Resolution of two fundamental issues in the dynamics of relativity and exposure of a real version of the emperor’s new clothes

Changbiao Wang∗ ShangGang Group, 10 Dover Road, North Haven, CT 06473, USA

In this paper, we aim to resolve two fundamental issues in the dynamics of relativity: (i) Under what condition, the time-column space integrals of a Lorentz four-tensor constitute a Lorentz four- vector, and (ii) under what condition, the time-element space integral of a Lorentz four-vector is a Lorentz scalar; namely two “conservation laws”, which are mispresented in traditional textbooks, and widely used in fundamental research, such as relativistic analysis of the momentum of light in a medium, and the proofs of the positive mass theorem in general relativity. To resolve issue (i), we have developed a generalized Lorentz four-vector theorem based on the principles of classical mathematical analysis, with a simplified analytic example given to illustrate how to transform a space integral from one inertial frame to another, and a strict mathematical derivation provided to confirm the effect of Lorentz contraction. We use this four-vector theorem to verify Møller’s theo- rem, and surprisingly find that Møller’s theorem is fundamentally wrong. We provide a corrected version of Møller’s theorem. We also use this four-vector theorem to analyze a plane light wave in a moving uniform medium, and find that the momentum and energy of Minkowski quasi-photon constitute a Lorentz four-vector and Planck constant is a Lorentz invariant. To resolve issue (ii), we have developed a generalized Lorentz scalar theorem. We use this theorem to verify the “invariant conservation law” in relativistic electrodynamics, and unexpectedly find that it is also fundamentally wrong. Thus the two “conservation laws” in traditional textbooks, which have magically attracted several generations of most outstanding scientists, turned out to be imaginary, just like the em- peror’s new clothes; creating a scientific myth in the modern theoretical and mathematical physics: Believing is seeing.

PACS numbers: 03.30.+p, 03.50.De, 04.20.-q, 11.30.-j Keywords: Conservation laws of relativity; Light momentum; Positive mass theorem; Gordon metric

I. INTRODUCTION resolve the electrostatic field problem [6]. In contrast to Laue’s theorem, Møller provided a four-vector theorem for a tensor that is required to be In the dynamics of relativity, the energy and momen- divergence-less, with a boundary condition imposed, but tum of a physical system is described by a Lorentz four- allowed to be time-dependent [3]. Møller’s theorem only tensor; such a tensor is usually called energy–momentum has a sufficient condition (instead of a sufficient and nec- tensor [1–3], stress tensor [4], stress–energy tensor [5, 6], essary condition) [13], but it is more attractive because or momentum–energy stress tensor [7]. If the tensor is the energy–momentum tensor for electromagnetic (EM) divergence-less, then the system is thought to be con- radiation fields varies with time [8–11]. It is widely rec- served [1–3], and it is a closed system [3]; thus the total ognized in the community that Møller’s theorem is abso- energy and momentum can be obtained by carrying out lutely rigorous so that this theorem has been widely used space integration of the time-column elements of the ten- in quantum electrodynamics [8] and relativistic analysis sor to constitute a Lorentz four-vector [1–5, 8–11]. of light momentum in a dielectric medium [9–11]. Mathematically speaking, if a tensor satisfies certain In this paper, we provide a generalized Lorentz four- conditions, the space integrals of the tensor’s time- vector theorem for a tensor which is not required to be column elements can form a Lorentz four-vector. For the time-independent and divergence-less, and on which no sake of convenience, we call such a mathematical state- boundary conditions are imposed. This theorem has a ment “four-vector theorem”. sufficient and necessary condition. We use this theo- Laue set up a four-vector theorem for a tensor that rem to verify Møller’s theorem, surprisingly finding that arXiv:1702.03200v24 [physics.gen-ph] 3 Dec 2020 is required to be time-independent [12]. Laue’s theorem Møller’s theorem is fundamentally wrong. only provides a sufficient condition (instead of a sufficient Like the four-vector theorem, a Lorentz scalar theo- and necessary condition), and it cannot be used to judge rem is a mathematical statement that under what con- the Lorentz property of the energy and momentum of ditions, the time-element space integral of a four-vector electrostatic fields. In a recent study, Laue’s theorem is is a Lorentz scalar. In Ref. [6], a scalar theorem for a improved to be a theorem that has a sufficient and nec- four-vector that is required to be time-independent is set essary condition, and it is successfully used to generally up, called “derivative von Laue’s theorem”, and it is suc- cessfully used to strictly resolve the invariance problem of total electric charge in relativistic electrodynamics. In this paper, we also provide a generalized Lorentz ∗Electronic address: changbiao [email protected] scalar theorem for a four-vector which is not required to 2 be time-independent and divergence-less, and on which Lorentz four-vector (Theorem 2), while the scalar theo- no boundary conditions are imposed. This theorem has rem provides a criterion to judge under what condition a sufficient and necessary condition. We use this scalar the space integral of the time-element of a four-vector is theorem to identify the validity of a well-known result in a Lorentz scalar (Theorem 3). The proofs of Theorem the dynamics of relativity that if a Lorentz four-vector 1 and Theorem 2 are very similar, and without loss of is divergence-less, then the time-element space integral generality, only the proof of Theorem 1 is given. of the four-vector is a Lorentz scalar [3, p. 168], namely Suppose that an inertial frame of X0Y 0Z0 moves uni- the “invariant conservation law” claimed by Weinberg [2, formly at βc relatively to the laboratory frame XYZ, p. 41]. We unexpectedly find that this widely-accepted where c is the vacuum light speed. The Lorentz trans- result is also fundamentally wrong. formation of time-space four-vector Xµ = (x, ct) is given The paper is organized as follows. In Sec. II, proofs are by [6, 7] given of Lorentz four-vector and scalar theorems based x0 = x + ξ(β · x)β − γβct, (1) on the principles of classical mathematical analysis, with 0 a simplified analytic example given to illustrate how to ct = γ(ct − β · x), (2) transform a space integral from one inertial frame to an- or conversely, given by other, and a strict mathematical derivation provided to 0 0 0 confirm the effect of Lorentz contraction. In Sec. III, x = x0 + ξ(β · x0)β − γβ ct0, (3) Møller’s theorem is proved to be incorrect, and a cor- ct = γ(ct0 − β0 · x0), (4) rected version of Møller’s theorem is provided. In Sec. IV, the “invariant conservation law” in relativistic electrody- where ξ ≡ (γ − 1)/β2 = γ2/(γ + 1), γ ≡ (1 − β2)−1/2, 0 ν namics is proved to be invalid; namely the current con- and β = −β. Note: Xµ = gµν X = (−x, ct), with µν tinuity equation ∇ · J + ∂ρ/∂t = 0 cannot be taken as gµν = g = diag(−1, −1, −1, +1) the Minkowski metric. the charge conservation law in relativity. In Sec. V, some According to the definition of tensors [3, p.108], if remarks and conclusions are given. In Appendix A it Ωµν (x, t) is a Lorentz four-tensor given in XYZ, where is demonstrated why the hyperplane differential-element µ, ν = 1, 2, 3, and 4, with the index 4 correspond- four-vector, introduced to transform space integrals be- ing to time component, then in X0Y 0Z0 the tensor tween Lorentz inertial frames in textbooks, contradicts Ω0µν x = x(x0, t0), t = t(x0, t0)
is obtained through the principles of mathematical analysis and the princi- “double” Lorentz transformation of Ωµν (x, t), given by ple of relativity. In Appendix B, as an application of Theorem 1 to Minkowski tensor for a plane light wave in 0µ 0ν 0µν ∂X ∂X λσ a moving uniform medium, the momentum–energy four- Ω (x, t) = Ω (x, t), (5) ∂Xλ ∂Xσ vector of the quasi-photon and the Lorentz invariance of 0µν  0 0 0 0  Planck constant are naturally derived. In Appendix C, Ω x = x(x , t ), t = t(x , t ) physical counterexamples of Thirring’s claims are pro- ∂X0µ ∂X0ν   vided. In Appendix D, an illustration is given of why = Ωλσ x = x(x0, t0), t = t(x0, t0) , (6) ∂Xλ ∂Xσ the proofs of the positive mass theorem in general rela- tivity are based on a flawed theoretical framework. In where ∂X0µ/∂Xλ and ∂X0ν /∂Xσ are obtained from Appendix E, the covariance of Gordon optical metric is Lorentz transformation Eqs. (1) and (2), while x = questioned. x(x0, t0) and t = t(x0, t0) denote Lorentz transformation Eqs. (3) and (4), respectively. Eq. (5) is the expression of Ω0µν observed in XYZ, and Eq. (6) is the expression of II. LORENTZ FOUR-VECTOR THEOREMS Ω0µν observed in X0Y 0Z0. AND SCALAR THEOREM Theorem 1. Suppose that Θµν (x, t) is an integrable In this section, proofs are given of Lorentz four-vector Lorentz four-tensor, defined in the domain V in the lab- and scalar theorems, and a simplified analytic example oratory frame XYZ, where µ, ν = 1, 2, 3, and 4, with the is provided to illustrate how to transform a space inte- index 4 corresponding to time component, and V includ- gral from one inertial frame to another and to explain ing its boundary is at rest in XYZ, namely any x ∈ V is why under time-space Lorentz transformation the effect independent of t. The space integrals of the time-column of Lorentz contraction, namely the relativistic effect of elements of the tensor in XYZ are defined as lengths of a rigid rod argued by Einstein according to Z the principle of relativity [20], is strictly supported by P µ = Θµ4(x, t)d3x. (7) the principles of classical mathematical analysis. V : t=const Four-vector theorems provide a criterion to judge un- The space integrals of time-column elements of the tensor der what condition the space integrals of the time-column 0 0 0 elements of a tensor constitute a Lorentz four-vector in X Y Z are defined as (Theorem 1) and under what condition the space inte- Z   P 0µ = Θ0µ4 x = x(x0, t0), t d3x0, (8) grals of the time-row elements of a tensor constitute a V 0: t, t0=const 3 where Consider the space integral transform from XYZ to   X0Y 0Z0, given by Θ0µ4 x = x(x0, t0), t Z b ∂X0µ ∂X04   I(t) = cos(x − ct)dx := Θλσ x = x(x0, t0), t . (9) ∂Xλ ∂Xσ a 0 0 Z xb [=b/γ−|β|ct ]  0 0  0 µ = cos γ(x + |β|ct ) − ct γdx The four-vector theorem states: P is a Lorentz four- 0 0 xa [=a/γ−|β|ct ] vector if and only if h i Z = sin(b − ct) − sin(a − ct) , Θµj(x, t)d3x = 0 V : t=const where the integration region [a, b] is fixed in XYZ, and for µ = 1, 2, 3, 4 and j = 1, 2, 3 (10) any x ∈ [a, b] is independent of t, while observed in X0Y 0Z0, according to the principle of relativity, the cor- holds. 0 0 responding integration region [xa, xb] must move at |β|c along the minus x-direction. There are a few main points to understand Theorem 1 In the above integral transform, a key problem to that should be noted, as follows. be solved is to determine which Lorentz transformation (i) The importance of the definition Eq. (9) should should be taken, Eq. (1) or Eq. (3), as shown below. 0µ be emphasized, otherwise the implication of P = b R 0µ4 3 0 R Θ d x is ambiguous, and we cannot set up the trans- (a) Why Eq. (3) is taken? From a cos(x − ct)dx in µ 0µ x0 formation between P and P . In Eq. (9), the space R b 0 0
0 0 0 0 XYZ to 0 cos γ(x +|β|ct )−ct γdx in X Y Z , λσ 0 0 xa variables x in Θ (x, t) are replaced by x = x(x , t ), R b namely the space Lorentz transformation Eq. (3), while t x in a cos(x − ct)dx must be replaced by λσ in Θ (x, t) is kept as it is. Note that in Eq. (7) for the x = γ(x0 + |β|ct0) [Eq. (3)], definition of P µ, the integration variables x = (x, y, z) are independent of t because the domain V is fixed in XYZ, instead of λσ which is the mathematical reason why x in Θ (x, t) in x = x0/γ + |β|ct derived from Eq. (9) should be replaced by x = x(x0, t0) [Eq. (3)], in- 0 stead of x = f(x0, t) derived from x0 = x0(x, t) given by x = γ(x − |β|ct) [Eq. (1)], 0 Eq. (1) where x are also functions of t in addition to x . because the (proper) integration region a ≤ x ≤ b is (ii) Observed in XYZ, like P µ, P 0µ is only depen- fixed in XYZ, and the region boundaries x = a and dent on t in general; confer Eq. (16). The quantity t0 x = b are independent of t, while x = x0/γ + |β|ct in the integrand Θ0µ4x = x(x0, t0), t
of P 0µ = P 0µ(t) = derived from x0 = γ(x−|β|ct) [Eq. (1)] is a function R 0µ4 0 0
3 0 of t in addition to x0. V 0: t, t0=const Θ x = x(x , t ), t d x is introduced as a constant parameter in the space integral transform (b) Differential element transformation. According to 0 0 0 0 0 0 from XYZ to X Y Z , and thus observed in X Y Z , (a) which is a strict mathematical rule, the differ- 0 0µ 0µ the boundary of V is moving so that P = P (t) does ential element transformation must be calculated 0 not contain t . from x = γ(x0 + |β|ct0) [Eq. (3)], with t0 taken as (iii) If Θλσ(x, t) is independent of t, then both P µ and a constant parameter, leading to dx = γdx0 where P 0µ are independent of t, namely they are constants. the Jacobian determinant ∂(x)/∂(x0) = ∂x/∂x0 = γ is taken into account. (iv) The symmetry (Θµν = Θνµ) and divergence-less µν 0 0 0 (∂ν Θ =0) are not required, and there are no boundary (c) Motion of region. Observed in X Y Z , the integra- µν 0 0 0 conditions imposed on Θ (x, t). tion region is given by xa ≤ x ≤ xb, and the region 0 0 0 0 boundaries xa [= a/γ−|β|ct ] and xb [= b/γ−|β|ct ] Analytical example. In order to better understand (i) are moving at a velocity of |β|c along the minus x- and (ii), let us take a simple one-dimensional example or x0-direction so that the integral does not contain to illustrate how to transform the space integral from t0 although the integrand cos γ(x0 + |β|ct0) − ct
γ one Lorentz inertial frame to another according to the contains t0. principles of classical mathematical analysis [14]. Suppose that X0Y 0Z0 moves at |β|c with respect to (d) Effect of Lorentz contraction. The expressions of XYZ along the positive x-direction. In such a case, the integration region boundaries space Lorentz transformations Eqs. (1) and (3) are, re- 0 0 xa = a/γ − |β|ct spectively, simplified into: 0 0 xb = b/γ − |β|ct x0 = γ(x − |β|ct), y0 = y, z0 = z for Eq. (1), 0 0 0 0 are governed by the mathematical rule (a), suggest- x = γ(x + |β|ct ), y = y , z = z for Eq. (3). 0 0 ing that xa and xb must be “measured” at the same 4

