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\title{Relativistic Thermodynamics and\\ the Classical Model
of the Electron}
\author{B.H. Lavenda\thanks{ Supported in part by MURST
60\% .}\\ Universit\`a di Camerino\\ I-62032 Camerino (MC)
Italy\\
E-mail:
lavenda@camserv.unicam.it}
\date{}
\begin{document}
\maketitle
\begin{abstract}
Einstein's famous relation between mass and energy is
interpreted in terms of the equivalence of the rate of
heating of a body and the rate of increase of its inertial
mass. In an adiabatic process, where the proper mass
remains constant, it is the heat content, and not the
energy, which is conserved because the pressure,
and not the volume, is Lorentz-invariant. There are two
categories of relativistic quantities: inertial and
thermodynamic ones, which are transformed into one another by
the work necessary to keep the inertial state in motion.
In a non-adiabatic process, the rate of heating is
Lorentz-invariant, which must always be greater than the
power that it generates.
\end{abstract}
\section{Introduction}
Einstein's famous relation \cite{Einstein}
\begin{equation}
\Delta E=c^2\Delta m \label{eq:Einstein}
\end{equation}
affirms that \lq\lq the mass of a body is a measure of its
\textit{energy} content.\rq\rq\ The example used by Einstein
involved the emission of radiation by a body which causes
it to cool. It is well-know that, under certain
circumstances, external forces can lead to the heating of a
body thereby increasing its rest mass. Electromagnetic
forces acting on a conductor produce Joule heating. Were
it not for the fact that the volume of the body is not
Lorentz-invariant, the change in energy
(\ref{eq:Einstein}) could be equated with the heat gained by
the body. In the transformation from a system at rest to one
moving at a constant velocity, the volume undergoes a
FitzGerald-Lorentz contraction so that the left-hand side of
(\ref{eq:Einstein}) is not the increase, or decrease, in the
quantity of heat gained, or lost, by the body. Rather, it is
the pressure that remains constant so the quantity of heat
gained by the body increases the heat content (enthalpy) of
the body. Then (\ref{eq:Einstein}) asserts that \lq\lq the
mass of a body is a measure of its \textit{heat}
content.\rq\rq\
\par
If $\Delta E$ were equal to the increase in inertial energy
it would transform as $\gamma:=1/\sqrt{1-\beta^2}$, where
$\beta:=u/c$, the ratio of the relative speed $u$ to that of
light in vacuum, $c$. Yet, Einstein \cite[Doc.
47]{Einstein} knew that the same quantity of heat is always
smaller when a system is in motion than when it is a rest,
while the kinetic energy of the system is increased by
the motion. Thus, if $\Delta E$ represents the increase in
kinetic energy, it transforms in the opposite manner to the
quantity of heat under a Lorentz transformation.
\par
Moreover, the discrepancy between the inertial
electromagnetic mass of an electron and the
electrostatic mass, defined as the energy of formation of
the spherical charge divided by $c^2$, has never been
cleared up in a totally satisfactory manner \cite{ADY}.
Einstein was probably unconcerned by the mere numerical
factor of
$\fourthirds$ by which the electrostatic mass differed from
the electromagnetic mass
\cite{Miller}. Lorentz's conjecture, that the origin of the
mass of an electron was purely electromagnetic, had to be
abandoned because there was something missing in the
expression of the energy that was of a non-electromagnetic
nature
\cite{Abraham}.\par The classic argument given by von Laue
\cite{Laue} is that since the system is not closed, the
energy-momentum vector does not transform like a 4-vector.
However, upon closing the system by considering the
existence of mechanical components to energies and momenta
of the electron---without violating Maxwell's equations
which are assumed to hold throughout the entire space---the
correct $\fourthirds$ factor was achieved in the expression
for the energy---but at a high expense. For this
procedure introduces a \lq negative\rq\ pressure, known as
Poincar\'e's stress, which is related to the binding
potential, or the work done by the internal binding forces as
the spherical charge distribution accelerates and contracts
to the shape of an oblate ellipsoid. This is the same as the
work done by a negative, constant pressure, or a positive,
constant tension, which in the words of Poincar\'e
is comparable
\lq\lq to a constant external pressure acting
on the deformable and compressible electron, whose work is
proportional to the variation of the volume of the
electron\rq\rq
\cite{Poincare}.\par Yet, it has always been the feeling
that the correct dependence of the mass upon velocity
should be achieved without introducing any model of the
electron, especially one of finite size. By adjoining
arbitrary factors to the total energy, so that the elusive
factor of $\fourthirds$ would be achieved, no longer meant
that one was dealing with an energy. Yet, the
(pre-Einstein) desire to maintain a mass-energy relationship
was stronger than the realization that a mass-heat content
relationship would give the $\fourthirds$ factor without
having to divide the energy up into electromagnetic and
mechanical contributions. This theme will be developed here,
leading to the result that the left-hand side of
(\ref{eq:Einstein}) is actually the increase in the heat
content due to heat that has been conveyed
to the particle to increase its inertial mass by an
amount $\Delta m$.\par The question arises why should the
heat content transform any differently from the energy?
Here, one must be careful to distinguish between inertial
and thermodynamic quantities. Planck \cite{Planck} showed
that the pressure, $p$, is Lorentz-invariant. Since the
proper volume,
$V$, decreases by a factor of
$\gamma^{-1}$ under a Lorentz transformation, the
compressional work,
$-p\,dV$, will also decrease by the same amount. And since
the product appears in the Euler relation for the
increase in the internal energy, both the internal energy and
the quantity of heat gained by the system will necessarily
decrease by a factor
$\gamma^{-1}$ under a Lorentz transformation. In
contradistinction, the proper mass of a point particle
increases by a factor of
$\gamma$, so that the energy in (\ref{eq:Einstein}) would
appear to increase by the same amount.
\par How then are inertial quantities related to
thermodynamic ones? A \lq strange result\rq, due a
relativistic dynamical effect, was first commented upon
by Einstein
\cite[Doc. 45]{Einstein}. In Einstein's words:
\begin{quotation}
If a rigid body on which originally no forces are acting is
subject to the influence of forces that do not impart
acceleration to the body, then these forces---observed from a
coordinate system that is moving relative to the
body---perform an amount of work $d E$ on the body that
depends only on the final distribution of forces and the
translation velocity.
\end{quotation}
The quantity
$dE$ that Einstein was referring appears as a tensor in
von Laue's \cite{Laue} relativistic formulation of elastic
materials, being the difference between the
\textit{absolute} and \textit{relative} stress tensors.
In scalar form, it can be the difference between the
total and internal energy densities.
Mathematically, it will give rise to a Lorentz-invariant,
while, physically, it represents the work necessary to
keep the system in a state of uniform motion.
