36.10 Relativistic Energy

Energy is our final topic in this chapter on relativity. Space, time, velocity, and momentum are changed by relativity, so it seems inevitable that we’ll need a new view of energy.

A photograph of liquid mercury collected at the bottom of a thin tube. — Mercury is a metal, so why is it a liquid down to temperatures of nearly –40°C? Surprisingly, because of relativity. Many of the heaviest elements in the periodic table have unexpected properties because their electrons are moving at a substantial fraction of the speed of light and thus exhibit relativistic effects. Chemical theories that include relativity are needed to explain the properties of the heaviest elements. In the case of mercury, relativistic effects alter the electron shells so that the outer electrons are much less capable of establishing bonds than is predicted by classical chemistry. Consequently, the bonds between mercury atoms are too weak for it to form a solid at room temperature.

In Newtonian mechanics, a particle’s kinetic energy $K = \frac{1}{2} m u^{2}$ can be written in terms of its momentum $p = m u$ as $K = p^{2} / 2 m$ . This suggests that a relativistic expression for energy will likely involve both the square of p and the particle’s mass. We also hope that energy will be conserved in relativity, so a reasonable starting point is with the one quantity we’ve found that is the same in all inertial reference frames: the spacetime interval s.

Let a particle of mass m move through distance $Δ x$ during a time interval $Δ t$ , as measured in reference frame S. The spacetime interval is

s^{2} = c^{2} {(Δ t)}^{2} - {(Δ x)}^{2} = invariant

We can turn this into an expression involving momentum if we multiply by ${(m / Δ τ)}^{2}$ , where $Δ τ$ is the proper time (i.e., the time measured by the particle). Doing so gives

{(m c)}^{2} {(\frac{Δ t}{Δ τ})}^{2} - {(\frac{m Δ x}{Δ τ})}^{2} = {(m c)}^{2} {(\frac{Δ t}{Δ τ})}^{2} - p^{2} = invariant

(36.37)

where we used $p = m (Δ x / Δ τ)$ from Equation 36.32.

Now $Δ t$ , the time interval in frame S, is related to the proper time by the time-dilation result $Δ t = γ_{p} Δ τ$ . With this change, Equation 36.37 becomes

(γ_{p} m c)^{2} - p^{2} = invariant

Finally, for reasons that will be clear in a minute, we multiply by $c^{2}$ , to get

(γ_{p} m c^{2})^{2} - {(p c)}^{2} = invariant

(36.38)

To say that the right side is an invariant means it has the same value in all inertial reference frames. We can easily determine the constant by evaluating it in the reference frame in which the particle is at rest. In that frame, where $p = 0$ and $γ_{p} = 1$ , we find that

(γ_{p} m c^{2})^{2} - {(p c)}^{2} = {(m c^{2})}^{2}

(36.39)

Let’s reflect on what this means before taking the next step. The spacetime interval s has the same value in all inertial reference frames. Equation 36.39 was derived from the definition of the spacetime interval; hence the quantity $m c^{2}$ is also an invariant having the same value in all inertial reference frames. In other words, if experimenters in frames S and $S'$ both make measurements on this particle of mass m, they will find that

{(γ_{p} m c^{2})}^{2} - {(p c)}^{2} = {(γ_{p}^{'} m c^{2})}^{2} - {(p' c)}^{2}

(36.40)

Experimenters in different reference frames measure different values for the momentum, but experimenters in all reference frames agree that momentum is a conserved quantity. Equations 36.39 and 36.40 suggest that the quantity $γ_{p} m c^{2}$ is also an important property of the particle, a property that changes along with p in just the right way to satisfy Equation 36.39. But what is this property?

The first clue comes from checking the units. $γ_{p}$ is dimensionless and c is a velocity, so $γ_{p} m c^{2}$ has the same units as the classical expression $\frac{1}{2} m v^{2}$ —namely, units of energy. For a second clue, let’s examine how $γ_{p} m c^{2}$ behaves in the low-velocity limit $u ≪ c$ . We can use the binomial approximation expression for $γ_{p}$ to find

γ_{p} m c^{2} = \frac{m c^{2}}{\sqrt{1 - u^{2} / c^{2}}} \approx (1 + \frac{1}{2} \frac{u^{2}}{c^{2}}) m c^{2} = m c^{2} + \frac{1}{2} m u^{2}

(36.41)

The second term, $\frac{1}{2} m u^{2}$ , is the low-velocity expression for the kinetic energy K. This is an energy associated with motion. But the first term suggests that the concept of energy is more complex than we originally thought. It appears that there is an inherent energy associated with mass itself.