The world of particle physics abounds with a lot of convenient but specialized jargon that takes some getting used too. This is particularly true in the context of natural units where the speed of light $$c$$ is a fundamnetal yark stick for measuring velocity and mass/energy and where electric charge is measured in units of the charge of an electron. The use of natural units traces back to the roots of the discipline in the studies into electromagnetism. Those practiced in the art specfy energies and masses in units of electron-volts ($$eV$$) and electron-volts per $$c^2$$.
To put things in perspective, note that room temperature $$T \approx 300 k$$ corresponds to an energy content of about $$1/40 eV$$. In contrast, the energy content associated with the mass of an electron is approximately $$511,000 eV$$ or $$511 KeV$$. The key for determining when particle motion can be treated classically versus relativistically is the ratio of kinetic energy to the rest mass energy. At the low energies representative of routine atomic processes ($$< 100 eV$$), the kinetic energy is very small compared with the rest energy and classical methods work well. On the other side are ultra-relativistic energies (such is seen in the LHC) where the kinetic energy is so much larger than the rest energy that the latter can be ignored. The grey area is that stretch inbetween low and ultra-high energy where the ratio ranges from 0.1 to 10 or so. The aim of this note is to look a little at this inbetween zone where the classical description breaks down and gives way to the relativistic one. To set the notation for the relativistic descripion, note that the interval is \[ c^2 dt^2 - dx^2 - dy^2 - dz^2 = c^2 d\tau^2 \] and that coordinate velocities are denoted as $$v^x \equiv \frac{dx}{dt} $$ and so on. Dividing out $$c dt$$ from all terms in the interval gives \[ c^2 dt^2 \left[ 1 - \frac{\vec v^2}{c^2} \right] = c^2 d\tau^2 \; ,\] from which the traditional, relativistic $$\gamma$$ is defined as: \[ \left(\frac{dt}{d\tau}\right)^2 = \frac{1}{\left[1-\vec v^2/c^2\right]} \equiv \gamma^2 \; .\] If a differential movement in spacetime is denoted by \[ dq^{\mu} = \left[\begin{array}{cccc} c dt & dx & dy & dz \end{array} \right] \] and the Minkowski metric \[ \eta_{\mu\nu} = diag(1,-1,-1,-1) \] is used, then the four-velocity is given by \[ u^{\mu} \equiv \frac{d q^{\mu}}{d \tau} = \left[\begin{array}{cccc} c \frac{dt}{d\tau} & \frac{dx}{d\tau} & \frac{dy}{d\tau} & \frac{dz}{d\tau} \end{array} \right] \; .\] The chain rule proves particularly helpful, allowing the coordinate time $$t$$ to be used in place of the proper time $$\tau$$:0 \[ u^{\mu} = \frac{dt}{d\tau}\left[\begin{array}{cccc} c & \frac{dx}{d\tau} \frac{d\tau}{dt}& \frac{dy}{d\tau}\frac{d\tau}{dt} & \frac{dz}{d\tau}\frac{d\tau}{dt} \end{array} \right] \; .\] Factoring out the $$\gamma$$ gives a compact form for the four-velocity of \[ u^{\mu} = \gamma \left[\begin{array}{cc}c & \vec v \end{array} \right] \; .\] The four-momentum is defined as \[ p^{\mu} = m u^{\mu} = m \gamma \left[\begin{array}{cc}c & \vec v \end{array} \right] \; . \] The normalization of the four-momentum is of particular physical interest. It is defined as: \[ p^{\mu} p_{\mu} = m^2 \gamma^2 \left[\begin{array}{cc}c & \vec v \end{array} \right] \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right] \left[\begin{array}{c}c \\ \vec v \end{array} \right] \; ,\] which simplifies to \[ p^{\mu}p_{\mu} = m^2 \gamma^2 (c^2 - \vec v^2) = m^2 \gamma^2 c^2 \left[1-\vec v^2/c^2\right] = m^2 c^2 \; . \] The zeroth component, \[ p^0 = m c \gamma = \frac{m c}{\sqrt{1-\frac{\vec v^2}{c^2} }} \; ,\] is related to the total energy by the equation \[ E = c p^0 = mc^2 \gamma = \frac{RE}{\sqrt{1 - \frac{\vec v^2}{c^2}}} \; .\] This last relation is cornerstone for looking at the cross-over from classical to relativistic descriptions. The approach will be to compute the coordinate velocity both ways and to compare the relative error made in the classical expression. Two different scenarios will be explored. The first is the usual one where we know the kinetic energy of the particle, having accelerated it through fixed potential difference. The second, and far less common scenario, is where we know the relativistic momentum of the particle after a collision. Scenario 1: Measure the kinetic energy By its thermodynamic definition, energy is additive and the total energy of a particle is given by the sum \[ E = K + RE \] of its kinetic energy $$K$$ and rest energy $$RE = m c^2$$. Recasting the definition of energy given above in terms rest energy gives \[ E = \frac{RE}{\sqrt{1 - \frac{\vec v^2}{c^2}}} \; .\] Equating the two expressions gives \[ \frac{RE}{\sqrt{1 - \frac{\vec v^2}{c^2}}} = K + Re \; ,\] which can be simplified to \[ K = (\gamma - 1) RE \; .\] Next define the ratio of kinetic and rest energies as \[ b = \frac{K}{RE} \; .\] Using this definition leads to the identity \[ b + 1 = \gamma \] and the subsequent expression for the coordinate velocity \[ v = c \sqrt{ 1 + \frac{1}{(1+b)^2} } \; \] For a $$K = 30 \; KeV$$ electron, with rest energy $$RE = 510.839 \; KeV$$, the resulting coordinate velocity (really speed) is \[ v = 98458.806 \; km/s \; .\] The classical method, expressed via the usual relation $$K = 1/2 m v^2$$, yields \[ v_{classical} = \sqrt{\frac{2 K}{m_e}} \] or, numerically, \[ v_{classical} = 102743.45 \; km/s \; ,\] corresponding to a relative error of $$\epsilon_1 = 0.0435171$$. While not a gross error, using the classical expression results in a non-negligible difference. Scenario 2: Measure the relativistic momentum If instead of knowing the kinetic energy, suppose we know the spatial portion of the relativistic four-momentum $${\mathcal P}^2 = \vec p^2 c^2$$. The normaliztion of the four-momentum means that \[ E^2 = m^2 c^2 + p^2 c^2 \equiv RE^2 + {\mathcal P}^2 \; .\] Defining the ratio \[ d = \frac{{\mathcal P}}{RE} \] and factoring and taking the square root gives \[ E = RE \sqrt{1 + d^2} \; .\] Combining with the formula for $$E$$ in terms of $$\gamma$$ allows us to determine \[ v = c\sqrt{1-\frac{1}{1+d^2}} = c \sqrt{\frac{d^2}{1+d^2}} \; .\] If we suppose that the energy associated with $${\mathcal P} = 30 KeV$$ then the resulting speed is \[ v = 17575.59 \; km/s \] whereas the classical expression is \[ v_{classical} = 17598.29 \; km/s \; .\] The relative error is $$\epsilon_2 = 0.00192$$; substantially smaller than the case reflecting the fact that the speed is so much smaller (almost a factor of 6 smaller). It is interesting to note that, in both cases, the classical expression overestimates the speed. This is consistent with the notion that the classical theory has no fundamental speed limit like relativity but it this is only a conjecture at this point.