Relativistic energy

Relativistic energy is a term used to refer to the energy of a body when taking relativistic effects into account. It is given by

E=p2c2+m2c4=γmc2E=\sqrt{ p^{2}c^{2}+m^{2}c^{4} }=\gamma mc^{2}

where pp is the momentum of the body, mm is its mass, cc is the speed of light and γ\gamma is the relativistic gamma. In the special case of an object with zero momentum, what energy is left is known as the rest (mass) energy of the body:

E0=mc2E_{0}=mc^{2}

More generally, this formula is known as the mass-energy equivalence.

Unlike classical mechanics where, barring potential energy, the energy would always be zero for an object with zero speed, in relativity there is always some energy for any massive object. As such, this rest energy must be of a special nature and we are thus left with three, not two, pieces of energy: potential energy, kinetic energy and rest energy. The relativistic energy is the sum of the last two. In fact, the kinetic energy of a body is

K=E−E0=E−mc2K=E-E_{0}=E-mc^{2}

The total relativistic energy of a system is a conserved quantity. This does not mean it is a relativistic invariant; conservation has nothing to do with frames of reference, it just means that the quantity remains the same before and after a process. Notably, this means that mass is not conserved, as it can "flow" into kinetic energy and viceversa so long the total relativistic energy is conserved.

From four-momentum

In the same way as classical momentum correlates quite directly with kinetic energy, so too does the four-momentum correlate with relativistic energy. Inspired by the fact that to get the classical energy we need to take the square norm of momentum (since E=∣p∣2/2mE=\lvert \mathbf{p} \rvert^{2}/2m), we'll take the norm of the four-momentum (we'll use ∣p∣≡p\lvert \mathbf{p} \rvert\equiv p for simplicity):

pμpμ=−p0p0+p⋅p=−(p0)2+p2=−γ2m2c2+γ2m2v2==m21−v2/c2(−c2+v2)=−m2c2c2−v2(c2−v2)=−m2c2\begin{align} p^{\mu}p_{\mu}&=-p^{0}p^{0}+\mathbf{p}\cdot \mathbf{p}=-(p^{0})^{2}+p^{2}=- \gamma ^{2}m^{2}c^{2}+ \gamma ^{2}m^{2}v^{2}= \\ &=\frac{m^{2}}{1- v^{2}/c^{2}}(-c^{2}+v^{2})=-\frac{m^{2}c^{2}}{c^{2}-v^{2}}(c^{2}-v^{2})=-m^{2}c^{2} \end{align}

But also

pμpμ=−E2c2+p2p^{\mu}p_{\mu}= -\frac{E^{2}}{c^{2}}+p^{2}

Equate the two and multiply by −c2-c^{2} to get

E2−p2c2=m2c4E^{2}-p^{2}c^{2}=m^{2}c^{4}

When reordered, we find what's known as the relativistic energy of a body:

E2=p2c2+m2c4\boxed{E^{2}=p^{2}c^{2}+m^{2}c^{4}}

In the simplest case of an object with no momentum, p2=0p^{2}=0, we find the well-known formula due to Einstein for rest energy (or rest mass energy):

E0=mc2\boxed{E_{0}=mc^{2}}

This is the energy an object holds exclusively due to its mass. This stands in contrast with kinetic energy KK, which is given by the motion. With this in mind, the relativistic energy is

E2=E02+K2E^{2}=E_{0}^{2}+K^{2}

We can also rewrite β\beta and γ\gamma as

γ=Emc2,β=pcE\gamma=\frac{E}{mc^{2}},\qquad\beta=\frac{pc}{E}

Connection to classical energy

We can see how the relativistic energy falls back to classical energy in the low velocity limit. If we extract the kinetic part of the energy

K=E−mc2=γmc2−mc2=mc2(γ−1)K=E-mc^{2}=\gamma mc^{2} -mc^{2}=mc^{2}(\gamma-1)

Expressing γ\gamma in binomial expansion for β≪1\beta\ll1 and taking the first order, we find

K≃(1+β22−1)=mc2β22=v22c2mc2=12mv2K\simeq\left(1+ \frac{\beta^{2}}{2} -1\right)=\frac{mc^{2}\beta^{2}}{2}=\frac{v^{2}}{2c^{2}}mc^{2}=\frac{1}{2}mv^{2}

which is the classical formula for kinetic energy.