time t0 physically, except for the case with βc = 0 Note that x = x(x0, t0) in Θλσx = x(x0, t0), t
denotes 0 0 0 0 0 0 ⇒ X Y Z coincides with XYZ, and xa (= a) and Eq. (3). By the change of variables x(x , t ) = x or 0 0 0 0 0 0 xb (= b) are independent of the times when they are (x , y , z ; t ) → (x, y, z) with t as a constant parameter, measured, respectively. Thus the effect of Lorentz from above Eq. (15) we obtain contraction or the relativistic effect of lengths of a 0µ 04 Z rigid rod, defined by 0µ ∂X ∂X 1 λσ 3 P = λ σ Θ (x, t)d x, (16) 0 0 ∂X ∂X γ V : t=const R b R xb 0 R xb 0 a dx = x0 (γdx ) = γ x0 dx a a where d3x = |∂(x, y, z)/∂(x0, y0, z0)|d3x0 = γd3x0 is 0 0 ⇐⇒ xb − xa = (b − a)/γ, employed, with the Jacobian determinant 0 0 0 is a strict and natural result from the principles of ∂(x, y, z)/∂(x , y , z ) = γ being explained as the effect b of Lorentz contraction physically (confer Appendix A). mathematical analysis, where R dx is the proper a Since t0 is introduced as a constant parameter in the x0 R b 0 0µ 0 length of the rod, and 0 dx is exactly the length change of variables, P is independent of t . xa 0 04 4 of the moving rod defined by Einstein, because xa From Eq. (16), with ∂X /∂X = γ and the definition x0 µ R µ4 3 0 R b 0 given by Eq. (7), P = V : t=const Θ (x, t)d x, taken and xb in x0 dx are the points at which “the two a into account, we have ends of the [moving] rod to be measured are located at a definite time [ at the same time t0 ]” [20]. ∂X0µ ∂X04 1 Z P 0µ = Θλj(x, t)d3x ∂Xλ ∂Xj γ Theorem 2. Suppose that Θµν (x, t) is an integrable V : t=const 0µ 04 Z Lorentz four-tensor, defined in the domain V in the lab- ∂X ∂X 1 λ4 3 + λ 4 Θ (x, t)d x oratory frame XYZ, where µ, ν = 1, 2, 3, and 4, with the ∂X ∂X γ V : t=const index 4 corresponding to time component, and V includ- ∂X0µ ∂X04 1 Z = Θλj(x, t)d3x ing its boundary is at rest in XYZ, namely any x ∈ V λ j ∂X ∂X γ V : t=const is independent of t. The space integrals of the time-row ∂X0µ Z elements of the tensor in XYZ are defined as + Θλ4(x, t)d3x ∂Xλ Z V : t=const ν 4ν 3 0µ 04 Z Π = Θ (x, t)d x. (11) ∂X ∂X 1 λj 3 V : t=const = λ j Θ (x, t)d x ∂X ∂X γ V : t=const The space integrals of time-row elements of the tensor in ∂X0µ X0Y 0Z0 are defined as + P λ, (with j = 1, 2, 3). (17) ∂Xλ Z   Π0ν = Θ04ν x = x(x0, t0), t d3x0, (12) µ V 0: t, t0=const If P is a Lorentz four-vector, then where 0µ 0µ ∂X λ   P = P (18) Θ04ν x = x(x0, t0), t ∂Xλ

04 0ν must hold. Inserting Eq. (18) into Eq. (17), we have ∂X ∂X λσ 0 0  := λ σ Θ x = x(x , t ), t . (13) ∂X ∂X ∂X0µ ∂X04 1 Z Θλj(x, t)d3x = 0, (19) The four-vector theorem states: Πν is a Lorentz four- λ j ∂X ∂X γ V : t=const vector if and only if Z 0µ λ
iν 3 where ∂X /∂X is the Lorentz transformation Θ (x, t)d x = 0 0µ λ
V : t=const matrix, with its determinant det ∂X /∂X = 1 [4, for ν = 1, 2, 3, 4 and i = 1, 2, 3 (14) p. 544]. With both sides of above Eq. (19) multiplied by γ, from Eq. (1) we have holds.

 2  1 + ξβx ξβxβy ξβxβz −γβx Proof of Theorem 1. From Eqs. (8) and (9) we have 2  ξβyβx 1 + ξβy ξβyβz −γβy   ξβ β ξβ β 1 + ξβ2 −γβ  Z    z x z y z z  P 0µ = Θ0µ4 x = x(x0, t0), t d3x0 −γβx −γβy −γβz γ V 0: t, t0=const

 a11 a12 a13     0  ∂X0µ ∂X04 Z   21 22 23 −γβx = Θλσ x = x(x0, t0), t d3x0. a a a 0 λ σ ×  31 32 33   −γβy  =   , (20) ∂X ∂X 0 0  a a a   0  V : t, t =const −γβ (15) a41 a42 a43 z 0 5

λj R λj 3 λ where a = V : t=const Θ (x, t)d x, with λ =1,2,3,4 where the space variables x in Λ (x, t) are replaced by and j =1,2,3. x = x(x0, t0), namely the space Lorentz transformation From above it is seen that Eq. (20)⇔Eq. (18) through Eq. (3), but t in Λλ(x, t) is kept as it is. Eq. (17) is valid. Thus for P µ to be a Lorentz four-vector, Making integration in Eq. (24) with respect to (x, y, z) the sufficient and necessary condition is given by over V in the laboratory frame, we have Z aλj = Θλj(x, t)d3x = 0 Z   V : t=const Λ04 x = x(x0, t0), t d3x V : t=const for λ = 1, 2, 3, 4 and j = 1, 2, 3. (21) 04 Z ∂X λ 0 0  3 = λ Λ x = x(x , t ), t d x. (25) The sufficiency of Eq. (21) is apparent because we ∂X V : t=const directly have Eq. (21)⇒Eq. (20)⇒Eq. (19)⇒Eq. (18) from Eq. (17). The necessity is based on the fact By the change of variables (x, y, z) → (x0, y0, z0; t0) with t0 that a four-vector must follow Lorentz rule between as a constant parameter in the left-hand side of Eq. (25), any two inertial frames, namely βc is arbitrary, while keeping the integrals of the right-hand side to be and thus aλj = 0 must hold for all λ and j, be- computed in XYZ frame, we obtain λ1 cause (βx 6= 0, βy = 0, βz = 0) ⇒ a = 0, λ2 Z   (βx = 0, βy 6= 0, βz = 0) ⇒ a = 0, and Λ04 x = x(x0, t0), t γd3x0 λ3 (βx = 0, βy = 0, βz 6= 0) ⇒ a = 0. Thus we V 0: t, t0=const finish the proof of the sufficiency and necessity. ∂X04 Z Theorem 3. Suppose that Λµ(x, t) = (Λ,Λ4) is an in- λ 3 = λ Λ (x, t)d x. (26) tegrable Lorentz four-vector, defined in the domain V in ∂X V : t=const the laboratory frame XYZ, where µ = 1, 2, 3, and 4, where d3x = |∂(x, y, z)/∂(x0, y0, z0)|d3x0 = γd3x0 is taken with the index 4 corresponding to time component, and into account, with ∂(x, y, z)/∂(x0, y0, z0) = γ the Jaco- V including its boundary is at rest in XYZ, namely any bian determinant. x ∈ V is independent of t. The Lorentz scalar theorem We define states: The time-element space integral Z   Φ0 = Λ04 x = x(x0, t0), t d3x0, (27) Z 0 0 Φ = Λ4(x, t)d3x (22) V : t , t=const V : t=const where Λ04x = x(x0, t0), t
is defined in Eq. (24). Since is a Lorentz scalar if and only if t0 is introduced as a constant parameter in the change of variables in the space integral, Φ0 does not contain t0 Z although the integrand Λ04x = x(x0, t0), t
in Eq. (27) Λi(x, t)d3x = 0 for i = 1, 2, 3 or contains t0. Thus with the both sides of Eq. (26) divided V : t=const by γ and then Eq. (27) inserted, we have Z Z 0 04 0 0  3 0 Λ(x, t)d3x = 0 (23) Φ = Λ x = x(x , t ), t d x 0 0 V : t=const V : t , t=const holds. 04 Z ∂X 1 λ 3 = λ Λ (x, t)d x R 4 3 ∂X γ V : t=const Proof. Corresponding to Φ = V : t=const Λ (x, t)d x given by Eq. (22), we first have to define Φ0 = R Λ04d3x0 in X0Y 0Z0, because the implication of Φ0 = R Λ04d3x0 it- 04 Z ∂X 1 4 3 04 0 0 = Λ (x, t)d x self is ambiguous before the dependence of Λ on x , t , ∂X4 γ and t is defined. For this end, from Lorentz transforma- V : t=const tion we have ∂X04 1 Z ∂X04 + Λi(x, t)d3x Λ04(x, t) = Λλ(x, t) ∂Xi γ ∂Xλ V : t=const (with i = 1, 2, 3)

04 04 0 0  ∂X λ 0 0  ⇒ Λ x = x(x , t ), t = λ Λ x = x(x , t ), t Z ∂X = Φ − β · Λ(x, t)d3x, (28) (24) V : t=const 6 where ∂X04/∂X4 = γ,(∂X04/∂Xi)Λi = −γβ·Λ, and the are not symmetric, although R 4 3 definition Φ = V : t=const Λ (x, t)d x given by Eq. (22) are employed. Λ04 = γ(Λ4 − β · Λ) and From Eq. (28) we obtain the sufficient and necessary Λ4 = γ(Λ04 − β0 · Λ0) condition for Φ = Φ0 (Lorentz scalar), given by are symmetric. This asymmetry comes from the fact that Z 0 0 0 0 Λi(x, t)d3x = 0 for i = 1, 2, 3 or V is fixed in XYZ, while V is moving in X Y Z . V : t=const

Z III. INVALIDITY OF MØLLER’S THEOREM Λ(x, t)d3x = 0. (29) V : t=const In this section, (i) Møller’s theorem is proved to be The sufficiency is apparent, while the necessity comes incorrect; (ii) based on Theorem 1, a counterexample of from the fact that β is arbitrary. Thus we complete the Møller’s theorem is given; and (iii) a corrected version of proof. Møller’s theorem is provided, with a detailed elucidation given of why the corrected Møller’s theorem only defines There are some main points to understand Theorem 3 a trivial zero four-vector for EM stress–energy tensor. that should be noted: Møller’s theorem. Suppose that Θµν (x, t) is an inte- (i) If Λµ(x, t) is independent of t, namely ∂Λµ/∂t ≡ 0, grable Lorentz four-tensor, defined in the domain V in then both Φ and Φ0 are constants. the laboratory frame XYZ, where µ, ν = 1, 2, 3, and µ (ii) The divergence-less (∂µΛ = 0) is not required, and 4, with the index 4 corresponding to time component, there are no boundary conditions imposed on Λµ(x, t). and V including its boundary is at rest in XYZ, namely any x ∈ V is independent of t. All the elements of the (iii) Asymmetry arising from resting V and moving tensor have first-order partial derivatives with respect to V 0. Directly from Eq. (2), we have time-space coordinates Xµ = (x, ct). Møller’s theorem µν µν states: If Θ (x, t) is divergence-less (∂ν Θ (x, t) = 0), Λ04 = γ(Λ4 − β · Λ) and Θµν (x, t) = 0 holds on the boundary of V for any time (−∞ < t < +∞)— zero boundary condition, then Z Z the time-column space integrals ⇒ Λ04d3x0 = γ(Λ4 − β · Λ)d3x0 0 0 Z V V P µ = Θµ4(x, t)d3x (30) V : t=const Z ⇒ Φ0 = Φ − β · Λd3x constitute a Lorentz four-vector [3, pp.166-169]. V with d3x0 = d3x/γ used, namely Eq. (28). Conversely, from Eq. (4) we have Proof. From Møller’s sufficient condition, we first demonstrate that the time-column space integrals, given by Eq. (30), are time-independent (∂P µ/∂t ≡ 0), then we Λ4 = γ(Λ04 − β0 · Λ0) prove that the sufficient condition is not enough to make Eq. (30) be a four-vector, and we conclude that Møller’s Z Z theorem is incorrect. 4 3 04 0 0 3 ⇒ Λ d x = γ(Λ − β · Λ )d x Since Θµν (x, t) = 0 holds on the boundary of V , using V V 3-dimensional Gauss’s divergence theorem we obtain Z Z 2 0 2 0 0 3 0 µi 3 ⇒ Φ = γ Φ − γ β · Λ d x ∂iΘ (x, t)d x = 0, with i = 1, 2, 3. (31) V 0 V : t=const with d3x = γd3x0 used. We find that Because the boundary of V is at rest in the laboratory Z frame, we have Φ0 = Φ − β · Λd3x and Z ∂ ∂ Z V ( ··· )d3x = ( ··· )d3x. (32) ∂t ∂t Z V : t=const V : t=const Φ = γ2Φ0 − γ2β0 · Λ0d3x0 µν 4 V 0 From ∂ν Θ (x, t) = 0, with Eq. (31), Eq. (32), and X = 7 ct taken into account, we have Thus like Eq. (17), we obtain a sufficient and necessary condition for constant P µ to be a Lorentz four-vector, Z µν 3 given below 0 = ∂ν Θ (x, t)d x V : t=const Z Z aλj = Θλj(x, t)d3x = 0 µi 3 V : t=const = ∂iΘ (x, t)d x for λ = 1, 2, 3, 4 and j = 1, 2, 3, (37) V : t=const Z µ4 3 + ∂4Θ (x, t)d x which is the same as Eq. (10). However Møller’s suf- ficient condition does not include this sufficient and V : t=const Z necessary condition, and accordingly, Møller’s theorem µ4 3 = ∂4Θ (x, t)d x is fundamentally wrong. Thus we finish the proof. V : t=const Counterexample of Møller’s theorem. To further convince ∂ Z = Θµ4(x, t)d3x. (33) readers, given below is a pure mathematical counterex- ∂(ct) ample to disprove Møller’s theorem based on Theorem V : t=const 1. As indicated in Sec. V later, this counterexample of Møller’s theorem is also the counterexample of Landau- Inserting Eq. (30) into above Eq. (33) yields Lifshitz and Weinberg’s versions of Laue’s theorem [6]. Suppose that there is a symmetric Lorentz four-tensor ∂P µ ∂ Z = Θµ4(x, t)d3x ≡ 0. (34) ∂t ∂t   V : t=const 0 0 0 f(x) 0 0 0 f(x) Aµν =   , µ R µ4 3  0 0 0 f(x)  Thus P = V : t=const Θ (x, t)d x is constant although the integrand Θµ4(x, t) may depend on t. However it f(x) f(x) f(x) −(ct)(fx + fy + fz) should be emphasized that (38)

