Introduction: The one-degree of freedom $E=m{C}^{\text{2}}$ mass energy equivalence proposed by Albert Einstein who claimed the speed of the light to be the same constant C in vacuum observed in every coordinate system. The derivation began with Henri A. Lorentz generalization of the Galileo transform for a spherical light source that travels with the speed of light C from the origin 

${x^{\prime }}^{\text{2}}+{y^{\prime }}^{\text{2}}+{z^{\prime }}^{\text{2}}-C^{\text{2}}{t^{\prime }}^{\text{2}}= x^{\text{2}}+y^{\text{2}}+z^{\text{2}}-C^{\text{2}} t^{\text{2}}=\text{0}$ ;

or, in one dimension x:

$y=y\text{'}; z=z\text{'}$ ;

${x^{\prime }}^2-C^2{t^{\prime }}^2= x^2-C^2 t^2=0$

Since the Galileo transform $x^{\prime }=x-vt;$   $t’=t$  are no longer valid. Now we wish to find a relationship with proportionality function k

$x’=k\left(x-vt\right)$

$x=k’\left(x’+vt’\right)$

$x’=\text{0 }leadstok\left(x-vt\right)=\text{0}$

$x=vt$

$x=k’\left(kx-kvt+vt’\right)$

which can be expressed as a function of x and t

$t^{\prime }=k\{t-x/v(\text{1}-\frac{\text{1}}{k{k}^{\prime }})\}$

$x^{\text{2}}-c^{\text{2}}{t}^{\text{2}}-k^{\text{2}}{(x-vt)}^{\text{2}}+c^{\text{2}}{k}^{\text{2 }}{\{t-\frac{x}{v}\left(\text{1}-\frac{\text{1}}{k{k}^{\prime }}\right))}^{\text{2}}=\text{0}$

or,

$\{1-k^2+k^2\frac{c^2}{v^2}{(1-\frac{1}{k{k}^{\prime }})}^2\}{x}^2+2\{k^{2}v- \frac{c^{2}k}{v}\left(1-\frac{1}{k{k}^{\prime }}\right)\}xt+\left\{c^2{k}^{2 }-c^2-v^2{k}^{2 }\right\}{t}^2=0$

This quadratic polynomial; in x & t can vanish identically only if all the coefficients vanish. We have derived the constant function

$k=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=k^{\prime }$

In summary, we have derived H.A. Lorentz transform

$k=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=k^{\prime }$

$y^{\prime }=y$

$z^{\prime }=z$

$t^{\prime }=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}$

Now we apply for the H.A. Lorentz transforms to integrate the force times the distance for the work done to become Einstein energy mass relationship.

Assume that a force F acting a mass m does nothing but accelerate it, we have:

$F=\frac{d\left(mv\right)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$

$\Delta E=\underset{0}{\overset{f}{{\displaystyle \int }}}Fds=\underset{0}{\overset{f}{{\displaystyle \int }}}\left(m\frac{dv}{dt}+v\frac{dm}{dt}\right)ds=\underset{0}{\overset{f}{{\displaystyle \int }}}mvdv+\underset{0}{\overset{f}{{\displaystyle \int }}}v.vdm=\frac{1}{2}\underset{0}{\overset{f}{{\displaystyle \int }}}md{v}^2+\underset{0}{\overset{f}{{\displaystyle \int }}}{v}^{2}dm$

From $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ we obtain $v^2=c^2\left(1-\frac{m_0^2}{m^2}\right)$ ; and $\frac{d\left(v^2\right)}{dm}=\frac{2m_o^2{C}^2}{m^2}$

$\Delta E=\underset{0}{\overset{f}{{\displaystyle \int }}}m\frac{m_0^2{C}^2}{m^3}dm+\underset{o}{\overset{f}{{\displaystyle \int }}}{C}^2\left(1-\frac{m_o^2}{m^2}\right)dm=\underset{0}{\overset{f}{{\displaystyle \int }}}{C}^2 dm;$

$\Delta E=C^2\Delta m$

$E=m{C}^2+ground state$

This Einstein relation has been verified by Taurus Star light passing the sun on May 29 1919 the photon energy and its mass equivalence has been bended by the solar mass attraction predicted by Eddington, Watson, and Dyson

$p=mv;m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}};E=m{C}^2$

I am indebted to my teaches: Chairman Blass, and Advisor Fr. Williams Nichols who have taught me to be simple and not any more simpler.

None.

Author declares that there are no conflicts of interest.

1. Blass GA. Theoretical Physics, Appleton-Centrury-Crofts, New York.1962:p.302, p.322, p.324.