Electromagnetic Angular Momentum of an Orbiting Charge

Abstract

The electric field of an orbiting charge or electron observed in the rotating frame takes on a circular trajectory with a maximum radius of $R=\frac{c}{\omega '}$. The resultant extended electromagnetic structure is used to derive the spin–orbit energy of the orbiting electron. A surprising result of the derived expression is that the orbital velocity has a specific value ($\beta _0 = \frac{1}{136.96}$) in close agreement ($99.94\%$) with the experimentally determined value for the fine structure constant ($\hbar =\frac{ke^2}{c}\frac{1}{\alpha _{\mathrm{exp}}} =\frac{ke^2}{c}\frac{1}{\beta _0} \rightarrow \alpha _{\mathrm{exp}}=\beta _0$). Furthermore, the derived spin–orbit expression does not include a g-factor (i.e $g=1$) which means that the Larmor and Thomas precessions are equal and opposite, resulting in zero net precession in the lab frame. These results suggest that the quantised fine structure constant and by extension Planck’s constant, are a natural extension of classical electromagnetism and conservation of energy in a rotating frame.

Appendices

Appendix 1: Synchronous Orbit

Electrons in orbital motion about a nucleus will synchronize their rotation so that their orientation is always facing towards the nucleus as shown in Fig. 6. Hence, the orbiting electron will appear in the lab frame to be spinning on its axis with the same frequency as the orbital frequency. This is evident by the observed spin–orbit interaction where orbital electrons align their spins with the spin of the nucleus (either parallel or antiparallel).

Fig. 6

Synchronised electron spin and orbital rotation [Wikipedia: Fictitious force]

An analogous effect occurs for astronomical bodies which is known as “tidal locking” where the gravitational gradient induces tidal friction leading to synchronous rotation of the orbiting body. For example, it is well known that the Moon always has the same side continuously facing the Earth. The time it takes for a body to become tidally locked is highly dependent on the orbital radius [45, 46]: $t_{\mathrm{lock}} \propto r_{\mathrm{orbit}}^5$. Therefore bodies with smaller orbital radii will exhibit synchronous rotation much more quickly and are less likely to be disrupted from external influences. Hence most moons are either observed or thought to be in synchronous rotation. Similarly, the electric field gradient of orbiting charges will also result in synchronous rotation of electrons in atoms.

The general equation for electromagnetic angular momentum is given by: $\mathbf {L}_{\mathrm{total}} = \mathbf {p}_{\mathrm{total}} \times \mathbf {r} = \int \partial m (\mathbf {v} \times \mathbf {r}) \partial V = \int \partial m \omega r^2 (\hat{\mathbf {v}} \times \hat{\mathbf {r}}) \partial V = \int \partial m \omega r^2 \hat{\mathbf {L}} \partial V$, where the electromagnetic field energy for each mass element is given by: $\partial m = \frac{\epsilon }{2c^2} E^2$. The contributions for the orbital and spin angular momentum can be determined separately using the Parallel axis theorem: $\mathbf {I}_{\mathrm{total}} = \mathbf {I}_{\mathrm{orbit}} + \mathbf {I}_{\mathrm{spin}}$ (as shown in Fig. 7).

Fig. 7

Parallel axis theorem separates the orbital and spin inertia/angular momentum

Appendix 2: Electromagnetic Particle Theory—A Brief Literature Summary

Electromagnetic mass theory has been used to account for the mass of a stationary charge ($m= \int u'_{\mathrm{E}} \partial V=\int \frac{1}{2} \epsilon _0 E'^2 \partial V$) [11,12,13,14,15]. The application of electromagnetic mass theory to electromagnetic momentum has been heavily discussed in the literature for over a century (for example see, Griffiths [22] or Janssen [47]). While Poynting’s vector can successfully describe electromagnetic momentum of a photon, it led to inconsistent results for a moving charge (“$\frac{4}{3}$ problem” [15, 16, 19, 22, 47]), rotating capacitor (“Trouton–Noble paradox” [48,49,50]), and a moving capacitor [51, 52]. The widely acknowledged solution to this apparent discrepancy has been given by Rohrlich [1, 16, 19,20,21,22] based on the Maxwell stress tensor is to rotate the hypersurface of the energy $P^0_{\mathrm{e}}=\int u \ \mathrm {d}^3x$, and momentum $P_{\mathrm{e}}=\int \mathbf {g} \ \mathrm {d}^3x$ (which do not form a 4-vector) by the vector $n^\mu =\gamma \left( 1,\beta \right)$. This modifies the energy and momentum components the Maxwell stress tensor which are not divergenceless in the presence of a charge and/or current to form a covariant 4-vector:

$$\begin{aligned} cP^0_{\mathrm{e}}= & {} \gamma ^2 \int \left( u-\mathbf {v} \cdot \mathbf {g}\right) \mathrm {d}^3x = \gamma ^2 \int \left( \mathbf {E}^2-\mathbf {B}^2\right) \mathrm {d}^3x = \gamma \int \mathbf {E'}^2\mathrm {d}^3x' , \nonumber \\ cP^i_{\mathrm{e}}= & {} \gamma ^2 \int \left( c g^i + T^{(M)}_{\mathrm{ij}} \beta ^i \right) \mathrm {d}^3x = \gamma ^2 \mathbf {\beta } \int \left( \mathbf {E}^2-\mathbf {B}^2\right) \mathrm {d}^3x = \gamma \mathbf {\beta } \int \mathbf {E'}^2\mathrm {d}^3x' \end{aligned}$$

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where $t'=$constant, $\mathrm {d}^3x' = \gamma \mathrm {d}^3x$, $u=\frac{1}{2} \epsilon _0 E^2+\frac{1}{2} \epsilon _0 c^2 B^2, \mathbf {g}=\frac{1}{c^2}\mathbf {S}=\epsilon _0 \left( \mathbf {E} \times \mathbf {B} \right)$,

and $T^M_{\mathrm{ij}}= \epsilon _0 \left[ E_{\mathrm{i}} E_{\mathrm{j}}+c^2B_{\mathrm{i}} B_{\mathrm{j}} - \frac{1}{2}\left( \mathbf {E} \cdot \mathbf {E} +c^2\mathbf {B} \cdot \mathbf {B}\right) \delta _{\mathrm{ij}}\right]$ is the Maxwell stress tensor.

A related approach by Butler [50, 53] instead uses a dual form of the electromagnetic field 4-vector $d\mathbf {G}_{\mu }=\left[ \gamma \left( \mathbf {E}^2-\mathbf {B}^2 \right) /2c^2\right] u_{\mu }$. Teukolsky [21] and Jackson [1] have stated that both Rohrlich’s and Butler’s solutions are both physically correct.

In summary, we can simply say that the electromagnetic momentum density: $\mathbf {p}_{\mathrm{v}} = \frac{u'_{\mathrm{E}}}{c^2} \mathbf {v}$ is invariant in accordance with relativity (ie $E'^2=E^2-c^2B^2 \rightarrow u_{\mathrm{total}}=u'_{\mathrm{E}}=\frac{1}{2}\epsilon _0 \left( E^2-c^2B^2\right)$), however the “retarded volume” over which electromagnetic field energy extends does transform ($V=\gamma V'$). Therefore the total electromagnetic momentum of a moving charge (or capacitor) is given by: $\mathbf {p} = \int \frac{u'_{\mathrm{E}}}{c^2} \mathbf {v} \mathrm {d}V = \gamma m' \mathbf {v}$ and is consistent with relativistic momentum [1, 16,17,18,19,20,21,22,23].

