# Relativistic Thermodynamics: A Modern 4-Vector Approach

Authors: J. Güémez
Journal: Physics Research International
 (2011)

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## Abstract

Using the Minkowski relativistic 4-vector formalism, based on Einstein's equation, and the relativistic thermodynamics asynchronous formulation (Grøn (1973)), the isothermal compression of an ideal gas is analyzed, considering an electromagnetic origin for forces applied to it. This treatment is similar to the description previously developed by Van Kampen (van Kampen (1969)) and Hamity (Hamity (1969)). In this relativistic framework Mechanics and Thermodynamics merge in the first law of relativistic thermodynamics expressed, using 4-vector notation, such as ΔUμ  =  Wμ  +  Qμ, in Lorentz covariant formulation, which, with the covariant formalism for electromagnetic forces, constitutes a complete Lorentz covariant formulation for classical physics.


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## Body

## 1. Introduction

During the 1960s and 1970s many physicists devoted considerable effort to finding the most adequate relativistic formulation of thermodynamics [1]. Yuen's 1970 paper [2] presents the state of the art on relativistic thermodynamics at this time. After the work by Van Kampen [3] and Hamity [4] introducing 4 vectors in thermodynamics and the clearly stated asynchronous formulation by Gamba [5], Cavalleri and Salgarelli [6], and Grøn [7] a consensual relativistic thermodynamics formalism should have been achieved. However, no agreement on the correct Relativistic thermodynamics was reached [8] ([9], pp. 303–305). Until recently, papers on this topic have been published [10], mainly on relativistic transformation of temperature [11, 12].LetZ be a composite body, which moves, in a reference frame S, under the action of k external (conservative and nonconservative) forces Fk=(Fxk,Fyk,Fzk), simultaneously applied during time interval dt, with resultant force ℱ=(∑kFk)=(ℱx,ℱy,ℱz), impulse ℐ=ℱdt, with nonzero work δWext≠0 (only conservative forces perform work) and that experiences a certain thermodynamic process, with internal energy variation dU≠0 and heat δQ≠0. In classical physics, the complete description of this process, expressed in Galilean covariant form, is given by [13]: (i) a vectorial equation (linear momentum-impulse equation) dp=ℐ:(1){dpxdpydpz}={∑kFxkdt∑kFykdt∑kFzkdt}
and (ii) a scalar equation (first law of thermodynamics or energy equation) [14]:(2)dKcm+dU=δWext+δQ.
From (1) the following equation can be obtained: (3)dKcm=F⋅dxcm,
or the center of mass equation [15], where dxcm is the displacement of the center of mass (cm) of Z and dKcm is its kinetic energy variation throughout the process.For an observer in frameSA in standard configuration with respect to frame S, with velocity V=(V,0,0) (Appendix A), it has the corresponding equations (4)dKcmA=F⋅dxcmA,dKcmA+dU=δWextA+δQ,
where the corresponding magnitudes are measured in SA, the mass, force, interval of time, impulse ℐ, linear momentum variation dp, internal energy dU, and heat δQ are Galilean invariants, and magnitude velocity vA=v-V, displacement dxcm=dxcmA-Vdt, kinetic energy dKcmA=dKcm-Vdp, and work δWextA=δWext-Vℐ, have their specific Galilean transformation [16].It is interesting to note that when forces applied toZ have an electromagnetic origin, with some force Fk obtained from “Lorentz force” equation Fk=q(E+v×B) (q electric charge, E electric field, and B magnetic field), the whole formalism is neither covariant under Galilean transformations (Lorentz force is not Galilean covariant [17]) nor covariant under Lorentz transformations (the previous thermodynamics formalism is not Lorentz covariant), in contradiction with Einstein's principle of inertia.After these considerations about Galilean relativistic thermodynamics, not compatible with electromagnetic interactions, it seems necessary to obtain a formalism for the first law of thermodynamics expressed according to the principles of the special theory of relativity, that is, Lorentzian relativistic thermodynamics, compatible with electromagnetic interactions. As a result of this, it will be possible to obtain a Lorentz covariant formalism for exercises in classical physics that include concepts of mechanics, thermodynamics, and electromagnetism.A modern view of a relativistic thermodynamics theory requires a clear definition of (i) the tensorial objects which characterize the equilibrium state of the system and of (ii) any tensorial object that characterizes the interaction of the system with its mechanical (work reservoir ([18], Chap. 3)) and thermal (heat reservoir ([18], pp. 89-90)) surroundings, with a prescription of the apparatus which measures it. The observables will depend, in general, on the physical system and on the observer (Appendix A), but the principle of relativity ensures that all inertial observers obtain equivalent descriptions of the same process. So, any relativistic formalism developed to describe a physical process must be according to this principle,  that is, it must be Lorentz covariant. This is the course chosen in this paper, in which we solve an exercise on the isothermal (nonquasistatic) compression of an ideal gas in the reference frame S0 in which the system is at rest and in a frame SA, in standard configuration to S0 (Appendix A), using the Minkowski 4-vectors—related through Lorentz transformations [19]—and a Lorentz covariant form for the first law of thermodynamics.The paper is arranged as follows. In Section2 the formalism, based on the principle of the inertia of energy (Einstein's equation) and on the asynchronous formulation, is developed. After that, in Section 2.3 the principle of similitude is enunciated, expressing the conditions under which the same equations can be used for an elementary particle and for a composite system. Section 3 presents the 4-vector energy function Uμ for different systems. The asynchronous formulation of 4-vector work Wμ is obtained in Section 4. In Section 5 thermal radiation 4-vector (heat) Qμ, based on photons, is introduced. In Section 6 the mathematical formulation of the relativistic thermodynamics first law is presented in Lorentz covariant form. In Section 7 the isothermal compression, by two pistons, of an ideal gas is solved by using the previously developed formalism in both frames S0, zero momentum frame, and SA, in standard configuration respect S0. Forces on pistons are described using an electromagnetic interaction, in its relativistic Lorentz covariant form. Finally, Section 8 proposes some conclusions regarding the possibility of solving exercises in classical physics in a complete Lorentz covariant form. Although we assume that the reader is familiar with the Minkowski 4-vector formalism, in Appendix A a brief review on 4-vectors and Lorentz transformation algebra is provided introducing the “metric tensor” gνμ and the “Lorentz transformation” ℒνμ(V) used in the paper [20].


## 2. Relativistic Thermodynamics Formalism

Relativistic thermodynamics formalism is developed in two steps: (i) Einstein’s equationE0=mc2, expressed as the principle of the inertia of energy, which allows us to obtain energy function U and the 4-vector energy function Uμ for a given system; (ii) the asynchronous formulation, that will allow us to obtain the work W performed by forces acting on a system and the 4-vector work Wμ. As a consequence, the principle of similitude can be formulated, according to which, and under very general circumstances, a composite system behaves as a whole in its interactions with its surroundings and equations for an elementary particle can be used with a composite, deformable system.


### 2.1. Inertia of Energy

It could be considered, in a broad sense, that the main goal of relativistic thermodynamics is to reach a unified description on point dynamics and extended-body dynamics [13].In order to ensure that an extended body behaves like a “single particle” interacting with its surroundings—work reservoirs or thermal bath—and so that it is physically meaningful to use Lorentz transformations, it is necessary that all forms of energy that make up the body contribute in the same way to its inertia [21]. These forms of energy must include those related with the mass of its constituent elementary particles, binding—nuclear, chemical, and so forth—energies (Figure 1), internal kinetic energy (see Section 7.1.1), electrostatic energy [22], and so forth, and energy of thermal radiation in equilibrium with matter inside the system [23] (see Section 5).Figure 1
An atom (A)—self-contained structure—is obtained from a nucleus (N), previously assembled from protons (p) and neutrons (n), and an ensemble of electrons (e). Nucleus inertiaℳN=UNc-2 decreases respect the inertia of its elementary particle components 𝒰N=6mp+4mn, due to the energy UN-𝒰N=-ŨN=-8hν′ released in its formation. Atom inertia ℳA=UAc-2 decreases with respect to the inertia of its component nucleus and electrons 𝒰A=6me+ℳN due to the energy UA-𝒰A=-ŨA=-8hν released in its formation.Einstein's EquationE0=mc2 for an extended body can be interpreted by relating its inertia—a body's reluctance to undergo a change in velocity [24]—with energy function [25]—energy content of the physical system or internal energy [26].Principle of the Inertia of Energy
for an extended body in complete equilibrium, any kind of energy inside the system, relativistically expressed in reference frameS0 in which the system as a whole is at rest, contributes to the energy functionU of the system [27]. Considering that all forms of energy are convertible between them [28] the inertiaℳ of a system [29] in equilibrium is ([30], p. 163)
(5)M=Uc-2.According to Einstein [31, 32] the inertia of a body changes with its content of energy [33] (Section 3).It is possible to define the inertiaℳ of a body (we prefer the term inertia, instead of mass [34], to avoid confusions when the system includes photons (Section 5)) as [35]: the inertia ℳ of a composite body equals the sum of its elementary particles mass (protons, neutrons, and electrons) m0: (6)m0=∑jmp+∑kmn+∑lme,
with energy 𝒰=m0c2, minus the minimum energy Ũ, divided by c2, necessary to separate its elementary particles so that they are far apart (Section 7.1.1): (7)M=(U-Ũ)c-2=m0-Ũc-2,
with U=𝒰-Ũ.


### 2.2. Asynchronous Formulation

For an extended, deformable body a relativistic theory cannot be directly formulated in an arbitrary inertial frame. It must be based on known prerelativistic descriptions. On the one hand, it seems necessary to maintain the classical concept that the resultant force on the body must be zero (zero total impulse) when the motion remains uniform and to assure that when no torque is applied to the system in a certain reference frame, no torque is applied to it in another frame [36]. On the other hand, in classical mechanics forces on an extended system are applied simultaneously. This simultaneity occurs in all inertial frames. In thermodynamics, heat is a kind of interchanged energy with (assumed implicitly) zero linear momentum.According to Cavalleri and Salgarelli, when forces on an extended, composite, body are applied, in order to develop a coherent formalism for relativistic thermodynamics, a privileged observer must exist, in reference frameS0, that performs experiments on the body that remains at rest (Figure 2) [6].Figure 2
Compression process in frameS0. A set of external forces Fk (k=1,2,3,4) are applied on an extended, deformable system during the same time interval Δt, as measured in frame S0, with zero total impulse and zero torque. The kth force has associated the displacement Δrk (r=(x,y)) and 3-vector velocity vk=Δrk/Δt. The center of mass (cm) of the system does not move during the process.According to Gamba [5]:“in theAsynchronous Formulation, observers in frames S0 and SA refer to the same experiment (the experiment performed in the privileged frame S0) and obtain its own physical magnitudes, expressed as 4-vectors. In this formulation both descriptions of the experiment are connected by true Lorentz transformations [19].”The observer inS0 takes an ideal surface, at rest, which delimits the system considered and measures energy, work or heat, interchanged through the surface during time interval Δt. An observer in SA obtains the same magnitudes by true Lorentz transformations, from the events considered by an observer in S0. The observer in SA does not perform a similar experiment to observer in S0 (synchronous formulation [37]), it just translates the experiment performed in S0 to its own physical magnitudes. Owing to the relativity of simultaneity, forces applied simultaneously in S0 will not be simultaneous in SA (asynchronous processes).The existence of frameS0 guarantees the correspondence between the relativistic and the classical descriptions; an equivalence necessary in the low velocity limit.In the asynchronous formulation, given the quantityAμ=Bμ+Cμ, where Bμ is defined for the event x1μ={x1,y1,z1,ct1} and Cμ is defined for the event x2μ={x2,y2,z2,ct2}, with x1μ≠x2μ, but with t2=t1, (8)Aμ(x1μ[t1],x2μ[t1])=Bμ(x1μ[t1])+Cμ(x2μ[t1]),
then quantity Aμ(x1μ[t1],x2μ[t1]) in S0 is the same as AAμ(x1Aμ[t1A],x2Aμ[t2A]) in SA when all subindex A quantities are obtained from the corresponding quantities in S0 through Lorentz transformations. Relativity gives rules to relate measurements made by observers in frame S0 to measurements made by observers in frame SA only if this definition is adopted [5].In the asynchronous formulation the 4-vector energy functionUμ is a time-like 4-vector in S0, with zero linear momentum components [38] (see Section 3): (9)Uμ={0,0,0,U}.
In frame SA, UAμ={cpA,0,0,EA} transforms under Lorentz transformation as UAμ=ℒνμ(V)Uν.For work due to external forces applied simultaneously in frameS0 with total zero impulse, the 4-vector work Wμ is a timelike 4-vector [1] (see Section 4): (10)Wμ={0,0,0,W}.As a generalization of this asynchronous formulation, in frameS0 every flux of energy through the frontier of the system as thermal radiation (heat) is exchanged with zero total impulse. Thus, heat is exchanged with zero linear momentum in frame S0 with a 4-vector Qμ related to thermal radiation exchange given by [3] (see Section 5) (11)Qμ={0,0,0,Q}.In frameSA the same Lorentz transformation ℒνμ(V) is common to UAμ, WAμ, with WAμ=ℒνμ(V)Wμ and to the 4-vector QAμ, that transforms as the energy (timelike) part of a (time-like) 4-vector, QAμ=ℒνμ(V)Qμ.


