Time dependent electromagnetic fields and 4-dimensional Stokes' theorem
aa r X i v : . [ phy s i c s . g e n - ph ] A ug Time dependent electromagnetic fields and 4-dimensional Stokes’theorem
Ryan Andosca ∗ Department of Physics, California State University Fresno, Fresno, CA 93740-8031, USA
Douglas Singleton † Department of Physics,California State University Fresno,Fresno, CA 93740-8031, USAandICTP South American Institute for Fundamental Research,UNESP - Univ. Estadual Paulista Rua Dr. Bento T. Ferraz 271,01140-070, S˜ao Paulo, SP, Brasil (Dated: September 1, 2016)
Abstract
Stokes’ theorem is central to many aspects of physics – electromagnetism, the Aharonov-Bohmeffect, and Wilson loops to name a few. However, the pedagogical examples and research workalmost exclusively focus on situations where the fields are time-independent so that one needonly deal with purely spatial line integrals ( e.g. H A · d x ) and purely spatial area integrals ( e.g. R ( ∇ × A ) · d a = R B · d a ). Here we address this gap by giving some explicit examples of howStokes’ theorem plays out with time-dependent fields in a full 4-dimensional spacetime context.We also discuss some unusual features of Stokes’ theorem with time-dependent fields related togauge transformations and non-simply connected topology. ∗ Electronic address: [email protected] † Electronic address: [email protected] . STOKES’ THEOREM IN 3 AND 4 DIMENSIONSA. 3D Stokes’ theorem Stokes’ theorem is used in many areas of physics, particularly in electricity and magnetismwhere it gives a connection between the electromagnetic potentials ( i.e. φ and A ) and thefields ( i.e. E and B ). Through Stokes’ theorem, the connection between the line integral ofthe 3-vector potential, A , and the area integral of the magnetic field, B , is I C A · d x = Z S ( ∇ × A ) · d a = Z S B · d a , (1)The subscripts C and S on the integrals indicate line and surface integrals respectively. Theclosed contour C is spanned by an infinite number of possible surfaces S . The contour hasa direction of traversal which is determined by the direction of d x , and this determines thedirection of the vector area, d a , of the surface, S , via the right-hand-rule (wrap the fingersof the right hand in the direction that the contour is traversed and the thumb points in thedirection of the vector area). This issue of the directionality of the area in Stokes’ theoremwill be important (but less familiar) when we move from a purely spatial area to a spacetime area. In the rest of the paper we will drop the subscripts C and S if there is no confusionas to whether the integral is a line or surface integral.The usual pedagogical examples of (1) involve time-independent 3-vector potentials andmagnetic fields. One common example in cylindrical coordinates is a solenoid of radius R ,with the axis of the solenoid and the magnetic field along the z -axis. The magnetic field forthis setup is B = B ˆ z for ρ < R and B = for ρ ≥ R , (2)where B is the constant value of the magnetic field. The vector potential has the followingform A = B ρ ϕ for ρ < R and A = B R ρ ˆ ϕ for ρ ≥ R . (3)For a contour that is a circle of radius ρ > R which goes around the solenoid, the area integralin (1) yields R B · d a = B πR . For the line integral one similarly finds H A · d x = B πR .Thus, for this setup, Stokes’ theorem works out to give I A · d x = B πR = Z B · d a . (4)2he above example is also the heart of the time-independent Aharonov-Bohm effect [1]where one performs the usual quantum mechanical two-slit experiment but with an infinitesolenoid, described by the above B and A , inserted between the slits of the experiment. Forexample, say one sends electrons at a two-slit set up. The electrons will form an interferencepattern on a screen placed down stream from the two slits due to the quantum mechanicalwave nature of the electrons. In this simple two-slit experiment the interference occursdue to the phase difference of the wavefunction coming from the two different slits, andthis comes from the path length difference from each slit to whatever point on the downstream screen one is interested in. The Aharonov-Bohm effect comes from placing an infinitesolenoid between the slits. Classically one does not expect any change since, classically,charged particles only respond to the magnetic field, via v c × B , not the vector potential.However quantum mechanically, due to minimal coupling, the electrons will pick up a phase, e ~ c R γ A · d x , when traveling along a contour γ . Now for some particular point on the screenthere will be two paths leading from each slit to that point – call these two paths γ and γ . Now in addition to the phase difference due to the path length difference there will bean additional phase difference coming from the line integrals of the vector potential namely e ~ c R γ A · d x − e ~ R γ A · d x = e ~ c H A · d x = e ~ c R B · d a . Thus one gets a phase shift of the standardinterference pattern of the two-slit experiment, which is given by e ~ c H A · d x = e ~ c R B · d a .This is the heart of the time-independent Aharonov-Bohm experiment – that one gets aphase shift in the interference pattern of the two-slit experiment despite the fact that theelectrons move in a region which if B field free, but where the vector potential, A , is non-zero. A fuller and more detailed account of the Aharonov-Bohm effect can be found insection 3.4 of reference [2]. Due to the close connection between Stokes’ theorem and theAharonov-Bohm effect we have in mind that the contours and surfaces discussed in thispaper in connection with Stokes’ theorem should be those associated with the