Lorentz Symmetry Group, Retardation and Energy Transformations in a Relativistic Engine

Abstract

In a previous paper, we have shown that Newton’s third law cannot strictly hold in a distributed system of which the different parts are at a finite distance from each other. This is due to the finite speed of signal propagation which cannot exceed the speed of light in vacuum, which in turn means that when summing the total force in the system the force does not add up to zero. This was demonstrated in a specific example of two current loops with time dependent currents, the above analysis led to suggestion of a relativistic engine. Since the system is effected by a total force for a finite period of time this means that the system acquires mechanical momentum and energy, the question then arises how can we accommodate the law of momentum and energy conservation. The subject of momentum conservation was discussed in a pervious paper, while preliminary results regarding energy conservation where discussed in some additional papers. Here we give a complete analysis of the exchange of energy between the mechanical part of the relativistic engine and the field part, the energy radiated from the relativistic engine is also discussed. We show that the relativistic engine effect on the energy is 4th-order in $\frac{1}{c}$ and no lower order relativistic engine effect on the energy exists.

1. Introduction

Special relativity is a theory of the structure of space-time. It dictates that the equations of any physical theory should be invariant under the Lorentz group of coordinate transformations. The theory was first introduced in Einstein’s famous 1905 paper: “On the Electrodynamics of Moving Bodies” [1]. This theory was a consequence of empiric observations and the laws of electromagnetism which were formulated in the middle of the nineteenth century by Maxwell in their famous four partial differential equations [2,3,4] which owe their current form to Oliver Heaviside [5]. One of the consequences of these equations is that an electromagnetic signal travels at the speed of light c, which led people to believe that light is an electromagnetic wave. This was later used by Albert Einstein [1,3,4] to formulate their special theory of relativity which postulates that the speed of light in vacuum c is the maximal allowed velocity in nature. According to the theory of relativity no object, message, signal (even if not electromagnetic) or field can travel faster than the speed of light in vacuum. Hence retardation, if someone at a distance R from me changes something I may not know about it for at least a retardation time of $\frac{R}{c}$. This means that action and its reaction cannot be generated at the same time because of the signal finite propagation speed.

Newton’s laws of motion are three physical laws that, together, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. The three laws of motion were first compiled by Isaac Newton in their Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), first published in 1687 [6,7]. We will only be interested in this paper in the third law which states: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

According to the third law, the total force in a system not affected by external forces is thus zero. This law has numerous experimental verifications and seems to be one of the corner stones of physics. However, in light of the previous discussion it is obvious that action and its reaction cannot be generated at the same time because of the finite speed of signal propagation, hence the third law is false in an exact sense although it can be true for most practical application due to the high speed of signal propagation. Thus, the total force cannot be null at a given time.

Most modern locomotive systems are based on two material parts each obtaining momentum, which is equal and opposite to the momentum gained by the second part. A typical example of this type of system is a rocket which sheds exhaust gas to propel itself. However, the above relativistic considerations suggest’s a new type of motor in which the system is not composed of two material bodies but of a material body and field. Ignoring the field a naive observer will see the material body gaining momentum created out of nothing, however, a knowledgeable observer will understand that the opposite amount of momentum is obtained by the field as was shown in [8]. Indeed Noether’s theorem dictates that any system possessing translational symmetry will conserve momentum and the total physical system containing matter and field is indeed symmetrical under translations, while every sub-system (either matter or field) is not. This was already noticed by Feynman [4]. Feynman describes two orthogonally moving charges, apparently violating Newton’s third law as the forces that the charges induce on each other do not cancel (last part of 26-2), this paradox is resolved in (27-6) in which it is shown that the momentum gained by the two charge system is balanced by the field momentum.

We will define a relativistic engine as a physical system in which its center of mass is in motion due to the interaction of its material parts. Those parts may be free to move with respect to each other or more practically held in a rigid frame. This is of no consequence as we will only be interested in the motion of the center of mass in this paper. We stress that a relativistic motor allows 3-axis motion (including vertical), it has no moving parts, it has zero fuel consumption (and thus zero carbon emission) and needs only electromagnetic energy as explained in the current paper which may be provided by solar panels. The relativistic engine is an ideal solution for space travel in which currently much of the space vehicle mass is devoted to fuel.

It should be noted that there is no Lorentz covariant definition for the relativistic center of mass, as pointed out by Pryce [9]. Indeed, the center of mass mentioned in our work is not Lorentz covariant, it is defined with respect to a specific frame of reference in which the relativistic engine is at rest (the laboratory frame of reference). Even after the engine is set into motion its velocity is small compared to the speed of light, and thus the Newtonian definition of the center of mass still holds approximately in that specific frame. One should not confuse the retardation speed of the electromagnetic field (which is equal to speed of light in vacuum) with the speed of the engine itself that may be quite modest. The device is denoted a relativistic engine because it takes advantage of the relativistic retardation effect not because it can reach relativistic speeds, in fact one can think of reasons why a macroscopic relativistic engine will not reach relativistic speed due to the high energy required. However, if one still would like to consider an engine moving at relativistic speeds then perhaps the center of mass concept should be replaced by something else, perhaps a relativistic center of momentum, however, this option is beyond the scope of the current paper.

To be sure, other systems can introduce retardation such as for two objects that are connected by a long spring, in which conservation of momentum and energy do not apply if one ignores the momentum and energy of the spring. This case along with other interesting cases are discussed in an excellent book “Dynamics of Particles and the Electromagnetic Field” [10], where more problems, and solutions, were treated than in the current paper, this includes the problem of initial conditions. This work is not about retardation in long springs but about retardation caused by the relativistic effect of a finite speed of signal propagation. Only in this later case, momentum and energy is transferred between field and material allowing the relativistic engine effect. This is quite different than transfer of momentum and energy between two material bodies such as two objects and a connecting long spring. Momentum and energy transfer between material bodies is well known and the center of mass of the two objects and the connecting long spring viewed as one system cannot move, if not initially in motion. However, this is rather different from the current case in which momentum and energy are transferred from the electromagnetic field to a material body. In the later example momentum and energy are transferred from one material body (the two objects) to another material body (the long spring) this is well known and perhaps of less interest than the transfer of momentum and energy from a material body to a field which is involved in the motion of the center of mass of the material body. The latter case suggests a new form of motor which is the relativistic engine.

In what follows, we will assume that the magnetization and polarization of the medium are small and therefore we neglect corrections to the Lorentz force suggested in [11]. In a paper by Griffiths and Heald [12], it was pointed out that strictly Coulomb’s law and the Biot-Savart law determine the electric and magnetic fields for static sources only. Time-dependent generalizations of these two laws introduced by Jefimenko [13] were used to explore the applicability of Coulomb and Biot-Savart outside the static domain.

In a previous paper, we used Jefimenko’s [3,13] equation to discuss the force between two current carrying coils [14]. This was later expanded to include the interaction between a current carrying loop and a permanent magnet [15,16], see Figure 1.

Since the system is affected by a total force for a finite period of time this means that the system acquires mechanical momentum and energy, the question then arises if we need to abandon the law of momentum and energy conservation. The subject of momentum conversation was discussed in [8]. In [17,18,19], some preliminary aspects of the exchange of energy between the mechanical part of the relativistic engine and the electromagnetic field were discussed. In particular it was shown that the electric energy expenditure is twice the kinetic energy gained by the relativistic motor. It was also shown how some energy may be radiated from the relativistic engine device if the coils are not configure properly. In this paper, we develop a methodology to deal with all aspects of the energy transformation in a relativistic engine.

