Impact on Mass of A Moving Charged Object

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:


𝐻=𝑣×𝐷

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:


𝐸=𝑄4𝜋𝜖𝑟2 ----(1)

Then the net magnetic field intensity becomes:

𝐻=𝑣×𝐷=𝑣𝐷sin𝜃=𝑣𝜖𝐸sin𝜃 ----(2)

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

𝐻=𝑣𝑄sin𝜃4𝜋𝑟2 ----(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=𝜇2|𝐻|2

Hence, 𝐾.𝐸=𝜇𝑣2𝑄2sin2𝜃32𝜋2𝑟4[using eqn(3))]

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

𝐾.𝐸=02𝜋0𝜋𝑎𝜇𝑣2𝑄2sin2𝜃32𝜋2𝑟4𝑟2sin𝜃𝑟𝜃𝜙

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

So,
𝐾.𝐸=𝜇𝑣2𝑄232𝜋202𝜋0𝜋𝑎sin3𝜃𝑟2𝑟𝜃𝜙=𝜇𝑣2𝑄2(2𝜋)32𝜋20𝜋𝑎sin3𝜃𝑟2𝑟𝜃=𝜇𝑣2𝑄28𝜋[1𝑟]𝑎0𝜋sin3𝜃𝜃=𝜇𝑣2𝑄28𝜋𝑎0𝜋14(sin3𝜃3sin𝜃)𝜃

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

𝐾.𝐸𝐸=𝜇4𝜋𝑄2𝑣23𝑎

Now if the mass of the object is m, the mechanical kinetic energy is given as:
𝐾.𝐸𝑀=12𝑚𝑣2

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:

𝐾.𝐸=𝐾.𝐸𝑀+𝐾.𝐸𝐸𝑜𝑟,12𝑀𝑣2=12𝑚𝑣2+𝜇4𝜋𝑄23𝑎𝑜𝑟,𝑀=𝑚+𝜇2𝜋𝑄23𝑎

Hence the apparent mass of the object appears to increase by the factor of 𝜇2𝜋𝑄23𝑎. 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=𝜇2𝐻2(1)

The electrostatic energy per unit volume is given as:

𝑃.𝐸=𝜖2𝐸2(2)

Total energy of the system shall remain constant, hence:

𝑃.𝐸+𝐾.𝐸=𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Taking derivative with respect to E:

𝐸(𝑃.𝐸)+𝐸(𝐾.𝐸)=𝐸(𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝑜𝑟,𝐸(𝜖2𝐸2)+𝐸(𝜇2𝐻2)=0
𝑜𝑟,𝜖𝐸+𝜇𝐻𝐸(𝑣×𝐷)=0[𝑎𝑠,𝐻=𝑣×𝐷]

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

𝐻𝑒𝑛𝑐𝑒,𝜖𝐸+𝜇𝜖𝐻𝐸(𝑣×𝐸)=0[𝑎𝑠,𝐷=𝜖𝐸]
𝑜𝑟,𝐸=𝐵𝐸(𝑣×𝐸)=0[𝑎𝑠,𝐵=𝜇𝐻]

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

We have:

𝑣=𝑣𝑥𝑖^+𝑣𝑦𝑗^+𝑣𝑧𝑘^,𝑎𝑛𝑑,𝐸=𝐸𝑥𝑖^+𝐸𝑦𝑗^+𝐸𝑧𝑘^
𝑆𝑜,𝑣×𝐸=(𝑣𝑦𝐸𝑧𝐸𝑦𝑣𝑧)𝑖^+(𝐸𝑥𝑣𝑧𝑣𝑥𝐸𝑧)𝑗^+(𝑣𝑥𝐸𝑦𝐸𝑥𝑣𝑦)𝑘^
𝐻𝑒𝑛𝑐𝑒,𝐵𝐸(𝑣×𝐸)=𝐵𝐸𝑥(𝑣×𝐸)𝑖^+𝐵𝐸𝑦(𝑣×𝐸)𝑗^+𝐵𝐸𝑧(𝑣×𝐸)𝑘^
=(𝐵𝑣𝑧𝐵𝑣𝑦)𝑖^+(𝐵𝑣𝑥𝐵𝑣𝑧)𝑗^+(𝐵𝑣𝑦𝐵𝑣𝑥)𝑘^=𝐵×𝑣
𝑆𝑜,𝐸1=𝐵×𝑣

