Why does relativistic kinetic energy (KE) equal to the total relativistic energy minus the rest mass energy?

Question

OK, I understand why $KE=\gamma mc^2-mc^2$, but why is it also equal to $E-E_0$?

Answer 1

I think the answer is that if we start with

$$KE = \gamma mc^2 - m c^2$$

then because $mc^2$ is the rest mass energy. $E_0$

$$ E_0 = m c^2 $$

and the total energy $E$ is given by

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

so substituting into your first equation gives

$$ KE = E - E_0 $$

does this help?

Answer 2

If you take the non-(special)-relativistic limit of that expression you find that $$\text{KE} \approx \left(\frac12mv^2 + mc^2\right) - mc^2$$ which is what you would expect from classical mechanics. The other way to see this is that, when a particle is at rest in a certain inertial frame of reference, its relativistic energy is given by its rest mass through $m_0c^2$. This represent a sort of frozen energy contained inside the body. Kinetic energy is then extra energy the body gains when it aquires a certain velocity.

Answer 3

Relativistic kinetic energy is defined as $E-E_0$ because this definition naturally follows from the more general definition of relativistic energy as $E = \gamma mc^2$ where $\gamma$ depends upon the velocity.

It then follows to define rest energy for $v=0, \gamma=1$ as $E_0 = mc^2$, and relativistic kinetic energy as the excess energy when its moving over the energy when it's not: $E-E_0 =\gamma mc^2 - mc^2$.

The more fundamental question to then ask is: why is relativistic energy defined as $E = \gamma mc^2$?

Energy is defined via Noether's theorem as that physical quantity that's conserved for a system whose physical laws don't change over time. For a single point particle of mass m

$$E =\gamma mc^2$$