The plan of this paper is as follows: First, we introduce the conservation of energy and momentum in a general electromagnetic system. Then we discuss the particular case of a simple relativistic engine made of two current loops of arbitrary geometry, in which we shall consider the mechanical momentum and energy gained by the engine. This will be followed by a general analysis of the energy transformations between the field and mechanical components in a relativistic engine. Which will be followed by a specific analysis of a relativistic engine taking into account the various energy contributions expanded in powers of $\frac{1}{c}$, in which c is the speed of light in vacuum. We show that the relativistic engine effect on the energy is 4th-order in $\frac{1}{c}$ and no lower order relativistic engine effect on the energy exists.

2. Energy and Momentum Conservation

Any system with space-time translational symmetry must conserve momentum and energy according to Noether’s theorem. In the case of a system with charge and current densities the energy-momentum conservation law takes the form [3]:

$$\int ({\partial }_{\alpha }{\Theta }^{\alpha \beta }+f^{\beta })d^{3}x=0$$

(1)

$x^{\alpha }$ are space-time coordinates such that $\alpha ,\beta \in \{0,1,2,3\}$, ${\partial }_{\alpha }$ is a partial derivative with respect to the four dimensional coordinates and Einstein’s summation convention is assumed, $d^{3}x$ is a volume element. In the above:

$${\Theta }^{\alpha \beta }=\left(\begin{array}{cc}{e}_{field}& \frac{1}{c}\overrightarrow{}{S}_p\\ \frac{1}{c}{\overrightarrow{S}}_p& -T_{ij}\end{array}\right)$$

(2)

and:

$$\int f^{\beta }{d}^{3}x=(\frac{d{E}_{mech}}{dt},\frac{d{\overrightarrow{P}}_{mech}}{dt}).$$

(3)

The various terms in the matrix appearing in Equation (2) are defined in terms of the electric field $\overrightarrow{E}$ and magnetic flux density $\overrightarrow{B}$ as follows. The field energy density $e_{field}$ is defined such that:

$$E_{field}\equiv \int e_{field}{d}^{3}x=\frac{{ϵ}_0}{2}\int \left({\overrightarrow{E}}^2+c^2{\overrightarrow{B}}^2\right)d^{3}x$$

(4)

where ${ϵ}_0$ is the vacuum permittivity (≃$8.85{10}^{-12}\mathrm{F}{\mathrm{m}}^{-1}$). Poynting’s vector is defined as:

$${\overrightarrow{S}}_p=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$$

(5)

where ${\mu }_0=4\pi {10}^{-7}$ is the vacuum magnetic permeability. $T_{ij}$ is the Maxwell stress tensor:

$$T_{ij}={ϵ}_0\left[E_i{E}_j+c^2{B}_i{B}_j-\frac{1}{2}({\overrightarrow{E}}^2+c^2{\overrightarrow{B}}^2){\delta }_{ij}\right]$$

(6)

$i,j\in \{1,2,3\}$ and ${\delta }_{ij}$ is Kronecker’s delta. Equation (3) contains the mechanical energy and momentum $E_{mech},{\overrightarrow{P}}_{mech}$ and the temporal derivative $\frac{d}{dt}$. Next we shall write the matrix Equation (1) in terms of the spatial and temporal components separately. The spatial components will yield the equation:

$$\frac{d{P}_{mechi}}{dt}+\frac{d{P}_{fieldi}}{dt}={\oint }_S{T}_{ij}{\widehat{n}}_{j}da$$

(7)

In the above, $P_{fieldi}$ is the i component of the field momentum of the system:

$${\overrightarrow{P}}_{field}={ϵ}_0\int \overrightarrow{E}\times \overrightarrow{B}{d}^{3}x$$

(8)

S is a closed surface encapsulating the volume in which the system is located, $\widehat{n}$ is a unit vector normal to the surface, $da$ is a surface element.

Equation (7) as proved in [3] is a precise statement of momentum conservation in a relativistic engine and from a pure point of view nothing else is needed, however, for the sake of concreteness a specific example for the two current loops relativistic engine was analyses in [8] and will not be repeated here.

To conclude this section we shall look at the zeroth component of Equation (1), which yields the energy conservation equation:

$$\frac{d{E}_{mech}}{dt}+\frac{d{E}_{field}}{dt}=-{\oint }_S{\overrightarrow{S}}_p·\widehat{n}da.$$

(9)

The derivative of the mechanical energy is the power needed to sustain the system and is given by [3]:

$$Power\equiv \frac{d{E}_{mech}}{dt}=\int d^{3}x\overrightarrow{J}·\overrightarrow{E}$$

(10)

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

$$\int d^{3}x\overrightarrow{J}·\overrightarrow{E}+\frac{d{E}_{field}}{dt}=-{\oint }_S{\overrightarrow{S}}_p·\widehat{n}da.$$

(11)

Or:

$$\int d^{3}x\overrightarrow{J}·\overrightarrow{E}=-\frac{d{E}_{field}}{dt}-{\oint }_S{\overrightarrow{S}}_p·\widehat{n}da.$$

(12)

In the following we will discuss the manifestation of energy conservation as described by Equation (12) for a relativistic engine.

3. The Case of Two Currents Loops of Arbitrary Geometry

Consider two wires having segments of length $d{\overrightarrow{l}}_1,d{\overrightarrow{l}}_2$ located at ${\overrightarrow{x}}_1,{\overrightarrow{x}}_2$, respectively, and carrying currents $I_1,I_2$ (see Figure 2).

According to [14] (Equation (38)), the force on loop ${}_2$ generated by loop ${}_1$ takes the form:

$${\overrightarrow{F}}_{21}=\frac{{\mu }_0}{4\pi }{I}_2\left(t\right)\sum _{n=0}^{\infty }\frac{I_1^{\left(n\right)}\left(t\right)}{n!}{(-\frac{1}{c})}^n(1-n)\oint \oint R_{12}^{n-3}{\overrightarrow{R}}_{12}(d{\overrightarrow{l}}_2·d{\overrightarrow{l}}_1)$$

(13)

in which ${\overrightarrow{R}}_{12}\equiv {\overrightarrow{x}}_1-{\overrightarrow{x}}_2$, $R_{12}\equiv \left|{\overrightarrow{R}}_{12}\right|$. $I_1\left(t\right)$ is written as a Taylor expansion in the time t and the terms $I^{\left(n\right)}\left(t\right)$ are the derivatives of order n with respect to the variable t. As in all expansions the above equation is valid only for a certain environment of t on the time axis which depends on the function $I_1\left(t\right)$, this environment shall be defined using the convergence radius $T_{max}$. That is Equation (13) is valid only in the domain $[t-T_{max},t+T_{max}]$. We note that there is no first order contribution to the force. Hence the next contribution to the force after the quasi-static term is second order. Let us define the dimensionless geometrical factor ${\overrightarrow{K}}_{21n}$ as:

$${\overrightarrow{K}}_{21n}=\frac{1}{h^n}\oint \oint R_{12}^{n-3}{\overrightarrow{R}}_{12}(d{\overrightarrow{l}}_2·d{\overrightarrow{l}}_1)=-{\overrightarrow{K}}_{12n}.$$