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:

𝐸2=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):

𝐸=(𝐵×𝑣)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 becomes𝑣=𝑣𝑧𝑘^

Now𝐵=𝜇𝐻=𝜇(𝑣×𝐷)=𝜇[𝑣𝑧𝑘^×(𝐷𝑥𝑖^+𝐷𝑦𝑗^+𝐷𝑧𝑘^)]=𝜇(𝑣𝑧𝐷𝑦𝑖^+𝑣𝑧𝐷𝑥𝑗^)(3)

Now the electric field intensity E becomes:

𝐸=(𝐵×𝑣)𝜙=[𝜇(𝑣𝑧𝐷𝑦𝑖^+𝑣𝑧𝐷𝑥𝑗^)×𝑣𝑧𝑘^]𝜙=𝜇(𝑣𝑧2𝐷𝑥𝑖^+𝑣𝑧2𝐷𝑦𝑗^)𝜙

Hence𝐷𝜖=𝜇(𝑣𝑧2𝐷𝑥𝑖^+𝑣𝑧2𝐷𝑦𝑗^)𝜙

Or,𝐷=𝜇𝜖(𝑣𝑧2𝐷𝑥𝑖^+𝑣𝑧2𝐷𝑦𝑗^)𝜖𝜙=𝑣𝑧2𝑐2(𝐷𝑥𝑖^+𝐷𝑦𝑗^)𝜖𝜙[𝜇𝜖=1𝑐2]

Now,(𝐷𝑥𝑖^+𝐷𝑦𝑗^+𝐷𝑧𝑘^)=𝑣𝑧2𝑐2(𝐷𝑥𝑖^+𝐷𝑦𝑗^)𝜖𝜙𝑜𝑟,𝐷𝑥(1𝑣𝑧2𝑐2)𝑖^+𝐷𝑦(1𝑣𝑧2𝑐2)𝑗^+𝐷𝑧𝑘^=𝜖𝜙

Splitting the above equation for D into components we get:

𝐷𝑥(1𝑣𝑧2𝑐2)=𝜖𝜙𝑥𝑜𝑟,𝐷𝑥=(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑥(4)
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦,𝐷𝑦=(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑦(5)
𝐴𝑛𝑑,𝐷𝑧=𝜖𝜙𝑧(6)

Putting the vector components together we get:

𝐷𝑥𝑖^+𝐷𝑦𝑗^+𝐷𝑧𝑘^=(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑥𝑖^(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑦𝑗^𝜖𝜙𝑧𝑘^
𝑜𝑟,𝐷=(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑥𝑖^(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑦𝑗^𝜖𝜙𝑧𝑘^

Now taking divergent both sides:

.𝐷=.[(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑥𝑖^(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑦𝑗^𝜖𝜙𝑧𝑘^]

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

.[(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑥𝑖^(𝑐2𝑐2𝑣𝑧2)𝜖𝜙𝑦𝑗^𝜖𝜙𝑧𝑘^]=0
𝑜𝑟(𝑐2𝑐2𝑣𝑧2)2𝜙𝑥2(𝑐2𝑐2𝑣𝑧2)2𝜙𝑦22𝜙𝑧2=0
𝑜𝑟2𝜙𝑥2+2𝜙𝑦2+(𝑐2𝑣𝑧2𝑐2)2𝜙𝑧2=0(7)

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

Lets put:

𝑧=𝑐𝑐2𝑣𝑧2𝑧

Using chain rule:𝜙𝑧=𝜙𝑧𝑧𝑧𝑜𝑟,2𝜙𝑧2=𝜙𝑧𝑧𝑧𝑧𝑧=2𝜙𝑧2(𝑧𝑧)2=(𝑐2𝑣𝑧2𝑐2)2𝜙𝑧2

Now eqn(7) becomes:2𝜙𝑥2+2𝜙𝑦2+2𝜙𝑧2=0

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

𝜙=𝐴𝑥2+𝑦2+𝑧2=𝐴𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2

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:

𝜙𝑥=𝐴𝑥(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32𝜙𝑦=𝐴𝑦(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32𝜙𝑧=𝑐2(𝑐2𝑣𝑧2)𝐴𝑧(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32

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

𝐷𝑥=𝑐2(𝑐2𝑣𝑧2)𝜖𝐴𝑥(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32(8)
𝐷𝑦=𝑐2(𝑐2𝑣𝑧2)𝜖𝐴𝑦(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32(9)
𝐷𝑧=𝑐2(𝑐2𝑣𝑧2)𝜖𝐴𝑧(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32(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.

𝐷.𝑛^𝑑𝑆=𝑄(11)

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

𝑛^=𝑥𝑎𝑖^+𝑦𝑎𝑗^+𝑧𝑎𝑘^

Now integral (11) becomes:

𝑄=𝜖𝐴𝑐2(𝑐2𝑣𝑧2)(𝑥𝑖^+𝑦𝑗^+𝑧𝑘^(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32).(𝑥𝑎𝑖^+𝑦𝑎𝑗^+𝑧𝑎𝑘^)𝑑𝑆=𝜖𝐴𝑐2𝑎(𝑐2𝑣𝑧2)𝑥2+𝑦2+𝑧2(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32𝑑𝑆=𝜖𝑎𝐴𝑐2(𝑐2𝑣𝑧2)𝑑𝑆(𝑥2+𝑦2+(𝑐2𝑐2𝑣𝑧2)𝑧2)32[𝑎𝑠,𝑥2+𝑦2+𝑧2=𝑎2]

Now for proper evaluation, we need to convert this to spherical coordinate.
So we have:𝑥=𝑟𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜃𝑦=𝑟𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃𝑧=𝑟𝑐𝑜𝑠𝜃𝑑𝑆=𝑟2𝑠𝑖𝑛𝜃𝑑𝜃𝑑𝜙

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:

𝑄=𝜖𝑎𝐴𝑐2(𝑐2𝑣𝑧2)0𝜋02𝜋𝑎2𝑠𝑖𝑛𝜃𝜃𝜙(𝑎2𝑐𝑜𝑠2𝜙𝑠𝑖𝑛2𝜃+𝑎2𝑠𝑖𝑛2𝜙𝑠𝑖𝑛2𝜃+(𝑐2𝑐2𝑣𝑧2)𝑎2𝑐𝑜𝑠2𝜃)32=𝜖𝐴𝑐2(𝑐2𝑣𝑧2)0𝜋02𝜋𝑠𝑖𝑛𝜃𝜃𝜙(𝑠𝑖𝑛2𝜃+(𝑐2𝑐2𝑣𝑧2)𝑐𝑜𝑠2𝜃)32=2𝜋𝜖𝐴𝑐2(𝑐2𝑣𝑧2)0𝜋𝑠𝑖𝑛𝜃𝜃(𝑠𝑖𝑛2𝜃+(𝑐2𝑐2𝑣𝑧2)𝑐𝑜𝑠2𝜃)32=2𝜋𝜖𝐴𝑐2(𝑐2𝑣𝑧2)0𝜋𝑠𝑖𝑛𝜃𝜃(1+(𝑣𝑧2𝑐2𝑣𝑧2)𝑐𝑜𝑠2𝜃)32

Now we solve this integral by substituting:𝑣𝑧𝑐2𝑣𝑧2𝑐𝑜𝑠𝜃=𝑡𝑎𝑛𝜓𝐻𝑒𝑛𝑐𝑒,𝑣𝑧𝑐2𝑣𝑧2𝑠𝑖𝑛𝜃𝜃=𝑠𝑒𝑐2𝜓𝜓

