Impact on Mass of A Moving Charged Object

Synopsis

When Speed << c
Impact on Mass of A Moving Charged Object – When Speed → c

When Speed << c

Till the end 19th century, when electrical science research were on peak, different pioneer came up with analysis that a charged body moving will gain mass.

J J Thomson was the first to come up with the idea of increase of mass of moving charged body when moving in space. Most electrical pioneers were mostly mathematical with very hotchpotch physical theory.

He made quantized electricity in to "physical" entity with once single concept called Faraday Tubes, which itself extends from the actual works of Faraday.

He rejected Maxwell's mathematical theory and took a physical approach, through empirically same as that of Maxwell. But different physical meaning.

He was the first to reject magnetic field, making it as a side effect of electric field in motion. In other words he brought physical unification of electricity and magnetism, unlike Maxwell's mathematical unification.

His visualization of mass and momentum is different than that of Newtonian. For example, it's assumed that mass is something resides inside an object. But according to Thomson, the mass is not just inside a charged object but rather extends throughout the space. Hence its motion leads to resistance hence appears like increase in apparent mass. This so-called relativistic mass increase has it's roots in Thomson, and Not Einstein.

He showed that magnetic field is due to motion of faraday tubes. Hence:

$\begin{aligned} \vec{H} = \vec{v} \times \vec{D} \end{aligned}$

He treated electric lines of force as not just abstract mathematical representations rather concrete physical reality as Faraday Tube of Induction. The displacement vector D is the abstract representation of number of excessive tubes passing per unit area in space between two points. And the net charge Q is the net effective Faraday tubes attached to a charged object.

So now you cannot have infinite number of lines of forces. Hence you cannot have fractional charge too, which electrolysis also confirms. Hence electricity is quantized in this manner, which can be traced to Thomson only.

Lets have a charged sphere with radius a moving with velocity v, with surface charge Q in the origin has the electric field intensity E at any point of space in spherical coordinate system. It's given by the expression:

$\begin{aligned} E = \frac{Q}{4 π ϵ r^{2}} \end{aligned} ----(1)$

Then the net magnetic field intensity becomes:

$\begin{aligned} \vec{H} & = \vec{v} \times \vec{D} \\ = v D \sin θ \\ = v ϵ E \sin θ \end{aligned} ----(2)$

Hence, from eqn(1) and eqn(2), we have:

$\begin{aligned} H = \frac{v Q \sin θ}{4 π r^{2}} \end{aligned} ----(3)$

The kinetic energy of the moving sphere is the net magnetic energy per unit volume in the system, which is given as:

$K . E = \frac{μ}{2} {| H |}^{2}$

Hence, $K . E = \frac{μ v^{2} Q^{2} \sin^{2} θ}{32 π^{2} r^{4}} [using eqn(3))]$

Now to get the total magnetic energy, we shall integrate the above expression over spherical coordinate system, as:

$\begin{aligned} K . E = \int_{0}^{2 π} \int_{0}^{π} \int_{a}^{\infty} \frac{μ v^{2} Q^{2} \sin^{2} θ}{32 π^{2} r^{4}} r^{2} \sin θ \partial r \partial θ \partial ϕ \end{aligned}$

Need to integrate over the entire space from the surface of sphere with radius a to infinity.

So,
$\begin{aligned} K . E & = \frac{μ v^{2} Q^{2}}{32 π^{2}} \int_{0}^{2 π} \int_{0}^{π} \int_{a}^{\infty} \frac{\sin^{3} θ}{r^{2}} \partial r \partial θ \partial ϕ \\ = \frac{μ v^{2} Q^{2} (2 π)}{32 π^{2}} \int_{0}^{π} \int_{a}^{\infty} \frac{\sin^{3} θ}{r^{2}} \partial r \partial θ \\ = \frac{μ v^{2} Q^{2}}{8 π} {[\frac{- 1}{r}]}_{a}^{\infty} \int_{0}^{π} \sin^{3} θ \partial θ \\ = \frac{μ v^{2} Q^{2}}{8 π a} \int_{0}^{π} \frac{1}{4} (\sin 3 θ - 3 \sin θ) \partial θ \end{aligned}$

Hence integrating we get the final electrical(magnetic) kinetic energy of the moving charge object as:

$\begin{aligned} K . E_{E} = \frac{μ}{4 π} \frac{Q^{2} v^{2}}{3 a} \end{aligned}$

Now if the mass of the object is m, the mechanical kinetic energy is given as:
$\begin{aligned} K . E_{M} = \frac{1}{2} m v^{2} \end{aligned}$

Now the total kinetic energy of the entire system is the sum of two energies, the mechanical and electrical.

If M is the apparent mass of the object on motion through the aether, the total kinetic energy is given by:

$\begin{aligned} K . E = K . E_{M} + K . E_{E} \\ o r, \frac{1}{2} M v^{2} = \frac{1}{2} m v^{2} + \frac{μ}{4 π} \frac{Q^{2}}{3 a} \\ o r, M = m + \frac{μ}{2 π} \frac{Q^{2}}{3 a} \end{aligned}$

Hence the apparent mass of the object appears to increase by the factor of $\frac{μ}{2 π} \frac{Q^{2}}{3 a}$ . This is for velocity if the speed is very less compared to that of speed of light. As we haven't taken the electric field intensity E distortion in higher speed, just magnetic field due to motion of tubes.