### 2.3. Principle of Similitude

The asynchronous formulation and the principle of the inertia of energy guarantee that the system can be described as a “single particle” [39] characterized by its energy function U or its inertia ℳ. These considerations permit us to enunciate [13] the following.Principle of Similitude
The mathematical expression for a physical law is the same when referred to an elementary particle, with tabulated massm, or when referred to a composite body, well characterized by its energy function U, and inertia ℳ=Uc-2.In the asynchronous formulation there is no difference between Lorentz transformations for an elementary particle and Lorentz transformations for an extended body, provided that the system is in equilibrium,  that is, energy functionU of the body is well defined. In frames like SA, in which the system has velocity V, differences between point dynamics and extended-body dynamics are due to the relativity of simultaneity [6],  that is, forces applied simultaneously in S0 but at different points of the body will not be simultaneous in SA.The principle of similitude has the following meaning. Physics equations, such as the Lorentz force equationF=q(E+v×B), Newton's second law of classical mechanics F=ma, or relativistic equations, such as E2=m2c4+c2p2 or p=(E/c2)v, are correct when they are applied to an elementary particle, with mass m and charge q, because every magnitude is well defined, for example, total energy E=γ(v)mc2, linear momentum p=γ(v)mv, and so forth, as well as the electric field E, the magnetic field B, and so forth, and forces applied are local forces, all of them applied at the same point. Similarly, a 4-vector, like Cμ=Aμ+Bμ or Cμ=c-1qℰνμvν, transforms between frames S0 and SA in standard configuration, by using the Lorentz transformation, ℒνμ(V), with CAμ=ℒνμ(V)Cν, and so forth, and where CAμ, and so forth, is in SA the same 4-vector Cμ in S0, because all of them are locally defined.When one wants to apply these equations to a process described on a composite, deformable body (e.g., a Ni atomic nucleus, a gas enclosed in a cylinder-piston system, a macroscopic chunk of Fe, etc.) and one wants to use the Lorentz transformation between reference inertial frames to transform 4-vectors, it is necessary to have previously ensured that the body behaves as a whole and that the principle of inertia of energy is satisfied. Because on an extended body different forces are applied at different points, it is necessary to ensure previously that there exists a reference frameS0 in which the center of mass does not move during the process. This goal is achieved when external forces are applied according to the asynchronous formulation and when the interval of time during which forces are applied on the mobile parts of the system is greater than the relaxation time of the system.Consider a gas enclosed in a cylinder-piston system. If the force on piston is applied in such a way that the velocity of the pistonvk is greater than the velocity of the sound in the gas vs, with a characteristic gas relaxation time tC given by tC≈L/vs, where L is a characteristic linear dimension of the system, then the system does not behave as a whole during time intervals Δt<tC because there are parts of it that do not feel the perturbation and so do not contribute to the inertia of the system. In this case the description of the process cannot be made according to the relativistic formalism to be developed here, the principle of similitude is not applicable and another formalism must be used to describe the process [40].When a process on a composite, extended body is carried out in such a way that the principle of inertia of energy is satisfied, the same set of equations valid for elementary particles can be used on the body.Consider a macroscopic body with well-defined energy functionU. In general, this energy function is temperature dependent U≡U(T) (see Section 7.1.1) (U dependence on volume will not be considered volume [41]) and also its inertia ℳ(T)=U(T)c-2, according to the principle of inertia of energy ([9], p. 289). When moving with velocity V (one-dimensional) in frame SA its linear momentum pA and total energy EA are given by (12)pA=γ(V)M(T)V,EA=γ(V)U(T),pA=EAc2V.
As previously noted, these results can be obtained from UAμ=ℒνμ(V)Uν. These equations constitute the generalization for an extended body of equations pA=γ(V)mV, EA=γ(V)mc2, and pA=(EA/c2)V for an elementary particle of mass m and velocity V. The total energy EA of the body can be expressed as (13)EA2=[U(T)]2+c2pA2=[M(T)]2c4+c2pA2.
Equation (13) is the generalization, in a thermodynamics context, of the equation E2=m2c4+c2p2 for an elementary particle.For an elementary particle massm and velocity V, the kinetic energy K is K=[γ(V)-1]mc2. The kinetic energy KA for an extended body, defined as KA=EA-U(T), is (14)KA=c2pA2EA+U(T)=[γ(V)-1]U(T)=[γ(V)-1]M(T)c2.
The kinetic energy of the body in frame S0, in which its linear momentum is null, is zero.


## 2.1. Inertia of Energy

It could be considered, in a broad sense, that the main goal of relativistic thermodynamics is to reach a unified description on point dynamics and extended-body dynamics [13].In order to ensure that an extended body behaves like a “single particle” interacting with its surroundings—work reservoirs or thermal bath—and so that it is physically meaningful to use Lorentz transformations, it is necessary that all forms of energy that make up the body contribute in the same way to its inertia [21]. These forms of energy must include those related with the mass of its constituent elementary particles, binding—nuclear, chemical, and so forth—energies (Figure 1), internal kinetic energy (see Section 7.1.1), electrostatic energy [22], and so forth, and energy of thermal radiation in equilibrium with matter inside the system [23] (see Section 5).Figure 1
An atom (A)—self-contained structure—is obtained from a nucleus (N), previously assembled from protons (p) and neutrons (n), and an ensemble of electrons (e). Nucleus inertiaℳN=UNc-2 decreases respect the inertia of its elementary particle components 𝒰N=6mp+4mn, due to the energy UN-𝒰N=-ŨN=-8hν′ released in its formation. Atom inertia ℳA=UAc-2 decreases with respect to the inertia of its component nucleus and electrons 𝒰A=6me+ℳN due to the energy UA-𝒰A=-ŨA=-8hν released in its formation.Einstein's EquationE0=mc2 for an extended body can be interpreted by relating its inertia—a body's reluctance to undergo a change in velocity [24]—with energy function [25]—energy content of the physical system or internal energy [26].Principle of the Inertia of Energy
for an extended body in complete equilibrium, any kind of energy inside the system, relativistically expressed in reference frameS0 in which the system as a whole is at rest, contributes to the energy functionU of the system [27]. Considering that all forms of energy are convertible between them [28] the inertiaℳ of a system [29] in equilibrium is ([30], p. 163)
(5)M=Uc-2.According to Einstein [31, 32] the inertia of a body changes with its content of energy [33] (Section 3).It is possible to define the inertiaℳ of a body (we prefer the term inertia, instead of mass [34], to avoid confusions when the system includes photons (Section 5)) as [35]: the inertia ℳ of a composite body equals the sum of its elementary particles mass (protons, neutrons, and electrons) m0: (6)m0=∑jmp+∑kmn+∑lme,
with energy 𝒰=m0c2, minus the minimum energy Ũ, divided by c2, necessary to separate its elementary particles so that they are far apart (Section 7.1.1): (7)M=(U-Ũ)c-2=m0-Ũc-2,
with U=𝒰-Ũ.


## 2.2. Asynchronous Formulation

For an extended, deformable body a relativistic theory cannot be directly formulated in an arbitrary inertial frame. It must be based on known prerelativistic descriptions. On the one hand, it seems necessary to maintain the classical concept that the resultant force on the body must be zero (zero total impulse) when the motion remains uniform and to assure that when no torque is applied to the system in a certain reference frame, no torque is applied to it in another frame [36]. On the other hand, in classical mechanics forces on an extended system are applied simultaneously. This simultaneity occurs in all inertial frames. In thermodynamics, heat is a kind of interchanged energy with (assumed implicitly) zero linear momentum.According to Cavalleri and Salgarelli, when forces on an extended, composite, body are applied, in order to develop a coherent formalism for relativistic thermodynamics, a privileged observer must exist, in reference frameS0, that performs experiments on the body that remains at rest (Figure 2) [6].Figure 2
Compression process in frameS0. A set of external forces Fk (k=1,2,3,4) are applied on an extended, deformable system during the same time interval Δt, as measured in frame S0, with zero total impulse and zero torque. The kth force has associated the displacement Δrk (r=(x,y)) and 3-vector velocity vk=Δrk/Δt. The center of mass (cm) of the system does not move during the process.According to Gamba [5]:“in theAsynchronous Formulation, observers in frames S0 and SA refer to the same experiment (the experiment performed in the privileged frame S0) and obtain its own physical magnitudes, expressed as 4-vectors. In this formulation both descriptions of the experiment are connected by true Lorentz transformations [19].”The observer inS0 takes an ideal surface, at rest, which delimits the system considered and measures energy, work or heat, interchanged through the surface during time interval Δt. An observer in SA obtains the same magnitudes by true Lorentz transformations, from the events considered by an observer in S0. The observer in SA does not perform a similar experiment to observer in S0 (synchronous formulation [37]), it just translates the experiment performed in S0 to its own physical magnitudes. Owing to the relativity of simultaneity, forces applied simultaneously in S0 will not be simultaneous in SA (asynchronous processes).The existence of frameS0 guarantees the correspondence between the relativistic and the classical descriptions; an equivalence necessary in the low velocity limit.In the asynchronous formulation, given the quantityAμ=Bμ+Cμ, where Bμ is defined for the event x1μ={x1,y1,z1,ct1} and Cμ is defined for the event x2μ={x2,y2,z2,ct2}, with x1μ≠x2μ, but with t2=t1, (8)Aμ(x1μ[t1],x2μ[t1])=Bμ(x1μ[t1])+Cμ(x2μ[t1]),
then quantity Aμ(x1μ[t1],x2μ[t1]) in S0 is the same as AAμ(x1Aμ[t1A],x2Aμ[t2A]) in SA when all subindex A quantities are obtained from the corresponding quantities in S0 through Lorentz transformations. Relativity gives rules to relate measurements made by observers in frame S0 to measurements made by observers in frame SA only if this definition is adopted [5].In the asynchronous formulation the 4-vector energy functionUμ is a time-like 4-vector in S0, with zero linear momentum components [38] (see Section 3): (9)Uμ={0,0,0,U}.
In frame SA, UAμ={cpA,0,0,EA} transforms under Lorentz transformation as UAμ=ℒνμ(V)Uν.For work due to external forces applied simultaneously in frameS0 with total zero impulse, the 4-vector work Wμ is a timelike 4-vector [1] (see Section 4): (10)Wμ={0,0,0,W}.As a generalization of this asynchronous formulation, in frameS0 every flux of energy through the frontier of the system as thermal radiation (heat) is exchanged with zero total impulse. Thus, heat is exchanged with zero linear momentum in frame S0 with a 4-vector Qμ related to thermal radiation exchange given by [3] (see Section 5) (11)Qμ={0,0,0,Q}.In frameSA the same Lorentz transformation ℒνμ(V) is common to UAμ, WAμ, with WAμ=ℒνμ(V)Wμ and to the 4-vector QAμ, that transforms as the energy (timelike) part of a (time-like) 4-vector, QAμ=ℒνμ(V)Qμ.