Now the above integral becomes:𝑄=2𝜋𝜖𝐴𝑐2(𝑐2𝑣𝑧2)𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑐2𝑣𝑧2𝑣𝑧𝑠𝑒𝑐2𝜓(1+𝑡𝑎𝑛2𝜓)32𝜓=2𝜋𝜖𝐴𝑐2𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑐𝑜𝑠𝜓𝜓=2𝜋𝜖𝐴𝑐2𝑣𝑧𝑐2𝑣𝑧2[𝑠𝑖𝑛𝜓]𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2=2𝜋𝜖𝐴𝑐2𝑣𝑧𝑐2𝑣𝑧2[𝑡𝑎𝑛𝜓1+𝑡𝑎𝑛2𝜓]𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2=2𝜋𝜖𝐴𝑐2𝑣𝑧𝑐2𝑣𝑧22𝑣𝑧𝑐2𝑣𝑧21+𝑣𝑧2𝑐2𝑣𝑧2=4𝜋𝜖𝐴𝑐𝑐2𝑣𝑧2

Now A becomes:𝐴=𝑄4𝜋𝜖𝑐2𝑣𝑧2𝑐

Now putting the value of A in scalar potential equation it becomes:𝜙=𝑄1𝑣𝑧2𝑐24𝜋𝜖𝑥2+𝑦2+(11𝑣𝑧2𝑐2)𝑧2(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:

𝜙=𝑄1𝑣𝑧2𝑐24𝜋𝜖𝑟𝑠𝑖𝑛2𝜃+(11𝑣𝑧2𝑐2)𝑐𝑜𝑠2𝜃

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

𝐷=𝑄4𝜋11𝑣𝑧2𝑐21(𝑥2+𝑦2+11𝑣𝑧2𝑐2𝑧2)32(𝑥𝑖^+𝑦𝑗^+𝑧𝑘^)(13)

In spherical coordinate, the D can be expressed as:𝐷=𝑄4𝜋11𝑣𝑧2𝑐21𝑟2(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)32(𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜃𝑟^+𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃𝜃^+𝑐𝑜𝑠𝜃𝜙^)

The above equation shows that the the faraday tubes are radial and the resultant polarization D varies inversely as𝐷1𝑟2(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)32

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 as𝐷.𝑛^1(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)32The 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=𝑄1𝑣𝑧2𝑐24𝜋𝜖𝑦2+(11𝑣2𝑐2)𝑥2𝑜𝑟,𝑦=±𝑄2𝛽2𝜙2(4𝜋𝜖)2𝛽2𝑥2[𝑤𝑒𝑟𝑒,𝛽=11𝑣2𝑐2]

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:𝐻=𝑄4𝜋11𝑣𝑧2𝑐2𝑣𝑧(𝑥2+𝑦2+11𝑣𝑧2𝑐2𝑧2)32(𝑦𝑖^+𝑥𝑗^)

The magnetic energy per unit volume is given from eqn(1) as𝜇2𝐻2=𝜇𝑄232𝜋21(1𝑣𝑧2𝑐2)𝑣𝑧2(𝑥2+𝑦2)(𝑥2+𝑦2+11𝑣𝑧2𝑐2𝑧2)3

Converting the kinetic energy per unit volume into spherical coordinate we get:𝜇𝑄232𝜋2𝑣𝑧2(1𝑣𝑧2𝑐2)𝑠𝑖𝑛2𝜃𝑟4(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)3

To calculate the total kinetic energy we need to integrate the above expression for the entire space after the sphere surface.𝐾.𝐸=𝜇𝑄232𝜋2𝑣𝑧2(1𝑣𝑧2𝑐2)02𝜋0𝜋𝑎𝑠𝑖𝑛2𝜃𝑟4(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)3𝑟2𝑠𝑖𝑛𝜃𝜃𝜙=𝜇𝑄216𝜋𝑣𝑧2(1𝑣𝑧2𝑐2)0𝜋𝑎𝑠𝑖𝑛3𝜃𝑟2(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)3𝜃=𝜇𝑄216𝜋𝑎𝑣𝑧2(1𝑣𝑧2𝑐2)0𝜋𝑠𝑖𝑛3𝜃(𝑠𝑖𝑛2𝜃+11𝑣𝑧2𝑐2𝑐𝑜𝑠2𝜃)3𝜃=𝜇𝑄216𝜋𝑎𝑣𝑧2(1𝑣𝑧2𝑐2)0𝜋𝑠𝑖𝑛3𝜃(1+𝑣𝑧2𝑐2𝑣𝑧2𝑐𝑜𝑠2𝜃)3𝜃

