrow{J_2}·\overrightarrow{J_1}+{\partial }_t^2\overrightarrow{J_1}·\overrightarrow{J_2}-\right.}\right.}\hfill \\ & & {\displaystyle \left.{\displaystyle \left.2{\partial }_t\overrightarrow{J_1}·{\partial }_t\overrightarrow{J_2}\right)}\right]}.\hfill \end{array}$$

(89)

After a few lines of calculations, it can now be shown that

$${\displaystyle \frac{d{E}_{field12volume}^{\left[4\right]}}{dt}}=-Powe{r}_{12}^{\left[4\right]},$$

(90)

in which $Powe{r}_{12}^{\left[4\right]}$ is the mechanical work done by the system and is given in Equation (81). For the surface term, we may only write the following:

$${\displaystyle \frac{d{E}_{field12surface}^{\left[4\right]}}{dt}}\approx -P{F}^{\left[4\right]},$$

(91)

in which $P{F}^{\left[4\right]}$ is the Poynting flux given in Equation (85). Although both sides of Equation (91) contain the same type of expressions, the numerical factors are only of the same order of magnitude—not identical. This is probably due to the crude approximations used in Section 4.5.2 and in the {appendices}. Notice, however, that this will not change the final results of the current paper.

4.5.4. Summary

The key results of the $n=4$ calculation are outlined as follows. First, we notice that in this order, we have, for the first time, all energy components: energy is stored and radiated and does mechanical work. This, however, is not relevant to the retarded field engine, as it involves higher derivatives of the field sources that are not needed for the functioning of the retarded field engine. The radiation terms do not balance exactly in the above calculation for the following reasons. First, in the Poynting flux, we keep only terms that are proportional to $R_{max}$, neglecting lower order terms. Second, in the integral calculation of the energy, we neglect corrections associated with a small change of boundary terms due to the change of variable (from $\overrightarrow{x}$ to $\overrightarrow{R_1}$), which considerably simplifies the calculation. A more accurate and detailed calculation should balance those terms as well; however, this has no baring on retarded field engine theory and, from this point of view, is redundant. The energy exchange is illustrated in Figure 2.

5. Discussion

In order to verify the validity of the energy balance equation at various orders of approximation, we systematically evaluate the mechanical work, field energy, and Poynting vector contributions corresponding to successive terms in the Taylor expansion of the electromagnetic fields. Each term, labeled by index n, represents a distinct order in the $1/c$ expansion and contributes differently to the total energy dynamics of the system. Below, we summarize the outcome of this analysis for $n=0$ through $n=4$, highlighting the nature of energy exchange and the role of radiation at each order. The results are also given in

Table 1. Table of energy balance results.

