# Newton’s Third Law in the Framework of Special Relativity for Charged Bodies

Shailendra Rajputa & Asher Yahaloma,b
a Ariel University, Kiryat Hamada POB 3, Ariel 40700, Israel
b Princeton University, Princeton, New Jersey 08543, USA
e-mails<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Newton’s third law states that any action is countered by a reaction of equal
magnitude but opposite direction. The total force in a system not affected by
external forces is thus zero. However, according to the principles of
relativity, a signal cannot propagate at speeds exceeding the speed of light.
Hence the action and reaction cannot be generated at the same time due to the
relativity of simultaneity. Thus, the total force cannot be null at a given
time. In a previous paper [1], we have shown that Newton’n third law cannot
strictly hold in a distributed system, where 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 the vacuum. A specific
example of two current loops with time dependent currents demonstrated that
the summing of the total force in the system does not add up to zero. This
analysis led to the suggestion of a relativistic engine [2, 3]. As the system
is affected by a total force for a finite period, the system acquires
mechanical momentum and energy. Now the question then arises how can we
accommodate the law of momentum and energy conservation. The subject of
momentum conversation was discussed in [4], while preliminary results
regarding energy conservation were discussed in [5, 6, 7]. Previous analysis
relied on the fact that the bodies were macroscopically natural, which means
that the number of electrons and ions is equal in every volume element. Here
we relax this assumption and study charged bodies, thus analyzing the
consequences on a possible electric relativistic engine.

PACS: 03.30.+p, 03.50.De
Keywords: Newton’s Third Law, Electromagnetism, Relativity

##  1 Introduction

Special relativity is a theory of the structure of space-time. It was
introduced in Einstein’s famous 1905 paper: ”On the Electrodynamics of Moving
Bodies” [8]. 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 his famous four partial differential
equations [9, 10, 11] which owe their current form to Oliver Heaviside [12].
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 [8, 10, 11] to
formulate his 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, any object, message, signal (even if not
electromagnetic), or field can not 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
simultaneously because of the signal finite propagation speed.

Newton’s laws of motion are three physical laws that, together, laid the
foundation for classical mechanics. These laws describe the relationship
between the body, acting forces, and motion of body in response to those
forces. The three laws of motion were first compiled by Isaac Newton in his
Philosophiae Naturalis Principia Mathematica (Mathematical Principles of
Natural Philosophy), first published in 1687 [13, 14]. In this paper, we are
interested only 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 applications due to the high speed of signal propagation. Thus
the total force cannot be null at a given time.

The locomotive systems of today are based on two material parts; each obtains
momentum that is equal and opposite to the momentum gained by the second part.
A typical example of this type of system is a rocket that sheds exhaust gas to
propel itself. However, the above relativistic considerations suggest 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 [4]. Indeed, Noether’s theorem dictates that any system
that possesses translational symmetry will conserve momentum. 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 [11]. 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.

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 [15]. In a paper by Griffiths & Heald [16] 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 [17] were used to explore the
applicability of Coulomb and Biot-Savart outside the static domain.

In a previous paper, we used Jefimenko’s [17, 10] equation to discuss the
force between two current carrying coils [1]. This was later expanded to
include the interaction between a current carrying loop and a permanent magnet
[2, 3]. Since the system is affected by a total force for a finite period,
hence 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 [4]. In [5, 6, 7] 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 might be radiated from the relativistic engine device if the coils are
not configured properly.

Previous analysis relied on the fact that the bodies were macroscopically
natural, which means that the number of electrons and ions is equal in every
volume element. Here we relax this assumption and study charged bodies, thus
analyse the consequences on a possible electric relativistic engine.

This paper will use Jefimenko s equation to discuss the force between two
charge carrying bodies. Two mathematical treatments will be given; in one we
consider an instantaneous action at a distance, here Newton’s third law hold,
and the total electromagnetic force will be shown to be equal to zero. Then we
consider the dynamic electromagnetic condition where the reaction to an action
cannot occur before having the action-generated information reach the affected
object, thus bringing about a non-zero resultant. We stress that the bodies
themselves are taken to be stationary, it’s only the charges in them that are
changing with time.

##  2 The Electro - Static Condition

Consider two bodies having volume elements $d^{3}x_{1},d^{3}x_{2}$ located at
$\vec{x}_{1},\vec{x}_{2}$ respectively and having static charge densities
$\rho_{1},\rho_{2}$ (see figure 1).

Figure 1: Two charged bodies.

The electro-static force applied by charged body $2$ on charged body $1$ is
given by Jackson [10]:

$\vec{F}_{12}=\frac{1}{4\pi\epsilon_{0}}\int\int\frac{d^{3}x_{1}d^{3}x_{2}\
\rho_{1}(\vec{x}_{1})\rho_{2}(\vec{x}_{2})\
\vec{x}_{12}}{|\vec{x}_{12}|^{3}}.$ (1)

where, $\vec{x}_{12}=\vec{x}_{1}-\vec{x}_{2}$ and
$\epsilon_{0}=8.85~{}10^{-12}$ $\rm Fm^{-1}$ is the vacuum permittivity. From
the above, it is easy to calculate the electro-static force applied by current
loop $1$ on charged body $2$ by changing indices $1\leftrightarrow 2$:

$\vec{F}_{21}=\frac{1}{4\pi\epsilon_{0}}\int\int\frac{d^{3}x_{1}d^{3}x_{2}\
\rho_{1}(\vec{x}_{1})\rho_{2}(\vec{x}_{2})\
\vec{x}_{21}}{|\vec{x}_{12}|^{3}}.$ (2)

However,

$\vec{x}_{21}=\vec{x}_{2}-\vec{x}_{1}=-(\vec{x}_{1}-\vec{x}_{2})=-\vec{x}_{12},\qquad|\vec{x}_{21}|=|\vec{x}_{12}|$
(3)

We obtain:

$\vec{F}_{21}=-\vec{F}_{12}.$ (4)

And the total force on two loop system is null:

$\vec{F}_{T}=\vec{F}_{12}+\vec{F}_{21}=0.$ (5)

The above result is independent of the geometry of the charged bodies and is
in complete agreement with Newton’s third law.

##  3 The Dynamic Electromagnetic Condition

Let us now consider the more general time dependent case. Maxwell’s equations
dictate that in this case one can not have a magnetic field without an
electric field and vice versa. Therefore we will consider both the electric
and magnetic parts of the Lorenz force $\vec{F}_{21}$. Let us suppose that the
electric field $\vec{E}$ and magnetic field $\vec{B}$ are created by charged
body $1$ and acts upon charged body $2$. Since a charged body may contain both
ions and free electrons we will have the Lorentz force:

$\vec{F}_{21}=\int d^{3}x_{2}\rho_{i2}(\vec{E}+\vec{v}_{i2}\times\vec{B})+\int
d^{3}x_{2}\rho_{e2}(\vec{E}+\vec{v}_{e2}\times\vec{B}).$ (6)

In the above, we integrate over the entire volume of charged body $2$.
$\rho_{i2}$ and $\rho_{e2}$ are the ion charge density and electron charge
density respectively, $\vec{v}_{i2}$ and $\vec{v}_{e2}$ are the ion velocity
field and electron velocity field respectively. The total charge density
amounts the sum of the ions charge density and free electrons charge density,
hence:

$\rho_{2}=\rho_{i2}+\rho_{e2}.$ (7)

Thus the electric terms in the above force equation cancel and we are left
with:

$\vec{F}_{21}=\int d^{3}x_{2}\rho_{2}\vec{E}+\int
d^{3}x_{2}\rho_{i2}\vec{v}_{i2}\times\vec{B}+\int
d^{3}x_{2}\rho_{e2}\vec{v}_{e2}\times\vec{B}.$ (8)

In the laboratory frame the ions being at rest we have: $\vec{v}_{i2}=0$.
Thus, we arrive at:

$\vec{F}_{21}=\int d^{3}x_{2}\rho_{2}\vec{E}+\int
d^{3}x_{2}\rho_{e2}\vec{v}_{e2}\times\vec{B}.$ (9)

Introducing the current density: $\vec{J}_{2}=\rho_{e2}\vec{v}_{e2}$, we
obtain:

$\vec{F}_{21}=\int
d^{3}x_{2}\left(\rho_{2}\vec{E}+\vec{J}_{2}\times\vec{B}\right).$ (10)