Impact on Mass of A Moving Charged Object – When Speed → c

Now we will take a much deeper and detailed analysis for a charged body on how it’s electric field distorts based on speed as first devised by J J Thomson.

The magnetic energy per unit volume of the system is given as:

𝐾 . 𝐸 = \frac{𝜇}{2} {∣ 𝐻 ∣}^{2} - - - - - - (1)

The electrostatic energy per unit volume is given as:

𝑃 . 𝐸 = \frac{𝜖}{2} {∣ 𝐸 ∣}^{2} - - - - - - (2)

Total energy of the system shall remain constant, hence:

𝑃 . 𝐸 + 𝐾 . 𝐸 = 𝐶 𝑜 𝑛 𝑠 𝑡 𝑎 𝑛 𝑡

Taking derivative with respect to E:

\frac{\partial}{\partial \vec{𝐸}} (𝑃 . 𝐸) + \frac{\partial}{\partial \vec{𝐸}} (𝐾 . 𝐸) = \frac{\partial}{\partial \vec{𝐸}} (𝐶 𝑜 𝑛 𝑠 𝑡 𝑎 𝑛 𝑡)

𝑜 𝑟, \frac{\partial}{\partial \vec{𝐸}} (\frac{𝜖}{2} {∣ 𝐸 ∣}^{2}) + \frac{\partial}{\partial \vec{𝐸}} (\frac{𝜇}{2} {∣ 𝐻 ∣}^{2}) = 0

𝑜 𝑟, - 𝜖 \vec{𝐸} + 𝜇 𝐻 \frac{\partial}{\partial \vec{𝐸}} (\vec{𝑣} \times \vec{𝐷}) = 0 [𝑎 𝑠, \vec{𝐻} = \vec{𝑣} \times \vec{𝐷}]

As, as kinetic energy is increasing, hence potential energy shall decrease. So the rate of change of potential energy shall be negative.

𝐻 𝑒 𝑛 𝑐 𝑒, - 𝜖 \vec{𝐸} + 𝜇 𝜖 𝐻 \frac{\partial}{\partial \vec{𝐸}} (\vec{𝑣} \times \vec{𝐸}) = 0 [𝑎 𝑠, \vec{𝐷} = 𝜖 \vec{𝐸}]

𝑜 𝑟, \vec{𝐸} = 𝐵 \frac{\partial}{\partial \vec{𝐸}} (\vec{𝑣} \times \vec{𝐸}) = 0 [𝑎 𝑠, \vec{𝐵} = 𝜇 \vec{𝐻}]

To find the derivative of v x E, lets write each vector in its component form and calculate the cross product.

We have:

\vec{𝑣} = 𝑣_{𝑥} \hat{𝑖} + 𝑣_{𝑦} \hat{𝑗} + 𝑣_{𝑧} \hat{𝑘}, 𝑎 𝑛 𝑑, \vec{𝐸} = 𝐸_{𝑥} \hat{𝑖} + 𝐸_{𝑦} \hat{𝑗} + 𝐸_{𝑧} \hat{𝑘}

𝑆 𝑜, \vec{𝑣} \times \vec{𝐸} = (𝑣_{𝑦} 𝐸_{𝑧} - 𝐸_{𝑦} 𝑣_{𝑧}) \hat{𝑖} + (𝐸_{𝑥} 𝑣_{𝑧} - 𝑣_{𝑥} 𝐸_{𝑧}) \hat{𝑗} + (𝑣_{𝑥} 𝐸_{𝑦} - 𝐸_{𝑥} 𝑣_{𝑦}) \hat{𝑘}

𝐻 𝑒 𝑛 𝑐 𝑒, 𝐵 \frac{\partial}{\partial \vec{𝐸}} (\vec{𝑣} \times \vec{𝐸}) = 𝐵 \frac{\partial}{\partial 𝐸_{𝑥}} (\vec{𝑣} \times \vec{𝐸}) \hat{𝑖} + 𝐵 \frac{\partial}{\partial 𝐸_{𝑦}} (\vec{𝑣} \times \vec{𝐸}) \hat{𝑗} + 𝐵 \frac{\partial}{\partial 𝐸_{𝑧}} (\vec{𝑣} \times \vec{𝐸}) \hat{𝑘}

\begin{aligned} = (𝐵 𝑣_{𝑧} - 𝐵 𝑣_{𝑦}) \hat{𝑖} + (𝐵 𝑣_{𝑥} - 𝐵 𝑣_{𝑧}) \hat{𝑗} + (𝐵 𝑣_{𝑦} - 𝐵 𝑣_{𝑥}) \hat{𝑘} \\ = \vec{𝐵} \times \vec{𝑣} \end{aligned}

\begin{array}{r} 𝑆 𝑜, \vec{𝐸_{1}} = \vec{𝐵} \times \vec{𝑣} \end{array}

This is the electric field intensity due to motional charge. Let’s call it E1 . Let’s E2 be the electric field in space due to the electric scalar potential Φ. Such that:

\vec{𝐸_{2}} = - \vec{\nabla} 𝜙

The scalar potential can be zero in closed loop, but not in our case.