n	Field Energy Derivative	Mechanical Power	Radiated Power
0	${\displaystyle \frac{d}{dt}}{\displaystyle \left[k\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \frac{{\rho }_1{\rho }_2}{R}}\right]}$	$-{\displaystyle \frac{d}{dt}}{\displaystyle \left[k\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \frac{{\rho }_1{\rho }_2}{R}}\right]}$	0
1	0	0	0
2	${\displaystyle \frac{k}{c^2}}{\displaystyle \frac{d}{dt}}{\displaystyle \left[\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{R\left({\partial }_t{\rho }_1{\partial }_t{\rho }_2\right)}{2}}+{\displaystyle \frac{\overrightarrow{J_1}·\overrightarrow{J_2}}{R}}\right)}\right]}$	$-{\displaystyle \frac{k}{c^2}}{\displaystyle \frac{d}{dt}}{\displaystyle \left[\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{R\left({\partial }_t{\rho }_1{\partial }_t{\rho }_2\right)}{2}}+{\displaystyle \frac{\overrightarrow{J_1}·\overrightarrow{J_2}}{R}}\right)}\right]}$	0
3	0	${\displaystyle \left({\displaystyle \frac{2k}{3c^3}}\right)}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left(\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·{\partial }_t^2\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)+\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)·{\partial }_t^2\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right)}$	$-{\displaystyle \left({\displaystyle \frac{2k}{3c^3}}\right)}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left(\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·{\partial }_t^2\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)+\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)·{\partial }_t^2\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right)}$
4	${\displaystyle \frac{k}{8c^4}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left[-{\displaystyle \frac{1}{3}}{\partial }_t^2{\rho }_1{\partial }_t^2{\rho }_2{\displaystyle \left(R^3+2R^2{R}_{max}\right)}+{\displaystyle \frac{1}{3}}{R}^3({\partial }_t^3{\rho }_2{\partial }_t{\rho }_1+{\partial }_t^3{\rho }_1{\partial }_t{\rho }_2)+\right.}$ ${\displaystyle \left.4{\partial }_t\overrightarrow{J_1}·{\partial }_t\overrightarrow{J_2}{\displaystyle \left(2R_{max}-R\right)}+4(R-R_{max})({\partial }_t^2\overrightarrow{J_2}·\overrightarrow{J_1}+{\partial }_t^2\overrightarrow{J_1}·\overrightarrow{J_2})\right]}$	${\displaystyle \frac{k}{2c^4}}\int d^3{x}_1{d}^3{x}_{2}R{\displaystyle \left[{\displaystyle \frac{1}{4}}{\displaystyle \left({\partial }_t^4{\rho }_1\left(\overrightarrow{x_1}\right)\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)-{\partial }_t^4{\rho }_2\left(\overrightarrow{x_2}\right)\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right)}·\overrightarrow{R}\right.}$ $-{\displaystyle \left.{\displaystyle \left({\partial }_t^3\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)+{\partial }_t^3\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)·\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)\right)}.\right]}$	${\displaystyle \frac{k{R}_{max}}{2c^4}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{1}{2}}{\displaystyle \left({\partial }_t^3\overrightarrow{J_1}·\overrightarrow{J_2}+{\partial }_t^3\overrightarrow{J_2}·\overrightarrow{J_1}\right)}-{\displaystyle \frac{1}{3}}{\displaystyle \left({\partial }_t\overrightarrow{J_1}·{\partial }_t^2\overrightarrow{J_2}+{\partial }_t\overrightarrow{J_2}·{\partial }_t^2\overrightarrow{J_1}\right)}\right)}$

.

For $n=0$, the zeroth-order energy equation is found to be satisfied. Mechanical work performed on or by the system leads to a corresponding increase or decrease in the field energy, as expected in the absence of any contribution from the Poynting vector. For $n=1$, we find that the first-order energy equation is trivially satisfied, that is, it reduces to an identity of $0=0$ under the condition in which the mechanical work, field energy, and Poynting contribution all vanish individually. Vanishing mechanical work implies that, to the first order in $1/c$, no mechanical energy is either supplied to or extracted from the relativistic engine. Furthermore, the first-order terms in the $1/c$ expansion do not contribute to either the field energy or the Poynting vector associated with the engine. In other words, in the first order, the electromagnetic fields neither stores energy nor transports it through radiation or absorption, indicating the absence of any real energy exchange. For $n=2$, we conclude that the second-order energy equation is, indeed, balanced: the power associated with mechanical work equals the negative time derivative of the field energy, as expected in the absence of any Poynting contribution. In the case of $n=3$, the absence of field energy implies that the electromagnetic field cannot store energy, leading to an energy balance strictly between radiation and mechanical work. Thus, incident radiation has the potential to drive charged particles to perform work. However, when the currents in the two subsystems are orthogonal, radiation is not emitted, and consequently, no work is done. It is also important to note that, if not properly configured, a relativistic motor may emit radiation. This behavior has been demonstrated in earlier studies involving uncharged relativistic motors [18]. For $n=4$, we conclude that the fourth-order energy balance holds. At this order, both the electric and magnetic-field energies are influenced by the mechanical work. At the order of $1/c^4$, the structure of our series expansion limits our ability to evaluate the radiation flux at infinity. Instead, we must restrict our analysis to the radiation flux through a spherical surface of radius $R_{max}$, beyond which the validity of our approximation breaks down. We also notice that for terms dependent on $R_{max}$, the approximation we use here is not sufficient to exactly balance the terms of the energy change and the Poynting flux, for which a better approximation is required.

Our motivation for this study is to understand the energy exchange in a relativistic motor; however, for this motor, one needs to consider only the first-order derivative of the charge density. On the other hand, $n\ge 3$ terms in the energy equation depend on higher derivatives. This problem will be resolved in the next section.

