Hidden but real: new relativistic "paradox" exposing the ubiquity of hidden momentum
HHidden but real: new relativistic “paradox” exposing the ubiquity of hiddenmomentum
Daniel A. Turolla Vanzella ∗ Instituto de F´ısica de S˜ao Carlos, Universidade de S˜ao Paulo,Caixa Postal 369, 13560-970, S˜ao Carlos, S˜ao Paulo, Brazil (Dated: October 7, 2020)The tight connection between mass and energy unveiled by Special Relativity, summarized bythe iconic formula E = mc , has revolutionized our understanding of nature and even shaped ourpolitical world over the past century through its military application. It is certainly one of the mostexhaustively-tested and well-known equations of modern science. Although we have become used toits most obvious implication — mass-energy equivalence —, it is surprising that one of its subtle —yet, inevitable — consequences is still a matter of confusion: the so-called hidden momentum . Oftenconsidered as a peculiar feature of specific systems or as an artifact to avoid paradoxal situations,here we present a new relativistic “paradox” which exposes the true nature and ubiquity of hiddenmomentum. We also show that hidden momentum can be forced to reveal itself through observableeffects, hopefully putting an end to decades of controversy about its reality. I. INTRODUCTION
Einstein’s iconic mass-energy relation, E = mc , is ar-guably the most famous formula in modern science. Itexpresses the equivalence between total mass m and to-tal energy E of a system ( c being the speed of lightin vacuum), with wide-ranging consequences: from theunattainability of the speed of light for massive objects,to particle production in high-energy accelerators; fromthe origin of the energy of stars — less than 0.1% of thestar’s mass, converted into radiation over its entire ex-istence —, to violent bursts of gravitational waves frommerging black holes — some of them sourced by sev-eral solar masses converted into energy in a fraction ofa second. Given the importance and generality of mass-energy equivalence, it may strike as a surprise that oneof its subtle — but inevitable — consequences is still amatter of confusion: the concept of hidden momentum [1]— here generalized as the (purely relativistic) part of to-tal momentum which is not encoded in the motion of thecenter of mass-energy (CME) of the system.In Newtonian mechanics, the total momentum P of a closed mechanical system — one which does not exchange matter with “the rest of the universe” — is always givenby P = M V cm , where M is the total mass of the systemand V cm is the velocity of its center of mass. This result,known as the center-of-mass theorem, holds true regard-less whether the Newtonian system is subject to externalforces or not. In contrast to that, a variety of relativis-tic systems possessing nonzero total momentum in therest frame of their CME has been identified over the pastdecades (see, e.g., Refs. [1–15]). Here, the term “rela-tivistic” does not necessarily mean that large velocitiesare involved, but rather that different inertial-frame de-scriptions are supposed to be Lorentz covariant instead ∗ Electronic address: [email protected] of Galilean covariant. This covariance constraint leadsto “unfamiliar” results (i.e., results inexistent in Newto-nian mechanics) even in the rest frame of the system –such as non-zero total momentum. Such rest-frame mo-mentum has been termed hidden momentum (HM) [1],which now seems to be somewhat unfortunate becauseapparently this has misled many to interpret its nature assomehow distinct from “regular” momentum — as we ar-gue here, from the relativity-theory perspective, it is not.Adding confusion to the story, all systems in which HMhad been identified, until now, involved interaction withelectromagnetic fields — where it even bears an interest-ing connection with the difference between canonical andkinematic electromagnetic momenta [15] — and/or mov-ing inner parts subject to some external force field. Thismasked the generic nature of HM as if it were an exoticfeature — undesired by some — of peculiar interactionlaws or specific systems.Here, we present a new relativistic “paradox” whichshows that this view is limited and that HM is ubiq-uitous in a relativistic world (i.e., a world supposed tobe covariant under Lorentz transformations). Moreover,the general definition of HM we propose, freeing its com-putation from the rest frame of the system, leads to aformula which explicitly shows its relation to asymmetricexchange/flow of energy. Finally, in order to conclusivelyshow that HM is as real as it could be, in the end we dis-cuss an observational consequence of its existence. Thepresent analysis is intended to put an end to decades ofconfusion about the nature and reality of the so-calledHM.The paper is organized as follows. In Sec. II, wepresent the new relativistic “paradox” involving a heat-conducting bar analyzed from different inertial-frameperspectives. In order to make the presentation clearer,textbook-level relativistic calculations which supportstatements made in this Section are described in Ap-pendix A. In Sec. III, we put the heat-conducting-bar(HCB) “paradox” in context with other known pseudo- a r X i v : . [ phy s i c s . c l a ss - ph ] O c t aradoxes, stressing their origin in our intuition basedon space and time as separate entities rather thanin inconsistencies with known theories. We also ar-gue that the HCB “paradox” is a close thermal ana-logue of another relativistic pseudo-paradox known as“Mansuripur’s paradox” [6]. In Sec. IV, we argue thatHM is just another inevitable consequence of Einstein’smass-energy relation E = mc , obtaining a general ex-pression for HM as dipole moment of energy-exchangerate (Subsec. IV A) and then applying this general resultto solve the HCB “paradox” (Subsec. IV B). In Sec. V,we show that, contrary to widespread belief expressed inthe literature, the existence of HM in a system can beobjectively tested. Finally, in Sec. VI, we present ourfinal comments and discussion. It is important to stressthat (i) the HCB pseudo-paradox (Sec. II) and (ii) thegeneral deduction of the HM formula (Subsec. IV A) areindependent presentations; the former is discussed onlybecause it evidentiates, in a concrete scenario, the genericnature of HM — which is the main point of this work. II. HEAT-CONDUCTING-BAR PARADOX
