Table Of ContentDo the standard expressions for the electromagnetic field-momentum need any
modifications?
∗
Ashok K. Singal
Astronomy and Astrophysics Division, Physical Research Laboratory, Navrangpura, Ahmedabad - 380 009, India
We investigate here thequestion raised in literature about the correct expression for theelectro-
magneticfield-momentum,especiallywhenstaticfieldsareinvolved. Forthisweexamineacoupleof
simplebutintriguingcases. Firstweconsiderasystem configurationinwhich electromagnetic field
6 momentum is present even though the system is static. We trace the electromagnetic momentum
1 to be present in the form of a continuous transport of electromagnetic energy from one part of the
0 system to another, without causing any net change in the energy of the system. In a second case
2 we show that the electromagnetic momentum is nil irrespective of whether the charged system is
stationaryorinmotion,eventhoughtheelectromagneticenergyispresentthroughout. Wedemon-
r
a strate that the conventional formulation of electromagnetic field-momentum describes the systems
M consistently without any real contradictions. Here we also make exposition of a curiosity where
electromagnetic energy decreases when the charged system gains velocity. Then we discuss the
4 more general question that has been raised – Are the conventional formulas for energy-momentum
1 of electromagnetic fields valid for all cases? Or more specifically, in the case of so-called “bound
fields,” do we need to change over to some modified definitions which have made their appearance
] even in standard text books? We show that in all cases it is only the conventional formulas which
h
leadtoresultsconsistentwiththerestofphysics,includingthespecialtheoryofrelativity,andthat
p
any proposed modifications are thussuperfluous.
-
s
s
a I. INTRODUCTION Equally intriguing is the case where electromagnetic
l
c momentum is nil in a charged system not only when it
s. Aquestionaboutthecorrectexpressionfortheelectro- is static but also when the system is in motion. This
c happens in spite of the fact that electromagnetic energy
magneticfieldmomentum,especiallyforstaticcases,has
i
s been raisedin the literature.1–3 The standardexpression is present in the system in either case, though it is even
y morebaffling that the energydecreaseswhen the system
for the electromagnetic field momentum is the volume
h integral,4,6 gainsmotion. Wetrackdownthepresenceofelectromag-
p
netic momentuminthese casesthat existsinthe formof
[
a continuous transport of electromagnetic energy from
2 P=ǫoZ E×Bdv. (1) one part of the system to another, without any net gain
v V of energy in the process. We also demonstrate how a
5
charged system could comprise less energy after it has
9 The expression is believed to be true even for static sys-
gathered motion from its initial state of rest.
4 temswherethesourcesoftheelectricandmagneticfields
6 may not be having any apparent motion, with the fields
0 therebybeing staticwithnotemporalchangeinthefield
. II. AN APPARITION OF MOMENTUM
1 values. Forinstance,inastaticsystemcomprisingapair
WITHOUT ANY APPARENT MOTION
0 of crossed electric and magnetic dipoles (p and m re-
6 spectively),thereisalinearelectromagneticmomentum,
1 P ∝ m×p, associated with the system. The puzzling In order to comprehend the occurrence of momentum
:
v questionthenis–whatreallyismovinginthisotherwise without motion we consider a specific example of two
i static system? The momentum is certainly not due to concentric spherical shells, one of radius a having a sur-
X
the drift velocities of chargesthat constitutes the steady face charge density proportional to cosθ, resulting in an
r
a current giving rise to m; after all that motion is present electric dipole moment p, and the second sphere of ra-
evenif the electric dipole p were absentand in that case dius b having a surface current density proportional to
there wouldnotbe anyelectromagneticmomentumPin sinθ,resultinginamagneticdipolemomentm. Thetwo
the system. Evenotherwise,any such momentum vector spheres are oriented such that p and m are perpendic-
due to drift velocities of conducting charges, when inte- ular to each other, which results in an electromagnetic
grated over a closed current loop, will have a nil value. linear momentum P = µ m×p/(4πb3) in this other-
o
Nor could the electric dipole p alone be the cause of any wisestaticsystem.1,7 Thesystemcouldbecalledstation-
momentum P. Electromagnetic field momentum makes arybecauseevenaconstantcurrentimpliesmovementof
an appearanceonly when both p and m are present and charges,howeverasfarastheelectricandmagneticfields
that too when they have at least some components mu- are concerned, there is no temporal change and in that
tually perpendicular. For further details and references sense it is a static system. Now should one change the
on the question of electromagnetic momentum we refer definition of electromagnetic momentum so as to make
the reader to the review article by Griffiths.3 it zero, or is there some unknown “hidden” momentum
2
FIG.1. Acurrentloop ABCD,carryingcurrentI,lyingpar- FIG. 2. The geometry for calculating electromagnetic mo-
alleltotheconstantelectricfieldEofachargedparallelplate mentumduetotransportofelectricenergyacrossthecircuit.