## 2.3. Principle of Similitude

The asynchronous formulation and the principle of the inertia of energy guarantee that the system can be described as a “single particle” [39] characterized by its energy function U or its inertia ℳ. These considerations permit us to enunciate [13] the following.Principle of Similitude
The mathematical expression for a physical law is the same when referred to an elementary particle, with tabulated massm, or when referred to a composite body, well characterized by its energy function U, and inertia ℳ=Uc-2.In the asynchronous formulation there is no difference between Lorentz transformations for an elementary particle and Lorentz transformations for an extended body, provided that the system is in equilibrium,  that is, energy functionU of the body is well defined. In frames like SA, in which the system has velocity V, differences between point dynamics and extended-body dynamics are due to the relativity of simultaneity [6],  that is, forces applied simultaneously in S0 but at different points of the body will not be simultaneous in SA.The principle of similitude has the following meaning. Physics equations, such as the Lorentz force equationF=q(E+v×B), Newton's second law of classical mechanics F=ma, or relativistic equations, such as E2=m2c4+c2p2 or p=(E/c2)v, are correct when they are applied to an elementary particle, with mass m and charge q, because every magnitude is well defined, for example, total energy E=γ(v)mc2, linear momentum p=γ(v)mv, and so forth, as well as the electric field E, the magnetic field B, and so forth, and forces applied are local forces, all of them applied at the same point. Similarly, a 4-vector, like Cμ=Aμ+Bμ or Cμ=c-1qℰνμvν, transforms between frames S0 and SA in standard configuration, by using the Lorentz transformation, ℒνμ(V), with CAμ=ℒνμ(V)Cν, and so forth, and where CAμ, and so forth, is in SA the same 4-vector Cμ in S0, because all of them are locally defined.When one wants to apply these equations to a process described on a composite, deformable body (e.g., a Ni atomic nucleus, a gas enclosed in a cylinder-piston system, a macroscopic chunk of Fe, etc.) and one wants to use the Lorentz transformation between reference inertial frames to transform 4-vectors, it is necessary to have previously ensured that the body behaves as a whole and that the principle of inertia of energy is satisfied. Because on an extended body different forces are applied at different points, it is necessary to ensure previously that there exists a reference frameS0 in which the center of mass does not move during the process. This goal is achieved when external forces are applied according to the asynchronous formulation and when the interval of time during which forces are applied on the mobile parts of the system is greater than the relaxation time of the system.Consider a gas enclosed in a cylinder-piston system. If the force on piston is applied in such a way that the velocity of the pistonvk is greater than the velocity of the sound in the gas vs, with a characteristic gas relaxation time tC given by tC≈L/vs, where L is a characteristic linear dimension of the system, then the system does not behave as a whole during time intervals Δt<tC because there are parts of it that do not feel the perturbation and so do not contribute to the inertia of the system. In this case the description of the process cannot be made according to the relativistic formalism to be developed here, the principle of similitude is not applicable and another formalism must be used to describe the process [40].When a process on a composite, extended body is carried out in such a way that the principle of inertia of energy is satisfied, the same set of equations valid for elementary particles can be used on the body.Consider a macroscopic body with well-defined energy functionU. In general, this energy function is temperature dependent U≡U(T) (see Section 7.1.1) (U dependence on volume will not be considered volume [41]) and also its inertia ℳ(T)=U(T)c-2, according to the principle of inertia of energy ([9], p. 289). When moving with velocity V (one-dimensional) in frame SA its linear momentum pA and total energy EA are given by (12)pA=γ(V)M(T)V,EA=γ(V)U(T),pA=EAc2V.
As previously noted, these results can be obtained from UAμ=ℒνμ(V)Uν. These equations constitute the generalization for an extended body of equations pA=γ(V)mV, EA=γ(V)mc2, and pA=(EA/c2)V for an elementary particle of mass m and velocity V. The total energy EA of the body can be expressed as (13)EA2=[U(T)]2+c2pA2=[M(T)]2c4+c2pA2.
Equation (13) is the generalization, in a thermodynamics context, of the equation E2=m2c4+c2p2 for an elementary particle.For an elementary particle massm and velocity V, the kinetic energy K is K=[γ(V)-1]mc2. The kinetic energy KA for an extended body, defined as KA=EA-U(T), is (14)KA=c2pA2EA+U(T)=[γ(V)-1]U(T)=[γ(V)-1]M(T)c2.
The kinetic energy of the body in frame S0, in which its linear momentum is null, is zero.


## 3. Four-Vector Energy Function

Energy functionU of a composite body is obtained from the energy function of its components (Section 2.1).  (1)
Universal constants (c, h (Planck), kB(Boltzmann), G, ϵ0, etc.) are relativistic invariants having the same value for all inertial observers in relative motion.(2)
For an elementary particle—proton, neutron, and electron—the inertia equals its tabulated mass—mp. mn, me, respectively.(3)
For a nucleus,ZAN, with Z protons and (A-Z) neutrons, its inertia ℳN equals the sum of the inertia of the elementary particles—with all elementary particles at infinite separation as initial arrangement—minus its binding energy (strong interaction) [42] ŨN divided by c2 (Figure 1):
(15)UN=[Zmp+(A-Z)mn]c2-|ŨN|,MN=UNc-2.(4)
For an atom, the inertiaℳA equals the sum of the inertia of its nucleus and electrons minus released energy ŨA (electromagnetic interaction) [43] divided by c2 (Figure 1):
(16)UA=UN+npmec2-|ŨA|,MA=UAc-2.
For instance, energy function u for a 4He atom (2 protons, 2 neutrons, and 2 electrons) [21] is given [31] by
(17)u=u0-(|ŨN|+|ŨA|),u0=2(mp+me)c2+2mnc2.(5)
For a molecule, formed byk atoms, the inertia ℳM is the sum of the inertia of its individual atoms minus the energy released when chemical bonds are formed [44] divided by c2:
(18)UM=∑kUAk-|ŨM|,MM=UMc-2.
Energy function UC of a composite, self-contained (stable) system is less than the sum of the energy function of its k constituents [45] 𝒰=∑kUk, UC<U.(6)
For a system of free noninteracting components [46] like a gas of He atoms, the inertia equals the sum of the total energy of components U=∑k(kk+uk)—kinetic energy and energy function of the kth component, respectively—divided by c2 (see Section 7.1.1).(7)
For thermal radiation (photons in a cavity with energy density proportional to fourth power of absolute temperature) filling a cavity [47] its total energy Up contributes to the total inertia of the system [48]. The thermal radiation emitted by a body can be described as radiation in a cavity [49] (see Section 5).As previously noted, in the zero-momentum frameS0 of a composite system with energy function U≡U(T) the 4-vector that denotes the state of the system is given by Uμ={0,0,0,U(T)}. For an observer in frame SA, 4-vector energy function UAμ={cpA,0,0,EA} is UAμ=ℒνμ(V)Uν, and one obtains (19)UAμ={-cγ(V)M(T)V,0,0,γ(V)U(T)},M(T)=U(T)c-2,pA=-γ(V)M(T)V,EA=γ(V)U(T),
according to the principle of inertia of energy (Section 7.2.1).Einstein Equation
for a completely isolated system that performs any kind of internal process, for example, annihilation or creation of particles, disintegration, inelastic collisions, and so forth, the inertia does not change along the process [13], according to the Principle of Inertia of Energy, with
(20)ΔUμ=0.


## 4. Four-Vector Work

In order to obtain a complete characterization of forces applied to a thermodynamical system (i.e., based on a fundamental interaction), we will describe forces as the interaction between an electric chargeqk located on the kth piston and a (static) electric field Ek=(Exk,Eyk,Ezk). This procedure guarantees a detailed description of forces and of its relativistic transformation between reference frames (Appendix B).Consider in frameS0 a set of k forces Fk=(Fxk,Fyk,Fzk)=q(Exk,Eyk,Ezk), with an electromagnetic origin, simultaneously applied, on different k pistons, on an extended body (Figure 3) during the same interval of time Δt, according to the previously discussed asynchronous formulation. Impulse Ik=(Ixk,Iyk,Izk) and work Wk for the kth force are given by (21)Ik=(Fxkdt,Fykdt,Fzkdt),Wk=Fk⋅drk=Fxkdxk+Fykdyk+Fzkdzk.Figure 3
Extended system, withk pistons (k=1,2,3,4), to which forces Fk=qEk are simultaneously applied during time interval Δt as measured in frame S0 by a set of synchronized clocks at rest (Figure 1). On the kth piston there is a clock that measures its proper time τk. The kth piston displaces drk during time interval dt, speed vk=drk/dt, and proper time interval dτk. Forces Fk have an electromagnetic origin: an electric field Ek, produced by a plane-parallel charged capacitor, interacts with a charge qk fixed to the kth piston.Thekth field Ek is represented by the 4×4×4-tensor Ekνμ: (22)εkνμ={000Exk000Eyk000EzkExkEykEzk0},
with the 4×4×4-tensor electromagnetic force Fkνμ: (23)Fkνμ=qεkνμ={000Fxk000Fyk000FzkFxkFykFzk0}.
The kth piston has a 4-vector displacement dxkμ and a 4-vector velocity vkμ:  (24)dxμ={dxk,dyk,dzk,cdt},vkμ=dxkμdτk=γ(vk){vxk,vyk,vzk,c},dtdτk=γ(vk),
where τk is the proper time of kth piston displacement.For thekth piston, two 4-vectors can be obtained: (i) the 4-vector Minkowski force Fkμ and (ii) the 4-vector work Wkμ.  (1)
The 4-vector Minkowski forceFkμ is given by ([20], Chap. 33)
(25)Fkμ=c-1Fkνμvkν=γ(vk){Fxk,Fyk,Fzk,c-1[Fk⋅vk]},
with Fk·vk=Fxkvxk+Fykvyk+Fzkvzk.(2)
The 4-vector workδWkμ is given by
(26)δWkμ=Fkνμdxkν={cFxkdt,cFykdt,cFzkdt,Fxkdxk+Fykdyk+Fzkdzk}={cIxk,cIyk,cIzk,Wk},
a 4-vector with units of energy. The 4-vector Fkμ can be obtained by deriving δWkμ in respect to proper time dτk of the kth piston as(27)δWkμdτk=dtdτkδWμdt=cFkμ.This obtention of the 4-vectorFkμ shows that δWkμ is a 4-vector itself (Appendix A). For a finite interval of time Δt, with constant force Fk, and 4-vector interval Δxkμ={Δxk,Δyk,Δzk,cΔt}, the 4-vector work Wkμ is (28)Wkμ=FkνμΔxkν={cFxkΔt,cFykΔt,cFzkΔt,FxkΔxk+FykΔyk+FzkΔzk}.For the set ofk forces simultaneously applied to the body at different pistons in frame S0 during the finite interval of time Δt (Figure 4), the total 4-vector “force-displacement product” (work) Wμ is the sum of the 4-vector Wkμ. The 4-vector total work Wμ is given by Wμ=∑kWkμ, with condition ∑kIk=0: (29)Wμ={0,0,0,W};W=∑kWk.(a) A gas contained in a cylinder is compressed under the action of two pistons, L, with electric charge+q fixed to it, and R, with electric charge -q fixed to it. On piston L, a force FL=qE is exerted by the electric field E and a force -(FL-δF) slightly smaller by the gas. On piston R a force FR=-qE is exerted by the electric field, an a force (FR-δF) by the gas. (b) Thus the gas is compressed under the action of force FL applied to a displacement ΔxL and a force FR applied to a displacement ΔxR. Both forces FL and FR are applied simultaneously during time interval Δt. Every piston acts as an intermediate agent between the work reservoir (electric field and battery to which the capacitor is connected) and the thermodynamics system (the gas).
(a)(b)In frameSA, WAμ=ℒνμ(V)Wν={cℐxA,0,0,WA}, with impulse ℐxA and “force-displacement product” WA being (30)IxA=-γ(V)(Wc-2)V,WA=γ(V)W.Adiabatic First Law ([18], Section 4.2)
A system that changes its energy function owing to forces applied to it, in an adiabatic process, is(31)ΔUμ=Wμ.