Consider the system depicted in Fig. 1(a), composedby a free bar connecting two thermal reservoirs at tem-peratures T and T — with, say, T ≥ T —, at rest in aninertial frame. In order to avoid unnecessary subtleties,we consider that (i) the stationary heat-flow regime hasbeen established, (ii) thermal contact between the barand each reservoir is symmetric (for instance, throughthe lateral surface of the bar), and (iii) the bar is coatedwith a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) en-sure that the CME of the bar stays at rest in the inertialrest frame of the reservoirs without the need for any me-chanical constraint; the heat-conducting bar in the sta-tionary regime is in static mechanical equilibrium. (Con-dition (iii) only serves to keep the system simple.) Fromthe rest-frame perspective, the effect of the reservoirs onthe bar is merely exchange of heat, with no net forcesor torques being applied. Let W > T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature difference T − T to be arbitrarily small by choosing bars with arbitrarilylarge thermal conductivities. Therefore, although unnec-essary, one can simplify further the setup considering themass-energy and temperature distributions along the barto be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the per-spective of another inertial frame, with respect to(w.r.t.) which the bar (and the whole system) moves withvelocity V perpendicular to itself — “moving frame”for short. Although it may sound odd at first, it fol-lows directly from Einstein’s Special Relativity that, in VV(a) (b) x ’ y ’ xy x ’ y ’ T T F - F D
FIG. 1: Heat-conducting-bar “paradox.” (a) A bar standsstill in an inertial frame, connecting two thermal reservoirs.In the stationary regime, the bar absorbs heat from the ther-mal reservoir at temperature T , at a rate W , and deliversheat at the same rate to the thermal reservoir at temperature T ≤ T . There are no net forces between the bar and thereservoirs. (b) The same situation observed from an inertialframe w.r.t. which the bar moves with velocity V perpen-dicular to itself. According to observers static in this lat-ter frame, the reservoirs apply opposite forces ± F = ± W V / c on the bar. Therefore, in this frame there exists a torque T = W ( V × D )/ c on the heat-conducting moving bar, where D is the spatial vector depicted in the figure. Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V (a) (b) f - f ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V FIG. 2: Spacetime depiction of the world-volume of the barand the energy-momentum exchange (given by the 4-forcedensities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along thetime direction, describing exchange of energy without net spa-tial forces. (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, inthis frame, there are force densities ± f acting on the bar. this new frame, the reservoirs apply opposite net forces ± F = ± W V / c at the moving-bar’s ends [see Fig. 1(b)].The proof of this fact is actually quite simple (a textbook-level exercise) and is explained in detail in Appendix A.In essence, due to Lorentz covariance, what is seen inthe rest frame as a pure exchange of energy (i.e., a 4-force density f a with only time component), correspondsto exchange of both energy and momentum according tothe “moving frame” (see Fig. 2, which is the spacetimedepiction of the bar in Fig. 1).Once the reader is convinced of the existence of suchforces, he/she promptly realizes that they lead to atorque on the heat-conducting moving bar, which (ne-glecting the spatial extension of the thermal contacts) is2iven by T = W ( V × D )/ c , (1)where D is the separation vector between the thermalcontacts (see Fig. 1); although the opposite forces haveno net effect on the total momentum of the bar as timepasses, they do change the bar’s angular momentum. Ifwe apply our Newtonian intuition — as is customarywhen arriving at relativistic “paradoxes” —, this torquew.r.t. the instantaneous CME position should try to ro-tate the bar. But this is obviously in conflict with thefact that in the reservoirs’ rest frame the bar is in staticmechanical equilibrium; there is absolutely no reason forrotation. We have stumbled on a (new) relativistic “para-dox.” III. NO REAL PARADOX
Relativistic “paradoxes” — more precisely, situationswhose descriptions from different inertial perspectives seem paradoxical when compared to each other — arenumerous and even serve as teaching tools in relativity.Rather than pointing to inconsistencies in fundamentaltheories, they reveal how our Newtonian perception ofspace and time as separate entities, instead of interwovenin an absolute four-dimensional spacetime, can be deceiv-ing. Their nature can be loosely classified as kinemati-cal — those which involve only time-interval and spatial-distance measurements — and dynamical — those whichinvolve forces. The twins’, the barn-pole, and the Bell’sspaceship “paradoxes” are well-known textbook samplesof the kinematical type — see, e.g., Ref. [16] —, whereasthe Trouton-Noble [17], the right-angle-lever [18], and thesubmarine [19–22] “paradoxes” are representative of thedynamical type. The heat-conducting-bar (HCB) “para-dox” presented above clearly fits into this latter class.Contrary to kinematical “paradoxes,” the dynamical onesare rarely addressed in relativity textbooks and introduc-tory courses. This may explain why many of them areunknown to nonrelativists or, when known, concepts in-volved in their resolution are seen with suspicion.In 2012, M. Mansuripur [6] analyzed in detail an in-genious dynamical “paradox” — previously discussed inRef. [23] — which, in a simplified but equivalent version,can be realized by a neutral magnet at rest in an inertialframe, where there exists a uniform (external) electricfield E perpendicular to the magnet’s magnetic dipolemoment m (see Fig. 3). In the magnet’s rest frame[Fig. 3(a)], the magnet “seems” oblivious to the presenceof the electric field — apart from induced polarization,which can be made negligible. However, looking at thesame system from another inertial frame, w.r.t. whichthe magnet moves with velocity V along the electric-field’s direction [Fig. 3(b)], the magnet now also bearsan electric dipole moment d = V × m / c — since m isultimately due to electric currents, not pairs of magneticmonopoles [24]. Thus, according to this inertial frame, there must exist a torque T = d × E = ( V ⋅ E ) m / c acting on the magnet, which would supposedly