paths andsurfaces of particles in an Aharonov-Bohm experiment.The magnetic field in (2) can also be obtained from a vector potential of the followingform A ′ = (cid:20) B ρ − B R ρ (cid:21) ˆ ϕ for ρ < R and A ′ = for ρ ≥ R . (5)This form of the vector potential is related to the original form given in (3) by the following3auge transformation A ′ → A + ∇ χ with χ = − B R ϕ χ is non-single valued and A ′ , for ρ < R , has a ρ singularity.These features (singular vector potential and non-single valued gauge function) indicate thatwhile the B field produced by the two different vector potentials in (3) and (5) is the samethe physical situation is different – for the vector potential (5) one has the original solenoidof radius R plus an idealized, infinitely thin solenoid placed along the symmetry axis with acurrent flowing in the opposite direction of the original solenoid. We will show shortly howto deal with this singularity in A ′ . One might naively conclude that Stokes’ theorem fails inthis new gauge. The area integral, R B · d a = B πR , is still the same since the B field is stillgiven by (2). However, for the circular contour with radius ρ > R , apparently H A ′ · d x = 0since A ′ = 0. The problem is the singularity at ρ = 0 in A ′ inside the solenoid. Due to this“puncture”, the space is said to have a non-simply connected topology. One can not spanthe simple circle contour with a surface that includes ρ = 0 since this point is no longer partof the space. To deal with this “puncture” at ρ = 0, we need to deform the simple circlecontour into the more complex contour in Fig. (1), which avoids ρ = 0. The outer circularcontour gives zero ( R outer A ′ · d x = 0) since A ′ = 0 for ρ > R . The line integrals for the tworadial segments cancel. The only non-zero contribution is from the inner circular contourwith an infinitesimal radius ǫ ≪ R . Since the inner contour is traversed in the oppositedirection from the outer contour, the inner line integral has a negative sign relative to theouter line integral. Putting it all together one finds that for the contour in Fig. (1). I A ′ · d x = Z inner A ′ · d x = − Z π (cid:20) B ǫ − B R ǫ (cid:21) ǫdϕ = B πR − B πǫ → B πR . (7)In the last step we have let ǫ →
0. This removes the singularity in A ′ at ρ = 0, and we find H A ′ · d x = R B · d a for the contour in Fig. (1), thus satisfying Stokes’ theorem. B. 4D Stokes’ theorem
The discussion of the previous subsection was in terms of the 3-vector potential. Since A is part of the 4-vector A µ = ( φ, A ), we can generalize the first expression in equation (1) as I A · d x → I A µ dx µ = − I φ cdt + I A · d x , (8)4 ε FIG. 1:
The modified contour for the gauge potential from (5) The inner circle has a radius ǫ ≪ R and the outer circle aradius ρ > R . This generalization was noted already in [1] and is discussed in more detail in references[2, 3]. Equation (8) involves both a spatial line integral as well as a time integral, thusmaking it ideal for time-dependent situations. In the section below we will make this laststatement more concrete by looking at the specific example of a time-dependent solenoidwith various contours. It should be noted that throughout the paper we use the ( − + ++)metric signature rather than the (+ − −− ) signature used in references [1–3]. Next, the5ight hand side of (1), which involves the magnetic field, can be generalized as [2, 3] Z B · d a → Z E · d x cdt + Z B · d a = 12 Z F µν dσ µν , (9)where in the last expression F µν = ∂ µ A ν − ∂ ν A µ is the Maxwell field strength tensor and dσ µν is an infinitesimal spacetime area. The signs of the area integrals of the fields of theintermediate expression in (9) can be checked using the expression for the contour integralsof the potentials in (8) in the following way: dropping the scalar potential part of (8) andthe electric area integral of (9) one recovers the usual result H A · d x = R B · d a ; in turndropping the 3-vector potential line integral of (8) and magnetic field area integral part of(9), one recovers the standard relationship φ = − R E · d x (inside the time integration).In the next section, we apply Stokes’ theorem to an explicit example which involves time-dependent fields and potentials and which uses the full four-vector potential, A µ = ( φ, A ). II. THE INFINITE, TIME-DEPENDENT SOLENOID
In this section we study the case of an infinite solenoid of radius R with a time-dependentmagnetic flux. We first consider a spacetime loop which does not enclose the solenoid. Forconcreteness and simplicity, we take the time dependence to be linear so that the 4-vectorpotential takes the form A µ = (cid:18) , , B tρ ϕ , (cid:19) for ρ < R and A µ = (cid:18) , , B tR ρ ˆ ϕ , (cid:19) for ρ ≥ R . (10)The scalar potential is zero ( i.e. φ = 0). This is similar to the expression given in (3) butwith B → B t . Note that in (10) B is a rate of magnetic field strength change, while in(3) B is just the magnetic field strength. The magnetic field connected with the vectorpotential in (10) is B = B t ˆ z for ρ < R and B = for ρ ≥ R , (11)and the electric field connected with (10) is E = − B ρ c ˆ ϕ for ρ < R and E = − B R ρc ˆ ϕ for ρ ≥ R . (12)The linear time dependence of the magnetic flux yields the above simple fields – A , B and E . The potential and fields in (10) (11) and (12) correspond to those given in [4] if the If one considers sinusoidal time dependence the magnetic field will be non-zero outside the solenoid andthe forms of both the electric and magnetic fields will involve Bessel and Neumann functions. [3] t = 0 whereas thepotentials and fields above are linearly increasing for all time, but as t becomes