## 5. Four-Vector Heat

Work is described in thermodynamics as oriented (nonrandom) internal energy transferred between a body and a work reservoir (Figure4). However, heat is described as random (or nondirected) internal energy transferred between two bodies at different temperatures [50]. Nondirected means “without linear momentum.”According to Rindler, in the special theory of relativity any transfer of energy, being equivalent to a transfer of inertia, necessarily involves momentum [51,p. 91]. This assessment is valid for all forms of radiation and must be valid for heat [52], whatever definition of heat is being used.The most direct argument on relativistic heat transformations is provided by Arzeliés [1]. Based on the principle of equivalence work-heat, this author assumes that relativistic heat transforms as relativistic work.The 4-vector heat,Qμ, is obtained in two steps. First, we obtain the 4-vector for thermal radiation (its frequency distribution fulfills Planck’s frequency distribution) enclosed in a cavity with walls at temperature T, and then the thermal radiation exchanged by a body as heat is described as thermal radiation in a cavity.In a generalization of the asynchronous formulation, we assume that in frameS0 (zero momentum frame) heat is emitted or absorbed with zero linear momentum ([3] p. 173). With the 4-vector Qμ given in frame S0 as Qμ={0,0,0,Q}, in frame SA, standard configuration, with QAμ={cpA,0,0,EA}, and QAμ=ℒνμ(V)Qν, a linear momentum associated to Q, must be ([4] p. 1746)(32)pA=-γ(V)Qc2V.
The relativistic linear momentum of heat in frame SA requires a physical interpretation—because of the contrast with no momentum for heat in classical thermodynamics [53]. In order to provide the relativistic interpretation of heat and the description of a thermal bath, we will describe thermal radiation as an ensemble of emitted photons enclosed in a cavity [54].A cavity with walls at temperatureT, measured with a gas thermometer at constant volume, and filled with photons that fit Planck’s frequency distribution—that is, thermal photons—constitutes a thermal bath. In frame S0 in which cavity walls are at rest, the total linear momentum of the photon ensemble is zero. In the monochromatic approximation [55] to Planck’s distribution, every photon has the same frequency ν, with ν(T)=AT (Wien's Law), where A is a constant (Figure 5).(a) Cavity with walls at temperatureT filled with thermal radiation (photons) frequency ν=ν(T) (monochromatic approximation). In frame S0 linear momentum for this ensemble of photons is zero pp=0 and its energy function is Up=Nhν. (b) In frame SA, the same ensemble of photons, with frequencies νL>ν, νR<ν (relativistic Doppler effect) and ν′ (aberration effect), has linear momentum ppA=γ(V)(Upc-1)V, according to the principle of inertia of energy, and total energy Ep=γ(V)Up.
(a)(b)The 4-vector waveωμ for a photon ([56], pp. 255–257) of wavelength λ and angular frequency ω=2π/𝒯, period 𝒯, that propagates in a direction given by the wave vector k, (33)k=(2πλcos⁡θ,2πλsin⁡θ,0),
is ([20], pp. 269-270) (34)ωμ={c2πλcos⁡θ,c2πλsin⁡θ,0,2πT}.For a givenrth photon, with frequency ν, λν=c, and moving in direction k=(cosθr,sinθr,0), there exists an energy 4-vector (ℏ=h/2π), (35)qrμ=ℏωrμ={c[hνc]cos⁡θr,c[hνc]sin⁡θr,0,hν}.
The norm of this 4-vector is ∥qrμ∥=0. An individual photon has no inertia.In frameS0, total linear momentum for the N photons inside the cavity at temperature T, pp, and its total energy Ep are given by (36)pxp=∑rhνccos⁡θr=0,pyp=∑rhνcsin⁡θr=0,Ep(T)=∑rhν=Nhν(T).
In the zero-momentum frame S0, total energy Ep(T) is the energy function Up(T) of the system. The 4-vector thermal radiation 𝒬μ, 𝒬μ=∑rqrμ is (37)Qμ={0,0,0,Up(T)},Up(T)=Nhν(T).
In frame SA, (38)QAμ=Lνμ(V)Qν={-cγ(V)Mp(T)V,0,0,γ(V)Up(T)},Mp(T)=Nhν(T)c-2.
The energy function Up(T) is the norm of the 4-vector ∥𝒬Aμ∥=Nhν(T). This photons ensemble has nonzero inertia [48] ℳp=Upc-2.Consider for a moment this cavity filled with thermal radiation containing one mole of atoms of a gas also. It is interesting to note that (i) photons of thermal radiation enclosed in a cavity, with Planck’s frequency distribution, the atoms of the gas, with its (ii) electrons distributed in electronic orbitals following Boltzmann's energy distribution, and (iii) atoms moving with Maxwell's (or Juttner distribution [57]) kinetic energy distribution, every distribution with the same parameter temperature T, contribute to the energy function and to the inertia [35] of the system. As previously discussed, energy function for an ensemble of atoms and energy function for an ensemble of thermal photons transform between inertial frames in the same way. Thus, thermal equilibrium at temperature T between matter and radiation is a relativistic invariant and every inertial observer will agree on that equilibrium (Figure 6).(a) A gas contained in a cylinder is compressed by a forceF applied to a piston inside a cavity (thermal resevoir) at temperature T (isothermal process). The centre of mass of the gas remains at rest in its initial and final equilibrium states. (b) During the compression process photons with frequency ν (monochromatic approximation) are emitted with zero linear momentum. (c) Heat Q emitted during the compression is characterized as the energy associated with the thermal radiation made up of all photons emitted contained into the cavity (with minus sign).
(a)(b)(c)After the obtention of a 4-vector𝒬μ for the contribution to its energy function by thermal radiation inside a cavity, it is necessary to characterize as a 4-vector heat Qμ the exchanged energy by a body as thermal photons.First of all, systems thermally interacting with each other cannot be in equilibrium if they are in relative motion [58]. In the Asynchronous Formulation generalization to heat, there exists a privileged frame S0 in which the system is at rest with respect to the thermal bath and in S0 thermal radiation (photons) is absorbed or emitted with zero total linear momentum.The energy absorbed, or emitted, by a body as thermal radiation (heat) throughout a process can be modeled as photons inside a cavity. A thermal system can absorb or emit photons through its frontier except in adiabatic processes. A photon emitted by a body, with frequencyν and direction u, contributes with -hνu/c to the linear momentum variation of the body and with -hν to the total energy variation of the body that emits it. A photon absorbed by a body, with frequency ν and direction u, contributes with +hνu/c to the linear momentum variation and with +hν to total energy increment of the system.Absorbed or emitted photons can be considered different phases in thermal equilibrium [59]. Thus, there is not “force-displacement product” (work) associated with emission or absorption of thermal radiation (photons).With the system and thermal bath mutually at rest, the ensemble of emitted photons (when the system is at higher temperature than thermal reservoir) is described as an ensemble of thermal photons in a cavity with zero total linear momentum, and so the ensemble of absorbed photons (when the system is at lower temperature than thermal reservoir).In frameS0, the 4-vector thermal radiation (heat) Qμ associated when the body emits (−) or absorbs (+) N photons with frequency ν(T) is given by (39)Qμ=-Qμ={0,0,0,∓Up(T)},
with Up(T)=ṄhνΔt, where Ṅ=dN/dt is the flux of photons (net number of photons exchanged in unit time) and N=ṄΔt is the net number of photons exchanged by the body during time interval Δt.For an observer in frameSA, QAμ={cpA,0,0,EA}, from QAμ=ℒνμ(V)Qν with linear momentum ppA and total energy EpA: (40)ppA=-γ(V)Mp(T)V;Mp(T)=Up(T)c-2,EpA=γ(V)Up(T)=γ(V)Mp(T)c2,
Physical interpretation of linear momentum for heat in frame SA will be obtained from relativistic Doppler and aberration effects applied to photons (see Section 7.2.3). The norm of QAμ is (41)‖QAμ‖=[EpA2-c2ppA2]1/2=Up(T)=Mp(T)c2,
with energy function Up(T) and inertia ℳp(T) relativistic invariants [60].If heat is defined as the total energy associated with the emitted (or absorbed) photonsas measured in the observer's frame, then QA=γ(V)Q, with Q=Up(T). If heat is defined as the emitted (or absorbed) energy carried by photons in frame S0 in which the interchange of photons with a thermal surrounding mutually at rest and zero linear momentum happens—as it is defined (implicitly) in classical thermodynamics—then Q is the norm of any 4-vector Qμ=∥QAμ∥ and it is a relativistic invariant. In any case, it is the 4-vector that possesses physical meaning, not its components.Heat
For a system that changes its energy function without forces applied to it, by heating, or cooling, in diathermal contact with a thermal bath, system and bath at mutual rest, is(42)ΔUμ=Qμ.


## 6. Relativistic Thermodynamics First Law

According to the generalized asynchronous formulation of relativistic thermodynamics, the description of a certain process on a composite, deformable system, and after the obtention of 4-vectors energy functionUμ, initial Uiμ and final Ufμ, work Wμ, and heat Qμ, is as follows.Relativistic Thermodynamics First Law
Mathematical: the relationship between variations in energy function of a system after a certain process, during which it interacts with a mechanical reservoir, with forces simultaneously applied to it during that process, and a thermal reservoir, system and reservoir mutually at rest, with thermal radiation interchanged by the system during the process, with every magnitude expressed as a 4-vector, is [4, 61]
(43)ΔUμ=Ufμ-Uiμ=Wμ+Qμ.No matter whether a system is self-contained or free (confined in a container), any energy, momentum,Uμ, Wμ or Qμ, is always a 4-vector provided that one performs a covariant summation at constant time (simultaneously) in the frame S0 in which the system is mutually at rest [2] (at least instantaneously) with its mechanical and thermal surroundings.In the frame referenceSA, the first law is expressed as  (44)ΔUAμ=UfAμ-UiAμ=WAμ+QAμ.
Every 4-vector in SA can be obtained by a Lorentz transformation for the corresponding 4-vector in S0(45)Lνμ(V)[Ufν-Uiν=Wν+Qν]⟶UfAμ-UiAμ=WAμ+QAμ.
This circumstance guarantees the first law Lorentz covariance.


## 7. Ideal Gas Isothermal Compression

A horizontal cylinder (Figure7), with thin metallic walls, section A, length L, containing 1 mol, NA atoms, of 4He gas, enclosed by two pistons, left (L) and right (R). We assume that helium behaves as an ideal gas, described by thermal equation of state P𝒱=NAkBT. The gas is in equilibrium under pressure Pi and at temperature T, volume 𝒱i, 𝒱i=RT/Pi. The limits of the system are the walls of the cylinder, considered diathermal. Pistons are considered adiabatic.Figure 7
Cylinder with gas, diathermal walls, closed by two pistons, R and L, inside a plane-parallel capacitor with charge surface densityσ. Laser (l), beam splitter (s), pistons of blocked mechanism (b) are used to assure that forces are applied simultaneously [62]. The gas in cylinder is compressed by forces acting on pistons. Force FL=F acts on piston L during time interval Δt and displacement ΔxL=Δx. Force FR=-F acts on piston R during time interval Δt and displacement ΔxR=-Δx. During the compression process, photons are emitted by cylinder walls with frequency ν and zero linear momentum in frame S0.


### 7.1. Compression in FrameS0

As privileged frameS0 we take the frame in which cylinder walls, plate parallel capacitor and thermal reservoir walls are at rest. During the compression process, forces on gas are applied simultaneously, during time interval Δt. Thermal radiation is interchanged with zero impulse in S0 and the gas center of mass remains at the same point, with initial and final zero velocity.


#### 7.1.1. Energy Function inS0

For simplicity, we assume that the atoms of He inside the cylinder possess only translational energy,  that is, all atoms are in its ground electronic state. In general, one can assume that gas velocity distribution is Juttner distribution [57] ([9], pp. 289–293). For simplicity, one assumes that atoms are randomly distributed inside the container and that every atom has the same translational energy,  that is, every He atom moves with the same speed v=v(T) (monokinetic approximation [50]), same modulus, but with different vectorial components v=(vx,vy,vz), v=|v|. In this approximation, v(T)=aT1/2. Constant a is obtained by imposing (46)k=[γ(v)-1]u=32kBT,
where k=[γ(v)-1]u is the kinetic energy of a He atom and u its energy function (Section 3).Linear momentumpj=(pxj,pyj,0) (for simplicity we assume x-y as movement of the atoms) and total energy Ej for the jth atom are (47)pxj=γ(v)mvx,pyj=γ(v)mvy,ej=γ(v)u,m=uc-2.
Initial total linear momentum pi=(pxi,pyi,0) and initial total energy (energy function ) Ui are given by (48)pxi=∑jpxj=γ(v)m∑kvxj=0,pyi=∑jpyj=γ(v)m∑kvyj=0,Ui=U(T)=γ(v)∑ju=NAγ(v)u.
The 4-vector initial energy function Uiμ is then  (49)Uiμ={0,0,0,Ui};Ui=γ(v)NAu.
Energy function Ui depends on temperature through temperature dependence on velocity v=v(T). In S0 (pi=0), total system energy ([63], Sec. 8.3) is, by definition, its energy function Ui=Ki+𝒰i, sum of the kinetic energy of helium atoms Ki=∑kkk=NA[γ(v)-1]u, and 𝒰i=∑ku=NAu.For an ideal gas in an isothermal process, energy function remains constant, as well as the temperature, and also He atom speed. The 4-vector final energy functionUfμ is then: (50)Ufμ={0,0,0,Uf},
with Uf=Ui=U(T).