∂ Z defined in the cubic domain V (−π ≤ x, y, z ≤ π), where Θµj(x, t)d3x = 0 ∂t f(x) = (sin x)2(sin y)2(sin z)2 is independent of time, V : t=const R 3 3 with V f(x)d x = π , and fx ≡ ∂f/∂x, fy ≡ ∂f/∂y, µν µν for j = 1, 2, 3 may not hold. (35) and fz ≡ ∂f/∂z. A is divergence-less (∂µA = 0 ⇔ µν µν νµ ∂ν A = 0 because of A = A ), and satisfies the µν From the divergence-less (∂ν Θ = 0) and the zero- Møller’s zero boundary condition: Aµν =0 holds on the boundary condition (Θµν = 0 on boundary), we have boundary x, y, z = ±π for −∞ < t < +∞. Thus Aµν achieved a conclusion that the time-column space inte- satisfies the sufficient condition of Møller’s theorem, and µ R µ4 3 grals P = V : t=const Θ (x, t)d x are time-independent constants. In what follows, we will show that the Z divergence-less and the zero-boundary condition is not M µ = Aµ4d3x = (π3, π3, π3, 0) (39) µ sufficient to make P be a four-vector. In other words, V : t=const Møller’s sufficient condition is not sufficient. From Eqs. (15)-(17) in the proof of Theorem 1, we have is supposed to be a Lorentz four-vector. However because Z   P 0µ = Θ0µ4 x = x(x0, t0), t d3x0 Z Z V 0: t, t0=const A41d3x = A42d3x V : t=const V : t=const

0µ 04 Z ∂X ∂X 1 λσ 3 Z = λ σ Θ (x, t)d x 43 3 ∂X ∂X γ V : t=const = A d x V : t=const

0µ 04 Z ∂X ∂X 1 λj 3 3 = λ j Θ (x, t)d x = π 6= 0, (40) ∂X ∂X γ V : t=const (allowed to be t-dependent) Aµν does not satisfy the sufficient and necessary condition Eq. (10) of Theorem 1, and accordingly, M µ = R Aµ4d3x is not a four-vector. Thus ∂X0µ V : t=const λ Møller’s theorem is disproved by this counterexample + λ P , (with j = 1, 2, 3). (36) ∂X based on Theorem 1. (t-independent) 8

The above counterexample shows that the sufficient Physically, the pre-assumption is extremely strong and condition of Møller’s theorem indeed does not includes severe, because it requires that (i) within the domain V , µν the sufficient and necessary condition Eq. (10) of The- there are no any sources (∂ν Te = 0), and (ii) the EM orem 1. Obviously, Møller’s theorem can be easily energy and Minkowski momentum never flow through corrected by adding the condition Eq. (10), as follows. the closed boundary of V for any time (E × H = 0 and D × B = 0 for −∞ < t < +∞). Thus this physical Corrected Møller’s theorem. Suppose that Θµν (x, t) system is never provided with any EM energy and mo- is an integrable Lorentz four-tensor, defined in the do- mentum. According to energy–momentum conservation main V in the laboratory frame XYZ, where µ, ν = 1, law, no EM fields can be supported within the domain 2, 3, and 4, with the index 4 corresponding to time com- V in such a case, leading to a zero field solution. Thus ponent, and V including its boundary is at rest in XYZ, the corrected Møller’s theorem only defines a trivial zero namely any x ∈ V is independent of t. It is assumed four-vector for an EM stress–energy tensor of a finite µν µν that Θ (x, t) is divergence-less (∂ν Θ (x, t) = 0), and closed physical system, even if this theorem is applicable. Θµν (x, t) = 0 holds on the boundary of V for any time (−∞ < t < +∞)— zero boundary condition. The cor- Example 2 for corrected Møller’s theorem. Nevertheless, rected Møller’s theorem states: The time-column space the corrected Møller’s theorem may define a non-zero integrals four-vector in general. As an example, consider the ten- Z sor given by P µ = Θµ4(x, t)d3x (41) V : t=const  0 0 0 0  0 0 0 0 Bµν (x, t) =   , (44) constitute a Lorentz four-vector if and only if  0 0 0 0  Z 0 0 0 f(x) Θµj(x, t)d3x = 0 V : t=const defined in the cubic domain V (−π ≤ x, y, z ≤ π), where 2 2 2 R 3 3 for µ = 1, 2, 3, 4 and j = 1, 2, 3. (42) f(x) = (sin x) (sin y) (sin z) , with V f(x)d x = π . µν µν B (x, t) is divergence-less (∂ν B = 0), and satisfies holds. the zero boundary condition: Bµν (x, t) = 0 on the boundary (x, y, z = ±π) for −∞ < t < +∞; thus the However we would like to indicate, by enumerating pre-assumption of corrected Møller’s theorem is satis- specific examples as follows, that the corrected Møller’s R µj 3 fied. On the other hand, V : t=const B (x, t)d x = 0 theorem has a limited application. holds for µ = 1, 2, 3, 4 and j = 1, 2, 3; thus Bµν (x, t) also satisfies the sufficient and necessary condition Example 1 for corrected Møller’s theorem. Consider Eq. (42) for the corrected Møller’s theorem. Accord- Minkowski EM stress–energy tensor for “a pure radia- R µ4 3 3 ingly, V : t=const B (x, t)d x = (0, 0, 0, π ) 6= 0 is a tion field in matter” [10], given by four-vector — the corrected Møller’s theorem may define a non-zero four-vector in general.  Tˇ cg  Teµν = (T µν )T , with T µν = M A , (43) cgM Wem Conclusion for corrected Møller’s theorem. In con- clusion, the corrected Møller’s theorem may define a µν µν µν where Te is the transpose of T , with ∂ν Te = non-zero four-vector in general; however, it only defines νµ ˇ
∂ν T = ∇ · TM + ∂gM/∂t, ∇ · (cgA) + ∂Wem/∂(ct) ; a trivial zero four-vector for an EM stress–energy tensor 2 gA = E×H/c is the Abraham momentum; gM = D×B of a finite closed physical system. Thus the application is the Minkowski momentum; Wem = 0.5(D·E+B·H) is of the theorem is limited. the EM energy density; and Tˇ M = −DE−BH+ˇI0.5(D· E+B·H) is the Minkowski stress tensor, with ˇI the unit Differences between three four-vector theorems. We have tensor [6]. We first assume that the corrected Møller’s three four-vector theorems: Theorem 1 and corrected theorem is applicable for this EM tensor. Then let us see Møller’s theorem (both presented in the present paper), what conclusion we can get. and generalized von Laue’s theorem (presented in The pre-assumption of corrected Møller’s theorem is Ref. [6]). For the convenience to compare, we write the tensor’s divergence-less plus a zero-boundary condi- down the generalized von Laue’s theorem from Ref. [6] tion. The zero-boundary condition requires that all the as follows. tensor elements be equal to zero on the boundary for any time (−∞ < t < +∞). Thus for the EM stress– Generalized von Laue’s theorem. Assume that energy tensor given by Eq. (43), the pre-assumption re- Θµν (x) is an integrable Lorentz four-tensor, defined in µν quires ∂ν Te = 0 holding within the finite domain V the domain V in the laboratory frame XYZ, where µ, ν of a physical system, and Poynting vector E × H = 0 = 1, 2, 3, and 4, with the index 4 corresponding to time and Minkowski momentum D × B = 0 holding on the component, V including its boundary is at rest in XYZ, boundary of V for any time (−∞ < t < +∞). and Θµν is independent of time (∂Θµν /∂t ≡ 0). The 9

µν µν generalized von Laue’s theorem states: The time-column- (i) ∂ν R = 0 holds but R (x, t) does not have a zero- µ R µ4 3 41 element space integrals P = V Θ (x)d x constitute boundary condition R (x, t) = π 6= 0 on the boundary: R µj 3
a Lorentz four-vector if and only if V Θ (x)d x = 0 x = π and −π ≤ y, z ≤ π, for example . Thus the cor- holds for all µ =1, 2, 3, 4 and j = 1, 2, 3. rected Møller’s theorem does not apply. (ii) ∂Rµν /∂t ≡ 0 does not hold, because of ∂R44/∂t = Between the corrected Møller’s theorem and the above −c 6= 0 . Thus the generalized von Laue’s theorem does generalized von Laue’s theorem, the difference is that not apply either. in the corrected Møller’s theorem, the divergence-less (iii) R Rµj(x, t)d3x = 0 for µ = 1, 2, 3, 4 (∂ Θµν = 0) plus a zero boundary condition (Θµν = 0 on V : t=const ν and j = 1, 2, 3 holds, satisfying the sufficient and nec- boundary) is taken as a pre-assumption, and Θµν (x, t) is essary condition Eq. (10) of Theorem 1. Thus N µ = allowed to be time-dependent, while in the generalized R µ4 3 R 3 µν V : t=const R (x, t)d x = V : t=const(x, 0, 0, −ct)d x = von Laue’s theorem, ∂Θ /∂t ≡ 0 is taken as a pre- 3 assumption, and Θµν (x, t) ≡ Θµν (x) is not allowed to be (0, 0, 0, −8π ct) is a Lorentz four-vector. time-dependent, but no boundary condition is required. From above example we see that Theorem 1 has a bet- Compared with the corrected Møller’s theorem and the ter adaptability. It is interesting to indicate that Theo- generalized von Laue’s theorem, Theorem 1 does not have rem 1 can be used to analyze the EM stress–energy tensor any pre-assumption; however, the three theorems have for a plane light wave in a dielectric medium, as shown the same definition P 0µ, as shown below. in Appendix B. From Eq. (36), we know that the same definition of P 0µ is used in both Theorem 1 and the corrected Møller’s theorem, given by IV. INVALIDITY OF WEINBERG’S CLAIM Z   P 0µ = Θ0µ4 x = x(x0, t0), t d3x0 V 0: t, t0=const In relativistic electrodynamics, there are two main- stream arguments for the Lorentz invariance of total elec- tric charge. One of them comes from an assumption that 0µ 04 Z ∂X ∂X 1 λσ 3 the total electric charge is an experimental invariant, as = λ σ Θ (x, t)d x. (45) ∂X ∂X γ V : t=const presented in the textbook by Jackson [4, p.555]; the other comes from a well-accepted “invariant conservation law” If Θµν (x, t) is independent of t, namely Θµν (x, t) ≡ that the divergence-less of current density four-vector Θµν (x), then the above Eq. (45) becomes makes the total charge be a Lorentz scalar, as claimed in the textbook by Weinberg [2, p.41]. In this section, by enumerating a counterexample we use Theorem 3 to Z   P 0µ = Θ0µ4 x = x(x0, t0) d3x0 disprove Weinberg’s claim. V 0: t, t0=const For a physical system defined in the domain V with S as its closed boundary, we will show that the divergence-less (∂ J µ = 0) of current density four-vector 0µ 04 Z µ ∂X ∂X 1 λσ 3 J µ = (J, cρ) plus a boundary zero-integral given by = λ σ Θ (x)d x. (46) ∂X ∂X γ V : t=const H S J(x, t) · dS = 0 makes the total charge Q in V be a time-independent constant; however, it is not enough to This is exactly the case of von Laue’s theorem presented make the constant be a Lorentz scalar. in Ref. [6], where Θ0µ4x = x(x0, t0)
is written as 0µ4 0 0 µ Θ (x , ct ), and t does not show up. Constant of total electric charge. From ∂µJ = 0 ⇒ ∇ · J + ∂ρ/∂t = 0, with Q = R ρd3x taken into account H V Adaptability of Theorem 1. Since Theorem 1 does not [21, p. 461], we have S J(x, t) · dS + dQ/dt = 0, where have a pre-assumption, it may have a better adaptability. R the exchangeability between ∂/∂t and V is employed so To show this, a specific example is given below. that R (∂ρ/∂t)d3x = dQ/dt holds, as usually presented Suppose that there is a symmetric Lorentz four-tensor V H in textbooks [2, p. 20] [41, p. 345]. If S J(x, t) · dS = 0 holds, then we have dQ/dt = 0 ⇒ Q = const . Phys-  0 0 0 x  ically, the current density J = ρu is a charge den- 0 0 0 0 sity flow, where u is the charge moving velocity, and Rµν (x, t) =   , (47) H  0 0 0 0  S J(x, t) · dS = 0 means that there is no net charge x 0 0 −ct flowing into or out of V . Thus the total electric charge Q in V is constant; however, it never means that this con- stant is a Lorentz invariant. (It is should be emphasized defined in the cubic domain V (−π ≤ x, y, z ≤ π) with µ H that only from ∂µJ = 0 without S J(x, t) · dS = 0 its boundary S (x, y, z = ±π). considered, one cannot derive Q = const in V .) From Eq. (47) we know that Why is the constant total charge Q in V in above, 10