Now, let us consider the coil that generates the magnetic field. The electric
and magnetic fields can be written as follows in terms of the vector and
scalar potentials [10]:

$\vec{E}=-\partial_{t}\vec{A}-\vec{\nabla}\Phi.$ (11)
$\vec{B}=\vec{\nabla}\times\vec{A}.$ (12)

Here, $\vec{\nabla}$ has the standard definition in vector analysis, $t$ is
time and $\partial_{t}$ is a partial derivative with respect to time. If the
field is generated by a charge density $\rho_{1}$ and current density
$\vec{J}_{1}$ in charged body $1$, we can solve for the scalar and vector
potentials and obtain the result [10]:

$\Phi(\vec{x}_{2})=\frac{1}{4\pi\epsilon_{0}}\int
d^{3}x_{1}\frac{\rho_{1}(\vec{x}_{1},t_{ret})}{R},\qquad\vec{R}\equiv\vec{x}_{12},\qquad
t_{ret}\equiv t-\frac{R}{c}.$ (13)
$\vec{A}(\vec{x}_{2})=\frac{\mu_{0}}{4\pi}\int
d^{3}x_{1}\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}.$ (14)

Here, $c=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$ is the speed of light in
vacuum. The above solutions satisfy the Lorentz gauge conditions:

$\vec{\nabla}\cdot\vec{A}+\frac{1}{c^{2}}~{}\partial_{t}\Phi=0$ (15)

Due to the conservation of charge:

$\vec{\nabla}\cdot\vec{J}+\partial_{t}\rho=0$ (16)

(See appendix A). Combining equation (14) with equation (12), we arrive at the
result:

$\vec{B}(\vec{x}_{2})=\vec{\nabla}_{\vec{x}_{2}}\times\vec{A}(\vec{x}_{2})=\frac{\mu_{0}}{4\pi}\int
d^{3}x_{1}\vec{\nabla}_{\vec{x}_{2}}\times\left(\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right).$
(17)

However, notice that111We use the notation
$\partial_{y}\equiv\frac{\partial}{\partial y}$.:

$\vec{\nabla}_{\vec{x}_{2}}\times\left(\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right)=\vec{\nabla}_{\vec{x}_{2}}R\times\partial_{R}\left(\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right).$
(18)

Since:

$\vec{\nabla}_{\vec{x}_{2}}R=-\frac{\vec{R}}{R}$ (19)

And:

$\partial_{R}\left(\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right)=-\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R^{2}}-\frac{\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})}{Rc}.$
(20)

Hence:

$\vec{\nabla}_{\vec{x}_{2}}\times\left(\frac{\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right)=\frac{\vec{R}}{R^{3}}\times\left(\vec{J}_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})\right).$
(21)

Inserting equation (21) into equation (17), we arrive at Jefimenko’s equations
[10, 17] for the magnetic field:

$\vec{B}(\vec{x}_{2})=\frac{\mu_{0}}{4\pi}\int
d^{3}x_{1}\frac{\vec{R}}{R^{3}}\times\left(\vec{J}_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})\right).$
(22)

For the electric field, we have two contributions according to equation (11),
one from the scalar potential and a second one from the vector potential:

$\vec{E}=\vec{E}_{a}+\vec{E}_{b},\qquad\vec{E}_{a}\equiv-\partial_{t}\vec{A},\qquad\vec{E}_{b}\equiv-\vec{\nabla}\Phi.$
(23)

Hence according to equation (14):

$\vec{E}_{a}(\vec{x}_{2})=-\frac{\mu_{0}}{4\pi}\int
d^{3}x_{1}\frac{\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}.$ (24)

And according to equation (13):

$\vec{E}_{b}(\vec{x}_{2})=-k\int
d^{3}x_{1}\vec{\nabla}_{\vec{x}_{2}}\left(\frac{\rho_{1}(\vec{x}_{1},t_{ret})}{R}\right),\qquad
k\equiv\frac{1}{4\pi\epsilon_{0}}\simeq 9\cdot 10^{9}.$ (25)

Above equation can also be written as:

$\vec{E}_{b}(\vec{x}_{2})=-k\int
d^{3}x_{1}\left[\frac{1}{R}\vec{\nabla}_{\vec{x}_{2}}\rho_{1}(\vec{x}_{1},t_{ret})+\rho_{1}(\vec{x}_{1},t_{ret})\vec{\nabla}_{\vec{x}_{2}}\frac{1}{R}\right],$
(26)

however:

$\vec{\nabla}_{\vec{x}_{2}}\frac{1}{R}=\frac{\hat{R}}{R^{2}},\qquad\hat{R}\equiv\frac{\vec{R}}{R}$
(27)

and also:

$\vec{\nabla}_{\vec{x}_{2}}\rho_{1}(\vec{x}_{1},t_{ret})=\vec{\nabla}_{\vec{x}_{2}}R~{}\left.\partial_{R}\rho_{1}(\vec{x}_{1},t_{ret})\right|_{\vec{x}_{1}}=\frac{\hat{R}}{c}\partial_{t}\rho_{1}(\vec{x}_{1},t_{ret}).$
(28)

It thus follows that:

$\vec{E}_{b}(\vec{x}_{2})=-k\int
d^{3}x_{1}\frac{\hat{R}}{R^{2}}\left[\rho_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\rho_{1}(\vec{x}_{1},t_{ret})\right].$
(29)

Adding $\vec{E}_{b}$ and $\vec{E}_{a}$ and taking into account that:

$\frac{\mu_{0}}{4\pi k}=\mu_{0}\epsilon_{0}=\frac{1}{c^{2}}$ (30)

We arrive at the Jefimenko’s expression [10, 17] for the electric field:

$\displaystyle\vec{E}(\vec{x}_{2})$ $\displaystyle=$ $\displaystyle-k\int
d^{3}x_{1}\frac{1}{R^{2}}\left[\left(\rho_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\rho_{1}(\vec{x}_{1},t_{ret})\right)\hat{R}\right.$
(31) $\displaystyle+$
$\displaystyle\left.\left(\frac{R}{c}\right)^{2}\frac{\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right].$

Inserting equation (22) and equation (31) into equation (10),we arrive at a
somewhat lengthy but straightforward expression:

$\displaystyle\vec{F}_{21}$ $\displaystyle=$ $\displaystyle-k\int\int
d^{3}x_{1}d^{3}x_{2}\frac{1}{R^{2}}$ (32)
$\displaystyle\left\\{\rho_{2}(\vec{x}_{2},t)\left[\left(\rho_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\rho_{1}(\vec{x}_{1},t_{ret})\right)\hat{R}+\left(\frac{R}{c}\right)^{2}\frac{\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})}{R}\right]\right.$
$\displaystyle\left.+\left(\frac{R}{c}\right)^{2}\left[\frac{\hat{R}}{R^{2}}\times\left(\vec{J}_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})\right)\right]\times\vec{J}_{2}(\vec{x}_{2},t)\right\\}$

###  3.1 The Quasi-Static Approximation

In the quasi-static approximation, it is assumed that $\tau=\frac{R}{c}$ is
small and can be neglected. Under the same approximation $t_{ret}\simeq t$.
Neglecting all terms of order $\tau$, equation (32) takes the Coulomb form:

$\vec{F}_{21}=-k\int\int d^{3}x_{1}d^{3}x_{2}\
\rho_{1}(\vec{x}_{1},t)\rho_{2}(\vec{x}_{2},t)\frac{\hat{R}}{R^{2}}.$ (33)

Above equation is similar to the equation (2) with slightly different
notation, but now the charge densities are time dependent and therefore not
static. Newton third law follows trivially using the same arguments as in
section 2. We notice that for vanishing charge densities equation (32) takes
the form:

$\displaystyle\vec{F}_{21}$ $\displaystyle=$ $\displaystyle-k\int\int
d^{3}x_{1}d^{3}x_{2}\frac{1}{R^{2}}$ (34)
$\displaystyle\left\\{\left(\frac{R}{c}\right)^{2}\left[\frac{\hat{R}}{R^{2}}\times\left(\vec{J}_{1}(\vec{x}_{1},t_{ret})+\left(\frac{R}{c}\right)\partial_{t}\vec{J}_{1}(\vec{x}_{1},t_{ret})\right)\right]\times\vec{J}_{2}(\vec{x}_{2},t)\right\\}.$

In this case the leading term is $O(\tau^{2})$, neglecting higher order terms
in $\tau$ this can be written as:

$\vec{F}_{21}=\frac{\mu_{0}}{4\pi}\int\int
d^{3}x_{1}d^{3}x_{2}\left\\{\vec{J}_{2}(\vec{x}_{2},t)\times\left[\frac{\hat{R}}{R^{2}}\times\vec{J}_{1}(\vec{x}_{1},t)\right]\right\\}.$
(35)

Or also as:

$\displaystyle\vec{F}_{21}$ $\displaystyle=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int\int
d^{3}x_{1}d^{3}x_{2}\left\\{\left(\vec{J}_{2}(\vec{x}_{2},t)\cdot\vec{J}_{1}(\vec{x}_{1},t)\right)\frac{\hat{R}}{R^{2}}\right.$
(36) $\displaystyle-$
$\displaystyle\left.\left(\vec{J}_{2}(\vec{x}_{2},t)\cdot\frac{\hat{R}}{R^{2}}\right)\vec{J}_{1}(\vec{x}_{1},t)\right\\}.$

Notice that according to equation (27):

$\vec{J}_{2}(\vec{x}_{2},t)\cdot\frac{\hat{R}}{R^{2}}=\vec{J}_{2}(\vec{x}_{2},t)\cdot\vec{\nabla}_{\vec{x}_{2}}\frac{1}{R}=\vec{\nabla}_{\vec{x}_{2}}\cdot\left(\frac{\vec{J}_{2}(\vec{x}_{2},t)}{R}\right)-\frac{1}{R}\vec{\nabla}_{\vec{x}_{2}}\cdot\vec{J}_{2}(\vec{x}_{2},t).$
(37)

However, since we assume that the charge density is null it follows from
equation (16) that $\vec{\nabla}_{\vec{x}_{2}}\cdot\vec{J}_{2}=0$ and:

$\vec{J}_{2}(\vec{x}_{2},t)\cdot\frac{\hat{R}}{R^{2}}=\vec{\nabla}_{\vec{x}_{2}}\cdot\left(\frac{\vec{J}_{2}(\vec{x}_{2},t)}{R}\right).$
(38)

Takin the volume integral of the left hand side and using Gauss theorem, we
obtain:

$\int d^{3}x_{2}~{}\vec{J}_{2}(\vec{x}_{2},t)\cdot\frac{\hat{R}}{R^{2}}=\oint
d\vec{S}_{2}\cdot\left(\frac{\vec{J}_{2}(\vec{x}_{2},t)}{R}\right),$ (39)

where $d\vec{S}_{2}$ and the integral is taken over the entire surface
encapsulating the volume. If the surface is take far from the system in which
the current density is non-vanishing it follows that the surface integral is
null and thus:

$\int d^{3}x_{2}~{}\vec{J}_{2}(\vec{x}_{2},t)\cdot\frac{\hat{R}}{R^{2}}=0.$
(40)

Thus equation (36) will take the form:

$\vec{F}_{21}=\frac{\mu_{0}}{4\pi}\int\int
d^{3}x_{1}d^{3}x_{2}\left(\vec{J}_{2}(\vec{x}_{2},t)\cdot\vec{J}_{1}(\vec{x}_{1},t)\right)\frac{\hat{R}}{R^{2}}.$
(41)

From which Newton’s third law follows trivially. We notice that this form is a
generalization of a similar result presented in [1] for the case of currents
that are restricted to flow on thin current loops. In the quasi-static
approximation, Newton’s third law holds regardless of the geometry of the
system involved, and the total force is null.

This should come as no surprise to the reader since, in the quasi-static
approximation, we totally neglect the time it takes a signal to propagate from
system $1$ to system $2$, assuming that this time is zero. Different results
are obtained when $\tau$ is not neglected, as we will see in the next
subsection.

###  3.2 The Case of a Finite $\tau$

Consider the charge density
$\rho(\vec{x}^{\prime},t_{ret})=\rho(\vec{x}^{\prime},t-\frac{R}{c})$, if
$\frac{R}{c}$ is small but not zero one can write a Taylor series expansion
around $t$ in the form:

$\rho(\vec{x}^{\prime},t_{ret})=\rho(\vec{x}^{\prime},t-\frac{R}{c})=\sum_{n=0}^{\infty}\frac{\partial^{n}_{t}\rho(\vec{x}^{\prime},t)}{n!}(-\frac{R}{c})^{n}.$
(42)

In the above $\partial^{n}_{t}\rho(\vec{x}^{\prime},t)$ is the partial
temporal derivative of order $n$ of $\rho(\vec{x}^{\prime},t)$. The same
expansion for the current density leads to the expression:

$\vec{J}(\vec{x}^{\prime},t_{ret})=\vec{J}(\vec{x}^{\prime},t-\frac{R}{c})=\sum_{n=0}^{\infty}\frac{\partial^{n}_{t}\vec{J}(\vec{x}^{\prime},t)}{n!}(-\frac{R}{c})^{n}.$
(43)

In the above $\partial^{n}_{t}\vec{J}(\vec{x}^{\prime},t)$ is the partial
temporal derivative of order $n$ of $\vec{J}(\vec{x}^{\prime},t)$. The above
expansions are valid only for a certain environment of $t$ on the time axis
which depends on the functions involved, this environment shall be defined
using the convergence radius $T_{max}$ which may be different for each
function. That is equation (42) and equation (43) is valid only in the domain
$[t-T_{max},t+T_{max}]$. As we expand in the delay time $\frac{R}{c}$ it
follows that the expansion is valid only for a limited range:

$R<R_{max}\equiv c~{}T_{max}.$ (44)

This means that basically we are dealing with a near field approximation,
however, as $c$ is a large number $R_{max}$ would be quite large for most
systems. Now inserting equation (42) into equation (13) we obtain:

$\displaystyle\Phi_{1}(\vec{x}_{2})$ $\displaystyle=$ $\displaystyle k\int
d^{3}x_{1}\frac{\rho_{1}(t_{ret})}{R}=k\sum_{n=0}^{\infty}\frac{1}{n!}\int
d^{3}x_{1}\partial^{n}_{t}\rho_{1}(\vec{x}^{\prime},t)\frac{1}{R}(-\frac{R}{c})^{n}$
(45) $\displaystyle=$ $\displaystyle
k\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n}_{t}\rho_{1}(\vec{x}_{1},t)R^{n-1}$

Similarly inserting equation (43) into equation (14) we obtain:

$\displaystyle\vec{A}_{1}(\vec{x}_{2})$ $\displaystyle=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int
d^{3}x_{1}\frac{\vec{J}_{1}(t_{ret})}{R}=\frac{\mu_{0}}{4\pi}\sum_{n=0}^{\infty}\frac{1}{n!}\int
d^{3}x_{1}\partial^{n}_{t}\vec{J}_{1}(\vec{x}^{\prime},t)\frac{1}{R}(-\frac{R}{c})^{n}$
(46) $\displaystyle=$
$\displaystyle\frac{\mu_{0}}{4\pi}\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-1}$

As
$\frac{\mu_{0}}{4\pi}=\frac{\mu_{0}\epsilon_{0}}{4\pi\epsilon_{0}}=\frac{k}{c^{2}}$
it follows that:

$\displaystyle\vec{A}_{1}(\vec{x}_{2})$ $\displaystyle=$ $\displaystyle
k\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{1}{c}\right)^{n+2}\int
d^{3}x_{1}\partial^{n}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-1}$ (47)
$\displaystyle=$ $\displaystyle
k\sum_{n=2}^{\infty}\frac{1}{(n-2)!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n-2}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-3}$

We are now at a position in which we can calculated the electric and magnetic
fields from the expansions given in equation (45) and equation (47). However,
before we proceed we introduce the following notation. Let $G^{[n]}$ be the
contribution of order $\frac{1}{c^{n}}$ to the quantity $G$, thus:

$G=\sum_{n=0}^{\infty}G^{[n]}$ (48)

Hence:

$\Phi_{1}^{[n]}(\vec{x}_{2})=\frac{k}{n!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n}_{t}\rho_{1}(\vec{x}_{1},t)R^{n-1}$ (49)
$\displaystyle\vec{A}_{1}^{[n]}(\vec{x}_{2})=\frac{k}{(n-2)!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n-2}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-3}\quad{\rm
for}~{}n>=2$ $\displaystyle\vec{A}_{1}^{[0]}=\vec{A}_{1}^{[1]}=0.$ (50)

It now follows from equation (23) that:

