A recent article in this journal by Narayan1 illustrated that magnetic forces do not obey Newton's third law of action and reaction via the example of two “point” electric dipoles whose moments $pi=p0,ieλt x̂i$ grow exponentially with time. He correctly found that $F12+F21+dPEM/dt=0,$ where $PEM=∫E×B dVol/4π c$ (in Gaussian units, with c as the speed of light in vacuum) is the electromagnetic-field momentum associated with the two dipoles.2 This is an example of a general result by Page and Adams3 published in this journal many years ago, in the so-called Darwin approximation4–6 that keeps terms only to order $v2/c2$ (which ignores electromagnetic radiation), where v is velocity. Narayan's demonstration is nice in that it holds to all orders of $v/c.$

However, Narayan mistakenly claimed that “there are no electromagnetic fields in the far field zone,” and hence, “no radiation is emitted” in his example, based on the appearance of his Eqs. (2) and (3). While those forms are convenient for later computations, they do not well display the characteristics of electromagnetic fields of a time-dependent dipole. This was the topic of the appendix to a paper in this journal by Berman,7,8 where the magnetic field of a time-dependent electric dipole (at the origin) was shown to be $B(r,t)=[p¨]× r̂/c2r+[ṗ]×r̂/cr2$ with $[p]=p(t−r/c)$. The term that varies as 1/r is a “radiation” field, nonzero except in the limit $t→−∞$ when the dipole moment is zero by construction. The radiation pattern of Narayan's two dipoles can be computed using Eq. (66.6) of Landau,5$dI/dΩ=cr2[Brad]2/4π,$ and Fig. 1 of Narayan, where $x̂1=x̂$ and $x̂2=ŷ,$ noting also that $r̂=sin θ cos φ x̂+sin θ sin φ ŷ+cos θ ẑ,$$dI/dΩ=λ4[(p1+p2)×r̂]2/4πc3=λ4[(p12+p22) cos2 θ+(p1 sin φ−p2 cos φ)2 sin2 θ]/4πc3.$ Then, the total intensity, integrated over solid angle is, $I=2λ4[p12+p22]/3c3.$9

We can also consider the radiated momentum, $dPrad/dt=∫dΩ d2Prad/dt dΩ,$ where $d2Prad/dt dΩ=(cr̂/4π) dI/dΩ.$10 Although $d2Prad/dt dΩ$ is nonzero, the total radiated momentum $dPrad/dt$ is zero, recalling the above form of $dI/dΩ$.

A variant of Narayan's example has been suggested by Onoochin,11 in which the time dependence of the dipoles is $pi=p0,itx̂i,$ and the corresponding charge and current densities are $ρi=−Iit(∂/∂xi)δ3(r−ri)$ and $Ji=Iidliδ3(r−ri)x̂i$ with $r1=0$ and $r2=a x̂.$ Such a configuration has been called “semistatic,”12 for which the electric and magnetic fields are given by the instantaneous Coulomb and Biot-Savart forms. There is no radiation in this variant. It follows that $F12+F21=−(I1dl1)(I2dl2) ŷ/a2c2=−dPEM/dt,$ more quickly than in Narayan's example.13

In a larger historical context, Ampère's insistence that magnetic forces obey Newton's third law earned him the sobriquet by Maxwell14 as the “Newton of electricity.” Ampère's authority held up acceptance of the “Lorentz” force law (stated obliquely by Maxwell in 186115,16) until efforts by Thomson17,18 and Heaviside19 in 1891 clarified that electromagnetic fields carry momentum as well as energy (following the first clear statement of the “Lorentz” force law by Heaviside20 in 1885).

In 1864, Maxwell discussed “electromagnetic momentum,”21 identifying this with Faraday's “electronic state” in Sec. 26, and clarifying in Sec. 57 that the electromagnetic momentum of charge q in an external vector potential A is qA(/c) (in the Coulomb gauge, as favored by Maxwell). This formulation suggests that electromagnetic momentum is a property of the charge, rather than of the electromagnetic field. That Maxwell's electromagnetic momentum is equivalent to the electromagnetic-field momentum of Thomson and Heaviside (the $PEM$ of this Letter) in quasistatic examples was first demonstrated by Thomson.22,23

Applying Maxwell's formulation, $PEM=Σ qjA(rj)/c,$ of electromagnetic momentum to the “semistatic” variant of Onoochin, we see that the (Coulomb-gauge) vector potential of dipole 1 is the same at the position of both charges of dipole 2, and so does not contribute to the momentum. The vector potential of dipole 2 along the x-axis is $A2(x)=I2dl2 ŷ/c|a−x|,$ so the electromagnetic momentum is $PEM=(q1dl1/c)dA2(0)/dx=(I1dl1)(I2dl2)t ŷ/a2c2,$ noting that $q1=I1t.$

1.
O.
Narayan
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Am. J. Phys.
89
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1033
1036
(
2021
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2.
We exclude the self-momentum of moving charges, and consider only the interaction field momentum.
3.
L.
Page
and
N. I.
, Jr.
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13
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141
147
(
1945
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4.
C. G.
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L. D.
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J. D.
Jackson
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P. R.
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76
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54
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9.
The electric-dipole radiation follows from Eq. (67.8) of Landau5 as $IE1=2[p¨2]/3c3=2λ4[(p1+p2)2]/3c3=2λ4[p12+p22]/3c3,$ which is the same as the total intensity.
11.
V.
Onoochin
, private communication.
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D. J.
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13.
For details, see the appendix of kirkmcd.princeton.edu/examples/narayan.pdf.
14.
J. C.
Maxwell
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J. C.
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17.
J. J.
Thomson
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19.
O.
Heaviside
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Philos. Trans. R. Soc.
183
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423
480
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20.
O.
Heaviside
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Electrician
14
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178
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J. C.
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512
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J. J.
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For discussion in this journal, see the appendix of
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