The total electric field intensity in space for the system is given as (E1 + E2):

\vec{𝐸} = (\vec{𝐵} \times \vec{𝑣}) - \vec{\nabla} 𝜙

Now lets say a charged object of surface charge Q and radius a moving with a velocity v in the direction of z-axis, for the sake of simplicity.

The velocity vector v becomes $\vec{𝑣} = 𝑣_{𝑧} \hat{𝑘}$

Now $\begin{aligned} \vec{𝐵} = 𝜇 \vec{𝐻} & = 𝜇 (\vec{𝑣} \times \vec{𝐷}) \\ = 𝜇 [𝑣_{𝑧} \hat{𝑘} \times (𝐷_{𝑥} \hat{𝑖} + 𝐷_{𝑦} \hat{𝑗} + 𝐷_{𝑧} \hat{𝑘})] \\ = 𝜇 (- 𝑣_{𝑧} 𝐷_{𝑦} \hat{𝑖} + 𝑣_{𝑧} 𝐷_{𝑥} \hat{𝑗}) - - - - - - (3) \end{aligned}$

Now the electric field intensity E becomes:

\begin{aligned} \vec{𝐸} & = (\vec{𝐵} \times \vec{𝑣}) - \vec{\nabla} 𝜙 \\ = [𝜇 (- 𝑣_{𝑧} 𝐷_{𝑦} \hat{𝑖} + 𝑣_{𝑧} 𝐷_{𝑥} \hat{𝑗}) \times 𝑣_{𝑧} \hat{𝑘}] - \vec{\nabla} 𝜙 \\ = 𝜇 (𝑣_{𝑧}^{2} 𝐷_{𝑥} \hat{𝑖} + 𝑣_{𝑧}^{2} 𝐷_{𝑦} \hat{𝑗}) - \vec{\nabla} 𝜙 \end{aligned}

Hence $\begin{aligned} \frac{\vec{𝐷}}{𝜖} & = 𝜇 (𝑣_{𝑧}^{2} 𝐷_{𝑥} \hat{𝑖} + 𝑣_{𝑧}^{2} 𝐷_{𝑦} \hat{𝑗}) - \vec{\nabla} 𝜙 \end{aligned}$

Or, $\begin{aligned} \vec{𝐷} & = 𝜇 𝜖 (𝑣_{𝑧}^{2} 𝐷_{𝑥} \hat{𝑖} + 𝑣_{𝑧}^{2} 𝐷_{𝑦} \hat{𝑗}) - 𝜖 \vec{\nabla} 𝜙 \\ = \frac{𝑣_{𝑧}^{2}}{𝑐^{2}} (𝐷_{𝑥} \hat{𝑖} + 𝐷_{𝑦} \hat{𝑗}) - 𝜖 \vec{\nabla} 𝜙 [𝜇 𝜖 = \frac{1}{𝑐^{2}}] \end{aligned}$

Now, $(𝐷_{𝑥} \hat{𝑖} + 𝐷_{𝑦} \hat{𝑗} + 𝐷_{𝑧} \hat{𝑘}) = \frac{𝑣_{𝑧}^{2}}{𝑐^{2}} (𝐷_{𝑥} \hat{𝑖} + 𝐷_{𝑦} \hat{𝑗}) - 𝜖 \vec{\nabla} 𝜙 𝑜 𝑟, 𝐷_{𝑥} (1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}) \hat{𝑖} + 𝐷_{𝑦} (1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}) \hat{𝑗} + 𝐷_{𝑧} \hat{𝑘} = - 𝜖 \vec{\nabla} 𝜙$

Splitting the above equation for D into components we get:

𝐷_{𝑥} (1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}) = - 𝜖 \frac{\partial 𝜙}{\partial 𝑥} 𝑜 𝑟, 𝐷_{𝑥} = - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑥} - - - (4)

𝑆 𝑖 𝑚 𝑖 𝑙 𝑎 𝑟 𝑙 𝑦, 𝐷_{𝑦} = - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑦} - - - (5)

𝐴 𝑛 𝑑, 𝐷_{𝑧} = - 𝜖 \frac{\partial 𝜙}{\partial 𝑧} - - - (6)

Putting the vector components together we get:

𝐷_{𝑥} \hat{𝑖} + 𝐷_{𝑦} \hat{𝑗} + 𝐷_{𝑧} \hat{𝑘} = - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑥} \hat{𝑖} - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑦} \hat{𝑗} - 𝜖 \frac{\partial 𝜙}{\partial 𝑧} \hat{𝑘}

𝑜 𝑟, \vec{𝐷} = - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑥} \hat{𝑖} - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑦} \hat{𝑗} - 𝜖 \frac{\partial 𝜙}{\partial 𝑧} \hat{𝑘}

Now taking divergent both sides:

\vec{\nabla} . \vec{𝐷} = \vec{\nabla} . [- (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑥} \hat{𝑖} - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑦} \hat{𝑗} - 𝜖 \frac{\partial 𝜙}{\partial 𝑧} \hat{𝑘}]

