# International Journal of Magnetics and Electromagnetism

(ISSN: 2631-5068)

Volume 6, Issue 1

Research Article

#### DOI: 10.35840/2631-5068/6526

# Angular Momentum Emission by a Rotating Dipole

Radi I Khrapko^{*}

### Table of Content

### Figures

**Figure 1:** Polarization of the electric field seen....

Polarization of the electric field seen by looking form different directions at a circular oscillator.

### References

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**Author Details**

Radi I Khrapko^{*}

Moscow Aviation Institute, Volokolamskoe Shosse 4, Russia

**Corresponding author**

Radi I Khrapko, Moscow Aviation Institute, Volokolamskoe Shosse 4, 125993, Moscow, Russia.

Accepted: December 09, 2020 | Published Online: December 11, 2020

Citation: Khrapko RI (2020) Angular Momentum Emission by a Rotating Dipole. Int J Magnetics Electromagnetism 6:026.

Copyright: © 2020 Khrapko RI. 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.

## Abstract

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.

## Keywords

Classical spin, Electrodynamics, Spin radiation

## PACS

75.10.Hk, 03.50.De

## Introduction

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

JH Poynting [2]: "If we put E for the energy in unit volume and G for the torque per unit area, we have $G=E\lambda /2\pi $".

This means that such radiation is Weyssenhoff's spin-fluid [3].

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].

$\underset{c}{{\rm Y}}{\text{}}^{\lambda \mu \nu}=-2{A}^{[\lambda}{\delta}_{\alpha}^{\mu ]}\frac{\partial L}{\partial ({\partial}_{\nu}{A}_{\alpha})}=-2{A}^{[\lambda}{F}^{\mu ]\nu}\text{(1)}$

Where $L=-{F}_{\mu \nu}{F}^{\mu \nu}/4$ is the free electromagnetic field Lagrangian, ${A}^{\lambda}$ is the vector potential, and ${F}_{\mu \nu}$ is the field-strength tensor. The local sense of a spin tensor is as follows. ${{\rm Y}}^{xyt}$ [J*s/m^{3}] is spin volume density, ${{\rm Y}}^{xyl}$ [J*s/m^{2}] 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].

${T}^{\mu \nu}=-{g}^{\mu \lambda}{F}_{\lambda \alpha}{F}^{\nu \alpha}+{g}^{\mu \nu}{F}_{\alpha \beta}{F}^{\alpha \beta}/4\text{(2)}$

The local sense of the energy-momentum tensor is as follows. ${T}^{xt}$ [N*s/m^{3}] is momentum volume density, ${T}^{tx}$ [kg/m^{2}*s] is mass-energy flux density. It means, e.g., $d{p}^{x}={T}^{xt}dV$ is the momentum in the volume $dV$.

Moment of momentum, e.g., $d{L}^{xy}=(x{T}^{yt}-y{T}^{xt})dV$ is the orbital angular momentum of the momentum contained in the volume $dV$. So, the total angular momentum possessed by the volume $dV$ is

$d{J}^{ik}=d{S}^{ik}+d{L}^{ik}=({{\rm Y}}^{ikt}+2{r}^{[i}{T}^{k]t})dV\text{(3)}$

The total torque per the area $d{a}_{l}$, i.e. angular momentum flux, is

$d{\tau}^{ik}=d\underset{S}{\tau}{\text{}}^{ik}+d{L}^{ik}/dt=({{\rm Y}}^{ikl}+2{r}^{[i}{T}^{k]l})d{a}_{l}\text{(4)}$

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 [28]:

"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

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

$E=\left[\frac{{\omega}^{2}(p{r}^{2}-(\mathrm{pr})r)}{4\pi {\epsilon}_{0}{c}^{2}{r}^{3}}+\frac{i\omega (p{r}^{2}-3(\mathrm{pr})r}{4\pi {\epsilon}_{0}c{r}^{4}}-\frac{(p{r}^{2}-3(\mathrm{pr})r}{4\pi {\epsilon}_{0}{r}^{5}}\right]\mathrm{exp}(ikr-i\omega t)\text{(5)}$