(14)

in the above h is some characteristic distance between the coils. In terms of ${\overrightarrow{K}}_{21n}$ we can write Equation (13) as:

$${\overrightarrow{F}}_{21}=\frac{{\mu }_0}{4\pi }{I}_2\left(t\right)\sum _{n=0}^{\infty }\frac{I_1^{\left(n\right)}\left(t\right)}{n!}{(-\frac{h}{c})}^n(1-n){\overrightarrow{K}}_{21n}.$$

(15)

The force due to coil 2 that acts on coil 1 is:

$${\overrightarrow{F}}_{12}=\frac{{\mu }_0}{4\pi }{I}_1\left(t\right)\sum _{n=0}^{\infty }\frac{I_2^{\left(n\right)}\left(t\right)}{n!}{(-\frac{h}{c})}^n(1-n){\overrightarrow{K}}_{12n}.$$

(16)

The total force on the system is thus:

$$\begin{array}{ccc}& & {\overrightarrow{F}}_T={\overrightarrow{F}}_{12}+{\overrightarrow{F}}_{21}=\hfill \\ & & \frac{{\mu }_0}{4\pi }\sum _{n=0}^{\infty }\frac{(1-n)}{n!}{(-\frac{h}{c})}^n{\overrightarrow{K}}_{12n}\left(I_1\left(t\right)I_2^{\left(n\right)}\left(t\right)-I_2\left(t\right)I_1^{\left(n\right)}\left(t\right)\right).\hfill \end{array}$$

(17)

We note that the quasi-static term $n=0$ does not contribute to the sum nor does the $n=1$ term. The fact that the retarded field “corrects” itself to first order in order to “mimic” a non retarded field was already noticed by Feynman [4]. Hence we can write:

$${\overrightarrow{F}}_T=\frac{{\mu }_0}{4\pi }\sum _{n=2}^{\infty }\frac{(1-n)}{n!}{(-\frac{h}{c})}^n{\overrightarrow{K}}_{12n}\left(I_1\left(t\right)I_2^{\left(n\right)}\left(t\right)-I_2\left(t\right)I_1^{\left(n\right)}\left(t\right)\right).$$

(18)

We conclude that in general Newton’s third law is not satisfied, taking the leading non-vanishing terms in the above sum we obtain:

$$\begin{array}{ccc}\hfill {\overrightarrow{F}}_T& \cong & -\frac{{\mu }_0}{8\pi }{\left(\frac{h}{c}\right)}^2{\overrightarrow{K}}_{122}\left(I_1\left(t\right)I_2^{\left(2\right)}\left(t\right)-I_2\left(t\right)I_1^{\left(2\right)}\left(t\right)\right).\hfill \end{array}$$

(19)

This result is correct up to a second order in $\frac{1}{c}$. Assuming that the total momentum of the system and the current derivatives are null at $t=0$, we obtain a mechanical linear momentum as follows:

$$\begin{array}{ccc}& & {\overrightarrow{P}}_{mech}={\int }_0^t{\overrightarrow{F}}_T\left(t^{\prime }\right)d{t}^{\prime }\cong \hfill \\ & & -\frac{{\mu }_0}{8\pi }{\left(\frac{h}{c}\right)}^2{\overrightarrow{K}}_{122}\left(I_1\left(t\right)I_2^{\left(1\right)}\left(t\right)-I_2\left(t\right)I_1^{\left(1\right)}\left(t\right)\right).\hfill \end{array}$$

(20)

For simplicity, we will from here on assume a direct current in loop 2 hence:

$${\overrightarrow{P}}_{mech}\cong \frac{{\mu }_0}{8\pi }{I}_1^{\left(1\right)}\left(t\right)I_2{\left(\frac{h}{c}\right)}^2{\overrightarrow{K}}_{122}\propto \frac{1}{c^2}.$$

(21)

For a calculation of ${\overrightarrow{K}}_{122}$ in particular geometries see [14,15,16]. We stress that the mechanical momentum given is of order $\frac{1}{c^2}$ and higher order terms are neglected. The kinetic mechanical energy associated with this momentum is:

$$E_{mech}=\frac{{\overrightarrow{P}}_{mech}^2}{2M}=\frac{1}{2}{\overrightarrow{P}}_{mech}·\overrightarrow{v}s.\propto \frac{1}{c^4}.$$

(22)

where M is the mass of the relativistic engine and:

$$\overrightarrow{v}s.=\frac{{\overrightarrow{P}}_{mech}}{M}\propto \frac{1}{c^2}$$

(23)

is the engine velocity. This indicates that unlike the conservation of momentum [8] which was independent of the mass and therefore of the velocity attained by the system, in the calculations of both the mechanical and electromagnetic energies the systems velocity and mass are of paramount importance. We also note that the expression for mechanical energy is of order $\frac{1}{c^4}$, lower order corrections do not exist and higher order corrections are neglected.

We define the relativistic engine effect as a motion of the center of mass of a system due to the interaction of its two subsystems, if there is no motion of the center of mass than the effect is null. It may be that the center of mass is moving due to the motion of each subsystem separately but then, according to the above definition this is not a relativistic engine effect but an effect caused by an external force.

3.1. Proposed Experiment

In the following, we calculate the factor ${\overrightarrow{K}}_{12n}$ for a geometry of two loops of unit radius and a unit distance placed one atop the other (see Figure 3).

According to Equation (14), we obtain by taking $h=1$ (in arbitrary units since ${\overrightarrow{K}}_{12n}$ is dimensionless).

$${\overrightarrow{K}}_{12n}=-\oint \oint R^{n-3}\overrightarrow{R}(d{\overrightarrow{l}}_2·d{\overrightarrow{l}}_1).$$

(24)

Every point on loop 1 and 2 has the coordinates ${\overrightarrow{x}}_1=$$(cos{\theta }_1,sin{\theta }_1,1)$ and ${\overrightarrow{x}}_2=(cos{\theta }_2,sin{\theta }_2,0)$, hence:

$$\overrightarrow{R}={\overrightarrow{x}}_{12}={\overrightarrow{x}}_1-{\overrightarrow{x}}_2=(cos{\theta }_1-cos{\theta }_2,sin{\theta }_1-sin{\theta }_2,1)$$

(25)

Furthermore, thus:

$$R^2=3-2cos({\theta }_1-{\theta }_2)$$

(26)

The line elements can be calculated as follows:

$$\begin{array}{ccc}\hfill d{\overrightarrow{l}}_1& =& d{\theta }_1{\widehat{\theta }}_1=d{\theta }_1(-sin{\theta }_1,cos{\theta }_1,0)\hfill \\ \hfill d{\overrightarrow{l}}_2& =& d{\theta }_2{\widehat{\theta }}_2=d{\theta }_2(-sin{\theta }_2,cos{\theta }_2,0)\hfill \end{array}$$

(27)

Hence:

$$d{\overrightarrow{l}}_1·d{\overrightarrow{l}}_2=d{\theta }_{1}d{\theta }_{2}cos({\theta }_1-{\theta }_2)$$

(28)