$\displaystyle\vec{E}_{1a}^{[n]}=-\partial_{t}\vec{A}_{1}^{[n]}=$
$\displaystyle-\frac{k}{(n-2)!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n-1}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-3}\quad{\rm
for}~{}n>=2$ $\displaystyle\vec{E}_{1a}^{[0]}=\vec{E}_{1a}^{[1]}=0.$ (51)

and

$\vec{E}_{1b}^{[n]}=-\vec{\nabla}_{\vec{x}_{2}}\Phi^{[n]}=-\frac{k}{n!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n}_{t}\rho_{1}(\vec{x}_{1},t)\vec{\nabla}_{\vec{x}_{2}}R^{n-1}.$
(52)

However, since:

$\vec{\nabla}_{\vec{x}_{2}}R^{n-1}=\vec{\nabla}_{\vec{x}_{2}}R~{}\partial_{R}R^{n-1}=-(n-1)R^{n-2}\hat{R}.$
(53)

it follows that:

$\vec{E}_{1b}^{[n]}=\frac{k(n-1)}{n!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\partial^{n}_{t}\rho_{1}(\vec{x}_{1},t)R^{n-2}\hat{R}.$ (54)

Zeroth order contribution comes only from the potential part of the electric
field and is the Coulomb contribution:

$\vec{E}_{1}^{[0]}=\vec{E}_{1b}^{[0]}=-k\int
d^{3}x_{1}\frac{\rho_{1}(\vec{x}_{1},t)}{R^{2}}\hat{R}$ (55)

We also deduce that $\vec{E}_{1b}^{[1]}=0$, hence:

$\vec{E}_{1}^{[1]}=\vec{E}_{1a}^{[1]}+\vec{E}_{1b}^{[1]}=0$ (56)

thus there is no first order correction to the electric field in a charged
system, first order corrections are also absent in an uncharged system [1].
The first term containing contributions from both the scalar and vector
potentials to the electric field is the second order term:

$\vec{E}_{1}^{[2]}=\vec{E}_{1a}^{[2]}+\vec{E}_{1b}^{[2]}=k\left(\frac{1}{c}\right)^{2}\int
d^{3}x_{1}\left[\frac{1}{2}\partial^{2}_{t}\rho_{1}(\vec{x}_{1},t)\hat{R}-\partial_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{-1}\right]$
(57)

As $\frac{1}{c}$ is quite small, it will suffice to consider contributions
till the second order. We now calculate the magnetic field using equation (17)
and equation (50) and obtain:

$\displaystyle\vec{B}_{1}^{[n]}(\vec{x}_{2})=\vec{\nabla}_{\vec{x}_{2}}\times\vec{A}_{1}^{[n]}(\vec{x}_{2})=$
$\displaystyle\frac{k}{(n-2)!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}\vec{\nabla}_{\vec{x}_{2}}\times\left[\partial^{n-2}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-3}\right]\quad{\rm
for}~{}n>=2$ $\displaystyle\vec{B}_{1}^{[0]}=\vec{B}_{1}^{[1]}=0.$ (58)

However:

$\vec{\nabla}_{\vec{x}_{2}}\times\left[\partial^{n-2}_{t}\vec{J}_{1}(\vec{x}_{1},t)R^{n-3}\right]=\vec{\nabla}_{\vec{x}_{2}}R^{n-3}\times\partial^{n-2}_{t}\vec{J}_{1}(\vec{x}_{1},t)$
(59)

and also:

$\vec{\nabla}_{\vec{x}_{2}}R^{n-3}=\vec{\nabla}_{\vec{x}_{2}}R~{}\partial_{R}R^{n-3}=(3-n)~{}R^{n-4}~{}\hat{R}$
(60)

Hence we may write:

$\displaystyle\vec{B}_{1}^{[n]}(\vec{x}_{2})=$ $\displaystyle
k\frac{(3-n)}{(n-2)!}\left(-\frac{1}{c}\right)^{n}\int
d^{3}x_{1}R^{n-4}~{}\hat{R}\times\partial^{n-2}_{t}\vec{J}_{1}(\vec{x}_{1},t)\quad{\rm
for}~{}n>=2.$ (61)

in particular:

$\vec{B}_{1}^{[2]}(\vec{x}_{2})=k\left(\frac{1}{c}\right)^{2}\int
d^{3}x_{1}R^{-2}~{}\hat{R}\times\vec{J}_{1}(\vec{x}_{1},t).$ (62)

Using the expressions of the electric and magnetic fields, we may now
calculate the force $n^{th}$ order contribution through equation (10):

$\vec{F}_{21}^{[n]}=\int
d^{3}x_{2}\left(\rho_{2}(\vec{x}_{2})\vec{E}_{1}^{[n]}(\vec{x}_{2})+\vec{J}_{2}(\vec{x}_{2})\times\vec{B}_{1}^{[n]}(\vec{x}_{2})\right).$
(63)

It follows that for the zeroth order we obtain:

$\vec{F}_{21}^{[0]}=\int
d^{3}x_{2}\rho_{2}(\vec{x}_{2})\vec{E}_{1}^{[0]}(\vec{x}_{2}).$ (64)

Inserting equation (55), we obtain Coulomb’s force:

$\vec{F}_{21}^{[0]}=-k\int\int
d^{3}x_{1}d^{3}x_{2}\rho_{1}(\vec{x}_{1})\rho_{2}(\vec{x}_{2})\frac{1}{R^{2}}\hat{R}=-\vec{F}_{12}^{[0]}.$
(65)

This type of force which is just the quasi static force satisfies Newton’s
third law (see section 3.1), hence the total force on the system is null:

$\vec{F}_{T}^{[0]}=\vec{F}_{21}^{[0]}+\vec{F}_{12}^{[0]}=0.$ (66)

The first order force in $\frac{1}{c}$ is null since the first order electric
and magnetic fields are null, thus:

$\vec{F}_{T}^{[1]}=\vec{F}_{21}^{[1]}=\vec{F}_{12}^{[1]}=0.$ (67)

We shall now proceed with the calculation of the second order force term, this
will suffice as $\frac{1}{c}$ is a rather small number. To do this we first
divide the force given in equation (63) into electric and magnetic terms:

$\displaystyle\vec{F}_{21}^{[2]}$ $\displaystyle=$
$\displaystyle\vec{F}_{21e}^{[2]}+\vec{F}_{21m}^{[2]}$
$\displaystyle\vec{F}_{21e}^{[2]}$ $\displaystyle\equiv$ $\displaystyle\int
d^{3}x_{2}~{}\rho_{2}(\vec{x}_{2})\vec{E}_{1}^{[2]}(\vec{x}_{2})$
$\displaystyle\vec{F}_{21m}^{[2]}$ $\displaystyle\equiv$ $\displaystyle\int
d^{3}x_{2}~{}\vec{J}_{2}(\vec{x}_{2})\times\vec{B}_{1}^{[2]}(\vec{x}_{2})$
(68)

Inserting equation (57) into equation (68), we readily obtain the electric
force:

$\vec{F}_{21e}^{[2]}=\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}~{}\left[\frac{1}{2}\rho_{2}\partial^{2}_{t}\rho_{1}\hat{R}-\rho_{2}\partial_{t}\vec{J}_{1}R^{-1}\right]$
(69)

The magnetic force can be obtained by inserting equation (62) into equation
(68):

$\displaystyle\vec{F}_{21m}^{[2]}$ $\displaystyle=$
$\displaystyle\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}~{}\vec{J}_{2}\times~{}\left(R^{-2}\hat{R}\times\vec{J}_{1}\right)$
(70) $\displaystyle=$ $\displaystyle\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}\left[\hat{R}\frac{\vec{J}_{1}\cdot\vec{J}_{2}}{R^{2}}-\vec{J}_{1}R^{-2}(\hat{R}\cdot\vec{J}_{2})\right]$

However,

$R^{-2}\hat{R}=\vec{\nabla}_{\vec{x}_{2}}R^{-1}$ (71)

it follows that:

$\vec{F}_{21m}^{[2]}=\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}\left[\hat{R}\frac{\vec{J}_{1}\cdot\vec{J}_{2}}{R^{2}}-\vec{J}_{1}\left((\vec{\nabla}_{\vec{x}_{2}}R^{-1})\cdot\vec{J}_{2}\right)\right]$
(72)

Let us look at the integral:

$\int d^{3}x_{2}(\vec{\nabla}_{\vec{x}_{2}}R^{-1})\cdot\vec{J}_{2}=\int
d^{3}x_{2}\left[\vec{\nabla}_{\vec{x}_{2}}\cdot\left(\frac{\vec{J}_{2}}{R}\right)-\frac{1}{R}\vec{\nabla}_{\vec{x}_{2}}\cdot\vec{J}_{2}\right]$
(73)

Using Gauss theorem and the continuity equation (16), we arrive at the
following expression:

$\int d^{3}x_{2}(\vec{\nabla}_{\vec{x}_{2}}R^{-1})\cdot\vec{J}_{2}=\oint
d\vec{S}_{2}\cdot\left(\frac{\vec{J}_{2}}{R}\right)+\int
d^{3}x_{2}\frac{1}{R}\partial_{t}\rho_{2}.$ (74)

The surface integral is performed over a surface encapsulating the volume of
the volume integral, if the volume integral is performed over all space the
surface is at infinity. Provided there are no currents at infinity:

$\int d^{3}x_{2}(\vec{\nabla}_{\vec{x}_{2}}R^{-1})\cdot\vec{J}_{2}=\int
d^{3}x_{2}\frac{1}{R}\partial_{t}\rho_{2}.$ (75)

Inserting the result given in equation (75) into equation (72), we arrive at
the following expression for the second order magnetic force:

$\vec{F}_{21m}^{[2]}=\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}\left[\hat{R}\frac{\vec{J}_{1}\cdot\vec{J}_{2}}{R^{2}}-\frac{\vec{J}_{1}}{R}\partial_{t}\rho_{2}\right].$
(76)

We remark that for the magnetic field, the second order is the lowest order
for the force as the zeroth and first order are null. Moreover, we observe
that the force is a sum of two parts, the first one satisfies Newton’s third
law (see section 3.1), and the second does not. We can now calculate the total
electromagnetic force by adding equation (69) with equation (76):

$\displaystyle\vec{F}_{21}^{[2]}$ $\displaystyle=$
$\displaystyle\vec{F}_{21e}^{[2]}+\vec{F}_{21m}^{[2]}$ (77)
$\displaystyle=\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}~{}\left[\frac{1}{2}\rho_{2}\partial^{2}_{t}\rho_{1}\hat{x}_{12}-\partial_{t}(\rho_{2}\vec{J}_{1})R^{-1}+\hat{x}_{12}\frac{\vec{J}_{1}\cdot\vec{J}_{2}}{R^{2}}\right]$

We now use the notation $\hat{R}=\hat{x}_{12}$ for clarity. From the above
expressions it is easy to calculate $F_{12}^{[2]}$ by exchanging the indices
$1$ and $2$:

$\displaystyle\vec{F}_{12}^{[2]}$ $\displaystyle=$ (78)
$\displaystyle\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}~{}\left[\frac{1}{2}\rho_{1}\partial^{2}_{t}\rho_{2}\hat{x}_{21}-\partial_{t}(\rho_{1}\vec{J}_{2})R^{-1}+\hat{x}_{21}\frac{\vec{J}_{1}\cdot\vec{J}_{2}}{R^{2}}\right]$

Combining equation (77) and equation (78) and taking into account that
$\hat{x}_{12}=-\hat{x}_{21}$, it follows that:

$\displaystyle\vec{F}_{T}^{[2]}$ $\displaystyle=$
$\displaystyle\vec{F}_{12}^{[2]}+\vec{F}_{21}^{[2]}=$ (79)
$\displaystyle\left(\frac{k}{c^{2}}\right)\int\int
d^{3}x_{1}d^{3}x_{2}~{}\left[\frac{1}{2}\left(\rho_{2}\partial^{2}_{t}\rho_{1}-\rho_{1}\partial^{2}_{t}\rho_{2}\right)\hat{R}-\partial_{t}(\rho_{1}\vec{J}_{2}+\rho_{2}\vec{J}_{1})R^{-1}\right]$

Now as $\frac{k}{c^{2}}=\frac{\mu_{0}}{4\pi}$ and since:

$\rho_{2}\partial^{2}_{t}\rho_{1}-\rho_{1}\partial^{2}_{t}\rho_{2}=\partial_{t}\left(\rho_{2}\partial_{t}\rho_{1}-\rho_{1}\partial_{t}\rho_{2}\right)$
(80)

It follows that:

$\vec{F}_{T}^{[2]}=\frac{\mu_{0}}{4\pi}\partial_{t}\int\int
d^{3}x_{1}d^{3}x_{2}~{}\left[\frac{1}{2}\left(\rho_{2}\partial_{t}\rho_{1}-\rho_{1}\partial_{t}\rho_{2}\right)\hat{R}-(\rho_{1}\vec{J}_{2}+\rho_{2}\vec{J}_{1})R^{-1}\right]$
(81)

In the following section, we describe some of the implications of the above
formula. We remark that in some fast changing systems the second order
correction will not suffice and higher order terms will be needed.

###  3.3 Some preliminary observations

According to Newton’s second law, a system with a non zero total force in its
center of mass, must have a change in its total linear momentum $\vec{P}(t)$:

$\vec{F}_{T}^{[2]}=\frac{d\vec{P}}{dt}$ (82)

Assuming that $\vec{P}(-\infty)=0$ and also that there are not current or
charge densities at $t=-\infty$, it follows from equation (81) that:

$\vec{P}(t)=\frac{\mu_{0}}{4\pi}\int\int
d^{3}x_{1}d^{3}x_{2}~{}\left[\frac{1}{2}\left(\rho_{2}\partial_{t}\rho_{1}-\rho_{1}\partial_{t}\rho_{2}\right)\hat{R}-(\rho_{1}\vec{J}_{2}+\rho_{2}\vec{J}_{1})R^{-1}\right]$
(83)

Comparing equation (83) with the momentum gain of a non charged relativistic
motor described by equation (64) of [4]:

$\vec{P}_{mech}\cong\frac{\mu_{0}}{8\pi}\partial_{t}I_{1}(t)I_{2}(\frac{h}{c})^{2}\vec{K}_{122},\qquad\vec{K}_{122}=-\frac{1}{h^{2}}\oint\oint\hat{R}(d\vec{l}_{2}\cdot
d\vec{l}_{1})$ (84)

where $h$ is a typical scale of the system, $I_{1}(t)$ and $I_{2}$ are the
currents flowing through two current loops and $d\vec{l}_{1},d\vec{l}_{2}$ are
the current loop line elements. We notice some major differences. First, we
notice a factor of $\left(\frac{h}{c}\right)^{2}$ in case of an uncharged
motor. As, for any practical system the scale $h$ is of the order of one, this
means that the charged relativistic motor is stronger than the uncharged motor
by a factor of $c^{2}\sim 10^{17}$ which is quite a considerable factor.
Second, we notice that for the uncharged motor, the current must be
continuously increased in order to maintain the momentum in the same
direction. Of course, one cannot do this for ever, hence the uncharged motor
is a type of a piston engine doing a periodic motion backward and forward and
can only produce motion forward by interacting with an external system (the
road). This is not the case for the charged relativistic motor. In fact we
obtain non vanishing momentum for stationary charge and current densities:

$\vec{P}(t)=-\frac{\mu_{0}}{4\pi}\int\int
d^{3}x_{1}d^{3}x_{2}~{}(\rho_{1}\vec{J}_{2}+\rho_{2}\vec{J}_{1})R^{-1}$ (85)

hence the charged relativistic motor can produce forward momentum without
interacting with any external system except the electromagnetic field. The
above expression can be somewhat simplified using the potential given in
equation (13) such that:

$\vec{P}(t)=-\frac{1}{c^{2}}\left[\int d^{3}x_{2}~{}\Phi_{1}\vec{J}_{2}+\int
d^{3}x_{1}~{}\Phi_{2}\vec{J}_{1}\right]$ (86)

in which we are reminded that $\frac{\mu_{0}}{4\pi
k}=\mu_{0}\epsilon_{0}=\frac{1}{c^{2}}$. Another observation is that in a
charged relativistic motor we do not need both subsystems to be charged, in
fact if $\rho_{2}=0$:

$\vec{P}(t)=-\frac{\mu_{0}}{4\pi}\int\int
d^{3}x_{1}d^{3}x_{2}~{}\rho_{1}\vec{J}_{2}R^{-1}=-\frac{1}{c^{2}}\int
d^{3}x_{2}~{}\Phi_{1}\vec{J}_{2}$ (87)

provided that the system has a non vanishing current density $\vec{J}_{2}$. We
underline that as in [4] the forward momentum gained by the mechanical system
will be balanced by a backward momentum gained by the electromagnetic system.
We shall now discuss the challenges in developing a charged relativistic
motor.

