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The Helmholtz Decomposition and the Coulomb Gauge Kirk T The Helmholtz Decomposition and the Coulomb Gauge Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (April 17, 2008; updated March 3, 2020) 1Problem Helmholtz showed in 1858 [1] (in a hydrodynamic context) that any vector field, say E,that vanishes suitably quickly at infinity can be decomposed as,1,2,3 E = Eirr + Erot, (1) 4 where the irrotational and rotational components Eirr and Erot obey, ∇ × Eirr =0, and ∇ · Erot =0. (2) For the case that E is the electric field, discuss the relation of the Helmholtz decomposition to use of the Coulomb gauge.5 2Solution The Helmholtz decomposition (1)-(2) is an artificial split of the vector field E into two parts, which parts have interesting mathematical properties. We recall that in electrodynamics the electric field E and the magnetic field B can be related to a scalar potential V and a vector potential A according to (in SI units), ∂ −∇V − A , E = ∂t (3) B = ∇ × A. (4) This results in another decomposition of the electric field E which might be different from that of Helmholtz. Here, we explore the relation between these two decompositions. We also recall that the potentials V and A are not unique, but can be redefined in a systematic way such that the fields E and B are invariant under such redefinition. A 1The essence of this decomposition was anticipated by Stokes (1849) in secs. 5-6 of [2]. 2For a review of the Helmholtz’ decomposition in nonelectromagnetic contexts, see [3]. 3Radiation fields, which fall off as 1/r at large distance r from their (bounded) source, do fall off sufficiently quickly for Helmholtz’ decomposition to apply, as reviewed in [4]. Doubts as to this were expressed in [5], but see [6]. See Appendix A for the Helmholtz decomposition of the electromagnetic fields of a Hertzian (“point”, oscillating) dipole, which illustrates that such a decomposition is readily made when radiation is present. 4The irrotational component is sometimes labeled “longitudinal” or “parallel”, and the rotational com- ponent is sometimes labeled “solenoidal” or “transverse”. 5Vector plane waves E ei(k·r−ωt) do not vanish “suitably quickly” at infinity, so care is required in applying the Helmholtz decomposition Eirr =(E · k) kˆ, Erot = E − Eirr of this mathematically useful, but physically unrealistic class of fields. See, for example, sec. 2.4.2 of [7]. 1 particular choice of the potentials is called a choice of gauge, and the relations (3)-(4) are said to be gauge invariant.6 Returning to Helmholtz’ decomposition, we note that he also showed how, ∇ · E(r ) ∇ × E(r ) Eirr(r)=−∇ dVol , and Erot(r)=∇ × dVol , (5) 4πR 4πR where R = |r − r|. Time does not appear in eq. (5), which indicates that the vector field E at some point r (and some time t) can be reconstructed from knowledge of its vector derivatives, ∇·E and ∇×E, over all space (at the same time t).7 The main historical significance of the Helmholtz decomposition (1) and (5) was in showing that Maxwell’s equations, which give prescriptions for the vector derivatives ∇ · E and ∇ × E, are mathematically sufficient to determine the field E.Since∇ · E = ρtotal/0 and ∇ × E = −∂B/∂t, the fields Eirr and Erot involve instantaneous action at a distance and should not be regarded as physically real. This illustrates how gauge invariance in necessary, but not sufficient, for electromagnetic fields to correspond to “reality”.8,9,10 6The gauge transformation A → A + ∇χ, V → V − ∂χ/∂(ct), leaves the fields E and B unchanged. A consequence of this is that when the vector potential is decomposed as A = Airr + Arot, the rotational part 2 is unchanged by transformations where ∇ χ = 0. That is, in such cases, Airr + Arot → (Airr + ∇χ)+Arot, where the term in parenthesis is the irrotational part of the transformed vector potential, so the rotational part, Arot, is unchanged by the gauge transformation. 2 On the other hand, if ∇ χ = 0 everywhere, then the gauge transformation is Airr + Arot → Airr +(Arot + ∇χ), which leaves the irrotational part of A unchanged. 7If the field E is known only within a finite volume V, bounded by a closed surface S, then the Helmholtz decomposition (1) and (5) becomes, ∇ · E(r ) nˆ · E(r ) Eirr(r)=−∇ dVol + dArea , (6) 4πR 4πR V S ∇ × E(r ) nˆ × E(r ) Erot(r)=∇ × dVol + dArea , (7) V 4πR S 4πR where nˆ is the inward unit normal vector on the surface S. That is, E = −∇V only if ∇ × E =0 inV,and nˆ × E =0 onS, (8) E = ∇ × A only if ∇ · E =0 inV,and nˆ · E =0 onS. (9) Neglect of the conditions on the surface S can lead to error, as remarked in [8, 9, 10, 11]. 