make itspin — in gross contradiction with the fact that in itsinertial rest frame the magnet stands still. Mansuripurconcludes that this contradiction is an “incontrovertibletheoretical evidence of the incompatibility of the Lorentzlaw [of force] with the fundamental tenets of special rel-ativity” [6]. ++ ++ −− −− m mdm VV EE(a) (b) F - F FIG. 3: Mansuripur’s “paradox.” (a) A magnet stands stillin an inertial frame, with magnetic dipole moment m per-pendicular to a uniform electric field E . (b) The same situa-tion observed from an inertial frame w.r.t. which the magnetmoves with velocity V parallel to the electric field. Now, themagnet also carries an electric dipole moment d , upon whichthe electric field exerts a torque. The HCB “paradox” presented earlier is a close ther-mal analogue of Mansuripur’s, with the thermal reser-voirs playing the role of the external electric field andthe heat-conducting bar substituting the magnet. Lessobvious is the analogue, in Mansuripur’s setup, of theheat exchange rate W between the bar and the reservoirs.But recalling that magnetism in materials is ultimatelydue to current densities j (even if quantum mechanical innature), one concludes that the magnet in its rest framedoes exchange energy with the external electric field ata rate, per volume, j ⋅ E : the magnet predominantly ab-sorbs (resp., delivers) energy from (resp., to) the externalfield where j ⋅ E > j ⋅ E < enforced by Lorentz covariance and, particularly,by E = mc . Certainly, no one would hold that Einstein’smass-energy relation is “incompatible with the funda-mental tenets of special relativity.” Therefore, there is nological reason for taking this stand regarding the Lorentzforce. IV. MASS-ENERGY EQUIVALENCE ANDHIDDEN MOMENTUM
Relativistic thermodynamics has its own history ofsubtleties and controversies. The most emblematic of3 igure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V (a) (b) f - f ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 ⌃ ⌃ ct ct Consider the system depicted in figure 1-(a), composed by a free bar connecting two thermalreservoirs at temperatures T and T – with, say, T T –, at rest in an inertial frame. In orderto avoid unnecessary subtleties, we consider that (i) the stationary heat-flow regime has been es-tablished, (ii) thermal contact between the bar and the reservoirs occurs only through the lateralwalls of the bar, and (iii) the bar is coated with a thermal insulator all over the parts which are notin contact with the reservoirs. Conditions (i) and (ii) ensure that the center of mass-energy (CME)of the bar stays at rest in the inertial rest frame of the reservoirs without the need of any externalforce; the heat-conducting bar in the stationary regime is in static mechanical equilibrium. (Condi-tion (iii) only serves to keep the system simple.) From the rest-frame perspective, the effect of thereservoirs on the bar is merely exchange of heat, with no net forces or torques being applied. Let W > represent the (constant) rate at which heat is exchanged between the bar and the reservoirs– flowing into (respectively, out of) the bar from (resp., to) the reservoir at temperature T (resp., T ). (Side note: for any given heat-exchange rate W , we can consider the temperature differ-ence T T to be arbitrarily small by choosing bars with arbitrarily large thermal conductivities.Therefore, although unnecessary, one can simplify further the setup considering the mass-energyand temperature distributions along the bar to be arbitrarily close to homogeneous.)Now, let us analyze the same setup from the perspective of another inertial frame, withrespect to (w.r.t.) which the bar (and the whole system) moves with velocity V perpendicular to2 Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V Figure 3
Microscopic modeling of energy-momentum transfer between reservoir andsystem. Although the force F obtained in figure 2 is independent of the details of thereservoir, it is instructive to reproduce this result using a microscopic modeling, wherethe reservoir is constituted by (massive or massless) particles colliding (inelastically) withthe system’s walls. (a) In the reservoir’s rest frame, the inelastic collisions transfer energyto/from the system without net exchange of momentum – due to isotropy; (b)
Seen fromanother inertial perspective, the inelastic collisions are no longer isotropic: transformingto this frame all momenta involved in depiction (a) , it is a textbook exercise to show thatpairs of collisions which are symmetric in the rest frame of the reservoir ( p = p )transfer momentum P = ( p + p ) = E V /c to the system when analyzed fromthe moving frame. Figure 4
Spacetime depiction of the world-volume of the bar and the energy-momentumexchange (given by the 4-force densities ± f a ) between the bar and the reservoirs. (a) Fromthe rest-frame perspective, f a has only component along the time direction, describing ex-change of energy without net spatial forces; (b) From the moving-frame perspective, the same f a clearly has nonvanishing spatial projection. Therefore, in this frame, there areforce densities ± f acting on the bar. f a Figure 5
Mansuripur’s “paradox.” (a)
A magnet stands still in an inertial frame, withmagnetic dipole moment m perpendicular to a uniform electric field E ; (b) The samesituation observed from an inertial frame w.r.t. which the magnet moves with velocity V FIG. 4: Four-dimensional representation of (the cross section z = f a ) where the elec-tric field favors the current density ( j ⋅ E >
