International Journal of Magnetics and Electromagnetism Int J Magnetics Electromagnetism 2631-5068 VIBGYOR Online Publishers 313 Kd Tower, Cotterells, Hemel Hempstead, Hertfordshire, England, HP1 1AU Angular Momentum Emission by a Rotating Dipole 10.35840/2631-5068/6526 Khrapko RI A new calculation confirms the presence of spin radiation along the axis of rotation of a dipole. This is further proof of the need to introduce the spin tensor into classical electrodynamics, along with the energy-momentum tensor. Research Article 6 1 OPEN ACCESS Angular Momentum Emission by a Rotating Dipole Radi I Khrapko Moscow Aviation Institute, Volokolamskoe Shosse 4, Russia Radi I Khrapko
Moscow Aviation Institute, Volokolamskoe Shosse 4, 125993, Moscow, Russia.
09 December 2020 11 December 2020 Khrapko RI 2020 Angular Momentum Emission by a Rotating Dipole Int J Magnetics Electromagnetism 2020 Khrapko RI, et al © This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

A new calculation confirms the presence of spin radiation along the axis of rotation of a dipole. This is further proof of the need to introduce the spin tensor into classical electrodynamics, along with the energy-momentum tensor.

75.10.Hk, 03.50.De

Circularly polarized electromagnetic radiation contains angular momentum in the form of the angular momentum density [1,2].

JH Poynting : "If we put E for the energy in unit volume and G for the torque per unit area, we have G=Eλ/2π".

This means that such radiation is Weyssenhoff's spin-fluid .

J Weyssenhoff: "By spin-fluid we mean a fluid each element of which possesses besides energy and linear momentum also a certain amount of angular momentum, proportional - just as energy and the linear momentum - to the volume of the element".

This is recorded in textbooks [4,5]. Since Emma Noether, this angular momentum has been described by the spin tensor density [6-8].

Υc​λμν=−2A[λδμ]α∂L∂(∂νAα)=−2A[λFμ]ν (1)

Where L=−FμνFμν/4 is the free electromagnetic field Lagrangian, Aλ is the vector potential, and Fμν is the field-strength tensor. The local sense of a spin tensor is as follows. Υxyt [J*s/m3] is spin volume density, Υxyl [J*s/m2] is spin flux density, i.e. torque per unit area (cf. J. H. Poynting). The spin tensor is used in the publications [9-20]. However, the spin tensor is ignored in works expressing the common point of view, e.g. [21-25].

Besides spin, any electromagnetic field contains mass-energy and momentum, which are described by the energy-momentum tensor [26,27].

Tμν=−gμλFλαFνα+gμνFαβFαβ/4 (2)

The local sense of the energy-momentum tensor is as follows. Txt [N*s/m3] is momentum volume density, Ttx [kg/m2*s] is mass-energy flux density. It means, e.g., dpx=TxtdV is the momentum in the volume dV.

Moment of momentum, e.g., dLxy=(xTyt−yTxt)dV is the orbital angular momentum of the momentum contained in the volume dV. So, the total angular momentum possessed by the volume dV is

dJik=dSik+dLik=(Υikt+2r[iTk]t)dV (3)

The total torque per the area dal, i.e. angular momentum flux, is

dτik=dτS​ik+dLik/dt=(Υikl+2r[iTk]l)dal (4)

It is important that spin is not associated with a moment of a linear momentum, or even with a motion of matter. Hehl writes about spin of an electron :

"The current density in Dirac's theory can be split into a convective part and a polarization part. The polarization part is determined by the spin distribution of the electron field. It should lead to no energy flux in the rest system of the electron because the genuine spin 'motion' take place only within a region of the order of the Compton wavelength of the electron".

Electromagnetic field of a rotating dipole p is well known [27,29,30].

E=[ω2(pr2−(pr)r)4πε0c2r3+iω(pr2−3(pr)r4πε0cr4−(pr2−3(pr)r4πε0r5]exp(ikr−iωt) (5)

H=[ω2r×p4πcr2+iωr×p4πr3]exp(ikr−iωt) (6)

The first terms of (5), (6) are proportional to 1/r and so represent radiation. This radiation is of circular polarization in the direction of the rotational axis, z-axis (see Figure 1 from ). Therefore this field contains the spin flux Υxyl. We calculate this spin flux per sphere r=Const in Section 3.

