aa r X i v : . [ phy s i c s . c l a ss - ph ] J a n The phase symmetryof classical electrodynamics
Serhii Samokhvalov ∗ Dniprovsk State Technical University, Ukraine ***
Abstract
The dynamic U (1) symmetry of classical electrodynamics without charges, similar to theknown phase symmetry of quantum mechanics, is analyzed. Using this symmetry, alternativeLagrangians of classical electrodynamics, similar to quantum-mechanical ones, were found,and their symmetry properties were investigated. ∗ e - mail : [email protected] Introduction
The most characteristic feature of quantum mechanics is the use of complex numbers to describestates, which leads to the existence of a compact phase of the wave function. The wave propertyof matter are connected with it. A notable peculiarity of the quantum-mechanical phase usage isthe existence of the phase symmetry (the requirement of the unitarity of the quantum theory).Besides the wave characteristics, the quantization of the particles quantity is the result ofthe phase compactness and their conservation is connected with the phase symmetry.Thus, the two main properties of quantum mechanics: wave-likeness and discreteness thereare due to phase symmetry, that requires the comprehension of the role of phase symmetry init to understand quantum-mechanical laws.We will look for such comprehension by comparing quantum mechanics with classical electro-dynamics, which, unlike quantum mechanics, was formulated from the very beginning as purelyreal, while at the same time having deep analogies with quantum mechanics.Greek indices take on values 0 , , ,
3, Latin indexes of the middle of the alphabet takeon values 1 , , diag ( g µν ) = ( − , + , + , +), for a completely anisymmetric tensor is accepted ε = 1. The wave function meets Schr¨odinger equation: i ~ ∂ t ψ = − ~ m △ ψ + V ψ. (1)If represent ψ = R exp ( iθ/ ~ ) = ρ (cid:18) cos θ ~ + i sin θ ~ (cid:19) , (2)where ρ = R = exp(2 τ ), complex equation (1) splits into two real equations (its real andimagine parts), i.e. its decomplexification takes place: Re : ∂ t θ = − m ( ∇ θ ) − V − Q, (3) Im : ∂ t ρ + ∇ (cid:18) ρ m ∇ θ (cid:19) = 0 , (4)2here Q = − ~ m △ RR = ~ m " (cid:18) ∇ ρρ (cid:19) − ∇ ρρ (5)– quantum-mechanical potential. That is it that makes a distinction between quantum-mechanicaland classical descriptions, and it’s caused by the superposition principle, i.e. by Schr¨odingerequation linearity (1). Equation (3) is Hamilton-Jacobi equation, and equation (4) is equationof particles transfer at velocity ~v = ∇ θ/m , i.e. ~p = ∇ θ .From equation (3) comes directly the equation of motion for a particle, if we take gradientfrom each of its parts ∂ t ~p + ~v ∇ ~p = −∇ U (6)(in Lagrangian coordinates d~p/dt = −∇ U ) and denote U = V + Q . So a particle is affected byquantum-mechanical force −∇ Q in addition to classical force −∇ V .If instead of ρ we use variable τ through which the quantum-mechanical potential is putdown like Q = − ~ m (cid:2) ( ∇ τ ) + ∇ τ (cid:3) , (7)equations (3) and (4) will become: ∂ t θ = − m ( ∇ θ ) − V + ~ m (cid:2) ( ∇ τ ) + ∇ τ (cid:3) , (8) ∂ t τ + 1 m ∇ θ ∇ τ = − m ∇ θ. (9)All the equations and definitions listed above are symmetrical in respect of phase transfor-mations: δθ = ~ δϕ. (10)With direct (algebraic) decomplexification ψ = ε + iχ Schr¨odinger