0) and past-directedotherwise ( j ⋅ E < ′ —therefore, no forces according to the magnet’s rest frame —while being nonzero and circulating on Σ — therefore, apply-ing a torque on the magnet according to the “moving frame.” them is probably the question of how temperature trans-forms from one inertial frame to another. It took about90 years for this to be recognized as an ill-posed question— hence, the conflicting answers given during this period(see Refs. [25, 26] and references therein). Fortunately,none of these subtleties — not even temperature transfor-mation — concerns us; the purpose of thermal reservoirsin the setup of Fig. 1 is only to guarantee an eventualstationary situation in the rest frame of the system.As mentioned earlier, several systems with nonvanish-ing total momentum in their rest frames have been foundand discussed in the literature (see, e.g., Refs. [1–5]) —including Mansuripur’s setup [7–11, 15]. All such systemsinvolved interaction with electromagnetic fields and/ormoving inner parts subject to some external force field,which led many to view it as a feature of peculiar inter-action laws or systems. Mansuripur, for instance, con-sidered HM to be an ad hoc addition to materials inter-acting with electromagnetic fields, with no justificationother than artificially avoiding paradoxal situations [12–14]. A better law of electromagnetic force, he reasoned,should be one which leads to no torque on the movingmagnet in an electric field — hence, doing away withHM. In this sense, the HCB “paradox” we discuss in thiswork is unique, for it does not depend on the inner detailsof the system (the bar and the heat/energy flow) and ofthe interaction with “the rest of the universe” (the ther-mal/energy reservoirs).The resolution of the HCB “paradox” — as well asMansuripur’s — consists in taking mass-energy equiva-lence to its ultimate consequences. As heat (i.e., energy)flows through the bar, it contributes to momentum invery much the same way as would a flow of matter. Infact, distinguishing contributions to the total momentumcoming from “different forms” of energy flows is quitecontrary to the spirit of relativity theory. Therefore, the total momentum of the bar in its rest frame [Fig. 1(a)]does not vanish — a purely relativistic effect. For thesame reason, according to the inertial frame w.r.t. whichthe bar moves with velocity V perpendicular to itself[Fig. 1(b)], there is a momentum contribution along thebar. Consequently, the bar’s total momentum P and theCME velocity V are misaligned; and dragging momen-tum P along a spatial direction which is not aligned to itinevitably leads to a time-varying angular momentum L (with d L / dt = V × P ) and, therefore, demands a torque —which, as we shall see below, is precisely the one suppliedby the thermal reservoirs in the moving frame. A. General treatment: hidden momentum asdipole moment of energy-exchange rate
Mass-energy equivalence has a very straightforwardconsequence which, nonetheless, is overlooked when ar-riving at dynamical relativistic “paradoxes:” the distinc-tion between closed and isolated systems, as usually madein Newtonian (i.e., Galilean-covariant) physics (includingnonrelativistic thermodynamics), has no absolute mean-ing in relativistic physics. While in Newtonian physicsa system can exchange energy without exchanging mass(i.e., it can be non-isolated but closed), this is obviouslyimpossible if mass and energy are the same physical quan-tity. In this sense, the relativistic dynamics of the heat-conducting bar depicted in Fig. 1 — which, in the New-tonian context, is a closed system for which the center-of-mass theorem would apply — is essentially the sameas that of a pipe segment carrying a steady fluid/particlecurrent, with fluid/particles entering the system at oneend of the pipe and leaving at the other — which, inthe Newtonian context, is an open system, for which thecenter-of-mass theorem does not apply. No one wouldobject that the pipe segment carrying a particle/fluidcurrent possesses nonzero momentum in the rest frameof its center of mass. Distinguishing this momentumfrom the one carried by the heat-conducting bar — or,for that matter, from the one carried by the magnet inMansuripur’s setup — is solely motivated by our Newto-nian view of the world.Although distinguishing HM from “regular” mo-mentum is artificial as far as relativistic dynamics isconcerned, it is useful, in order to demystify it further,to pin down the elements which compel our Newtonianintuition to make such a distinction. Basically, allinstances of HM involve systems where there is a clearnotion of a velocity field v of their “constituents”— usually taken to be particles or fluid elements —and an associated nonnegative (not identically null) number density n which, together, satisfy the continuityequation ∂ t n + ∇ ⋅ ( n v ) = all inertial frames. Putin spacetime language: these systems in which HM canbe identified posses, associated to their constituents,a time-like, future-directed 4-vector field n a — the4-current number density, whose components in inertial4artesian coordinates read n µ = ( nc, n v ) — satisfyingthe tensorial equation ∇ a n a =
0. The key point is thatthe existence of such a 4-current number density allowsus to extend to the relativistic context, in a consistentmanner, the notion of “closed systems:”
Definition:
A system will be said to be closed if one ofthe following holds:(a) The system is isolated — i.e., its extended (see below) stress-energy-momentum tensor¯ T ab satisfies ∇ a ¯ T ab ≡ n a (as definedabove) whose extension ¯ n a (see below) satis-fies ∇ a ¯ n a ≡ T ab...cd... with support supp (T ) ,its extension ¯ T ab...cd... is the tensor field defined overthe whole spacetime which coincides with T ab...cd... in supp (T ) and is zero otherwise. This is a mere tech-nicality, useful when treating systems which are,themselves, part of larger ones. Due to possiblediscontinuities in ¯ T ab...cd... , the equations above are tobe taken in the distributional sense.)The definition above clearly recovers, in the Newtonianregime, the notion of closed systems as those which do not exchange matter/mass, for in this case the mass densityitself satisfies the continuity equation and can be takento be n up to a multiplicative constant. This fact will beused later when we restrict attention to closed systems.Let S be a system with stress-energy-momentum ten-sor whose components in inertial Cartesian coordinates {( ct, x )} are given by T µν S . We assume that at each in-stant t , T µν S goes to zero sufficiently fast at spatial infinityso that the manipulations and integrals which follow be-low are well defined. The center of mass-energy (CME)of S is given by X cme ∶= M c ∫ d x ¯ T S x , (2)where M = ∫ d x ¯ T S / c is the (possibly time-dependent)total mass of the system. (All integrations are carriedover the entire spatial sections t = constant; hence theuse of ¯ T µν S instead of T µν S .) Multiplying Eq. (2) by M and taking the time derivative, we get: M V cme = − dMdt X cme + c ∫ d x ∂ ¯ T S x = c ∫ d x ∂ ¯ T S ( x − X cme ) , (3)where V cme ∶= d X cme / dt is the velocity of the CME of S .The fact that system S is not necessarily isolatedmeans that ∂ µ ¯ T µν S = f ν , where f µ is the 4-force den-sity acting on S — in particular, f is related to energyexchange rate W through d x f = dW / c . Substituting ∂ ¯ T S = f − ∂ j ¯ T j S into Eq. (3) leads to: M V cme − c ∫ d x f ( x − X cme ) = − c ∫ d x ∂ j ¯ T j S ( x − X cme )= − c ∫ d x ∂ j [ ¯ T j S ( x − X cme )] + c ∫ d x ¯ T j S ∂ j x = ∫ d x p = P , (4)where ( p ) j = ¯ T j S / c are the components of the momentumdensity of the system.So far, very little has been imposed on the system S .In fact, the only assumptions are that the integrals aboveconverge and that the surface term coming from the firstintegral in the right-hand side (r.h.s.) of the second lineof Eq. (4) vanishes at spatial infinity. But now, we re-strict attention to closed systems, as defined earlier. Thereason is that for closed systems, the second term in theleft-hand side (l.h.s.) of Eq. (4) is purely relativistic, sincein the Newtonian regime n is proportional to mass den-sity and, therefore, f / c ∝ ∇ a ¯ n a ≡
