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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?

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I see nothing wrong in this procedure. Where do you see it not to be allowed?

The only objection raised by your post is calling "total mass" the "relativistic mass", the use of which is nowadays discouraged as stated by John Rennie in his answer to your question you link to.

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  • $\begingroup$ 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 $? $\endgroup$ – user104372 Apr 21 '16 at 10:54
  • $\begingroup$ it works for all speeds. If v is much smaller than c, it will approximate conventional kinetic energy calculation. $\endgroup$ – Peter R Apr 21 '16 at 13:11
  • $\begingroup$ The rest mass energyof the electron is .511 MeV and the lorentz factor is 2 for .866c so your calculation looks correct. The preferred formula is a more accurate way to define the total energy. In simple cases and if you know what you are doing, subtracting the rest mass energy from the total relativistc energy works fine. $\endgroup$ – Peter R Apr 21 '16 at 15:52
  • $\begingroup$ 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. $\endgroup$ – L. Levrel Apr 21 '16 at 19:08

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