#### 7.1.2. Work inS0

Forces on pistons are described as produced by the interaction of an electric charge with an electromagnetic field (Figure7). Static electric positive +q and negative -q charges are located on right and left pistons, respectively. The whole device, gas plus pistons, is located inside the homogeneous electric field E produced by a plane-parallel capacitor ([64], Chap. 13) with charge surface density σ.In frameS0 the capacitor is at rest. A horizontal electric field Ex is created inside the capacitor. For this (uniform) electric field the potential 4-vector Φμ={Ax,Ay,Az,ϕ}, where A=(Ax,Ay,Az) and ϕ are the vector potential and the scalar potential, respectively, is given by the contravariant 4-vector [65]: (51)Φμ={0,0,0,-Exx},Ex=σ2ϵ0.The electromagnetic4×4-tensor ℰνμ—double contravariant—is given by ([51], Section 42) (52)Eνμ=∂Φν∂xμ-∂Φμ∂xν.
For the horizontal plane-parallel capacitor the 4×4-tensor ℰνμ is (53)Eνμ={000Ex00000000-Ex000}.
The mixed 4×4-tensor ℰνμ is obtained as (54)Eνμ=gνξEξμ={000-Ex00000000-Ex000}.Initially, pistons are locked by a blocked mechanism (b in Figure7). A laser (l in Figure 7) and a beam splitter (s in Figure 7) that is located just between the pistons, are used to release simultaneously both pistons. When the laser is turned on the split beams will arrive at the blocked mechanism of both left and right pistons and, at time t=0, are unlocked allowing electric field charges interaction simultaneously on both L and R pistons [62].When pistons are released, the Minkowski forceFkμ on kth piston is (55)Fkμ=qEνμvkμ,vkμ=dxkνdτk,
where q is the electric charge on piston, ℰνμ is the 4×4-tensor electromagnetic field, and vkμ and τk are the kth piston 4-vector velocity and proper time, respectively [66].For the left piston (subindex L), displacementΔxL=Δx and velocity vL=ΔxL/Δt=v, 4-vector ΔxLμ and 4-vector velocity vLμ, are respectively (56)ΔxLμ={Δx,0,0,cΔt},vLμ=γ(v){v,0,0,c}.
For the right piston (R), displacement ΔxR=-Δx and velocity vR=ΔxR/Δt=-v, 4-vector ΔxRμ and 4-vector velocity vRμ, are respectively (57)ΔxRμ={-Δx,0,0,cΔt},vRμ=γ(v){-v,0,0,c}.The Minkowski forces due to electromagnetic interaction are:(58)FLμ=-qEνμvLν=γ(v){+cqEx,0,0,qExv},FRμ=+qEνμvRν=γ(v){-cqEx,0,0,qExv}.
With forces acting on pistons, FL=(qEx,0,0) and FR=(-qEx,0,0), 4-vectors work are, respectively, (59)WLμ=-qEνμΔxLν={+cqExΔt,0,0,qExΔx},WRμ=+qEνμΔxRν={-cqExΔt,0,0,qExΔx}.
The total 4-vector work Wμ is then  (60)Wμ=WLμ+WRμ={0,0,0,2qExΔx}.


#### 7.1.3. Heat inS0

During (slow) gas compression,N=ṄΔt photons with frequency ν(T) are emitted, with zero total linear momentum. Photons are emitted through the horizontal walls of the cylinder, N/2 photons are emitted in direction θ+=π/2 and N/2 photons in direction θ-=-π/2. Total linear momentum for this ensemble of photons pp=(pxp,pyp,0) is (61)pxp=0,pyp=N2hνsin⁡θ++N2hνsin⁡θ-=0.
Total energy of these emitted photons E=Nhν, with pp=0, is its energy function Up: (62)Up=Nhν.
According to the principle of inertia of energy, these N photons have inertia [63] ℳp=Upc-2. The 4-vector for heat Qμ (thermal radiation emitted from the point of view of the system) is then given by  (63)Qμ={0,0,0,-ṄhνΔt}.


#### 7.1.4. First Law inS0

From first lawUfμ-Uiμ=Wμ+Qμ, one obtains 0=2qExΔx-ṄhνΔt or (64)ṄhνΔt=2qExΔx,Ṅhν=2qExv.
The configurational work done on the gas provides the energy emitted as heat.


### 7.2. Compression in FrameSA

An observer in frameSA obtains the corresponding 4-vector UiAμ, UfAμ, WAμ, and QAμ by measuring different magnitudes—displacements ΔxA, time intervals ΔtA, velocities vA, forces FA, photon frequency νA, and so forth—in its own frame (Figure 8). With first law in frame SA expressed as UfAμ-UiAμ=WAμ+QAμ, the Lorentz invariance of this equation assures that the same result as in S0,  that is, ṄhνΔt=2qEΔx, is obtained.Figure 8
Isothermal compression given in Figure7, described from the point of view of an observer in frame SA, standard configuration, with velocity V with respect to frame S0. In frame SA forces are not applied simultaneously and heat carries linear momentum.


#### 7.2.1. Energy Function inSA

Let there be one mol of He atoms moving inside the cylinder. In frameS0, with zero total linear momentum, velocities of atoms are measured simultaneously. During the same interval of time Δt, displacement Δrj=(Δxj,Δyj,0) for the jth atom is measured and its 3-vector velocity is given by vk=(Δrj/Δt)=(vx,vy,0). In order to ensure that in frame S0 total linear momentum is zero, for every atom j moving with velocity vj=(vxj,vyj,0) there must exist another atom n moving with opposite velocity vn=(vxn,vyn,0), such that vxj=-vxn and vyj=-vyn.The velocityvA=(vxsA,vysA,0) of the sth atom is given, in terms of its velocity v=(vxs,vys,0) measured in S0 and velocity V=(V,0,0) of frame SA with respect to frame S0, by the equation ([51], Section 12) (65)vsA=1γ(V)[1-vxsV/c2](γ(V)[vxs-V],vys,0),
with the useful relations ([67], p. 69) (66)γ(vsA)=γ(v)γ(V)[1-vxsVc2],γ(vsA)vxsA=γ(v)γ(V)(vxs-V),γ(vsA)vysA=γ(v)vys.For every pairj-n of opposite atoms, total momentum and total energy are easily obtained using the previous transformations: (67)px(j+n)A=γ(vjA)(uc-2)vxjA+γ(vrA)(uc-2)vxnA=γ(V)[2γ(v)uc-2]V,py(j+n)A=γ(vjA)(uc-2)vyjA+γ(vrA)(uc-2)vynA=0,
and total energy is given by (68)EjA=γ(vjA)u=γ(v)γ(V)[1-vxVc2]u,EnA=γ(vnA)u=γ(v)γ(V)[1+vxVc2]u,E(j+n)A=EjA+EnA=γ(V)[2γ(v)u].
For the N/2 total pairs of opposite atoms, one has(69)UiAμ={-cγ(V)[γ(v)NAuc2]V,0,0,γ(V)[γ(v)NAu]}={-cγ(V)MV,0,0,γ(V)U(T)},M=U(T)c-2
This is the same result obtained from Lorentz transformation [68] on the 4-vector energy function in S0, given in (49), UiAμ=ℒνμUiν.A similar description for 4-vector final energy functionUfAμ, with UfAμ=ℒνμUfν, is(70)UfAμ{-cγ(V)MV,0,0,γ(V)U(T)}.


#### 7.2.2. Work inSA

By considering the locked-unlocked piston set, laser beam, splitter, blocked mechanism described previously in frameS0, it is evident that forces that are simultaneously applied in S0 are not simultaneous in frame SA (Figure 8) [1].To obtain the 4-vectorWAμ in SA, relativistic transformations of time intervals, spatial displacements, and forces must be used.In SA, 4-vector displacements are (71)ΔxRAμ=Lνμ(V)ΔxRμ={γ(V)[-Δx-VΔt],0,0,cγ(V)[Δt+Vc2Δx]},ΔxLAμ=Lνμ(V)ΔxLμ={γ(V)[+Δx-VΔt],0,0,cγ(V)[Δt-Vc2Δx]}.
Spatial displacements ΔxLA and ΔxRA associated with forces FLA and FRA, respectively, are different in SA as well as time intervals: ΔtLA≠ΔtRA [1].It is assumed that 4-vector forces acting on extended bodies are transformed in the same way as 4-vector forces acting on point particles [69, 70]. Force FA=(FxA,FyA,FzA) measured with respect to SA is given, in terms of force F=(Fx,Fy,Fz) measured with respect to S0 and the velocity V=(V,0,0) of frame SA with respect to frame S0, by [71] (72)FxA=Fx-(V/c2)(Fxvx+Fyvy)1-vxV/c2,FyA=γ-1(V)Fy1-vxV/c2.
For horizontal forces (Fy=Fz=0) and Fx=±F, (73)FLA=qEx,FRA=-qEx.
Total impulse ℐA and work WA in frame SA are given by (74)IA=FLAΔtLA+FRAΔtRA=-γ(V)[2qExΔxc-2]V,WA=FLAΔxLA+FRAΔxRA=γ(V)[2qExΔx],
with (75)WAμ={cIA,0,0,WA}.
This is the same result for WAμ obtained by using Lorentz transformation on 4-vector work in S0 given by (60), WAμ=ℒνμ(V)WAν.The same result is obtained if one considers relativistic transformation of electric and magnetic fields ([72], Section 10.5).The4×4-tensor electromagnetic field in SA, ℰAνμ, can be obtained as (Appendix A) ([20], p. 281) (76)EAνμ=Lνμ(V)EAνμL+νμ(V)={000-Ex00000000-Ex000}.
The electromagnetic field does not change in frame SA with respect to the field in S0 (no magnetic field appears in SA). The 4-vector work WLAμ and WRAμ can by obtained by applying its definition in SA: (77)WLAμ=qEAνμΔxLAν={cqExγ(V)[Δt-(Vc2)Δx],0,0,qγ(V)×Ex[Δx-VΔt][Δt-(Vc2)Δx]},WRAμ=-qEAνμΔxRAν={-cqEx[Δt+(Vc2)Δx],0,0,-q×Ex[-Δx-VΔt][Δt+(Vc2)Δx]}.InSA total impulse, (78)IA=-cγ(V)[2qEΔxc2]V,
is not zero and total work is (79)WA=γ(V)[2qEΔx].
The 4-vector work WAμ in SA is then  (80)WAμ=WLAμ+WRAμ={-cγ(V)[2qEΔxc2]V,0,0,γ(V)[2qEΔx]},
a result previously obtained.


#### 7.2.3. Heat inSA

From relativistic Doppler effect, frequencyνA in frame SA for a photon emitted in frame S0 with frequency ν and direction (cosθ,sinθ,0) is given by (81)νA=γ(V)[1-β(V)cos⁡θ]ν.
The relativistic aberration effect [73] indicates that photon direction (cosθA,sinθA,0) as measured in SA is (82)cos⁡θA=cos⁡θ-β(V)1-β(V)cos⁡θ,sin⁡θA=γ-1(V)sin⁡θ1-β(V)cos⁡θ.In frameS0, for a photon jth emitted with angle θj=+π/2 there is another photon nth emitted with angle θn=-π/2 (both with frequency ν, in order to assure zero total linear momentum for emitted photons). In frame SA photons are emitted with frequency νA=γ(V)ν, higher than frequency measured in S0 (transverse Doppler effect). Photons in SA are emitted with angles larger than π/2 (in absolute value) (Figure 8).In frameSA, total linear momentum and total energy are easy to obtain for this pair of opposite emitted photons (in S0) and for the N/2 pairs of emitted photons pairs, total linear momentum and energy are (83)pxpA=-cγ(V)[Nhνc-2]V,pypA=0,EpA=γ(V)Nhν=γ(V)ṄhνΔt.
For the ensemble of emitted photons the 4-vector is (84)Qpμ={-cγ(V)MpV,0,0,γ(V)Nhν},Mp=Nhνc-2.
In frame SA, N/2 are emitted with angles ±θA, with zero linear momentum in y direction. These photons carry linear momentum in direction -x and its total linear momentum is pxA=-γ(V)(hν/c2)V.The 4-vector heat emitted by the bodyQAμ=-Qpμ is (85)QAμ={c[γ(V)Nhνc2]V,0,0,-γ(V)Nhν}.
This result was obtained from Lorentz transformation on the 4-vector heat in S0, given by (63), QAμ=ℒνμ(V)Qν. An ensemble of photons with energy ṄhνΔt and zero linear momentum, has an inertia Nhνc-2 associated. The ensemble of N thermal photons does not transform between frames like a photon but like an elementary particle.Total energy for photonsEpA as measured in SA and total energy for photons Up in S0 are related as  (86)EpA=γ(V)Up.
The norm ∥QAμ∥ of the 4-vector heat QAμ is an invariant, with (87)‖QAμ‖=Up=ṄhνΔt.