equation (1) splits intotwo equations: Re : ~ ∂ t χ = − ~ m △ ε − V ε, (11)3 m : ~ ∂ t ε = − ~ m △ χ + V χ. (12)Phase symmetry for equations (11) and (12) is put down as: ∂ε = − χδϕ, δχ = εδϕ. (13) Maxwell equations in space without charges are:1 c ∂ t ~H = −∇ × ~E, ∇ · ~H = 0 , (14)1 c ∂ t ~E = ∇ × ~H, ∇ · ~E = 0 . (15)Equations (14) are the first pair and equations (15) are the second pair of Maxwell equations.Equations(14) are consequence of expression ~E and ~H through electromagnetic potentials (Ja-cobi identities).Equations (14) and (15) are symmetric with respect to Rainich transformations [1], whichform group U (1) R : δ ~E = − ~Hδϕ, δ ~H = ~Eδϕ, (16)which are the analogue of phase transformations (13) in quantum mechanics. This symmetry iscompletely dynamic and it doesn’t exist in the presence of charges (though symmetry (13) takeplace also with the presence of the potential V ). Transformations (16) confuses the first andthe second pairs of Maxwell equations, i.e. they confuses Lagrangian equation for Maxwell’sLagrangian and Jacobi identity for F µν .This symmetry lets to put down Maxwell equations in complex form: ic ∂ t ~ξ = ∇ × ~ξ, (17) ∇ · ~ξ = 0 , (18)4here ~ξ = ~E + i ~H . If equation (17) is written in components: ic ∂ t ξ i − ε ijk ∂ j ξ k = 0 , (19)it look a lot like Dirac equation.Let’s also introduce ~η = ~ξ ∗ = ~E − i ~H . Then ~ξ · ~η = E + H , (20) i ( ~ξ × ~η ) = 2( ~E × ~H ) . (21)Vector ~η obeys the equations: ic ∂ t ~η = −∇ × ~η, ∇ · ~η = 0 . (22)Rainich transformations (16) for fields ~ξ and ~η have such form: δ~ξ = i~ξδϕ, δ~η = − i~ηδϕ, (23)or for finite transformations: ~ξ ′ = e iϕ ~ξ, ~η ′ = − e − iϕ ~η. (24)Multiplying (17) by ~η , and (22) by ~ξ , and adding them, we get: ∂ t ( ~ξ · ~η ) + ∇ · h ic ( ~ξ × ~η ) i = 0 . (25)Energy density and Umov-Pointing vector ω = 18 π ( ~ξ · ~η ) = 18 π ( E + H ) , (26) ~S = ic π ( ~ξ × ~η ) = c π ( ~E + ~H ) (27)let equation (25) be written in the form of the equation: ∂ t ω + ∇ · ~S = 0 . (28)5or any substance with density ρ under its transfer velocity we have to take vector ~v thatbelongs to the transfer equation: ∂ t ρ + ∇ · ( ρ~v ) = 0 . (29)Thus, the speed of propagation of electromagnetic energy is determined as follows: ~v = ~S/ω = 2 c ~E × ~HE + H (30)and we have v = c only in the case of electromagnetic wave when ~E ⊥ ~H and E = H . In othercases v = c . For example, for electrostatics ~H = 0 and ~v = 0. This is effect of mass.Let the constant electric field ~ε be superimposed upon an electromagnetic wave ~E , ~H (sucha solution is physical, i.e. it definitely obeys Maxwell equations). Then we get from (30): ~v = ~S/ω = 2 c ~E × ~H + ~ε × ~HE + H + 2 ~ε · ~E + ε . (31)Let’s assume that ε is small and ~x = ~ε/E . From (31) with the accuracy to the first order on x we get: ~v = c [(1 − ~x · ~n E ) ~n S + ~x × ~n H ] , (32)where ~n E and ~n H are unit vectors along vectors ~E , ~H and ~n S = ~n E × ~n H . From (32) comes thatthe velocity of the an electromagnetic wave spreading in the constant electric field fluctuateswith the wave frequency not only in the value (the first term in (32)) but also in the direction(the last term). On the analogy with (2) let’s put: ~ξ = J / (cid:18) ~n E cos θ ~ + i~n H sin θ ~ (cid:19) , (33)i.e. E = J / cos θ ~ , H = J / sin θ ~ , (34)6 = E + H . Phase θ corresponds