On electromagnetic momentum of an electric dipole in a magnetic field
aa r X i v : . [ phy s i c s . c l a ss - ph ] S e p On electromagnetic momentum of an electric dipole in a magneticfield
Ben Yu-Kuang Hu ∗ Department of Physics, University of Akron, Akron, OH 44325-4001 (Dated: August 16, 2018)
Abstract
The total linear electromagnetic field momentum P em of a stationary electric dipole p in a staticmagnetic field B is considered. The expression P em = B × p , which has previously been impliedto hold in all static magnetic field situations, is not valid in general. The contribution of theelectromagnetic momentum of the fringing fields of the dipole is discussed. It is shown that wheneither the static magnetic field or the electric dipole moment is changed, the mechanical impulseon the system equals − ∆ P em , and hidden momentum does not need to be invoked in order toconserve total momentum. . INTRODUCTION In classical electromagnetism, the electric ( E ) and magnetic ( B ) fields store linear mo-mentum in the form of electromagnetic momentum density ǫ ( E × B ) [SI units are usedthroughout this paper]. The total electromagnetic momentum of a system is the integral ofthe electromagnetic momentum density, P em = ǫ Z E ( r ) × B ( r ) d r . (1)It is therefore possible for static electric and magnetic fields to store electromagnetic mo-mentum (although the total momentum of the system must zero, as is discussed later). Toillustrate the consequences of the electromagnetic momentum on a specific system, severalauthors have studied the electromagnetic momentum due to electric dipoles in the presenceof static magnetic fields. In this paper, we consider two aspects of electromagnetic momentum stored in the fieldsdue to a stationary electric dipole in a static magnetic field. The first is the expression forthe electromagnetic momentum stored in the system, which has been reported to be (seeRefs. 3–5 and Ref. 6) P em = 12 B × p , (2)where p is the electric dipole and B is the magnetic field at the position of dipole. Eq. (2) wasderived for stationary electric dipoles in locally uniform static magnetic fields producedby a long uniform solenoid and a spinning sphere with a uniform surface charge density.Ref. 6 inadvertently omitted to mention that the result was derived for locally uniformfields, implying that it was true in general. Given that the electromagnetic momentum fora stationary magnetic moment m in a static (not necessarily uniform) electric field is P em = 1 c E × m , (3)where E is the electric field at the position of the magnetic moment and c is the speedof light, it seems plausible that there should be an equivalent expression for a stationaryelectric dipole in a static, non-uniform magnetic field. However, the results Eqs. (2) and (3)are incompatible with each other, as illustrated in the following example.Assume there is a magnetic moment m = m ˆ z at the origin and an electric dipole p = p ˆ x at R ˆ x (where carets indicate unit vectors), as shown in Fig. 1. The magnetic field at a2isplacement r = 0 from the magnetic moment is (see e.g. , Ref. 1, p. 255) B = µ πr [3( m · ˆ r )ˆ r − m ] . (4)The electric field at a displacement r from the electric dipole is obtained by replacing m by p and µ by ǫ − in the above equation. The magnetic field at R ˆ x due to the magneticdipole is B ( r = ˆ x R ) = − µ m/ (4 πR ) ˆ z , and the electric field at the origin due to the electricdipole is E ( ) = p/ (2 πǫ R ) ˆ x . These together with Eqs. (2), (3) and c − = µ ǫ giveFrom Eq. (2): P em = − µ mpπR ˆ y ; (5a)From Eq. (3): P em = − µ mpπR ˆ y . (5b)Since P em evaluated using Eqs. (2) and (3) do not agree, at least one of these expressionsis not valid in general. It turns out that it is Eq. (2) that is not in general valid, evenin cases where the magnetic field is locally uniform, and P em cannot in general be writtensolely in terms of p and B at the position of the dipole. In Section II of this paper, we giveseveral expressions for P em for a stationary electric dipole in a static magnetic field, all ofwhich depend (explicitly or implicitly) on the electric current configuration that generatesthe magnetic field. This is followed by examples and a discussion regarding the reasonsbehind the differences in total amount of P em for different current configurations.Section III deals with another aspect of an electric dipole in a magnetic field: the impulseimparted to the system when either the electric dipole moment or magnetic field is reducedto zero. A recent paper by Babson et al. seemed to imply (although it was not theirintent ) that it is necessary to take into account the hidden momentum in the system inorder for the total momentum to be conserved. We show that in these systems, the lossof electromagnetic momentum is always equal to the mechanical impulse imparted to thesystem, and it is not necessary to invoke the presence of hidden momentum to conserve thetotal momentum.Section IV contains a discussion of the results of this paper. II. P em FOR A STATIONARY ELECTRIC DIPOLE IN A STATIC MAGNETICFIELD
The total electromagnetic momentum P em for the case in which the electric and mag-netic fields E and B are due to stationary charges distribution ρ ( r ) and steady current3istributions J ( r ) which are local (do not extend to infinity), can also be expressed as P em = 1 c Z V ( r ) J ( r ) d r , (6a)where V ( r ) is the scalar potential in the Coulomb gauge, V ( r ) = 14 πǫ Z ρ ( r ′ ) | r − r ′ | d r ′ . (6b)In Appendix A, the derivation of Eq. (6a) is reproduced, and circumstances in which thelocality of J can be relaxed are discussed.The electromagnetic momentum can alternatively be expressed as P em = Z ρ ( r ) A ( r ) d r . (7a)where A ( r ) is the vector potential in the Coulomb gauge ( ∇ · A = 0) for a static currentsource, A ( r ) = µ π Z J ( r ′ ) | r − r ′ | d r ′ . (7b)It is easy to see that both expressions Eqs. (6a) and (7a) for P em are equivalent, by substi-tuting Eq. (6b) into (6a) and Eq. (7b) into (7a), and using c − = ǫ µ .First, we reconfirm that Eq. (3) is valid for a magnetic dipole in the presence of a staticelectric field. The vector potential for a magnetic dipole at r m in the Coulomb gauge is A = µ m × ( r − r m ) / (4 π | r − r m | ), so Eq. (7a) gives P em = µ Z d r ρ ( r ) m × ( r − r m )4 π | r − r m | = µ ǫ (cid:20) − Z d r ρ ( r ) ( r m − r )4 πǫ | r − r m | (cid:21) × m , (8)which yields Eq. (3) since the term in the square parentheses is the electric field at theposition of the magnetic dipole and µ ǫ = c − . A. Expressions for P em for stationary electric dipole in a static magnetic field We now present four expressions for the total electromagnetic field momentum for anelectric dipole p in the presence of a static current density J ( r ) which produces a staticmagnetic field B ( r ) and a corresponding vector potential in the Coulomb gauge A ( r ). Inthese expressions, the gradients and the magnetic fields are evaluated at the position r of4he electric dipole. The expressions are P em = ( p · ∇ ) A ( r ) (9a)= − µ π Z [ p · ( r − r ′ )] J ( r ′ ) | r − r ′ | d r ′ (9b)= B × p + ∇ ( p · A ) (9c)= B × p − µ π Z ( r − r ′ ) [ p · J ( r ′ )] | r − r ′ | d r ′ . (9d)Expression (9a), which was given in Ref. 13, can be derived by taking a point dipole p ata position r to be the limit of point charges − q at r and q at r + l in which | l | ≡ l → q → ∞ , with the product l q = p being finite. Using this in Eq. (7a) and the expansion A ( r + l ) ≈ A ( r ) + ( l · ∇ ) A ( r ), results in P em = lim l → ql → p q [ A ( r + l ) − A ( r )] = lim l → ql → p ( q l · ∇ ) A ,which gives expression (9a).Combining Eq. (9a) with Eq. (7b) and using the relationship (where ∇ is the gradientwith respect the variable r ) ∇ | r − r ′ | = − r − r ′ | r − r ′ | , (10)[or alternatively using the scalar potential for a point dipole, V ( r ) = p · r / (4 πǫ r ), inEq. (6a)] yields expression (9b).Expression (9c) is obtained by using the vector identity [see, e.g. , Refs. 1 or 2] ∇ ( p · A ) = p × ( ∇ × A ) + A × ( ∇ × p ) + ( p · ∇ ) A + ( A · ∇ ) p , together with ∇ × A = B and p nothaving any spatial dependence, gives expression (9c). Finally, using Eq. (7b) in Eq. (9c) andutilizing Eq. (10) yields expression (9d). Expressions (9c) and (9d) show that, in additionto the B × p term that is analogous to the E × m /c for a magnetic dipole in an electricfield, there is an additional term which is dependent on the details of the current distributionwhich does not vanish in general. Thus, the electromagnetic momentum of an electric dipolein the magnetic field cannot be expressed solely as a function of the dipole moment and thelocal magnetic field.The differences between the forms of the expressions for P em for a magnetic dipole in anelectric field and for an electric dipole in a magnetic field are due to the differences betweenthe sources of static electric and magnetic field. Static magnetic fields are caused by electriccurrents and static electric fields are caused by electric charges. Therefore, despite the factthat the magnetic and electric fields for magnetic and electric dipoles have the same form5away from the dipoles themselves) the expressions for P em are different. For more details,see Ref. 13. B. P em for various cases of electric dipoles in a magnetic field We use the expressions Eqs. (9a) – (9d) to evaluate P em for a static electric dipole in astatic magnetic field. We obtain the correct result for the magnetic and dipole configurationshown in Fig. 1. We then consider cases of an electric dipole in a uniform magnetic fieldproduced by a spinning uniformly charged spherical shell and a uniform circular solenoid,where we reproduce the result B × p . We see then that in the case of a uniform fieldproduced by electric currents in two parallel plates, P em is not B × p . This is followed bya discussion of the reasons for the differences in P e m in these cases.
1. Configuration shown in Fig. 1
We now re-evaluate P em for the example shown in Fig. 1. The vector potential for amagnetic dipole m = m ˆ z at the origin in the Coulomb gauge, in spherical and cartesiancoordinates, is A m ( r ) = µ m × r πr = µ m sin θ πr ˆ φ = µ m π − y ˆ x + x ˆ y ( x + y + z ) / . (11)Therefore, for an electric dipole p = p ˆ x at position R ˆ x [or in spherical coordinates, r = R , θ = π/ φ = 0] in the presence of the magnetic dipole at the origin, Eq. (9a) gives P em = p (cid:20) ∂∂r (cid:18) µ m sin θ πr (cid:19)(cid:21) r = Rθ = π/ ˆ φ = − µ mp πR ˆ φ , (12)which agrees with the result for P em using Eq. (3), since ˆ φ = ˆ y when φ = 0. In thisparticular case, P em = 2 B × p .Alternatively, we can also use Eq. (9c). In this case, it is more convenient to us theCartesian-coordinate expression for A m ( r ), which gives[ ∇ ( p · A m )] ( R, , = µ mp π (cid:20) ∇ − y ( x + y + z ) / (cid:21) ( R, , = − µ mp πR ˆ y = B × p . (13)Using this in Eq. (9c) also gives P em = 2 B × p .6 . Constant magnetic field inside uniformly charged spinning spherical shell McDonald , Gsponer and Babson et al. evaluated P em for a capacitor or electric dipole p , in a uniformly charged spinning spherical shell (or equivalently a uniformly magnetizedsphere) producing a uniform field B inside the sphere. In all cases, they obtained the result P em = B × p . We will see that this result is reproduced by the expressions given in theprevious section, and is in fact independent of the position of the dipole (or equivalentlycapacitor) within the sphere.The vector potential in the Coulomb gauge is A in ( r ) = µ Rσ ( ω × r ) and this givesa uniform magnetic field inside the sphere of B = µ σR ω . Thus, for this case the vectorpotential inside the sphere can be written as A in ( r ) = 12 B × r . (14)Since p · A in = 12 p · ( B × r ) = − r · ( B × p ), Eq. (9c) gives P em = B × p + 12 ∇ ( p · ( B × r )) = 12 B × p , (15)independent of the position of the electric dipole inside the sphere. By linear superposition,this result holds for any distribution of dipoles within the sphere. This explains why in Ref. 3obtained result Eq. (15) for a the electromagnetic momentum of a spherical-cap capacitorinside a uniformly magnetized sphere (which is equivalent to a rotating uniformly chargedspherical surface), where p is the total dipole moment of the charged capacitor. In fact, solong as the capacitor is wholly contained within the sphere, the capacitor can any shape andthe result would still hold.
3. Constant magnetic field created by a cylindrical solenoid
We now consider the case of an electric dipole in a region of constant magnetic fieldcreated by an infinite cylindrical solenoid of radius R that is centerd along the z -axis. Thevector potential in the Coulomb gauge in cylindrical coordinates is A ( r ) = Bs ˆ φ = B × r if s R,BR s ˆ φ , if s > R (16)7here s is the distance from the z -axis. These give B = B ˆ z inside the solenoid and B = 0outside. We now evaluate P em for a dipole inside and outside the solenoid. Dipole within the solenoid : Since the vector potential inside the cylinder is the same formas A in ( r ) for the rotating sphere, the calculation also yields P em = B × p . As in thecase of the dipole within the sphere, this result is independent of the position of the dipoleinside the solenoid, and therefore even a macroscopic object, such as a capacitor, with atotal dipole moment p within the solenoid is also P em = B × p .At this stage, one might surmise that even though P em = B × p does not hold fornon-uniform magnetic fields, perhaps it is always true for uniform magnetic fields. The nextand subsequent examples show that this is not the case. Dipole outside the solenoid : For a dipole outside the solenoid, the electromagnetic mo-mentum is P em = ( p · ∇ ) A = BR (cid:18) p s ∂∂s + p φ s ∂∂φ + p z ∂∂z (cid:19) ˆ φ ( φ ) s = − BR s (cid:16) p s ˆ φ + p φ ˆ s (cid:17) , (17)where ∂ ˆ φ /dφ = − ˆ s has been used. Since the magnetic field outside the solenoid is zero,clearly P em = B × p . In Appendix B, this result is reproduced by direct integration of theelectromagnetic momentum density.