µ R 3 resulting from ∂µJ = 0 ⇒ ∇ · J = −∂ρ/∂t, not a cludes the sufficient and necessary condition Λd x = 0 Lorentz invariant? That is because Q in V in frame of Theorem 3. XYZ and Q0 in V 0 in frame X0Y 0Z0 do not refer to The current density four-vector J µ = (J, cρ) and the same total charge in the same volume physically, the above counterexample W µ = (W,W 4) are all where X0Y 0Z0 moves at βc 6= 0 relatively to XYZ. divergence-less, while the Lorentz property of their This conclusion directly comes from the fact that, the time-element space integrals does not depends on the R 3 volume V in V (∂ρ/∂t)d x is fixed in XYZ, and the divergence-less. Now let us take a look of four-vectors 0 R 0 0 3 0 0 0 0 volume V in V 0 (∂ρ /∂t )d x is fixed in X Y Z so that are not divergence-less, and see what difference they R 0 R may have. that ∂/∂t and V , and ∂/∂t and V 0 are exchangeable, respectively, as shown in Eq. (32), in order to make both Example 1. Consider a four-vector, given by Γ µ = R (∂ρ/∂t)d3x = dQ/dt and R (∂ρ0/∂t0)d3x0 = dQ0/dt0 V V 0 (Γ,Γ 4) = (sin x sin y sin z, 0, 0, 0) defined in the cu- hold in general. However according to Einstein’s relativ- bic domain V (−π ≤ x, y, z ≤ π), with ∂ Γ µ = ity [20], V fixed in XYZ is moving observed in X0Y 0Z0. µ 0 R 0 0 3 0 0 0 0 cos x sin y sin z 6= 0 holding except for some individual Now V in V 0 (∂ρ /∂t )d x is at rest in X Y Z , and R 3 0 discrete points, and V : t=const Γd x = (0, 0, 0) = 0 hold- thus V and V do not denote the same volume, and Q R 4 3 0 ing. According to Theorem 3, Φ = V : t=const Γ d x (= and Q do not denote the same total charge. Therefore, 0 µ 0) is a Lorentz scalar, because Φ = Φ − β · the continuity equation ∂µJ = 0 cannot be taken as R 3 µ V : t=const Γd x = Φ. Put it simply, for ∂µΓ 6= 0, the “invariant [charge] conservation law” in relativity R 4 3 [2, p. 40], which can be better understood from the Γ d x is a Lorentz scalar. mathematical counterexample given below. Example 2. Consider a four-vector, given by U µ = (U,U 4) = (sin2 x sin2 y sin2 z, 0, 0, 0) defined in the cu- Counterexample of Weinberg’s claim. Why is ∂ J µ = 0 µ µ bic domain V (−π ≤ x, y, z ≤ π), with ∂µU = not a sufficient condition to make Q = const be a Lorentz 2 cos x sin x sin2 y sin2 z 6= 0 holding except for some scalar, even additionally plus a zero-boundary condition individual discrete points, and R Ud3x = J µ = (J, cρ) = 0 on S ⇒ H J · dS = 0? To understand V : t=const S (π3, 0, 0) 6= 0 holding. According to Theorem 3, Φ = this, consider a mathematical four-vector, given by R 4 3 V : t=const U d x (= 0) is not a Lorentz scalar, because   0 R 3 3 0 µ 4 Φ = Φ − β · V : t=const Ud x = Φ − βxπ ⇒ Φ 6= Φ for W (x.t) = (W,W ) = f(x), 0, 0, −(ct)fx (48) µ R 4 3 any βx 6= 0. Put it simply, for ∂µU 6= 0, U d x is not defined in the cubic domain V (−π ≤ x, y, z ≤ π), a Lorentz scalar.
4 where W = f(x), 0, 0 , W = −(ct)fx, and f(x) = From above Example 1 and Example 2, we know that 2 2 2 R 3 3 µ R 4 3 µ R 4 3 (sin x) (sin y) (sin z) , with V f(x)d x = π . W (x, t) Γ d x is a Lorentz scalar for ∂µΓ 6= 0, and U d x µ µ satisfies the zero-boundary condition, namely W = is not a Lorentz scalar for ∂µU 6= 0. We have known 4 R 4 3 µ (W,W ) = 0 holds on the boundary S (x, y, z = ±π). that J d x is a Lorentz scalar for ∂µJ = 0 [6], while µ µ W is divergence-less, namely ∂µW = 0, and in addi- the counterexample of Weinberg’s claim tells us that µ H R 4 3 µ tion, W has a zero-boundary condition ⇒ S W · dS = W d x is not a Lorentz scalar for ∂µW = 0. Thus µ 0 holds. According to Weinberg’s claim, ∂µW = 0 we can generally conclude that whether the time-element makes R W 4d3x be a Lorentz scalar. space integral of a four-vector is a Lorentz scalar has R 3 3 However because of V : t=const Wd x = (π , 0, 0) 6= 0, nothing to do with the divergence-less property of the W µ does not satisfy the sufficient and necessary condition four-vector. Eq. (23) of Theorem 3. Thus according to Theorem 3, The invariance problem of total electric charge has R 4 3 R 3 Φ = V : t=const W d x = V : t=const −(ct)fxd x (= 0) been resolved by using “derivative von Laue’s theorem” is not a Lorentz scalar. To better understand this, from in Ref. [6], which indicates that the invariance comes from Eq. (28) we have two facts: (a) J µ is a four-vector and (b) the moving ve- Z locity of any charged particles is less than vacuum light Φ0 = Φ − β · Wd3x speed. This strict theoretical result removes the assump- V : t=const tion that the total charge is an experimental invariant [4, Z 3 p.555]. = Φ − βx f(x)d x V The difference between the derivative von Laue’s the- 3 orem [6] and Theorem 3 is that the derivative von Laue’s = Φ − βxπ , (49) theorem has a pre-assumption of ∂Λµ/∂t ≡ 0, namely Λµ = (Λ,Λ4) is not allowed to be time-dependent, while 0 and Φ = Φ cannot hold for any βx 6= 0. Thus Weinberg’s Theorem 3 does not. For example, we also can use the µ claim is disproved, namely ∂µW = 0 is not a sufficient derivative von Laue’s theorem [6] to analyze the four- condition to make R W 4d3x be a scalar. vector Γ µ = (Γ,Γ 4) = (sin x sin y sin z, 0, 0, 0) discussed µ µ In the above counterexample, ∂µW = 0 holds but above because ∂Γ /∂t ≡ 0 holds, but we cannot use it to R Wd3x = 0 does not hold. Thus in general, the suffi- analyze W µ(x, t) given by Eq. (48), because ∂W µ/∂t ≡ 0 µ cient condition ∂µΛ = 0 of Weinberg’s claim does not in- does not hold. Thus Theorem 3 has a better adaptability. 11

µν V. REMARKS AND CONCLUSIONS include the divergence-less of a tensor (∂ν Θ (x, t) = 0), but they do not include or derive the sufficient and nec- In this paper, we have developed Lorentz four-vector essary condition Eq. (10). Thus it is not appropriate that µν theorems (Theorem 1 for column four-vector and Theo- ∂ν Θ (x, t) = 0 is recognized to be “conservation Law” rem 2 for row four-vector; they are essentially the same) [18, p. 310] or to “guarantee conservation of the total 4- and Lorentz scalar theorem (Theorem 3). Based on The- momentum” [21, p. 443] in traditional textbooks. orem 1, we find that the well-established Møller’s theo- The most convincing way to disprove a mathematical rem is fundamentally wrong, and we provided a corrected conjecture is to provide its counterexample. The coun- version of Møller’s theorem (see Sec. III). Based on The- terexample Eq. (38) for Møller’s version of Laue’s theo- orem 3, we disproved Weinberg’s claim, and obtained a rem, Aµν , is also a counterexample of Landau-Lifshitz general conclusion for the Lorentz property of a four- and Weinberg’s versions of Laue’s theorem, because Aµν µν vector’s time-element space integral (see Sec. IV). is both divergence-less (∂ν A = 0) and symmetric We have shown that the sufficient condition of Møller’s (Aµν = Aνµ), while the Landau-Lifshitz version takes theorem makes the time-column space integrals of a ten- the divergence-less of a tensor as a sufficient condition, sor be time-independent constants; however, it is not and the Weinberg’s version takes the divergence-less plus a sufficient condition to make the integrals constitute a symmetry of a tensor as a sufficient condition. In other a Lorentz four-vector. The corrected Møller’s theorem words, Aµν given by Eq. (38) is a common mathematical has a limited application; especially for Minkowski EM counterexample to disprove Møller’s, Landau-Lifshitz, stress–energy tensor, the corrected Møller’s theorem only and Weinberg’s versions of Laue’s theorem which are defines a trivial zero four-vector. all mathematical conjectures, independent of physics al- We have shown that there are three four-vector the- though originating from physics, because all physical im- orems: (a) generalized von Laue’s theorem; (b) cor- plications have already been turned into mathematical rected Møller’s theorem; and (c) Theorem 1. The gen- descriptions. eralized von Laue’s theorem, presented in Ref. [6], has a There is a well-known result in the dynamics of rel- pre-assumption that tensor Θµν is required to be time- ativity that if a Lorentz four-vector is divergence-less, independent (∂Θµν /∂t ≡ 0). The corrected Møller’s then the time-element space integral of the four-vector theorem, provided in the present paper, also has a pre- is a Lorentz scalar; for example, Weinberg claims that assumption that tensor Θµν is required to be divergence- “for any conserved four-vector”, namely for any four- µν µ µ less (∂ν Θ = 0) and required to satisfy a zero boundary vector J = (J, cρ) that satisfies equation ∂µJ = 0, condition (Θµν = 0 on boundary) but Θµν is allowed to cQ = R J 4d3x = R cρd3x “defines a time-independent be time-dependent. Compared with the generalized von scalar” [2, p.41], and Møller also claims a proof of such Laue’s theorem and corrected Møller’s theorem, Theorem a result in his textbook [3, p.168]. However in the 1 does not have any pre-assumption, while the three the- present paper, we have shown based on Theorem 3 in orems have the same sufficient and necessary condition. Sec. IV that this well-known result is not correct. We Thus Theorem 1 has a better adaptability, as shown by have found a general conclusion for the Lorentz prop- a specific example described by Eq. (47) in Sec. III. erty of a four-vector’s time-element space integral, stat- However it should be noted that, just because the ing that whether the time-element space integral of a generalized von Laue’s theorem has a pre-assumption four-vector is a Lorentz scalar has nothing to do with of ∂Θµν (x, t)/∂t ≡ 0 (but no boundary condition re- the four-vector’s divergence-less property. Accordingly, quired) and the corrected Møller’s theorem has a pre- it is not appropriate for the current continuity equa- µν µ assumption of ∂ν Θ (x, t) = 0 plus Θ(x, t) = 0 on tion ∂µJ (x, t) = 0 to be called “invariant conservation boundary (zero boundary condition), the four-vector Law” [2, p. 40] or “the law of charge conservation” [21, µ R µ4 3 p. 559, p. 443] in textbooks. P = V : t=const Θ (x, t)d x defined by the two theo- rems is time-independent (∂P µ/∂t ≡ 0). Thus the gen- As presented in traditional textbooks, the local con- eralized von Laue’s theorem and the corrected Møller’s servation law of energy–momentum in general relativity, µν theorem can be taken as “conservation laws” in a tradi- given by T ;ν = 0 [22, p. 298], is covariantly generalized µν µν tional sense. from the conservation law T ,ν = ∂ν T = 0 in spe- We also have shown that there are three incorrect cial relativity [1, p. 83] [2, p. 45] [26]. Thus the validity four-vector theorems: (i) Møller’s theorem, which is also of the latter is a necessary condition for the validity of called “Møller’s version of Laue’s theorem” in Ref. [6]; the former. This law is often used for relativistic analysis (ii) Landau-Lifshitz version of Laue’s theorem; and (iii) of the Abraham-Minkowski debate on the momentum of Weinberg’s version of Laue’s theorem. Møller’s version is light in a medium [9–11], and it is also thought to play disproved in the present paper by taking the mathemat- an important role in gravitation theory [21, p.132], be- ical tensor Eq. (38) as a counterexample, while Landau- cause “the GR [general relativity] field equations should Lifshitz and Weinberg’s versions are disproved in Ref. [6] be consistent with energy and momentum conservation, µν by taking the EM tensor of a charged metal sphere in T ;ν = 0” [22, p. 299]. This law is so well-established free space as a counterexample. All the sufficient condi- that often no citations are given for its origin in research tions of the three disproved versions of Laue’s theorem articles [27–30]. However in fact, the traditional conser- 12