$H=\left[\frac{{\omega}^{2}r\times p}{4\pi c{r}^{2}}+\frac{i\omega r\times p}{4\pi {r}^{3}}\right]\mathrm{exp}(ikr-i\omega t)\text{(6)}$

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 [31]). Therefore this field contains the spin flux ${{\rm Y}}^{xyl}$. 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 $d{a}_{l}$ of the sphere $r=Const$. $d{L}^{ik}/dt=2{r}^{[i}{T}^{k]l}d{a}_{l}=0$. 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 $d{a}_{l}=\{0,0,d{a}_{z}\}$, $E=\{{E}_{x},0,0\},\text{\hspace{0.33em}}H=\{0,{H}_{y},0\},\text{\hspace{0.33em}}r=\{0,0,z\}$, i.e. ${F}_{tx},\text{\hspace{0.33em}}{F}^{tx},\text{\hspace{0.33em}}{F}_{xz},\text{\hspace{0.33em}}{F}^{xz}$ are not equal to zero only. Using this coordinates we find according to (2): ${T}^{xz}=-{g}^{xx}{F}_{x\alpha}{F}^{z\alpha}=0,\text{\hspace{0.33em}}\text{\hspace{0.33em}}{T}^{yz}=-{g}^{yy}{F}_{y\alpha}{F}^{z\alpha}=0$. 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):

$d{L}^{ik}/dtd\Omega ={\omega}^{3}{p}^{2}{\mathrm{sin}}^{2}\theta /16{\pi}^{2}{\epsilon}_{0}{c}^{3}\text{(7)}$

Where, $d\Omega =\mathrm{sin}\theta d\theta d\varphi $. 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 Radiation by a Rotating Dipole

Spin radiated by the first terms field was calculated in [15] using the spin volume density ${{\rm Y}}^{xyt}$ on the assumption that this density is moving at the speed of light. Here the spin flux density ${{\rm Y}}^{xyl}$ is used. This is more naturally.

Using

$E=\frac{{\omega}^{2}(p{r}^{2}-(\mathrm{pr})r)}{4\pi {\epsilon}_{0}{c}^{2}{r}^{3}}\mathrm{exp}(ikr-i\omega t),\text{\hspace{1em}}H=\frac{{\omega}^{2}r\times p}{4\pi c{r}^{2}}\mathrm{exp}(ikr-i\omega t)\text{,}{p}_{x}=p,\text{\hspace{1em}}{p}_{y}=ip\text{(8)}$

yields

${E}_{x}={F}_{tx}=\frac{{\omega}^{2}p({r}^{2}-{x}^{2}-ixy)}{4\pi {\epsilon}_{o}{c}^{2}{r}^{3}}\text{,}{E}_{y}={F}_{ty}=\frac{{\omega}^{2}p(i{r}^{2}-xy-i{y}^{2})}{4\pi {\epsilon}_{o}{c}^{2}{r}^{3}},\text{}{E}_{z}={F}_{tz}=\frac{-{\omega}^{2}p(zx+izy)}{4\pi {\epsilon}_{o}{c}^{2}{r}^{3}}\text{(9)}$

${H}_{x}={F}^{zy}=\frac{-i{\omega}^{2}pz}{4\pi c{r}^{2}},\text{}{H}_{y}={F}^{xz}=\frac{{\omega}^{2}pz}{4\pi c{r}^{2}},\text{}{H}_{z}={F}^{yx}=\frac{{\omega}^{2}p(ix-y)}{4\pi c{r}^{2}}\text{(10)}$