large theresults of [4] yield those given above after units conversion.The 3-vector potential in (10) can be “redistributed” to form a new scalar and 3-vectorpotential, which gives the same electric and magnetic fields. The new 4-vector potential is A ′ µ = ( φ ′ , A ′ ) = (cid:18) B R ϕ c , , (cid:20) B tρ − B R t ρ (cid:21) ˆ ϕ , (cid:19) for ρ < R (13)and A ′ µ = ( φ ′ , A ′ ) = (cid:18) B R ϕ c , , , (cid:19) for ρ ≥ R , (14)The two forms of the potentials for the time-dependent solenoid are related by a gaugetransformation given by A ′ µ → A µ + ∂ µ χ with χ = − B R ϕt A ′ has a ρ singularity and thegauge function, χ , is non-single valued.We begin by first evaluating H A µ dx µ for these two gauges as given in equations (10) and(13) (14), on the closed spacetime path shown in figure (2). A. Spacetime loop integral for the 4-vector potential from (10)
The evaluation of the loop integral H A µ dx µ for the potential in (10) is split into 6 seg-ments, as shown in Fig. (2). Since in the end we want to connect our discussion of thetime-dependent Stokes’ theorem with the time-dependent Aharonov-Bohm effect [5], we willtake our paths to be those traversed by a particle and thus the path lengths will be param-eterized in terms of the particle’s velocity. We will take the linear speed of the particle tobe the same along every path. In general Stokes’ theorem could apply to paths which arenot physically realizable paths for a particle. For example, even for the time-dependent caseone could take a circular path around the solenoid at one instant in time. In this case onewould just get H A · d x = R B · d a but the circular path would not be one that a real particlecould travel. In addition, one would not touch on the time-dependent nature of this casewhich, is one of the central points of this article.Stokes’ theorem with time-independent fields has closed spatial loops when evaluating H A · d x . Similarly, the time-dependent case has closed spacetime loops. To create a closed7 IG. 2:
Closed loop path outside a solenoid with a time-changing magnetic flux. spacetime loop we consider two particles which begin at the same spacetime point, moveapart and then come back together. We then time reverse the path of one of the particlesand add this to the result of the path of the other particle. This is also the procedure for theAharonov-Bohm effect with time-independent fields. The paths of the two moving particlesare shown, with paths 1, 2 and 3 for one particle and paths 4, 5 and 6 for the other particle.The arrows indicate the real directions of travel of each particle, starting from the middle ofthe inner arc at radius ρ and ending at the middle of the outer arc at radius ρ . To obtaina closed spacetime path we apply the time reversal operator, T [ ... ], to the results of the lineintegrals for paths 4, 5 and 6. The operation of T [ ... ] takes t → − t . The results of applying T [ ... ] to other physical quantities can be found in section 6.10 of reference. [6] The oddnessof A under time reversal ( i.e. T [ A ] = − A ) and the evenness of x under time reversal ( i.e. T [ x ] = x ) implies that T [ R A · d x ] = − R A · d x . Thus the operation of T [ ... ] in segments4, 5 and 6 has the effect of multiplying the results of these line integrals by −
1. To get theclosed spacetime loop we add the time reversed paths 4, 5 and 6 to the results from paths1, 2 and 3. Paths 1, 3, 4 and 6 cover an angle of ϕ /
2, with 1 and 3 going between 0 and ϕ /
2, and with 4 and 6 going between 0 and − ϕ /
2. The linear speed of the particles istaken to be constant throughout so that the angular speed along the inner paths, 1 and 4, islarger than the angular speed along the outer paths, 3 and 6. The detailed definitions andcalculations for each segment are given in Appendix I. Using these results we find that the8losed spacetime loop is I A µ dx µ = (cid:18)Z + Z + Z + T (cid:20)Z + Z + Z (cid:21)(cid:19) A · d x = − B R ϕ ω (cid:20) ρ ρ + 4( ρ − ρ ) ρ ϕ + 1 (cid:21) . (16)In (16) ω is the angular speed at which paths 1 and 4 from Fig. (2) are traversed; ρ is thedistance from the center of the solenoid to the inner paths 1 or 4; ρ is the distance fromthe center of the solenoid to the outer paths 3 and 6. It is important to note for later that,when viewed from above, the direction of traversal of the closed loop is clockwise. B. Spacetime loop integral for the 4-vector potential from (13) and (14)
The evaluation of the loop integral for the 4-potential in (13) now just involves thescalar potential H A µ dx µ → − H φ ′ cdt since A ′ = 0 outside the solenoid. The details of thecalculation for the 6 segments can be found in Appendix II. Collecting together the resultsfor the time-reversed paths 4, 5, and 6 (which, as in the previous case, changes the sign forthese integrals as given in Appendix II) and adding these to the results from paths 1, 2 and3 gives the closed spacetime loop result for the potential in this gauge as I A ′ µ dx µ = − I φ ′ cdt = − (cid:18)Z + Z + Z + T (cid:20)Z + Z + Z (cid:21)(cid:19) φ ′ cdt = − B R ϕ ω (cid:20) ρ ρ + 4( ρ − ρ ) ρ ϕ + 1 (cid:21) . (17)Comparing (16) with (17) we see that the two different gauges give the same result for H A µ dx µ , as is expected for this gauge invariant quantity. We next calculate the spacetimearea integral of the fields i.e. R F µν dσ µν = R E · d x cdt + R B · d a . C. Spacetime area integral for the fields from (11) and (12)