µ µν vation laws including both ∂µΛ = 0 and ∂ν T = 0 [21, These problems never got any attention in the proofs of p. 443] can be directly disproved by simple physical coun- the positive mass theorem by Schoen and Yau in 1979 [32] terexamples, as shown in Appendix C, although they are and 1981 [39], Witten [33] and Nester [34] in 1981, Parker claimed to have been already proved with the use of the and Taubes in 1982 [35], and Gibbons and coworkers in modern language of exterior calculus [19, p. 318]. Thus 1983 [36]. All the proofs are based on a flawed theoret- clarifying the two conservation laws in the present paper ical framework set up by Arnowitt, Deser, and Misner has a general significance. [37, 38], where the total energy–momentum P µ in an In conclusion, we have generally resolved two funda- asymptotically flat spacetime is required to “obey the µν mental issues in the dynamics of relativity: (a) Un- differential conservation law T ,µ = 0 [of the energy– der what condition, the time-column space integrals of momentum tensor T µν ]” and required to “transform as a Lorentz four-tensor constitute a Lorentz four-vector a four-vector” [38]. However in fact, the conservation µν µ (Theorem 1), and (b) under what condition, the time- law T ,µ = 0 cannot guarantee that P is a four-vector, element space integral of a Lorentz four-vector is a as shown in Appendix D. Apparently, this problem is Lorentz scalar (Theorem 3). Both Theorem 1 and The- ignored in all the proofs [32–36, 39]. Moreover, the equa- ν µν µν µν orem 3 have their own sufficient and necessary condi- tions Tµ ;ν = 0, T ;µ = 0, T ;ν = 0, or T ,ν = 0 have tions, which have nothing to do with the divergence-less been always claimed as conservation law in later litera- µν µ µν νµ (∂ν Θ = 0 or ∂µΛ = 0), symmetry (Θ = Θ ), and ture, such as in Ref. [28] by Ratra and Peebles in 1988, boundary conditions. This point is especially important. Ref. [27] by Witten in 1991, Ref. [22, p. 299] by Rindler in For example, from Theorem 1 we can directly judge that 2006, and Ref. [29] by Biˇc´akand Schmidt in 2016. From Møller’s theorem is incorrect, because the sufficient con- this one can see that the community in the field of rela- dition of Møller’s theorem does not include the sufficient tivity has never become aware of these problems before and necessary condition of Theorem 1, as shown in coun- the publications of [6, 7]. terexample Eq. (38). A similar argument is applicable to Recently, Bojowald criticizes that the constant wave the “invariant conservation law” claimed by Weinberg [2, four-vector Kµ in Appendix C does not constitute p. 40], as shown in counterexample Eq. (48). a counterexample to the traditional conservation law µ R 4 3 As a practical application, we have used Theorem 1 ∂µΛ = 0 ⇒ Λ d x = invariant, because “it does to analyze Minkowski EM tensor for a plane light wave not obey the usual boundary conditions of conservation in a moving medium with Einstein’s light–quantum hy- laws, which state that the integrated current must vanish pothesis taken into account, and we find that the four- at timelike boundaries or at infinity” [40]. Apparently, momentum of quasi-photon and the Lorentz invariance Bojowald’s criticism has no basis at all, because the con- µ of Planck constant can be naturally derived (see Ap- servation law ∂µΛ = 0, as claimed in [2, p. 40][19][21, pendix B). Einstein’s light–quantum hypothesis is the ba- p. 443], does not require Λµ = 0 holding on bound- sis of Bohr frequency condition of radiation from atomic aries. A typical example is the current density four- transitions, the invariance of Planck constant is an im- vector J µ = (J, cρ) for a sphere of perfect conductor plicit postulate in Dirac relativistic quantum mechanics with a radius of R (6= 0) and a total static charge of [7], and the existence of four-momentum of the quasi- Q (6= 0) in free space. Observed in the sphere-rest frame, photon is required by momentum–energy conservation the current density is given by J = 0 holding everywhere, law in Einstein-box thought experiment [25]. Thus this and according to the basic properties for a perfect con- natural derivation is compatible with the quantum the- ductor, we know that any net charge resides on the sur- ory and the momentum–energy conservation law, and the face [41, p. 97], with the charge density given by ρ = result obtained strongly supports the Minkowski tensor Qδ(r − R)/(4πR2), where δ(x) is Dirac delta function. as being the most qualified momentum–energy tensor for The current four-vector J µ satisfies the invariant conser- µ R 4 3 descriptions of light–matter interactions in the frame of vation law ∂µJ = 0 with cQ = J d x as a Lorentz the principle of relativity. invariant, as claimed by Weinberg [2, p. 40]; however, 4 2 In relativity, charge conservation law refers to that the J = cρ = cQδ(r −R)/(4πR ) 6= 0 holds on the spherical total charge is a time- and frame-independent constant [2, boundary r = R. From this elegant example of classi- µ p. 41], as demonstrated in Eq. (A6) of Appendix A, while cal electrodynamics — four-current J not vanishing on energy-momentum conservation law refers to that the to- boundary for charged metallic sphere in free space — tal energy and momentum constitute a covariant and one can see that Bojowald’s criticism is exactly based on time-independent four-vector [2, p. 46], as demonstrated his groundless speculation, and it is not justified. Thus µ for a plane light wave in Appendix B. The problems of K 6= 0 on boundaries does not lose its qualification to be µ ∂ Λµ = 0 and ∂ Θµν = 0 as being conservation laws in a counterexample; namely, the wave four-vector K with µ ν µ R 4 3 relativity were first discovered in [6, 7], and they are ex- ∂µK = 0 and K d x 6= invariant in Appendix C is plicitly illustrated and generally resolved in the present a completely qualified counterexample to the traditional µ R 4 3 paper. One might argue that in the establishment of conservation law ∂µΛ = 0 ⇒ Λ d x = invariant. the positive mass theorems in general relativity, rigorous Bojowald also criticizes that instead of from Theorem mathematics has been used and loopholes possibly left in 1 developed in the present paper, the total momentum– earlier works do not exist. Unfortunately, this is not true. energy four-vector should be constructed from a covari- 13 ant combination of a stress–energy tensor and the hyper- proper element multiplied by its four-velocity normal- plane differential-element (HDE) four-vector [40], as pre- ized to light speed c) is constructed to define the change sented in literature such as the textbook by Jackson [4, of variable formula. In the laboratory frame XYZ, the p. 757]. However as shown in Appendix A, the HDE four- HDE four-vector is given by vector is apparently in contradiction with the principle of relativity and the principles of classical mathematical dSµ = (dS, dS4) analysis. Thus this criticism by Bojowald is not justified = (0, 1)d3x = (0, 0, 0, 1)d3x. (A2) either, coming from his negligence of the self-consistency of a physical theory, and his lack of a good knowledge In the frame X0Y 0Z0 moving at βc with respect to XYZ, about how to transform a space integral from one iner- the HDE four-vector is given by tial frame to another according to the change of variables theorem in mathematical analysis [14]. dS0µ = (dS0, dS04) = (γ0β0, γ0)d3x, (A3)

where (γ0β0, γ0) is the normalized four-velocity of V 0 Appendix A: How to transform space integrals (fixed in XYZ) moving at β c = −βc observed in between Lorentz inertial frames? X0Y 0Z0. With γ0 = γ taken into account we have

dS04 = d3x0 = γd3x = γdS4 (A4) In special relativity, the hyperplane differential- element (HDE) four-vector is often used to transform — the change of variable formula for Technique-II. In triple (space) integrals from one Lorentz inertial frame above Eqs. (A2) and (A3), dSµ = (0, 1)d3x means that to another [4, p. 757]. In this Appendix, we will show d3x is fixed in XYZ, while dS0µ = (β0, 1)d3x0 means that that the HDE four-vector itself is not consistent with the d3x0 = γd3x moves at β0c = −βc in X0Y 0Z0. Note that principles of classical mathematical analysis and the prin- d3x0/γ (= d3x) is an invariant in the Technique-II. ciple of relativity although it is Lorentz covariant (confer In special relativity developed by Einstein [20], a mov- Appendix E for the definition of covariance). ing object Lorentz-contracts in the direction of motion, Under time-space Lorentz transformation, there are while keeping in the same the dimensions of the object in two techniques used to transform space (triple) integrals all the transverse directions. Now d3x is fixed in XYZ, between inertial frames. The first technique is based on and it is moving observed in X0Y 0Z0. Thus according the change of variables theorem, as presented in mathe- to Einstein, d3x0, denoting the size of d3x observed in matical analysis [14], and used in Laue’s original paper X0Y 0Z0, is given by to derive Laue’s theorem [12], and also used to develop my theory in the present paper and the previous work [6]; d3x d3x0 = , (A5) called Technique-I for convenience. In this technique, the γ differential element transformation formula or the change of variable formula from the laboratory frame XYZ to which is exactly the same as Eq. (A1) derived from the the frame X0Y 0Z0 is given by change of variables theorem [14]. That is because the change of variables theorem is the strict mathematical

3 ∂(x, y, z) 3 0 3 0 basis of the effect of Lorentz contraction under time-space d x = d x = γd x , (A1) ∂(x0, y0, z0) Lorentz transformation, as shown in the Analytical exam- ple in Sec. II. where X0Y 0Z0 moves at βc with respect to XYZ, while However according to Eq. (A4), Technique-II requires XYZ moves at β0c = −βc with respect to X0Y 0Z0; d3x is d3x0 = γd3x; thus Technique-II contradicts both the the proper differential element fixed in XYZ, while d3x0 change of variables theorem in mathematical analysis and is the corresponding element moving at β0c = −βc in and the effect of Lorentz contraction in Einstein’s special X0Y 0Z0; Jacobian determinant ∂(x, y, z)/∂(x0, y0, z0) = γ relativity. More seriously, Technique-II directly contra- with γ = (1 − β2)−1/2 = (1 − β02)−1/2 = γ0 is calcu- dicts Lorentz invariance of total charge; in other words, lated from Eq. (3), as illustrated in the Analytical exam- if Technique-II were used, then a non-zero total charge ple in Sec. II. Note that γd3x0 (=moving element d3x0 would not be a Lorentz invariant, which is shown below. multiplied by its relativistic factor γ0 = γ) is a Lorentz Total charge Q = invariant in Technique-I. First we invariant in the Technique-I. show that Lorentz invariance of total charge is always The second technique is based on the HDE (hyper- valid in Technique-I. Without loss of generality, we as- plane differential-element) four-vector, as presented in sume that a charge distribution is created by charged some respected textbooks, such as the book by Jack- particles which move at the same velocity, otherwise it son [4, p. 757]; called Technique-II for convenience. In can be treated discretely, as shown in Ref. [6]. Accord- Technique-II, the integral domain V is assumed to be at ing to special relativity, there must exist an inertial frame rest in the laboratory frame, and then similar to the clas- XYZ where the charged particles are at rest. Thus in the sical particle’s four-momentum = proper mass multiplied charge-rest frame XYZ (taken as the laboratory frame), R 3 by its four-velocity, a differential-element four-vector (= the total charge can be formulated as Q = V ρ(x)d x, 14 where ρ(x) with ∂ρ/∂t ≡ 0 is the charge distribution, V charge Q is always a Lorentz invariant in Technique-I; is the volume at rest, and Q is the (time-independent) both cases have nothing to do with the boundary condi- total charge in V . In such a case, all charged particles tions of J µ = (J, cρ). are stationary and frozen in V so that no current exists Conclusion. From above analysis we can conclude that (J = 0). Observed in a frame X0Y 0Z0 moving at βc 6= 0 Technique-II is based on the HDE four-vector, as pre- with respect to the charge-rest frame XYZ, the volume sented in [4, p. 757], and it has three flaws: (i) contra- V 0 is moving at β0c = −βc, but there are no charged par- dicting the change of variables theorem in mathemati- ticles crossing through the boundary of V 0 although the cal analysis [14]; (ii) contradicting the effect of Lorentz current J0 = −γβcρ 6= 0 holds. In Technique-I, as shown contraction in special relativity; and (iii) contradicting in Eq. (A5), the change of variable formula is given by the Lorentz invariance of total charge. To put it simply, d3x0 = d3x/γ. From this, with cρ0 = γ(cρ − β · J) and Technique-II follows neither the principles of mathemat- J = 0 taken into account, we have ical analysis nor the principle of relativity. Thus the Z Lorentz covariant hyperplane differential-element four- cQ0 : = (cρ0)d3x0 vector Eq. (A3) is only a mathematical four-vector, in- V 0 stead of a physical four-vector [24]. Z = γ(cρ − β · J)(d3x/γ) Two subtle, but apparently fundamental issues. In V analysis of the Lorentz invariance of total charge in a Z 3 volume, a subtle and important issue is about how to = (cρ)d x =: cQ define the volume. If there are charged particles crossing V through the boundary of the volume, the total charge in = invariant. (A6) the volume may not be Lorentz invariant [15], possibly Thus we finish the proof that the total charge is always leading to a doubt of the completeness of Maxwell EM a Lorentz invariant in Technique-I. theory [16]. Thus the correct volume is supposed to be Total charge Q = non-invariant in Technique-II. Now the one that moves at the same velocity as that of the we show why Technique-II contradicts the Lorentz in- charge, as argued above. Another subtle issue is how to variance of total charge. In the laboratory frame XYZ correctly understand the definition of total charge. For (charge-rest frame), the four-current density is given by example, by analysis of an infinite straight charged wire, µ 4 4 R 3 J = (J,J ) with J = 0 and J = cρ, and the total Bili`cpuzzled that the standard definition Q = V ρd x R 3 R 4 and the so-called covariant definition Q = R J µdS (in charge Q is defined by cQ = V (cρ)d x = V J dS4. Ob- µ served in the frame X0Y 0Z0 moving at βc with respect units with c = 1) are not equivalent in general [15]; now to the laboratory frame XYZ, we have J 0µ = (J0,J 04) we know that the problem turned out to be here: the 0 4 04 0 4 R µ R 0µ 0 with J = −γβJ and J = cρ = γJ , and the total transform of triple integral J dSµ = J dSµ contra- 0 0 R 0 3 0 R 04 0 dicts the change of variables theorem in mathematical charge Q is defined by cQ = V 0 (cρ )d x = V 0 J dS4. According to Eq. (A2) and Eq. (A3) of Technique-II, we analysis [14], as shown in Eqs. (A1) and (A4). 3 have dSµ = (−dS, dS4), with dS = 0 and dS4 = d x; and 0 0 0 0 dSµ = (−dS , dS4) = (γβ, γ)dS4, with dS = −γβdS4 0 3 0 and dS4 = γdS4 = d x . From this we have Appendix B: Natural derivation of four-momentum of quasi-photon and invariance of Planck constant Z Z Z 4 µ 0µ 0 J dS4 = J dSµ = J dSµ V V V 0 Quasi-photons are the carriers of the momentum and Z Z 04 0 0 0 energy of the light in a medium, and they are descriptions = J dS4 − J · dS V 0 V 0 of the macroscopic average of light-matter microscopic in- Z Z teractions [17]. Lorentz invariance of Planck constant is 04 0 4 = J dS4 − (−γβJ ) · (−γβdS4) an implicit postulate in Dirac relativistic quantum the- V 0 V Z Z ory [7]. In this appendix, we will apply Theorem 1 to 04 0 2 4 = J dS4 − (γβ) J dS4 Minkowski EM tensor for a plane light wave in a moving V 0 V uniform medium, and find that the momentum–energy four-vector (four-momentum) of the quasi-photon and Z Z  2 4 04 0 the Lorentz invariance of Planck constant can be nat- ⇒ 1 + (γβ) J dS4 = J dS4 V V 0 urally derived. ⇒ γ2cQ = cQ0 ⇒ Q 6= Q0 (A7) For a plane light wave propagating in a moving non- dispersive, isotropic, uniform medium, observed in the medium-rest frame, Minkowski quasi-photons character- if Q 6= 0 holds, where β 6= 0 is assumed ⇒ γ > 1. Thus izing the light-matter interactions move along the wave we finish the proof that a non-zero total charge (Q 6= 0) vector kw at the speed c/n, with n (> 1) the refractive in- is not a Lorentz invariant in Technique-II. dex. In such a case, there is a photon-rest frame moving So far we have shown that the total charge Q 6= 0 is at the quasi-photon velocity relatively to the medium- never a Lorentz invariant in Technique-II, while the total rest frame [7, 17]. Observed in the photon-rest frame 15

