Kinetic energy of a relativistic massive point charge in a constant and uniform electric field \(\frac{mv^2}{2}=\frac{mc^2}{2}(1-e^{-2\frac{\Delta Uq}{mc^2}})\) The kinetic energy is derived under the assumption that Coulomb's force is proportional to \(\ 1-v^2/c^2\) For simplicity, I took the case when there is no initial velocity. Accelerated particle is emitted from the origin Derivation: From Newton's Second Law: \(m\frac{dv}{dt}=qE(1-v^2/c^2)\) Here \(E\) - the electric field intensity We integrate: \(\frac{dv}{(1-v^2/c^2)}=\frac{qE}{mc}dt\) \(\frac{v}{c}=th(\frac{qE}{mc}t)\) [1] Once again, we integrate \(\frac{dx}{dt}=c\ th(\frac{qE}{mc}t)\) \(x=\frac{mc^2}{qE}\ ln\{ch(\frac{qE}{mc}t)\}\) Next we do recalculate the electric field intensity into a potential difference \(\Delta U=xE\) \(\Delta Uq=mc^2\ ln\{ch(\frac{qE}{mc}t)\}\) \(e^{\frac{\Delta Uq}{mc^2}}=ch(\frac{qE}{mc}t)\) We use the equation [1], to eliminate the time \(\frac{mv^2}{2}=\frac{mc^2}{2}th^2(\frac{qE}{mc}t)=\frac{mc^2}{2}\frac{e^{2\frac{\Delta Uq}{mc^2}}-1}{e^{2\frac{\Delta Uq}{mc^2}}}=\frac{mc^2}{2}(1-e^{-2\frac{\Delta Uq}{mc^2}})\)

Why \(F=qE(1-v^2/c^2)\)? If the relativistic increase of mass does not exist, then the relativistic increase of energy does not exist also. Derivation: A mass-spectrometers display for us: either a relativistic increases of mass exist, or Lorentz's force disappears. But any relativistic increases of mass do not exist! Hence: Lorentz's force disappears. Hence: Coulomb's force disappears also. My formula reflects this fact.

Can be \(F=qE\sqrt{1-v^2/c^2}\)? No! Derivation: \(\frac{dv}{\sqrt{1-v^2/c^2}}=\frac{qE}{mc}dt\) \(\frac{v}{c}=sin(\frac{qE}{mc}t)\) [1] \(\frac{dx}{dt}=c\ sin(\frac{qE}{mc}t)\) \(x=\frac{mc^2}{qE}(1-cos(\frac{qE}{mc}t))\) \(\Delta Uq=mc^2(1-cos(\frac{qE}{mc}t))\) \(cos(\frac{qE}{mc}t)=1-\frac{\Delta Uq}{mc^2}\) \(\frac{mv^2}{2}=\frac{mc^2}{2}sin^2(\frac{qE}{mc}t)=\frac{mc^2}{2}(1-(1-\frac{\Delta Uq}{mc^2})^2)\) If \(\Delta U=\frac{mc^2}{q}\), then \(v=c\). Electrons reaches velocity of light at energies 50 keV. That (obviously) contrary to experiment.

Lorentz force: \(F=qvB(1-v^2/c^2)\) Coulomb force: \(F=qE(1-v^2/c^2)\) Cyclotron radius: \(R=\frac{mv}{qB(1-v^2/c^2)}\)

A real (longitudinal) dimensions of SRT associated with visual dimensions so: \(\Delta x_r=\frac{\Delta x_v}{\sqrt{1-v_v^2/c^2}}\) In Master Theory: \(\Delta x_r=\frac{\Delta x_v}{1-v_v^2/c^2}\) Hence the visual speed and related real speed (in Master Theory) as follows: \(v_r=\frac{v_v}{1-v_v^2/c^2}\) \(v_v=\frac{2v_r}{\sqrt{1+v_r^2/c^2}+1}\) If the rocket is uniformly accelerated moving (astronauts to do measuring acceleration by mass on a spring), then: \(v_r=at\) \(v_v=\frac{2at}{\sqrt{1+(at)^2/c^2}+1}\)

I think that the transformation of visual coordinates and visual time in the real-world coordinates and in real time in Master Theory should look like this: \(x_r=\frac{x_v+v_vt_v}{1-v_v^2/c^2}\) \(y_r=\frac{y_v}{\sqrt{1-v_v^2/c^2}}\) \(z_r=\frac{z_v}{\sqrt{1-v_v^2/c^2}}\) \(t_r=t_v+v_vx_v/c^2\)

Looking at the starry sky, we realize that the light is up to us (maybe) thousands of years. We see the stars and galaxies are not in the real coordinates, and - in visual coordinates. It is this fact and showing the Lorentz Transformations to us. PRapping some fool said Lorentz Transformations show real-world coordinates, and all this nonsense began to repeat after him.