capacitor.
transportedper unit time along s, the vector joining the
that cancels the electromagnetic momentum to make it current elements in side CD to that in AB, implying a
nil in a static system. In fact where does the electro- momentum,
magneticmomentumitselflieinsuchstatic systems? As
I I
theword“momentum”conjuresupavisionofsomekind E·dls =E×da , (2)
c2 c2
of motion, where, if any, is such motion in the system
corresponding to the electromagnetic momentum? where da is the area across the loop contained between
To understand that, we first examine a simpler case these two current elements. It is certainly not implied
of a small rectangular loop carrying a constant current here that necessarily a one-to-one energy transfer takes
I = nev and lying in a static electric field E, assumed place between current elements placed symmetrically in
to be constant and parallel to a side of the current loop. twooppositesidesofthecircuit,thisprocedure,asimple
Here n is the number of charges per unit length of the book-keeping,isadoptedmerelytomakethecalculations
circuit,eistheelectricchargeofconductingchargedpar- easybytakingadvantageofthesymmetryofthesystem.
ticles, and v is their drift velocity. On any small current We integrate over the whole loop to find the total linear
element dl of the loop, the electric field does work at a momentum to be,
rate nevdl·E = Idl·E. We notice that in one arm of
E×aI E×m
thelooptheelectricfieldEisdoingapositiveworkwhile = , (3)
c2 c2
in the other parallel arm the work done is negative, and
no work being done in the other two sides. Thus while wherem=aI isthemagneticdipolemomentofthecur-
there may be a gain of energy in one arm of the loop, rent loop. It should be noted that the electric current
there is an equal loss in the opposite arm; effectively a could be due to either positive or/and negative charges,
continuous exchange of electrical energy takes place be- making no difference in the final result. We could re-
tween opposite parts of the circuit because of the pres- place the rectangular shape of the loop with a polygon
ence of electrical field. Though in this process there is with larger number of sides and in the limit even with a
no net change in the total energy content of the system, circular one without changing any of our essential argu-
nevertheless there is a linear momentum associatedwith ments; in fact we could have any irregular shaped loop
this transport of energy across the loop. With respect as it can be always realizedby a superpositionof a suffi-
to Fig. (1), the electric field pumps energy into arm AB ciently large number of regular shaped loops. Moreover
and which is getting drained in arm CD, thus effectively Evectornowneednotnecessarilybeonlyintheplaneof
electric field gives rise to a transport of electrical energy the current loop. This therefore is a fairly general result
from side CD, where it disappears, to side AB where it as long as the current loop is small enough for E to be
appears. It is basically this transport of electric energy considered constant over its extent.
across the circuit that constitutes electromagnetic mo- The system of course remains stationary as there will
mentum in an otherwise static case. beanequalandoppositemomentumassociatedwiththe
Theelectromagneticmomentumcanbecalculatedeas- stabilizing forces (irrespective of the detailed model of
ily if we first consider a pair of current elements placed these forces of constraint) that keep the current in the
symmetrically in the two arms of the loop (Fig. 2). The loop everywhere constant (as we assumed it to begin
power being fed into the circuit element of length dl in with) in spite of the effects of the electric field E on the
arm AB through electric field is I E·dl, while a simi- loop (see Section V).
lar amount of power is being drained from an element It is now straight forward to apply this result to the
dl in arm CD. Effectively an energy IE · dl is being two spherical shells,1,7 comprising a pair of electric and
3
magnetic dipole, oriented perpendicular to each other.