XYZ (taken as the laboratory frame here), there are energy transport. However in fact, it is not the real some fantastic EM phenomena to take place. energy velocity in general. The real energy velocity is the phase velocity, required by both Fermat’s principle 1. The EM fields E = 0, H = 0, D = D0 cos Ψ, and and energy conservation law which are physical postu- B = B0 cos Ψ hold, where D0 6= 0 and B0 6= 0 lates independent of Maxwell equations [23]. That is with D0 ⊥ B0 are the constant amplitudes, leading 0 0 0 0 why β c = (E × H )/Wem is called “photon apparent to EM energy density Wem = 0.5(D·E+B·H) = 0 2 velocity” in Ref. [7]. and Abraham momentum gA = E × H/c = 0, but According to the Lorentz transformation of the wave the Minkowski momentum gM = D × B 6= 0. µ four-vector K = (kw, ω/c), with ω = 0 in XYZ taken 2. The wave angular frequency ω = 0 and the wave into account, we have phase Ψ = (ωt − kw · x) = (−kw · x) hold, with the 0 0 0 0 0 0 ω = γ(ω − kw · β c) ⇒ ω = kw · β c. (B8) wave vector kw 6= 0, leading to all the EM fields behaving as static fields [17]. As mentioned above, both Fermat’s principle and en- ergy conservation law require the EM energy to propa- The Minkowski EM stress–energy tensor T µν is given e gate at the phase velocity [7], given by by Eq. (43), with Tˇ M = −DE − BH + ˇIWem = 0 and cg = 0 holding ⇒ the holding of R Tˇ d3x = 0 0 A V : t=const M 0 ω kw 0 0 0 R 3 β c = ⇒ ω = k · β c. (B9) 0 and cgAd x = 0 ⇒ the holding of ph 0 0 w ph V : t=const |kw| |kw| R µj 3 V : t=const Te d x = 0 for µ = 1, 2, 3, 4 and j = 1, 2, 3. According to the sufficient and necessary con- From Eq. (B8) and Eq. (B9), we have µν µν dition Eq. (10) of Theorem 1, Te -time-column (= T - k0 · β0 = k0 · β0 . (B10) time-row) space integrals w w ph Z   µ Wem(Ψ) 3 0 0 P = c gM(Ψ), d x (B1) As indicated, the volume V in Eq. (B2) moves at β c. c 0 0 V : t=const If the same volume V moves at βphc, we label it as V 0 , called “light volume” [7]. Observed in the photon- constitute a Lorentz four-vector, which is time- light rest frame XYZ, the volume V is at rest, and the light independent because of all the EM fields behaving as volume Vlight is also at rest due to ω = 0 ⇒ β c = 0; being static in the photon-rest frame ⇒ ∂Teµν /∂t ≡ 0, ph
namely Vlight = V . including ∂ gM,Wem/c /∂t ≡ 0. 0 0 0 0 0 0 0 Observed in X Y Z , from Eq. (B10) we find that Ψ = Observed in the inertial frame X Y Z moving rela- 0 0 0 0 0 0 0 0 (ω t − kw · x ) takes the same value for x = x0 + β ct tively to the photon-rest frame XYZ at a velocity of βc, 0 0 0 0 0 0 0 0 with x ∈ V and for x = x0 + βphct with x ∈ Vlight. we have 0 0 Thus with V in Eq. (B2) replaced by Vlight, we obtain an Z  0 0  0µ 0 0 Wem(Ψ ) 3 0 equivalent expression, given by P = c gM(Ψ ), d x , (B2) V 0: t0=const c Z  0 0  where the phase is given by Ψ0 = (ω0t0 − k0 · x0), and V 0 0µ 0 0 Wem(Ψ ) 3 0 w P = c gM(Ψ ), d x . 0 0 0 c is moving at β c = −βc. Vlight: t =const From the Lorentz transformation of EM field-strength (B11) tensors [7, see Eqs. (3) and (4) there], with E = 0 and H = 0 taken into account we have The above Eq. (B11) is a time-independent and Lorentz

0 covariant four-vector. E = γβ × Bc, (B3) So far, with Theorem 1 applied to Minkowski EM ten- H0 = −γβ × Dc, (B4) sor we have arrived at a conclusion that for a plane light D0 = γD − ξ(β · D)β, (B5) wave in a medium, the total Minkowski EM momen- tum and energy contained in a light volume constitute B0 = γB − ξ(β · B)β, (B6) a Lorentz four-vector. where γ = (1 − β2)−1/2 and ξ = (γ − 1)/β2. From above The above result implies that observed in any inertial Eqs. (B3)-(B6), we obtain frame, all photons in a light volume are moving at the same velocity as that of the light volume and no pho- E0 × H0 tons cross the boundary of the volume; in other words, = −βc = β0c, (B7) W 0 observed at the same time in all inertial frames, respec- em tively, the photons in the light volume are the same pho- 0 called the energy velocity traditionally, where Wem = tons. This Lorentz property of light volume was first 0.5(D0 · E0 + B0 · H0) = D0 · E0 = B0 · H0 is the EM formulated in Ref. [7, see Eq. (50) there], directly based 0 0 0 0 energy density. Because (E × H )/Wem = β c given on the Lorentz covariance of EM field-strength tensors by above Eq. (B7) is the moving velocity of the vol- (instead of Minkowski EM tensor). This makes sense, be- ume V 0, seemingly it indeed looks like the velocity of cause Minkowski tensor is a covariant combination of the 16

EM field-strength tensors [17, footnote 7], and in princi- The answer is no, because this plane light wave is a ple, all physical results obtained from Minkowski tensor non-trivial solution of Maxwell equations (non-zero field are already embodied in the EM field-strength tensors solution), and the pre-assumption of corrected Møller’s [24]. theorem cannot be satisfied, namely the divergence-less µν νµ ˇ Especially, if there is only one photon contained in the ∂ν Te = ∂ν T = ∇ · TM + ∂gM/∂t, ∇ · (cgA) + light volume V 0 and Einstein light-quantum hypothe-
light ∂Wem/∂(ct) = 0 is fulfilled [6], but the zero-boundary sis is taken into account, namely condition (Teµν = 0 on the boundary of V ) cannot be Z 14 24 34 0 3 0 0 fulfilled; for example, (Te , Te , Te ) = cgM = cD × B = Wemd x = ~ω , (B12) 2 0 0 cD × B cos (−k · x) = 0 cannot hold on the whole Vlight: t =const 0 0 w closed boundary surface of a finite 3D domain V 6= 0 then because cos2(−k · x) = 0 only appears on the discrete Z Z w 0 3 0 0 0 3 0 planes with kw · x = (2l + 1)π/2, where l is an arbitrary gMd x = (D × B )d x 0 0 0 0 integer. Vlight: t =const Vlight: t =const Z 0 0 0 3 0 Question 3: Is Theorem 1 applicable for identifying = (Wem/ω )kwd x the Lorentz property of light momentum and energy for V 0 : t0=const light a plane light wave in free space? The answer is no. In free space, for a non-trivial plane wave observed in 0 = ~kw (B13) any given inertial frame XYZ, the EM fields can be writ- ten as (E, H, D, B) = (E , H ,  E , µ H ) cos Ψ, where is the momentum of the photon in V 0 , where the 0 0 0 0 0 0 light E0 6= 0 and H0 6= 0 with E0 ⊥ H0 are the constant Lorentz invariant expression D0 × B0 = (W 0 /ω0)k0 em w amplitudes, Ψ = (ωt − kw · x) is the wave phase, kw is used [7, see Eq. (37) there], and is the reduced ~ with |kw| = ω/c 6= 0 is the wave vector, and 0 and Planck constant. Inserting Eq. (B12) and Eq. (B13) into µ0 the vacuum permittivity and permeability constants. Eq. (B11), we find that Abraham and Minkowski momentums are equal, namely 0 0 0µ 0µ g = g = E × H/c2 (6= 0). (~kw, ~ω /c) = ~K (= P /c) (B14) A M According to Theorem 1, the sufficient and necessary 0µ 0 0 is the photon’s four-momentum, where K = (kw, ω /c) condition for the plane wave can be written as is a known (wave) four-vector. Since ~K0µ and K0µ 0µ 0ν Z are both four-vectors, ~(gµν K X ) = scalar and µj 3 0µ 0ν Te (x, t)d x = 0 (gµν K X ) = scalar must hold, where gµν is the V : t=const Minkowski metric and X0ν is the time-space four-vector. 0µ 0ν for µ = 1, 2, 3, 4 and j = 1, 2, 3, (B15) Note that (gµν K X ) is the scalar of phase, and it can be any real number. Thus the Planck constant must be ~ where the volume V is at rest in XYZ, and Teµν is given a Lorentz invariant. This conclusion was first obtained by Eq. (43). in Ref. [7] directly from two EM field-strength tensors. From Eq. (43) we have Conclusion. From above it is seen that the four- Z momentum of quasi-photon in a medium and the in- 41 42 43 3 variance of Planck constant are naturally obtained by (Te , Te , Te ) d x V : t=const applying Theorem 1 to Minkowski tensor, with taken Z 3 into account Einstein light-quantum hypothesis that is = cgAd x 6= 0 (B16) the basis of Bohr frequency condition of atomic transi- V : t=const tions in quantum theory. On the other hand, as shown in Einstein-box thought experiment [25], momentum– for a finite volume V (6= 0). energy conservation law requires the quasi-photon to From above Eq. (B16) we find that the sufficient and have a four-momentum. Thus the Minkowski tensor is necessary condition Eq. (B15) cannot hold. Thus we con- compatible with the quantum theory of atomic light radi- clude that for a plane light wave in free space, the to- ation and the momentum–energy conservation law, and it tal (Minkowsi = Abraham) momentum and energy con- is the correct momentum–energy tensor for descriptions tained in a given volume V that is at rest in an iner- of light–matter interactions. tial frame cannot constitute a Lorentz four-vector. How- ever, it does not mean that the momentum and energy Question 1: Is the generalized von Laue’s theorem of light cannot constitute a four-vector, because the pho- [6] applicable for identifying the Lorentz property of tons in free space cannot be stopped in a given volume light momentum and energy for a plane light wave in a V that is resting in an inertial frame in terms of Ein- medium? The answer is yes, because its pre-assumption µν stein’s hypothesis of constancy of light speed. From this ∂Te /∂t ≡ 0 is satisfied, as shown above. it follows that always there are photons crossing through Question 2: Is the corrected Møller’s theorem applica- the boundary of V , and flowing into and out of V , and ble for identifying the Lorentz property of light momen- thus the photons in V are not the same photons ob- tum and energy for a plane light wave in a medium? served for different time. (Note that the photon density 17