The evaluation of the spacetime area integral R F µν dσ µν , for the fields in (11) and (12)on the spacetime area implied by Fig. (2), reduces to R E · d x cdt since B = 0 outside thesolenoid. The detailed calculations for R E · d x cdt are given in Appendix III.We need to combine the results for the spacetime areas associated with the segments 1,2, and 3 with the time reversed spacetime areas associated with the time-reverse segments9or 4, 5 and 6. From [6], E and x are even under time reversal ( i.e. do not change sign)whereas t is odd ( i.e. changes sign). Thus applying T to R E · d x cdt changes the sign T [ R E · d x cdt ] = − R E · d x cdt . This means that the time-reversed spacetime areas associatedwith segments 4, 5, and 6 are equivalent to the spacetime areas associated with segments 1,2, and 3. With all this in mind the total spacetime area integral is Z E · d x cdt = (cid:18)Z + Z + Z + T (cid:20)Z + Z + Z (cid:21)(cid:19) E · d x cdt = − B R ϕ ω (cid:20) ρ ρ + 4( ρ − ρ ) ρ ϕ + 1 (cid:21) (18)We see that this result agrees with the spacetime line integral of the 4-vector potential from(16) or (17). Thus we find that, for this case, the time-dependent 4D version of Stokes’theorem is satisfied. In the next subsection we examine the case in which the path enclosesthe solenoid and therefore the spacetime area has a magnetic field contribution. D. 4D Stokes’ Theorem for a path enclosing the solenoid
We now consider a closed spacetime loop that encloses the solenoid as shown in Fig. (3).We will use the the results of the preceding subsections and the appendices to perform thecalculations. To enclose the solenoid with a spacetime path we eliminate paths 2, 3, 5 and6 and then extend paths 1 and 4 around to form a closed loop.For the form of the 4-vector potential given in (10), the closed spacetime loop integral, H A µ dx µ , can be obtained using the line integrals of paths 1 and 4 (given by equations (34)and (37) respectively) with ϕ = 2 π Z A · d x = B R π ω and Z A · d x = − B R π ω . (19)From (19), one can obtain I A µ dx µ = + (cid:18)Z A · d x + T Z A · d x (cid:19) = B R π ω . (20)For this gauge H A µ dx µ = − H φ cdt + H A · d x → H A · d x so only the 3-vector potentialcontributes. Also H A · d x = R A · d x − R A · d x since the direction of path 4 must betime-reversed to obtain a closed spacetime loop. It is worth noting that, when viewed fromabove, the path closes in a counterclockwise sense. This is the reverse of the closed path10 IG. 3:
Closed loop path that encloses the solenoid with the time-changing magnetic flux. This closed loop is obtained fromthe loop in figure (2) by discarding paths 2, 3, 5 and 6 and by extending paths 1 and 4 using ϕ = 2 π . ε ₄ ε ₁ FIG. 4:
The spacetime loop for the case when the 4-vector potentials are given by (13) and (14). The singularity at ρ = 0 isexcised by the small loop of radius ǫ ≪ R . The initial direction of integration of the different paths are shown. To get a closedspacetime loop we time reverse path 4, path l and path ǫ . in Fig. (2). This will have important consequences when we discuss the “direction” of thespacetime area associated with the spacetime contours.Next we examine the form of the 4-vector potential given in (13) and (14), obtained fromthe form of the 4-vector potential given in (10) by the gauge transformation (15). Due tothe ρ singularity for the ρ < R ρ = 0 needs to be removed using a contour similar to the one used in time-independent case(see Fig. (1)). The contour that we use now is shown in Fig. (4), with arrows indicatingthe direction of travel prior to time reversal of the paths 4, l and ǫ . We have contributions12rom the scalar potential only from paths 1 and 4 on the outer loop. The results from paths1 and 4 (given by (40) and (43) respectively) with ϕ = 2 π are Z φ ′ cdt = B R π ω and Z φ ′ cdt = − B R π ω . (21)Next we need to take into account the small interior circular path which we take to be ofradius ǫ ≪ R . Since φ ′ = B R ϕ c does not depend on ρ , the results for the interior paths willbe the same as those in (21) giving Z ǫ φ ′ cdt = B R π ω and Z ǫ φ ′ cdt = − B R π ω . (22)The subscripts ǫ , ǫ on the integrals above indicate that these are the interior half circlepaths of radius ǫ ≪ R corresponding to the outer paths 1 and 4 respectively. We haveassumed the traversal of the inner circle is at the same angular velocity, ω , as the outercircle. We will see below that this is the only value for the angular velocity that is able toexcise the singularity.At this point we calculate the scalar potential contribution to H A ′ µ dx µ . From (21) (22),we get I φ ′ cdt = Z φ ′ cdt + Z ǫ φ ′ cdt + Z l φ ′ cdt + T (cid:20)Z φ ′ cdt + Z ǫ φ ′ cdt + Z l φ ′ cdt (cid:21) = B R π ω − B R π ω + Z l φ ′ cdt + B R π ω − B R π ω − Z l φ ′ cdt = 0 (23)Although we did not explicitly calculate R l φ ′ cdt and R l φ ′ cdt , it is clear that they are thesame in magnitude and cancel after we apply time reversal.The only non-zero and uncanceled contribution to H A ′ µ dx µ comes from the interior 3-vector part of (13). The − B R t ρ ˆ ϕ piece gives the same result as (19) since the extra negativein this part of the 3-potential is balanced by the fact that the inner circular path is traversedin the opposite direction as the outer circle after time reversal. So we have Z ǫ A ′ · d x = B R π ω − B ǫ π ω and Z ǫ A ′ · d x = − B R π ω + B ǫ π ω . (24)There is an additional contribution from the B ρt ˆ ϕ term in the 3-potential relative to (19),but in the limit ǫ → I A ′ µ dx µ = + (cid:18)Z ǫ A ′ · d x + T Z ǫ A ′ · d x (cid:19) = B R π ω − B ǫ π ω → B R π ω , (25)13here at the end we have taken ǫ →