\par
\section{Lorentz Invariants and Relative Stresses}
Consider the
Lorentz transformation of the energy-matter tensor
\[
\mathbf{T}=\left(\begin{array}{cc}
\mbox{\boldmath{$\tau$}} & i\mbox{\boldmath{$g$}}c\\
i\mbox{\boldmath{$g$}}c & \varepsilon
\end{array}
\right), \]
where $\mbox{\boldmath{$\tau$}}$ is the stress,
$\mbox{\boldmath{$g$}}$ the momentum density, and
$\varepsilon$ the energy density. If we consider the
transformation of the energy-matter tensor from a rest
frame, $O$, to a moving one, $O^{\prime}$, at a constant
relative velocity
$\mathbf{u}$, the Lorentz transformation is
\[
\mathbf{T}^{\prime}_{ik}=\alpha_{il}
\alpha_{km}\; \mathbf{T}_{lm},
\]
where
\[
\mbox{\boldmath$\alpha$}=\left(
\begin{array}{cc}
\gamma & i\beta\gamma\\
-i\beta\gamma & \gamma
\end{array}
\right). \]
As a consequence, the momentum density, stress, and energy
density transform as
\begin{equation}
\mathbf{g}^{\prime}=\gamma^2\frac{\mathbf{u}}{c^2}
\left(\varepsilon-\tau
\right), \label{eq:g}
\end{equation}
\begin{equation}
\tau^{\prime}=\gamma^2\left(\tau-\beta^2\varepsilon
\right) \label{eq:tau}
\end{equation}
and
\begin{equation}
\varepsilon^{\prime}=\gamma^2\left(\varepsilon-
\beta^2\tau
\right),
\label{eq:e}
\end{equation}
respectively, since $\mathbf{g}=0$ in
$O$. These equations can be combined to give the
\textit{relative} stress tensor \cite{Laue}
\begin{equation}
\mbox{\boldmath$\bar{\tau}$}:=
\mbox{\boldmath$\tau$}^{\prime}
+\mathbf{u}
\otimes\mathbf{g}^{\prime}=\mbox{\boldmath$\tau$},
\label{eq:tau-bis}
\end{equation}
and the
\textit{relative} energy density
\begin{equation}
\bar{\varepsilon}:=\varepsilon^{\prime}-\mathbf{u}\cdot
\mathbf{g}^{\prime}=\varepsilon\ge0.
\label{eq:e-bis}
\end{equation}
$\mbox{\boldmath{$\tau$}}^{\prime}$ is the absolute stress
in the moving frame, and $\otimes$ is the direct product.
The advantage of working with relative stresses and
densities is that they are Lorentz-invariants. The same
relations apply to the Maxwell stress tensor in a frame
moving with velocity
$\mathbf{u}$.\par In electromagnetism, the momentum density
$\mathbf{g}$ is the Poynting vector divided by
$c^2$,
$\mathbf{g}=(\mathbf{E}\times\mathbf{H})/c$, and the
electromagnetic energy density
$\varepsilon=\half\left(E^2+H^2\right)$,
where
$\mathbf{E}$ and $\mathbf{H}$, are the electric and
magnetic field strengths, respectively. The relative
velocity for a transversal plane wave,
$\mathbf{u}=c(\mathbf{E}\times\mathbf{H})/E^2$, is the
condition that the transformed magnetic field vanishes.
The electromagnetic analog to the Lorentz-invariant
$\bar{\varepsilon}$ is
\[\half\left(E^2+H^2\right)-
\frac{\left(\mathbf{E}\times
\mathbf{H}\right)^2}{E^2}=\half\left(E^2-H^2\right)\ge0,
\]
since
$\mathbf{E}\cdot\mathbf{H}=0$. The relative speed
$u=cH/E\le c$. If the fields were not
orthogonal, then the second invariant quantity would be the
scalar
$\mathbf{E}\cdot\mathbf{H}$. For a light wave in vacuum, both
Lorentz-invariants vanish because $H=E$ and the two fields
are orthogonal to one another.\par For the moment, let us
consider the frame $O^{\prime}$ as the rest frame and the
frame $O$ moving at a velocity $-\mathbf{u}$ relative to
$O^{\prime}$. The electric and magnetic field strengths in
frames
$O$ and
$O^{\prime}$ are related by
\begin{eqnarray}
\mathbf{E}^{\prime} & = &\gamma\left(\mathbf{E}
+(\mbox{\boldmath{$\beta$}}\times\mathbf{H})\right)\nonumber\\
\mathbf{H}^{\prime} & = & \gamma\left(
\mathbf{H}-(\mbox{\boldmath{$\beta$}}
\times\mathbf{E})\right)=0,
\label{eq:Lorentz}
\end{eqnarray}
where $\mbox{\boldmath{$\beta$}}=\mathbf{u}/c$. The
second relation is the Biot-Savart law. The
Maxwell stress tensor in $O^{\prime}$ is
\begin{equation}
\mbox{\boldmath{$\tau$}}^{\prime}=\gamma^{-2}\left(
\mathbf{E}\mathbf{E}-\half
E^2\mathbf{I}\right),\label{eq:Maxwell}
\end{equation}
where $\mathbf{I}$ is the unit tensor. From the second
equation in (\ref{eq:Lorentz}) we have
\[\mathbf{E}(\mbox{\boldmath{$\beta$}}\times\mathbf{H})=-
\beta^2\mathbf{E}\mathbf{E}.\]
With the aid of the tensor equality
\[\mathbf{E}(\mbox{\boldmath{$\beta$}}\times\mathbf{H})
=\mbox{\boldmath{$\beta$}}(\mathbf{E}\times\mathbf{H})+
\mathbf{H}(\mbox{\boldmath{$\beta$}}\times\mathbf{E})-
\mbox{\boldmath{$\beta$}}\cdot(\mathbf{E}\times\mathbf{H})
\mathbf{I},\]
the transformed absolute Maxwell stress tensor in $O$ is
\begin{eqnarray}
\mbox{\boldmath{$\tau$}}
& = & \mathbf{E}\mathbf{E}-\half E^2\mathbf{I}+
\mbox{\boldmath{$\beta$}}(\mathbf{E}\times\mathbf{H})
+\mathbf{H}(\mbox{\boldmath{$\beta$}}\times\mathbf{E})
-\mbox{\boldmath{$\beta$}}\cdot(\mathbf{E}\times\mathbf{H})
\mathbf{I}-\half\mathbf{E}\cdot(\mbox{\boldmath{$\beta$}}
\times\mathbf{H})\mathbf{I}\nonumber\\
& = &\mathbf{E}\mathbf{E}-\half E^2\mathbf{I}+
\mbox{\boldmath{$\beta$}}(\mathbf{E}\times\mathbf{H})+
\mathbf{H}\mathbf{H}-\mbox{\boldmath{$\beta$}}\cdot
(\mathbf{E}\times\mathbf{H})\mathbf{I}-\half
\mathbf{E}\cdot(\mathbf{E}\times\mbox{\boldmath{$\beta$}})
\cdot\mathbf{H}\mathbf{I}\nonumber\\
& = & \mathbf{E}\mathbf{E}+\mathbf{H}\mathbf{H}
-\half\left(E^2+H^2\right)\mathbf{I}+
\mbox{\boldmath{$\beta$}}(\mathbf{E}\times\mathbf{H}),
\label{eq:Maxwell-bis}
\end{eqnarray}
where we have used
$\mbox{\boldmath{$\beta$}}\cdot(\mathbf{E}\times\mathbf{H})=
\mathbf{H}\cdot(\mbox{\boldmath{$\beta$}}\times\mathbf{E})$.
Since $\mathbf{g}^{\prime}=0$, we are dealing with the
inverse Lorentz transformation, which amounts to changing the
sign of the relative velocity, $\mathbf{u}$, in all the
previous formulas. Hence, (\ref{eq:Maxwell-bis}) is
precisely (\ref{eq:tau-bis}), with the relative velocity
sign reversed,
\begin{equation}
\mbox{\boldmath{$\bar{\tau}$}}=\mathbf{E}\mathbf{E}+
\mathbf{H}\mathbf{H}-\half\left(E^2+H^2\right)\mathbf{I}
\label{eq:Maxwell-tris}
\end{equation}
is the relative Maxwell stress tensor in $O$, and
\[\mathbf{u}\otimes\mathbf{g}=\mbox{\boldmath{$\beta$}}\left(
\mathbf{E}\times\mathbf{H}\right),\]
is the difference between the absolute stress,
(\ref{eq:Maxwell-bis}) and the
relative stress (\ref{eq:Maxwell-tris}).