4. Two parallel plates with counter-propagating currents
Assume that there are a pair thin conducting plates are in the x – y plane and they are at z = ± a . The plate on the top carries a uniform current density j in the + x direction andthe plate at the bottom carries the an equal magnitude of current in the − x direction, asshown in Fig. 2. The calculation of the vector potential using Eq. (7b), is analogous to thatof the scalar potential from parallel plate capacitors with a uniform charge distribution, andit gives a vector potential of A ( r ) = Bz ˆ x , if − a < z < a, ± Ba ˆ x , if z ≷ a, (18)where B = µ j , and a magnetic field of B ˆ y for | z | < a and B = 0 for | z | > a .8he P em for an electric dipole outside the plates is zero, since A is constant. For anelectric dipole p in between the plates, using Eq. (9a), the total electromagnetic momentumis P em = (cid:18) p x ∂∂x + p y ∂∂y + p z ∂∂z (cid:19) Bz ˆ x = p z B ˆ x . (19)Thus, only the component of p perpendicular to the plates contributes to P em . This providesan example where there is a non-zero uniform magnetic field, but P em = B × p .Note that in this case, even though the currents and hence the magnetic fields extend toinfinity, the expression Eq. (7a) is convergent and the surface term in Eq. (A1) in AppendixA goes to zero in the limit when the volume of integration goes to infinity. Therefore,Eq. (9a) [which is based on Eq. (7a)] can be used to obtain P em . In Appendix C, the aboveresult for P em is also obtained by evaluating ǫ R d r E × B directly.To circumvent any problems which might be associated with infinite currents and fields,one can assume that the current-carrying plates are large but finite, and the top and bottomplates are connected at the ends where the currents flow to and from. In this case, therewill be contributions to P em from the fringing fields, but these can be made arbitrarily smallby increasing the size of the plates. For readers who are skeptical that the above exampleis consequence of the unbounded currents and magnetic fields, which can sometimes giveill-defined physical quantities, an example of P em = B × p for purely local currents inshown in the Appendix D. C. Which gauge of A ( r ) should be used? Eqs. (9a) and (9c) indicate that P em for an electric dipole in a constant magnetic fieldis dependent on form of the vector potential A ( r ) at the dipole. In the case of a constantmagnetic field B = B ˆ y , both A ( r ) = B ˆ y × r = B ( z ˆ x − x ˆ z ) and A ( r ) = Bz ˆ x (or for thatmatter, A = B [(1 − α ) z ˆ x − αx ˆ z ] for any constant α ) satisfy ∇ × A = B ˆ y and ∇ · A = 0.As shown above, the value of P em depends on which A ( r ) is used. So, which of these is thecorrect one to use? The Helmholtz theorem states that the appropriate A in the Coulombgauge ( ∇ · A = 0) is determined by B = ∇ × A over all space. Provided that B goes to zerosufficiently fast as r → ∞ , A ( r ) is determined by the current density J ( r ) by Eq. (7b). Thisexpression gives A = B ˆ y × r for the interior regions of uniform cylindrical solenoids andspinning spheres, and A = ˆ x Bz for interior region of parallel plates with constant current9ensities. D. Contribution of the fringing electric fields of a capacitor to P em A capacitor can be modeled as a superposition of point dipoles. Since the P em is linearin E , which in turn is linear in p , the results for P em for a point dipole are equally validfor a capacitor. For cylindrical solenoids and spinning spheres of uniform surface charge,the factor of in the result P em = B × p implies P em = P capacitor , where P capacitor is theelectromagnetic momentum that is stored in between the capacitor plates. Obviously, thefringing electric fields of the capacitor contain a contribution of − P capacitor . On the otherhand, for a constant field generated by a “solenoid” consisting of parallel conducing plateswith counter-propagating currents, if the capacitor plates are oriented in the same way as thecurrent-carrying plates, the P em = P capacitor , but if the capacitor plates are perpendicularto the plates carrying the current, P em = 0. What causes these differences in the totalelectromagnetic momentum?Consider the cases shown in Fig. 3(a) – (c), in which the capacitors can be approximatedas an electric dipole moment is the positive z -direction, and the magnetic field is in thepositive y -direction. In these cases, the y - and z -components of the total electromagneticmomentum P em are zero, so to determine the contribution to P em , we need only considerthe z -component of the electric field due to the dipole (since the electromagnetic momentumdensity is ǫ E × B , and here B is in the y -direction.) Fig. 3(d) indicates the regions where E z is positive and where it is negative around the electric dipole. Since the z -component ofthe electric field for a dipole at the origin is E z ( r ) = p πǫ r (3 cos θ − − p ǫ δ ( r ), E z > − (1 / √ ≈ o with respect to the positive andnegative z -axes, as indicated by the shaded regions in Fig. 3(d). When B in the positive y -direction, these regions will give a negative contribution to P em ,x , while conversely, theregions within approximately 35 o of the “equator” of the dipole, the electric field gives apositive contribution to P em ,x .In the fringing field regions around the capacitors where the magnetic field exists, thevolume of the regions where E z > E z <
0) increases (decreases) in the configurationsshown in Figs. 3(b), (a) and (c), respectively. This results in a corresponding increase inthe contribution of the fringing fields, in that order. In Fig. 3(b), apparently the fringing10eld contributions to P em of the E z > E z < E z > E z < III. MOMENTUM IMPARTED TO THE SYSTEM WHEN CURRENT ORDIPOLE IS CHANGED
In this section, we attempt to clarify the discussion in Babson et al. on the role of hiddenmomentum in the conservation of total momentum of a system consisting of an electricdipole in a magnetic field, when the electromagnetic momentum is changed. It is shown herethat the mechanical impulse imparted to the system due to either a change in the currentproducing the magnetic field or the electric dipole is equal to the loss of electromagneticmomentum of the system, and hence it is not necessary to invoke hidden momentum toconserve the total momentum. A. Proof that impulse imparted to the system equals − ∆ P em
1. Current is changed
