Irodov 1.373: Where Motion Doubles a Particle's Energy

Problem Irodov 1.373

A particle of rest mass $m_0$ moves with a speed $v$ at which its kinetic energy comes out exactly equal to its rest energy $m_0c^2$ . Find that speed.

Given

$m_0$
$T = m_0c^2$
$\beta = v/c$
$c = 3.00\times10^8$ m/s

Find

The speed

v

at which the kinetic energy reaches the rest energy

Solution

Step 1Turn the condition into a condition on the Lorentz factor

The kinetic energy of a moving particle is everything its total energy carries beyond the rest energy, so writing the total energy as $\gamma m_0c^2$ with $\gamma = (1-\beta^2)^{-1/2}$ gives the kinetic energy in the same terms.

T = \gamma m_0c^2-m_0c^2 = (\gamma-1)\,m_0c^2

Setting this equal to the rest energy and dividing both sides by $m_0c^2$ removes the mass from the problem entirely.

\begin{align*}(\gamma-1)\,m_0c^2 &= m_0c^2 \\ \gamma-1 &= 1\end{align*}

so the required state is the one whose Lorentz factor is two.

\gamma = 2

(1)

Step 2Invert the Lorentz factor for the speed

The definition of $\gamma$ is a statement about $\sqrt{1-\beta^2}$ , so it inverts by squaring and solving for $\beta^2$ , and the inversion is worth doing once for a general $\gamma$ rather than for the value at hand.

\begin{aligned} \gamma &= \frac{1}{\sqrt{1-\beta^2}} \\ 1-\beta^2 &= \frac{1}{\gamma^2} \\ \beta^2 &= 1-\frac{1}{\gamma^2} \end{aligned} \tag{2}

The speed is $c$ times the positive root of this, so the whole problem now rests on the single number $\gamma$ of (1).

Step 3Put the number in

Substituting the required $\gamma = 2$ of (1) into (2), one factor at a time:

\frac{1}{\gamma^2} = \frac14, \qquad 1-\frac{1}{\gamma^2} = \frac34, \qquad \sqrt{\frac34} = 0.866

Multiplying that fraction of $c$ by $c = 3.00\times10^8$ m/s gives the speed itself.

v = c\sqrt{1-\frac{1}{\gamma^2}} = \frac{\sqrt3}{2}\,c = 2.6\times10^8\ \text{m/s}

(3)

Kinetic energy in units of the rest energy, against speed. The gold curve is the relativistic γ-1, the royal one the classical β^2/2. Gold reaches 1 at the marked abscissa 0.87, where the classical curve stands at 0.38. — Kinetic energy in units of the rest energy, against speed. The gold curve is the relativistic $\gamma-1$ , the royal one the classical $\beta^2/2$ . Gold reaches $1$ at the marked abscissa $0.87$ , where the classical curve stands at $0.38$ .

Check

Run the speed back through the condition (1). At $\beta = \sqrt3/2$ the quantity $1-\beta^2$ is $1/4$ , its square root is $1/2$ , and the Lorentz factor is its reciprocal, $\gamma = 2$ ; then $T = (\gamma-1)m_0c^2$ is $m_0c^2$ , the rest energy, as required. The classical formula would have answered this question with $\frac{1}{2}m_0v^2 = m_0c^2$ , that is $v = c\sqrt2$ , a speed no particle can have, so the relativistic treatment is not a refinement here but the only one that can answer at all.

Remark

At this speed the particle’s total energy in the laboratory frame is $2m_0c^2$ , half of it rest energy and half of it kinetic: below this speed most of a particle’s energy is rest energy, above it most is kinetic, so $\sqrt3\,c/2$ is the crossover. Two other readings of the same $\gamma = 2$ , both in that same laboratory frame: the particle’s relativistic momentum is twice the Newtonian $m_0v$ , and one second of its own proper time corresponds to two seconds of laboratory time. The kinetic energy this costs is $0.51$ MeV for an electron and $938$ MeV for a proton, so the very same speed is routine bench physics for the one and accelerator work for the other.