To solve the above integral, let’s set:𝑐𝑜𝑠𝜃=𝑢𝐻𝑒𝑛𝑐𝑒,𝑠𝑖𝑛𝜃𝜃=𝑢

Now the integral becomes:𝐾.𝐸=𝜇𝑄216𝜋𝑎𝑣𝑧2(1𝑣𝑧2𝑐2)11𝑢21(1+𝑣𝑧2𝑐2𝑣𝑧2𝑢2)3𝑢

Now again set:𝑣𝑧𝑐2𝑣𝑧2𝑢=𝑡𝑎𝑛𝜓𝑠𝑜,𝑢=𝑐2𝑣𝑧2𝑣𝑧𝑠𝑒𝑐2𝜓𝜓

Now the integral reduced to:𝐾.𝐸=𝜇𝑄216𝜋𝑎1(1𝑣𝑧2𝑐2)𝑐2𝑣𝑧2𝑣𝑧𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2((𝑐2𝑣𝑧2)𝑠𝑖𝑛2𝜓𝑐𝑜𝑠2𝜓𝑣𝑧2𝑐𝑜𝑠4𝜓)𝜓=𝜇𝑄216𝜋𝑎1(1𝑣𝑧2𝑐2)𝑐2𝑣𝑧2𝑣𝑧𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2(𝑐2𝑠𝑖𝑛2𝜓𝑐𝑜𝑠2𝜓𝑣𝑧2𝑐𝑜𝑠2𝜓)𝜓=𝜇𝑄216𝜋𝑎𝑣𝑧𝑐2𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2(𝑐2𝑣𝑧2𝑠𝑖𝑛2𝜓𝑐𝑜𝑠2𝜓𝑐𝑜𝑠2𝜓)𝜓=𝜇𝑄216𝜋𝑎𝑣𝑧𝑐2𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2(𝑐2𝑣𝑧218(1𝑐𝑜𝑠4𝜓)12(1+𝑐𝑜𝑠2𝜓))𝜓=𝜇𝑄216𝜋𝑎𝑣𝑧𝑐2𝑐2𝑣𝑧2[𝑐2𝑣𝑧218(𝜓𝑠𝑖𝑛4𝜓4)12(𝜓+𝑠𝑖𝑛2𝜓2)]𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2=𝜇𝑄216𝜋𝑎𝑣𝑧𝑐2𝑐2𝑣𝑧2[𝑐28𝑣𝑧2𝜓𝑐28𝑣𝑧2𝑠𝑖𝑛4𝜓4𝜓2𝑠𝑖𝑛2𝜓4]𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2

By substituting:𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2=𝑣2the above equation reduced to:

𝐾.𝐸=𝜇𝑄216𝜋𝑎𝑣𝑧𝑐2𝑐2𝑣𝑧2[𝑣2(1𝑐24𝑣𝑧2)+12𝑠𝑖𝑛2𝑣2(1+𝑐24𝑣𝑧2𝑐𝑜𝑠2𝑣2)]

So now the total kinetic energy of the system is the sum of mechanical and magnetic energy, which is:𝐾.𝐸=𝐾.𝐸𝑀+𝐾.𝐸𝐸12𝑀𝑣𝑧2=12𝑚𝑣𝑧2+𝜇𝑄216𝜋𝑎𝑣𝑧𝑐2𝑐2𝑣𝑧2{𝑣2(1𝑐24𝑣𝑧2)+12𝑠𝑖𝑛2𝑣2(1+𝑐24𝑣𝑧2𝑐𝑜𝑠2𝑣2)}=12𝑣𝑧2[𝑚+𝜇𝑄28𝜋𝑎𝑐2𝑣𝑧𝑐2𝑣𝑧2{𝑣2(1𝑐24𝑣𝑧2)+12𝑠𝑖𝑛2𝑣2(1+𝑐24𝑣𝑧2𝑐𝑜𝑠2𝑣2)}]𝑜𝑟,𝑀=𝑚+𝜇𝑄28𝜋𝑎𝑐2𝑣𝑧𝑐2𝑣𝑧2{𝑣2(1𝑐24𝑣𝑧2)+12𝑠𝑖𝑛2𝑣2(1+𝑐24𝑣𝑧2𝑐𝑜𝑠2𝑣2)}

