Electromagnetic approach to find the origin of mass, electromagnetic mass theory

Fig 1

Considering a simple uniformly surface-charged massless sphere with radius $a$ and total charge $q$ (see Figure 1), energy and momentum of an electromagnetic field can be acquired:
\begin{align}
    E_\text{EM} &=\frac{1}{2}\int \mathbf{E} \cdot \mathbf{D} \; d^3x = \frac{1}{2}\frac{q^2}{4\pi\epsilon_0}\frac{1}{a}, \\ \textbf{p}_\text{EM} &=\int \mathbf{E}\times \mathbf{H} \; d^3x = \frac{2}{3} \frac{q^2}{4\pi\epsilon_0} \frac{\textbf{v}}{ac^2}.
\end{align}
The Lorentz boost transformation of $E_\text{EM}$ makes no change under the low-speed approximation ($v \ll c$). The discrepancy between momentum-derived mass and energy-derived mass appears: 
\begin{equation}
    m_\text{EM}^{(p)} = \frac{\textbf{p}_\text{EM}}{\textbf{v}_\text{EM}} = \frac{2}{3} \frac{q^2}{4\pi\epsilon_0 ac^2}, \;\; m_\text{EM}^{(E)}=\frac{E_\text{EM}}{c^2} =  \frac{1}{2} \frac{q^2}{4\pi\epsilon_0 ac^2}
\end{equation}
which was well-known ``$4/3$-paradox,'' since their relation is $m_\text{EM}^{(p)}=\frac{4}{3}m_\text{EM}^{(E)}$. It seemed to disprove the MEE for the electromagnetic field. This paradox was solved in many papers (Fermi 1922, Dirac1938, Schwinger 1983, Rohrlich 1960, Medina 2006) in that the factor of $4/3$ was compensated for by a non-electromagnetic contribution to make consistent with $E_\text{EM}=m_\text{EM}c^2$. The Lorentz boost of rest-frame energy-momentum tensor $T_{\mu\nu}$ alongside the x-axis (velocity direction) gives additional mechanical momentum inside the sphere:
\begin{equation}\tag{1}
    p_x = \gamma(v)^2 \int_\text{out} v \frac{u_\text{EM}}{c^2} \; d^3x + \gamma(v)^2 \int_\text{in} v \frac{T_{xx}}{c^2} \;d^3x \simeq \frac{2}{3} \frac{q^2}{4\pi\epsilon_0} \frac{\boldsymbol{v}}{ac^2} - \frac{1}{6} \frac{q^2}{4\pi\epsilon_0} \frac{\boldsymbol{v}}{ac^2} = \frac{1}{2} \frac{q^2}{4\pi\epsilon_0} \frac{\boldsymbol{v}}{ac^2},
\end{equation}
resulting in $m_\text{EM}^{(p)}=m_\text{EM}^{(E)}$.

 

 

Fig 2

Next let us show $m_\text{EM}^{(E)}=m_\text{EM}^{(F)}$. I now bring up an acceleration. When a volume-charged sphere is being accelerated, differential forces acting on each charge element becomes angle dependent (Figure 2R) due to their influence by retarded time (Figure 2L). All these differential forces being integrated over the sphere creates a kind of self-force. Let the state of being infinitely spread out and not interacting with each fragmented charge be chosen as a reference state. Then the self-energy to collect all these fragments is 
\begin{equation}\tag{1}
    E_\text{self} = \frac{1}{8\pi\epsilon_0}\int\int \frac{\sigma^2}{\left|\textbf{x}_1-\textbf{x}_2\right|} \; d\tau_1 \; d\tau_2,
\end{equation}
where $\sigma$ represents the uniform charge density. The self-energy to accelerate this sphere is calculated ($\textbf{x}=\textbf{x}_1-\textbf{x}_2$) after some lengthy algebra as
\begin{equation}\tag{2}
    \textbf{F}_\text{self} = \frac{1}{4\pi\epsilon_0} \int\int \left( \frac{\textbf{x}}{\left|\textbf{x}\right|^3} + \frac{\textbf{a} \cdot \textbf{x}}{c^2\left|\textbf{x}\right|^3}\textbf{x} - \frac{\textbf{a}}{c^2 \left| \textbf{x} \right|}\right) \left(1+\frac{\textbf{a}\cdot\textbf{x}}{2c^2} \right) \sigma^2 \; d\tau_1 \; d\tau_2  \simeq -\frac{\textbf{a}}{8\pi\epsilon_0 c^2} \int\int \frac{\sigma^2}{\left|\textbf{x}_1-\textbf{x}_2\right|} \; d\tau_1 \; d\tau_2,
\end{equation}
resulting in $m_\text{EM}^{(E)}=m_\text{EM}^{(F)}$ when $v\ll c$.