When the current is changed the vector potential changes by ∆ A ( r ), which induces achange in the electric field E = − ∂ A /∂t (the Faraday effect). The momentum imparted toa charge distribution ρ ( r ) as the current changes is ~ I ρ = Z dt Z d r ρ ( r ) E ( r , t ) = Z d r ρ ( r ) (cid:20) − Z dt ∂ A ∂t (cid:21) = − Z d r ρ ( r ) ∆ A ( r ) = − ∆ P em , (20)where the last equality comes from Eq. (7a).Thus, when the magnitude of the current ischanged, the change in the electromagnetic momentum is equal and opposite to the impulsegiven to the charges. Since this is true for any arbitrary distribution of charges, it is alsotrue for the momentum imparted on the dipole.11 . Electric dipole moment is changed
When the dipole moment is changed, there is an electric current density J dip due to thetransfer of charge within the dipole which experiences a Lorentz force F dip = J dip × B . Inaddition, J dip itself produces a magnetic field, B dip , which imparts a Lorentz force on thecurrent in the solenoid. Let us examine the impulses exerted by these forces and comparethese to the change in the electromagnetic momentum.Assuming the dipole is at the origin, the current distribution due to the change in thedipole moment is J dip ( r , t ) = ˙ p ( t ) δ ( r ) . (21)The impulse on the dipole, which is equal to the momentum imparted on the dipole, is givenby ~ I dip = Z dt Z J dip ( r , t ) × B ( r ) d r = Z dt ˙ p ( t ) × B ( ) = ∆ p × B ( ) . (22)The current J dip ( t ) produces a magnetic field at position r ′ and time t of B dip ( r ′ , t ) = − µ π (cid:20) r ′ × [ ˙ p + ( r ′ /c )¨ p ] r ′ (cid:21) ret , (23)where “ret” indicates that ˙ p and ¨ p are evaluated at retarded time t ret = t − | r ′ | /c . Theimpulse imparted on the current in the solenoid is ~ I sol = Z dt Z J ( r ′ ) × B dip ( r ′ , t ) d r ′ = − µ π Z J ( r ′ ) × (cid:18) r ′ × ∆ p r ′ (cid:19) d r ′ = µ π Z (cid:20) − r ′ ( J ( r ′ ) · ∆ p ) r ′ + ∆ p (cid:18) J ( r ′ ) · r ′ r ′ (cid:19)(cid:21) d r ′ , (24)where the identity A × ( B × C ) = B ( A · C ) − C ( A · B ) has been used in the last equality.(The ¨ p term in B dip ( r , t ) does not contribute because ∆ ˙ p = 0, since p is constant beforeand after it is changed.) The second term in the last expression in Eq. (24) is zero because Z J ( r ′ ) · r ′ r ′ d r ′ = − Z J ( r ′ ) · ∇ ′ (cid:18) r ′ (cid:19) d r ′ = Z r ′ ∇ ′ · J ( r ′ ) − ∇ ′ · (cid:18) J ( r ′ ) r ′ (cid:19) = 0 (25)12ince ∇ ′ · J ( r ′ ) = 0 for static currents, and from the divergence theorem, R ∇ ′ · (cid:0) J ( r ′ ) /r ′ (cid:1) d r ′ = R S J ( r ′ ) /r ′ · d a ′ = 0, because S is a surface at infinity and J ( r ′ ) is be localized. Therefore, ~ I sol = − µ π Z r ′ ( J ( r ′ ) · ∆ p ) r ′ d r ′ . (26)This implies that the total impulse given to the system due to the change ∆ p in the dipolemoment at the origin is ~ I dip + ~ I sol = ∆ p × B ( ) − µ π Z r ′ ( J ( r ′ ) · ∆ p ) r ′ d r ′ . (27)But from Eq. (9d), this is exactly the negative of the change in the electromagnetic momen-tum for an electric dipole at r = 0, so ~ I dip + ~ I sol + ∆ P em = 0 , (28)which shows that the momentum imparted to the system when the electric dipole momentchanges is equal to the change in field electromagnetic momentum. B. Examples
We now examine several cases of electric dipoles in magnetic fields created by differentsources, specifically by a magnetic dipole, and by cylindrical, spherical and parallel-platesolenoids. In each case, we show explicitly that when either the electric dipole or themagnetic-field-producing current is changed, the impulse on the system is equal to thechange in electromagnetic momentum.
1. Magnetic and electric dipole at arbitrary orientation and displacement
For an arbitrary placement of an magnetic moment m and an electric dipole p displacedby r = 0 from the magnetic moment, the electromagnetic momentum is P em = µ πr (cid:18) ( m × p ) − r ( p · r )( m × r ) (cid:19) . (29)Let us assume that m is at the origin and p is at r . Magnetic dipole changed – When the magnetic dipole is changed by ∆ m , the change inthe electromagnetic momentum from Eq. (29) is∆ P em = µ πr (cid:18) (∆ m × p ) − r ( p · r )(∆ m × r ) (cid:19) . (30)13et us assume, for notational simplicity, that the change in the dipole is quasi-static, so thatone can ignore retardation effects and the ¨ m term. (The derivation can easily be extendedto include non-quasi-static changes.) This change produces an electric field E m . dip ( r , t ) = µ π r × ˙ m r . (31)The force on the electric dipole is ∇ ( p · E m . dip ), so the momentum imparted on the electricdipole due to a change in the magnetic moment is ~ I = Z ∇ ( p · E m . dip ) dt = µ π ∇ (cid:16) p · r r (cid:17) × ∆ m = µ π (cid:20) p r − p · r ) r r (cid:21) × ∆ m . (32)Eqs. (30) and (32) give ∆ P em + ~ I = 0.It is easy to generalize the above from a point electric dipole to an arbitrary distributionof static charges ρ ( r ). The electric field E dip ( r , t ) induced by the change in m imparts animpulse on ρ ( r ) of (using Eq. (31), Coulomb’s law for the electric field and µ ǫ = c − ) ~ I = Z d r ρ ( r ) (cid:20)Z dt E dip ( r , t ) (cid:21) = Z d r µ π (cid:20) ρ ( r ) r × ∆ m r (cid:21) = − c E × ∆m , (33)where E is electric field at the origin; i.e. , the position of the magnetic moment. Since∆ P em = c − E × ∆ m , this implies that ~ I + ∆ P em = 0 when m is changed for an arbitrarydistribution of static charges. Electric dipole changed – If the electric dipole is changed by ∆ p , the change in the elec-tromagnetic momentum from Eq. (29) is∆ P em = µ πr (cid:18) ( m × ∆ p ) − r (∆ p · r )( m × r ) (cid:19) . (34)The impulse of the magnetic field on the charge current in the electric dipole, using Eq. (22)and Eq. (4) for the field of the magnetic dipole, is ~ I e . dip . = ∆ p × B = µ πr (cid:20) m · r )(∆ p × r ) r − ∆ p × m (cid:21) . (35)In addition, the current of the electric dipole at r , in the quasi-static limit, induces a magneticfield at r ′ of B e . dip ( r ′ , t ) = − µ π ( r ′ − r ) × ˙ p | r ′ − r | . (36)14he impulse on the magnetic dipole is ~ I m . dip . = Z ∇ ′ [ m · B e . dip ( r ′ , t )] r ′ =0 dt = − µ π ∇ (cid:18) m · [( r ′ − r ) × ∆ p ] | r ′ − r | (cid:19) r ′ =0 = − µ π ∇ ′ (cid:18) ( r ′ − r ) · (∆ p × m ) | r ′ − r | (cid:19) r ′ =0 = − µ π (cid:20) ∆ p × m r − r · (∆ p × m ) r r (cid:21) . (37)Therefore, the total impulse on the system is ~ I = ~ I e . dip . + ~ I m . dip . = µ πr (cid:2) m · r )(∆ p × r ) + 2( m × ∆ p ) r + 3 r · (∆ p × m ) r (cid:3) = µ πr (cid:2) − ( m × ∆ p ) r + 3(∆ p · r )( r × m ) (cid:3) (38)where in the last equality we use the identity ( m · r )(∆ p × r ) + ( r · r )( m × ∆ p ) + (∆ p · r )( r × m ) + [ r · (∆ p × m )] r = 0 . (39)Eqs. (34) and (38) give ∆ P em + ~ I = 0 .