8 The forms (5) are not the only possible representations of Eirr and Erot. For example, we could write, ∇ · E(r ) ∇ × E(r ) Eirr(r)=−∇ dVol + C , and Erot(r)=∇ × dVol + ∇χ , (10) 4πR 4πR for any constant C and any differentiable scalar function χ, without changing the values of Eirr and Erot. Also, since [∇ × E(r)]/R = ∇ × [E(r)/R]+∇ × E(r)/R,ifE falls off sufficiently quickly at large R we can rewrite Erot as, E(r ) Erot(r)=∇ × ∇ × dVol . (11) 4πR 9See Appendix A for an application of the Helmholtz decomposition to Hertzian dipole radiation 10The Helmholtz decomposition leads to interesting interpretations of the momentum and angular mo- mentum associated with electromagnetic fields [7]. 2 The Helmholtz decomposition (1) and (5) can be rewritten as, E = −∇V + ∇ × F, (12) where, ∇ · E(r) ∇ × E(r) V (r)= dVol, and F(r)= dVol. (13) 4πR 4πR It is consistent with usual nomenclature to call V a scalar potential and F a vector potential. That is, Helmholtz decomposition lends itself to an interpretation of fields as related to derivatives of potentials. When the vector field E is the electric field, it also obeys Maxwell’s equations, two of which are (in SI units and for media where the permittivity is 0), ρ ∂B ∇ · E = , and ∇ × E = − , (14) 0 ∂t where ρ is the electric charge density and B is the magnetic field. If we insert these physics relations into eq. (13), we find, ρ(r) V r d , ( )= π R Vol (15) 4 0 ∂ B(r) F(r)=− dVol. (16) ∂t 4πR The scalar potential (15) is calculated from the instantaneous charge density, which is exactly the prescription (44) of the Coulomb gauge. That is, Helmholtz + Maxwell implies use of the Coulomb-gauge prescription for the scalar potential. However, eq. (16) for the vector potential F does not appear to be that of the usual procedures associated with the Coulomb gauge. Comparing eqs. (12)-(13) and (16), we see that we can introduce another vector potential A which obeys, ∂ ∇ × − A , F = ∂t (17) such that, B(r) A(r)=∇ × dVol, (18) 4πR and, ∂ −∇V − A , E = ∂t (3) which is the usual way the electric field is related to a scalar potential V and a vector potential A. Note also that eq. (18) obeys the Coulomb gauge condition (40) that ∇ · A =0.11 Thus, the Helmholtz decomposition (1) and (5) of the electric field E is equiva- lent to the decomposition (3) in terms of a scalar and a vector potential, provided those potentials are calculated in the Coulomb gauge.12 11See, for example, sec. 3 of [12]. 12 The fields Eirr = −∇V − ∂Airr/∂(ct)andErot = −∂Arot/∂(ct) can be deduced from scalar potential (C) V and vector potential A = Airr + Arot in any gauge, but only in the Coulomb gauge is Airr =0suchthat (C) (C) the Helmholtz decomposition has the simple form Eirr = −∇V and Erot = −∂A /∂t. 3 Using various vector calculus identities, we have, ∇ × B(r ) d ∇ 1 × B(r ) d − ∇ 1 × B(r ) d A(r)= πR Vol = R π Vol = R π Vol 4 4 4 ∇ × B(r ) d ∇ × B(r ) d = πR Vol + πR Vol 4 4 ∇ × B(r) B(r) ∇ × B(r) = dVol − dArea × = dVol, (19) 4πR 4πR 4πR provided B vanishes sufficiently quickly at infinity. In view of the Maxwell equation ∇·B =0, we recognize eq. (19) as the Helmholtz decomposition B = ∇ × A for the magnetic field.13 We can go further by invoking the Maxwell equation, 1 ∂E ∇ × B = μ J + , (23) 0 c2 ∂t where J is the current density vector, the medium is assumed to have permeability μ0,and c is the speed of light, so that, μ J(r) ∂ E(r) A(r)= 0 dVol + dVol. (24) 4π R ∂t 4πc2R This is not a useful prescription for calculation of the vector potential, because the second term of eq. (24) requires us to know E(r)/c2 to be able to calculate E(r).14 But, c2 is a big number, so E/c2 is only a “small” correction, and perhaps can be ignored.15 If we do so, 13We can verify the consistency of eqs. (18) and (19) by taking the curl of the latter. For this, we note that, ∇ × B(r) 1 1 ∇ × = −(∇ × B(r)) × ∇ =(∇ × B(r)) × ∇ . (20) 4πR 4πR 4πR The i-component of this is, ki i,j,kjlm(∂lBm)∂k(1/4πR)=δlm(∂l Bm)∂k(1/4πR)=(∂kBi)∂k(1/4πR) − (∂iBk)∂k(1/4πR) 2 = ∂k[Bi∂k(1/4πR)] − Bi∇ (1/4πR) − ∂k[(1/4πR) ∂iBk]+(1/4πR) ∂i∇ · B 3 = Bi(r ) δ (r − r )+∂k[Bi∂k(1/4πR) − (1/4πR) ∂iBk]. (21) The volume integral of this gives B(r) plus a surface integral that vanishes if the magnetic field falls off sufficiently quickly at large distances. That is, ∇ × A = B for the vector potentials given by eqs.
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