 

 

 

 

Detailed proof of (1) & (2)

Consider the radius vector of a sphere $\boldsymbol{r}$. When the vector $\boldsymbol{r}$ is affected by the anisotropic gravitation $\boldsymbol{g}$, the average proper velocity of light also affected to become

\begin{equation}
c_\text{avg} = c\left(1+\frac{\textbf{g}\cdot\textbf{r}}{2c^2} \right).
\end{equation}

For small gravity $\textbf{g}\cdot\textbf{r} \ll c^2$, since $r=ct$, the proper radius vector to reach a point around the sphere can be expressed as

\begin{equation}
r_\text{avg}^{-1}\simeq r^{-1} \left(1-\frac{\textbf{g}\cdot\textbf{r}}{2c^2} \right).
\end{equation}

From the straight forward application to situation of Figure 2, with the equivalence principle $\textbf{a}=-\textbf{g}$, the anisotropic volume element can be derived:

\begin{equation}
d\tau^{(a)}=\left.d\tau_\text{avg}\right|_{\textbf{a}=-\textbf{g}}=d\tau \left(1+\frac{\textbf{a}\cdot\textbf{r}}{2c^2} \right).
\end{equation}

Also I get the expression for the infinitesimal scalar potential considering $\textbf{a}=-\textbf{g}$

\begin{equation}
d\phi^{(a)} = \left. \frac{1}{4\pi\epsilon_0} \frac{\sigma \;d\tau_\text{avg}}{r_\text{avg}}\right|_{\textbf{a}=-\textbf{g}}=\frac{1}{4\pi\epsilon_0}\frac{\sigma\;d\tau}{r}\left(1+\frac{\textbf{a}\cdot\textbf{r}}{2c^2} \right)^2,
\end{equation}

and upto the leading order,

\begin{equation}
d\phi^{(a)} = \frac{1}{4\pi\epsilon_0} \frac{\sigma\;d\tau}{r} \left(1+\frac{\textbf{a}\cdot\textbf{r}}{c^2} \right).
\end{equation}

Thus the self-force that the electromagnetic field exerts upon the other volume charge element is

\begin{align*}
\textbf{F}_\text{self} &= -\int \sigma\;d\tau_2^{(a)}\nabla \int d\phi^{(a)}_1 \\ &= \frac{1}{4\pi\epsilon_0} \int\int \left( \frac{\textbf{x}}{\left|\textbf{x}\right|^3} + \frac{\textbf{a} \cdot \textbf{x}}{c^2\left|\textbf{x}\right|^3}\textbf{x} - \frac{\textbf{a}}{c^2 \left| \textbf{x} \right|}\right) \left(1+\frac{\textbf{a}\cdot\textbf{x}}{2c^2} \right) \sigma^2 \; d\tau_1 \; d\tau_2,
\end{align*}

where $\textbf{x}=\textbf{r}_1-\textbf{r}_2$. Here I have derived the left part of (2). Similarly, one also can derive (1) as

\begin{equation}
E_\text{self} = \frac{1}{8\pi\epsilon_0}\int\int \frac{\sigma^2}{\left|\textbf{x}_1-\textbf{x}_2\right|} \; d\tau_1 \; d\tau_2.
\end{equation}

 

Now let us shift our attention to the problem of $m_\text{EM}^{(F)}=m_\text{EM}^{(E)}$. The expression (2) can be written within the leading order of $c^{-2}$ as

\begin{equation}
    \textbf{F}_\text{self} = \frac{1}{8\pi\epsilon_0} \int\int \left( \frac{\textbf{x}}{\left|\textbf{x}\right|^3} + \frac{3}{2}\frac{\textbf{a} \cdot \textbf{x}}{c^2\left|\textbf{x}\right|^3}\textbf{x} - \frac{\textbf{a}}{c^2 \left| \textbf{x} \right|}\right) \sigma^2 \; d\tau_1 \; d\tau_2.
\end{equation}