\par
Henceforth, we will consider the frame
$O^{\prime}$ to be moving at a relative velocity with respect
to
$O$ which we assume to be at rest. A system is said to be
{\em closed} when
the stress tensor $\mbox{\boldmath{$\tau$}}=0$ everywhere in
the body at rest
\cite{Moller}. In such a closed system,
the energy densities in the two frames are related by
$\varepsilon^{\prime}=\gamma^2\varepsilon$, and integrating
(\ref{eq:e-bis}) over the volume
$V$ leads to
\begin{equation}\bar{E}^{\prime}=E^{\prime}
-\mathbf{u}\cdot\mathbf{G}^{\prime}=
\frac{E}{\sqrt{1-\beta^2}}-\frac{\beta^2E}
{\sqrt{1-\beta^2}}=E\sqrt{1-\beta^2}\ge0, \label{eq:clock}
\end{equation} since the volume undergoes a
FitzGerald-Lorentz contraction in
$O^{\prime}$,
\begin{equation}
V^{\prime}=V\sqrt{1-\beta^2}. \label{eq:Fitz}
\end{equation} Therefore, in the closed
system, the proper energy, E, transforms like the
frequency of a clock, which is diminished by the motion,
while the total energy, $E^{\prime}=E/\sqrt{1-\beta^2}$, is
increased by the motion.\par Early in his career, de Broglie
\cite{deBroglie} was struck with the paradoxical difference
between
transverse Doppler effect, in which the frequency
is decreased by $\omega\sqrt{1-\beta^2}$, and the
relativistic variation of the frequency
$\omega/\sqrt{1-\beta^2}$, which is increased
by the motion. According to de Broglie, the latter frequency
is associated with the wave that is caused by mass in motion.
When a stationary clock with frequency
$\omega$ is set in motion, it will travel a distance $u\,dt$
in time
$dt$. For an observer who sees the clock go by at a velocity
$u$, the phase change is
$\omega\left(dt-u\,dt/v_{\rm p}\right)$, where
$v_{\rm p}=c/\beta$ is the phase velocity. During the same
time interval, another observer moves a distance $u\,dt$, and
experiences a phase change of
$\omega\sqrt{1-\beta^2}\,dt$. Since no distinction can be
made between the two phase variations,
\begin{equation}\frac{\omega}{\sqrt{1-\beta^2}}\left(
1-\frac{u}{v_{\rm
p}}\right)\,dt=\omega\sqrt{1-\beta^2}\,dt.
\label{eq:clock-bis}
\end{equation}
The momentum arising from the motion is
\[G^{\prime}=\frac{1}{v_{\rm
p}}\frac{\hbar\omega}{\sqrt{1-\beta^2}} \]
and
\[uG^{\prime}=\beta^2\frac{\hbar\omega}{\sqrt{1-\beta^2}}\]
is the work that is required to set the clock in motion,
where $\hbar$ is Planck's constant divided by $2\pi$.
Hence, (\ref{eq:clock-bis}) is precisely
(\ref{eq:clock}).\par In the presence of stresses acting on
the system,
$E^{\prime}$ no longer represents the total energy, which in
certain circumstances can even become negative
\cite[164]{Moller}. However, even in the presence of a
spatial tensor in the rest frame
$\mbox{\boldmath{$\tau$}}$, the relative energy
\begin{eqnarray}
\bar{E}^{\prime} & = &
\int\,\varepsilon^{\prime}\,dV^{\prime}-\int\,\mathbf{u}
\cdot\mathbf{g}^{\prime}\,dV^{\prime}
\nonumber\\
& = & \frac{E-\left(\mathbf{u}\cdot\int
\mbox{\boldmath{$\tau$}}dV\cdot\mathbf{u}\right)/c^2}
{\sqrt{1-\beta^2}}-\beta^2\frac{E-\left(\mathbf{u}\cdot
\int\mbox{\boldmath{$\tau$}}dV\cdot\mathbf{u}\right)/u^2}
{\sqrt{1-\beta^2}}\nonumber\\
& = &
E\sqrt{1-\beta^2},
\label{eq:Ebar}
\end{eqnarray}
is still independent of the
stresses acting on the system, and is positive semi-definite,
thus representing a true energy.
\par Specializing to the case where all non-diagonal
components of the stress tensor
vanish, the relative stress tensor
coincides with the negative pressure
\begin{equation}
\bar{\tau}_{ik}=-p^{\prime}\,\delta_{ik}=-p\delta_{ik},
\label{eq:p}
\end{equation}
where $\delta_{ik}$ is the Kronecker delta function.
The pressure $p$ is Lorentz-invariant \cite{Planck}. The
(positive) work done on the body to compress it by an amount
$dV^{\prime}$ is obtained from (\ref{eq:tau-bis}) as
\begin{equation}
-p^{\prime}dV^{\prime}=\tau^{\prime}
dV^{\prime}+\mathbf{u}\cdot d\mathbf{G}^{\prime}
=-\frac{p\,dV+\beta^2dE}
{\sqrt{1-\beta^2}}+
\beta^2\frac{dE+p\,dV}{\sqrt{1-
\beta^2}}=-p\,dV\sqrt{1-\beta^2}. \label{eq:pV}
\end{equation}
The volume $dV^{\prime}$ undergoes a FitzGerald-Lorentz
contraction in accordance with (\ref{eq:Fitz}).
\par
Adding (\ref{eq:pV}) to (\ref{eq:Ebar}) gives the
heat content
\begin{equation}
\bar{H}^{\prime}=\bar{E}^{\prime}+pV^{\prime}=
\left(E+pV\right)\sqrt{1-\beta^2}.
\label{eq:Hbar}
\end{equation}
If heat is conveyed to a body then the amount of heat that
is absorbed is equal to the increase in the internal
energy---provided the volume remains constant during the
process. However, the volume undergoes a FitzGerald-Lorentz
contraction so that the quantity of heat absorbed cannot be
equal to the increase in the internal energy. Rather, the
process occurs at constant pressure, since the pressure is
Lorentz-invariant. Consequently, the amount of heat
transmitted to the body is equal to the heat content
(\ref{eq:Hbar}).\par The relative motion of $O^{\prime}$ with
respect to $O$ can be likened to a Joule-Thomson process.
In a Joule-Thomson process, a gas is steadily transferred
from one compartment to another through a porous plug. The
adjective \lq steadily\rq\ implies that the pressure of the
two compartments is always the same. The invariance of the
pressure is maintained by constantly moving pistons on
both sides of the partition. Since the system is
isolated adiabatically, the change in the internal energy,
$\Delta E$ is equal to the difference in the work to move
the gas out of one compartment and into the other,
$-p\Delta V$. Hence,
the heat content will be invariant during the process,
$\Delta H=\Delta E+p\Delta V=0$.
\par Integrating the momentum transformation law
(\ref{eq:g}) over the same volume results in
\begin{equation}
\mathbf{G}^{\prime}=\int\,\mathbf{g}^{\prime}\,dV^{\prime}=
\frac{\mathbf{u}}{c^2}\frac{\left(E+pV\right)}
{\sqrt{1-\beta^2}},
\label{eq:G}
\end{equation}
or in terms of the relative energy,
\begin{equation}
\mathbf{G}^{\prime}=
\frac{\mathbf{u}}{c^2}\frac{\bar{E}^{\prime}
+pV^{\prime}}{1-\beta^2}.
\label{eq:G-bis}
\end{equation}
The momentum density $\mathbf{g}$ is not Lorentz-invariant.