to the action for photons and at Rainich transformations(23) obtains the addition: δθ = ~ δϕ, (35)which completely coincides with (10).Now our goal is in phase decomplexification of dynamic (with derivativs with respect to t ) Maxwell equations (17) by analogy with decomplexification of Schr¨odinger equations. As aresult we get: Re : − ∂ t J sin θ ~ ~n H − J ~ cos θ ~ ∂ t θ~n H − J sin θ ~ ∂ t ~n H = c (cid:26) cos θ ~ ( ∇ J × ~n E ) − J ~ sin θ ~ ( ∇ θ × ~n E ) + 2 J cos θ ~ ( ∇ × ~n E ) (cid:27) , (36) Im : ∂ t J cos θ ~ ~n E − J ~ sin θ ~ ∂ t θ~n E + 2 J cos θ ~ ∂ t ~n E = c (cid:26) sin θ ~ ( ∇ J × ~n H ) + 2 J ~ cos θ ~ ( ∇ θ × ~n H ) + 2 J sin θ ~ ( ∇ × ~n H ) (cid:27) . (37)Let’s multiply (36) by cos θ ~ ~n H , and (37) by sin θ ~ ~n E and add. As a result we get the analogueof Hamilton-Jacobi equation (3) for photons: ∂ t θ + ~ J ∇ · (cid:18) J c cos (cid:18) θ ~ (cid:19) ~n S (cid:19) + ~ c ~n H · ( ∇ × ~n E ) + ~n E · ( ∇ × ~n H )] = 0 . (38)If multiply (36) by − sin θ ~ ~n H , and (37) by cos θ ~ ~n E and add, we’ll get photons transfer equation: ∂ t J + ∇ · (cid:18) J c sin (cid:18) θ ~ (cid:19) ~n S (cid:19) = 0 , (39)which is the same as (28). Equations (38) and (39) do not exhaust equations (36) and (37).Other ratios can be obtained from them. Let’s further suppose c = 1, and also instead of index t let’s write index 0.7axwell’s Lagrangian of electromagnetic field L M = − π F µν F µν = 18 π ( E − H ) = 18 π Reξ (40)leads to second pair of Maxwell equations (15) ∂ ν F µν = 0 (41)only provided we express F µν through potentials A µ : F µν = ∂ µ A ν − ∂ ν A µ , (42)which we should vary. The first pair of Maxwell equations (14): ∂ ν ˜ F µν = 0 , (43)where ˜ F µν = 12 ε µνρτ F ρτ , (44)is Jacobi identity - the consequence of (42).In spite of U (1) R -invariance of Maxwell equations, Lagrangian L M is not U (1) R -invariant,though obviously Lorentz-invariant. Transformations (16) really lead to the addition: δL M = − π ~E · ~Hδϕ. (45)Let us try to find the U (1) R -invariant Lagrangian of electrodynamics without worrying aboutits Lorentz invariance. At that let’s use the analogy of dynamic Maxwell and Dirac equations,and write down Lagrangian of electrodynamics, which similar to Dirac’s Lagrangian: L D = − l π h ~H · ( ∂ ~E − ∇ × ~H ) − ~E · ( ∂ ~H + ∇ × ~E ) i . (46)where l is the constant of the length dimension.Variational derivatives of L D with respect to ~E and ~H give the dynamic part of Maxwellequations both the first (14) and the second (15) pair: δL D δ ~E = l π ( ∂ ~H + ∇ × ~E ) = 0 , (47)8 L D δ ~H = l π ( ∂ ~E − ∇ × ~H ) = 0 . (48)Lagrangian L D don’t give zero divergences for ~E and ~H as Lagrange equations.Lagrangian L D is invariant with respect to transformations (16) ( U (1) R - invariant), thoughobviously is not invariant with respect to Lorenz transformations. U (1) R -invariance allows us to write down L D in the complex form (with respect to ~ξ and ~η ): L D = − l π h ~η · ( i∂ ~ξ − ∇ × ~ξ ) − ~ξ · ( i∂ ~η + ∇ × ~η ) i . (49)Independent variation on ~ξ and ~η obviously leads to equation (17) and the first of theequations (22). U (1) R - invariance in notation (49) becomes utterly evident. L D At the Lorenz transformations with parameters ϑ i (angles) and v i (velocities), we have: δE i = δϑ k ε kij E j − δv k ε kij H j , (50) δH