0. This expresses the well-known fact that in Newtonian mechanics, the to-tal momentum of an arbitrary closed system (isolated ornot) is completely encoded in the motion of its center ofmass and its total mass — the center-of-mass theorem.In relativity theory, on the other hand, we see that asym-metric (w.r.t. the CME) exchanges of energy between S and “the rest of the universe” contribute to the total mo-mentum of the system; now, P cannot be assessed onlyby keeping track of the system’s mass-energy distribu-tion. This motivates us to define the “hidden” part ofthe total momentum of the system as P h ∶= P − M V cme ,5hich can then be calculated by: P h = − c ∫ d x f ( x − X cme ) = − c ∫ dW ( x − X cme ) . (5)In words: the hidden momentum of a closed systemis given by (minus 1 / c times) the dipole moment(w.r.t. X cme ) of its energy-exchange rate. Notice thatour definition not only frees HM from being identifiedonly in the rest frame of the system (where V cme = ),but also shows that HM does not depend on the innerdetails of the system; it does not depend on the natureof T ab S (electromagnetic, thermal, mechanical, etc.), butonly on how it fails to be conserved.[31] B. Hidden momentum in a heat-conducting bar
Applying the definition given in Eq. (5), or its discreteversion P h = − c ∑ j ( x j − X cme ) W j , (6)to the system depicted in Fig. 1, we promptly obtain: P h = W D / c . (7)Therefore, in the situation depicted in Fig. 1(b), drag-ging the total momentum P = M V + P h of the bar ata constant velocity V leads to an angular momentum L which changes at a rate d L dt = V × P = V × P h = W V × D / c . Comparing this result with the torque given in Eq. (1),provided by the forces ± F , we see that everything fitsperfectly: the torque provided by the forces seen fromthe moving frame is exactly the one needed to keep thespinless bar in uniform motion.Obviously, Eq. (5) can also be applied to the magnetdepicted in Fig. 3 [7]. Recalling that j = ∇ × M , where M is the magnet’s magnetization, we have: P h = − c ∫ d x ( j ⋅ E ) ( x − X cme )= − c ∫ d x [∇ ⋅ ( M × E )] ( x − X cme )= c ∫ d x ( M × E ) = c (∫ d x M ) × E = c m × E . (8)Therefore, in situation depicted in Fig. 3(b), the totalangular momentum L of the magnet changes at a rate: d L dt = V × P = V × P h = ( V ⋅ E ) m / c , which matches exactly the torque applied on the magnetaccording to the “moving frame” — see Sec. III. V. HIDDEN BUT REAL... AND OBSERVABLE
By now, we hope we have convinced the reader thatHM, far from being a peculiar property of specific sys-tems or interaction laws, is simply a legitimate relativis-tic contribution to total momentum coming from energyflows in a closed, non-isolated system — a distinction mo-tivated solely by our Newtonian intuition. In this sense,not only it is ubiquitous in a relativistic world, but alsoit is as real as any other form of momentum. In fact,we present below a final conclusive evidence which cor-roborates this view: we show that HM can be convertedinto “regular” momentum and, as such, its existence hasobservable consequences.Consider the same system depicted in Fig. 1(a) in sta-tionary regime of heat flow. Suppose, now, that thethermal contacts with both reservoirs are suddenly inter-rupted simultaneously in the bar’s frame (e.g., by someclever preprogrammed mechanism attached to the baritself which shields the contacts with some thermal insu-lator). Since this interruption can be performed withoutany external forces applied on the bar, its total momen-tum will not change in the process. However, once thethermal contacts are interrupted, the bar is isolated and,as such, the center-of-mass theorem must hold. This im-plies that the bar’s initial total momentum, which wascompletely “hidden,” now has to manifest itself as mo-tion of the bar’s CME: V cme = P h M = W D M c , where M is the mass of the bar in situation depictedin Fig. 1(a). (Note that if, in preparing the setup repre-sented in Fig. 1(a), we had not held the bar fixed until thestationary heat-flow regime had been established, the barmight have acquired a velocity in the opposite directionin order to keep its total momentum zero — in case ther-mal contacts are established with no net external forcesacting on the bar in its rest frame. This is the reasonfor assumption (i) made in Sec. II.) Although this veloc-ity is probably too small in realistic situations, it consti-tutes evidence that HM is not some imaginary conceptwith no objective existence. (Obviously, HM containedin the “rest of the universe” will also be converted intomotion of the CME of this latter, but since the “rest-of-the-universe” mass is supposedly much larger than thatof the system of interest, such effect may be completelyneglected.)The same conclusion holds for the magnet representedin Fig. 3: if the electric field is removed without exertingnet forces on the magnet — which may be difficult dueto possibly inhomogeneous magnetic fields generated inthe process —, the magnet would acquire a velocity V = P h M = m × E M c , with M being the magnet’s mass. This may be seen assomewhat similar to the Richardson–Einstein–de Haas6ffect [28, 29], but now for linear instead of angularvelocity/momentum. (In this hypothetical scenario, itseems reasonable to conjecture that the opposite mo-mentum of the “rest of the universe” may become man-ifest through emission of electromagnetic waves.) Sinceboth M and m , in ideal situations, scale with the mag-net’s volume, we can make an order-of-magnitude es-timation for V ∶= ∥ V ∥ using Bohr’s magneton, m ∶=∥ m ∥ ∼ µ B ∼ − GeV / T, and the proton’s mass, M ∼ M p ∼ / c : V ∼ [ E /( / m )] × − nm / s,where E ∶= ∥ E ∥ . As expected, the typical velocities forreal magnets in realistic external electric fields are ex-tremely small. However, one might try to amplify thiseffect by using systems in which M scales with size slowerthan does m — as in solenoids, in the macroscopic sce-nario, or Rydberg atoms, in the atomic realm. VI. DISCUSSION