#### 7.2.4. First Law inSA

From the first law inSA, UfAμ-UiAμ=WAμ+QAμ and the 4-vectors given in (69),(80),(85), one obtains (88)0=-cγ(V)[2qEΔxc2]V+cγ(V)[Nhνc2]V,0=γ(V)[2qEΔx]-γ(V)[Nhν].
These two equations are redundant, and one obtains (89)Nhν=2qEΔx.
This is a result previously obtained in frame S0.In frameSA, total energy EA, linear momentum pA, and energy function U remain constant during the compression process—these magnitudes remain constant in frame S0 too. In frame SA the set of forces applied to the gas produces a net impulse —in contrast with the net zero impulse in frame S0—and the ensemble of photons emitted during the process carries linear momentum—in contrast with the net zero momentum for emitted photons in S0. Impulse and work provided by external forces on the gas, represented by the 4-vector WAμ, are transmitted to the ensemble of emitted photons, represented by QAμ. When photons are emitted, the gas gets a (positive) linear momentum due to this emission of thermal radiation that compensates for the (negative) linear momentum provided by forces applied on it, with a result of total zero linear momentum variation. Similarly, energy carried for photons is provided by the work done by the forces. This transformation, QAμ=-WAμ, is the relativistic generalization for the complete transformation of work W into heat Q for an isothermal process on an ideal gas, U=U(T), with ΔU=0. The description of this process in frame S0 is the usual description in classical thermodynamics, with energy associated with heat but with no linear momentum.


## 7.1. Compression in FrameS0

As privileged frameS0 we take the frame in which cylinder walls, plate parallel capacitor and thermal reservoir walls are at rest. During the compression process, forces on gas are applied simultaneously, during time interval Δt. Thermal radiation is interchanged with zero impulse in S0 and the gas center of mass remains at the same point, with initial and final zero velocity.


### 7.1.1. Energy Function inS0

For simplicity, we assume that the atoms of He inside the cylinder possess only translational energy,  that is, all atoms are in its ground electronic state. In general, one can assume that gas velocity distribution is Juttner distribution [57] ([9], pp. 289–293). For simplicity, one assumes that atoms are randomly distributed inside the container and that every atom has the same translational energy,  that is, every He atom moves with the same speed v=v(T) (monokinetic approximation [50]), same modulus, but with different vectorial components v=(vx,vy,vz), v=|v|. In this approximation, v(T)=aT1/2. Constant a is obtained by imposing (46)k=[γ(v)-1]u=32kBT,
where k=[γ(v)-1]u is the kinetic energy of a He atom and u its energy function (Section 3).Linear momentumpj=(pxj,pyj,0) (for simplicity we assume x-y as movement of the atoms) and total energy Ej for the jth atom are (47)pxj=γ(v)mvx,pyj=γ(v)mvy,ej=γ(v)u,m=uc-2.
Initial total linear momentum pi=(pxi,pyi,0) and initial total energy (energy function ) Ui are given by (48)pxi=∑jpxj=γ(v)m∑kvxj=0,pyi=∑jpyj=γ(v)m∑kvyj=0,Ui=U(T)=γ(v)∑ju=NAγ(v)u.
The 4-vector initial energy function Uiμ is then  (49)Uiμ={0,0,0,Ui};Ui=γ(v)NAu.
Energy function Ui depends on temperature through temperature dependence on velocity v=v(T). In S0 (pi=0), total system energy ([63], Sec. 8.3) is, by definition, its energy function Ui=Ki+𝒰i, sum of the kinetic energy of helium atoms Ki=∑kkk=NA[γ(v)-1]u, and 𝒰i=∑ku=NAu.For an ideal gas in an isothermal process, energy function remains constant, as well as the temperature, and also He atom speed. The 4-vector final energy functionUfμ is then: (50)Ufμ={0,0,0,Uf},
with Uf=Ui=U(T).


### 7.1.2. Work inS0

Forces on pistons are described as produced by the interaction of an electric charge with an electromagnetic field (Figure7). Static electric positive +q and negative -q charges are located on right and left pistons, respectively. The whole device, gas plus pistons, is located inside the homogeneous electric field E produced by a plane-parallel capacitor ([64], Chap. 13) with charge surface density σ.In frameS0 the capacitor is at rest. A horizontal electric field Ex is created inside the capacitor. For this (uniform) electric field the potential 4-vector Φμ={Ax,Ay,Az,ϕ}, where A=(Ax,Ay,Az) and ϕ are the vector potential and the scalar potential, respectively, is given by the contravariant 4-vector [65]: (51)Φμ={0,0,0,-Exx},Ex=σ2ϵ0.The electromagnetic4×4-tensor ℰνμ—double contravariant—is given by ([51], Section 42) (52)Eνμ=∂Φν∂xμ-∂Φμ∂xν.
For the horizontal plane-parallel capacitor the 4×4-tensor ℰνμ is (53)Eνμ={000Ex00000000-Ex000}.
The mixed 4×4-tensor ℰνμ is obtained as (54)Eνμ=gνξEξμ={000-Ex00000000-Ex000}.Initially, pistons are locked by a blocked mechanism (b in Figure7). A laser (l in Figure 7) and a beam splitter (s in Figure 7) that is located just between the pistons, are used to release simultaneously both pistons. When the laser is turned on the split beams will arrive at the blocked mechanism of both left and right pistons and, at time t=0, are unlocked allowing electric field charges interaction simultaneously on both L and R pistons [62].When pistons are released, the Minkowski forceFkμ on kth piston is (55)Fkμ=qEνμvkμ,vkμ=dxkνdτk,
where q is the electric charge on piston, ℰνμ is the 4×4-tensor electromagnetic field, and vkμ and τk are the kth piston 4-vector velocity and proper time, respectively [66].For the left piston (subindex L), displacementΔxL=Δx and velocity vL=ΔxL/Δt=v, 4-vector ΔxLμ and 4-vector velocity vLμ, are respectively (56)ΔxLμ={Δx,0,0,cΔt},vLμ=γ(v){v,0,0,c}.
For the right piston (R), displacement ΔxR=-Δx and velocity vR=ΔxR/Δt=-v, 4-vector ΔxRμ and 4-vector velocity vRμ, are respectively (57)ΔxRμ={-Δx,0,0,cΔt},vRμ=γ(v){-v,0,0,c}.The Minkowski forces due to electromagnetic interaction are:(58)FLμ=-qEνμvLν=γ(v){+cqEx,0,0,qExv},FRμ=+qEνμvRν=γ(v){-cqEx,0,0,qExv}.
With forces acting on pistons, FL=(qEx,0,0) and FR=(-qEx,0,0), 4-vectors work are, respectively, (59)WLμ=-qEνμΔxLν={+cqExΔt,0,0,qExΔx},WRμ=+qEνμΔxRν={-cqExΔt,0,0,qExΔx}.
The total 4-vector work Wμ is then  (60)Wμ=WLμ+WRμ={0,0,0,2qExΔx}.


### 7.1.3. Heat inS0

During (slow) gas compression,N=ṄΔt photons with frequency ν(T) are emitted, with zero total linear momentum. Photons are emitted through the horizontal walls of the cylinder, N/2 photons are emitted in direction θ+=π/2 and N/2 photons in direction θ-=-π/2. Total linear momentum for this ensemble of photons pp=(pxp,pyp,0) is (61)pxp=0,pyp=N2hνsin⁡θ++N2hνsin⁡θ-=0.
Total energy of these emitted photons E=Nhν, with pp=0, is its energy function Up: (62)Up=Nhν.
According to the principle of inertia of energy, these N photons have inertia [63] ℳp=Upc-2. The 4-vector for heat Qμ (thermal radiation emitted from the point of view of the system) is then given by  (63)Qμ={0,0,0,-ṄhνΔt}.


### 7.1.4. First Law inS0

From first lawUfμ-Uiμ=Wμ+Qμ, one obtains 0=2qExΔx-ṄhνΔt or (64)ṄhνΔt=2qExΔx,Ṅhν=2qExv.
The configurational work done on the gas provides the energy emitted as heat.


## 7.1.1. Energy Function inS0

For simplicity, we assume that the atoms of He inside the cylinder possess only translational energy,  that is, all atoms are in its ground electronic state. In general, one can assume that gas velocity distribution is Juttner distribution [57] ([9], pp. 289–293). For simplicity, one assumes that atoms are randomly distributed inside the container and that every atom has the same translational energy,  that is, every He atom moves with the same speed v=v(T) (monokinetic approximation [50]), same modulus, but with different vectorial components v=(vx,vy,vz), v=|v|. In this approximation, v(T)=aT1/2. Constant a is obtained by imposing (46)k=[γ(v)-1]u=32kBT,
where k=[γ(v)-1]u is the kinetic energy of a He atom and u its energy function (Section 3).Linear momentumpj=(pxj,pyj,0) (for simplicity we assume x-y as movement of the atoms) and total energy Ej for the jth atom are (47)pxj=γ(v)mvx,pyj=γ(v)mvy,ej=γ(v)u,m=uc-2.
Initial total linear momentum pi=(pxi,pyi,0) and initial total energy (energy function ) Ui are given by (48)pxi=∑jpxj=γ(v)m∑kvxj=0,pyi=∑jpyj=γ(v)m∑kvyj=0,Ui=U(T)=γ(v)∑ju=NAγ(v)u.
The 4-vector initial energy function Uiμ is then  (49)Uiμ={0,0,0,Ui};Ui=γ(v)NAu.
Energy function Ui depends on temperature through temperature dependence on velocity v=v(T). In S0 (pi=0), total system energy ([63], Sec. 8.3) is, by definition, its energy function Ui=Ki+𝒰i, sum of the kinetic energy of helium atoms Ki=∑kkk=NA[γ(v)-1]u, and 𝒰i=∑ku=NAu.For an ideal gas in an isothermal process, energy function remains constant, as well as the temperature, and also He atom speed. The 4-vector final energy functionUfμ is then: (50)Ufμ={0,0,0,Uf},
with Uf=Ui=U(T).


## 7.1.2. Work inS0

Forces on pistons are described as produced by the interaction of an electric charge with an electromagnetic field (Figure7). Static electric positive +q and negative -q charges are located on right and left pistons, respectively. The whole device, gas plus pistons, is located inside the homogeneous electric field E produced by a plane-parallel capacitor ([64], Chap. 13) with charge surface density σ.In frameS0 the capacitor is at rest. A horizontal electric field Ex is created inside the capacitor. For this (uniform) electric field the potential 4-vector Φμ={Ax,Ay,Az,ϕ}, where A=(Ax,Ay,Az) and ϕ are the vector potential and the scalar potential, respectively, is given by the contravariant 4-vector [65]: (51)Φμ={0,0,0,-Exx},Ex=σ2ϵ0.The electromagnetic4×4-tensor ℰνμ—double contravariant—is given by ([51], Section 42) (52)Eνμ=∂Φν∂xμ-∂Φμ∂xν.
For the horizontal plane-parallel capacitor the 4×4-tensor ℰνμ is (53)Eνμ={000Ex00000000-Ex000}.
The mixed 4×4-tensor ℰνμ is obtained as (54)Eνμ=gνξEξμ={000-Ex00000000-Ex000}.Initially, pistons are locked by a blocked mechanism (b in Figure7). A laser (l in Figure 7) and a beam splitter (s in Figure 7) that is located just between the pistons, are used to release simultaneously both pistons. When the laser is turned on the split beams will arrive at the blocked mechanism of both left and right pistons and, at time t=0, are unlocked allowing electric field charges interaction simultaneously on both L and R pistons [62].When pistons are released, the Minkowski forceFkμ on kth piston is (55)Fkμ=qEνμvkμ,vkμ=dxkνdτk,
where q is the electric charge on piston, ℰνμ is the 4×4-tensor electromagnetic field, and vkμ and τk are the kth piston 4-vector velocity and proper time, respectively [66].For the left piston (subindex L), displacementΔxL=Δx and velocity vL=ΔxL/Δt=v, 4-vector ΔxLμ and 4-vector velocity vLμ, are respectively (56)ΔxLμ={Δx,0,0,cΔt},vLμ=γ(v){v,0,0,c}.
For the right piston (R), displacement ΔxR=-Δx and velocity vR=ΔxR/Δt=-v, 4-vector ΔxRμ and 4-vector velocity vRμ, are respectively (57)ΔxRμ={-Δx,0,0,cΔt},vRμ=γ(v){-v,0,0,c}.The Minkowski forces due to electromagnetic interaction are:(58)FLμ=-qEνμvLν=γ(v){+cqEx,0,0,qExv},FRμ=+qEνμvRν=γ(v){-cqEx,0,0,qExv}.
With forces acting on pistons, FL=(qEx,0,0) and FR=(-qEx,0,0), 4-vectors work are, respectively, (59)WLμ=-qEνμΔxLν={+cqExΔt,0,0,qExΔx},WRμ=+qEνμΔxRν={-cqExΔt,0,0,qExΔx}.
The total 4-vector work Wμ is then  (60)Wμ=WLμ+WRμ={0,0,0,2qExΔx}.