Inserting Equations (28), (26) and (25) into Equation (14) we arrive at:

$${\overrightarrow{K}}_{12n}=-{\int }_0^{2\pi }d{\theta }_2{\int }_0^{2\pi }d{\theta }_{1}cos({\theta }_1-{\theta }_2){\left(\sqrt{3-2cos({\theta }_1-{\theta }_2)}\right)}^{n-3}\overrightarrow{R}({\theta }_1,{\theta }_2).$$

(29)

We now make a change of variable in the above integral ${\theta }^{\prime }={\theta }_1-{\theta }_2$, in terms of the new variable:

$${\overrightarrow{K}}_{12n}=-{\int }_0^{2\pi }d{\theta }_2{\int }_{-{\theta }_2}^{2\pi -{\theta }_2}d{\theta }^{\prime }cos\left({\theta }^{\prime }\right){\left(\sqrt{3-2cos\left({\theta }^{\prime }\right)}\right)}^{n-3}\overrightarrow{R}({\theta }^{\prime }+{\theta }_2,{\theta }_2).$$

(30)

In which:

$$\begin{array}{ccc}\hfill \overrightarrow{R}({\theta }^{\prime }+{\theta }_2,{\theta }_2)& =& ((cos{\theta }^{\prime }-1)cos{\theta }_2-sin{\theta }^{\prime }sin{\theta }_2,\hfill \\ & & (cos{\theta }^{\prime }-1)sin{\theta }_2+sin{\theta }^{\prime }cos{\theta }_2,1)\hfill \end{array}$$

(31)

The integrand is periodic with respect to ${\theta }^{\prime }$ with a period of $2\pi$, we thus write Equation (30) as:

$${\overrightarrow{K}}_{12n}=-{\int }_0^{2\pi }d{\theta }_2{\int }_0^{2\pi }d{\theta }^{\prime }cos\left({\theta }^{\prime }\right){\left(\sqrt{3-2cos\left({\theta }^{\prime }\right)}\right)}^{n-3}\overrightarrow{R}({\theta }^{\prime }+{\theta }_2,{\theta }_2).$$

(32)

Noticing that:

$${\int }_0^{2\pi }cos{\theta }_{2}d{\theta }_2={\int }_0^{2\pi }sin{\theta }_{2}d{\theta }_2=0$$

(33)

We arrive at the equation:

$${\overrightarrow{K}}_{12n}=\left(0,0,-2\pi {\int }_0^{2\pi }d{\theta }^{\prime }cos\left({\theta }^{\prime }\right){\left(\sqrt{3-2cos\left({\theta }^{\prime }\right)}\right)}^{n-3}\right).$$

(34)

Vanishing of the x and y components of the force is expected due to the symmetry. The z components can be calculated analytically, for even values of n one needs elliptic functions, for example:

$$\begin{array}{ccc}\hfill K_{z122}& =& 4\pi \left(Ee\right(-4)-3Ke(-4\left)\right)\simeq -4.94,\hfill \\ \hfill Ee\left(m\right)& \equiv & {\int }_0^{\frac{\pi }{2}}d\theta \sqrt{1-m{sin}^2\theta }\hfill \\ \hfill Ke\left(m\right)& \equiv & {\int }_0^{\frac{\pi }{2}}d\theta \frac{1}{\sqrt{1-m{sin}^2\theta }}\hfill \end{array}$$

(35)

K values for $n\le 6$ are given in

Table 1. Table of the geometric factor K values.

n	${\mathit{K}}_{\mathit{z}12\mathit{n}}$
0	−7.18
1	−6.74
2	−4.94
3	0
4	11.97
5	39.48
6	101.03

.

Finally notice that for multiple loops we have:

$$K_{z12n}\sim N_1{N}_2.$$

(36)

where $N_1$ and $N_2$ are the number of turns in coil 1 and coil 2, respectively.

The Problem of Current Switching

We will address the problem of achieving constant force which is of interest for locomotive applications. A constant force may be achieved by having a direct current in one loop $I_1\left(t\right)=\overline{I_1}$ and a current of uniform second derivative on the other $I_2\left(t\right)=\frac{1}{2}\overline{I_2}\frac{t^2}{{\tau }_c^2}$. In this case the accelerating force will be according to Equation (19):

$$F_{Tz}\cong -\frac{{\mu }_0}{8\pi }{\left(\frac{h}{c}\right)}^2\frac{1}{{\tau }_c^2}{K}_{z122}\overline{I_1}\overline{I_2}$$

(37)

Assuming the case of circular loops of the previous subsection we obtain from Equation (35):

$$F_{Tz}\cong 4.94\frac{{\mu }_0}{8\pi }{N}_1{N}_2{\left(\frac{h}{c}\right)}^2\frac{1}{{\tau }_c^2}\overline{I_1}\overline{I_2}$$

(38)

For the values given in

Table 2. The choice of parameters for the force calculation.

Parameter	Value
$N_1$	1000
$N_2$	1000
$\overline{I_1}$	100 A
$\overline{I_2}$	100 A
h	0.1 m
$\frac{h}{c}$	0.3 ns
${\tau }_c$	10 ns

$F_{Tz}\cong 2.74$ Newton. The switching time may represent some difficulty which one can overcome with advanced enough switching technology, perhaps using low resistivity superconducting materials. Another possibility is using numerous modular solid-state devices each with fast switching and small current such that an appreciable amount of cumulative forcing will result. In any case increasing the current for ever is not possible, but probably can be maintained sufficiently long to allow experimental verification. Changing periodically the current is of course possible but in this case one is led back to the standard piston going up and down as in an ordinary car motor and this type of motion needs to transfer momentum to the road in order to convert circular to linear motion.

4. Field Energy

Consider two sub-systems denoted system ${}_1$ and system ${}_2$ which are far apart such that their interaction is negligible. In this case, Equation (12) is correct for each sub-system separately, that is:

$$\int d^{3}x{\overrightarrow{J}}_1·{\overrightarrow{E}}_1=-\frac{d{E}_{field1}}{dt}-{\oint }_S{\overrightarrow{S}}_{p1}·\widehat{n}da.$$

(39)

$$\int d^{3}x{\overrightarrow{J}}_2·{\overrightarrow{E}}_2=-\frac{d{E}_{field2}}{dt}-{\oint }_S{\overrightarrow{S}}_{p2}·\widehat{n}da.$$

(40)

Next we will put the two loops closer together such that they may interact but without modifying the charge and the current densities of each of the subsystems. The total fields of the combined system are:

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

(41)

Since both the field energy Equation (4) and Poynting’s vector Equation (5) are quadratic in the fields the following result is obtained:

$$\begin{array}{ccc}\hfill E_{field}& \equiv & \frac{{ϵ}_0}{2}\int \left({\overrightarrow{E}}^2+c^2{\overrightarrow{B}}^2\right)d^{3}x=E_{field1}+E_{field2}+E_{field12}\hfill \\ \hfill E_{field1}& \equiv & E_{Efield1}+E_{Mfield1}\equiv \frac{{ϵ}_0}{2}\int \left({\overrightarrow{E}}_1^2+c^2{\overrightarrow{B}}_1^2\right)d^{3}x\hfill \\ \hfill E_{field2}& \equiv & E_{Efield2}+E_{Mfield2}\equiv \frac{{ϵ}_0}{2}\int \left({\overrightarrow{E}}_2^2+c^2{\overrightarrow{B}}_2^2\right)d^{3}x\hfill \\ \hfill E_{field12}& \equiv & E_{Efield12}+E_{Mfield12}\hfill \\ & \equiv & {ϵ}_0\int \left({\overrightarrow{E}}_1·{\overrightarrow{E}}_2+c^2{\overrightarrow{B}}_1·{\overrightarrow{B}}_2\right)d^{3}x\hfill \end{array}$$