We conclude this section by listing the limiting assumptions of our analysis:

• Our analysis is within the framework of classical electromagnetism; this means that we exclude extremely high electric and magnetic fields, which are beyond the Schwinger limit and for which a proper physical treatment will require quantum electrodynamics.
• We assume that each of the interacting subsystems (subsystems 1 and 2) is small enough such that the retardation effects can be neglected within each system. The precise meaning of “small enough” for slow, non-relativistic systems was developed in [19] and has to do with the size (R) of each subsystem satisfying $R<{\displaystyle \frac{2c^2}{a}}$, in which a is the acceleration of any microscopic constituent of those systems. The abovementioned paper concentrates on gravitational systems, but the analysis should be approximately relevant to electrodynamic systems as well.
• We assume that the subsystems have the same charge and current densities with respect to the engine frame but not necessarily with respect to the laboratory frame. The engine may move and even accelerate with respect to the laboratory frame.

6. The Source of the Charged Retarded Field Engine Energy

So far, our discussion has been quite general. We have calculated the energy balance terms for various orders of $1/c^n$ up to $n=4$. We have not paid specific attention to the characteristics of the relativistic engine, which we consider now. Suppose we have two static charge distributions consisting of a retarded field engine and suppose one induces a time-dependent change in one of the charge distributions generating momentum according to Equation (8), after which the charge distribution becomes static again, as shown in Figure 5.

Notice that at this point, we have two static charge distributions in the engine’s frame but not in the laboratory frame. In the laboratory frame, the two subsystems connected to the engine are now moving with a velocity of ${\overrightarrow{v}}_s$ (see Equation (10)), which is the velocity of the engine. The engine has gained kinetic energy, as represented by Equation (9). We now inquire regarding the source of this energy.

Let $\overrightarrow{x}$ be the location vector of a point in the laboratory (inertial) frame and let ${\overrightarrow{x}}_c\left(t\right)$ be the location of the engine in that frame, as shown in Figure 6.

It follows that the location of the same point with respect to the engine is expressed as follows:

$${\overrightarrow{x}}^{\prime }=\overrightarrow{x}-{\overrightarrow{x}}_c\left(t\right),\Rightarrow \overrightarrow{x}={\overrightarrow{x}}^{\prime }+{\overrightarrow{x}}_c\left(t\right).$$

(92)

It also follows that

$${\overrightarrow{v}}_s={\displaystyle \frac{d{\overrightarrow{x}}_c\left(t\right)}{dt}}.$$

(93)

The charge density can be expressed in either the laboratory (inertial) frame or in the engine frame. In the first case, it is designated by $\rho$, and in the second, it is designated by ${\rho }_m$; these two functions are related as follows:

$$\rho (\overrightarrow{x},t)={\rho }_m({\overrightarrow{x}}^{\prime },t)={\rho }_m(\overrightarrow{x}-{\overrightarrow{x}}_c\left(t\right),t).$$

(94)

In the case in which the charge distribution is “static” in the engine frame, this reduces to the following:

$$\rho (\overrightarrow{x},t)={\rho }_m(\overrightarrow{x}-{\overrightarrow{x}}_c\left(t\right)).$$

(95)

Similarly, the partial temporal derivative can be calculated as follows:

$${\partial }_t\rho (\overrightarrow{x},t){|}_{\overrightarrow{x}}={\partial }_t{\rho }_m(\overrightarrow{x}-{\overrightarrow{x}}_c\left(t\right),t){{|}_{\overrightarrow{x}}=-{\overrightarrow{v}}_s·\overrightarrow{\nabla }{\rho }_m+{\partial }_t{\rho }_m|}_{{\overrightarrow{x}}^{\prime }}.$$

(96)

For the case in which the charge density is “static” in the engine frame, this reduces to the following:

$${\partial }_t\rho (\overrightarrow{x},t){|}_{\overrightarrow{x}}=-{\overrightarrow{v}}_s·\overrightarrow{\nabla }{\rho }_m=-\overrightarrow{\nabla }·{\displaystyle \left(\rho {\overrightarrow{v}}_s\right)}.$$

(97)

Taking into account the continuity (Equation (A9)), we notice that although no current densities were initially assumed in the system, it now follows that, from the laboratory point of view, the engine now carries current density of the following form:

$$\overrightarrow{J}=\rho {\overrightarrow{v}}_s$$

(98)