i = δϑ k ε kij H j − δv k ε kij E j , (51)or in the complex form: δξ i = ( δϑ k + iδv k ) ε kij ξ j , (52) δη i = ( δϑ k + iδv k ) ε kij η j . (53)Beside this: δ∂ = − δv i ∂ i , (54) δ∂ m = δv m ∂ + δϑ k ε kmi ∂ i . (55)9et’s bring into use designations: R s = i∂ ξ s − ε smn ∂ m ξ n = 4 πl δL D δη s , (56) D = ∂ n ξ n , (57)relativeistic transformations of which are: δR s = δϑ k ε ksj R j − iδv s D, (58) δD = − iδv k R k . (59)With the use of R s we can write down: L D = l π Re ( η s R s ) , δL D = l π Re [ δ ( η s R s )] . (60)But: δ ( η s R s ) = iδv m η m ( ε mnk R k − g mn D ) (61)and when accomplishing Maxwell equations R k = 0 , D = 0 , (62) δL D disappears. It’s interesting, that lorenzinvariance of electrodynamics, that is based on theLagrangian L D , demands D = 0 as the extra condition, but D = 0 isn’t Lagrange equation for L D .Let’s write δL D with help of ~E and ~H :4 πδL D = δv m l (cid:2) H n ( ε mnk R Rek − g mn D Re ) − E n ( ε mnk R Imk − g mn D Im ) (cid:3) , (63)where R Res = − ∂ H s − ε smn ∂ m E n , R Ims = ∂ E s − ε smn ∂ m H n , (64)10 Re = ∂ n E n , D Im = ∂ n H n . (65)Substituting (64) and (65) in (63), we get:8 πδL D = δv m l [ − ε mnk ( H n ∂ H k + E n ∂ E k ) + ( H k ∂ m E k − E k ∂ m H k )+( E k ∂ k H m − H k ∂ k E m ) + ( E m ∂ k H k − H m ∂ k E k )] . (66)As we see, δL D doesn’t transform into divergence on the level of the fields ~E and ~H .Both Lagrangians are invariant with respect to space-time translations [2]: δx µ = t µ , (67) δx = A µ t µ , (68)( x - is the fifth coordinate), that lead to the such transformations of the fields: δA µ = − t ν F νµ , (69) δE i = − t ν ∂ ν E i , (70) δH i = − t ν ∂ ν H i . (71) Let’s find currents whose conservation is ensured by the considered invariants of both La-grangians.If the Lagrangian is symmetric with respect to transformations: δx µ = ε a X µa , (72) δq A = ε a α Aa , (73)11hen, according to Noether’s theorem, currents J µa = − ∂L∂ ( ∂ µ q A ) α Aa − LX µa . (74)are conserved on the extremals.For the Lagrangian L M in case of translations (67) and (69) we obtain: α Aa ∼ − F νµ , X µa ∼ δ µa (75)and as a conserved current acts Maxwell’s energy-momentum tensor T µν = 14 π (cid:18) − F ρµ F ρν + 14 δ µν F ρτ F µτ (cid:19) , (76)for which T coincide with ω (26), and T i with S i (27).For the Lagrangian L D in case of translations (67), (70) and (71), we obtain: α Aa ∼ − ∂ ν E i , − ∂ ν H i ; X Aa ∼ δ Aa . (77)Then the density of the ”energy” for the Lagrangian L D is: T = l π n ~E · [ ∇ × ~E ] + ~H · [ ∇ × ~H ] o . (78)On shell: ∇ × ~E = − ∂ ~H, ∇ × ~H = ∂ ~E (79)and instead of (78) we have: T = − l π n ~E · ∂ ~H − ~H · ∂ ~E o , (80)that practically coincides with density of energy for Dirac’s field and, apparently, doesn’t co-incide with (26). For example, for electrostatic field, (80) gives 0. Due to the fact that inthe electromagnetic wave ~E ⊥ ~H , for the flat-polarized wave with (80) we also get 0. Very odd”energy”!For the density of the ”momentum” for the Lagrangian L D we have: T i ~e i = l π n [ ~E × ∂ ~E ] + [ ~H × ∂ ~H ] o . (81)12r on shell (79) T i ~e i = l π n [ ~E × [ ∇ × ~H ]] − [ ~H × [ ∇ × ~E ]] o . (82)It’s also very odd quantity. For the flat-polarized electromagnetic wave with (81) we get T i = 0.Now let’s consider Rainich current, that is Noether current of the