\par
Under a Lorentz transformation, the energy transforms as
\cite[Doc. 45]{Einstein}
\begin{equation}
E^{\prime}=\bar{E}^{\prime}+\mathbf{u}\cdot
\mathbf{G}^{\prime}=\frac{E+\beta^2pV}
{\sqrt{1-\beta^2}}.
\label{eq:Eprime}
\end{equation}
The second term in (\ref{eq:Eprime}) is the \lq strange
result\rq\ referred to by Einstein in the Introduction.
However, merely by adding $pV$ to both sides of
(\ref{eq:Eprime}) converts the total energy into the total
heat content
\begin{equation}
H^{\prime}=E^{\prime}+pV^{\prime}=\frac{E
+pV}
{\sqrt{1-\beta^2}},
\label{eq:H}
\end{equation}
which transforms \textit{inversely} to the heat content
defined in (\ref{eq:Hbar}). Equation (\ref{eq:Eprime})
provides a whole new outlook on Einstein's
\lq strange result\rq\ \cite{Einstein}. According to
Einstein, even though there are no forces acting in $O$, a
change in its internal state, in
$O$, gives rise to an additional kinetic energy in
$O^{\prime}$. For a system at rest, the total energy
coincides with the internal energy, which is a function of
the entropy and volume. For a state in motion, the
change in the total energy,
\begin{equation}
dE^{\prime}=d\bar{E}^{\prime}+\mathbf{u}\cdot
d\mathbf{G}^{\prime},\label{eq:internal}
\end{equation}shows that the total energy is a function of
the momentum, in addition to the other thermodynamic
extensive variables. Moreover, the second term in
(\ref{eq:internal}) has the pivotal role of transforming
products of conjugate thermodynamic variables that decrease
by an amount
$\gamma^{-1}$ into inertial ones that increase as
$\gamma$ under a Lorentz transformation. Whereas
thermodynamic quantities are bounded, as
$\beta\rightarrow1$, inertial quantities are unbounded. Since
only differences in thermodynamic quantities can be
measured, the rest energy can always be subtracted out
giving the correct nonrelativistic limit.
\par
\section{The Second Law}
In the frame $O^{\prime}$, in relative motion with
respect to $O$, the first law is given by
\begin{equation}
dQ^{\prime}=dE^{\prime}-dW^{\prime}. \label{eq:first}
\end{equation}
It is conventional to write the work done on the
system as
\cite{Pauli}
\begin{equation}
dW^{\prime}=-p\,dV^{\prime}+\mathbf{u}\cdot
d\mathbf{G}^{\prime},
\label{eq:W}
\end{equation}
where the first term is the compressional work if
$dV^{\prime}<0$, and the second term is the work done in
the transformation from $O$ to $O^{\prime}$.
$dQ^{\prime}$ is the amount of heat
transferred to the system. Taken by itself, equation
(\ref{eq:first}) would lead one to identify $E^{\prime}$ with
the internal energy in the $O^{\prime}$ frame. However,
when (\ref{eq:W}) is introduced into (\ref{eq:first}) we
obtain the difference
$dE^{\prime}-\mathbf{u}\cdot d\mathbf{G}^{\prime}$, which,
according to (\ref{eq:internal}) is the change in internal
energy, $d\bar{E}^{\prime}$, in $O^{\prime}$. Neither
$\varepsilon^{\prime}$ nor
$\mathbf{u}\cdot\mathbf{g}^{\prime}$ is Lorentz-invariant,
only their difference is,
$\bar{\varepsilon}$. This is analogous to the
thermodynamic statement that neither the work nor the heat
conveyed to the system are exact differentials, only their
sum is, the internal energy.
\par Introducing (\ref{eq:W}) into (\ref{eq:first}) results
in
\begin{equation}
dQ^{\prime}=d\bar{E}^{\prime}+p\,dV^{\prime}=(d\bar{E}+p\,dV)
\sqrt{1-\beta^2}
=dQ\sqrt{1-\beta^2}. \label{eq:Q}
\end{equation}
Neither the total energy $E^{\prime}$ nor the momentum
$\mathbf{G}^{\prime}$ transform in a characteristic way
under a Lorentz transform; only the difference of the former
and the scalar product of the latter with $\mathbf{u}$
transform as the volume. The quantity of heat conveyed to
a body in motion is less than the quantity of heat conveyed
to the same body at rest.\par Were the internal energy to
be confused with the total relativistic energy
$E^{\prime}$ in (\ref{eq:first}), then
\begin{eqnarray}
dQ^{\prime} & = & dE^{\prime}+p\,dV^{\prime}=
d\bar{E}^{\prime}+p\,dV^{\prime}+\mathbf{u}\cdot
d\mathbf{G}^{\prime}\nonumber\\ & = &
\frac{dE+pdV}
{\sqrt{1-\beta^2}}=\frac{dQ}{\sqrt{1
-\beta^2}}. \label{eq:Q-wrong}
\end{eqnarray}
This is the result that would be obtained by equating the
change in the total relativistic energy,
$dE^{\prime}=dQ^{\prime}$, where
$dE^{\prime}=\gamma dE$ with the amount of heat
absorbed
\cite{Arzelies}. Apart from confusing the
total energy with the internal energy, equating the change in
the internal energy to the heat absorbed requires the volume
to remain constant, which it obviously cannot under a
Lorentz transformation.
\par
Not all extensive thermodynamic variables transform like the
volume. The number particles in the system and the number of
micro-complexions corresponding to a single macroscopic
state, or its logarithm, must be the same in all inertial
frames. This implies that the number of particles and the
entropy must be Lorentz-invariant,
\[
S^{\prime}=S. \]
At equilibrium, there are two equivalent representations:
entropy and internal energy. Since the entropy is invariant
while the internal energy is not, there is a
symmetry breaking between the internal energy and entropy
representations. The absolute temperature
$T^{\prime}$ is the integrating denominator of the heat
transferred,
\begin{equation}
\frac{dQ^{\prime}}{T^{\prime}}=dS^{\prime},
\label{eq:second}
\end{equation} and according to the second law it requires
the temperature to transform as
\cite{Planck}
\begin{equation}
T^{\prime}=T\sqrt{1-\beta^2}. \label{eq:T}
\end{equation}
This led Einstein \cite[Doc. 47]{Einstein} to conclude that
the temperature in a moving system is always lower than the
temperature of the reference system at rest. The product $TS$
decreases by a factor of
$\gamma^{-1}$ under a Lorentz transformation, like the proper
energy and volume. But, because the entropy is
Lorentz-invariant, the conjugate intensive variable must
undergo a FitzGerald-Lorentz contraction. This is in full
accord with the thermodynamic definition of temperature
\begin{equation}
\frac{1}{T^{\prime}}=\left(\frac{\partial
S}{\partial\bar{E}^{\prime}}\right)_{V^{\prime}}
=\left(\frac{\partial
S} {\partial
E}\right)_V\frac{1}{\sqrt{1-\beta^2}}=\frac{1}
{T\sqrt{1-\beta^2}},
\label{eq:T-bis}
\end{equation}
provided the \textit{internal} energy $\bar{E}^{\prime}$
transforms according to (\ref{eq:Ebar}). Contrarily, if
(\ref{eq:Q-wrong}) were valid, the temperature would
increase with the velocity,
\[
T^{\prime}=\frac{T}{\sqrt{1-\beta^2}}, \]
requiring that we equate the energy
in (\ref{eq:T-bis}) with the total relativistic energy.