2. Electric dipole on the axis of a cylindrical solenoid and at the center of a rotating chargedspherical shell
The cases in which an electric dipole is on the axis of a cylindrical solenoid and at thecenter of a rotating charged uniform spherical shell were considered by Babson et al. . Inboth cases, P em = B × p . Furthermore the impulse imparted on the system when thedipole strength or the magnetic field is changed is also the same in both cases. Magnetic field changed – When the current that is the source of the magnetic field ischanged by ∆ B , a straightforward generalization of the calculation given in Babson et al. shows that the impulse given to the dipole is − ∆ B × p . Since ∆ P em = ∆ B × p , thisimplies ∆ P em + ~ I = 0. Electric dipole changed – When the electric dipole is changed by ∆ p , a straightforwardgeneralization of the calculation of Ref. 5 shows that the impulse to the dipole is ~ I dip = − B × ∆ p and the impulse to the solenoid is ~ I sol = B × ∆ p , giving a total impulse of ~ I = ~ I dip + ~ I sol = − B × ∆ p . Again, ∆ P em + ~ I = 0.15 . Electric dipole in a “solenoid” of parallel counter-propagating currents In the case where the magnetic field is created by currents in parallel conducting platesin the configuration shown in Fig. 2 and considered in Sec. II B 4, the electromagnetic mo-mentum is given by Eq. (19).
Magnitude of current in the solenoid is changed – When the current density is changedby ∆ j = ± ∆ j ˆ x in the plate at z = ± a , the magnetic field is changed by ∆ B = ∆ B ˆ y inbetween the plates, and therefore Eq.(19) yields∆ P em = p z ∆ B ˆ x . (40)The change in the magnetic field induces an electric field, which can be deduced by Amperianloops. From the symmetry of the problem, the electric field must be in the ˆ x direction, andindependent of the x -coordinate. Taking rectangular amperian loops which have a length L in the x -direction and d in the z -direction, the area through the loop is L d and the changein flux ∆
BLd is equal to R E · d l around the loop. This shows that R [ E x ( z ) − E x ( z + d )] dt =∆ Bd . Thus, the momentum of a dipole inside the solenoid | z | < a , has momentum impartedof the form ~ I = Z q [ E x ( z + d ) − E x ( z )] dt ˆ x = − qd ∆ B ˆ x = − ∆ B p z ˆ x . (41)This gives ∆ P em + ~ I = 0. Electric dipole changed – When the electric dipole changes, Eq.(19) gives∆ P em = ∆ p z B ˆ x . (42)The change in the p results in a current density j = δ ( r − r dip ) ˙ p (where r dip is the positionof the dipole) and R ˙ p dt = ∆ p . As in previous instances considered, when p changes, thereare two impulses on the system. First, there is the Lorentz force of the magnetic field B onthe current in the dipole, which gives an impulse of ~ I dip = Z dt Z d r j × B = Z dt ˙ p × B = ∆ p × B . (43)Then, there is the Lorentz force of the magnetic field induced by the change in the dipole(see Eq. (35)) on the currents in the parallel-plate solenoid. We treat each of the threedifferent components of the dipole, p = p x ˆ x , p y ˆ y and p z ˆ z , individually.16 p z : When the z -component of p is changes by ∆ p z , the impulse on the the dipole due tothe magnetic field is ~ I dip = ∆ p z B (ˆ z × ˆ y ) = − ∆ p z B ˆ x . The magnetic fields inducedby the current in the dipole are in the azimuthal direction and symmetric with respectto the azimuthal angle, so by symmetry the total impulse ~ I sol due to these magneticfields on the currents at z = ± a integrates to zero. Hence, ~ I = ~ I sol + ~ I dip = − ∆ p z B ˆ x .∆ p y : When the y -component of p is changed by ∆ p y , the impulse due to the magneticfield on the current in the dipole ~ I dip = ∆ p × B = 0. By symmetry, the magneticfields produced by the change in the electric dipole has x and z components. The x -component of the magnetic field does not impart an impulse on the currents at z = ± a , and the impulses from the z -components cancel due to symmetry. Therefore, ~ I = ~ I sol + ~ I dip = 0.∆ p x : For ∆ p = ∆ p x ˆ x , the impulse on the current in the dipole is ~ I dip . = p × B = ∆ p x B (ˆ x × ˆ y ) = ∆ p x B ˆ z . If we let r ′ be the position with respect to the dipole, then the currentin the dipole produces a magnetic field B e . dip ( r ′ , t ) given by Eq. (36) [with r = 0 inthat equation]. The current at z = + a has a surface current density of j = j ˆ x , so theforce per unit area is j ˆ x × B e . dip ( r ′ , t ). The impulse on the entire sheet is ~ I sol , + a = Z dx ′ Z dy ′ Z dt j × B e . dip ( r ′ , t )= µ π Z dx ′ Z dy ′ j ˆ x × (ˆ x ∆ p × r ′ ) r ′ = µ j ∆ p π Z dx ′ Z dy ′ ˆ x (ˆ x · r ′ ) − r ′ r ′ = − µ j ∆ p π Z dx ′ Z dy ′ y ′ ˆ y + z ′ ˆ z r ′ , (44)where z ′ > z -coordinate of the current sheet at z = + a relative to theelectric dipole. The ˆ y component vanishes because the integrand is anti-symmetricwith respect to y ′ . Changing variables of the integral R dx ′ R dy ′ to 2 π R ds ′ s ′ (takingadvantage of azimuthal symmetry of the integrand) gives, using a change of variables ζ = s ′ /z ′ , ~ I sol , + a = − µ j ∆ p Z ∞ ds ′ s ′ z ′ ( z ′ + s ′ ) / ˆ x = − B ∆ p Z ∞ dζ ζ (1 + ζ ) / ˆ x = − B ∆ p x x , (45)17here we have used B = µ j . For the current in the plane at the z = − a , thecalculation is identical except that the the current is j is in the opposite direction,and z ′ < ζ = s ′ / | z ′ | = − s ′ / ( − z ′ ) introduces a negativesign. These two sign changes negate each other, so ~ I sol , − a = − B ∆ p ˆ x , and the totalimpulse on the solenoid is ~ I sol = ~ I sol , + a + ~ I sol , − a = − B ∆ p x ˆ x . (46)Therefore ~ I sol + ~ I dip = 0.Comparing these results for ~ I caused by ∆ p x , ∆ p y and ∆ p z with the change in the fieldmomentum Eq. (42), we see that ∆ P em + ~ I = 0 for ∆ p in any direction. IV. DISCUSSION
An electric dipole in a static magnetic field and a magnetic dipole in a static electric fieldboth generally give rise to a non-zero electromagnetic field momentum P em = ǫ R d r E × B .While the P em for a point magnetic dipole in a static electric field can be expressed interms of magnetic moment and the local electric field at the position of the moment, namelyEq. (3), the same cannot be said for a point electric dipole in a static magnetic field; Eq. (2)is not true in general, even for locally uniform magnetic fields. To determine P em for a pointelectric dipole in a static magnetic field, one needs to know the full current distributionwhich produces the magnetic field, or the appropriate derivatives of the magnetic vectorpotential in the Coulomb gauge at the position of the electric dipole; see Eqs. (9a) – (9d).The difference between electric dipoles in static magnetic fields and magnetic dipoles instatic electric fields is due the difference in the sources of static electric and magnetic fields;static electric fields are produced by charges whereas