Although Mansuripur’s speculation on alternative lawsof electromagnetic force is a valid inquiry — which canonly be definitely settled by experiments —, the genericnature of the HCB “paradox” — with electromagnetismand moving inner parts playing no explicit essential role— shows that the existence of torques acting on spin-less, uniformly-moving objects is a ubiquitous featureof relativistic dynamics. As stated earlier, this torque( T = V cme × P ) is responsible for translating the CME ofthe system (with velocity V cme ) along a direction whichis not aligned to its total momentum ( P ) — which ex-poses the existence of HM. The generalized definition ofHM as P h ∶= P − M V cme , proposed in Subsec. IV A — which makes sense not onlyin the rest frame of the system (where V cme = ) —,led to a formula relating HM with asymmetric (w.r.t. thesystem’s CME) exchange of energy with “the rest of theuniverse:” P h = − c ∑ j ( x j − X cme ) W j , (9)where X cme is the CME position and x j is the positionwhere energy exchange occurs at a rate W j ( W j >
0, ifenergy enters the system; W j <
0, if energy leaves the sys-tem). The interpretation is simple: this asymmetry leadsto energy flows in the system which, regardless their na-ture, contribute to momentum in very much the sameway as do matter flows — thanks to mass-energy equiv-alence. As stressed earlier, distinguishing contributionsto the total momentum coming from “different forms” ofenergy flows is quite contrary to the spirit of relativitytheory — reason why a covariant, observer-independentdefinition of HM does not (and cannot) exist. Notwith-standing, although artificial and not strictly necessary, thinking in terms of HM may help our Newtonian intu-ition to spot effects which might pass unnoticed otherwise— as the one discussed in Sec. V.Obviously, only experiments can decide on the correct-ness of candidate laws of Nature. For instance, whetheror not HM is in fact present in a magnet subject to anexternal electric field strongly depends on the ultimateorigin of its magnetic dipole moment [15]. Here, in orderto arrive at the well-known Eq. (8), we have adopted thestandard view that any magnetic moment is ultimatelydue to electric currents, even if of a quantum nature.Notwithstanding, Eq. (9) is equally valid whatever is themicroscopic modeling of magnetic dipoles. Be it as itmay, the point is that aiming at substituting a law offorce solely on the basis that it leads to HM is a mis-guided effort. As made explicit by the HCB “paradox”and Eq. (9), HM is simply an inevitable consequence of E = mc when seen from an arbitrary inertial frame.Moreover, its existence can, in principle, be tested byforcing it to reveal itself as motion of the CME of thesystem — as shown in Sec. V. Looking at this the otherway around, measuring the amount of HM in atomic-size“magnets” subject to external fields may end up beinguseful for investigating or confirming the ultimate natureof elementary magnetic moments. Acknowledgments
The author thanks Kirk T. McDonald, VladimirHnizdo, David J. Griffiths, and Massud Mansuripur fordiscussions which motivated — not necessarily implyingagreement with — this work. In particular, we thankKirk T. McDonald for calling attention to Ref. [27]. Theauthor also thanks Barnab´as Deme for discussions re-garding the similarity between the effect discussed inSec. V and the Richardson–Einstein–de Haas effect.
Appendix A: Energy-momentum transfer in inelasticcollisions and force exchange with thermal reservoirs
Consider the symmetric process depicted in Fig. 5(a),where two identical particles with opposite momenta( p ′ = − p ′ ) and vanishing total angular momentum col-lide with a surface at rest. We shall allow the collisionsto be inelastic, each delivering an energy ∆ E ′ / P ′ = −( ∆ p ′ + ∆ p ′ ) = ). In Fig. 5(b), the same process is depicted as seen from an inertial framew.r.t. which the surface moves with velocity V . Obvi-ously, the whole process is determined from its descrip-tion above; all one has to do is to Lorentz transform theprimed quantities to this new frame. By doing so —which is a textbook exercise —, one realizes that the mo-mentum exchanges between the particles and the surfaceare no longer symmetric (∆ p ≠ − ∆ p ) and that a net7 a) (b) Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 equivalence. Unless one is willing to consider that magnetism in materials is not ultimately due toelectric current densities ( j ) – which exchange energy with electric fields E at a rate density j · E –,hidden momentum is inevitable in Mansuripur’s setup. Aiming at substituting a law of force solelyon the basis of it leading to HM is a misguided effort. Methods
Here, we present the calculations which support all the conclusions drawn in this article.
Force on a moving object in contact with a thermal reservoir in its rest frame.
We begin byjustifying P = E = 0 P = E V /c E = E Hidden momentum in general.
Let A be a system characterized by the energy-momentum tensorwhose components in an inertial Cartesian coordinates { ( t, x ) } are given by T µ⌫ A . For technicalsimplicity, we assume that at each instant t , T µ⌫ A = 0 only in a bounded spatial region (i.e., T µ⌫ A has compact spatial support). If @ µ T µ⌫ A = 0 everywhere, then A is said to be isolated .6 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 equivalence. Unless one is willing to consider that magnetism in materials is not ultimately due toelectric current densities ( j ) – which exchange energy with electric fields E at a rate density j · E –,hidden momentum is inevitable in Mansuripur’s setup. Aiming at substituting a law of force solelyon the basis of it leading to HM is a misguided effort. Methods
Here, we present the calculations which support all the conclusions drawn in this article.
Force on a moving object in contact with a thermal reservoir in its rest frame.
We begin byjustifying P = E = 0 P = E V /c E = E Hidden momentum in general.