(42)

$$\begin{array}{ccc}\hfill {\overrightarrow{S}}_p& \equiv & \frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}={\overrightarrow{S}}_{p1}+{\overrightarrow{S}}_{p2}+{\overrightarrow{S}}_{p12}\hfill \\ \hfill {\overrightarrow{S}}_{p1}& \equiv & \frac{1}{{\mu }_0}{\overrightarrow{E}}_1\times {\overrightarrow{B}}_1\hfill \\ \hfill {\overrightarrow{S}}_{p2}& \equiv & \frac{1}{{\mu }_0}{\overrightarrow{E}}_2\times {\overrightarrow{B}}_2\hfill \\ \hfill {\overrightarrow{S}}_{p12}& \equiv & \frac{1}{{\mu }_0}\left({\overrightarrow{E}}_1\times {\overrightarrow{B}}_2+{\overrightarrow{E}}_2\times {\overrightarrow{B}}_1\right)\hfill \end{array}$$

(43)

The power invested in the combined system in bilinear in the current flux density and the electric field according to Equation (10). This will lead to the following expression:

$$\begin{array}{ccc}\hfill Power& =& \int d^{3}x\overrightarrow{J}·\overrightarrow{E}=Powe{r}_1+Powe{r}_2+Powe{r}_{12}\hfill \\ \hfill Powe{r}_1& \equiv & \int d^{3}x{\overrightarrow{J}}_1·{\overrightarrow{E}}_1\hfill \\ \hfill Powe{r}_2& \equiv & \int d^{3}x{\overrightarrow{J}}_2·{\overrightarrow{E}}_2\hfill \\ \hfill Powe{r}_{12}& \equiv & \int d^{3}x\left({\overrightarrow{J}}_1·{\overrightarrow{E}}_2+{\overrightarrow{J}}_2·{\overrightarrow{E}}_1\right)\hfill \end{array}$$

(44)

Subtracting from Equation (12) the expressions given in Equations (39) and (40):

$$\begin{array}{c}Power-Powe{r}_1-Powe{r}_2\hfill \\ \equiv \frac{d(E_{field}-E_{field1}-E_{field2})}{dt}-{\oint }_S\left({\overrightarrow{S}}_p-{\overrightarrow{S}}_{p1}-{\overrightarrow{S}}_{p2}\right)·\widehat{n}da.\hfill \end{array}$$

(45)

taking into account Equations (42)–(44) we arrive at:

$$Powe{r}_{12}=-\frac{d{E}_{field12}}{dt}-{\oint }_S{\overrightarrow{S}}_{p12}·\widehat{n}da.$$

(46)

5. Field Energy of a System of Two Current Loops

We now study the case of two loops of arbitrary geometry as described in Section 3. However, now we assume two time dependencies. One due to the intrinsic time dependence of the current in the loop and another is due to its motion as part of the relativistic engine. Thus, the current density is:

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

(47)

in which ${\overrightarrow{J}}^{\prime }(\overrightarrow{x},t)$ is the current density in the moving frame of the relativistic engine and $\frac{d{\overrightarrow{x}}_c\left(t\right)}{dt}=\overrightarrow{v}s.\left(t\right)$ is the velocity of the relativistic engine. The vector potential is [3]:

$$\overrightarrow{A}(\overrightarrow{x},t)=\frac{{\mu }_0}{4\pi }\int d^3{x}^{\prime }\frac{\overrightarrow{J}({\overrightarrow{x}}^{\prime },t_{ret})}{R},\overrightarrow{R}\equiv \overrightarrow{x}-{\overrightarrow{x}}^{\prime },t_{ret}\equiv t-\frac{R}{c}.$$

(48)

Which can be written as a power series in $-\frac{R}{c}$ in the form:

$$\begin{array}{ccc}\hfill \overrightarrow{A}(\overrightarrow{x},t)& =& \frac{{\mu }_0}{4\pi }\sum _{n=0}^{\infty }\frac{1}{n!}\int d^3{x}^{\prime }\frac{1}{R}{(-\frac{R}{c})}^n\frac{d^n}{d{t}^n}\overrightarrow{J}({\overrightarrow{x}}^{\prime },t)\hfill \\ \hfill & =& \frac{{\mu }_0}{4\pi }\sum _{n=0}^{\infty }\frac{1}{n!}\frac{d^n}{d{t}^n}\int d^3{x}^{\prime }\frac{1}{R}{(-\frac{R}{c})}^n\overrightarrow{J}({\overrightarrow{x}}^{\prime },t)\hfill \\ \hfill & =& \frac{{\mu }_0}{4\pi }\sum _{n=0}^{\infty }\frac{1}{n!}\frac{d^n}{d{t}^n}\int d^3{x}^{\prime }\frac{1}{R}{(-\frac{R}{c})}^n{\overrightarrow{J}}^{\prime }({\overrightarrow{x}}^{\prime }-{\overrightarrow{x}}_c\left(t\right),t),\hfill \end{array}$$

(49)

As in the case of Equation (13) this expansion is only valid in the domain $[t-T_{max},t+T_{max}]$ in which $T_{max}$ being the “convergence radius” of the above series. Hence this is only valid for distances satisfying $\frac{R}{c}<T_{max}$ or:

$$R<c{T}_{max}\equiv R_{max}$$

(50)

since c is a big number we will occasionally use Equation (49) for infinite distances while remembering that although $R_{max}$ may be big it is not infinite. Let us introduce a comoving integration variable: $\tilde{\overrightarrow{x}}={\overrightarrow{x}}^{\prime }-{\overrightarrow{x}}_c\left(t\right),{\overrightarrow{x}}^{\prime }=\tilde{\overrightarrow{x}}+{\overrightarrow{x}}_c\left(t\right)$ such that: $R\left(t\right)=|{\overrightarrow{x}}^{\prime }-\overrightarrow{x}|=|\tilde{\overrightarrow{x}}+{\overrightarrow{x}}_c\left(t\right)-\overrightarrow{x}|$ in such a comoving coordinate system we have:

$$\overrightarrow{A}(\overrightarrow{x},t)=\frac{{\mu }_0}{4\pi }\sum _{n=0}^{\infty }\frac{1}{n!}\frac{d^n}{d{t}^n}\int d^3\tilde{x}\frac{1}{R\left(t\right)}{(-\frac{R\left(t\right)}{c})}^n{\overrightarrow{J}}^{\prime }(\tilde{\overrightarrow{x}},t),$$

(51)

For a thin and uniform current loop this can be written as:

$$\overrightarrow{A}(\overrightarrow{x},t)=\frac{{\mu }_0}{4\pi }\sum _{n=0}^{\infty }\frac{1}{n!}\frac{d^n}{d{t}^n}\left[I\left(t\right)\oint d\tilde{\overrightarrow{l}}\frac{1}{R\left(t\right)}{(-\frac{R\left(t\right)}{c})}^n\right].$$

(52)