The magnetic dipole moment m comprises a uniformly
charged spherical shell of radius, say, b rotating at a
constant rate1. The external electric field of the elec-
tric dipole moment p can be written as E = [3(p.ˆr)ˆr−
p]/(4πǫ r3).1,7 Ignoringtheradialcomponentofthefield
o
which does no work on conducting charges in the spher-
ical shell, we find the rest of electric field component
on the shell surface to be a constant −p/(4πǫ b3). Di-
o
viding the spherical shell into a large number of circu-
lar current loops and then summing over them we ar-
rive at the total electromagnetic linear momentum of
the system as µ m×p/(4πb3), the result obtained us-
o
ingtheconventionalformulafortheelectromagneticfield
momentum.1,7
III. THE CASE OF MISSING MOMENTUM IN
A MOVING CHARGED SYSTEM FIG. 3. A charged parallel plate capacitor stationary in its
rest-frameS′ whichismovingwithavelocityV with respect
to thelab-frame S.
We shall now show that it is equally possible to have
a charged system which has a nil electromagnetic mo-
mentum irrespectiveofwhether the systemisstatic or is
from plate 2 to plate 1 through the force of attraction,
in motion. Consider again the last example of a parallel
which forms a part of the total electromagnetic momen-
plate capacitor, sans the current loop. However now the
tum of the system. One can easily visualize the effect of
capacitor is moving parallel to the plate separation with
work being done by these electromagnetic forces, if we
avelocityV inthe lab-frameSwhileS′ istherestframe
consider the capacitor plates at this stage not clamped
of the charge (Fig. (2)).
together and thus free to move under these forces of at-
Let Q and −Q be the total charges on the capacitor
traction. Then as seen in the lab-frame S, plate 1 will
plates, each of area A, these quantities being invariants
gain a velocity higher than V while the velocity of plate
between S′ and S. The electrostatic field, E =Q/(Aǫ ),
o 2 will reduce below V, implying a transport of energy
a constant in the region between the plates, is also the atarate(Q2/2ǫ A)V,throughelectricattraction,across
o
same in either frame. Similarly the mutual force of at-
lengthlofthecapacitorsystem(i.e.,fromplate2toplate
traction between plates, F = Q2/(2ǫ A), as well as the
o 1). The resulting momentum is in a direction opposite
energydensity,ǫ E2/2=Q2/(2ǫ A2),oftheelectrostatic
o o to V.
field do not change between two frames.
Therefore the total electromagnetic momentum of the
Let the separation between the two plates be l in the system is,
lab-frame S, then there will be an electric field energy
U = Q2l/(2ǫoA), as calculated in S. However, the mag- P =UV − Q2 Vl = 0. (4)
netic fieldis zerothroughout,therefore the field momen- em c2 2ǫ Ac2
o
tum is zero, which is rather perplexing since the system
with an energy U and moving with a velocity V is ex- Thus we see that to get a consistent picture of the elec-
pected to possess a momentum UV/c2 in the lab-frame tromagnetic field momentum, one needs to track down
S. Why is the momentum nil for a moving system? and properly account for the contribution of the electro-
Actually due to the force of attraction between the magneticforcesduringthemotionofthesystem,thecon-
plates, there is a negative momentum component (i.e., tributions that might not be so apparent in many cases.
in a direction opposite to the motion of the capacitor),
due to the energy flow caused by the force of attraction
IV. AN INTRIGUING CASE OF ENERGY
between the moving plates. This excess momentum is
DECREASING WITH MOTION!
overandabovethe usual relativistic momentumformula
UV/c2. This extra momentum is due to the fact that
there is a continuous flow of energy into the system at The separations of the two plates, l and l′ in frames
its plate 1 due to the work being done at a rate F21·V, S and S′ are related, due to the Lorentz contraction, by
due to F21, the force of attraction on plate 1. At the l=l′/γ. Therefore the electric energies U =Q2l/(2ǫoA)
same time an equal amount of energy is flowing out of and U′ = Q2l′/(2ǫ A), obtained from the volume inte-
o
the system due to F12 at plate 2. Although there is no gralsof the energy density inS andS′, imply U =U′/γ.
net increase of the system energy due to these forces, Thisappearstobe atvariancewith the usualrelativistic
there is a continuous transport of energy taking place transformationformulaswhereoneexpectstheenergyof
4
the moving system in lab-frame S to be instead higher will continue till the electric field giving rise to this non-
by a factor γ. How come the energy has decreased with uniform current is cancelled in the circuit. Of course
motion? in that case we will no longer find an electromagnetic
To understandthis, consider a case where the charged momentum in the system, but that may not come as a
capacitor,initiallyatrestinthelab-frameSwithaplate surprisesincewewouldnothaveanelectricfieldeitherin
separation l and energy U = Q2l/(2ǫ A), is given a theregionofm. Thereforeinordertostudythemomen-
o
push to acquire a velocity V so as to come to rest in tum arising from E×m, we need to maintain a uniform
S′. In frame S′ the capacitor was initially moving with electric current in the circuit using some constraints.