static magnetic fields are produced bycharge currents.It has also been shown that when either the electric dipole moment or the current produc-ing the magnetic field is changed the mechanical impulse on the system ~ I is equal to the lossof electromagnetic momentum for an electric dipole in the magnetic field. The motivationfor showing this comes from Babson et al. , who stated in the conclusion section that “thereis no reason why this impulse should equal the momentum originally stored in the fields” inthese situations, thus unintentionally inferring that it is necessary to take into account the18idden momentum in the system in order for the total momentum to be conserved. Hid-den momentum is a form of mechanical momentum that associated with internally movingparts in the presence of a potential gradient, and is a relativistic effect. As shown in Ref. 5,for a stationary electric dipole in a static magnetic field, the hidden momentum must beincluded for the total linear momentum to vanish, as is required since the“center-of-energy”is stationary. However, there is no need to invoke hidden momentum in order to complywith the law of conservation of momentum when electromagnetic momentum is changed. It is interesting to note that, for a stationary electric dipole in a static magnetic field, wheneither the electric dipole moment or the current producing the magnetic field is changed, theovert mechanical momentum ( i.e. , the non-hidden component of the mechanical momentum)is conserved, even when there is non-zero electromagnetic momentum initially present in thesystem. By Newton’s second law, the change in the mechanical momentum is equal to theimpulse on the system; i.e. , ∆ P overt + ∆ P hidden = ~ I . Since ~ I = − ∆ P em , this implies that∆ P overt = − ∆[ P em + P hidden ]. But since P em = − P hidden in a static system, this gives∆ P overt = 0, independent of what the ∆ P em is.For example, considering a stationary electric dipole in the presence of a static magneticfield produced by a solenoid, if the current in the solenoid remains constant and dipolecompletely discharges, the dipole acquires momentum B × p , and the solenoid always endsup with a momentum of − B × p , regardless of what the initial P em is. To illustrate this,consider the cases shown in Figs. 3(a) – (c). In Fig. 4(a), the hidden momentum that isinitially contained in the solenoid is − B × p . When the dipole is discharged, the impulsereceived by the solenoid from the magnetic field generated by the discharging dipole is − B × p . In addition, the hidden momentum of − B × p that is initially in the solenoidis transformed into overt momentum when the dipole discharges, giving the solenoid a totalovert momentum of − B × p . In Fig. 4(b), when the dipole is discharged, the total impulsereceived by the solenoid currents is zero, but the solenoid had a hidden momentum of − B × p that is transformed into an overt mechanical momentum. In Fig. 4(c), the total impulsereceived by the solenoid currents is − B × p , but the solenoid had no hidden momentum. Onthe other hand, if the dipole remains constant and the current in solenoid is turned off, the thedipole acquires an overt momentum that is equal to the initial electromagnetic momentum,and the solenoid acquires an overt momentum that is equal to the initial hidden momentumin the solenoid, which is equal to the negative of the initial electromagnetic momentum.19hus, in all cases, the total change in the overt momentum of the dipole and solenoid iszero. Acknowledgments
The hospitality of Prof. Antti-Pekka Jauho at Aalto University, Helsinki, Finland, andProf. Christopher Stanton at the University of Florida, Gainesville, where some of thiswork was performed, and useful correspondence with Prof. David J. Griffiths are gratefullyacknowledged. The author also gratefully acknowledges an anonymous referee for usefulsuggestions that have improved the presentation of this manuscript.
Appendix A: Derivation of Eq. (6a)
To see under what conditions Eq. (6a) holds, let us first rederive this result for a finitevolume V , with surface S . For static electric and magnetic fields, P em = ǫ Z V d r E × B = − ǫ Z V d r ∇ V × B = ǫ Z V d r [ V ( ∇ × B ) − ∇ × ( V B )] = 1 c Z V d r V J − ǫ Z S d a V (ˆ n × B ) . (A1)where we have used E = −∇ V , ∇ × ( V B ) = ∇ V × B + V ∇ × B , ∇ × B = µ J , µ ǫ = c − ,and R V d r ∇× A = R S ˆ n × A , where ˆ n is the unit vector perpendicular to the surface. Letting V → ∞ , one obtains Eq. (6a), provided that the last term in Eq. (A1), ǫ R S d a V (ˆ n × B ) → S → ∞ . This condition is satisfied for an electric dipole in a uniform “solenoid” ofconsisting of infinite parallel plates, since as the radius r of the surface the radius of S increase, the integrand V (ˆ n × B ) ∼ r − whereas area of S where the integrand is non-zeroincreases as r . Appendix B: Derivation of Eq. (17) by direct integration of the electromagneticmomentum density
We first derive the electromagnetic momentum for a point charge that is outside a uniformcylindrical solenoid. The electromagnetic momentum of a point charge q that is a perpen-dicular distance s from the an infinitely long, infinitesimally small solenoid along the z -axis20arrying magnetic flux Φ m can be obtained by integrating the electromagnetic momentumover the length of the solenoid. The result is, in cylindrical coordinates, is P em = q Φ m πs ˆ φ . (B1)The P em for the charge q a distance s from the axis of of a solenoid of finite radius R ( < s )with uniform magnetic field B can be obtained by using the above result and integratingover the cross section of the solenoid. If we integrate along a shell of radius r and width dr , as shown in Fig. 4, by symmetry, the P em must be in the azimuthal direction. Themagnitude of the electromagnetic momentum due to this shell is dP em = qB ( r dr )2 π Z π cos ψr ′ dθ (B2)where r ′ , θ and ψ are shown in Fig. 4. (The factor cos ψ arises because we are taking theazimuthal component of dP em .) Using the law of cosines, which gives r = r ′ + s − sr ′ cos ψ and r ′ = r + s − rs cos θ , the integral can be evaluated to give dP em = qBr dr πs (cid:18) π + Z π s − r r + s − rs cos θ dθ (cid:19) = qBr drs . (B3)The total momentum is obtained by integrating from r = 0 to r = R , yielding P em = qBR s ˆ φ . (B4)Consider now a dipole which consists of two charges q and − q separated by a distance l ,in the limit l → ql → p . For p in the z -direction, the contributions to P em from thetwo charges in the dipole cancel each other, so P em = 0. For p = p s ˆ s , given by charges − q and q at s ˆ s and ( s + l )ˆ s respectively, Eq. (B1) yields P em = BR φ " lim l → ql → ps (cid:18) − qs + qs + l (cid:19) = BR φ lim ql → p s (cid:20) − qls (cid:21) = − BR s p s ˆ φ . (B5)For p = p φ ˆ φ , given by charges q and − q are at the