Using $A=-{\displaystyle \int Edt=-iE/\omega}$ yields

${A}_{x}=\frac{\omega p(-i{r}^{2}+i{x}^{2}-xy)}{4\pi {\epsilon}_{o}{c}^{2}{r}^{3}},\text{}{A}_{y}=\frac{\omega p({r}^{2}+ixy-{y}^{2})}{4\pi {\epsilon}_{o}{c}^{2}{r}^{3}},\text{}{A}_{z}=\frac{\omega p(izx-zy)}{4\pi {\epsilon}_{o}{c}^{2}{r}^{3}}\text{(11)}$

Accordingly to ${Y}^{\lambda \mu \nu}=-2{A}^{[\lambda}{F}^{\mu ]\nu}$, we have

${{\rm Y}}^{xyx}=-\frac{\Re}{2}\left\{{\overline{A}}^{x}{F}^{yx}\right\}=\frac{{\omega}^{3}{z}^{2}x}{32{\pi}^{2}{\epsilon}_{0}{c}^{3}{r}^{5}},\text{}{{\rm Y}}^{xyy}=\frac{\Re}{2}\left\{{\overline{A}}^{y}{F}^{xy}\right\}=\frac{{\omega}^{3}{z}^{2}y}{32{\pi}^{2}{\epsilon}_{0}{c}^{3}{r}^{5}},$

${{\rm Y}}^{xyz}=-\frac{\Re}{2}\{{\overline{A}}^{x}{F}^{yz}-{\overline{A}}^{y}{F}^{xz}\}=\frac{{\omega}^{3}({r}^{2}+{z}^{2})z}{32{\pi}^{2}{\epsilon}_{0}{c}^{3}{r}^{5}}\text{(12)}$

Because of $d\underset{S}{\tau}{\text{}}^{ik}={{\rm Y}}^{ikl}d{a}_{l}$, we need the Cartesian coordinates of elements of the sphere $r=Const$, which spherical coordinates are $d{a}_{v}=\{d{a}_{r}=d\theta d\varphi ,\text{\hspace{0.33em}}d{a}_{\theta}=0,\text{\hspace{0.33em}}d{a}_{\varphi}=0\}$. The transformation coefficients are;

$\frac{\partial r}{\partial x}=\frac{x}{r},\text{\hspace{1em}}\frac{\partial r}{\partial y}=\frac{y}{r},\text{\hspace{1em}}\frac{\partial r}{\partial z}=\frac{z}{r}$, and $\sqrt{g}={r}^{2}\mathrm{sin}\theta $. So we have

$d{a}_{l}=\{d{a}_{x}=x\mathrm{sin}\theta d\theta d\varphi r,\text{\hspace{0.33em}}d{a}_{y}=yr\mathrm{sin}\theta d\theta d\varphi ,\text{\hspace{0.33em}}d{a}_{z}=zr\mathrm{sin}\theta d\theta d\varphi \},\text{and}$

$d\underset{S}{\tau}{\text{}}^{xy}={{\rm Y}}^{xyl}d{a}_{l}={{\rm Y}}^{xyx}d{a}_{x}+{{\rm Y}}^{xyy}d{a}_{y}+{{\rm Y}}^{xyz}d{a}_{z}$

$=\frac{{\omega}^{3}{p}^{2}({z}^{2}{x}^{2}+{z}^{2}{y}^{2}+{r}^{2}{z}^{2}+{z}^{4})}{32{\pi}^{2}{\epsilon}_{0}{c}^{3}{r}^{4}}\mathrm{sin}\theta d\theta d\varphi =\frac{{\omega}^{3}{p}^{2}}{16{\pi}^{2}{\epsilon}_{0}{c}^{3}}{\mathrm{cos}}^{2}\theta \mathrm{sin}\theta d\theta d\varphi \text{(13)}$

This result, $d\underset{S}{\tau}{\text{}}^{xy}/d\Omega =\frac{{\omega}^{3}{p}^{2}}{16{\pi}^{2}{\epsilon}_{0}{c}^{3}}{\mathrm{cos}}^{2}\theta $, is coincided with Ref. [15]. The angular distribution of the spin radiation is represent in Figure 3.

## Conclusion

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?" [38] (was submitted on 07 October, 1999).