\par
Combining the first, (\ref{eq:first}), and second,
(\ref{eq:second}), laws we get
\begin{equation}
T^{\prime}\,dS=dE^{\prime}+p\,dV^{\prime}-\mathbf{u}\cdot
d\mathbf{G}^{\prime}.
\label{eq:Euler}
\end{equation}
By confusing the total energy with the internal energy,
the entropy acquires a dependence upon the momentum, or if
the mass were constant, upon the velocity itself. This would
destroy the invariance of the entropy since it would not be
the same in all inertial frames of reference. Yet, if we
were to insist that $E^{\prime}$ in the differential form of
Euler's relation, (\ref{eq:Euler}), were the total energy,
the difference between
(\ref{eq:Eprime}) and the scalar product of the relative
velocity with the momentum (\ref{eq:G}) would give
\[\frac{E+\beta^2p\,V}{\sqrt{1-\beta^2}}-\beta^2\frac{
E+p\,V}{\sqrt{1-\beta^2}}=\bar{E}^{\prime}=
E\sqrt{1-\beta^2},\]
analogous to (\ref{eq:Ebar}), which is the true internal
energy. This eliminates any possible dependence of the
entropy upon the momentum because (\ref{eq:Euler}) reduces to
\begin{equation}
T^{\prime}\,dS=d\bar{E}^{\prime}+p\,dV^{\prime}.
\label{eq:Euler-bis}
\end{equation} Hence,
\textit{the relative energy,
$\bar{E}^{\prime}$, must be identified as the thermodynamic
internal energy}. And since the right-hand side of
(\ref{eq:Euler-bis}) transforms according to (\ref{eq:Ebar})
and (\ref{eq:pV}), the temperature will transform according
to (\ref{eq:T}).
\section{The Elusive Factor of $\frac{4}{3}$}
The vanishing of
the trace of the energy-momentum tensor in three dimensional
space gives the equation of state \cite{Laue}
\begin{equation}
p=\third\varepsilon. \label{eq:Laue}
\end{equation}
If this equation of state is characteristic of the electron,
it would necessarily assign to it a zero rest mass.
On the contrary, the equation of state (\ref{eq:Laue})
does not characterize the electrons, but, rather, the
electromagnetic interaction between them \cite{Landau}.
Moreover, the equation of state must be valid in any frame
and this identifies the internal energy density
as the Lorentz-invariant. In a system in relative motion, the
equation of state (\ref{eq:Laue}) is
\begin{equation}
p=\third\bar{\varepsilon},\label{eq:Laue-bis}
\end{equation}
and not
$3p=\varepsilon^{\prime}=\bar{\varepsilon}+ug^{\prime}$. It
is important to observe that the total energy density in any
Lorentz frame is equal to
$\gamma^2\varepsilon$ because the volume decreases by the
factor of
$\gamma^{-1}$ under a Lorentz transformation, and, at the
same time, the total energy increases by a factor of
$\gamma$. Therefore, the total energy density increases by
$\gamma^2$ on going from the rest frame to any other
inertial frame. If the energy density in (\ref{eq:Laue-bis})
were identified as the total energy density, it would
contradict the fact that the pressure is
Lorentz-invariant.
\par
According to (\ref{eq:G}), the transverse mass is still
given by
\begin{equation}
m_{\perp}=\frac{G^{\prime}}{u}=\fourthirds
\frac{E/c^2}{\sqrt{1-\beta^2}}.
\label{eq:transverse}
\end{equation}
The reason that (\ref{eq:transverse}) gives the
correct result is that the equation of state (\ref{eq:Laue})
was used in (\ref{eq:G}). Only in a state of motion is there
a distinction between the total and internal energies.
Even had we invoked the equation of state
(\ref{eq:Laue-bis}) in (\ref{eq:G-bis}) we would have still
obtained the same result because of (\ref{eq:Ebar}).\par
The electron's total energy resulting its
self-electromagnetic fields is given by (\ref{eq:Eprime}),
where \begin{equation}
E=\half\int\,\mathbf{E}\cdot\mathbf{E}\,dV=
\eighth\int_{r_0}^\infty\,\frac{e^2}{r^2}\,dr
=\eighth\frac{e^2}{r_0}=m_{\rm es}c^2 \label{eq:m-es}
\end{equation}
and $m_{\rm es}$ is the electrostatic mass.
This is the energy of formation of a spherical charge, where
$r_0$ is the \lq classical\rq\ radius of the electron. The
momentum (\ref{eq:transverse}) gives the correct expression
for the electromagnetic (transverse) mass,
$G^{\prime}/u=\fourthirds m_{\rm es}$, but this will not
coincide with the (\ref{eq:Eprime}), divided by $c^2$,
unless the term
$\third\gamma m_{\rm es}\beta^2$ is replaced by $\third\gamma
m_{\rm es}$ \cite{Poincare}. Quite ingeniously, the
momentum and energy were split into field \lq\lq $f$\rq\rq\
and mechanical
\lq\lq $m$\rq\rq\ components
\[G^{\prime}=G^{f \prime}+G^{m \prime}\;\;\;\;\;\;\;\;\; {\rm and}
\;\;\;\;\;\;\;\;\;E^{\prime}=E^{f
\prime}+E^{m \prime} .\]
Because $G^{f\,\prime}$ gave the correct result,
$G^{m\,\prime}$ was required to vanish. But because of the
discrepancy in the numerical value of the mass, both energy
components were necessary in order to obtain the desired
expression for the total energy. The mechanical component
of the energy was assumed to have the value
$E^{m}=\third m_{\rm es}c^2=-p^mV$. The pressure $p^{m}$ was
added to the field pressure,
\[p=p^f+p^m,\]
so as annul
the total pressure $p$ through the balance $p^f=-p^m$. The
negative pressure
$p^m$ was referred to as Poincar\'e's stress \cite{Lorentz}.
Then, since the inertial component satisfies the same Lorentz
transformation
\[
E^{m
\prime}=\gamma\left(E^{m}+\beta^2p^mV\right)
\]
as the field energy (\ref{eq:Eprime}), the energy related
to the Poincar\'e stress was found to be
\begin{equation}E^{m \prime}=\third m_{\rm
es}c^2\sqrt{1-\beta^2}.
\label{eq:Poincare}
\end{equation}
This was the non-electromagnetic contribution to the energy
that was required in order to make the energy expression,
\[E^{\prime}=\frac{m_{\rm es}c^2+\beta^2\third m_{\rm es}c^2}
{\sqrt{1-\beta^2}}+\third m_{\rm es}c^2\sqrt{1-\beta^2},\]
coincide with the expression for the momentum. Even
today, (\ref{eq:Poincare}) is associated with the work
done by the internal binding forces as the spherical
charge distribution accelerates and contracts
\cite{ADY}. The binding energy is proportional to the
volume, and this explains the factor $\gamma^{-1}$ in
(\ref{eq:Poincare}). The non-electromagnetic origin of
(\ref{eq:Poincare}) supposedly sounded the death knell for
the advocates of a purely
electromagnetic origin of the mass of an electron
\cite{Miller}.\par If the problem has anything more than
historical interest, it can be resolved much more simply.
Transferring our attention from the total energy to the total
heat content (\ref{eq:H}), which is conserved in an adiabatic
process, and evaluating it with
the equation of state (\ref{eq:Laue}) give
\begin{equation}
H^{\prime}=E^{\prime}+pV^{\prime}=
\left(E+pV\right)\sqrt{1-\beta^2}+
\beta^2\frac{E+pV}{\sqrt{1-\beta^2}}
=\fourthirds\frac{E}{\sqrt{1-\beta^2}},
\label{eq:H-bis}
\end{equation}
where the proper energy $E$ is given by (\ref{eq:m-es}).