This extended electromagnetic particle model is used in this paper to estimate the spin–orbit interaction energy and the resultant velocity/momentum of the orbital electron. The mass element we use in our calculations is:

$$\begin{aligned} \partial m= u'_{\mathrm{E}} \partial V= \frac{1}{2} \epsilon _0 E'^2 \gamma \partial V' \end{aligned}$$

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Furthermore, it is conjectured that in the electron’s frame of reference there is only an electric field, and that there is no spin, no magnetic field and no magnetic moment (See Appendix 3). Whereas in the lab frame, there will be spin, a magnetic field and a magnetic moment. As stated above the net electromagnetic field density will be the same as that in the electron’s frame of reference (ie net electromagnetic field density is invariant: $E'^2=E^2-c^2B^2 \rightarrow u_{\mathrm{total}}=u'_{\mathrm{E}}=\frac{1}{2}\epsilon _0 \left( E^2-c^2B^2\right)$). If the magnetic field or magnetic moment were to somehow have an additional effect on the electromagnetic mass, it would be of the order of $B \approx \beta \times E$. Hence $\frac{u_{\mathrm{B}}}{u_{\mathrm{E}}} \approx \beta ^2 \approx \frac{1}{137^2} \approx 0.0000533$ which would be a 2nd or 3rd order effect to the accuracy of the fine structure constant value estimated in this paper.

Appendix 3: Mass of an Electron: Electromagnetic $+$ Non-electromagnetic

The total mass of an electron is considered to have both electromagnetic and perhaps also non-electromagnetic mass contributions. The electromagnetic contribution is commonly given by integrating the electromagnetic contributions over all space extending from $r=\infty$ to the spherical radius: $r=a$ on which the charge of the electron is evenly distributed: $m_{\mathrm{elec}}= \int ^\infty _a \frac{1}{c^2} u'_{\mathrm{E}} \partial V=\int ^\infty _a \frac{\epsilon _0}{2 c^2} E'^2 \partial V$ [11,12,13,14,15].

The possible non-electromagnetic contribution refers to all “nonelectrical” components of the electron (sometimes referred to as “mechanical” mass) that includes forces which keep the electron together and not fly apart from the electric repulsive forces. The extra nonelectrical forces are also known collectively by the more elegant name of “Poincarè stresses”. Hence these “nonelectrical” components will be located within the spherical volume of the electric charge. Hence the “the Poincarè stresses” will be confined to a maximum radius of: $r \le a$. $m_{\mathrm{non-elec}}= \int ^a_0 F_{\mathrm{Poincar}\grave{\mathrm{e}}} \partial V$ So collectively the total mass is given by:

$$\begin{aligned} m_{\mathrm{total}}= & {} \int ^\infty _a \frac{\epsilon _0}{2 c^2} E'^2 \partial V + \int ^a_0 \frac{1}{c^2} F_{\mathrm{Poincar}\grave{\mathrm{e}}} \partial V \nonumber \\= & {} m_{\mathrm{elec}} + m_{\mathrm{non-elec}} \end{aligned}$$

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For such an electron in an orbit, the total orbital angular momentum will be proportional to the total mass: $m_{\mathrm{total}}$:

$$\begin{aligned} \mathbf {L}_{\mathrm{orbit}}= & {} \int \partial m_{\mathrm{total}} \omega r_0^2 \partial V \nonumber \\= & {} \left( m_{\mathrm{elec}} + m_{\mathrm{non-elec}}\right) \omega r_0^2 \end{aligned}$$

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However for such an electron spinning about its axis, the spin angular momentum will be dominated by the electromagnetic mass inertia which extends from radius of the spherical charge to infinity: $r: a \rightarrow \infty$, while the nonelectrical mass inertia contribution is confined to the radius of the spherical charge $r \le a$. Therefore the total spin angular momentum is given by:

$$\begin{aligned} \mathbf {S} =&\int \partial m_{\mathrm{total}} \omega r^2 \partial V \nonumber \\ =&\int ^\infty _a \frac{\epsilon _0}{2 c^2} E'^2 \omega r^2 \partial V + \int ^a_0 \frac{1}{c^2} F_{\mathrm{Poincar}\grave{\mathrm{e}}} \omega r^2 \partial V \nonumber \\ \approxeq&\int ^\infty _a \frac{\epsilon _0}{2c^2} E'^2 \omega r^2 \partial V \end{aligned}$$

(42)

Hence, it is the extended electromagnetic structure that will dominate the rotational inertia and subsequently the spin–orbit contribution.