2 2 Np = Wem/(~ω) = [0E0 /(~ω)] cos Ψ is a wave, depen- Appendix C: Physical counterexamples dent on time and space locations.) On the other hand, of Thirring’s claims because of the relativity of simultaneity, photons may cross through the boundary of V at the same time in The most effective and convincing way to disprove a one frame, but these photons cannot cross through the claim is to give its counterexample. In this appendix, we boundary of V at the same time in other frames; thus will provide two physical counterexamples of the claims leading to a result that the photons in V are not the made by Thirring. same photons observed in different frames. That is why In his book [19], with the help of exterior calculus the total momentum and energy of the photons contained Thirring claims: in V cannot constitute a four-vector in such a case. µ R 4 3 Therefore, Theorem 1 only can be used to identify the (i) ∂µΛ = 0 makes Λ d x be a Lorentz scalar Lorentz property of the total momentum and energy of (namely “invariant conservation Law” claimed by Wein- materials or particles, which are moving at a velocity less berg [2, p. 40]); than the vacuum light speed c so that there is a material- µν R µ4 3 (ii) ∂ν T = 0 makes T d x be a Lorentz four- rest or particle-rest inertial frame, such as in the case for vector (namely, Landau-Lifshitz version of Laue’s theo- a plane light wave in a dielectric medium shown above, rem [6]). where Minkowski quasi-photon propagates at a velocity of c/n < c [17]. (Note: If a momentum–energy tensor In both claims (i) and (ii), no boundary conditions are is contributed by materials or particles which move at required. However, claim (i) can be directly disproved different velocities individually, then the tensor should by a simple counterexample Λµ = Kµ, and claim (ii) can be discretized so that each of the discretized tensors is be directly disproved by a simple counterexample T µν = µ ν µ contributed by the materials or particles which move at K K , where K = (kw, ω/c) is the wave four-vector for the same velocity, just like in the proof of the Lorentz a plane wave in free space, first shown by Einstein [20], invariance of total charge given in Ref. [6]). with kw the wave vector, ω (6= 0) the angular frequency, It should be noted that in free space, the invariance of and |kw| = ω/c holding. This is illustrated as follows. Planck constant ~ and the covariance of wave four-vector µ µ ν µ Apparently, ∂µK = 0 and ∂ν (K K ) = 0 are both K have been already proved in [7] and [20], respectively. valid because Kµ is independent of space and time vari- Thus the four-momentum of photon in free space, equal 0 0 0 µ ables (x, t). Suppose that X Y Z frame moves with re- to ~K , is a solved problem theoretically. spect to the laboratory frame XYZ at a velocity of βc Further specific explanation: Why is along the wave vector kw. Observed in XYZ, we have K4 = ω/c and Z Z E × H cg d3x = d3x A c Z ω Z V : t=const V : t=const K4d3x = d3x, (C1) V c V 6= 0 where the integral domain V is fixed in XYZ. Observed always valid for a finite V 6= 0 for a plane wave in free in X0Y 0Z0, we have K04 = ω0/c and space? For a non-trivial plane wave in free space, observed in Z Z ω0 K04d3x0 = d3x0, (C2) any inertial frames, the power flow or Poynting vector V 0 V 0 c

2 E × H = E0 × H0 cos Ψ 6≡ 0 where the integral domain V 0 is moving at β0c = −βc. ⇒ E0 × H0 6= 0 From the Lorentz transformation of Kµ, we obtain the Doppler frequency shift [20], given by holds; otherwise, there are no energy flowing and no wave. 0 R 2 3 ω ω  On the other hand, we have V : t=const cos Ψd x > = γ − β · kw 0 holding; thus leading to the holding of c c R 3 cgAd x 6= 0 for a plane wave in free V : t=const ω space. = γ(1 − |β|) , (C3) R 2 3 c Note: V : t=const cos Ψd x > 0 comes from the fact that V 6= 0 is a finite 3D volume, and cos2 Ψ ≥ 0 2 holds with the zero points only appearing on discrete where γ = (1 − β )−1/2 is the time dilation factor, and planes, and thus there must exist a smaller volume β k kw ⇒ β · kw = |β| × |kw| = |β|ω/c is employed. V ∗ ⊂ V , where cos2 Ψ > 0 exactly holds ⇒ the hold- As shown in Eq. (A5) of Appendix A, the change of R 2 3 R 2 3 3 0 3 ing of V : t=const cos Ψd x ≥ V ∗: t=const cos Ψd x > 0. variable formula in such a case is given by d x = d x/γ. Inserting Eq. (C3) into Eq. (C2), with d3x0 = d3x/γ and 18

Eq. (C1) taken into account, we have condition Eq. (23) of Theorem 3 satisfied, ⇒ R J 4d3x = scalar, as shown in Eq. (A6) of Appendix A. However Z Z 0 µ 04 3 0 ω 3 0 for K = (kw, ω/c), there is no such a frame where K d x = d x R 3 R 4 3 V 0 V 0 c kw = 0 ⇒ kwd x = 0 holds; thus K d x is not a ω Z d3x scalar. Why is there no such a frame for kw = 0? As = γ(1 − |β|) c γ we know, the photon momentum–energy four-vector is V given by Kµ = ( k , ω/c), with the Planck constant. ω Z ~ ~ w ~ ~ = (1 − |β|) d3x According to Einstein’s hypothesis of constancy of light c V speed, there is no photon-rest frame in free space, and Z the photon momentum k 6= 0 ⇒ k 6= 0 holds in any = (1 − |β|) K4d3x. (C4) ~ w w V frames. Note that in above Eq. (C4), the integrand K04 satisfies the definition Eq. (24) in the proof of Theorem 3. Appendix D: Why is the theoretical framework for From above Eq. (C4) we have the positive mass theorem flawed? Z Z K04d3x0 6= K4d3x for βc 6= 0. (C5) In general relativity, the metric gµν are solutions of V 0 V Einstein’s field equations for a given energy–momentum tensor T µν that causes space to curve [21, p. 5]. Accord- R 4 3 Thus we conclude that V K d x is not a Lorentz scalar, ing to Arnowitt, Deser and Misner [37], the definition of and Thirring’s claim (i) is disproved. the total energy–momentum P µ is given by the volume On the other hand, we have integral of the components of T 0µ, namely Z Z Z Z T µ4d3x = KµK4d3x = Kµ K4d3x. (C6) P µ := T 0µd3x, (D1) V V V

µ R 4 3 which can be expressed as surface integrals through Ein- In above Eq. (C6), K is a four-vector, but V K d x is not a Lorentz scalar; thus their product stein’s field equations and Gauss’s theorem, and where P 0 = E is the total energy [38]. Z  Z In the proofs of the positive mass theorem [33, 35, 39], µ 4 3 µ4 3 (K ) × K d x = T d x (C7) the total energy follows the definition in [37, 38] [21, V V p. 462], given by must not be a four-vector, and Thirring’s claim (ii) is 1 Z disproved as well. E = (g − g )d2Sj, (D2) 16π jk,k kk,j Proof by contradiction that Eq. (C7) is not a four- vector. If Eq. (C7) were a four-vector, then where the integral is evaluated over a closed surface in the asymptotically flat region surrounding the source of Z  gravitation. µ ν 4 3 (gµν K X ) K d x = scalar (C8) Arnowitt, Deser and Misner claim that because of the V µν µ conservation law T ,µ = 0, “P should transform as would hold, where Xν = (x, ct) is the time-space four- a four-vector” [38], which is clearly endorsed by Nester vector, and [34]. In the textbook by Misner, Thorne and Wheeler [21, p. 462], it is also emphasized that the conservation µ ν µν µν µ R 0µ 3 µ (gµν K X ) = (ωt − kw · x) = scalar (C9) law T ,µ = 0 (T ,ν = 0) makes P = T d x (P = R T µ0d3x) be a four-vector. is the scalar of phase, with gµν = diag(−1, −1, −1, +1) Unfortunately, as shown by the counterexample in Ap- the Minkowski metric. Thus from Eq. (C8) and pendix C, T µν = 0 cannot guarantee that P µ is a R 4 3
,µ Eq. (C9) it follows that V K d x must be a four-vector. Thus the Arnowitt-Deser-Misner theoreti- R 4 3
scalar, but V K d x is not a scalar according to cal framework used for the proofs of the positive mass Eq. (C5). This contradiction indicates that Eq. (C7), theorem is flawed. R µ4 3 µ R 4 3
V T d x=K V K d x , cannot be a Lorentz four- Nevertheless, one might argue that in the framework µν vector. of the positive mass theorem, in addition to T ,µ = 0 An interesting question: Why is R K4d3x not a scalar there is another important requirement, called dominant µ R 4 3 energy condition [39], reading: for the wave four-vector K = (kw, ω/c) while J d x is µ a scalar for the current density four-vector J = (J, cρ)? T 00 ≥ |T µν | (D3) That is because the moving velocity of any charged par- µν 00 µν ticle is less than the vacuum light speed c, and there for each µ, ν [30, 35]; and both T ,µ = 0 and T ≥ |T | is a particle-rest frame where the particle current J = together make P µ = R T 0µd3x be a four-vector (in above 0 ⇒ R Jd3x = 0 holds, with the sufficient and necessary Arnowitt-Deser-Misner practice µ, ν = 0, 1, 2, 3 used). 19

However this is not true, because the counterexample Lagrange formalism, must be confirmed by the Maxwell equations [4, p. 598]. Thus any justified results that can- µν µ ν T = K K (D4) not be derived from Maxwell equations could constitute a challenge of the completeness of classical electromagnetic in Appendix C itself satisfies the dominant energy con- theory. dition It is well-established that Maxwell equations are co- T 44 ≥ |T µν |, (D5) variant under Lorentz transformation of two EM field- strength tensors [24]. This covariance has three proper- namely ties:

K4K4 ≥ |KµKν | (D6) (a) Maxwell equations keep invariant in mathematical form in all Lorentz inertial frames. or (b) All physical quantities appearing in the equations (ω/c)2 ≥ |KµKν |, (D7) have the same physical definitions in all frames. (c) Maxwell equations in one frame do not include any µ µ where K = (kw, ω/c) with |kw| = ω/c ≥ |K | (here physical quantities defined in other inertial frames. Rindler’s practice µ, ν = 1, 2, 3, 4 [22, p. 138] used). µν The above three properties do not contain each other. To That is to say, although the conservation law T ,ν = 0 µν µ ν put it simply, Maxwell equations are frame-independent or ∂ν T = ∂ν (K K ) = 0 and the dominant energy condition T 44 ≥ |T µν | or K4K4 ≥ |KµKν | are both sat- under Lorentz transformation of two EM field-strength isfied, tensors. Another typical example for the covariance is the Z four-momentum of a massive particle, given by P µ : = T µ4d3x Z µ 4 3 µ = K K d x m0U (Non-covariant form) Z  = K4d3x Kµ (D8) m = U µ (Covariant form in relativity) γ is not a four-vector, because R K4d3x
is not a scalar,  mc2  = mv, (Covariant form in 3D space) (E1) as shown in Eq. (C5) of Appendix C. c From above we can see that all the proofs of the posi- tive mass theorem [32–36, 39] are based on a flawed the- in a general XYZ frame, where m0 is the particle’s mass oretical framework, and thus the validity of the theorem in the particle-rest frame — rest (proper) mass, U µ = itself could be called into question. γ(v, c) is the four-velocity, m (= γm0) is the mass, γ = (1 − v2/c2)−1/2, and v is the particle’s velocity. µ The definition of m0U is in a non-covariant form. Appendix E: Is Gordon optical metric covariant? µ m0U has the same mathematical form and physical def- inition in all inertial frames, and complies with Proper- How to define the covariance of physical quanti- ties (a) and (b), but it is not consistent with Property ties/equations is one of the most important issues in op- µ (c), because m0U includes m0 defined in the particle- tics of moving media. Actually, it is a long-lasting un- µ rest frame, and thus m0U cannot provide the frame- solved fundamental problem in the theory of relativity. In independent definitions of momentum and mass in a gen- this appendix, we take Maxwell equations and the four- µ eral frame. For example, m0U defines m0 as the parti- momentum of a massive particle as examples to give the µ cle’s mass in the particle-rest frame, but m0U does not definition of covariance, and indicate why the covariance define the particles’s mass m in a general frame. of Gordon metric tensor [42] is indeterminate. Properties (b) and (c) require that the definitions of In principle, the definitions of physical quantities used particle’s momentum and mass must be the same in all to construct a covariant theory of light propagation in inertial frames and they do not include any quantities moving media are supposed to be given in a general in- defined in other frames, namely ertial frame, otherwise the covariance of the constructed theory is ambiguous. At least, one cannot identify the Momentum = Mass × Velocity = mv. (E2) covariance of the quantity without a definition given in a general frame. Thus we have the mass transformation m = γm0. As a first principle, Maxwell equations are the physi- Obviously, the definition of four-momentum cal basis for descriptions of macro electromagnetic phe- (m/γ)U µ = (mv, mc2/c) is frame-independent un- nomena, and all other alternative approaches, including der its Lorentz transformation. Especially, like Maxwell 20 equations, it does not include any quantities of other rest frame). Obviously, this is not consistent with Prop- frames; thus this definition complies with Property (c), erty (c). In their works [43, 44], Leonhardt and Piwnicki in addition to Properties (a) and (b). are supposed to provide the definition of the refractive Note that (m/γ)U µ is a covariant form in relativity, index in a general frame in order to remove this in- with (m/γ) an invariant and U µ the four-velocity of the consistency and make it expressed in a covariant form, particle in a general XYZ frame, while (mv, mc2/c) is just like Eq. (E2) leading to the frame-independent four- a covariant form in 3D-space, with mv the momentum momentum (m/γ)U µ = (mv, mc2/c) shown above; un- and mc2 the energy of the particle in a general XYZ fortunately, the authors failed to do so. Thus the covari- frame; just like Maxwell equations have the two forms of ance of Γ µν cannot be identified, namely the covariance covariance (confer [17, Footnote 7 there]). of Gordon optical metric Γ µν is ambiguous, depending Unfortunately, Property (c) has been usually neglected on how to define the refractive index in a general inertial in constructing a tensor in the community. A typical ex- frame. ample is Gordon optical metric tensor [42], which is gen- It should be indicated that Gordon defined the eralized to describe the effective gravitational field for a (proper) refractive index n0 by assuming that the nonuniformly moving medium in the works by Leonhardt medium is isotropic, observed in the medium-rest frame and Piwnicki [43, 44]. The Gordon metric in a general [42]. However when a medium is moving, it becomes XYZ frame reads: anisotropic in general [45], and Gordon’s way to define n0 is not valid to define the refractive index in a general µν µν 2 µ ν Γ = g + (n0 − 1)u u , (E3) inertial frame. It also should be indicated that the way to define the µν where g = diag(−1, −1, −1, +1) is the Minkowski met- refractive index in a general frame could essentially re- µ µ ric, u = U /c = γ(β, 1) with β = v/c is the normalized vise the descriptions of light propagation. For example, four-velocity of the moving medium (namely the medium in Wang’s theory where the refractive index in a uniform moves at v = βc with respect to XYZ), and n0 is the medium is defined in a general inertial frame, propaga- refractive index in the medium-rest frame. tion of light complies with Fermat’s principle in all iner- µν µν 2 µ ν From Eq. (E3), we find that Γ = g + (n0 − 1)u u tial frames [7], while in Leonhardt’s theory, it does only is given in a general frame XYZ, but it includes n0 “in the special case of a medium at rest” [46]. which is a quantity defined in another frame (medium-