We are now at a position to calculate the work done by the internal electric field in such a system. This is done by using Equation (54), which describes the time derivative of this work. We remind the reader that we have shown this work to be exactly equal to the amount of energy change in the electromagnetic field (see Equation (58)). Thus,

$$E_{mech}=-{\displaystyle \frac{k}{c^2}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{R{\displaystyle \left({\partial }_t{\rho }_1\left(\overrightarrow{x_1}\right){\partial }_t{\rho }_2\left(\overrightarrow{x_2}\right)\right)}}{2}}+{\displaystyle \frac{\overrightarrow{J_1}\left(\overrightarrow{x_1}\right)·\overrightarrow{J_2}\left(\overrightarrow{x_2}\right)}{R}}\right)}.$$

(99)

This can be written in the following form:

$$E_{mech}=-{\displaystyle \frac{k}{2c^2}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{R}{2}}{\displaystyle \left({\partial }_t{\rho }_1{\partial }_t{\rho }_2+{\partial }_t{\rho }_1{\partial }_t{\rho }_2\right)}+{\displaystyle \frac{\overrightarrow{J_1}·\overrightarrow{J_2}+\overrightarrow{J_1}·\overrightarrow{J_2}}{R}}\right)}.$$

(100)

Taking into account Equations (97) and (98), it follows that

$$E_{mech}={\displaystyle \frac{k}{2c^2}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{R}{2}}{\displaystyle \left({\partial }_t{\rho }_1{\overrightarrow{v}}_s·\overrightarrow{\nabla }{\rho }_2+{\overrightarrow{v}}_s·\overrightarrow{\nabla }{\rho }_1{\partial }_t{\rho }_2\right)}-{\displaystyle \frac{\overrightarrow{J_1}·{\overrightarrow{v}}_s{\rho }_2+{\rho }_1{\overrightarrow{v}}_s·\overrightarrow{J_2}}{R}}\right)}.$$

(101)

Notice that

$$\int d^3{x}_{2}R{\overrightarrow{v}}_s·{\overrightarrow{\nabla }}_2{\rho }_2=\int d^3{x}_2{\displaystyle \left[{\overrightarrow{\nabla }}_2·\left(R{\rho }_2{\overrightarrow{v}}_s\right)-{\rho }_2{\overrightarrow{v}}_s·{\overrightarrow{\nabla }}_{2}R\right]}.$$

(102)

Using the Gaussian theorem and Equation (A34), this can be written in the following form:

$$\int d^3{x}_{2}R{\overrightarrow{v}}_s·{\overrightarrow{\nabla }}_2{\rho }_2=\oint d{\overrightarrow{S}}_2·\left(R{\rho }_2{\overrightarrow{v}}_s\right)+\int d^3{x}_2{\rho }_2{\overrightarrow{v}}_s·\widehat{R}=\int d^3{x}_2{\rho }_2{\overrightarrow{v}}_s·\widehat{R},$$

(103)

in which we assume, as usual, that there is no charge on a far encapsulating surface. Similarly,

$$\int d^3{x}_{1}R{\overrightarrow{v}}_s·{\overrightarrow{\nabla }}_1{\rho }_1=-\int d^3{x}_1{\rho }_1{\overrightarrow{v}}_s·\widehat{R}.$$

(104)

Plugging Equations (103) and (104) into Equation (101) it follows that

$$E_{mech}={\displaystyle \frac{1}{2}}{\overrightarrow{v}}_s·{\displaystyle \frac{k}{c^2}}\int \int d^3{x}_1{d}^3{x}_2{\displaystyle \left({\displaystyle \frac{\widehat{R}}{2}}{\displaystyle \left({\partial }_t{\rho }_1{\rho }_2-{\rho }_1{\partial }_t{\rho }_2\right)}-{\displaystyle \frac{\overrightarrow{J_1}{\rho }_2+{\rho }_1\overrightarrow{J_2}}{R}}\right)}={\displaystyle \frac{1}{2}}{\overrightarrow{v}}_s·\overrightarrow{P}=E_k.$$

(105)

The linear momentum is given by Equation (8), and the kinetic energy was defined in Equation (9). Hence, the work done by the retarded electromagnetic field is converted into the kinetic energy of the retarded field engine. We use an $n=2$ expression for the work, but since ${\overrightarrow{v}}_s$ is proportional to ${\displaystyle \frac{1}{c^2}}$, it turns out to be proportional to, ${\displaystyle \frac{1}{c^4}}$, as previously discussed, thereby resolving the dilemma stated in the end of the previous section.