same distance s from the z -axis but withazimuthal angles that are different by an angle ∆ φ = l/s , Eq. (B1) yields P em = BR s lim l → ql → pφ h q ˆ φ ( φ + ∆ φ ) − q ˆ φ ( φ ) i = BR s lim ql → p φ " q ls ∂ ˆ φ ∂φ = − BR s p φ ˆ s . (B6)21 ppendix C: Alternative derivation for P em due to electric dipole in between anti-parallel current sheets The result Eq. (19) for the momentum of an electric dipole in a constant magnetic fieldcreated by conducting parallel plates can also be obtained by integrating ǫ E × B over allspace. To do this, we divide the volume integral in between the parallel current-carryingplates into an integral across the entire x – y plane, followed by an integral from z = − a to z = + a ; i.e. , P em = ǫ Z a − a dz ′ E ( z ′ ) × B ˆ y ; (C1a)where E ( z ′ ) = Z ∞−∞ dx Z ∞−∞ dy E ( x, y, z ′ ) . (C1b)is the integral of the electric field over the x – y plane at z = z ′ . Below, we evaluate E ( z ′ ) forphysical dipoles p = q l , with a finite q and l , and then take the point dipole limit q → ∞ , l → ql → p in the end. We show here that E ( z ′ ) is zero for any physical dipole exceptwhen plane at z = z ′ lies in between the positive and negative charges of the dipole.By symmetry, the component E ( z ′ ) along the surface of the plane due to a staticpoint charge is be zero. (For example, for a point charge on the z -axis, E x,y ( x, y, z ′ ) = − E x,y ( − x, − y, z ′ ), so when E x or E y are integrated over the x – y , the contributions from x, y and − x, − y cancel each other.) By linear superposition, this is also true for electric fieldsdue to static dipoles. Therefore, the only possible non-zero component is the component of E that is perpendicular to the surface, which is electric flux through the surface. By Gauss’law, the magnitude of the electric field flux through the plane due to a point charge q is | q | / (2 ǫ ), independent of the distance of the charge from the plane. (This is because halfof the flux lines that emanate from the charge will pass through the plane – see Fig. 5(a).When both the charges of a dipole are on the same side of the plane, the contributions fromthe positive and negative charges cancel. The only case when E ( z ′ ) is non-zero is whenthe plane straddles the charges; see Fig. 5(b)). So, if the dipole consists of a charge − q at( x , y , z ) and q at ( x + l x , y + l y , z + l z ), then E ( z ′ ) is equal to − q/ǫ ˆ z for z < z ′ < z + l z ,and zero otherwise. Therefore, P em = ǫ Z z + l z z dz ′ (cid:18) − qBǫ (cid:19) ˆ z × ˆ y = ql z B ˆ x = p z B ˆ x , (C2)reproducing Eq. (19) 22 ppendix D: Example of P em = B × p for purely local currents Current and charge configurations that extend to infinity can at times yield ill-definedresults for quantities such as the electromagnetic field momentum. To show that the resultsdescribed in Sec. II B are not artifacts of the unbounded currents used in the examples,in this appendix a case is presented in which P em = B × p even when only purely localcurrents and charges are present.Consider a spinning sphere with a uniform surface charge, as in Refs. 5 and 3 and inSection II B 2, together with a torus which contains magnetic field. The magnetic field inthe torus is caused by a surface current on the torus, and hence both current and fieldare local. In the limit where the torus is infinitesimally small, the vector potential of themagnetic of the torus in the Coulomb gauge has the same form as the magnetic field of apoint dipole; namely, in spherical coordinates (with the z -axis perpendicular to the planeof the torus) A tor ,r = 2 Cr cos θ (D1a) A tor ,θ = Cr sin θ (D1b) C = µ V tor I π (D1c)where V tor is the volume of the torus and I is the current times number of windings of thewire around the torus.The spinning sphere creates a uniform magnetic field B s inside the sphere, and the elec-tromagnetic momentum due to the electric dipole and the sphere is P em , s = B s × p . Forsimplicity, let the dipole be at a distance R from the torus and on the axis that passesthrough the torus, as shown in Fig. 6. The contribution to the electromagnetic momentumof the magnetic field inside the torus and the electric field of the dipole p = p r ˆ r + p θ ˆ θ (where the origin is taken to be the position of the torus), is P em , tor = CR − ( − p r ˆ r + p θ ˆ θ ),by Eq. (9a). Therefore, the total electromagnetic momentum is P em , s + P em , tor = B s × p ,in general. ∗ Electronic address: [email protected] D. J. Griffiths,
Introduction to Electrodynamics (Prentice-Hall, New Jersey, 2010), 4th ed. J. D. Jacskon,
Classical Electrodynamics (Wiley, New York, 1999), 3rd ed. K. T. Mcdonald, “Electromagnetic momentum of a capacitor in a uniform magnetic field,” < ∼ mcdonald/examples/cap momentum.pdf > (June 18, 2006; updatedSeptember 4, 2014; 14 pp). Andre Gsponer, “On the electromagnetic momentum of static charge and steady current distri-butions,” Eur. J. Phys. , 1021–1042 (2007). David Babson, Stephen P. Reynolds, Robin Bjorkquist and David J. Griffiths, “Hidden momen-tum, field momentum, and electromagnetic impulse,” Am. J. Phys. , 862–833 (2009). David J. Griffiths, “Resource Letter EM-1: Electromagnetic Momentum,” Am. J. Phys. ,7–18 (2012). W. H. Furry, “Examples of Momentum Distrubtion in the Electromagnetic Field and Matter,”Am. J. Phys. , 621–636 (1969). This was also pointed out in J. Franklin, “The electromagnetic momentum of static charge-current distributions,” Am. J. Phys. , 869 (2014). It should be noted that the present authordoes not agree with the main conclusions of Franklin’s paper. See also D. J. Griffiths and V.Hnizdo, “Comment on ‘The electromagnetic momentum of static-charge distributions,’ by Jer-rold Franklin [Am. J. Phys. 82, 869875 (2014)],” Am. J. Phys. , 279 (2015), and J. Franklin,“Response to ‘Comment on “The electromagnetic momentum of static charge-current distribu-tions” [Am. J. Phys. 83, 279 (2015)]’ ”, Am. J. Phys. , 280 (2015). David J. Griffiths, private communication. For a review of hidden momentum, see Ref. 6 and references therein. J. J. Thomson,
Electricity and Matter , (Charles Scribner’s Sons, New York, 1904) pp. 30–33. M. G. Calkin, “Linear Momentum of Quasistatic Electromagnetic Fields,” Am. J. Phys. ,921–925, (1966). David J. Griffiths, “Dipoles at rest,” Am. J. Phys. , 979–987 (1992). David J. Griffiths, “The field of a uniformly polarized object,” Am. J. Phys., , 187 (1992). See e.g. , Ref. 1, p. 247. See, e.g. , John Michael Finn,