transformations (16) forthe Lagrangian L D . In such case: α Aa ∼ − H i , E i ; X µa = 0 (83)and current is: J = l ω = l π ( E + H ) , (84) J i = l S i = l π [ ~E × ~H ] i . (85)It’s interesting, that ”phase” current for the Lagrangian L D coincides (up to the factor l )with ”temporal” current for the Lagrangian L M . What does it mean? Maybe, synchronousmoving by phase happens when moving by time?If quantity of photons is connected with the U (1) R -symmetry, then the formulas (84) and (85)give the ”quantum-mechanical” interpretation of the light intensity. In such case the Lagrangian L D is better than L M . The transition from L M to L D may correspond to ”quantization”.Let’s note, that when ∂ ~E ∼ − hν ~H and ∂ ~H ∼ hν ~E (which corresponds to the synchronousphase transformation during evolution over time), from (80) and (81) we obtain: T ∼ hνω, T i ∼ hνS i . (86)If ω is proportional to the quantity of the photons, and S i is proportional to the density of theirflow, then obtained formulas lead to the quantum interpretation of the energy as a quantity thatis proportional to hν , and of the momentum as a quantity that is proportional to h/λ .Let’s note, that Rainich phase transformations (16) and translations along the fifth coordi-nate δx = t , (87)13hich are associated with the conservation of charges, in this formalism are not related. Maybe,it takes place only for uncharged electromagnetic field, where quantity of particles (photons)isn’t equal to quantity of charges, as for electrons.At rotations (50), (51): α Aa ∼ ε kij E j , ε kij H j ; X µa = 0 , (88)so the current is: M k = l S k = l π [ ~E × ~H ] k , (89) M lk = l π h δ lk ( E + H ) − E l E k − H l H k i , (90)and the conservation law is: ∂ S k + ∂ k ω − π ( E l ∂ l E k + H l ∂ l H k ) = 0 (91)(when getting it was used that ∂ k E k = 0, ∂ k H k = 0).At boosts (50), (51): α Aa ∼ − ε kij H j , ε kij E j , X µa = 0 , (92)and current is: L k = 0 , (93) L lk = − l π ( E l H k − H l E k ) . (94)In such case conservation law leads to identity: E l ∂ l E k = H l ∂ l H k (95)(when getting it was also used that ∂ k E k = 0, ∂ k H k = 0).14 Relativistic generalization of L D The necessity of relativistic generalization of L D consists of the following. First, the Lagrangian L D (46) has a lower time index 0, that ensures the fact, that it’s Rainich current is a zerocomponent of ordinary Maxwell’s energy-momentum tensor (76). Second, not all of Maxwellequations (but only dynamical) are Lagrange equations for L D . And third, it is desirable to findthe ”reasons” of phase symmetry. It, maybe, would allow to solve the mystery of electron withit’s spin, mass, etc.Firstly, let’s write down formulas: F i = E i , H i = 12 ε ijk F jk , F jk = ε ijk H k , (96)˜ F µν = 12 ε µνρτ F ρτ , (97)˜ F i = H i , E i = − ε ijk ˜ F jk , ˜ F ij = − ε ijk E k , (98)Lagrangian (46): L D = l π h ~H · ( ∂ ~E − ∇ × ~H ) − ~E · ( ∂ ~H + ∇ × ~E ) i . (99)is a zero component of vector Lagrangian: L Dρ = l π ( ˜ F ρν ∂ µ F νµ − F ρν ∂ µ ˜ F νµ ) . (100)Independent variation of vector Lagrangian L Dρ with respect to F µν and ˜ F µν lead to pair ofequations: δL Dρ δF µν = − l π h g ρµ ∂ σ ˜ F νσ + ∂ ν ˜ F ρµ − ( µ < −− > ν ) i = 0 , (101) δL Dρ δ ˜ F µν = l π [ g ρµ ∂ σ F νσ + ∂ ν F ρµ − ( µ < −− > ν )] = 0 . (102)Summing by ρ and µ , we get first and