0. Thus by excising the singularity that exists in thisgauge we find that the results for the closed spacetime integrals given in (20) and (25) agree.We note that the interior circular path and the two paths in the ˆ ρ direction are not phys-ical paths that the particles traverse, but are simply artifacts used to excise the singularityat ρ = 0. Additionally, as we remarked above, we needed to arbitrarily take the traversalof the inner circle at the same angular velocity as the outer circle so that the singularitywould be removed. This arbitrariness is absent from the time-independent case. In the time-dependent case, the “strength” of the ρ singularity is ∝ B R t , and thus changes linearly intime.We now calculate the right hand side of Stokes’ theorem – the field version of the spacetimearea integral, R F µν dσ µν = R E · d x cdt + R B · d a . In this case, we use the two spacetimesurfaces shown in Fig. (5), both of which span the spacetime path used in evaluating H A µ dx µ .For the spacetime area that is the side of the spacetime cylinder – the right side of Fig.(5) – only the electric field will contribute i.e. R F µν dσ µν → R E · d x cdt . We can obtainthis electric piece using our results from (46) and (49) with ϕ = 2 π , but with one subtlety:we need to reverse the signs relative to those given in (46) and (49) so that we have Z E · d x cdt = + B R π ω ; Z E · d x cdt = − B R π ω . (26)The reason for the sign reversal is as follows: when we form a closed spacetime loop fromthe two contours in Fig. (3), the direction of the spacetime loop is counterclockwise whenthe solenoid is viewed from above. As previously mentioned, this is the reverse of thepaths in Fig. (2). For purely spatial examples of Stokes’ theorem, reversing the direction oftraversal for the path reverses the direction of the 3-vector area via the right-hand-rule. Here,although there is no equivalent right-hand-rule for the spacetime area, we nevertheless needto reverse the spacetime area “direction” when the closed spacetime path is traversed in theopposite direction. This might be viewed as an extension of the right-hand-rule to spacetimepaths and areas. One can obtain this result on the “direction” of the spacetime area lessheuristically using differential forms and wedge product notation. Reviewing differentialforms and wedge product notation is outside the scope of this article, but the interestedreader can find nice expositions on this in the textbooks by Frankel [7], Felsager [8] andRyder. [2] To get the spacetime area contribution associated with the closed spacetime14 yx t yx FIG. 5:
Two of the many possible spacetime areas associated with the spacetime loop, H A µ dx µ . The spacetime area on theleft hand side ( i.e. the slanted top of the spacetime cylinder) gets a contribution only from the magnetic field. The spacetimearea on the right hand side ( i.e. the side of the spacetime cylinder) gets a contribution only from the electric field. R R E · d x cdt since E and d x are even under time reversal, but dt is odd. We then add thetime reversed result for path 4 to the result for area 1 which gives Z E · d x dt = (cid:18)Z + T Z (cid:19) E · d x cdt = B R π ω . (27)This result agrees with the results for the loop integral for the 4-vector potentials in the twodifferent gauges, given in (20) and (25).We now calculate R F µν dσ µν for the spacetime area on the left hand side of Fig. (5) i.e. the slanted “top” of the spacetime cylinder. For this spacetime area, only the magneticfield contributes i.e. R F µν dσ µν → R B · d a . For the previous case, shown in Fig. (2), ourspacetime loop enclosed a region where B = 0. Now from the right hand side of Fig. (5)we can see that the spacetime area will have B = 0 so we expect R B · d a = 0. For thespacetime area associated with path 1 we have Z π Z R B · ˆ z ρdρdϕ = B R ω Z π ϕdϕ = B R π ω , (28)where we have used t = ϕ/ω to write the magnetic field magnitude as B t → B ϕ/ω . Notethat here the area vector points along the positive z-axis d a = ˆ z ρdρdϕ . For the spacetimearea associated with segment 4 we have Z − π Z R B · ( − ˆ z ) ρdρdϕ = − B R ω Z − π ϕdϕ = − B R π ω . (29)Again we have used t = ϕ/ω and dϕ = ωdt . Here the area vector points along the negativez-axis ( d a = − ˆ z ρdρdϕ ) since the path is traversed in a clockwise direction. To get thespacetime area associated with the closed spacetime loop, we add the result of (28) to thetime reversed result of (29) Z B · d a = Z B · d a + T (cid:20)Z B · d a (cid:21) = Z B · d a − Z B · d a = B R π ω . (30)This result agrees with the results for the loop integrals for the 4-vector potentials in thetwo different gauges given in (20) and (25), and agrees with the result for the spacetime areaintegral from (27). This is reminiscent of the charging capacitor demonstration of Maxwell’sdisplacement current. [9] In this example, there is a circular loop that encloses a wire whichis charging up a capacitor. The closed loop integral of the magnetic field ( i.e. H B · d x )gives B ∝ I ( t ) ρ ˆ ϕ . If the surface chosen to span this loop cuts through the wire then I ( t )16s the current due to charges. However, the surface can be chosen so that it goes betweenthe capacitor plates where one has no current from charges. In this region one does havea time-changing electric field and so the contribution to the magnetic field loop integralnow comes from the displacement current. Here, the loop integral of the vector potentialis related to the spacetime area integral of the electric field in one case, and related to thespacetime area integral of the magnetic field in the other case. III. SUMMARY AND CONCLUSIONS
In this work, we have given explicit examples of Stokes’ theorem in the case where thereexists time-dependent fields ( i.e.
4D Stokes’ theorem). There are many examples of Stokes’theorem as it applies to time-independent fields ( i.e.
