Before moving on further in the study of spacetime, for the purpose of note we look onto the derivation of the formula $E=mc^2$.
We know that, Force is defined as the rate of change of momentum i.e., $F=\frac {d}{dt}(mv)$
According to the theory of relativity, both mass and velocity are a variable quantity. We therefore, can rewrite the above as follows,
$F=\frac {d}{dt}(mv)$ = $m\frac {dv}{dt}$+$v\frac {dm}{dt}$
Let the force $F$ displace the body through a distance $dx$. Then, the increase in the kinectic energy $dE_k$ of the body is equal to the work done $F.dx$
Hence, $dE_k= Fdx= (m\frac {dv}{dt}+v\frac {dm}{dt})dx$
$=(mv)dv + v^2 dm$
Now, according to the law of mass variation with velocity, we have;
$m=\gamma m_0$ , where $\gamma$ represent the factor $1/\sqrt{1-\frac{v^2}{c^2}}$
taking the differential of the mass and the velocity in the equation we have the following relation,
$c^2dm=(mv)dv+v^2dm$
substituting, the value of in the differential of the kinectic energy, we have as follows
$dE_k=c^2dm$
Integrating the above differential from $ m_0$ to $m $ i.e, $\int_{m_0}^{m}c^2dm$, we have
$E_k=mc^2-m_0c^2$ or,
rearranging, $E_k+m_0c^2=E_t=mc^2$
This represent the total energy in the system, $E=mc^2$
We know that, Force is defined as the rate of change of momentum i.e., $F=\frac {d}{dt}(mv)$
According to the theory of relativity, both mass and velocity are a variable quantity. We therefore, can rewrite the above as follows,
$F=\frac {d}{dt}(mv)$ = $m\frac {dv}{dt}$+$v\frac {dm}{dt}$
Let the force $F$ displace the body through a distance $dx$. Then, the increase in the kinectic energy $dE_k$ of the body is equal to the work done $F.dx$
Hence, $dE_k= Fdx= (m\frac {dv}{dt}+v\frac {dm}{dt})dx$
$=(mv)dv + v^2 dm$
Now, according to the law of mass variation with velocity, we have;
$m=\gamma m_0$ , where $\gamma$ represent the factor $1/\sqrt{1-\frac{v^2}{c^2}}$
taking the differential of the mass and the velocity in the equation we have the following relation,
$c^2dm=(mv)dv+v^2dm$
substituting, the value of in the differential of the kinectic energy, we have as follows
$dE_k=c^2dm$
Integrating the above differential from $ m_0$ to $m $ i.e, $\int_{m_0}^{m}c^2dm$, we have
$E_k=mc^2-m_0c^2$ or,
rearranging, $E_k+m_0c^2=E_t=mc^2$
This represent the total energy in the system, $E=mc^2$
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