Dividing through by the volume $V^{\prime}$ gives
\begin{equation}
\frac{h^{\prime}}{c^2}=\frac{\bar{h}+\mathbf{u}
\cdot\mathbf{g}^{\prime}}{c^2}=
\gamma^2\varrho_{\rm em},
\label{eq:hprime}
\end{equation}
where
\[\mathbf{g}^{\prime}=\gamma^2\varrho_{\rm
em}\mathbf{u}\]
is the momentum density, and
\begin{equation}
\varrho_{\rm
em}=\frac{\varepsilon+p}{c^2}=\frac{\bar{h}}{c^2}=
\fourthirds\frac{\varepsilon}{c^2}\label{eq:varrho}
\end{equation}
is Lorentz-invariant mass density. The electromagnetic mass
density, (\ref{eq:varrho}) is precisely Planck's
\cite{Planck} generalization.
\par Planck referred to
(\ref{eq:varrho}) as the
\lq\lq law of inertia of energy\rq\rq; the corresponding
momentum density
$\mathbf{u}\,\bar{h}/c^2$ was in the same direction as
Poynting's vector, $\mathbf{u}\,\bar{h}$, for the flow of
energy. It is unexplainable why Planck failed to realize
that mass should be related to a conserved quantity in an
adiabatic process. Perhaps he wanted to salvage the
mass-energy correspondence all
costs since their densities were off by a mere numerical
factor in (\ref{eq:varrho}). This elusive
$\fourthirds$ factor bears witness to the fact that the
heat content is proportional to the relativistic mass, where
the $\third$ factor comes from the equation of state
(\ref{eq:Laue}). It was therefore incorrect
to associate the mass of an electron with $E^{\prime}/c^2$,
since
its density is not Lorentz-invariant like the mass density
(\ref{eq:varrho}). It was also incorrect, and unnecessary,
to divide the energy and momentum into electromagnetic and
inertial contributions. Energy does not transform into
energy under a Lorentz transformation. An additional
factor, (\ref{eq:Poincare}), was added to make sure that it
did. The presence of the term
$pV^{\prime}$ in the heat content (\ref{eq:H}) means that a
particle cannot be compressed to a point particle.
The relativistic electromagnetic mass should have been
associated with the heat content $\bar{H}^{\prime}/c^2$,
rather than the total energy,
$E^{\prime}/c^2$, because the latter is not conserved
in an adiabatic process in which the volume is not
constant.\par
\section{Relativistic Generalization of the First Law}
If the momentum is introduced as the
derivative of the \textit{total} heat content,
(\ref{eq:H}),
\[\frac{\partial H^{\prime}}{\partial
u_k}=G^{\prime}_k,\]
with respect to the velocity, then the Legendre transform,
\begin{equation}
H^{\prime}-u_k\frac{\partial H^{\prime}}{\partial
u_k}=H^{\prime}-\mathbf{u}\cdot\mathbf{G}^{\prime}=
(E+pV)\sqrt{1-\beta^2} =
\bar{H}^{\prime}, \label{eq:Legendre}
\end{equation}
introduces the heat content (\ref{eq:Hbar}), as its dual.
The velocity at which energy is propagated,
\begin{equation}
\mathbf{u}=\frac{\mathbf{G}^{\prime}}{H^{\prime}/c^2},
\label{eq:group}
\end{equation}
is in the direction of the Poynting vector.
Introducing this velocity into the Legendre
transform (\ref{eq:Legendre}) allows us to cast the latter
in the form
\begin{equation}
\mathbf{G}^{\prime}\cdot\mathbf{G}^{\prime}-
\frac{H^{\prime\,2}}{c^2}=-\frac{H^{\prime\,2}}{c^2}
\left(1-\beta^2\right)=-\frac{H^2}{c^2}=
\mbox{invariant},
\label{eq:G-inv}
\end{equation}
because $H^{\prime}=\gamma^2\bar{H}^{\prime}$, according to
(\ref{eq:Hbar}) and (\ref{eq:H}). This is the relativistic
thermodynamic Hamilton-Jacobi equation. The heat content
is the invariant corresponding to the rest mass in the
usual relativistic Hamilton-Jacobi equation. The invariance
of (\ref{eq:G-inv}) shows that
\begin{equation}G_i^{\prime}=\left(\mathbf{G}^{\prime},
\frac{i}{c}H^{\prime}\right)
\label{eq:4-G}
\end{equation}
is a 4-vector. The 4-momentum (\ref{eq:4-G}) is a
time-like 4-vector.
\par
Electromagnetic forces acting on a particle can
produce changes in its rest mass through Joule heating,
thereby increasing the rest mass of the particle. According
to (\ref{eq:Hbar}) and (\ref{eq:G-bis}), the total rate of
change of the heat content is given by
\[
\frac{d \bar{H}^{\prime}}{d\tau}
+\mathbf{u}\cdot
\frac{d\mathbf{G}^{\prime}}{d\tau}=\dot{Q}^{\prime}+
\frac{\beta^2}{1-\beta^2}\dot{Q}^{\prime}=
\frac{\dot{Q}^{\prime}}{1-\beta^2}=\frac{dH^{\prime}}{d\tau},
\] where
$\dot{Q}^{\prime}=
d\left(\bar{E}^{\prime}+pV^{\prime}\right)/d\tau$
is the rate of heating at constant pressure in $O^{\prime}$.
The heat content decreases as
$\gamma^{-1}$,
\[ \bar{H}^{\prime}=H\sqrt{1-\beta^2},\]
under a Lorentz transformation, whereas the total heat
content increases by a factor of $\gamma$,
\[H^{\prime}=\frac{H}{\sqrt{1-\beta^2}}.\]
The former is a thermodynamic quantity, behaving as the
proper volume under a Lorentz transformation, while the
latter is a inertial quantity, transforming as the proper
mass of a point particle.\par However,
since the proper time is
\begin{equation}
\tau=\sqrt{1-\beta^2}\,t, \label{eq:time}
\end{equation}
the rates of heating
are the same in $O$ and $O^{\prime}$,
\begin{equation}
\dot{Q}^{\prime}=\frac{d\bar{H}^{\prime}}{d\tau}=
\frac{d\bar{H}}{dt}=\dot{Q}. \label{eq:Q-dot}
\end{equation}
The equality of the total rates of heating, as measured in
the two inertial frames, $O$ and $O^{\prime}$, is due to the
compensation of the FitzGerald-Lorentz contraction of the
volume (\ref{eq:Fitz}) and the dilation of time
(\ref{eq:time}). Since the rates of heating are the same in
the two frames, while the quantity of heat conveyed is less
in the moving frame than the one at rest, some of the heat
is being converted into work in $O^{\prime}$. That work is
necessary to keep the body in a state of uniform motion
[\textit{vid}. (\ref{eq:W-bis})].
\par
Consider an infinitesimal proper volume element, $\delta V$,
such as the volume occupied by a particle.
The rate at which heat is being conveyed to the particle in
$O$ is
$\dot{Q}=\dot{q}\delta V$, where $\dot{q}$ is the rate of heating per unit volume,
while, in $O^{\prime}$, it is
$\dot{q}^{\prime}\delta V^{\prime}$. On the strength of the
invariance of the rates of heating (\ref{eq:Q-dot}) and the
FitzGerald-Lorentz contraction, (\ref{eq:Fitz}), of the
proper volume $\delta V$, it follows that the density of
rate of heating in $O^{\prime}$ is greater than in $O$,
\begin{equation}
\dot{q}^{\prime}=\frac{\dot{q}}{\sqrt{1-\beta^2}}.