3D Stokes’ theorem), but these areperhaps the first explicit worked out examples of Stokes’ theorem with time-dependentfields. In fact, there are some claims that Stokes’ theorem does not apply to time-dependentfields (see the first footnote on page 305 and the first paragraph on page 312 of reference[10]). Despite these assertions, we have found that Stokes’ theorem can be applied to time-dependent cases if care is taken in how the spacetime loop is closed. In particular weinvestigated an infinite solenoid with a linearly increasing magnetic flux. There we showed H A µ dx µ = R F µν dσ µν for the spacetime loop closed outside the solenoid – see Fig. (2).We also showed that H A ′ µ dx µ , for the closed spacetime loop from Fig. (2), remained thesame for a gauge transformed 4-potential as given in (13) (14), demonstrating the gaugeinvariance of H A µ dx µ .We also investigated the more subtle case when the spacetime loop enclosed the solenoid.Here we also calculated H A µ dx µ in the two gauges given in (10) and (13) (14). For the4-vector potential (obtained with the multi-valued gauge function χ ) given in (13) (14), wehad to excise the singularity at ρ = 0 via the contour given in Fig. (4). In terms of thespacetime area integral of the fields we carried out the calculation using the two spacetimeareas given in Fig. (5): (i) the side of the spacetime cylinder where only the electric fieldcontributed; (ii) the slanted top of the spacetime cylinder where only the magnetic fieldcontributed. The result of R F µν dσ µν came either from the side spacetime area or theslanted top spacetime area. This is similar to the standard example used to demonstrateMaxwell’s displacement current. 17he time-dependent Stokes’ theorem is closely related to the time dependent Aharonov-Bohm effect and the the discussion in this article has been closely guided by this connection.The spacetime contours from Figs. (2) (3) were taken to be those that a real particlecould traverse since this is what occurs in the Aharonov-Bohm effect. Although much workhas been done on the time-independent Aharonov-Bohm effect, much less has been donein regard to the time-dependent Aharonov-Bohm effect. There are only two experimentsthat we have found which have been done on the time-dependent Aharonov-Bohm effect –one accidental experiment [11] and one purposeful experiment. [12] Two theoretical papers[13, 14] were written in an attempt to explain the surprising non-result of the accidentalexperiment of Marton et al. [11] More recently, there have been some theoretical papers[5, 15, 16] dealing with the time-dependent Aharonov-Bohm effect. Still, aside from the twoexperiments in [11, 12] (both of which gave unclear results), there is little in the way ofexperimental results for the time-dependent Aharonov-Bohm effect.As a final comment we note that the electric field associated with the solenoid, given in(12), is of the form one would expect for a current of magnetic charges – just as a currentof electric charges produces a magnetic field of the form B ∝ ρ ˆ ϕ , so too a current ofmagnetic charges would produce an electric field of the form E ∝ ρ ˆ ϕ . This connection tomagnetic charge also offers another example of a non-single valued gauge transformation.The following 3-vector potential (we now use spherical polar coordinates r, θ, ϕ , rather thanthe cylindrical coordinates, ρ, ϕ, z ) yields a monopole magnetic field B = ∇ × A = g ˆ r /r A monopole = g (1 − cos θ ) r sin θ ˆ ϕ . (31)The vector potential in (31) is single valued, but it has the usual Dirac string singularitypathology along the negative z-axis i.e. θ = π . One can also obtain a magnetic monopolefield from A monopole = − g (1+cos θ ) r sin θ ˆ ϕ which has a Dirac string singularity along the positivez-axis i.e. θ = 0. These two forms of the monopole 3-vector potential are related by thegauge transformation A → A − ∇ χ with χ = 2 gϕ . In this case the gauge function χ is non-single valued, but the two forms of the gauge potential A monopole are single valued. Thus,this is not exactly like the 4-potentials and gauge transformation for the time-dependentsolenoid case, given in (13) (14) and (15), where both the potentials and gauge functionwere non-single valued.It is easy to see that one can also get a magnetic monopole field from the following18lternative 3-vector potential [3, 17, 18] A monopole = − gϕ sin θr ˆ θ , (32)which does not have the Dirac string singularity of (31), but is non-single valued due tothe ϕ dependence of A θ . The two vector potentials in (31) and (32) are related by a gaugetransformation of the form A µ → A µ + ∂ µ χ ; χ = − g (1 − cos θ ) ϕ (33)Here we see that both the gauge transformation function χ in (33) and the 3-vectorgauge potential (32) are non-single valued, which then is similar to the situation for thetime-dependent solenoid as given in (13) , (14), (15). Acknowledgments:
DS is supported by a 2015-2016 Fulbright Scholars Grant to Braziland by grant Φ . Appendix I