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Ann. Phys. 322, 891 (1905), English translation: On the the local conservation of energy and momentum”. electrodynamics of moving bodies. [31] G. Arfken, Mathematical Methods for Physicists (Or- [21] C. W. Misner, K. S. Thorne, and A. Wheeler, Gravitation lando, Academic Press, 1985), 3rd edition. (W. H. Freeman and Company, San Francisco, 1973). [32] R. Schoen and S. T. Yau, On the proof of the [22] W. Rindler, Relativity: Special, General, and Cosmo- positive mass conjecture in general relativity, Com- logical (Oxford, NY, 2006), 2nd edition. In his book mun. Math. Phys. 65, 45 (1979). Rindler argues that in special relativity, four-current den- [33] E. Witten, A new proof of the positive energy theorem, sity J µ = (J, cρ) satisfies charge conservation equation Commun. Math. Phys. 80, 381 (1981). µ µ J ,µ(= ∂µJ ) = 0 (on p. 142 of the book), and charge [34] J. M. Nester, A new gravitational energy expression with Q = R (J 4/c)d3x
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Material to help reading

Dear Readers, In my paper, I think that I have generally resolved two fundamental issues in the dynamics of relativity: (a) Under what condition, the time-column space integrals of a Lorentz four-tensor constitute a Lorentz four-vector (Theorem 1), and (b) under what condition, the time-element space integral of a Lorentz four-vector is a Lorentz scalar (Theorem 3). Both Theorem 1 and Theorem 3 have their own sufficient and necessary conditions, which have nothing to do with the divergence-less, symmetry, and boundary conditions. This point is especially important. For example, from Theorem 1 we can directly judge that Møller’s theorem is incorrect, because the sufficient condition of Møller’s theorem does not include the sufficient and necessary condition of Theorem 1, as shown in counterexample Eq. (38). A similar argument is applicable to the “invariant conservation law” claimed by Weinberg [2, p.40], as shown in counterexample Eq. (48).

µ Why is the constant total charge Q in V, resulting from ∂ µ J = 0 ⇒ ∇ ⋅ J = − ∂ρ ∂t , not a Lorentz invariant? As shown in Sec. IV, that is because Q in V in frame XYZ and Q′ in V ′ in frame X ′Y′Z′ do not refer to the same total charge in the same volume physically, where X ′Y′Z′ moves at βc ≠ 0 relatively to XYZ . This conclusion directly comes from the ∂ρ ∂ 3 ′ fact that, the volume V in ∫ V ( t)d x is fixed in XYZ , and the volume V in ∂ρ ∂ 3 ′ ′ ′ ∂ ∂ ∂ ∂ ∫ V ' ( ' t')d x' is fixed in X Y Z so that t and ∫V , and t ' and ∫V ' are exchangeable, 3 3 respectively, in order to make both ∫V (∂ρ ∂t)d x = dQ dt and ∫V ' (∂ρ' ∂t')d x' = dQ' dt' hold in general. However according to Einstein’s relativity [20], V fixed in XYZ is moving ′ ′ ′ ′ ∂ρ ∂ 3 ′ ′ ′ ′ observed in X Y Z . Now V in ∫ V ' ( ' t')d x' is at rest in X Y Z , and thus V and V do not denote the same volume, and Q and Q′ do not denote the same total charge. µ Therefore, the current continuity equation ∂ µ J = 0 cannot be taken as the “invariant [charge] conservation law” in relativity [2, p.40].

As a practical application, I have used Theorem 1 to analyze Minkowski EM tensor for a plane light wave in a moving medium with Einstein’s light-quantum hypothesis taken into account, finding that the four-momentum of quasi-photon and the Lorentz invariance of Planck constant can be naturally derived (see Appendix B).

In relativity, charge conservation law refers to that the total charge is a time- and frame- independent constant [2, p. 41], while energy-momentum conservation law refers to that the total energy and momentum constitute a covariant and time-independent four-vector µ µν [2, p. 46]. The problems of ∂ µ Λ = 0 and ∂ν Θ = 0 as being conservation laws in relativity were first discovered in [6,7], and they are explicitly illustrated and generally resolved in the present paper. These problems never got any attention in literature before the publications of [6,7], such as in the proofs of the positive mass theorem by Schoen and Yau [39] and by Witten [33], and in the textbook of Rindler [22].

My manuscript has been greatly and carefully revised according to the referee reports from Physical Review D.

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Referee A did not provide any specific criticisms and comments about novelty and originality, and whether analysis and calculations support claimed conclusions.

Referee B has two main points which constitute thought-provoking challenges over the novelty and originality of my research work, and took me a lot of time to understand.

Point (1). Referee B argues that Rindler’s book has already solved the problems about the traditional conservations laws. Point (2). Referee B argues that in the proofs of the positive mass theorem, “rigorous mathematics has been used and loopholes possibly left in earlier works do not exist.”

By careful analysis, I find that Rindler did not realize the problems of conservation laws, not to speak of solving them, while the positive mass theorem itself is based on the µν incorrect conservation law T ,µ = 0 , which is not just a “loophole” but an error that is not fixable.

I think that I have clarified the above two points in the revised manuscript, and all detailed explanations are attached in Response to Referee B. Thank you for your precious time to review my paper.

Sincerely, Changbiao Wang

Attached: Response to Referee B

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Response to Referee B

Response to Point (1). Referee B argues that Rindler’s book has already solved the problems about the traditional conservations laws. I am sorry, I don’t think so, which is shown below. (i) Invariance of total charge in a closed physical system is the charge conservation law, which means that the total charge is a time-independent and frame-independent constant. It is a fundamental issue in the dynamics of relativity. In his book, Prof. Rindler argues that in special relativity, four-current density J µ = (J,cρ) satisfies µ µ “conservation of charge” equation J ,µ (= ∂ µ J ) = 0 , copied below

W. Rindler, Relativity (Oxford, NY, 2006), p. 142

and the charge q [ = ∫ (J 4 / c)d 3 x ] is “[Lorentz] invariant from frame to frame”, copied below

W. Rindler, Relativity (Oxford, NY, 2006), p. 140

(Note that any charged particle takes up a finite space.) However the “conservation of charge” µ equation ∂ µ J = 0 is not a sufficient condition for q to be “invariant from frame to frame”. Apparently, Prof. Rindler did not realize this problem.

µ In Appendix C of my paper, a counterexample is given to show why ∂ µ J = 0 is not a sufficient condition for q to be “invariant from frame to frame”. The counterexample is the wave four-vector µ K = (k w ,ω c)

for a plane wave in free space, first shown by Einstein, with k w the wave vector, ω (≠ 0) the angular µ µ frequency, and k w = ω c holding. Like the four-current density J = (J,cρ) satisfying ∂ µ J = 0 , the wave four-vector K µ also satisfies µ ∂ µ K = 0 ,

4 3 3 but K d x = (ω c)∫V d x is not a Lorentz invariant. Why? This is physically explained in Appendix ∫V C, copied below.

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See: Appendix C of my paper

µ From above, we can see that Prof. Rindler did not realize the problem of J ,µ = 0 as being “conservation of charge”, not to speak of solving this problem. Therefore, in his book Prof. Rindler did not provide a proper formulation of the charge conservation law in the frame of relativity. (ii) The covariance of total four-momentum in a closed physical system is the energy-momentum conservation law, which means that the total energy and momentum constitute a covariant and time- independent four-vector for a closed physical system. It is also a fundamental issue in the dynamics of relativity.

µν µν In his book, Prof. Rindler argues that T ,ν (= ∂ν T ) = 0 is the energy-momentum four conservation equations for a closed physical system in special relativity, copied below

W. Rindler, Relativity (Oxford, NY, 2006), p. 298

however, there is no definition that is provided for what is the total four-momentum of the system.

Since Prof. Rindler did not provide the definition of total four-momentum for the energy-momentum µν µν tensor T that satisfies the conservation equation ∂ν T = 0 , Prof. Rindler did not provide a proper formulation of the covariance of total four-momentum for a closed system — the energy-momentum conservation law in the frame of relativity.

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Conclusion to Point (1) In his book, µ (i) Prof. Rindler did not realize that the “conservation of charge” equation ∂ µ J = 0 is not a sufficient condition for q = ∫ (J 4 / c)d 3 x to be “invariant from frame to frame”; (ii) Prof. Rindler did not provide the definition of total four-momentum based on the energy- momentum tensor T µν for a closed system. Thus the issues about the invariance of total charge and the covariance of total four-momentum for a closed system are not appropriately formulated. In other words, Prof. Rindler did not solve the problems µ µν of ∂ µ J = 0 and ∂ν T = 0 as being conservation laws. In my revised manuscript, I added Ref. [22] to answer this comment by Referee B.

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Response to Point (2). Referee B argues that in the proofs of the positive mass theorem, “rigorous mathematics has been used and loopholes possibly left in earlier works do not exist.”

I am sorry, I don’t think so, because in the proofs of the positive mass theorem, the traditional µν conservation law ∂ µT = 0 is employed. In other words, the proofs are based on a flawed theoretical framework, which is shown below.

The theoretical framework used for the proofs are shown in Arnowitt-Deser-Misner framework (part 1) Arnowitt-Deser-Misner framework (part 2) Arnowitt-Deser-Misner framework (part 3) Arnowitt-Deser-Misner framework (part 4)

µν The problem of the conservation law ∂ µT = 0 never got any attention in the proofs of the positive mass theorem by Schoen and Yau in 1979 and in 1981, Witten and Nester in 1981, Parker and Taubes in 1982, and Gibbons and coworkers in 1983. All the proofs are based on a flawed theoretical framework set up by Arnowitt, Deser, and Misner, where the total energy-momentum P µ in an asymptotically flat µν spacetime is required to “obey the differential conservation law T ,µ = 0 [of the energy-momentum µν tensor T ]” and required to “transform as a four-vector”; see Arnowitt-Deser-Misner framework (part µν µ 2). However in fact, the conservation law T ,µ = 0 cannot guarantee that P is a four-vector, as shown in Appendix D. Apparently, this problem is ignored in all the proofs.

Conclusion to Point (2) µν The proofs of the positive mass theorem are based on a flawed theoretical framework where ∂ µT = 0 is taken to be the energy-momentum conservation law, and this problem has never got any attention in the community.

To answer this comment by Referee B, I greatly revised Sec. V and added Appendix D.

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Arnowitt-Deser-Misner framework (part 1)

R. Arnowitt, S. Deser, and C. W. Misner, Phys. Rev. 118, 1100 (1960) Arnowitt-Deser-Misner framework (part 1): According to Arnowitt, Deser and Misner, the definition of the total energy-momentum P µ is given by the volume integral of the components of 0µ µ 0µ 3 T , namely P = ∫T d x , which can be expressed as surface integrals through Einstein’s field equations and Gauss’s theorem, and where P0 = E is the total energy.

Arnowitt-Deser-Misner framework (part 2)

R. Arnowitt, S. Deser, and C. W. Misner, Phys. Rev. 122, 997 (1961) Arnowitt-Deser-Misner framework (part 2): According to Arnowitt, Deser and Misner, because of µν µ the conservation law T ,µ = 0 , P should transform as a four-vector, namely it is a four-vector.

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Arnowitt-Deser-Misner framework (part 3)

Misner, Thorne, and Wheeler, Gravitation (W. H. Freeman and Company, SF, 1973), p. 443 µ Arnowitt-Deser-Misner framework (part 3): In this textbook, it is claimed that ∂ µ J = 0 makes 0 3 µν µ µ 0 3 Q = ∫ J d x be an invariant, and ∂ν T = 0 makes P = ∫T d x be a constant four-vector.

Arnowitt-Deser-Misner framework (part 4)

Misner, Thorne, and Wheeler, Gravitation (W. H. Freeman and Company, SF, 1973), p. 462 µ µ Arnowitt-Deser-Misner framework (part 4): Here in this textbook, P = ∫T 0d 3 x is, once again, µν µν clearly claimed as a four-vector because of the source conservation laws ∂ν T = 0 ( T is the source term of Einstein’s field equations). The total energy expression P0 = E is exactly the one used for the proofs of the positive mass theorem by Schoen and Yau, Witten, and Parker and Taubes.

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Schoen-Yau proof

Schoen and Yau, Commun. Math. Phys. 79, 231-260 (1981) Schoen-Yau proof: The total mass definition is taken from the one in Arnowitt-Deser-Misner framework (part 4). This theorem has been proved in their “previous paper in the important case” [Commun. Math. Phys. 65, 45 (1979)]. The two papers takes the same Arnowitt-Deser-Misner framework.

Witten’s proof

Witten, Commun. Math. Phys. 80, 381 (1981) Witten’s proof: The total mass definition is taken from the one in Arnowitt-Deser-Misner framework (part 4). Nester [Phys. Lett. A 83, 241 (1981)], Parker and Taubes [Commun. Math. Phys. 84, 223 (1982)], and Gibbons, Hawking, Horowitz, and Perry [Commun. Math. Phys. 88, 295 (1983)] follow Witten’s work, using Arnowitt-Deser-Misner framework.

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