As we are dealing with field in space, hence no charge is established there, so the divergence of D should be 0. Hence:

\vec{\nabla} . [- (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑥} \hat{𝑖} - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝜖 \frac{\partial 𝜙}{\partial 𝑦} \hat{𝑗} - 𝜖 \frac{\partial 𝜙}{\partial 𝑧} \hat{𝑘}] = 0

𝑜 𝑟 - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) \frac{\partial^{2} 𝜙}{\partial 𝑥^{2}} - (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) \frac{\partial^{2} 𝜙}{\partial 𝑦^{2}} - \frac{\partial^{2} 𝜙}{\partial 𝑧^{2}} = 0

𝑜 𝑟 \frac{\partial^{2} 𝜙}{\partial 𝑥^{2}} + \frac{\partial^{2} 𝜙}{\partial 𝑦^{2}} + (\frac{𝑐^{2} - 𝑣_{𝑧}^{2}}{𝑐^{2}}) \frac{\partial^{2} 𝜙}{\partial 𝑧^{2}} = 0 - - - (7)

Now to solve this equation, we have to do coordinate transform of Z axis.

Lets put:

𝑧^{'} = \frac{𝑐}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} 𝑧

Using chain rule: $\frac{\partial 𝜙}{\partial 𝑧^{'}} = \frac{\partial 𝜙}{\partial 𝑧} \frac{\partial 𝑧}{\partial 𝑧^{'}} \begin{aligned} 𝑜 𝑟, \frac{\partial^{2} 𝜙}{\partial 𝑧^{' 2}} & = \frac{\partial 𝜙}{\partial 𝑧} \frac{\partial 𝑧}{\partial 𝑧^{'}} \frac{\partial 𝑧}{\partial 𝑧^{'}} \\ = \frac{\partial^{2} 𝜙}{\partial 𝑧^{2}} (\frac{\partial 𝑧}{\partial 𝑧^{'}})^{2} \\ = (\frac{𝑐^{2} - 𝑣_{𝑧}^{2}}{𝑐^{2}}) \frac{\partial^{2} 𝜙}{\partial 𝑧^{2}} \end{aligned}$

Now eqn(7) becomes: $\frac{\partial^{2} 𝜙}{\partial 𝑥^{2}} + \frac{\partial^{2} 𝜙}{\partial 𝑦^{2}} + \frac{\partial^{2} 𝜙}{\partial 𝑧^{' 2}} = 0$

Now this is Laplace equation for potential. Following is the standard solution for electric potential:

\begin{aligned} 𝜙 & = \frac{𝐴}{\sqrt{𝑥^{2} + 𝑦^{2} + 𝑧^{' 2}}} \\ = \frac{𝐴}{\sqrt{𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2}}} \end{aligned}

To find the constant A, we need to first find gradient of the scalar potential Φ and find the Displacement D, and then integrate the Displacement over entire surface of the sphere and equate with the net charge Q as per Gauss law.

The gradient of scalar potential, ∇.Φ is:

\frac{\partial 𝜙}{\partial 𝑥} = \frac{- 𝐴 𝑥}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}}

\frac{\partial 𝜙}{\partial 𝑦} = \frac{- 𝐴 𝑦}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}}

\frac{\partial 𝜙}{\partial 𝑧} = \frac{𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \frac{- 𝐴 𝑧}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}}

Now putting the value of components of gradient of potential in eqn(4), (5) and (6), we get:

𝐷_{𝑥} = \frac{𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \frac{𝜖 𝐴 𝑥}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}} - - - (8)

𝐷_{𝑦} = \frac{𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \frac{𝜖 𝐴 𝑦}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}} - - - (9)

𝐷_{𝑧} = \frac{𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \frac{𝜖 𝐴 𝑧}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}} - - - (10)

Now according to Gauss law of electrostatics, the total electric displacement across a surface of a charged object is always the amount of charge contained.

∯ \vec{𝐷} . \hat{𝑛} 𝑑 𝑆 = 𝑄 - - - (11)

The normal unit vector n for the sphere with radius a is defined as:

\hat{𝑛} = \frac{𝑥}{𝑎} \hat{𝑖} + \frac{𝑦}{𝑎} \hat{𝑗} + \frac{𝑧}{𝑎} \hat{𝑘}

Now integral (11) becomes:

\begin{aligned} 𝑄 & = \frac{𝜖 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} ∯ (\frac{𝑥 \hat{𝑖} + 𝑦 \hat{𝑗} + 𝑧 \hat{𝑘}}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}}) . (\frac{𝑥}{𝑎} \hat{𝑖} + \frac{𝑦}{𝑎} \hat{𝑗} + \frac{𝑧}{𝑎} \hat{𝑘}) 𝑑 𝑆 \\ = \frac{𝜖 𝐴 𝑐^{2}}{𝑎 (𝑐^{2} - 𝑣_{𝑧}^{2})} ∯ \frac{𝑥^{2} + 𝑦^{2} + 𝑧^{2}}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}} 𝑑 𝑆 \\ = \frac{𝜖 𝑎 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} ∯ \frac{𝑑 𝑆}{(𝑥^{2} + 𝑦^{2} + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑧^{2})^{\frac{3}{2}}} [𝑎 𝑠, 𝑥^{2} + 𝑦^{2} + 𝑧^{2} = 𝑎^{2}] \end{aligned}