Classical Mechanics (Infinity Science, Hingham, MA, 2008) p. 80. From the partial differential equation point of view, the equations ∇ · A = 0, ∇ × A = B and ∇ × B = µ J are equivalent to ∇ A = − µ J ; i.e. , each component of A satisfies Pois-son’s equation. When the magnetic field is produced by surface currents the appropriate A is btained by matching boundary conditions across each surface current in the same way thatthe scalar potential must be matched across surface charges. Each component of the magneticvector potential must be continuous across the boundary, and for a surface current density K , ∂ A k , in /∂r ⊥ − ∂ A k , out /∂r ⊥ = µ K . See e.g. , Section 5.4.2 in Ref. 1. In fact, this argument was initially used to identify R ρ ( r ) A ( r ) d r (for A ( r ) in the Coulombgauge) as the electromagnetic momentum. See Ref. 12. This can be derived in the following way. Assume that there is a charge q at l ˆ z and − q at theorigin, and the current flows from one charge to another along an infinitesimally small wire alongthe z -axis. The current density is J ( r ) = ˙ q ˆ z δ ( x ) δ ( y )[ θ ( l − z ) θ ( z )] = ( ˙ ql )ˆ z δ ( x ) δ ( y )[ θ ( l − z ) θ ( z )] /l .Let l →
0. Then, [ θ ( l − z ) θ ( z )] /l → δ ( z ) and ˙ ql ˆ z = ˙ p , so J = ˙ p δ ( r ). David J. Grfiiths, “Dynamic dipoles,” Am. J. Phys. , 867–872 (2011). Proof: Applying r · to the left hand side of Eq (39) and using r · (∆ p × r ) = ( r × m ) · r = 0 gives( r · r )[( m × ∆ p ) · r ] + [ r · (∆ p × m )]( r · r ) = 0 (D2)Applying r × to the left hand side of Eq. ((39) and using A × ( B × C ) = B ( A · C ) − C ( A · B ),gives ( m · r )[ r × (∆ p × r )] + ( r · r )[ r × ( m × ∆ p )] + (∆ p · r )[ r × ( r × m )]= ( m · r )[( r · r )∆ p − (∆ p · r ) r ] + ( r · r )[( r · ∆ p ) m − ( r · m )∆ p ]+ (∆ p · r )[( m · r ) r − ( r · r ) m ] = 0 . Since r · W = 0 and r × W = 0 ⇔ W = 0 (for r = 0), this proves Eq. (39). (The identitycan also be confirmed by the “brute force” method of setting ∆ p = ∆ p ˆ x , m = m x ˆ x + m y ˆ y and r = x ˆ x + y ˆ y + z ˆ z , calculating the left hand side of Eq. (39) component by component andshowing that each is zero.). The reader might draw the incorrect inference if the term “this impulse,” which refers to theimpulse received by “some element(s) in the system,” is interpreted to mean the total impulsereceived by system. Sidney Coleman and J. H. Van Vleck, “Origin of ‘Hidden Momentum Forces’ on Magnets,”Phys. Rev. , 1370–1375 (1968). In fact, in Ref. 5, Table I would have been correct if the the magnetic field was created bycounter-propagating currents in parallel plates which are oriented in the same direction asthe capacitor plates. (In this case, there is no net impulse on the currents when the dipole ischarges.) Table II is inconsistent because it does not include the impulse of the magnetic fieldproduced by the current of the discharging capacitor on the solenoid. When that is included,and the middle entry is relabelled “Momentum delivered to capacitor and solenoid,” all entriesin the right hand column on that table would be − BQ ˆ x . The fact that the total ∆ P overt = 0 can also be seen the in tables IV and V in Ref. 5, which showthe momentum transferred to the electric dipole and solenoid or spinning charged sphere wheneither the dipole moment vanishes or the current stops. The label p hid (for hidden momentum)in the second and third rows of these tables is somewhat misleading, because there is no longerany hidden momentum once the dipole discharges completely or the current stops. The hiddenmomentum would have been transformed to overt mechanical momentum. Thus, the entries inthe middle two columns of the second and third rows of those tables give the overt mechanicalmomenta of the dipole and sphere/solenoid after the dipole discharges or the current stops.Note that in all cases the final total P overt is zero. Ben Yu-Kuang Hu, “Introducing electromagnetic momentum,” Eur. J. Phys. , 873-881 (2012). N. J. Carron, “On the fields of a torus and the role of the vector potential,” Am. J. Phys. ,717–729 (1995). IG. 1: Configuration of electric ( p ) at the origin and magnetic ( m ) dipole at R ˆ x which demon-strates that Eqs. 2 and 3 give inconsistent results for P em . The dotted line labeled E is the electricfield due to p at m , and the dotted line labeled B is the magnetic field due to m at p .FIG. 2: Electric dipole in a uniform magnetic field produced by currents in parallel plates. Theplates are in the x – y plane at z = ± a . The current in the top (bottom) layer is in the positive(negative) x -direction, which produces a magnetic field in the positive y -direction. The lateralextent of the plates is assumed to be much, much larger than the separation 2 a of the plates. IG. 3: Capacitors in the uniform magnetic fields produced by (a) infinite cylindrical solenoidwith uniform current density around the circumference (b) infinite parallel plates with currents inopposite directions, and perpendicular the dipole of the capacitor and (c) same as (b) but withcurrents densities parallel and anti-parallel to the electric dipole moment of the capacitor. In thesethree cases, the electromagnetic momenta P em is (a) B × p , (b) B × p and (c) 0, where p isthe electric dipole moment of the capacitor and B is the magnetic field. Figure (d) shows theregions around an electric dipole that is oriented in the positive z -direction where the z -componentof the electric field of the dipole is positive (shaded regions) and negative (white regions). Thus,when the magnetic field is in the positive y -direction ( i.e. , into the paper), the x -component of theelectromagnetic momentum is negative in the shaded regions and positive in the white regions. IG. 4: Geometry of integration to obtain the electromagnetic momentum of a point charge outsidea uniform solenoid. IG. 5: (a) Electric flux lines from a positive point charge piercing an infinite plane. Since theplane is infinite, all the electric field lines with a component that is directed towards the plane willpierce the plane. This implies that half the electric field lines that emanate from the point chargewill pierce the plane. Therefore, from Gauss’ law, for an infinite plane (and a point charge q whichis not on the plane itself), the flux through the plane R E · d r = q/ǫ . When there are two chargesof equal magnitude and opposite sign on the same side of the plane, the the contributions of the+ and − charges cancel each other and the electric flux through the plane is zero,. (b) Electricflux lines for an infinite plane in between a positive (black dot) and negative (white dot) chargesof equal magnitude. The contribution for each charge to the flux is equal, and therefore the totalflux is q/ǫ . IG. 6: A situation where the electric current does not extend to infinity, and P em = B × p , where B is the locally uniform magnetic field around electric dipole p . The electric dipole (indicated bythe arrow) is in a uniform magnetic field created by a spinning sphere with a uniform surfacecharge, and is a distance R away from and on the axis of a torus that contains magnetic flux.away from and on the axis of a torus that contains magnetic flux.