Let A be a system characterized by the energy-momentum tensorwhose components in an inertial Cartesian coordinates { ( t, x ) } are given by T µ⌫ A . For technicalsimplicity, we assume that at each instant t , T µ⌫ A = 0 only in a bounded spatial region (i.e., T µ⌫ A has compact spatial support). If @ µ T µ⌫ A = 0 everywhere, then A is said to be isolated .6 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E = E ) The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 x ’ y ’ Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 V xy x ’ y ’ V Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7Eq. (1) by M and taking the time derivative, we get: M V cme = dMdt X cme + 1 c Z d x @ T A x = 1 c Z d x @ T A ( x X cme ) (2)where V cme := d X cme /dt is the velocity of the center of mass-energy of A . p (in)1 p (out)1 p (in)2 p (out)2 p p ( E > The fact that system A may interact with another system means that @ µ T µ⌫ A = f ⌫ , where f µ is the 4-force density acting on A . Substituting @ T A = f @ j T j A into equation (2) leads to: M V cme c Z d x f ( x X cme ) = c Z d x @ j T j A ( x X cme )= c Z d x @ j [ T j A ( x X cme )] + 1 c Z d x T j A @ j x = Z d x p = P , (3)where ( p ) j = T j A /c are the components of the momentum density of the system. Note that thesecond term in the left-hand side of equation (3) is purely relativistic, since mass conservation inNewtonian mechanics – which follows from covariance of Newton’s second law under Galileantransformations – implies f /c = 0 in the corresponding limit. This expresses the well-known factthat in Newtonian mechanics, the total momentum of an arbitrary system is completely encodedin the motion its center of mass ( V cme ) and its total mass ( M ). In relativistic mechanics, on theother hand, asymmetric (w.r.t. the center of mass-energy) exchanges of energy between A and “therest of the universe” contribute to the total momentum of the system; now, P cannot be assessedonly by keeping track of the motion of the system’s mass-energy distribution. This motivates us to7 FIG. 5: (a) Symmetric inelastic collisions of two identicalparticles with a surface at rest. The symmetry of the setupleads to no net momentum transfer to the surface. (b) Thesame process analyzed from an inertial frame w.r.t. whichthe surface is moving with velocity V . In this new frame,the collisions are no longer symmetric and a net momentum∆ P = ∆ E V / c is transferred to the surface, where ∆ E isthe energy delivered into the surface during the process. As-suming processes like this occurring at a constant rate, a netforce given by F = W V / c would be exerted on the surface,where W is the rate at which energy is delivered into the sur-face. By modeling a thermal reservoir as an isotropic bathof particles, this implies that an object at rest in a thermalreservoir is subject to a net force F = W V / c when seen froma reference frame w.r.t. which the whole system (object andreservoir) is moving — with W being the rate at which en-ergy (heat) is absorbed by the object. Note that this forceis exactly the one needed to keep an object with increasingmass, dM / dt = W / c , in uniform motion. momentum ∆ P = ∆ E V / c is transferred to the surface,where ∆ E = γ ∆ E ′ is the net energy delivered into thesurface in this frame ( γ is the Lorentz factor). The im-plication is clear: inelastic collisions which are symmetricin the rest frame of the surface exert a net force on thesurface when analyzed from inertial frames w.r.t. whichthe surface is moving.Modeling a thermal reservoir as an isotropic bath ofparticles, the result above inevitably leads to the con-clusion that an object static (and symmetrically im-mersed [32]) in a thermal reservoir is subject to a (purelyrelativistic) net force F = W V / c when seen from an in-ertial frame w.r.t. which the whole system (object andreservoir) is moving with velocity V , with W being therate at which energy (heat) is absorbed by the object.Although this may sound odd at first, it becomes quiteobvious when one realizes the need of an external forcein order to keep the constant velocity of an object with increasing rest energy (i.e., rest mass). In fact, the ex-istence of this relativistic force F = W V / c can be in-ferred from this more general argument, independent ofmicroscopic modeling of the reservoir (see Fig. 6). Theimportance of the microscopic collisional model is explic-itly showing that the existence of such a force does not depend on the fate of the absorbed energy ∆ E — forinstance, whether it is accumulated in the object or con-stantly drained to sustain a heat flow (as in Fig. 1) [33].Therefore, although, strictly speaking, the situations de-picted in Fig. 1 and Figs. 5 and 6 represent differentsystems, the origin of the forces seen according to the (a) (b) x ’ y ’ xy V equivalence. Unless one is willing to consider that magnetism in materials is not ultimately due toelectric current densities ( j ) – which exchange energy with electric fields E at a rate density j · E –,hidden momentum is inevitable in Mansuripur’s setup. Aiming at substituting a law of force solelyon the basis of it leading to HM is a misguided effort. Methods
Here, we present the calculations which support all the conclusions drawn in this article.
Force on a moving object in contact with a thermal reservoir in its rest frame.
We begin byjustifying P = E = 0 P = E V /c E = E Hidden momentum in general.
Let A be a system characterized by the energy-momentum tensorwhose components in an inertial Cartesian coordinates { ( t, x ) } are given by T µ⌫ A . For technicalsimplicity, we assume that at each instant t , T µ⌫ A = 0 only in a bounded spatial region (i.e., T µ⌫ A has compact spatial support). If @ µ T µ⌫ A = 0 everywhere, then A is said to be isolated .6 equivalence. Unless one is willing to consider that magnetism in materials is not ultimately due toelectric current densities ( j ) – which exchange energy with electric fields E at a rate density j · E –,hidden momentum is inevitable in Mansuripur’s setup. Aiming at substituting a law of force solelyon the basis of it leading to HM is a misguided effort. Methods
Here, we present the calculations which support all the conclusions drawn in this article.
Force on a moving object in contact with a thermal reservoir in its rest frame.
We begin byjustifying P = E = 0 P = E V /c E = E Hidden momentum in general.
Let A be a system characterized by the energy-momentum tensorwhose components in an inertial Cartesian coordinates { ( t, x ) } are given by T µ⌫ A . For technicalsimplicity, we assume that at each instant t , T µ⌫ A = 0 only in a bounded spatial region (i.e., T µ⌫ A has compact spatial support). If @ µ T µ⌫ A = 0 everywhere, then A is said to be isolated .6 equivalence. Unless one is willing to consider that magnetism in materials is not ultimately due toelectric current densities ( j ) – which exchange energy with electric fields E at a rate density j · E –,hidden momentum is inevitable in Mansuripur’s setup. Aiming at substituting a law of force solelyon the basis of it leading to HM is a misguided effort. Methods
Here, we present the calculations which support all the conclusions drawn in this article.