Now for proper evaluation, we need to convert this to spherical coordinate.
So we have: $𝑥 = 𝑟 𝑐 𝑜 𝑠 𝜙 𝑠 𝑖 𝑛 𝜃 𝑦 = 𝑟 𝑠 𝑖 𝑛 𝜙 𝑠 𝑖 𝑛 𝜃 𝑧 = 𝑟 𝑐 𝑜 𝑠 𝜃 𝑑 𝑆 = 𝑟^{2} 𝑠 𝑖 𝑛 𝜃 𝑑 𝜃 𝑑 𝜙$

The variable r should be set to radius a as we are not integrating over volume.

So now putting the above expression into the integral we get:

\begin{aligned} 𝑄 & = \frac{𝜖 𝑎 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \int_{0}^{𝜋} \int_{0}^{2 𝜋} \frac{𝑎^{2} 𝑠 𝑖 𝑛 𝜃 \partial 𝜃 \partial 𝜙}{(𝑎^{2} 𝑐 𝑜 𝑠^{2} 𝜙 𝑠 𝑖 𝑛^{2} 𝜃 + 𝑎^{2} 𝑠 𝑖 𝑛^{2} 𝜙 𝑠 𝑖 𝑛^{2} 𝜃 + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑎^{2} 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}} \\ = \frac{𝜖 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \int_{0}^{𝜋} \int_{0}^{2 𝜋} \frac{𝑠 𝑖 𝑛 𝜃 \partial 𝜃 \partial 𝜙}{(𝑠 𝑖 𝑛^{2} 𝜃 + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}} \\ = \frac{2 𝜋 𝜖 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \int_{0}^{𝜋} \frac{𝑠 𝑖 𝑛 𝜃 \partial 𝜃}{(𝑠 𝑖 𝑛^{2} 𝜃 + (\frac{𝑐^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}} \\ = \frac{2 𝜋 𝜖 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \int_{0}^{𝜋} \frac{𝑠 𝑖 𝑛 𝜃 \partial 𝜃}{(1 + (\frac{𝑣_{𝑧}^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}) 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}} \end{aligned}

Now we solve this integral by substituting: $\frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} 𝑐 𝑜 𝑠 𝜃 = 𝑡 𝑎 𝑛 𝜓 𝐻 𝑒 𝑛 𝑐 𝑒, \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} 𝑠 𝑖 𝑛 𝜃 \partial 𝜃 = 𝑠 𝑒 𝑐^{2} 𝜓 \partial 𝜓$

Now the above integral becomes: $\begin{aligned} 𝑄 & = \frac{2 𝜋 𝜖 𝐴 𝑐^{2}}{(𝑐^{2} - 𝑣_{𝑧}^{2})} \int_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} \frac{- \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}{𝑣_{𝑧}} \frac{𝑠 𝑒 𝑐^{2} 𝜓}{(1 + 𝑡 𝑎 𝑛^{2} 𝜓)^{\frac{3}{2}}} \partial 𝜓 \\ = \frac{- 2 𝜋 𝜖 𝐴 𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} \int_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} 𝑐 𝑜 𝑠 𝜓 \partial 𝜓 \\ = \frac{- 2 𝜋 𝜖 𝐴 𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {[𝑠 𝑖 𝑛 𝜓]}_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} \\ = \frac{- 2 𝜋 𝜖 𝐴 𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {[\frac{𝑡 𝑎 𝑛 𝜓}{\sqrt{1 + 𝑡 𝑎 𝑛^{2} 𝜓}}]}_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} \\ = \frac{- 2 𝜋 𝜖 𝐴 𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} \frac{\frac{- 2 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}{\sqrt{1 + \frac{𝑣_{𝑧}^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}}}} \\ = \frac{4 𝜋 𝜖 𝐴 𝑐}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} \end{aligned}$

Now A becomes: $𝐴 = \frac{𝑄}{4 𝜋 𝜖} \frac{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}{𝑐}$

Now putting the value of A in scalar potential equation it becomes: $\begin{aligned} 𝜙 & = \frac{𝑄 \sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}}{4 𝜋 𝜖 \sqrt{𝑥^{2} + 𝑦^{2} + (\frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}) 𝑧^{2}}} \end{aligned} - - - (12)$

You can see when the speed approaches speed of light the factor vz/c becomes prevalent and the potential becomes undefined at that speed.