second pair of Maxwell equations (43), (41): ∂ σ ˜ F νσ = 0 , (103)15 σ F νσ = 0 , (104)which is the full system of Maxwell equations (besides dynamical equations also equality tozero of divergences ~E and ~H ). Are there additional equations in equations (101) and (102) inaddition to Maxwell equations? In any case the multiplication of (101) and (102) by ε ρτµν againleads to equations (104) and (103) (in reverse order). So each of the equations (101) or (102)are enough to get the full system of Maxwell equations (103) and (104).If not to consider F µν and ˜ F µν as independent and to use the formula (97), we’ll get: δL Dρ δF µν = l π h ε ρτµν ∂ σ F τσ − g ρµ ∂ σ ˜ F νσ + g ρν ∂ σ ˜ F µσ − ∂ ν ˜ F ρµ + ∂ µ ˜ F ρτ + ε στµν ∂ τ F σρ i = 0 , (105)in particular, from extra, spatial part of vector Lagrangian L Dm , we have: δL Dm δF ij = l π h ε mij ∂ σ F σ − g mi ∂ σ ˜ F jσ + g mj ∂ σ ˜ F iσ − ∂ j ˜ F mi + ∂ i ˜ F mj + ε kij ∂ k F m − ε kij ∂ F mk i = 0 , (106) δL Dm δF i = l π h ε mik ∂ σ F kσ − g mi ∂ l ˜ F l − ∂ i ˜ F m + ∂ ˜ F mi − ε ikl ∂ l F mk i = 0 . (107)We know, that variation of L D is enough to get only dynamical part of Maxwell equations.However, variation of spatial part L Dm is enough to get the full system of Maxwell equations.Really: ε mij δL Dm δF ij = 2 l π ∂ k E k = 0 , (108) δ mi δL Dm δF ij = − l π ∂ σ ˜ F jσ = l π ( ∂ H j + ε jim ∂ i E m ) = 0 , (109) δ mi δL Dm δF i = − l π ∂ k H k = 0 , (110) ε mij δL Dm δF i = l π ∂ σ F jσ = − l π ( ∂ E j − ε jim ∂ i H m ) = 0 . (111)16 Conserved currents of vector Lagrangian
At translations δF µν = − t α ∂ α F µν , (112) δ ˜ F µν = − t α ∂ α ˜ F µν , (113) α Aa ∼ − ∂ ν F µν , − ∂ ν ˜ F µν , X µa ∼ δ µa , (114)and in consequence of translation invariance of the vector Lagrangian, current T µρα = l π ( ˜ F ρτ ∂ α F τµ − F ρτ ∂ α ˜ F τµ ) − l π δ µα ( ˜ F ρτ ∂ σ F τσ − F ρτ ∂ σ ˜ F τσ ) . (115)is conserved: ∂ µ T µρα = 0. On shell the last term disappears, and the conservation equation of”energy-momentum” for L Dρ takes the form: ∂ µ ˜ F ρτ ∂ α F τµ = ∂ µ F ρτ ∂ α ˜ F τµ . (116)Phase transformations: δF µν = − ˜ F µν δϕ, (117) δ ˜ F µν = F µν δϕ, (118)that generalize transformations (16), lead to the current: J µρ = l π (cid:16) ˜ F ρτ ˜ F τµ + F ρτ F τµ (cid:17) = l π (cid:18) F ρτ F τµ + 14 δ µρ F αβ F αβ (cid:19) , (119)which coincides (up to the factor l ) with the energy-momentum tensor (76) for the Maxwell’sLagrangian.Let us consider the Lorentz transformation of L Dρ . With them: δF µν = ω σµ F σν + ω σν F µσ , (120)17 ˜ F µν = ω σµ ˜ F σν + ω σν ˜ F µσ , (121) δL Dρ = ω σρ L Dρ . (122)It follows from (122) that since on shell L Dρ = 0, we have δL Dρ = 0.
10 Conclusion
It remains unclear how the Rainich phase θ relates to the ordinary quantum-mechanical phaseof the Hilbert space of quantum electrodynamics.As to role of a phase in the quantum mechanics, maybe it exists a variant of quantummechanics, which has only dynamic phase symmetry, as Maxwell’s electrodynamics, with thepossibility of expression of independent components of ψ -function through potentials.It remains mysterious the existence of vector Lagrangians and the possibility of getting thesame physical quantities from different Lagrangians as a result of different symmetries. It wouldbe worth to examine this question in general case. References [1]