a velocity −V with a plate separation l/γ and energy
One way to invoke such forces of constraint could be
U = Q2l/(2ǫ Aγ). In lab-frame S both plates of the
o to adopt a simple model of current, with charges spread
capacitor were given the push simultaneously, thus both
with equal spacing on a conveyor belt going around in a
plates continueto movewith a velocityV withthe same
circuitlikeABCDofFig. (1),constrainedtomovewitha
plate separation l afterwards. In frame S′, it will ap-
uniformspeedalongitspathsoastomaintainaconstant
pear that the capacitor plates initially moving with ve-
and uniform electric current. As the force on charges
locity −V are being stopped so as to come to rest in S′. due to the imposed field E is parallel to the drift veloc-
However,likeinthe famouspole andbarnproblem,8 the ity v on one side of the circuit, the field does a positive
stopping of the plates will not be a simultaneous event
workonthatsideandtherewillbeanequalbutnegative
in S′, and the plate 2 will come to rest before plate 1 by
work done on the opposite side of the circuit. Thus the
a time interval ∆t′ = γlV/c2, thus plate 1 will continue
agency, in order to keep a constant velocity of the belt
tomovewithavelocity−Vduringthe time∆t′ andwill
andtherebymaintainauniformcurrent,willgainenergy
thus be a distance ∆l′ = γlV2/c2 farther from plate 2,
on arm AB of the circuit but will impart energy on arm
before it also comes to rest in S′. Therefore the plate
CD.Thustheagencydrainsenergyononesideofthecir-
separation in S′ will become (l/γ)+∆l′ =lγ =l′. Also
cuit and feeds equal amount of energy on the other side,
duringthetime∆t′ themovementofplate1willdowork
thus a mechanical transport of energy from the side AB
against the force of attraction on it and that will add to
tothesideCDiscontinuouslytakingplace,andtherefore
the energy of the system which will add to the electro-
a mechanical momentum exists in the circuit. Such has
magnetic field energy of the system ∆U = F γlV2/c2
21 sometimes been termed as a “hidden momentum”, this
making the net electromagnetic energy of the system in name is perhaps unfortunate1,3, as it is not any more
the rest frame S′ as,
hidden than the electromagnetic momentum itself which
is also in the form of electromagnetic energy flow from
Q2l Q2lγV2 Q2lγ
+ = =Uγ. (5) one part of the system to another. As this momentum
2ǫ Aγ 2ǫ A c2 2ǫ A
o o o is from the transport of mechanical energy due to the
Thus energy in the rest frame S′ is γ times higher than forces of constraint that maintain a steady current in
the energy in the lab-frame S with respect to which the the system, in spite of the applied electric field, it would
capacitor system is moving with a velocity V. This of be termed better as “mechanical momentum” to distin-
guish it from the electromagnetic momentum. Another
course is consistent with the energy calculated from the
good example of the appearance of mechanical momen-
standard formula for energy in electromagnetic fields in
tum in the system when electric current is constrained
the system for each reference frame.
to be uniform across the circuit is given in Griffiths3,6.
The momentum due to the electromagnetic origin and
V. MECHANICAL MOMENTUM DUE TO THE that due to the mechanical cause are equal and opposite
STABILIZING FORCES and thus cancel each other making the net momentum
(electromagnetic plus mechanical) of this static system
nil. Thus we see that even though the total momentum
There are also forces of stabilization in all such cases.
ofthis staticsystemis zero(thatfits the name “static”),
We do not specify the ultimate nature of these forces as
the pure electromagnetic part of the momentum is not
that may not have any relevance for the problem under
zero. We get into complications when we try to describe
discussion. All we know or care for is that these keep
thetotalmomentumofthesystemonapureelectromag-
the system in equilibrium by providing stabilizing forces
netic basis.
equal and opposite to any unbalanced electromagnetic
forces in the system. In the same way, the separation between the capaci-
For instance, in the case of the currentloop ofSection tor plates in Fig. (2) is kept fixed in a given frame, in
II,ifauniformcurrentisnotmaintainedusingsomesta- spiteofthemutualforceofattractionbetweentheplates,
bilizing forces or constraints then a higher value of elec- thentheremustbesomeforcesofconstraint,e.g.,anon-
tric current in arm AB and a lower current in arm CD conductingrodkeepingtheplatesapart(ormaybesome
(due to the imposed electric field E) will soon result in clamps holding the plates fixed in their positions). Irre-
piling of positive charges on arm BC and a deficiency of spective of the nature or details of these forces of con-
similarchargesin armDA (orequivalently anaccumula- straint, without any ambiguity, on each plate they must
tion of negative charges on arm DA), and this accruing beequalandoppositetotheforceofelectricalattraction.