7. Conclusions

A relativistic engine is not a self-sustaining device, nor is it in perpetuum mobile; it requires an input of energy to function. This energy is drawn from the electromagnetic field, reducing its energy in the process. In this study, we explore the validity of energy conservation in relation to a charged retarded engine. This research focuses on how energy is conserved in charged retarded systems, proposing an exciting new way to power spacecraft. By using electromagnetic fields instead of traditional rocket fuel, this technology could help spaceships travel faster and farther while using much less energy.

The practical applications of the proposed “retarded field engine” are described in several papers. In [13], we describe a relativistic engine based on a charged capacitor/electret (system 1) and a coil carrying current (system 2) (see Figure 7). In this case, the capacitor is charged for a brief moment, after which both currents and charges are stationary, the force and momentum are in the direction of the current in the upper part of the figure.

In [20], we describe a relativistic engine based on two wires carrying periodic currents (see Figure 8). In this case, the upper wire is system 1, and the lower wire is system 2. The force is generated in the y direction, which is perpendicular to both wires. The current in each wire propagates along the wire.

In [14], we describe a relativistic engine composed of a permanent magnet (system 1) and a capacitor (system 2) (see Figure 9). In this case, the capacitor is charged for a brief moment, after which both (magnetic) currents and charges are stationary. The force and momentum are in the direction of the magnetic current in the upper part of the figure.

The major obstacles to obtaining a useful retarded field engine have to do with the phenomenon of dielectric breakdown, which limits the amount of charge density that can be accumulated before a discharge occurs and the amount of current density that a wire can carry, even if it is superconducting. Modest figures that are relevant to the configuration described in Figure 7 are presented in

Table 2. Maximal momentum gained by a relativistic motor for three cases of parameters. We assume an extreme case of charge density of $\sigma =3.7\times {10}^{-3}$ Coulomb/${\mathrm{m}}^2$ and current density pf $J_0=5\times {10}^7$ Ampere/${\mathrm{m}}^2$.

	Car	Rocket Engine Size	Giant Cube	Units
a	6	200	1000	m
b	2	10	1000	m
d	1	10	1000	m
w	0.2	0.4	0.4	m
$P_T$	0.3	868	$3.1\times {10}^7$	kg m/s

.

Obviously, one needs a very sensitive and precise experimental setup to measure such effects. On the atomic scale, the figures are quite different; for example, one can reach a volume charge density of

$$\overline{\rho }={\displaystyle \frac{e}{\frac{4}{3}\pi a_0^3}}\simeq 2.6\times {10}^{11}\mathrm{Coulomb}/{\mathrm{m}}^3,$$

(106)

in which $e\simeq 1.6\times {10}^{-19}$ Coulomb is the electron charge and $a_0\simeq 0.53\times {10}^{-10}\mathrm{m}$ is Bohr’s radius. The surface charge density of an electron in a hydrogen atom has an order of magnitude of

$$\overline{\sigma }={\displaystyle \frac{e}{4\pi a_0^2}}\simeq 4.5\mathrm{Coulomb}/{\mathrm{m}}^2.$$

(107)

This is a thousand times larger than what is available for macroscopic charge densities. The order of magnitude of the current density is

$$\overline{J}={\displaystyle \frac{e\hslash }{m_e{a}_0^4}}\simeq 2.3\times {10}^{18}\mathrm{Ampere}/{\mathrm{m}}^2=2.3\times {10}^{14}\mathrm{Ampere}/{\mathrm{cm}}^2$$

(108)

This current density is ten orders of magnitude larger than what can be achieved at a macroscopic scale. This motivates suggestions for implementations involving material design [21,22] (see, for example, Figure 10 and Figure 11).

The above examples explicitly highlight the practical implications of this paper’s findings. The authors welcome potential experimental demonstrations of any of the above configurations. They also direct readers to possible future work in the fields of material science and solid-state physics, in which it is hoped that materials that are relevant to the retarded field engine could be produced.