\label{eq:q-dot}
\end{equation}
Multiplying both sides of (\ref{eq:q-dot}) by the proper
time interval gives
\begin{equation}
q=q^{\prime}, \label{eq:q-Lorentz}
\end{equation}
showing that the density of heat is Lorentz-invariant. This
was to be expected from the transformation property of the
heat (\ref{eq:Q}), and the FitzGerald-Lorentz contraction
(\ref{eq:Fitz}) of the volume.
\par
Since the specific volume
$1/\delta V$ satisfies the continuity equation, the particle
volume will satisfy
\begin{equation}
\frac{d}{dt}\delta V=\delta
V\,\mbox{\boldmath{$\nabla$}}\cdot\mathbf{u}.
\label{eq:V}
\end{equation}
The is an expression for the conservation of matter, as
distinct from that of energy \cite{Eckart}. Writing
$\bar{H}=\bar{h}\delta V$ for the heat content in
(\ref{eq:Q-dot}), and using (\ref{eq:V}), we obtain the
balance equation
\begin{equation}
\frac{d\bar{h}}{dt}+\bar{h}\mbox{\boldmath{$\nabla$}}
\cdot\mathbf{u}=\frac{\partial\bar{h}}{\partial
t}+\mbox{\boldmath{$\nabla$}}
\cdot\bar{h}\mathbf{u}=\dot{q},
\label{eq:h}
\end{equation}
for the heat content density.
Introducing the mass density, according to
(\ref{eq:varrho}), in (\ref{eq:h}) gives \cite[eqn (227), p.
135]{Moller}
\begin{equation}
\frac{\partial\varrho_{\rm em}}{\partial t}+
\mbox{\boldmath{$\nabla$}}\cdot\varrho_{\rm em}\mathbf{u}=
\frac{\dot{q}}{c^2}. \label{eq:Moller}
\end{equation}
Multiplying both sides of (\ref{eq:Moller}) by $\delta V$,
integrating by parts, and taking into account (\ref{eq:V}),
result in
\begin{equation}
\frac{d}{dt}\delta m_{\rm em}=\frac{\delta\dot{Q}}{c^2}.
\label{eq:new}
\end{equation}
No longer can the distinction be made between the canonical
and grand canonical ensembles, which asserts that heat can be
transferred without matter, but matter cannot be transferred
without heat. Expression (\ref{eq:new}) places heat and
matter on the same level through the equivalence of the rate
of heating in the rest frame and the rate of increase of the
rest mass.
\par
Although this relation appears in M{\o}ller \cite[eqn (64),
p. 107]{Moller}, he still refers to it as a consequence of
(\ref{eq:Einstein}).\footnote{M{\o}ller writes the equation
of motion in the covariant form
\[\frac{d}{d\tau}\delta G_i=\delta F_i,\]
where $\tau$ is the proper time interval, and the
4-momentum $\delta G_i=\delta m_{\rm em}U_i$ in our
notation. Because $\tau$ is invariant, it guarantees that the
two 4-vectors in the equation will transform in the same way
in every system of inertia. Yet, when he arrives at
(\ref{eq:new}),
$\tau$ has now become \lq\lq identical with the time in the
rest system.\rq\rq\ The latter is the correct identification
of
$\tau$.} However, which energy in (\ref{eq:Einstein}) is
is not specified. Rather, (\ref{eq:new})
reinterprets (\ref{eq:Einstein}) insofar as it asserts that
the only way the rest mass can increase is through
the rate of heating. Moreover, mass conservation, $\delta
m_{\rm em}=\mbox{constant}$, corresponds to the conservation
of the heat content, $H=\mbox{constant}$, corresponding to an
adiabatic process. The energy cannot be conserved in all
systems of inertia because the volume does not remain
constant in all inertial frames.\par If (\ref{eq:Moller}) is
now multiplied by
$\delta V^{\prime}$, and an integration by
parts is performed, there results
\begin{equation}
\frac{d}{d\tau}\delta m^{\prime}_{\rm
em}=\frac{\dot{Q}^{\prime}}{c^2}. \label{eq:new-bis}
\end{equation}
Based on the invariance of the rates of heating in the two
inertial frames, the mass also undergoes a
FitzGerald-Lorentz contraction
\begin{equation}
\delta m_{\rm em}^{\prime}=\delta m_{\rm em}\sqrt{1-\beta^2}.
\label{eq:m-Lorentz}
\end{equation}
An immediate consequence of (\ref{eq:m-Lorentz}) is that
the electromagnetic mass density $\varrho_{\rm em}$ is an
invariant. This is implied by the invariance of the
pressure and the internal energy density in
(\ref{eq:varrho}). Because mass occupies volume, the
FitzGerald-Lorentz contraction of mass in motion
(\ref{eq:m-Lorentz}) is a natural consequence. That mass in
motion would behave in this manner, viewed from an observer
in the rest frame, would be a direct confirmation of a
finite volume of the electron. However, direct measurement
would not be easy since it is dwarfed by the work done by
to keep the particle in a state of steady motion in the
expression for the total heat content
\begin{equation}\delta H^{\prime}=\delta\bar{H}^{\prime}+
\mathbf{u}\cdot\delta\mathbf{G}^{\prime}
=\delta m_{\rm em}c^2\sqrt{1-\beta^2}+\frac{\delta m_{\rm em}
u^2}{\sqrt{1-\beta^2}}=\frac{\delta m_{\rm em}c^2}{\sqrt{
1-\beta^2}}.\label{eq:Hprime}
\end{equation}
\par
In summary, thermodynamic quantities decrease by a factor of
$\gamma^{-1}$ under a Lorentz transformation, and have
Lorentz-invariant densities, while inertial quantities
increase by a factor of $\gamma$. However, mass belongs to
both categories: Because of its inertia, the mass increases
by a factor of
$\gamma$ under a Lorentz transformation, while because mass
occupies volume, it will decrease by a factor of
$\gamma^{-1}$ under the same transformation. For a free
particle, the combined effect (\ref{eq:H}) shows that the
inertial contribution prevails.\footnote{An exception would
be an electron bound to a lattice, where its motion
bears little resemblance to that of a free particle. In
such a system, the first term in (\ref{eq:Hprime}) would
prevail, giving rise to the well-known energy band structure
in solids. Chopping up space-time into discrete units of the
Compton wavelength and the Compton wavelength divided by
$c$, respectively, has the effect of introducing a lattice
structure for free particle motion giving a bounded energy
spectrum identical to (\ref{eq:m-Lorentz})
\cite{Lav01}. }
\par
\par The transfer of heat gives rise to a reactive force
which maintains the kinematical constraint that the moving
frame be inertial. This can be shown by defining a
4-velocity
\begin{equation}
U_i=\left(\frac{\mathbf{u}}{\sqrt{1-\beta^2}},
i\frac{c}{\sqrt{1-\beta^2}}\right), \label{eq:4-U}
\end{equation}
and a 4-force
\begin{equation}
\delta
F_i^{\prime}=\left(\frac{\delta\mathbf{F}^{\prime}}
{\sqrt{1-\beta^2}},
\frac{i}{c}\frac{\delta\dot{Q}^{\prime}}{\sqrt{1-\beta^2}}\right),
\label{eq:4-F}
\end{equation}
acting on a particle of volume $\delta V^{\prime}$ in the
frame $O^{\prime}$. It is important to bear in mind that the
force
$\delta\mathbf{F}^{\prime}$ has a non-mechanical origin;
it arises when ever heat is conveyed to the particle
through convection or heat radiation.\par The equation of
motion is
\begin{equation}
U_i\frac{d}{d\tau}\delta m_{\rm em}^{\prime}+
\delta m_{\rm em}^{\prime}\frac{dU_i}{d\tau}=\delta
F_i^{\prime}.
