# Could we find relativistic energy in a simple way?

I asked a couple of questions on mass(-energy) increasing with speed, but there is still a very simple aspect I cannot understand, I hope you can give a simple and direct answer:

The formula to find the relativistic energy of a particle is :$$E^2 = p^2c^2 + m^2c^4$$ which is derived from the original Lorentz formula for mass:$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ which gives the value of the total mass including the rest mass and, subtracting the latter, we get the kinetic energy (net increase of total energy): $$E_k=m_0c^2*\left[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1\right]$$

If this is correct we know that when $v = 0.866c$ the total mass is twice the rest mass and $E_k$ is equivalent to one rest mass, in the case of an electron 0.511 MeV.

My question is: why is it not suggested (or allowed) to find the increased energy (the kinetic energy) in such a simple way? If an electron's speed is $0.866 c$, its total mass is 2 electron masses, one is rest mass and one is $E_k$: $\gamma= 2$, and $2m_0-m_0 = 1 m_0$. Because of the equivalence of mass-energy, the relativistic energy of the particle is therefore: 0.511 MeV. Conversely if we give an electron an acceleration of 0.511 MeV we know right away that its mass will double.

Is this procedure wrong in any way?

• So, you are saying that it is always correct to deduce the relativistic KE simply multiplying rest mass ( in this case .511 MeV) by $\gamma -1$ instead of using the more complicated formula usually suggested : $E^2 = p^2c^2 + m^2c^4$? – user104372 Apr 21 '16 at 10:54
• Beware! $E≠E_k$. $E$ is the total energy, $E_k=E-mc^2$ is the kinetic energy. $\sqrt{p^2c^2+m^2c^4}$ is $E$, not $E_k$. Note that the square root is a pain to work with, so when one studies inelastic collisions (e.g. Compton diffusion), one makes sure to have $E^2$ show up! In inelastic collisions mass is not conserved, so kinetic energy neither, and you have no choice but to consider total energy. – L. Levrel Apr 21 '16 at 19:08