5
Therefore their contributions to the energy and momen- and momentum in S are given as,4
tum of the system must also be equal and opposite to
ǫ
that calculated above for the electrical forces. U = oγ2 (E2−c2B2)dv. (8)
For instance, there must be a mechanical momentum 2 ZV
in the rod holding the two capacitor plates apart. This
is because as the system moves, as seen the lab-frame in
ǫ γ2V
S, work is being done on the holding-rodby plate 1 at a P= o (E2−c2B2)dv. (9)
rate F21·V, and an equal amount of work −F12·V is 2 c2 ZV
being done by the rod on plate 2. Although there is no
Actually this question was discussed previously at
net increaseof the energy of the roddue to this, there is
length12,13 where it is explicitly shown for the spherical
a continuous transport of energy taking place from the
model of a charged particle that when we consider the
left-end of the rod to its right-end, and this energy-flow
contributions of all electromagnetic forces (which may
through the rod is nothing but a mechanicalmomentum
not always be so obvious, and that according to us is
Pme =(Q2Vl/(2ǫoAc2). primarily the reason for so much confusion in this sub-
The total momentum of the system will be given by
ject) to the energy momentum of the considered sys-
the momentum Pem in the electromagnetic fields (which tem we invariably arrive at the result derived “indepen-
could represent only the pure electromagnetic part of a
dently” from conventional electromagnetic field energy-
system)plusPme,themechanicalmomentumassociated momentum formulas. Thereby a natural explanation for
with the forces of constraintas calculatedabove. There-
the famous but intriguing factor of 4/3 in the inertial
fore the total (electromagnetic plus mechanical) momen-
electromagnetic mass of a charge, emerges without any
tum of the system is given by
ambiguityandthereisnocompellingreasonforadopting
achangeintheconventionalformulas. Itisnotevensim-
V
Ptotal =Pem+Pme =Uc2. (6) poflyvaiewmaatstesroomfemtiemreescoandvveonciaetnecdeionrtahneinlitdeivraidtuurael.’1s4pOoinnet
can put forward the following additional arguments in
Theeffectofsuchstabilizingforceshasbeenshowntobe
essential for a successful explanation9 of the null results support of the conventional formulas vis-a-vis the modi-
of the famous Trouton-Noble experiment.10 fications as referred to above.
1. The total energy-momentum of a charged system is
the energy-momentum of its electromagnetic parts plus
the energy-momentum associated with the other (stabi-
VI. DISCUSSION
lizing, non-electromagnetic!) forces of constraint. Since
the electromagnetic fields could represent the energy-
Throughacoupleofsimpleexamples,wedemonstrated momentum associated with only the electromagnetic
above how the unbalanced electromagnetic forces in a phenomenon (after all expressions for E and B make
system make additional contribution to the energy mo- no reference and hence contain no information about
mentumofthesystem. Thestandardformulasforenergy these “other” forces), the energy-momentum associated
momentum of the electromagnetic fields of a system al- with the stabilizing forces could not be represented by
readycontainsallthese electromagneticcontributions to these. Therefore a modification in definition of energy-
the electromagnetic energy momentum. This brings us momentum in electromagnetic fields to include contri-
to the more general question raised by Romer:1 Are the bution of these “hidden” non-electromagnetic forces is
conventionalformulasforenergyandmomentumofelec- certainly not appropriate.
tromagnetic fields valid for all cases or do we need to 2. Ingeneral,electromagneticfieldsrepresentvaluesat
change over, at least in the case of “bound fields,” to space-time events and these fields themselves are not as-
some modified definitions11, which have made their ap- signed any velocities unlike a localized material particle.
pearance even in standard text books?4,5 To get an idea Thereforetochooseoneparticularreferenceframe(asort
ofthesemodifieddefinitions,weconsiderasimplesystem of “rest-frame” for the fields) on a preferential basis to
in which an inertial frame, say S′ can be found in which calculate the volume integral of electromagnetic energy-
all charges are at rest. Then there is no magnetic field momentum is not proper. In fact it will be against the
andonlyanelectricfieldexistsinframeS′. Thenenergy very spirit of relativity.