The potential can be expressed in spherical coordinate as:

\begin{aligned} 𝜙 & = \frac{𝑄 \sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}}{4 𝜋 𝜖 𝑟 \sqrt{𝑠 𝑖 𝑛^{2} 𝜃 + (\frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}) 𝑐 𝑜 𝑠^{2} 𝜃}} \end{aligned}

Now putting the value of A in eqn(8), (9), (10) we get D as:

\vec{𝐷} = \frac{𝑄}{4 𝜋} \frac{1}{\sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}} \frac{1}{(𝑥^{2} + 𝑦^{2} + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑧^{2})^{\frac{3}{2}}} (𝑥 \hat{𝑖} + 𝑦 \hat{𝑗} + 𝑧 \hat{𝑘}) - - - (13)

In spherical coordinate, the D can be expressed as: $\vec{𝐷} = \frac{𝑄}{4 𝜋} \frac{1}{\sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}} \frac{1}{𝑟^{2} (𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}} (𝑐 𝑜 𝑠 𝜙 𝑠 𝑖 𝑛 𝜃 \hat{𝑟} + 𝑠 𝑖 𝑛 𝜙 𝑠 𝑖 𝑛 𝜃 \hat{𝜃} + 𝑐 𝑜 𝑠 𝜃 \hat{𝜙})$

The above equation shows that the the faraday tubes are radial and the resultant polarization D varies inversely as $𝐷 \propto \frac{1}{𝑟^{2} (𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}}$

The result shows that the polarization is greatest when θ=π/2, and least when θ=0. The faraday tubes thus leave the poles of the sphere and tend to crowd at the equator. These is due to the tendencies of tubes to set themselves at right angles to the direction in which they are moving.

The surface charge density is given by multiplying the displacement with normal vector, which is proportional as

\vec{𝐷} . \hat{𝑛} \propto \frac{1}{(𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{\frac{3}{2}}}

The surface charge density is maximum at equator and minimum at poles.

To plot the potential and displacement, we can convert eqn(12) of potential into two dimensional equation for the sake of simplicity in visualization. We can replace z as x as we can visualize moving in x direction and remove the actual x part. Also the equation is with respect to moving frame. To get the proper field picture, we can transform the x-axis into the stationary from using x → x – vt. Then we can ploy y as:

𝜙 = \frac{𝑄 \sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}}{4 𝜋 𝜖 \sqrt{𝑦^{2} + (\frac{1}{1 - \frac{𝑣^{2}}{𝑐^{2}}}) 𝑥^{2}}} 𝑜 𝑟, 𝑦 = \pm \sqrt{\frac{𝑄^{2}}{𝛽^{2} 𝜙^{2} (4 𝜋 𝜖)^{2}} - 𝛽^{2} 𝑥^{2}} [𝑤 ℎ 𝑒 𝑟 𝑒, 𝛽 = \frac{1}{\sqrt{1 - \frac{𝑣^{2}}{𝑐^{2}}}}]

Following are the plot of potential (in gray) and displacement (in blue) for the the charged object when the v =0 and when v = 0.8c.

You can clearly see when the speed approach that of light, the field starts to distort as they don’t get enough time to realign, hence compressing the aether. Following is a short visualization with animation.

The magnetic field intensity H can be now derived from D using eqn(3), as: $\vec{𝐻} = \frac{𝑄}{4 𝜋} \frac{1}{\sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}} \frac{𝑣_{𝑧}}{(𝑥^{2} + 𝑦^{2} + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑧^{2})^{\frac{3}{2}}} (- 𝑦 \hat{𝑖} + 𝑥 \hat{𝑗})$

The magnetic energy per unit volume is given from eqn(1) as $\frac{𝜇}{2} {∣ 𝐻 ∣}^{2} = \frac{𝜇 𝑄^{2}}{32 𝜋^{2}} \frac{1}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \frac{𝑣_{𝑧}^{2} (𝑥^{2} + 𝑦^{2})}{(𝑥^{2} + 𝑦^{2} + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑧^{2})^{3}}$

Converting the kinetic energy per unit volume into spherical coordinate we get: $\frac{𝜇 𝑄^{2}}{32 𝜋^{2}} \frac{𝑣_{𝑧}^{2}}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \frac{𝑠 𝑖 𝑛^{2} 𝜃}{𝑟^{4} (𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{3}}$

To calculate the total kinetic energy we need to integrate the above expression for the entire space after the sphere surface. $\begin{aligned} 𝐾 . 𝐸 & = \frac{𝜇 𝑄^{2}}{32 𝜋^{2}} \frac{𝑣_{𝑧}^{2}}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \int_{0}^{2 𝜋} \int_{0}^{𝜋} \int_{𝑎}^{\infty} \frac{𝑠 𝑖 𝑛^{2} 𝜃}{𝑟^{4} (𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{3}} 𝑟^{2} 𝑠 𝑖 𝑛 𝜃 \partial 𝜃 \partial 𝜙 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋} \frac{𝑣_{𝑧}^{2}}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \int_{0}^{𝜋} \int_{𝑎}^{\infty} \frac{𝑠 𝑖 𝑛^{3} 𝜃}{𝑟^{2} (𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{3}} \partial 𝜃 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧}^{2}}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \int_{0}^{𝜋} \frac{𝑠 𝑖 𝑛^{3} 𝜃}{(𝑠 𝑖 𝑛^{2} 𝜃 + \frac{1}{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}} 𝑐 𝑜 𝑠^{2} 𝜃)^{3}} \partial 𝜃 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧}^{2}}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \int_{0}^{𝜋} \frac{𝑠 𝑖 𝑛^{3} 𝜃}{(1 + \frac{𝑣_{𝑧}^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}} 𝑐 𝑜 𝑠^{2} 𝜃)^{3}} \partial 𝜃 \end{aligned}$