## 7.1.3. Heat inS0

During (slow) gas compression,N=ṄΔt photons with frequency ν(T) are emitted, with zero total linear momentum. Photons are emitted through the horizontal walls of the cylinder, N/2 photons are emitted in direction θ+=π/2 and N/2 photons in direction θ-=-π/2. Total linear momentum for this ensemble of photons pp=(pxp,pyp,0) is (61)pxp=0,pyp=N2hνsin⁡θ++N2hνsin⁡θ-=0.
Total energy of these emitted photons E=Nhν, with pp=0, is its energy function Up: (62)Up=Nhν.
According to the principle of inertia of energy, these N photons have inertia [63] ℳp=Upc-2. The 4-vector for heat Qμ (thermal radiation emitted from the point of view of the system) is then given by  (63)Qμ={0,0,0,-ṄhνΔt}.


## 7.1.4. First Law inS0

From first lawUfμ-Uiμ=Wμ+Qμ, one obtains 0=2qExΔx-ṄhνΔt or (64)ṄhνΔt=2qExΔx,Ṅhν=2qExv.
The configurational work done on the gas provides the energy emitted as heat.


## 7.2. Compression in FrameSA

An observer in frameSA obtains the corresponding 4-vector UiAμ, UfAμ, WAμ, and QAμ by measuring different magnitudes—displacements ΔxA, time intervals ΔtA, velocities vA, forces FA, photon frequency νA, and so forth—in its own frame (Figure 8). With first law in frame SA expressed as UfAμ-UiAμ=WAμ+QAμ, the Lorentz invariance of this equation assures that the same result as in S0,  that is, ṄhνΔt=2qEΔx, is obtained.Figure 8
Isothermal compression given in Figure7, described from the point of view of an observer in frame SA, standard configuration, with velocity V with respect to frame S0. In frame SA forces are not applied simultaneously and heat carries linear momentum.


### 7.2.1. Energy Function inSA

Let there be one mol of He atoms moving inside the cylinder. In frameS0, with zero total linear momentum, velocities of atoms are measured simultaneously. During the same interval of time Δt, displacement Δrj=(Δxj,Δyj,0) for the jth atom is measured and its 3-vector velocity is given by vk=(Δrj/Δt)=(vx,vy,0). In order to ensure that in frame S0 total linear momentum is zero, for every atom j moving with velocity vj=(vxj,vyj,0) there must exist another atom n moving with opposite velocity vn=(vxn,vyn,0), such that vxj=-vxn and vyj=-vyn.The velocityvA=(vxsA,vysA,0) of the sth atom is given, in terms of its velocity v=(vxs,vys,0) measured in S0 and velocity V=(V,0,0) of frame SA with respect to frame S0, by the equation ([51], Section 12) (65)vsA=1γ(V)[1-vxsV/c2](γ(V)[vxs-V],vys,0),
with the useful relations ([67], p. 69) (66)γ(vsA)=γ(v)γ(V)[1-vxsVc2],γ(vsA)vxsA=γ(v)γ(V)(vxs-V),γ(vsA)vysA=γ(v)vys.For every pairj-n of opposite atoms, total momentum and total energy are easily obtained using the previous transformations: (67)px(j+n)A=γ(vjA)(uc-2)vxjA+γ(vrA)(uc-2)vxnA=γ(V)[2γ(v)uc-2]V,py(j+n)A=γ(vjA)(uc-2)vyjA+γ(vrA)(uc-2)vynA=0,
and total energy is given by (68)EjA=γ(vjA)u=γ(v)γ(V)[1-vxVc2]u,EnA=γ(vnA)u=γ(v)γ(V)[1+vxVc2]u,E(j+n)A=EjA+EnA=γ(V)[2γ(v)u].
For the N/2 total pairs of opposite atoms, one has(69)UiAμ={-cγ(V)[γ(v)NAuc2]V,0,0,γ(V)[γ(v)NAu]}={-cγ(V)MV,0,0,γ(V)U(T)},M=U(T)c-2
This is the same result obtained from Lorentz transformation [68] on the 4-vector energy function in S0, given in (49), UiAμ=ℒνμUiν.A similar description for 4-vector final energy functionUfAμ, with UfAμ=ℒνμUfν, is(70)UfAμ{-cγ(V)MV,0,0,γ(V)U(T)}.


### 7.2.2. Work inSA

By considering the locked-unlocked piston set, laser beam, splitter, blocked mechanism described previously in frameS0, it is evident that forces that are simultaneously applied in S0 are not simultaneous in frame SA (Figure 8) [1].To obtain the 4-vectorWAμ in SA, relativistic transformations of time intervals, spatial displacements, and forces must be used.In SA, 4-vector displacements are (71)ΔxRAμ=Lνμ(V)ΔxRμ={γ(V)[-Δx-VΔt],0,0,cγ(V)[Δt+Vc2Δx]},ΔxLAμ=Lνμ(V)ΔxLμ={γ(V)[+Δx-VΔt],0,0,cγ(V)[Δt-Vc2Δx]}.
Spatial displacements ΔxLA and ΔxRA associated with forces FLA and FRA, respectively, are different in SA as well as time intervals: ΔtLA≠ΔtRA [1].It is assumed that 4-vector forces acting on extended bodies are transformed in the same way as 4-vector forces acting on point particles [69, 70]. Force FA=(FxA,FyA,FzA) measured with respect to SA is given, in terms of force F=(Fx,Fy,Fz) measured with respect to S0 and the velocity V=(V,0,0) of frame SA with respect to frame S0, by [71] (72)FxA=Fx-(V/c2)(Fxvx+Fyvy)1-vxV/c2,FyA=γ-1(V)Fy1-vxV/c2.
For horizontal forces (Fy=Fz=0) and Fx=±F, (73)FLA=qEx,FRA=-qEx.
Total impulse ℐA and work WA in frame SA are given by (74)IA=FLAΔtLA+FRAΔtRA=-γ(V)[2qExΔxc-2]V,WA=FLAΔxLA+FRAΔxRA=γ(V)[2qExΔx],
with (75)WAμ={cIA,0,0,WA}.
This is the same result for WAμ obtained by using Lorentz transformation on 4-vector work in S0 given by (60), WAμ=ℒνμ(V)WAν.The same result is obtained if one considers relativistic transformation of electric and magnetic fields ([72], Section 10.5).The4×4-tensor electromagnetic field in SA, ℰAνμ, can be obtained as (Appendix A) ([20], p. 281) (76)EAνμ=Lνμ(V)EAνμL+νμ(V)={000-Ex00000000-Ex000}.
The electromagnetic field does not change in frame SA with respect to the field in S0 (no magnetic field appears in SA). The 4-vector work WLAμ and WRAμ can by obtained by applying its definition in SA: (77)WLAμ=qEAνμΔxLAν={cqExγ(V)[Δt-(Vc2)Δx],0,0,qγ(V)×Ex[Δx-VΔt][Δt-(Vc2)Δx]},WRAμ=-qEAνμΔxRAν={-cqEx[Δt+(Vc2)Δx],0,0,-q×Ex[-Δx-VΔt][Δt+(Vc2)Δx]}.InSA total impulse, (78)IA=-cγ(V)[2qEΔxc2]V,
is not zero and total work is (79)WA=γ(V)[2qEΔx].
The 4-vector work WAμ in SA is then  (80)WAμ=WLAμ+WRAμ={-cγ(V)[2qEΔxc2]V,0,0,γ(V)[2qEΔx]},
a result previously obtained.


### 7.2.3. Heat inSA

From relativistic Doppler effect, frequencyνA in frame SA for a photon emitted in frame S0 with frequency ν and direction (cosθ,sinθ,0) is given by (81)νA=γ(V)[1-β(V)cos⁡θ]ν.
The relativistic aberration effect [73] indicates that photon direction (cosθA,sinθA,0) as measured in SA is (82)cos⁡θA=cos⁡θ-β(V)1-β(V)cos⁡θ,sin⁡θA=γ-1(V)sin⁡θ1-β(V)cos⁡θ.In frameS0, for a photon jth emitted with angle θj=+π/2 there is another photon nth emitted with angle θn=-π/2 (both with frequency ν, in order to assure zero total linear momentum for emitted photons). In frame SA photons are emitted with frequency νA=γ(V)ν, higher than frequency measured in S0 (transverse Doppler effect). Photons in SA are emitted with angles larger than π/2 (in absolute value) (Figure 8).In frameSA, total linear momentum and total energy are easy to obtain for this pair of opposite emitted photons (in S0) and for the N/2 pairs of emitted photons pairs, total linear momentum and energy are (83)pxpA=-cγ(V)[Nhνc-2]V,pypA=0,EpA=γ(V)Nhν=γ(V)ṄhνΔt.
For the ensemble of emitted photons the 4-vector is (84)Qpμ={-cγ(V)MpV,0,0,γ(V)Nhν},Mp=Nhνc-2.
In frame SA, N/2 are emitted with angles ±θA, with zero linear momentum in y direction. These photons carry linear momentum in direction -x and its total linear momentum is pxA=-γ(V)(hν/c2)V.The 4-vector heat emitted by the bodyQAμ=-Qpμ is (85)QAμ={c[γ(V)Nhνc2]V,0,0,-γ(V)Nhν}.
This result was obtained from Lorentz transformation on the 4-vector heat in S0, given by (63), QAμ=ℒνμ(V)Qν. An ensemble of photons with energy ṄhνΔt and zero linear momentum, has an inertia Nhνc-2 associated. The ensemble of N thermal photons does not transform between frames like a photon but like an elementary particle.Total energy for photonsEpA as measured in SA and total energy for photons Up in S0 are related as  (86)EpA=γ(V)Up.
The norm ∥QAμ∥ of the 4-vector heat QAμ is an invariant, with (87)‖QAμ‖=Up=ṄhνΔt.


### 7.2.4. First Law inSA

From the first law inSA, UfAμ-UiAμ=WAμ+QAμ and the 4-vectors given in (69),(80),(85), one obtains (88)0=-cγ(V)[2qEΔxc2]V+cγ(V)[Nhνc2]V,0=γ(V)[2qEΔx]-γ(V)[Nhν].
These two equations are redundant, and one obtains (89)Nhν=2qEΔx.
This is a result previously obtained in frame S0.In frameSA, total energy EA, linear momentum pA, and energy function U remain constant during the compression process—these magnitudes remain constant in frame S0 too. In frame SA the set of forces applied to the gas produces a net impulse —in contrast with the net zero impulse in frame S0—and the ensemble of photons emitted during the process carries linear momentum—in contrast with the net zero momentum for emitted photons in S0. Impulse and work provided by external forces on the gas, represented by the 4-vector WAμ, are transmitted to the ensemble of emitted photons, represented by QAμ. When photons are emitted, the gas gets a (positive) linear momentum due to this emission of thermal radiation that compensates for the (negative) linear momentum provided by forces applied on it, with a result of total zero linear momentum variation. Similarly, energy carried for photons is provided by the work done by the forces. This transformation, QAμ=-WAμ, is the relativistic generalization for the complete transformation of work W into heat Q for an isothermal process on an ideal gas, U=U(T), with ΔU=0. The description of this process in frame S0 is the usual description in classical thermodynamics, with energy associated with heat but with no linear momentum.