Let us define:

$$\begin{array}{ccc}\hfill {\overrightarrow{A}}^{\left(n\right)}(\overrightarrow{x},t)& =& \frac{{\mu }_0}{4\pi n!}\frac{d^n}{d{t}^n}\left[I\left(t\right)\oint d\tilde{\overrightarrow{l}}\frac{1}{R\left(t\right)}{(-\frac{R\left(t\right)}{c})}^n\right]\hfill \\ & =& {(-1)}^n\frac{{\mu }_0}{4\pi n!c^n}\frac{d^n}{d{t}^n}\left[I\left(t\right)\oint d\tilde{\overrightarrow{l}}R{\left(t\right)}^{n-1}\right],\hfill \end{array}$$

(53)

in terms of ${\overrightarrow{A}}^{\left(n\right)}$ we may write:

$$\overrightarrow{A}(\overrightarrow{x},t)=\sum _{n=0}^{\infty }{\overrightarrow{A}}^{\left(n\right)}(\overrightarrow{x},t).$$

(54)

We notice ${\overrightarrow{A}}^{\left(n\right)}$ is at least of order $\frac{1}{c^n}$ but may contain higher order terms. The reason for this is the fact that ${\overrightarrow{A}}^{\left(n\right)}$ contains temporal derivatives of $R\left(t\right)$ which satisfy the equation:

$$\frac{dR\left(t\right)}{dt}=-\frac{\overrightarrow{R}\left(t\right)}{R\left(t\right)}·\overrightarrow{v}s.=-\widehat{R}·\overrightarrow{v},\widehat{R}\equiv \frac{\overrightarrow{R}\left(t\right)}{R\left(t\right)}$$

(55)

and $\overrightarrow{v}$ is proportional to $\frac{1}{c^2}$ for a relativistic engine according to Equation (23). Thus, ${\overrightarrow{A}}^{\left(n\right)}$ may contain terms which are up to order of ${\left(\frac{1}{c}\right)}^{3n-2}$. We also notice that the zeroth order terms takes the form:

$${\overrightarrow{A}}^{\left(0\right)}(\overrightarrow{x},t)=\frac{{\mu }_0}{4\pi }I\left(t\right)\oint d\tilde{\overrightarrow{l}}\frac{1}{R\left(t\right)},$$

(56)

which is just the quasi static approximation to $\overrightarrow{A}$. It also easy to see that the first order term:

$${\overrightarrow{A}}^{\left(1\right)}(\overrightarrow{x},t)=0$$

(57)

as $\oint d\tilde{\overrightarrow{l}}=0$. As ${\overrightarrow{A}}^{\left(n\right)}$ contains terms in various orders of $\frac{1}{c}$, and we will be interested in analyzing the energy problem for a definite order of $\frac{1}{c}$, we introduce the square bracket notation $G^{\left[n\right]}$ to denote a quantity of order ${\left(\frac{1}{c}\right)}^n$. Obviously, generally ${\overrightarrow{A}}^{\left[n\right]}\ne {\overrightarrow{A}}^{\left(n\right)}$. The electric field $\overrightarrow{E}$ can be calculated (in the gauge $\mathsf{\Phi }=0$) as [3]:

$$\overrightarrow{E}=-{\partial }_t\overrightarrow{A}.$$

(58)

We shall define for computational convenience:

$${\overrightarrow{E}}^{\left(n\right)}\equiv -{\partial }_t{\overrightarrow{A}}^{\left(n\right)}.$$

(59)

This means that ${\overrightarrow{E}}^{\left[n\right]}\ne {\overrightarrow{E}}^{\left(n\right)}$ will contain terms coming from ${\overrightarrow{A}}^{\left[n\right]}$ but also from ${\overrightarrow{A}}^{[n-2]}$ as the derivative will create terms which have an additional ${\left(\frac{1}{c}\right)}^2$ factor. As for the magnetic field $\overrightarrow{B}$ which can be calculated as [3]:

$$\overrightarrow{B}=\overrightarrow{\nabla }\times \overrightarrow{A}$$

(60)

it is easy to see that ${\overrightarrow{B}}^{\left[n\right]}$ is the same order as ${\overrightarrow{A}}^{\left[n\right]}$ and:

$${\overrightarrow{B}}^{\left[n\right]}=\overrightarrow{\nabla }\times {\overrightarrow{A}}^{\left[n\right]}$$

(61)

In what follows we will use the above expressions to analyzed the energy transformation order by order starting from $n=0$ and up to $n=4$ which according to Equation (22) is the most relevant for analyzing the energy transfer between the field and the mechanical components of a relativistic engine.

5.1. Normalization

To avoid carrying the factor $\frac{{\mu }_0}{4\pi }$ in numerous calculations we shall define:

$${\overrightarrow{A}}^{\prime }\equiv \overrightarrow{A}\frac{4\pi }{{\mu }_0}\Rightarrow \overrightarrow{A}={\overrightarrow{A}}^{\prime }\frac{{\mu }_0}{4\pi }.$$

(62)

Furthermore, hence:

$${\overrightarrow{E}}^{\prime }\equiv \overrightarrow{E}\frac{4\pi }{{\mu }_0}\Rightarrow \overrightarrow{E}={\overrightarrow{E}}^{\prime }\frac{{\mu }_0}{4\pi },{\overrightarrow{B}}^{\prime }\equiv \overrightarrow{B}\frac{4\pi }{{\mu }_0}\Rightarrow \overrightarrow{B}={\overrightarrow{B}}^{\prime }\frac{{\mu }_0}{4\pi }$$

(63)

The field energy (see Equation (42)) can be calculated in terms of the new variables as:

$$E_{field12}=\frac{{\mu }_0}{16{\pi }^2}\int \left(\frac{1}{c^2}{\overrightarrow{E}}_1^{\prime }·{\overrightarrow{E}}_2^{\prime }+{\overrightarrow{B}}_1^{\prime }·{\overrightarrow{B}}_2^{\prime }\right)d^{3}x$$

(64)

where we are reminded that $c^2=\frac{1}{{\mu }_0{ϵ}_0}$, the Pointing vector (Equation (43)) will take the following form in terms of the new variables:

$${\overrightarrow{S}}_{p12}=\frac{{\mu }_0}{16{\pi }^2}\left({\overrightarrow{E}}_1^{\prime }\times {\overrightarrow{B}}_2^{\prime }+{\overrightarrow{E}}_2^{\prime }\times {\overrightarrow{B}}_1^{\prime }\right).$$

(65)

Finally the mechanical work which is represented by Equation (44) will take the form:

$$Powe{r}_{12}=\frac{{\mu }_0}{4\pi }\int d^{3}x\left({\overrightarrow{J}}_1·{\overrightarrow{E}}_2^{\prime }+{\overrightarrow{J}}_2·{\overrightarrow{E}}_1^{\prime }\right).$$

(66)

We introduce the following normalization:

$$\begin{array}{ccc}\hfill {\overrightarrow{S}}_{p12}^{\prime }& \equiv & \frac{16{\pi }^2}{{\mu }_0}{\overrightarrow{S}}_{p12}\Rightarrow {\overrightarrow{S}}_{p12}=\frac{{\mu }_0}{16{\pi }^2}{\overrightarrow{S}}_{p12}^{\prime }\hfill \\ \hfill {\overrightarrow{S}}_{p12}^{\prime }& =& \left({\overrightarrow{E}}_1^{\prime }\times {\overrightarrow{B}}_2^{\prime }+{\overrightarrow{E}}_2^{\prime }\times {\overrightarrow{B}}_1^{\prime }\right).\hfill \end{array}$$