U′ is given by the volume integral, 3. Even if the modified definition might work for a
single charged particle or when all the charged particles
U′ = ǫo E′2dv′. (7) in the system being considered are moving together as
2 ZV′ a bunch with the same velocity vector, so that they all
could have a common rest-frame,such a modified defini-
Theninthelab-frameS,whichismovingwithavelocity tiondoesnotworkwhentherearemorethanoneparticle,
V with respect to S′, the electromagnetic field energy with different particles having different velocity vectors
andmomentum in the modified definitions5 are givenby with respect to each other as seen in any inertial frame.
U = γU′ and P = γVU′/c2. Accordingly field energy In such a case there can be no common rest-frame for
6
the particles and their so-called bound fields. For each electromagnetic origin and that due to the mechanical
such particle the fields have to be treated separately. In cause are equal and opposite and thus cancel each other
other words, what parts of the electromagnetic fields at making the net momentum of this static system nil (as
any event belong to what particle need to be first rec- in Section II). Now if we do modify the electromagnetic
ognized and then broken into contributions of each such momentumdefinitionto makeitzero(or evensomewhat
particleseparately. Thiswaythecontributiontotheelec- different from the standarddefinition) in a static system
tromagneticfieldenergy-momentumhastobecalculated like we discussed here, then we will end up with a net
separately for each particle, as each particle contributes momentum in a static system because a modification in
differently in different reference frames. This is so be- the definitionof the electromagneticmomentum willnot
cause each such contribution has to be evaluated with change the contribution of the mechanical momentum,
respectto the rest-frameofthe individualparticleasthe unless of course we modify the definition of the mechan-
Lorentz factor γ enters into the definition of the such ical momentum formula too. That will open a Pandoras
“bound” electromagnetic fields (viz. Eqs. (8) and (9)) box. It is thus best to stick to the standard definition
and has to be then integrated over all particles. That moresoasnorealcontradictionresultsfromtheirusage.
means algebraic sum of the electric and magnetic fields In our opinion time has come to abandon these mod-
(as components of vectors or a tensor) of all contribu- ified definitions as these do not seem to serve any use-
tions at that event do not suffice to calculate energy- ful purpose except creating confusion in the subject. It
momentum contributions of electromagnetic fields. In is only the standard definitions which are “correct” and
fact such a resultant field will not comply with the prin- should only be retained in text-books.
ciple of linear superposition and from the field values VII. CONCLUSIONS
in one inertial frame one cannot transform to another
frame using the standardfield transformationlaws. One We examined a couple of simple electromagnetic sys-
might wonder whether such a modified definition serves tems,firstonepossessingmomentumwithoutanyappar-
any useful purpose at all. entmotionandasecondonewithmotionbutwithoutany
4. Tohavetwoseparatedefinitions forso-called“free” momentum. Weshowedthatthismysteriouselectromag-
and “bound” fields is in itself an unnecessary compli- neticmomentaexistsintheformofatransportofelectri-
cation where one single definition has been giving the calenergytakingplacefromonepartofthesystemtoan-
correct and consistent results and been doing fine in all other, all within the system. We resolvedthese puzzling,
cases. apparentparadoxes,withintheconventionalformulation
5. If we do modify the definition of electromagnetic of the electromagnetic field energy-momentum, thereby
field momentum for such static cases (in the strict mod- obviating the need for modifications in the definition of
ified definition as say, in Ref. [4], so that the electro- standard formulation that have been advocated in the
magnetic field momentum is zero (or at least different) literature. We demonstrated that the standard formu-
in static cases, then we end up in bigger problems as lation suffices to explain all electromagnetic phenomena
we will have to modify definition of momentum even in andthatitisthe standardformulationonlywhichyields
mechanical cases so as to conserve momentum in elec- result consistent with the rest of physics, especially with
tromechanical phenomenon. The momentum due to the the special theory of relativity,.
∗
[email protected] 1992).
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6 D. J. Griffiths, Introduction to electrodynamics, 3rd ed. 25 1605-1620 (1992).
(Prentice, NewJersey, 1999). 13 T. H.Boyer, “Classical modelof theelectron and thedef-
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Poyntingvector in static fields” Am.J. Phys.35, 153-156 25, 3246-3250 (1982).
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