To solve the above integral, let’s set: $𝑐 𝑜 𝑠 𝜃 = 𝑢 𝐻 𝑒 𝑛 𝑐 𝑒, 𝑠 𝑖 𝑛 𝜃 \partial 𝜃 = - \partial 𝑢$

Now the integral becomes: $\begin{aligned} 𝐾 . 𝐸 & = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧}^{2}}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \int_{1}^{- 1} \frac{𝑢^{2} - 1}{(1 + \frac{𝑣_{𝑧}^{2}}{𝑐^{2} - 𝑣_{𝑧}^{2}} 𝑢^{2})^{3}} \partial 𝑢 \end{aligned}$

Now again set: $\begin{array}{r} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} 𝑢 = 𝑡 𝑎 𝑛 𝜓 \\ 𝑠 𝑜, \partial 𝑢 = \frac{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}{𝑣_{𝑧}} 𝑠 𝑒 𝑐^{2} 𝜓 \partial 𝜓 \end{array}$

Now the integral reduced to: $\begin{aligned} 𝐾 . 𝐸 & = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{1}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \frac{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}{𝑣_{𝑧}} \int_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} ((𝑐^{2} - 𝑣_{𝑧}^{2}) 𝑠 𝑖 𝑛^{2} 𝜓 𝑐 𝑜 𝑠^{2} 𝜓 - 𝑣_{𝑧}^{2} 𝑐 𝑜 𝑠^{4} 𝜓) \partial 𝜓 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{1}{(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})} \frac{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}{𝑣_{𝑧}} \int_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} (𝑐^{2} 𝑠 𝑖 𝑛^{2} 𝜓 𝑐 𝑜 𝑠^{2} 𝜓 - 𝑣_{𝑧}^{2} 𝑐 𝑜 𝑠^{2} 𝜓) \partial 𝜓 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧} 𝑐^{2}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} \int_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} (\frac{𝑐^{2}}{𝑣_{𝑧}^{2}} 𝑠 𝑖 𝑛^{2} 𝜓 𝑐 𝑜 𝑠^{2} 𝜓 - 𝑐 𝑜 𝑠^{2} 𝜓) \partial 𝜓 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧} 𝑐^{2}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} \int_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} (\frac{𝑐^{2}}{𝑣_{𝑧}^{2}} \frac{1}{8} (1 - 𝑐 𝑜 𝑠 4 𝜓) - \frac{1}{2} (1 + 𝑐 𝑜 𝑠 2 𝜓)) \partial 𝜓 \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧} 𝑐^{2}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {[\frac{𝑐^{2}}{𝑣_{𝑧}^{2}} \frac{1}{8} (𝜓 - \frac{𝑠 𝑖 𝑛 4 𝜓}{4}) - \frac{1}{2} (𝜓 + \frac{𝑠 𝑖 𝑛 2 𝜓}{2})]}_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} \\ = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧} 𝑐^{2}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {[\frac{𝑐^{2}}{8 𝑣_{𝑧}^{2}} 𝜓 - \frac{𝑐^{2}}{8 𝑣_{𝑧}^{2}} \frac{𝑠 𝑖 𝑛 4 𝜓}{4} - \frac{𝜓}{2} - \frac{𝑠 𝑖 𝑛 2 𝜓}{4}]}_{𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}}^{𝑡 𝑎 𝑛^{- 1} \frac{- 𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}} \end{aligned}$

By substituting: $𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} = 𝑣_{2}$ the above equation reduced to:

𝐾 . 𝐸 = \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧} 𝑐^{2}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} [𝑣_{2} (1 - \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{1}{2} 𝑠 𝑖 𝑛 2 𝑣_{2} (1 + \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}} 𝑐 𝑜 𝑠 2 𝑣_{2})]

So now the total kinetic energy of the system is the sum of mechanical and magnetic energy, which is: $𝐾 . 𝐸 = 𝐾 . 𝐸_{𝑀} + 𝐾 . 𝐸_{𝐸} \begin{aligned} \frac{1}{2} 𝑀 𝑣_{𝑧}^{2} & = \frac{1}{2} 𝑚 𝑣_{𝑧}^{2} + \frac{𝜇 𝑄^{2}}{16 𝜋 𝑎} \frac{𝑣_{𝑧} 𝑐^{2}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {𝑣_{2} (1 - \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{1}{2} 𝑠 𝑖 𝑛 2 𝑣_{2} (1 + \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}} 𝑐 𝑜 𝑠 2 𝑣_{2})} \\ = \frac{1}{2} 𝑣_{𝑧}^{2} [𝑚 + \frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {𝑣_{2} (1 - \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{1}{2} 𝑠 𝑖 𝑛 2 𝑣_{2} (1 + \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}} 𝑐 𝑜 𝑠 2 𝑣_{2})}] \end{aligned} 𝑜 𝑟, 𝑀 = 𝑚 + \frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {𝑣_{2} (1 - \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{1}{2} 𝑠 𝑖 𝑛 2 𝑣_{2} (1 + \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}} 𝑐 𝑜 𝑠 2 𝑣_{2})}$

