How can we define total energy and rest energy in relativistic energy equation?

I have learned that, derivation of mass-energy equivalence leads to $K=\gamma mc^2 -mc^2$ and it means that $\gamma mc^2$ is total energy and $mc^2$ is rest energy. But doesn't this equation just for $\gamma mc^2 -mc^2$ as a whole, not what $\gamma mc^2$ is and what $mc^2$ is? Is there an additional method to define exactly what each is? Or is there a nicer way to explain mass-energy equivalence?

I guess it depends on what you think is nicer. I will provide a derivation that I think is nicer, starting from the action principle.

Recall, that in classical mechanics the action $S$ is equal to the time integral of the Lagrangian $L$ and it is used to derive the equations of motion.

In non-relativistic classical mechanics, the action for a free particle is $$ S_{fNR} = \int dt L_{fNR}\; $$ where $$ L_{fNR} = \frac{1}{2}mv^2\;.\tag{0} $$

Recall further that shifting the Lagrangian by a constant does not change the equations of motion.

In relativistic classical mechanics, we seek an action $S_R$ that is Lorentz invariant. For a free particle we have one convenient invariant, the length of the particle's world $\int ds=\int c d\tau$, where $\tau$ is the proper time. So, we presume that $S$ is proportional to that length: $$ S_{fR} = \alpha \int ds\tag{1}\;, $$ where $\alpha$ is an undetermined constant.

Rewriting Eq. (1) via $ds=\sqrt{ds^2}=\sqrt{c^2dt^2 - d\vec x^2}$, we have $$ S_{fR} = \alpha \int dt \sqrt{c^2 - \vec v^2} $$ $$ S_{fR}= \int dt c\alpha\sqrt{1-\frac{v^2}{c^2}} $$ $$ =\int dt L_{fR}\;, $$ where $$ L_{fR}=c\alpha\sqrt{1-\frac{v^2}{c^2}}\;.\tag{2} $$

We now want to make a connection to the non-relativistic limit. This is the small $v/c$ limit. In this limit we find $$ L_{fR} \underbrace{\to}_{\frac{v}{c}\to 0} c\alpha(1 -\frac{1}{2}\frac{v^2}{c^2}+\ldots)\;, $$ and to make this agree with the non-relativistic expression of Eq. (0), we have to set $\alpha = -m c$.

Thus, returning to Eq. (2), we have $$ L_{fR} = -mc^2\sqrt{1-\frac{v^2}{c^2}}\;.\tag{3} $$

And we define the momentum in the usual way as $$ \vec p_R \equiv \frac{\partial L_R}{\partial \vec v} $$ $$ =-mc^2\frac{1}{2}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{-2\vec v}{c^2} $$ $$ =m\gamma \vec v\;, $$ where $\gamma = 1/\sqrt{1-v^2/c^2}$;.

And we define the (hamiltonian) energy in the usual way as $$ E_{fR} = \vec p\cdot \vec v - L_{fR} $$ $$ =m\gamma v^2 + mc^2/\gamma = mc^2 \gamma\;. $$

Thus, we have $E_{fR} = mc^2 \gamma$, which can again be expanded to find that, in the non-relativistic limit, $$ E_{fR}\to mc^2 + \frac{1}{2}mv^2 + \ldots\;, $$ so, the rest energy (the energy when the velocity is zero) is seen to be a constant $mc^2$. Thus we write $$ E_{fR} = K + mc^2\;, $$ where $m$ is the invariant rest mass and where $K=(\gamma - 1)mc^2$ has the correct non-relativistic limit of $\frac{1}{2}mv^2$.

I think one argument would be to put:

$$ F(v) = \gamma(v) m c^2, \ \ C = mc^2, $$

then you have by assumption that:

$$ F(v) - C = E(v) - R, $$

where $E(v)$ is the total energy and $R$ is the "true" rest energy of the body.

Now if we also define a constant $\kappa=C-R$, this immediately gives:

$$ F(v) = E(v) + C - R = E(v) + \kappa $$

So the proposed value for the total energy $F(v)$, differs from $E(v)$ by a numerical constant. But we know that an addition of a constant to the total energy is not physically meaningful, hence practically this specific part of the query is not physically meaningful.

Therefore ultimately the question comes down to whether the specific physical constant $C=mc^2$, with $m$ being the invariant mass, is or is not the true rest energy, here denoted by $R$. This can be tested experimentally, where indeed we find that $C=R=mc^2$ holds good. Once we determine that, it also becomes very natural to define the total energy as $\gamma m c^2$.

Using 4-vectors, particles are described by their 4-momenta, $\tilde p$.

Using the (+,-,-,-)-signature convention with natural units... but restricting to the (1+1)-case for simplicity...

The invariant-mass of a particle is given by the square-root of the Minkowski-dot-product of the 4-momentum with itself: $$m=\sqrt{\tilde p\cdot \tilde p}$$ [As in intro physics, where $\vec A$ has magnitude $A=\sqrt{\vec A\cdot\vec A}$ (leaving off the arrowhead), we could have called our 4-momentum $\tilde m$ (instead of $\tilde p$).]

Given a timelike-observer (using 4-velocity $\hat t$ [a unit timelike-vector tangent to the observer worldline]),

• the time-component of the particle 4-momentum is the "[total] relativistic energy of the particle as measured by $\hat t$" $$E=\hat t \cdot \tilde p = m\cosh\theta =\gamma m,$$ which, in the case when $m>0$, could be written in terms of the mass $m$ and the rapidity $\theta$ (the Minkowski-angle between $\hat t\cdot \tilde p$, future-timelike vectors when $m>0$).
• the spatial-component of the particle 4-momentum is the "relativistic momentum of the particle as measured by $\hat t$" $$P=\hat t_\perp \cdot \tilde p = m\sinh\theta = \beta\gamma m$$
• with $$\frac{P}{E}=\tanh\theta=\beta=\frac{v}{c}.$$ ($E$ and $P$ are essentially the legs of a Minkowski-right-triangle with hypotenuse $\tilde p$ and legs parallel and Minkowski-perpendicular to $\hat t$.

When $m>0$, the 4-momentum $\tilde p$ is timelike and can be expressed in terms of $m$ and $\theta$-relative-to-$\hat t$. With $\theta$ between $\hat t$ and $\tilde p$, then $E$ is the adjacent-side and $P$ is the opposite side in this Minkowski-right-triangle.)

For $m>0$, when $P=0$, then $E$ coincides with $m$. So, $m$ could be called the rest-energy.

Then, the kinetic energy for a massive particle ($m>0$) is then often formulated as the difference--the part of the total-relativisitic-energy that is not-the-rest-energy: $$T=E-m=\gamma m-m=(\gamma-1)m.$$

Certainly, the action approach gives more motivation for these results.

Given the 4-momentum formulation, one can solve problems with an energy-momentum diagram.

• See my answer to Momentum diagram for two colliding Particles for examples and links to other examples.

• See my answer to How is the time-component of the spacetime interval in a spacetime diagram related to the time component of the energy-momentum 4 vector? for ideas related to relativistic-kinetic-energy.