\label{eq:U-dot}
\end{equation}
Projecting (\ref{eq:U-dot}) in the direction of $U_i$, we
get
\begin{equation}
-c^2\frac{d}{d\tau}\delta m_{\rm em}^{\prime}=
\frac{\mathbf{u}\cdot\delta\mathbf{F}^{\prime}-
\delta\dot{Q}^{\prime}}{1-\beta^2}=-\delta\dot{Q} ,
\label{eq:Q-inv}
\end{equation}
by noting that $U_iU_i=-c^2$, and $U_i\,dU_i/d\tau=0$. The
last equality in (\ref{eq:Q-inv}) follows from
(\ref{eq:new-bis}), and the invariance of the rate of
heating (\ref{eq:Q-dot}).
\textit{The rate of heating must always be greater than the
power which it generates},
\[
\delta\dot{Q}^{\prime}>\mathbf{u}\cdot
\delta\mathbf{F}^{\prime}.
\]
This inequality can be taken as a kinetic
statement of the second law, which prohibits the 4-force
(\ref{eq:4-F}) from ever becoming space-like. Since the force
$\delta\mathbf{F}^{\prime}$ arises as a consequence of
heating, and is not an external force which can act on the
system in the absence of heat absorption, the power which
it generates can never exceed the rate at which the
system is being heated.\par Solving (\ref{eq:Q-inv}) for the
power results in
\begin{equation}
\mathbf{u}\cdot\delta\mathbf{F}^{\prime}=\beta^2
\delta\dot{Q}^{\prime},
\label{eq:power}
\end{equation}
testifying to the fact that the rate of heating is always
greater than the power that it produces. It is easily
verified by introducing (\ref{eq:new-bis}) into the equation
of motion (\ref{eq:U-dot}), and rearranging to give
\begin{equation}\delta m^{\prime}_{\rm
em}\frac{dU_i}{d\tau} =\delta
F_i^{\prime}-\frac{\delta\dot{Q}^{\prime}}{c^2}U_i.
\label{eq:U-dot-bis}
\end{equation}
Equation (\ref{eq:U-dot-bis}) is comparable with the
usual relativistic equation of motion
\begin{equation}m\frac{d\mathbf{u}}{d\tau}=
\mathbf{F}^{\prime}-\frac{\mathbf{u}}{c^2}
\left(\mathbf{F}^{\prime} \cdot
\mathbf{u}\right)\label{eq:mechanical}
\end{equation}
where $\mathbf{F}^{\prime}$ is the vector force acting
on a particle of rest mass $m$. The second term in
(\ref{eq:mechanical}) arises from the mass dependence upon
the velocity, and it is analogous to the second term
in (\ref{eq:U-dot-bis}), which is due to the rate of
increase of the proper mass through heating. Both of
these components of the force are in the direction
of the motion and can, consequently, generate power.
Scalar multiplication of (\ref{eq:U-dot-bis}) by
$U_i$ gives precisely
(\ref{eq:power}) because it is equivalent to the condition
$\delta F^{\prime}_iU_i+\delta\dot{Q}^{\prime}=0$.\par This
last condition enables us to write the equation of motion
(\ref{eq:U-dot-bis}) as
\[\delta m_{\rm em}^{\prime}\frac{dU_i}{d\tau}
=\delta\mathcal{F}_i^{\prime},\]
where
\begin{equation}\delta\mathcal{F}_i^{\prime}=\delta
F_i^{\prime}+U_i\frac{\delta
F^{\prime}_kU_k}{c^2}=\left(\frac{\delta\mathbf{F}^{\prime}-
\mathbf{u}\,\delta\dot{Q}^{\prime}/c^2}
{\sqrt{1-\beta^2}},0\right)\label{eq:reactive}
\end{equation}
is the \lq reactive\rq\ force, analogous to a genuine
mechanical 3-force. It is the only force present in the
absence of a mechanical force, and it satisfies the
condition
$\delta\mathcal{F}^{\prime}_iU_i=0$, on account of
(\ref{eq:power}). The fourth component of the 4-force
usually expresses the power produced by mechanical
forces, and the fourth component of the equation of motion
is a statement of energy conservation. In the case of the
4-force (\ref{eq:4-F}), the fourth equation of motion
(\ref{eq:new}) is a statement of the first law,
\textit{i.e.}, the balance of energy. This balance of
energy is incorporated into the first three equations of
motion in the reactive force (\ref{eq:reactive}).\par
Consequently, the force necessary to keep the particle in a
uniform state of motion is \cite[p. 107]{Moller}
\begin{equation}\delta\mathbf{F}^{\prime}=
\frac{\delta\dot{Q}^{\prime}}{c^2}\mathbf{u}.
\label{eq:F}
\end{equation}
It is
well-known that the component of the force parallel to the
relative motion is an invariant
\cite[p. 74]{Moller}. A property of any 4-vector is that it
forms an invariant with itself. If
(\ref{eq:4-F}) is a 4-force, then it must follow that
\[\delta
F_i^{\prime}\delta
F^{\prime}_i=\gamma^2\left(\delta\mathbf{F}^{\prime}\cdot\delta\mathbf{F}^{\prime}
-\frac{\delta\dot{Q}^{\prime\,2}}{c^2}\right)=
-\frac{\delta\dot{Q}^{\prime\,2}}{c^2}=\mbox{invariant},\]
where use has been made of (\ref{eq:F}).
The 4-force is time-like, and it confirms our
previous result, (\ref{eq:Q-dot}), that the rate of heating
is, indeed, an invariant.
\par Multiplying (\ref{eq:power}) by the time interval
$\Delta\tau$ in which heat has been conveyed to the particle
gives the work done to keep the particle in a state of
uniform motion,
\begin{equation}
\Delta W^{\prime}=-\beta^2\Delta Q^{\prime}, \label{eq:W-bis}
\end{equation}
where $\Delta Q^{\prime}=\dot{Q}^{\prime}\Delta\tau$
represents the total amount of heat transmitted to the
particle in the proper time interval $\Delta\tau$. Since
work has been done, the quantity of heat that
is converted into internal energy will be diminished by a
factor of $\gamma^{-2}$.
\par
Any 4-vector not only forms an invariant with itself, but it
does so with any other 4-vector. In an adiabatic
process
\begin{equation}\delta
G_i^{\prime}U_i=\frac{\delta\mathbf{G}^{\prime}\cdot
\mathbf{u}-\delta H^{\prime}}{\sqrt{1-\beta^2}}=-\frac{
\delta\bar{H}^{\prime}}{\sqrt{1-\beta^2}}=-\delta H=
\mbox{invariant},\label{eq:GU}
\end{equation}
while in any irreversible process in which heating occurs
\begin{equation}\delta F_i^{\prime}U_i=
\frac{\delta\mathbf{F}^{\prime}\cdot\mathbf{u}-
\delta\dot{Q}^{\prime}}{1-\beta^2}=-\delta\dot{Q}=
\mbox{invariant}.\label{eq:FU}
\end{equation}
The latter is precisely (\ref{eq:Q-inv}). The condition of
adiabaticity (\ref{eq:GU}), $\delta H=\mbox{constant}$, is
analogous to Newton's first law,
$\mathbf{G}=\mbox{constant}$. Likewise, (\ref{eq:FU}),
$d\delta H/dt=\delta\dot{Q}$, is analogous to Newton's
second law, $d\mathbf{G}/dt=\mathbf{F}$. \par
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\end{document}