In this appendix we carry out the details of the 6 line integrals for H A µ dx µ with the 4-vector potential from (10). For path 1, the vector potential from (10) is A = B tR ρ ˆ ϕ since theradius on this path is ρ . The infinitesimal path length element is d x = ρ dϕ ˆ ϕ . The path istraversed at a constant angular speed ω so that we have the relationship ϕ = ωt → dϕ = ωdt .The particle starts at ϕ = 0 and t = 0 and ends at ϕ = ϕ at t = ϕ ω , yielding Z A · d x = B R Z ϕ / tdϕ = B R ω Z ϕ / ω tdt = B R ϕ ω (34)For path 2 we have A = B tR ρ ˆ ϕ (now ρ varies) and d x = dρ ˆ ρ . Thus A · d x = 0 and weget no contribution to the loop integral from this line segment. We have taken the velocityalong path 2 to be the same as the velocity along path 1 (namely ρ ω ) and the distancetraveled is ρ − ρ (the distance from line segment 1 to line segment 3). Thus for path 2 we19ave Z A · d x ∝ Z ˆ ϕ · ˆ ρ dρ = 0 . (35)Although the line integral for path 2 is zero, time passes in the traversal of the path. Thiswill have an effect on the result for line segment 3. The amount of time that passes duringthe traversal of path 2 is ∆ t = ρ − ρ ρ ω . For path 3 from (10) we have A = B tR ρ ˆ ϕ , since theradius on this path is ρ . The infinitesimal length element is d x = − ρ dϕ ˆ ϕ . The path istraversed at a constant angular speed ω ′ = ω ρ ρ (the angular speed is slower) but this ensuresthat the linear speed along paths 1, 2 and 3 is the same . We now have the relationship ϕ = ω ′ t → dϕ = ω ′ dt . The particle starts at ϕ = ϕ and t i = ϕ ω + ρ − ρ ρ ω (this offset time ρ − ρ ρ ω is connected with the traversal of path 2). The particle ends at ϕ = 0 at t f = t i + ϕ ω ′ ,yielding Z A · d x = − B R Z t f t i tdϕ = − B R ω ′ Z t f t i tdt = − B R ω ′ t (cid:12)(cid:12)(cid:12)(cid:12) t f t i = − B R ϕ ω (cid:20) ρ ρ + 4( ρ − ρ ) ρ ϕ + 2 (cid:21) (36)Next we calculate R A · d x for line segments 4, 5 and 6. The integral R A · d x is the sameas R A · d x except d x = − ρ dϕ ˆ ϕ which changes the final result by a sign Z A · d x = − B R ϕ ω (37)The integral R A · d x , like R A · d x is zero since Z A · d x ∝ Z ˆ ϕ · ˆ ρ dρ = 0 . (38)Again time ∆ t = ρ − ρ ρ ω passes during the traversal of path 5, affecting the result of path 6.Finally, path 6 is similar to path 3 except d x = ρ dϕ ˆ ϕ which changes the final overall sign.Similar to path 3, path 6 is traversed at a constant angular speed ω ′ = ω ρ ρ which ensuresthat the linear speed along path 6 and path 4 are the same (in fact the speed along all sixsegments is taken to be the same). As before, we have the relationship ϕ = ω ′ t → dϕ = ω ′ dt . Here the requirement that the linear speed be the same along all line segments is a convenience. However,since one of the applications of our analysis is to the Aharonov-Bohm effect where one wants to eliminateor minimize the external forces on the particle tracing out the spacetime path we take the speed to beconstant Of course at the bends in the paths there will be forces but these can be thought of as thebending forces due to crystalline diffraction such as in the experiment in [11] ϕ = − ϕ and t i = ϕ ω + ρ − ρ ρ ω and ends at ϕ = 0 and t f = t i + ϕ ω ′ ,yielding Z A · d x = B R ϕ ω (cid:20) ρ ρ + 4( ρ − ρ ) ρ ϕ + 2 (cid:21) (39)Notice that the result for path 6 is equivalent to the negative of path 3. Appendix II
In this appendix we carry out the details of the 6 line segment integrals for H A µ dx µ forthe 4-vector potential from (13). For path 1 we have Z φ ′ cdt = B R Z ϕ / ω ϕdt = B R ω Z ϕ / ω tdt = B R ϕ ω , (40)where we use ϕ = ωt . For path 2, the scalar potential is constant φ ′ = B R ϕ and the timeto traverse path 2 is, as in Appendix I, ∆ t = ρ − ρ ρ ω . So for path 2 we have Z φ ′ cdt = B R ϕ Z ∆ t dt = B R ϕ (cid:18) ρ − ρ ρ ω (cid:19) . (41)For path 3 we use the relationship ϕ = ϕ − ω ′ t . The integral is Z φ ′ cdt = B R Z ϕ / ω ′ ϕdt = B R (cid:18) ϕ t − ω ′ t (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) ϕ / ω ′ = B R ϕ ρ ωρ . (42)Next we calculate R φ ′ cdt for line segments 4, 5, and 6. Line segment 4 is similar to path1 except that ϕ = − ωt which then changes the sign of the result Z φ ′ cdt = − B R ϕ ω . (43)Path 5 is similar to path 2 except now the scalar potential takes a different constant value φ ′ = − B R ϕ . Z φ ′ cdt = − B R ϕ (cid:18) ρ − ρ ρ ω (cid:19) . (44)Path 6 is similar to path 3 except now we have ϕ = − ϕ + ω ′ t . The integral is Z φ ′ cdt = − B R ϕ ρ ωρ . (45) Appendix III
21n this appendix we carry out details of the spacetime area integral of the electric andmagnetic fields for the spacetime loop given in Fig. 2For the spacetime area connected with path 1, the electric field and infinitesimal lineelement are E = − B R cρ ˆ ϕ and d x = ρ dϕ ˆ ϕ respectively, so Z E · d x cdt = − Z ϕ / ω dt Z ωt B R dϕ = − Z ϕ / ω (cid:18) B R ωt (cid:19) dt = − B R ϕ ω . (46)For the spacetime area connected with path 2, the electric field is E = − B R cρ ˆ ϕ – now theradial coordinate ρ is not fixed at ρ = ρ but rather runs from ρ to ρ . The infinitesimalline element is d x = ρdϕ ˆ ϕ so that for each ρ along path 2 (with ρ ≤ ρ ≤ ρ ) the dϕ integration runs from ϕ = 0 to ϕ = ϕ /