## 7.2.1. Energy Function inSA

Let there be one mol of He atoms moving inside the cylinder. In frameS0, with zero total linear momentum, velocities of atoms are measured simultaneously. During the same interval of time Δt, displacement Δrj=(Δxj,Δyj,0) for the jth atom is measured and its 3-vector velocity is given by vk=(Δrj/Δt)=(vx,vy,0). In order to ensure that in frame S0 total linear momentum is zero, for every atom j moving with velocity vj=(vxj,vyj,0) there must exist another atom n moving with opposite velocity vn=(vxn,vyn,0), such that vxj=-vxn and vyj=-vyn.The velocityvA=(vxsA,vysA,0) of the sth atom is given, in terms of its velocity v=(vxs,vys,0) measured in S0 and velocity V=(V,0,0) of frame SA with respect to frame S0, by the equation ([51], Section 12) (65)vsA=1γ(V)[1-vxsV/c2](γ(V)[vxs-V],vys,0),
with the useful relations ([67], p. 69) (66)γ(vsA)=γ(v)γ(V)[1-vxsVc2],γ(vsA)vxsA=γ(v)γ(V)(vxs-V),γ(vsA)vysA=γ(v)vys.For every pairj-n of opposite atoms, total momentum and total energy are easily obtained using the previous transformations: (67)px(j+n)A=γ(vjA)(uc-2)vxjA+γ(vrA)(uc-2)vxnA=γ(V)[2γ(v)uc-2]V,py(j+n)A=γ(vjA)(uc-2)vyjA+γ(vrA)(uc-2)vynA=0,
and total energy is given by (68)EjA=γ(vjA)u=γ(v)γ(V)[1-vxVc2]u,EnA=γ(vnA)u=γ(v)γ(V)[1+vxVc2]u,E(j+n)A=EjA+EnA=γ(V)[2γ(v)u].
For the N/2 total pairs of opposite atoms, one has(69)UiAμ={-cγ(V)[γ(v)NAuc2]V,0,0,γ(V)[γ(v)NAu]}={-cγ(V)MV,0,0,γ(V)U(T)},M=U(T)c-2
This is the same result obtained from Lorentz transformation [68] on the 4-vector energy function in S0, given in (49), UiAμ=ℒνμUiν.A similar description for 4-vector final energy functionUfAμ, with UfAμ=ℒνμUfν, is(70)UfAμ{-cγ(V)MV,0,0,γ(V)U(T)}.


## 7.2.2. Work inSA

By considering the locked-unlocked piston set, laser beam, splitter, blocked mechanism described previously in frameS0, it is evident that forces that are simultaneously applied in S0 are not simultaneous in frame SA (Figure 8) [1].To obtain the 4-vectorWAμ in SA, relativistic transformations of time intervals, spatial displacements, and forces must be used.In SA, 4-vector displacements are (71)ΔxRAμ=Lνμ(V)ΔxRμ={γ(V)[-Δx-VΔt],0,0,cγ(V)[Δt+Vc2Δx]},ΔxLAμ=Lνμ(V)ΔxLμ={γ(V)[+Δx-VΔt],0,0,cγ(V)[Δt-Vc2Δx]}.
Spatial displacements ΔxLA and ΔxRA associated with forces FLA and FRA, respectively, are different in SA as well as time intervals: ΔtLA≠ΔtRA [1].It is assumed that 4-vector forces acting on extended bodies are transformed in the same way as 4-vector forces acting on point particles [69, 70]. Force FA=(FxA,FyA,FzA) measured with respect to SA is given, in terms of force F=(Fx,Fy,Fz) measured with respect to S0 and the velocity V=(V,0,0) of frame SA with respect to frame S0, by [71] (72)FxA=Fx-(V/c2)(Fxvx+Fyvy)1-vxV/c2,FyA=γ-1(V)Fy1-vxV/c2.
For horizontal forces (Fy=Fz=0) and Fx=±F, (73)FLA=qEx,FRA=-qEx.
Total impulse ℐA and work WA in frame SA are given by (74)IA=FLAΔtLA+FRAΔtRA=-γ(V)[2qExΔxc-2]V,WA=FLAΔxLA+FRAΔxRA=γ(V)[2qExΔx],
with (75)WAμ={cIA,0,0,WA}.
This is the same result for WAμ obtained by using Lorentz transformation on 4-vector work in S0 given by (60), WAμ=ℒνμ(V)WAν.The same result is obtained if one considers relativistic transformation of electric and magnetic fields ([72], Section 10.5).The4×4-tensor electromagnetic field in SA, ℰAνμ, can be obtained as (Appendix A) ([20], p. 281) (76)EAνμ=Lνμ(V)EAνμL+νμ(V)={000-Ex00000000-Ex000}.
The electromagnetic field does not change in frame SA with respect to the field in S0 (no magnetic field appears in SA). The 4-vector work WLAμ and WRAμ can by obtained by applying its definition in SA: (77)WLAμ=qEAνμΔxLAν={cqExγ(V)[Δt-(Vc2)Δx],0,0,qγ(V)×Ex[Δx-VΔt][Δt-(Vc2)Δx]},WRAμ=-qEAνμΔxRAν={-cqEx[Δt+(Vc2)Δx],0,0,-q×Ex[-Δx-VΔt][Δt+(Vc2)Δx]}.InSA total impulse, (78)IA=-cγ(V)[2qEΔxc2]V,
is not zero and total work is (79)WA=γ(V)[2qEΔx].
The 4-vector work WAμ in SA is then  (80)WAμ=WLAμ+WRAμ={-cγ(V)[2qEΔxc2]V,0,0,γ(V)[2qEΔx]},
a result previously obtained.


## 7.2.3. Heat inSA

From relativistic Doppler effect, frequencyνA in frame SA for a photon emitted in frame S0 with frequency ν and direction (cosθ,sinθ,0) is given by (81)νA=γ(V)[1-β(V)cos⁡θ]ν.
The relativistic aberration effect [73] indicates that photon direction (cosθA,sinθA,0) as measured in SA is (82)cos⁡θA=cos⁡θ-β(V)1-β(V)cos⁡θ,sin⁡θA=γ-1(V)sin⁡θ1-β(V)cos⁡θ.In frameS0, for a photon jth emitted with angle θj=+π/2 there is another photon nth emitted with angle θn=-π/2 (both with frequency ν, in order to assure zero total linear momentum for emitted photons). In frame SA photons are emitted with frequency νA=γ(V)ν, higher than frequency measured in S0 (transverse Doppler effect). Photons in SA are emitted with angles larger than π/2 (in absolute value) (Figure 8).In frameSA, total linear momentum and total energy are easy to obtain for this pair of opposite emitted photons (in S0) and for the N/2 pairs of emitted photons pairs, total linear momentum and energy are (83)pxpA=-cγ(V)[Nhνc-2]V,pypA=0,EpA=γ(V)Nhν=γ(V)ṄhνΔt.
For the ensemble of emitted photons the 4-vector is (84)Qpμ={-cγ(V)MpV,0,0,γ(V)Nhν},Mp=Nhνc-2.
In frame SA, N/2 are emitted with angles ±θA, with zero linear momentum in y direction. These photons carry linear momentum in direction -x and its total linear momentum is pxA=-γ(V)(hν/c2)V.The 4-vector heat emitted by the bodyQAμ=-Qpμ is (85)QAμ={c[γ(V)Nhνc2]V,0,0,-γ(V)Nhν}.
This result was obtained from Lorentz transformation on the 4-vector heat in S0, given by (63), QAμ=ℒνμ(V)Qν. An ensemble of photons with energy ṄhνΔt and zero linear momentum, has an inertia Nhνc-2 associated. The ensemble of N thermal photons does not transform between frames like a photon but like an elementary particle.Total energy for photonsEpA as measured in SA and total energy for photons Up in S0 are related as  (86)EpA=γ(V)Up.
The norm ∥QAμ∥ of the 4-vector heat QAμ is an invariant, with (87)‖QAμ‖=Up=ṄhνΔt.


## 7.2.4. First Law inSA

From the first law inSA, UfAμ-UiAμ=WAμ+QAμ and the 4-vectors given in (69),(80),(85), one obtains (88)0=-cγ(V)[2qEΔxc2]V+cγ(V)[Nhνc2]V,0=γ(V)[2qEΔx]-γ(V)[Nhν].
These two equations are redundant, and one obtains (89)Nhν=2qEΔx.
This is a result previously obtained in frame S0.In frameSA, total energy EA, linear momentum pA, and energy function U remain constant during the compression process—these magnitudes remain constant in frame S0 too. In frame SA the set of forces applied to the gas produces a net impulse —in contrast with the net zero impulse in frame S0—and the ensemble of photons emitted during the process carries linear momentum—in contrast with the net zero momentum for emitted photons in S0. Impulse and work provided by external forces on the gas, represented by the 4-vector WAμ, are transmitted to the ensemble of emitted photons, represented by QAμ. When photons are emitted, the gas gets a (positive) linear momentum due to this emission of thermal radiation that compensates for the (negative) linear momentum provided by forces applied on it, with a result of total zero linear momentum variation. Similarly, energy carried for photons is provided by the work done by the forces. This transformation, QAμ=-WAμ, is the relativistic generalization for the complete transformation of work W into heat Q for an isothermal process on an ideal gas, U=U(T), with ΔU=0. The description of this process in frame S0 is the usual description in classical thermodynamics, with energy associated with heat but with no linear momentum.


## 8. Conclusions

A coherent development of modern relativistic thermodynamics requires (i) a guarantee that the system behaves according to the principle of the inertia of energy,  that is, forces are applied in such a way that the system behaves as a whole [29] and (ii) that the experiment in frame S0 is performed in such a way that equations for elementary (point) particles can be applied to the extended thermodynamic system (principle of similitude). When this goal is achieved, in the asynchronous formulation formalism, the Minkowski 4-vector calculus in special relativity can be used for nonlocal (extended bodies) as well as for local (elementary particles) 4-vector quantities.For the description of the process performed by bodyZ (Section 1) using the special theory of Relativity with the Minkowski 4-vector formalism, (1)-(2) merge in the first law of relativistic thermodynamics: (90)dUμ=δWμ+δQμ.
According to the asynchronous formulation, in frame S0 body Z is instantaneously at rest ([74], p. 41), with vi=0, and vf=v, v=(vx,vy,vz), v=|v|,—with equations (91){γ(v)Mfvxγ(v)Mfvyγ(v)Mfvzγ(v)Uf-Ui}={∑kFxkdt∑kFykdt∑kFzkdtδW}+{000δQ},
with Fk conservative forces simultaneously applied to Z and with (92)δW=∑k(Fk⋅vk)dt,δQ=Ṅhνdt,Mi=Uic-2,Mf=Ufc-2.
where Ṅ is the flux of photons between Z and its thermal reservoir. Thus, the coherence of the developed formalism is based on the principle of the inertia of energy, with the Lorentz transformation, that guarantees that any kind of energy, U, W, or Q, that contributes to the temporal (energy) component of a 4-vector in frame S0, contributes with the inertia ℳ=Uc-2 to the spatial (linear momentum) component of the body in frame SA.If every forceFk acting on Z has its origin in the interaction of an electric charge q with an electromagnetic field, with a 4×4-tensor ℰkνμ, the 4-vector Minkowski force, Fkμ, is given by (93)Fkμ=qcEkνμvkμ=γ(vk){Fxk,Fyk,Fzk,c-1Fk⋅vk},
and the corresponding 4-vector work (infinitesimal), δWkμ, is (94)δWkμ=qEkνμdxkμ={cFxkdt,cFykdt,cFzkdt,Fk⋅dxk},
with (95)cFkμ=δWkμdτk
and with (96)δWμ=∑kδWkμ=c∑kFkμdτk,dτk=γ-1(vk)dt.In conclusion, the formulation of the first law of relativistic thermodynamics using Minkowski 4-vector formalism, introducing 4-vectorUμ and 4-vector Qμ, and considering an electromagnetic origin for the 4-vector work Wμ, allows us to solve exercises in classical physics, including concepts of mechanics, thermodynamics, and electromagnetism, in a complete Lorentz covariant formalism.



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*Source: 387351-2011-10-01.xml*