At the same time this radiation contains no orbital angular momentum flux per elements dal of the sphere r=Const. dLik/dt=2r[iTk]ldal=0. Really, the first terms fields E & H are orthogonal to each other and to the vector r. So, in any point, we can enter local Cartesian coordinates such that dal={0,0,daz}, E={Ex,0,0}, H={0,Hy,0}, r={0,0,z}, i.e. Ftx, Ftx, Fxz, Fxz are not equal to zero only. Using this coordinates we find according to (2): Txz=−gxxFxαFzα=0,  Tyz=−gyyFyαFzα=0. So the orbital angular momentum is not radiated.

The second terms field of (5), (6) contains the orbital angular momentum flux, or torque, per the sphere r=Const. In Refs [32-37], spherical coordinates were used, and the angular distribution of the torque was obtained (see Figure 2):

dLik/dtdΩ=ω3p2sin2θ/16π2ε0c3 (7)

Where, dΩ=sinθdθdϕ. This torque is located in the neighborhood of the plane of rotation where the polarization is near linear. This torque is not radiated. This torque is like a static torque that someone can apply (Figure 2).

Spin radiated by the first terms field was calculated in  using the spin volume density Υxyt on the assumption that this density is moving at the speed of light. Here the spin flux density Υxyl is used. This is more naturally.

Using

E=ω2(pr2−(pr)r)4πε0c2r3exp(ikr−iωt), H=ω2r×p4πcr2exp(ikr−iωt), px=p, py=ip (8)

yields

Ex=Ftx=ω2p(r2−x2−ixy)4πεoc2r3, Ey=Fty=ω2p(ir2−xy−iy2)4πεoc2r3, Ez=Ftz=−ω2p(zx+izy)4πεoc2r3 (9)

Hx=Fzy=−iω2pz4πcr2, Hy=Fxz=ω2pz4πcr2, Hz=Fyx=ω2p(ix−y)4πcr2 (10)

Using A=−∫Edt=−iE/ω yields

Ax=ωp(−ir2+ix2−xy)4πεoc2r3, Ay=ωp(r2+ixy−y2)4πεoc2r3, Az=ωp(izx−zy)4πεoc2r3 (11)

Accordingly to Yλμν=−2A[λFμ]ν, we have

Υxyx=−R2{A¯¯¯xFyx}=ω3z2x32π2ε0c3r5, Υxyy=R2{A¯¯¯yFxy}=ω3z2y32π2ε0c3r5,

Υxyz=−R2{A¯¯¯xFyz−A¯¯¯yFxz}=ω3(r2+z2)z32π2ε0c3r5 (12)

Because of dτS​ik=Υikldal, we need the Cartesian coordinates of elements of the sphere r=Const, which spherical coordinates are dav={dar=dθdϕ, daθ=0, daϕ=0}. The transformation coefficients are;

∂r∂x=xr, ∂r∂y=yr, ∂r∂z=zr, and g√=r2sinθ. So we have

dal={dax=xsinθdθdϕr, day=yrsinθdθdϕ, daz=zrsinθdθdϕ}, and

dτS​xy=Υxyldal=Υxyxdax+Υxyyday+Υxyzdaz

=ω3p2(z2x2+z2y2+r2z2+z4)32π2ε0c3r4sinθdθdϕ=ω3p216π2ε0c3cos2θsinθdθdϕ (13)

This result, dτS​xy/dΩ=ω3p216π2ε0c3cos2θ, is coincided with Ref. . The angular distribution of the spin radiation is represent in Figure 3.

A rotating electric dipole emits angular momentum flux of two types: (i) Spin flux, which is directed mainly along the axis of rotation and determined by the spin tensor, and (ii) Orbital angular momentum flux determined by the energy-momentum tensor. The spin flux is not recognized by nowadays electrodynamics.

I am eternally grateful to Professor Robert Romer for the courageous publication of my question: "Does a plane wave really not carry spin?"  (was submitted on 07 October, 1999).

Polarization of the electric field seen by looking form different directions at a circular oscillator. https://www.vibgyorpublishers.org/content/ijme/ijme-6-026-001.gif Torque distribution. https://www.vibgyorpublishers.org/content/ijme/ijme-6-026-002.gif Spin flux distribution. https://www.vibgyorpublishers.org/content/ijme/ijme-6-026-003.gif