(67)

$$\begin{array}{ccc}\hfill E_{field12}^{\prime }& \equiv & \frac{16{\pi }^2}{{\mu }_0}{E}_{field12}\Rightarrow E_{field12}=\frac{{\mu }_0}{16{\pi }^2}{E}_{field12}^{\prime }\hfill \\ \hfill E_{field12}^{\prime }& =& \int \left(\frac{1}{c^2}{\overrightarrow{E}}_1^{\prime }·{\overrightarrow{E}}_2^{\prime }+{\overrightarrow{B}}_1^{\prime }·{\overrightarrow{B}}_2^{\prime }\right)d^{3}x\hfill \end{array}$$

(68)

$$\begin{array}{ccc}\hfill Powe{r}_{12}^{\prime }& \equiv & \frac{16{\pi }^2}{{\mu }_0}Powe{r}_{12}\Rightarrow Powe{r}_{12}=\frac{{\mu }_0}{16{\pi }^2}Powe{r}_{12}^{\prime }\hfill \\ \hfill Powe{r}_{12}^{\prime }& =& 4\pi \int d^{3}x\left({\overrightarrow{J}}_1·{\overrightarrow{E}}_2^{\prime }+{\overrightarrow{J}}_2·{\overrightarrow{E}}_1^{\prime }\right).\hfill \end{array}$$

(69)

In terms of the normalized quantities we may write the relativistic engine energy transformation Equation (46) as:

$$Powe{r}_{12}^{\prime }=-\frac{d{E}_{field12}^{\prime }}{dt}-{\oint }_S{\overrightarrow{S}}_{p12}^{\prime }·\widehat{n}da.$$

(70)

This can be analyzed order by order in terms of powers of $\frac{1}{c}$ hence:

$$Powe{r}_{12}^{\prime \left[n\right]}=-\frac{d{E}_{field12}^{\prime \left[n\right]}}{dt}-{\oint }_S{\overrightarrow{S}}_{p12}^{\prime \left[n\right]}·\widehat{n}da.$$

(71)

We notice that because of the pre-factor $\frac{1}{c^2}$ of the electric field contribution in Equation (68) this term will not contribute to the field energy for the lowest orders: $n=0,1$. We also notice that since we are considering a case of two loop currents Equation (69) takes the form:

$$Powe{r}_{12}^{\prime }=4\pi \left(I_1\left(t\right)\oint d\widetilde{\overrightarrow{l_1}}·{\overrightarrow{E}}_2^{\prime }\left({\overrightarrow{x}}_1\right)+I_2\left(t\right)\oint d\widetilde{\overrightarrow{l_2}}·{\overrightarrow{E}}_1^{\prime }\left({\overrightarrow{x}}_2\right)\right).$$

(72)

We are now ready to analyzed the field contribution order by order. In what follow we give the main results the reader who is interested in the details is referred to Appendix A.

5.2. $n=0$

For $n=0$ we conclude that the energy equation of order zero is indeed balanced. Mechanical work invested or extracted in the system results in increase or decrease in the field energy accordingly. To be more specific the magnetic field energy is affected by the mechanical work. The power related to the mechanical work is:

$$Powe{r}_{12}^{\left[0\right]}=-{\partial }_t{I}_1\left(t\right)I_2{M}_{12}^{\left[0\right]}$$

(73)

in which:

$$M_{12}^{\left[0\right]}\equiv \frac{{\mu }_0}{4\pi }\oint d{\overrightarrow{l}}_1·\oint d{\overrightarrow{l}}_2\frac{1}{|{\overrightarrow{x}}_1-{\overrightarrow{x}}_2|}$$

(74)

and this is equal to minus the derivative of the field energy:

$$E_{field12}^{\left[0\right]}=I_1\left(t\right)I_2{M}_{12}^{\left[0\right]}.$$

(75)

which is what is expected from Equation (71) for the case that there is no Poynting contribution. We underline that those contributions are not related the relativistic engine effect as none of the terms depends on the engine velocity $\overrightarrow{v}$, and thus the above expression will be valid even if the engine is infinitely massive and no motion occurs.

5.3. $n=1$

For $n=1$ we conclude that the energy equation of order one is indeed balanced in a trivial way. Equation (71) is satisfied in the sense that $0=0$. From now on we will disregard in our calculations any field term of order one, as their contribution to any expression is obviously null.

5.4. $n=2$

We conclude that the energy equation of the second order is indeed balanced. Mechanical work invested or extracted in the system results in increase or decrease in the field energy accordingly. To be more specific the magnetic field energy is affected by the mechanical work. The power related to the mechanical work is:

$$Powe{r}_{12}^{\left[2\right]}=-\frac{{\mu }_0}{8\pi c^2}\frac{d^3{I}_1\left(t\right)}{d{t}^3}{I}_2\oint \oint d\overrightarrow{l_2}·d\overrightarrow{l_2}{R}_{12}$$

(76)

and this is equal to minus the derivative of the field energy:

$$E_{field12}^{\left[2\right]}=\frac{{\mu }_0}{8\pi c^2}\frac{d^2{I}_1\left(t\right)}{d{t}^2}{I}_2\oint \oint (d\overrightarrow{l_1}·d\overrightarrow{l_2})R_{12}.$$

(77)

which is what is expected for the case that there is no Poynting contribution. We underline that those contributions are not related the relativistic engine effect as non of the terms depends on the engine velocity $\overrightarrow{v}$, and thus the above expression will be valid even if the engine is infinitely massive and no motion occurs. We do not expect any relativistic engine contributions for orders smaller than $\frac{1}{c^4}$. For a phasor current of frequency $\omega$ we obtain a relativistic correction to the classical mutual inductance which is important for large systems with high frequency.

$$M_{12}^{\left[2\right]}\equiv -\frac{{\mu }_0{\omega }^2}{8\pi c^2}\oint d{\overrightarrow{l}}_1·\oint d{\overrightarrow{l}}_2{R}_{12}$$

(78)

5.5. $n=3$

We conclude that the energy of the third order is indeed balanced. Mechanical work invested or extracted in the system results in increase or decrease in the field energy accordingly. To be more specific the magnetic field energy is affected by the mechanical work. The power related to the mechanical work is:

$$Powe{r}_{12}^{\left[3\right]}=\frac{{\mu }_0}{24\pi c^3}\frac{d^4{I}_1\left(t\right)}{d{t}^4}{I}_2\oint \oint d\overrightarrow{l_1}·d\overrightarrow{l_2}{R}_{12}^2$$

(79)

and this is equal to minus the derivative of the volume field energy:

$$E_{fieldV12}^{\left[3\right]}=-\frac{{\mu }_0}{24\pi c^3}\frac{d^3{I}_1\left(t\right)}{d{t}^3}{I}_2\oint \oint (d\overrightarrow{l_1}·d\overrightarrow{l_2})R_{12}^2.$$

(80)

However, for the third order in $\frac{1}{c}$ there is also a surface contribution to the field energy:

$$E_{fieldS12}^{\left[3\right]}=\frac{{\mu }_0}{9\pi c^3}{I}_2\frac{d^3{I}_1\left(t\right)}{d{t}^3}\overrightarrow{A}{r}_1·\overrightarrow{A}{r}_2.$$