This system has the interchanging symmetry $\boldsymbol{x}_1 \leftrightarrow \boldsymbol{x}_2$, yielding $\boldsymbol{x}\to -\boldsymbol{x}$ and invariant $\boldsymbol{F}_\text{self}$. Exchange-odd term thus has no contribution when integrated over the sphere as

\begin{equation}
\textbf{F}_\text{self} = \frac{1}{8\pi\epsilon_0} \int\int \left(\frac{3}{2}\frac{\textbf{a} \cdot \textbf{x}}{c^2\left|\textbf{x}\right|^3}\textbf{x} - \frac{\textbf{a}}{c^2 \left| \textbf{x} \right|}\right) \sigma^2 \; d\tau_1 \; d\tau_2.
\end{equation}

Using the decomposition $\textbf{x}=\textbf{x}_\parallel + \textbf{x}_\perp$ such that $\textbf{x}_\perp \perp \textbf{a}$, then, in the second term, $\left(\textbf{a} \cdot \textbf{x}\right)\textbf{x}=\left(\textbf{a} \cdot \textbf{x}_\parallel\right)\textbf{x}_\parallel + \left(\textbf{a} \cdot \textbf{x}_\parallel\right)\textbf{x}_\perp$. Integrating over the

sphere, the rotated contribution that the vector $\textbf{x}$ is rotated $180^{\circ}$ about $\textbf{a}$ perfectly cancels out to leave only the parallel term as

\begin{equation}
\textbf{F}_\text{self} = \frac{1}{8\pi\epsilon_0} \int\int \left(\frac{3}{2}\frac{\textbf{a} \cdot \textbf{x}_\parallel} {c^2\left|\textbf{x}\right|^3}\textbf{x}_\parallel - \frac{\textbf{a}}{c^2 \left| \textbf{x} \right|}\right) \sigma^2 \; d\tau_1 \; d\tau_2.
\end{equation}

Since the magnitude of $\left( \textbf{a} \cdot \textbf{x}_\parallel \right) \textbf{x}_\parallel$ is $\left| \left( \textbf{a} \cdot \textbf{x}_\parallel \right) \textbf{x}_\parallel\right| = ax^2\cos^2\theta$ where $\theta$ is the intersection angle between $\textbf{a}$ and $\textbf{x}$, one can write $\left( \textbf{a} \cdot \textbf{x}_\parallel \right) \textbf{x}_\parallel = \textbf{a} \left( \hat{\textbf{e}}_\textbf{a} \cdot \textbf{x} \right)^2$. Then the first term can be rewritten as ($\hat{\textbf{e}}_\textbf{x}=\textbf{x}/|\textbf{x}|$ and $\hat{\textbf{e}}_\textbf{a}=\textbf{a}/|\textbf{a}|$)

\begin{equation}\tag{3}
\int\int \frac{\textbf{a} \cdot \textbf{x}_\parallel}{|\textbf{x}|^3} \textbf{x}_\parallel\; d\tau_1 \; d\tau_2 = \int\int \frac{\left(\hat{\textbf{e}}_\textbf{x} \cdot \hat{\textbf{e}}_\textbf{a}\right)^2}{x}\textbf{a} \; d\tau_1\;d\tau_2.
\end{equation}

For the spherical charge distribution, all directions are indistinguishable. Therefore one can extract the average directional integral out from (3) as

\begin{align}
    \int\int \frac{\left(\hat{\textbf{e}}_\textbf{x} \cdot \hat{\textbf{e}}_\textbf{a}\right)^2}{x}\textbf{a} \; d\tau_1\;d\tau_2 &= \frac{1}{4\pi}\int\int \frac{\textbf{a} d\tau_1 \; d\tau_2}{x}\int \left( \hat{\textbf{e}}_\textbf{x} \cdot \hat{\textbf{e}}_\textbf{a}\right)^2 d\Omega \\
    &=\frac{1}{4\pi}\int\int \frac{\textbf{a} d\tau_1 \; d\tau_2}{x}\int \cos^2 \theta \; d\Omega\\
    &=\frac{1}{3}\int\int \frac{\textbf{a} d\tau_1 \; d\tau_2}{x}.
\end{align}

Thus I get the consistent result:

\begin{equation}
\textbf{F}_\text{self} = -\frac{\textbf{a}}{8\pi\epsilon_0 c^2} \int\int \frac{\sigma^2}{\left|\textbf{x}_1-\textbf{x}_2\right|} \; d\tau_1 \; d\tau_2 = -\frac{E_\text{self}}{c^2}\textbf{a}.
\end{equation}