###  3.4 Charge density

The amount of charge that can be accumulated in a given volume or surface is
limited by the phenomena of electrical breakdown in which the surrounding
medium is separated into electron and ions and becomes a plasma. As the
surrounding becomes conductive a discharge occurs and the charge density is
reduced. The dielectric strength $E_{max}$ of air is 3 MV/m [18], for high
vacuum 20-40 MV/m [19] and for diamond 2000 MV/m [20]. For an infinite surface
the surface density $\sigma$ is:

$\sigma=2\epsilon E<\sigma_{max}=2\epsilon
E_{max},\qquad\epsilon=\epsilon_{r}\epsilon_{0}$ (88)

in which $\epsilon_{r}$ is the relative susceptibility. For air
$\sigma_{max}\simeq 53~{}\mu C/m^{2}$. To estimate the amount of charge $Q$
which one can maintain in a given volume we notice that for a spherical
symmetric charge ball we obtain at a distance $r$ the radial field $E$:

$E=\frac{kQ}{r^{2}}$ (89)

the stronger field is on the ball itself, that is at $r=r_{s}$ we must have:

$\frac{kQ}{r_{s}^{2}}<E_{m}\Rightarrow Q<Q_{m}=\frac{1}{k}r_{s}^{2}E_{m}$ (90)

For a typical size of $1$ m, the maximal charge is $3.3~{}10^{-4}$ C. Hence
regardless if we have surface charge or a volume charge the maximal charge
scales as the square of the dimension of the system , that is as $h^{2}$. A
possible approach to increase the available charge density is to use electret.
Fluorinated parylene (Parylene HT, SCS) offers excellent surface charge
density of $3.7~{}{\rm mC/m^{2}}$ found for a $7.3~{}{\rm\mu m}$ film by Hsi-
wen and Yu-Chong [21], this material has a dielectric strength of 204.58 MV/m.
However, as the thickness of the material grows the charge density is reduced.

###  3.5 Current density

The amount of current a device can generate dependent on its voltage and
internal resistance. Provided that the external impedance is not significant,
the resulting currents are denoted as short currents be the order of a few
thousand amperes for a standard domestic electrical installation and as high
as hundreds of thousands of amperes in large industrial power systems. If the
current is flowing through a metal conductor, then heat is generated due to
the finite resistivity of the conductor. Large currents require a thick
conductor to avoid excessive heating. This problem can be circumvented using a
superconductor, although this will require cooling to extremely low
temperatures, making the system quite cumbersome. But, even a superconductor
has a critical current density and superconduction properties losses above
that. Jung, S. G. et al. [22] have reported critical current densities as high
as $5~{}{\rm kA/cm^{2}}$. Coil windings will enable to reuse the current, and
the number of windings in a given area is also critical to the performance of
the system. Proximity of the current to the charge will also affect the amount
of generated momenta equation (87) . However, putting a conductor too close to
the charge may result in discharge, hence a balance should be striked.

###  3.6 Scalability

It is clear from equation (87) that the larger the relativistic engine the
more powerful it is. However, since the charge surface density is limited the
momenta generate will scale as the second power of the size of the charged
subsystem and as the third power of the size of the current carrying system.

##  4 An Example

Consider an electret of thickness $d$ and area of $a\times b$, the electret is
put inside a tight superconductive winded coil (see figures 2 and 3).

Figure 2: A relativistic engine. Figure 3: A cross section of the relativistic
engine.

As the current in the $x$ direction is close to the negative charge it will
make a considerable contribution to the momentum equation equation (87) while
the return current which is far away from the negative charge will make a
relatively small contribution. The situation is reversed with regard to the
positive charge, here the return current will make the main contribution.
Hence the total momentum contribution is doubled with respect to the current
flowing in the $x$ direction and interacting with negative charge alone. The
current flowing in the $y$ direction will not contribute to the momentum as
the contribution from the current flowing upwards is exactly balanced by the
contribution of the current flowing downwards. Let us look at the negative
charge layer in figure 3, we shall model this as a surface charged layer in
the plane $y=0$ with a surface charge density $\sigma$ such that:

$\rho_{1}(x_{1},y_{1},z_{1})=-\sigma\delta(y_{1})\left\\{\begin{array}[]{cc}1&-\frac{a}{2}<x_{1}<\frac{a}{2},\quad-\frac{b}{2}<z_{1}<\frac{b}{2}\\\
0&{\rm else}\end{array}\right.$ (91)

in the above $\delta(y_{1})$ is a Kronecker delta.

$\vec{J}_{2}(x_{2},y_{2},z_{2})=\hat{x}J_{0}w\delta(y_{2}+\Delta)\left\\{\begin{array}[]{cc}1&-\frac{a}{2}<x_{2}<\frac{a}{2},\quad-\frac{b}{2}<z_{2}<\frac{b}{2}\\\
0&{\rm else}\end{array}\right.$ (92)

where $J_{0}$ is the current density, $w$ is the width of the winding,
$\hat{x}$ is a unit vector in the $x$ direction and $\Delta$ is the distance
of the current plane from the negative charge plane. Obviously their is a
difference between the $\Delta$ of the current: $\Delta_{1}$ and the return
current: $\Delta_{2}$. The integrated current flowing through the system is:

$I=J_{0}\ w\ b.$ (93)

However, the current flowing through each single wire depends on the winding
number $N_{w}$:

$I_{c}=\frac{I}{N_{w}}$ (94)

Plugging equation (91) and equation (92) into equation (87), we arrive at the
following momentum equation:

$\vec{P}=\frac{\mu_{0}}{4\pi}\sigma
J_{0}w\hat{x}\int_{-\frac{a}{2}}^{+\frac{a}{2}}dx_{1}\int_{-\frac{b}{2}}^{+\frac{b}{2}}dz_{1}\int_{-\frac{a}{2}}^{+\frac{a}{2}}dx_{2}\int_{-\frac{b}{2}}^{+\frac{b}{2}}dz_{2}\frac{1}{R}.$
(95)

Such that:

$R=\sqrt{(x_{1}-x_{2})^{2}+(z_{1}-z_{2})^{2}+\Delta^{2}}.$ (96)

Defining the dimensionless variables:

$x^{\prime}_{1}=\frac{x_{1}}{a},\quad z^{\prime}_{1}=\frac{z_{1}}{b},\quad
x^{\prime}_{2}=\frac{x_{2}}{a},\quad z^{\prime}_{2}=\frac{z_{2}}{b},\quad
R^{\prime}=\frac{R}{\Delta},\quad a^{\prime}=\frac{a}{\Delta},\quad
b^{\prime}=\frac{b}{\Delta}.$ (97)

We may write:

$\vec{P}=\frac{\mu_{0}}{4\pi}\sigma
J_{0}wab^{2}\tilde{\Lambda}(a^{\prime},b^{\prime})\hat{x}.$ (98)

The function $\tilde{\Lambda}(a^{\prime},b^{\prime})$ is dimensionless and
depends on the dimensionless quantities $a^{\prime},b^{\prime}$. It can be
evaluated using the quadruple integral:

$\tilde{\Lambda}(a^{\prime},b^{\prime})\equiv\int_{-\frac{1}{2}}^{+\frac{1}{2}}dx^{\prime}_{1}\int_{-\frac{1}{2}}^{+\frac{1}{2}}dz^{\prime}_{1}\int_{-\frac{1}{2}}^{+\frac{1}{2}}dx^{\prime}_{2}\int_{-\frac{1}{2}}^{+\frac{1}{2}}dz^{\prime}_{2}\frac{a^{\prime}}{R^{\prime}}.$
(99)

In which:

$R^{\prime}=\sqrt{a^{\prime 2}(x^{\prime}_{1}-x^{\prime}_{2})^{2}+b^{\prime
2}(z^{\prime}_{1}-z^{\prime}_{2})^{2}+1}.$ (100)

We shall attempt to perform at least part of this integration analytically.
For this, we shall define the auxiliary function:

$\displaystyle\tilde{\Gamma}(a^{\prime},b^{\prime},x^{\prime}_{2},z^{\prime}_{2})$
$\displaystyle\equiv$
$\displaystyle\int_{-\frac{1}{2}}^{+\frac{1}{2}}dx^{\prime}_{1}\int_{-\frac{1}{2}}^{+\frac{1}{2}}dz^{\prime}_{1}\frac{a^{\prime}}{R^{\prime}}$
$\displaystyle\Rightarrow\tilde{\Lambda}(a^{\prime},b^{\prime})$
$\displaystyle=$
$\displaystyle\int_{-\frac{1}{2}}^{+\frac{1}{2}}dx^{\prime}_{2}\int_{-\frac{1}{2}}^{+\frac{1}{2}}dz^{\prime}_{2}\tilde{\Gamma}(a^{\prime},b^{\prime},x^{\prime}_{2},z^{\prime}_{2}).$
(101)

We shall now make a change of variables:

$x=x^{\prime}_{1}-x^{\prime}_{2},\qquad z=z^{\prime}_{1}-z^{\prime}_{2},$
(102)

such that:

$\displaystyle\tilde{\Gamma}(a^{\prime},b^{\prime},x^{\prime}_{2},z^{\prime}_{2})$
$\displaystyle=$
$\displaystyle\int_{-\frac{1}{2}-x^{\prime}_{2}}^{+\frac{1}{2}-x^{\prime}_{2}}dx\int_{-\frac{1}{2}-z^{\prime}_{2}}^{+\frac{1}{2}-z^{\prime}_{2}}dz\frac{a^{\prime}}{R^{\prime}}$
$\displaystyle R^{\prime}$ $\displaystyle=$ $\displaystyle\sqrt{a^{\prime
2}x^{2}+b^{\prime 2}z^{2}+1}.$ (103)

Finally, we define:

$\displaystyle\Psi(a^{\prime},b^{\prime},x,z^{\prime}_{2})$
$\displaystyle\equiv$
$\displaystyle\int_{-\frac{1}{2}-z^{\prime}_{2}}^{+\frac{1}{2}-z^{\prime}_{2}}dz\frac{a^{\prime}}{R^{\prime}}$
$\displaystyle\Rightarrow\tilde{\Gamma}(a^{\prime},b^{\prime},x^{\prime}_{2},z^{\prime}_{2})$
$\displaystyle=$
$\displaystyle\int_{-\frac{1}{2}-x^{\prime}_{2}}^{+\frac{1}{2}-x^{\prime}_{2}}dx\Psi(a^{\prime},b^{\prime},x,z^{\prime}_{2}).$
(104)

The function $\Psi$ can be integrated analytically by introducing the new
variable:

$z^{\prime\prime}=\frac{b}{a}\frac{z}{\sqrt{x^{2}+a^{\prime-2}}}$ (105)

in terms of which:

$\displaystyle\Psi(a^{\prime},b^{\prime},x,z^{\prime}_{2})$ $\displaystyle=$
$\displaystyle\frac{a}{b}\int_{z^{\prime\prime}_{1}}^{z^{\prime\prime}_{2}}dz^{\prime\prime}\frac{1}{\sqrt{1+z^{\prime\prime
2}}}=\frac{a}{b}\left[{\rm arcsinh}(z^{\prime\prime}_{2})-{\rm
arcsinh}(z^{\prime\prime}_{1})\right]$ $\displaystyle z^{\prime\prime}_{1}$
$\displaystyle=$
$\displaystyle\frac{b}{a}\left(\frac{-\frac{1}{2}-z^{\prime}_{2}}{\sqrt{x^{2}+a^{\prime-2}}}\right),\qquad
z^{\prime\prime}_{2}=\frac{b}{a}\left(\frac{+\frac{1}{2}-z^{\prime}_{2}}{\sqrt{x^{2}+a^{\prime-2}}}\right).$
(106)

Unfortunately the calculation of $\tilde{\Gamma}$ and $\tilde{\Lambda}$ can
only be done numerically.

For the case $b^{\prime}=a^{\prime}$ the function $\tilde{\Lambda}$ is a
single variable function depicted in figure 4.

Figure 4: The function $\tilde{\Lambda}$ for the case $b^{\prime}=a^{\prime}$.

It can be seen that for $a^{\prime}=0$ the function is null, while for
$a^{\prime}=\infty$ it approaches $2.973\sim 3$. Hence if the ratio of size of
the engine to the winding width is about $20$ and $d\sim a$ one does not loss
a significant amount of force & momentum due to the return current. However,
looking at the case $b^{\prime}\neq a^{\prime}$ depicted in figure 5, we a see
an obvious advantage for a slender relativistic motor in the direction of
motion.

Figure 5: The function $\tilde{\Lambda}$ for the case $b^{\prime}\neq
a^{\prime}$, the color code is shown to the right of the contour plot.

We also point out that the return current interacts symmetrically with the
positive charge in a beneficial way, as both the sign of the charge and the
direction of the current are reversed (see figure 3) this doubles the momentum
gain such that according to equation (98) we have:

$\vec{P}_{T}=\frac{\mu_{0}}{2\pi}\sigma
J_{0}wab^{2}\left(\tilde{\Lambda}(\frac{a}{\Delta_{1}},\frac{b}{\Delta_{1}})-\tilde{\Lambda}(\frac{a}{\Delta_{2}},\frac{b}{\Delta_{2}})\right)\hat{x}.$
(107)

The quantities $\Delta_{1}$ and $\Delta_{2}$ are defined through figure 3 and
we thus expect:

$\Delta_{1}=\frac{w}{2},\qquad\Delta_{2}=\frac{w}{2}+d.$ (108)

The force produced by the relativistic engine depends on the rise time of the
current which can be increased gently or abruptly:

$\vec{F}_{T}=\frac{d\vec{P}_{T}}{dt}=\frac{\mu_{0}}{2\pi}\sigma\frac{dJ_{0}}{dt}wab^{2}\left(\tilde{\Lambda}(\frac{a}{\Delta_{1}},\frac{b}{\Delta_{1}})-\tilde{\Lambda}(\frac{a}{\Delta_{2}},\frac{b}{\Delta_{2}})\right)\hat{x}.$
(109)

We choose three different configurations of a relativistic motor, one is a
size of standard car and its parameters are described in the first column of
table 1 the other describes a considerably larger motor which may be suitable
for space travel is given in the second column of table 1, the third column
describes a cube of immense dimensions.

| car | rocket size engine | 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 | $3.1\ 10^{7}$ | kg m/s
Table 1: Maximal momentum gained by a relativistic motor for three cases of
parameters. We assume an extreme case of charge density $\sigma=3.7\ 10^{-3}$
Coulomb/$\rm m^{2}$ (see section 3.4), and current density $J_{0}=5\ 10^{7}$
Ampere/$\rm m^{2}$ (see section 3.5).

We see that despite the extreme parameters the maximum momenta gained are
quite modest,this is a direct result of the phenomena of dielectric breakdown.

##  5 The nano relativistic motor

In the previous section we have shown that intrinsic parameter limitations,
especially dielectric breakdown lead to somewhat modest amount of momenta that
can be gained by relativistic motor. However, it seams that in the microscopic
domain this limitation is rather relaxed. For example if consider ionic
crystals such as the prevalent table salt: $Na^{+}Cl^{-}$. This salt
solidifies to a face centered cubic lattice in which the lattice constant is
$l=564.02$ pm. Taking for example the $100$ plane of this lattice (see figure
6) we can see that the charge density is periodical in which in each half unit
cell we have a surface charge density of $\pm 2~{}{\rm Coulomb/m^{2}}$. This
is of course thousand time larger than available macroscopic charge densities
(see section 3.4), however, on the macroscopic average this leads to a null
charge density.

Figure 6: The $100$ plane of a table salt $Na^{+}Cl^{-}$ lattice, blue circles
stand for Sodium positive ions and yellow circles stand for Chlorine negative
ions.