(81)

in which:

$$\overrightarrow{A}r\equiv \frac{1}{2}\oint \overrightarrow{x}\times d\overrightarrow{l}$$

(82)

such that the total field energy is:

$$E_{field12}^{\left[3\right]}=E_{fieldV12}^{\left[3\right]}+E_{fieldS12}^{\left[3\right]}$$

(83)

The change in the field energy through the surface terms results in radiation as described by the Poynting flux:

$${\oint }_S{\overrightarrow{S}}_{p12}^{\left[3\right]}·\widehat{n}da=-\frac{{\mu }_0}{9\pi c^3}{I}_2\frac{d^4{I}_1\left(t\right)}{d{t}^4}\overrightarrow{A}{r}_1·\overrightarrow{A}{r}_2.$$

(84)

curiously this flux can be avoided by configuring the loops to be orthogonal to each other [19]. We underline that third order contributions are not related the relativistic engine effect as non of the terms depends on the engine velocity $\overrightarrow{v}$, and thus the above expression will be valid even if the engine is infinitely massive and no motion occurs. We do not expect any relativistic engine contributions for orders smaller than $\frac{1}{c^4}$. For a phasor current of frequency $\omega$ the result indicates a resistive relativistic correction to the classical mutual inductance which is important for large systems with high frequency.

$$M_{12}^{\left[3\right]}\equiv \frac{j{\mu }_0{\omega }^3}{24\pi c^3}\oint d{\overrightarrow{l}}_1·\oint d{\overrightarrow{l}}_2{R}_{12}^2$$

(85)

5.6. $n=4$

We conclude that the energy of the fourth order is indeed balanced. Mechanical work invested or extracted in the system results in increase or decrease in the field energy accordingly. In the fourth order both electric and magnetic field energies are affected by the mechanical work. The power related to the mechanical work is:

$$Powe{r}_{12}^{\left[4\right]}=-\frac{{\mu }_0}{96\pi c^2}\frac{d^5{I}_1\left(t\right)}{d{t}^5}{I}_2\oint \oint (d\overrightarrow{l_1}·d\overrightarrow{l_2})R_{21}^3+6\frac{d{E}_{mech}}{dt}.$$

(86)

it contains both work done by the mutual inductance and on the relativistic engine. This is equal to minus the derivative of the volume field energy:

$$E_{fieldV12}^{\left[4\right]}=-6E_{mech}+\frac{{\mu }_0}{96\pi c^4}\frac{d^4{I}_1\left(t\right)}{d{t}^4}{I}_2\oint \oint (d\overrightarrow{l_1}·d\overrightarrow{l_2})R_{12}^3.$$

(87)

The total field energy is s made of an electric part:

$$E_{Efield12}^{\left[4\right]}=-2E_{mech}.$$

(88)

and a magnetic part:

$$E_{MfieldV12}^{\left[4\right]}=-4E_{mech}+\frac{{\mu }_0}{96\pi c^4}\frac{d^4{I}_1\left(t\right)}{d{t}^4}{I}_2\oint \oint (d\overrightarrow{l_1}·d\overrightarrow{l_2})R_{12}^3.$$

(89)

Moreover, for the fourth order in $\frac{1}{c}$ there is also a surface contribution to the field energy which satisfies:

$$\frac{d{E}_{fieldS12}^{\prime \left[4\right]}}{dt}=-\int da\widehat{n}·({\overrightarrow{E}}_{1mui}^{\prime \left[4\right]}\times {\overrightarrow{B}}_2^{\prime \left[0\right]})=-{\oint }_S{\overrightarrow{S}}_{p12}^{\left[4\right]}·\widehat{n}da$$

(90)

such that the total field energy is:

$$E_{field12}^{\left[4\right]}=E_{fieldV12}^{\left[4\right]}+E_{fieldS12}^{\left[4\right]}$$

(91)

The change in the field energy through the surface terms results in radiation as described by the Poynting flux:

$$\begin{array}{ccc}& & {\oint }_S{\overrightarrow{S}}_{p12}^{\left[4\right]}·\widehat{n}da=\frac{{\mu }_0}{{\left(4\pi \right)}^2}{\oint }_S{\overrightarrow{S}}_{p12}^{\prime \left[4\right]}·\widehat{n}da=-\frac{{\mu }_0}{480\pi c^4}\frac{d^5{I}_1\left(t\right)}{d{t}^5}{I}_2{R}_{max}\hfill \\ & & \oint \oint \left(7(d{\overrightarrow{l}}_1·d{\overrightarrow{l}}_2)({\overrightarrow{x}}_1·{\overrightarrow{x}}_2)+2(d{\overrightarrow{l}}_1·{\overrightarrow{x}}_2)(d{\overrightarrow{l}}_2·{\overrightarrow{x}}_1)\right).\hfill \end{array}$$

(92)

We underline that fourth order contributions are the only one that are related to the relativistic engine effect as some of the terms depends on the engine velocity $\overrightarrow{v}$. If the engine is infinitely massive and no motion occurs, we are left with the mutual inductance correction terms and radiation terms which involve higher order derivatives. For a phasor current of frequency $\omega$ we obtain a relativistic correction to the classical mutual inductance which is important for large systems with high frequency.

$$M_{12}^{\left[4\right]}\equiv \frac{{\mu }_0{\omega }^4}{96\pi c^4}\oint d{\overrightarrow{l}}_1·\oint d{\overrightarrow{l}}_2{R}_{12}^3$$

(93)

6. Conclusions

A relativistic engine is not a “perpetuum mobile” it requires energy to operate. The energy needed for its operation comes at the expanse of the electromagnetic field energy. Moreover, we have shown that the total energy required is six times the mechanical energy obtained by the engine as energy must be invested also in driving the needed current for its operation through the loops (see Equations (A182) and (A195)). Two times comes at the expense of the electric field energy and four times at the expense of the magnetic field energy. Notice that we have not taken resistive losses into account but if the coils are not superconductive this should be taken into account as well. As we collected all the terms up to and including $\frac{1}{c^4}$ we have encountered for most of the time terms that can be thought of as relativistic corrections to the mutual inductance formula and are not connected in any way to the relativistic engine effect. For order $\frac{1}{c^3}$ there are also radiation losses which may be avoided by cleverly constructing the loop coils orthogonal to each other. For order $\frac{1}{c^4}$ the nature of our series expansion prevents us from evaluating the radiation flux at infinity and we must suffice with the radiation flux over a sphere of radius $R_{max}$ which is the distance after which our approximation becomes invalid.

In this work we have only dealt with the energy exchange due to the interaction of two loops but of course even a single loop looses energy due to radiation. Future works will consider other relativistic concepts such as an electric (rather than magnetic) relativistic engine. We will also be interested in studying the relativistic ramifications of small body moving in a large structure generating an “external field”. In such a body we expect additional contribution to the main “classical” force which are due to the relativistic retardation.

Finally we remark that although an energy of $6E_{mech}$ seems excessive and inefficient, it is highly efficient with respect to other types of engines which are purely electromagnetic. For example, to reach a momentum p using a photon engine one needs an energy of $E_p=pc$ while for a relativistic engine an energy of $E_r=3pvs.$ will suffice. The ratio is $\frac{E_p}{E_r}=\frac{c}{3v}$ which is a huge number for non relativistic speeds.