2. The integration over dt runs from t = 0 to t = ρ − ρ ρ ω which corresponds to moving from ρ to ρ at a speed of ρ ω . The dt integrationin combination with the dϕ integration sweeps out the spacetime area. Thus for path 2 thespacetime area integral is Z E · d x cdt = − Z ρ − ρ ρ ω dt Z ϕ / B R dϕ = − Z ρ − ρ ρ ω (cid:18) B R ϕ (cid:19) dt = − B R ϕ ρ ω ( ρ − ρ ) . (47)For the spacetime area connected with path 3 the electric field is E = − B R cρ ˆ ϕ and theinfinitesimal line element is d x = ρ dϕ ˆ ϕ . The dϕ integration goes from ϕ = 0 to ϕ = ϕ − ω ′ t (the angular velocity is ω ′ = ρ ρ ω so that the linear speed is the same on each path). The dt integration runs from t = 0 to t = ϕ / ω ′ . Thus the spacetime integral connected withpath 3 is Z E · d x cdt = − Z ϕ / ω ′ dt Z ϕ − ω ′ t B R dϕ = − Z ϕ / ω ′ B R (cid:16) ϕ − ω ′ t (cid:17) dt = − B R ϕ ρ ωρ . (48)The spacetime area integration connected with path 4 is similar to the integration connectedwith path 1, except here the limits on the dϕ integration run from ϕ = 0 to ϕ = − ωt , yielding Z E · d x cdt = − Z ϕ / ω dt Z − ωt B R dϕ = Z ϕ / ω (cid:18) B R ωt (cid:19) dt = B R ϕ ω . (49) Note that the spacetime area integration connected with path 2 runs along dϕ . It is not along the linearpath 2 which would be an integration along dρ . R E · d x cdt = − R E · d x cdt . The spacetime areaintegration connected with path 5 is similar to the integration connected with path 2, exceptthe limits on the dϕ integration run from ϕ = 0 to ϕ = − ϕ /
2, yielding Z E · d x cdt = − Z ρ − ρ ρ ω dt Z − ϕ / B R dϕ = Z ρ − ρ ρ ω (cid:18) B R ϕ (cid:19) dt = B R ϕ ρ ω ( ρ − ρ ) . (50)Note that comparing (47) and (50) illustrated that R E · d x cdt = − R E · d x cdt . Finally,the spacetime area integration connected with path 6 is similar to the integration connectedwith path 3, except here the limits on the dϕ integration run from ϕ = 0 to ϕ = ω ′ t − ϕ / Z E · d x cdt = − Z ϕ / ω ′ dt Z ω ′ t − ϕ B R dϕ = − Z ϕ / ω ′ B R (cid:16) ω ′ t − ϕ (cid:17) dt = B R ϕ ρ ωρ . (51)Again note that comparing (48) and (51) illustrates that R E · d x cdt = − R E · d x cdt .As a final comment, the “direction” of the spacetime area associated with the closedspacetime paths of Fig. (2) or Fig. (3) does not have a well known right hand rule as is thecase for purely spatial contours and areas. The directionality that we have chosen above forthe spacetime area is taken so that the area integral of the fields in (18) agrees with the resultsof the contour integrals of the two vector potentials given in (16) and (17) for the closedcontour given in Fig. (2) ( i.e. this choice ensures that Stokes’ theorem works out for Fig.(2)). In contrast, for the path which encloses the solenoid in Fig. (3) the direction in whichthe path is traversed, and thus the spacetime “area” direction, is reversed relative to thepath in Fig (2). Thus the spacetime area associated with the contour in Fig. (3) must havethe opposite sign from the spacetime area which comes from the contour in Fig. (2). Thisagain ensures that Stokes’ theorem works out for the contour which encloses the solenoid.One can regard the procedure described above, with the determining of the “direction”of the spacetime area based on the direction in which the closed spacetime contour closes(clockwise or counterclockwise) as an extension of the right hand rule to spacetime contoursand surfaces. A more rigorous way to determine the direction of the area for mixed spatial23nd time surfaces is given through the use of differential forms [2, 7, 8]. [1] Y. Aharonov and D. Bohm, “Significance of electromagnetic potentials in the quantum the-ory”, Phys. Rev. , 485-491 (1959).[2] L. Ryder, Quantum Field Theory nd edition, (Cambridge University Press, Cambridge UK1996).[3] J. Macdougall and D. Singleton, “Stokes’ theorem, gauge symmetry and the time-dependentAharonov-Bohm effect”, J. Math. Phys. , 042101 (2014).[4] T.A. Abbott and D.J. Griffiths, “Acceleration without radiation”, Am. J. Phys. , 241-244 (2013).[6] J.D. Jackson, Classical Electrodynamics , 3 rd edition, section 6.10 (John Wiley & Sons Inc.,New York).[7] T. Frankel, The Geometry of Physics: An Introduction , 3 rd edition (Cambridge UniversityPress, Cambridge UK 2012).[8] B. Felsager, Geometry, Particles, and Fields , (Springer-Verlag Press, New York, NY 1997).[9] D. J. Griffiths
Introduction to Electrodynamics , 3 rd edition, section 7.3 (Prentice Hall, UpperSaddle River, NJ 1999).[10] R. A. Brown and D. Home, “Locality and Causality in Time-Dependent Aharonov-BohmInterference”, Nuovo Cimento B , 303-316 (1992).[11] L. Marton, J. A. Simpson, and J. A. Suddeth, “An Electron Interferometer”, Rev. Sci. Instr. , 1099-1104 (1954).[12] A. N. Ageev, S. Yu. Davydov, and A. G. Chirkov, “Magnetic Aharonov-Bohm Effect underTime-Dependent Vector Potential”, Technical Phys. Letts. , 392-393 (2000).[13] F.G. Werner and D. Brill, “Significance of Electromagnetic Potentials in the Quantum Theoryin the Interpretation of Electron Interferometer Fringe Observations”, Phys. Rev. Letts., ,344-347 (1960).[14] J. Macdougall, D. Singleton, and E. C. Vagenas, “Revisiting the Marton, Simpson, and Sud-deth experimental confirmation of the AharonovBohm effect”, Phys. Lett. A , 1689-1692 , 4319-4325 (1992).[16] M. Bright, D. Singleton and A. Yoshida, “Aharonov-Bohm phase for an electromagnetic wavebackground”, Eur. Phys. J. C , 446 (2015).[17] G. B. Arfken, H-J. Weber: “Mathematical Methods for Physicists” (Harcourt/AcademicPress), 5th Edition, page 130.[18] R.K. Ghosh and P.B. Pal, “A non-singular potential for the Dirac monopole”, Phys. Lett. B , 387-390 (2003)., 387-390 (2003).