Irodov 3.375: When Kinetic Energy Equals Rest Energy

Problem Irodov 3.375

An electron, initially at rest, is set moving by a uniform electric field of strength $E = 10$ kV/cm. How long after the start does its kinetic energy become equal to its rest energy?

The electron e at rest in the uniform electric field E, seen side on. — The electron $e$ at rest in the uniform electric field $\vect E$ , seen side on.

Given

$E = 10$ kV/cm
$v(0) = 0$
kinetic energy equal to the rest energy at the moment sought

Find

the time

t

from the start

Solution

Step 1Momentum grows uniformly

The force on the electron, $eE$ in magnitude, is constant and along the field line, and the electron starts from rest, so its momentum grows linearly in time:

\frac{dp}{dt} = eE \quad\Longrightarrow\quad p = eEt \tag{1}

Step 2The momentum when the kinetic energy is

T

The total energy is $T + mc^2$ , and it is tied to the momentum by the invariant relation between energy and momentum. Solving that relation for $pc$ :

\begin{aligned} (T + mc^2)^2 &= (pc)^2 + (mc^2)^2, \\ (pc)^2 &= T^2 + 2Tmc^2 \end{aligned}

pc = \sqrt{T(T + 2mc^2)}

(2)

Step 3The time

Solving (1) for the time, and writing it through $pc$ so that (2) can go in:

t = \frac{p}{eE} = \frac{pc}{ceE}

Numerically, $T = mc^2$ makes the root $\sqrt{3}\,mc^2$ , and the field is $1.0\times10^6$ V/m:

t = \frac{\sqrt{3}\,mc}{eE} = \frac{\sqrt{3}\,(0.911\times10^{-30})(3.00\times10^{8})}{(1.60\times10^{-19})(1.0\times10^{6})} \text{ s}

t = \frac{\sqrt{T(T + 2mc^2)}}{ceE} \approx 3.0 \text{ ns}

(3)

Check

Units: $mc/(eE)$ is a momentum over a force, a time. Consistency: the work $eEx$ equals $T$ , so the electron has moved $x = mc^2/(eE) = 0.511$ m, an average speed of $x/t = 0.58\,c$ ; its final speed is $v = pc^2/(T + mc^2) = (\sqrt{3}/2)\,c = 0.87\,c$ , comfortably above the average for a particle that starts from rest, and below $c$ .

Remark

Newtonian mechanics would have the electron reach $c$ after only $mc/(eE) = 1.7$ ns. In fact $p = eEt$ grows without limit while the speed $v = c\,eEt/\sqrt{(mc)^2 + (eEt)^2}$ only creeps toward $c$ : a constant force keeps adding momentum and energy, and almost none of it shows up as speed once $eEt \gg mc$ .