To see how we can circumvent this unfortunate reality we will investigate
equation (87), calculating the spatial Fourier transform of the scalar
potential and the current density such that:

$\displaystyle\Phi_{1}(\vec{k})$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\Phi_{1}(\vec{x})e^{-i\vec{k}\cdot\vec{x}}d^{3}x$
$\displaystyle\vec{J}_{2}(\vec{k})$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\vec{J}_{2}(\vec{x})e^{-i\vec{k}\cdot\vec{x}}d^{3}x.$
(110)

Using the theorem of Parseval [23], we may now write equation (87) in the
form:

$\vec{P}(t)=-\frac{1}{c^{2}(2\pi)^{3}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}d^{3}k~{}\Phi_{1}^{*}(\vec{k})\vec{J}_{2}(\vec{k})$
(111)

In this form it is obvious, that a microscopic distribution of periodic charge
density will be beneficial if it is a accompanied by a current density
distribution of the same period. How can we obtain such a microscopic charge
density distribution? To this end we remind the reader that microscopic
currents are associated with the electronic motion and electronic spin. The
magnetization $\vec{M}$ is related to the magnetization current $\vec{J}_{M}$
by the formulae [2, 3]:

$\vec{J}_{M}\equiv\vec{\nabla}\times\vec{M}.$ (112)

We may replace in equation (86), equation (87) and equation (111) $\vec{J}$ by
$\vec{J}_{M}$ to obtain a similar effect. Furthermore, the magnetization
$\vec{M}$ is related to a the microscopic dipole moments $\vec{m}_{i}$
through:

$\vec{M}\equiv\frac{1}{V}\sum_{i}\vec{m}_{i}.$ (113)

in which we sum over all dipoles and divide by the sample volume $V$. In known
magnetic materials such as iron the magnetic dipole moments are related to the
spin configuration of the atom. In $\alpha$-iron the spins of the two unpaired
electrons in each atom generally align with the spins of its neighbors. This
happens because the orbitals of those two electrons ($d_{z^{2}}$ and
$d_{x^{2}-y^{2}}$) do not point toward neighboring atoms in the lattice, and
therefore are not involved in metallic bonding. Hence the ideal configuration
for a relativistic motor will involve and ionic lattice in which one species
of atom (say the positively charged) will involve free spins that can be
manipulated by an external magnetic field thus creating a relativistic engine
effect which can be easily manipulated in three axis. The details of such an
engine are beyond the scope of the current paper.

##  6 Conclusion

In this paper, we have shown that, in general, Newton’s third law is not
compatible with the principles of special relativity and the total force on a
two charged body system is not zero. Still, momentum is conserved if one takes
the field momentum into account, and the same is true for energy.

The main results of this paper are given by equation (82) and equation (83),
which describe the total relativistic force for charged bodies. Not all
configurations of charged relativistic motors were analyzed here due to space
limitations. We have concentrated our efforts on a system in which one part is
charged and the second part carries current. It is noticed that the charge
relativistic motor is many orders of magnitude powerful than an uncharged
relativistic motor. Still, it is shown that the limitations of dielectric
breakdown and current density lead to a rather modest momentum generated by
the engine on the macroscopic level.

Those limitations can be somewhat relaxed at the atomic level, as the surface
charge density for typical ionic crystals is three orders of magnitude larger
than what can be achieved at the macroscopic level. Thus future work will
concentrate on the development of nano relativistic engines. Other open
subjects are exploring additional possible charged relativistic configurations
as suggested by equation (83) and studying in detail momentum and energy
transformation from the electromagnetic to the mechanical elements of the
motor.

Appendices

##  Appendix A The consistency of the gauge condition

Set $\vec{x}=\vec{x}_{2}$, $\vec{x}^{\prime}=\vec{x}_{1}$ and
$R=|\vec{x}^{\prime}-\vec{x}|$, thus equation (14) takes the form:

$\vec{A}(\vec{x},t)=\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\frac{\vec{J}(\vec{x}^{\prime},t_{ret})}{R}.$ (114)

in which we suppressed the subscript of $\vec{J}$. Taking a divergence of both
sides of the above equation we obtain:

$\displaystyle\vec{\nabla}\cdot\vec{A}$ $\displaystyle=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\vec{\nabla}\cdot\left(\frac{\vec{J}(\vec{x}^{\prime},t_{ret})}{R}\right)$
(115) $\displaystyle=$ $\displaystyle\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\left(\frac{\vec{\nabla}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})}{R}+\vec{J}(\vec{x}^{\prime},t_{ret})\cdot\vec{\nabla}\frac{1}{R}\right).$

However:

$\vec{\nabla}\frac{1}{R}=-\vec{\nabla}^{\prime}\frac{1}{R}$ (116)

Hence:

$\vec{\nabla}\cdot\vec{A}=\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\left(\frac{\vec{\nabla}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})}{R}-\vec{J}(\vec{x}^{\prime},t_{ret})\cdot\vec{\nabla}^{\prime}\frac{1}{R}\right).$
(117)

This leads to:

$\vec{\nabla}\cdot\vec{A}=\frac{\mu_{0}}{4\pi}\left(\int
d^{3}x^{\prime}\frac{\vec{\nabla}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})+\vec{\nabla}^{\prime}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})}{R}-\oint
d\vec{S}\cdot\frac{\vec{J}(\vec{x}^{\prime},t_{ret})}{R}\right).$ (118)

in which we have used Gauss theorem. The surface integral is taken over a
closed surface encapsulating the volume of integration. However, the volume
can be infinite and thus the surface is taken at infinity at which we assume
there are no current densities. Hence:

$\vec{\nabla}\cdot\vec{A}=\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\frac{\vec{\nabla}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})+\vec{\nabla}^{\prime}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})}{R}.$
(119)

now $\vec{J}(\vec{x}^{\prime},t_{ret})$ depends on $\vec{x}^{\prime}$ both
directly and through $t_{ret}$, thus:

$\vec{\nabla}^{\prime}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})=\vec{\nabla}^{\prime}|_{t_{ret}}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})+\partial_{t_{ret}}\vec{J}(\vec{x}^{\prime},t_{ret})\cdot\vec{\nabla}^{\prime}t_{ret}.$
(120)

$\vec{\nabla}^{\prime}|_{t_{ret}}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})$ means
taking a divergence with respect to $\vec{x}^{\prime}$ but leaving $t_{ret}$
constant. However:

$\vec{\nabla}^{\prime}t_{ret}=\vec{\nabla}^{\prime}R~{}\partial_{R}t_{ret}=-\vec{\nabla}R~{}\partial_{R}t_{ret}=-\vec{\nabla}t_{ret}$
(121)

Hence:

$\vec{\nabla}^{\prime}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})=\vec{\nabla}^{\prime}|_{t_{ret}}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})-\partial_{t_{ret}}\vec{J}(\vec{x}^{\prime},t_{ret})\cdot\vec{\nabla}t_{ret}.$
(122)

We notice that:

$\vec{\nabla}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})=\partial_{t_{ret}}\vec{J}(\vec{x}^{\prime},t_{ret})\cdot\vec{\nabla}t_{ret},$
(123)

it now follows:

$\vec{\nabla}^{\prime}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})=\vec{\nabla}^{\prime}|_{t_{ret}}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})-\vec{\nabla}\cdot\vec{J}(\vec{x}^{\prime},t_{ret}).$
(124)

Inserting equation (124) into equation (119) leads to:

$\vec{\nabla}\cdot\vec{A}=\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\frac{\vec{\nabla}^{\prime}|_{t_{ret}}\cdot\vec{J}(\vec{x}^{\prime},t_{ret})}{R}.$
(125)

Let us now look at the scalar potential given in equation (13), this can be
written using the current notation as:

$\Phi=\frac{1}{4\pi\epsilon_{0}}\int
d^{3}x^{\prime}\frac{\rho(\vec{x}^{\prime},t_{ret})}{R}.$ (126)

Taking a partial temporal derivative of $\Phi$ and multiplying by
$\frac{1}{c^{2}}=\epsilon_{0}\mu_{0}$ we arrive at:

$\frac{1}{c^{2}}\partial_{t}\Phi=\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\frac{\partial_{t}\rho(\vec{x}^{\prime},t_{ret})}{R}.$ (127)

Combining equation (127) with equation (125) it follows that:

$\frac{1}{c^{2}}\partial_{t}\Phi+\vec{\nabla}\cdot\vec{A}=\frac{\mu_{0}}{4\pi}\int
d^{3}x^{\prime}\left.\frac{\partial_{t}\rho(\vec{x}^{\prime},t)+\vec{\nabla}^{\prime}\cdot\vec{J}(\vec{x}^{\prime},t)}{R}\right|_{t=t_{ret}}.$
(128)

However, due to the charge conservation equation (16) we obtain:

$\frac{1}{c^{2}}\partial_{t}\Phi+\vec{\nabla}\cdot\vec{A}=0.$ (129)

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