Force on a moving object in contact with a thermal reservoir in its rest frame.
We begin byjustifying P = E = 0 P = E V /c E = E Hidden momentum in general.
Let A be a system characterized by the energy-momentum tensorwhose components in an inertial Cartesian coordinates { ( t, x ) } are given by T µ⌫ A . For technicalsimplicity, we assume that at each instant t , T µ⌫ A = 0 only in a bounded spatial region (i.e., T µ⌫ A has compact spatial support). If @ µ T µ⌫ A = 0 everywhere, then A is said to be isolated .6 equivalence. Unless one is willing to consider that magnetism in materials is not ultimately due toelectric current densities ( j ) – which exchange energy with electric fields E at a rate density j · E –,hidden momentum is inevitable in Mansuripur’s setup. Aiming at substituting a law of force solelyon the basis of it leading to HM is a misguided effort. Methods
Here, we present the calculations which support all the conclusions drawn in this article.
Force on a moving object in contact with a thermal reservoir in its rest frame.
We begin byjustifying P = E = 0 P = E V /c E = E Hidden momentum in general.
Let A be a system characterized by the energy-momentum tensorwhose components in an inertial Cartesian coordinates { ( t, x ) } are given by T µ⌫ A . For technicalsimplicity, we assume that at each instant t , T µ⌫ A = 0 only in a bounded spatial region (i.e., T µ⌫ A has compact spatial support). If @ µ T µ⌫ A = 0 everywhere, then A is said to be isolated .6 V x ’ y ’ F FIG. 6: Force applied on a moving object by a comoving ther-mal reservoir. (a) An object at rest in the reservoir’s frame ex-changes an amount ∆ E ′ of energy in a time interval ∆ t ′ , withno momentum transfer (due to the symmetry of the reser-voir in its rest frame). According to mass-energy equivalence,this corresponds to a (rest-)mass variation ∆ M ′ = ∆ E ′ / c .(b) The same situation seen from another reference frame: avariation ∆ M = γ ∆ M ′ in the object’s mass, at constant ve-locity V , corresponds to a momentum transfer ∆ P = ∆ M V = γ ∆ E ′ V / c from the reservoir, in a time interval ∆ t = γ ∆ t ′ .Therefore, in this frame, the reservoir must (and does) applya net force F = ∆ P / ∆ t = ∆ E ′ V /( c ∆ t ′ ) = W V / c on theobject, where W = ∆ E ′ / ∆ t ′ = ∆ E / ∆ t is the energy exchangerate. Causality/locality ensures that this final result cannotdepend on whether the energy exchange ∆ E ′ is accumulatedin the object or if it is used to sustain a stationary heat flow,as in Fig. 1. “moving frame” is the same in all of them. For an inter-esting quantum microscopic scenario where this force isalso needed to conciliate different inertial-frame descrip-tions, see Ref. [30]. [1] W. Shockley and R. P. James, “Try Simplest Cases”Discovery of “Hidden Momentum” Forces on “MagneticCurrents.” , Phys. Rev. Lett. , 876 (1967).[2] I. Brevik, Experiments in phenomenological electrody-namics and the electromagnetic energy-momentum ten-sor , Phys. Rep. , 133-201 (1979).[3] L. Vaidman, Torque and force on a magnetic dipole , Am.J. Phys. , 978–983 (1990).[4] V. Hnizdo, Conservation of linear and angular momen- tum and the interaction of a moving charge with a mag-netic dipole , Am. J. Phys. , 242–246 (1992).[5] V. Hnizdo, Hidden mechanical momentum and the fieldmomentum in stationary electromagnetic and gravita-tional systems , Am. J. Phys. , 515–518 (1997).[6] M. Mansuripur, Trouble with the Lorentz Law of Force:Incompatibility with Special Relativity and MomentumConservation , Phys. Rev. Lett. , 193901 (2012).[7] D. A. T. Vanzella,
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Willa decaying atom feel a friction force? , Phys. Rev. Lett. , 053601 (2017).[31] After acceptance of this manuscript for publication,Ref. [27] came to our attention, where “internal momen-tum,” defined in a similar manner as HM here, is pro-posed to substitute the latter. We essentially agree withthe key points of Ref. [27], whose attempt to demys-tify HM is aligned with ours. Notwithstanding, Ref. [27]still limits its discussion to an electrodynamic system,restricting energy exchanges to mechanical work: f = f ⋅ v / c . In particular, its definition of “internal momen-tum” [following from its Eq. (5)] would vanish for thescenario depicted in Fig. 1(a), in conflict with both ourEq. (7) and the observable effect predicted in Sec. V. TheHCB “paradox,” the precise characterization of systemswhich bear HM, and the observational consequence ofHM which we discuss in what follows constitute impor-tant new ingredients to showing that HM (or “internalmomentum”) is a generic relativistic feature — as we hadalready claimed in Ref. [7].[32] If the immersion is not symmetric, there may appear anadditional contribution related to the fact that the netforce may not be zero in the rest frame of the system.[33] The same conclusion can be reached in the more gen-eral argument presented in Fig. 6 by invoking localityand causality: the force exerted by the reservoir at theboundary of the object when the energy is exchangedcannot depend on the later use of this energy.relativistic feature — as we hadalready claimed in Ref. [7].[32] If the immersion is not symmetric, there may appear anadditional contribution related to the fact that the netforce may not be zero in the rest frame of the system.[33] The same conclusion can be reached in the more gen-eral argument presented in Fig. 6 by invoking localityand causality: the force exerted by the reservoir at theboundary of the object when the energy is exchangedcannot depend on the later use of this energy.