So the mass of the object is increased by amount: $\frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {𝑣_{2} (1 - \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{1}{2} 𝑠 𝑖 𝑛 2 𝑣_{2} (1 + \frac{𝑐^{2}}{4 𝑣_{𝑧}^{2}} 𝑐 𝑜 𝑠 2 𝑣_{2})}$

By evaluating the above extra mass quantity by converting factors sin2v2 and cos2v2 in terms of tan and evaluate it using the value of v2, we get:

\frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} {𝑡 𝑎 𝑛^{- 1} (\frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}) (\frac{4 𝑣_{𝑧}^{2} - 𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{(2 𝑣_{𝑧}^{2} + 𝑐^{2})}{4 𝑣_{𝑧} 𝑐^{2}} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}}

You can see if velocity approaches to that of speed of light (vz = c), the mass tends to increase to infinity. So it’s velocity will remain constant. So its impossible to increase the velocity of a charged boy more than that of the speed of light in this scenario.

If speed of the object is very very less than that of light,

\begin{aligned} the factor 𝑡 𝑎 𝑛^{- 1} \frac{𝑣_{𝑧}}{\sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} can be approximated as: \\ ≃ 𝑣_{𝑧} (𝑐^{2} - 𝑣_{𝑧}^{2})^{\frac{- 1}{2}} - \frac{𝑣_{𝑧}^{3}}{3} (𝑐^{2} - 𝑣_{𝑧}^{2})^{\frac{- 1}{6}} [Using Maclaurin Series] \\ ≃ \frac{𝑣_{𝑧}}{𝑐} {(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})}^{- \frac{1}{2}} - \frac{𝑣_{𝑧}^{3}}{3 𝑐^{3}} {(1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}})}^{- \frac{1}{6}} \\ ≃ \frac{𝑣_{𝑧}}{𝑐} - \frac{𝑣_{𝑧}^{3}}{3 𝑐^{3}} (1 + \frac{𝑣_{𝑧}^{2}}{6 𝑐^{2}}) [dropping the factor \frac{𝑣_{𝑧}^{2}}{𝑐^{2}} in the first term and using Binomial expansion for the second] \end{aligned}

So the increased mass quantity becomes $\frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{𝑐^{2}}{𝑣_{𝑧} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}} [{\frac{𝑣_{𝑧}}{𝑐} - \frac{𝑣_{𝑧}^{3}}{3 𝑐^{3}} (1 + \frac{𝑣_{𝑧}^{2}}{6 𝑐^{2}})} (\frac{4 𝑣_{𝑧}^{2} - 𝑐^{2}}{4 𝑣_{𝑧}^{2}}) + \frac{(2 𝑣_{𝑧}^{2} + 𝑐^{2})}{4 𝑣_{𝑧} 𝑐^{2}} \sqrt{𝑐^{2} - 𝑣_{𝑧}^{2}}] \begin{aligned} = \frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{1}{4 𝑣_{𝑧}^{2} \sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}} [(1 - \frac{𝑣_{𝑧}^{2}}{3 𝑐^{2}} - \frac{𝑣_{𝑧}^{4}}{18 𝑐^{4}}) (4 𝑣_{𝑧}^{2} - 𝑐^{2}) + (2 𝑣_{𝑧}^{2} + 𝑐^{2}) \sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}] \\ ≃ \frac{𝜇 𝑄^{2}}{8 𝜋 𝑎} \frac{1}{4 𝑣_{𝑧}^{2}} [4 𝑣_{𝑧}^{2} - \frac{4 𝑣_{𝑧}^{4}}{3 𝑐^{2}} - \frac{2 𝑣_{𝑧}^{6}}{9 𝑐^{4}} - 𝑐^{2} + \frac{𝑣_{𝑧}^{2}}{3} + \frac{𝑣_{𝑧}^{4}}{18 𝑐^{2}} + 2 𝑣_{𝑧}^{2} + 𝑐^{2}] [By dropping the negligible factor \sqrt{1 - \frac{𝑣_{𝑧}^{2}}{𝑐^{2}}}] \\ = \frac{𝜇}{2 𝜋} \frac{𝑄^{2}}{𝑎} \frac{1}{16} [\frac{19}{3} - \frac{69 𝑣_{𝑧}^{2}}{54 𝑐^{2}} - \frac{2 𝑣_{𝑧}^{4}}{9 𝑐^{4}}] \\ ≃ \frac{𝜇}{2 𝜋} \frac{𝑄^{2}}{3 𝑎} \frac{19}{16} [Dropping the negligible last two terms] \\ ≃ \frac{𝜇}{2 𝜋} \frac{𝑄^{2}}{3 𝑎} [As the factor \frac{19}{16} = 1.18 is small and will be getting smaller if more terms of approximation is used. So dropped] \end{aligned}$

This is the same factor which we obtained in our pervious analysis without considering any field distortion.

So does this prove that speed of light is the maximum limit to any charged bodies? And what about neutral bodies?

Well that’s not the case. The limit of speed of light is for very limiting condition like the one mentioned here. There are many ways to overcome it. Unfortunately this very specific case for electrodynamics of moving bodies is being misinterpreted and generalized into erroneous theories like relativity without any physical medium.

Will write about it in future post.

source: parthasarathimishra.com

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