So the mass of the object is increased by amount:𝜇𝑄28𝜋𝑎𝑐2𝑣𝑧𝑐2𝑣𝑧2{𝑣2(1𝑐24𝑣𝑧2)+12𝑠𝑖𝑛2𝑣2(1+𝑐24𝑣𝑧2𝑐𝑜𝑠2𝑣2)}

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:

𝜇𝑄28𝜋𝑎𝑐2𝑣𝑧𝑐2𝑣𝑧2{𝑡𝑎𝑛1(𝑣𝑧𝑐2𝑣𝑧2)(4𝑣𝑧2𝑐24𝑣𝑧2)+(2𝑣𝑧2+𝑐2)4𝑣𝑧𝑐2𝑐2𝑣𝑧2}

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,the factor𝑡𝑎𝑛1𝑣𝑧𝑐2𝑣𝑧2can be approximated as:𝑣𝑧(𝑐2𝑣𝑧2)12𝑣𝑧33(𝑐2𝑣𝑧2)16[Using Maclaurin Series]𝑣𝑧𝑐(1𝑣𝑧2𝑐2)12𝑣𝑧33𝑐3(1𝑣𝑧2𝑐2)16𝑣𝑧𝑐𝑣𝑧33𝑐3(1+𝑣𝑧26𝑐2)[dropping the factor 𝑣𝑧2𝑐2 in the first term and using Binomial expansion for the second]

So the increased mass quantity becomes𝜇𝑄28𝜋𝑎𝑐2𝑣𝑧𝑐2𝑣𝑧2[{𝑣𝑧𝑐𝑣𝑧33𝑐3(1+𝑣𝑧26𝑐2)}(4𝑣𝑧2𝑐24𝑣𝑧2)+(2𝑣𝑧2+𝑐2)4𝑣𝑧𝑐2𝑐2𝑣𝑧2]=𝜇𝑄28𝜋𝑎14𝑣𝑧21𝑣𝑧2𝑐2[(1𝑣𝑧23𝑐2𝑣𝑧418𝑐4)(4𝑣𝑧2𝑐2)+(2𝑣𝑧2+𝑐2)1𝑣𝑧2𝑐2]𝜇𝑄28𝜋𝑎14𝑣𝑧2[4𝑣𝑧24𝑣𝑧43𝑐22𝑣𝑧69𝑐4𝑐2+𝑣𝑧23+𝑣𝑧418𝑐2+2𝑣𝑧2+𝑐2][By dropping the negligible factor 1𝑣𝑧2𝑐2 ]=𝜇2𝜋𝑄2𝑎116[19369𝑣𝑧254𝑐22𝑣𝑧49𝑐4]𝜇2𝜋𝑄23𝑎1916[Dropping the negligible last two terms]𝜇2𝜋𝑄23𝑎[As the factor 1916=1.18 is small and will be getting smaller if more terms of approximation is used. So dropped]

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

Learn more: Ether Technology
 
Revealed At Last... Must watch!
Free Energy Magnetic Generator and synthesizes many other technologies imbued with Nikola Tesla's technological identity

✔ Nikola Tesla’s method of magnifying electric power by neutralizing the magnetic counter-forces in an electric generator

Generates Energy-On-Demand: 👉 Free Energy Will Change Our World Forever

✔ Combination of induction motor and alternator 
✔ Combine generators with induction motors - self-powered generators with rotary motion
✔ Various methods of generating high power immobile generators

✔ Or maybe called Overunity for the system. Mother Nature doesn't care about people calling or naming phenomena. Overunity/Free Energy, Zero Point Energy (ZPE) are just a few different words

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