Irodov 1.374: How Slow Is Slow Enough – Amit & Deepali Sinha Foundation

Problem Irodov 1.374

The speed of a particle of rest mass $m_0$ is to be obtained from its kinetic energy $T$ using the classical relation $T = \frac{1}{2}m_0v^2$ instead of the relativistic one. Find the values of the ratio of kinetic energy to rest energy, $T/m_0c^2$ , for which the speed so obtained differs from the true speed by a relative error smaller than $\varepsilon = 0.010$ .

Given

$m_0$
$T$
$\varepsilon = 0.010$
$\tau = T/m_0c^2$

Find

The values of

\tau

for which the classical speed is in error by less than

\varepsilon

Solution

Step 1The two speeds the same kinetic energy implies

Write $\tau = T/m_0c^2$ for the ratio the answer is to be expressed in. The classical formula and the relativistic one assign different speeds to one and the same kinetic energy, so both are needed before they can be compared. Solving $T = \frac{1}{2}m_0v^2$ for the speed and pulling one factor of $c$ out of the root,

v_{\text{cl}} = \sqrt{\frac{2T}{m_0}} = c\sqrt{\frac{2T}{m_0c^2}} = c\sqrt{2\tau}

For the true speed, the relativistic kinetic energy $T = (\gamma-1)m_0c^2$ fixes the Lorentz factor, and inverting it gives $\beta^2$ .

\begin{align*}\gamma &= 1+\tau \\ \beta^2 &= 1-\frac{1}{\gamma^2} \\ &= \frac{(1+\tau)^2-1}{(1+\tau)^2} \\ &= \frac{\tau(2+\tau)}{(1+\tau)^2}\end{align*}

v = c\,\frac{\sqrt{\tau(2+\tau)}}{1+\tau}

(1)

Step 2The relative error in closed form

The relative error is the excess of the classical speed over the true one, measured against the true one, so it is a ratio of the two speeds less one. Dividing $v_{\text{cl}}$ by (1), the factors of $c$ and one factor of $\sqrt\tau$ cancel.

\begin{align*}\frac{v_{\text{cl}}}{v} &= \sqrt{2\tau}\cdot\frac{1+\tau}{\sqrt{\tau(2+\tau)}} \\ &= (1+\tau)\sqrt{\frac{2}{2+\tau}}\end{align*}

Subtracting one gives the error itself, exactly and with no approximation yet made.

\delta = \frac{v_{\text{cl}}-v}{v} = (1+\tau)\sqrt{\frac{2}{2+\tau}}-1

(2)

Step 3Expand for a small ratio and impose the bound

The bound asked for is a small one, so the crossing must occur at a small $\tau$ , and (2) may be expanded about $\tau = 0$ . Writing the root as a power and using $(1+u)^{-1/2} = 1-u/2+\ldots$ with $u = \tau/2$ ,

\begin{align*}\sqrt{\frac{2}{2+\tau}} &= \left(1+\frac{\tau}{2}\right)^{-1/2} \\ &= 1-\frac{\tau}{4}+O(\tau^2)\end{align*}

Multiplying by the prefactor $1+\tau$ and keeping terms to first order,

\begin{align*}\delta &= (1+\tau)\left(1-\frac{\tau}{4}\right)-1+O(\tau^2) \\ &= 1+\tau-\frac{\tau}{4}-1+O(\tau^2) \\ &= \frac{3}{4}\tau+O(\tau^2)\end{align*}

so the error grows three quarters as fast as the energy ratio. Requiring $\delta < \varepsilon$ therefore caps $\tau$ .

\frac{T}{m_0c^2} < \frac{4}{3}\varepsilon \approx 0.013

(3)

The exact relative error against the energy ratio τ. It rises from zero and crosses the royal line δ = ε = 0.010 at τ = 0.013; the admissible values of τ lie to the left of the marked crossing. — The exact relative error (2) against the energy ratio $\tau$ . It rises from zero and crosses the royal line $\delta = \varepsilon = 0.010$ at $\tau = 0.013$ ; the admissible values of $\tau$ lie to the left of the marked crossing.

Check

Test the bound against the exact error (2), which the expansion never entered. At $\tau = \frac43\varepsilon = 0.01333$ it gives $\delta = 0.00997$ , just inside the permitted $0.010$ , and at $\tau = 0.015$ it gives $0.0112$ , outside it; the exact crossing sits at $\tau = 0.01337$ . The first-order bound is therefore right to about three parts in a thousand, and it errs on the safe side for a reason worth naming: the terms the truncation drops are negative, $\delta = 3\tau/4-5\tau^2/32+\ldots$ , so the retained $3\tau/4$ overstates the true error and the bound admits slightly fewer values of $\tau$ than the exact condition would.

Remark

Read as a speed rather than as an energy ratio, the bound is remarkably generous: putting $\tau = 0.013$ back into (1) gives $v = 0.16c$ , about $4.8\times10^7$ m/s, so the classical formula is good to one percent up to a sixth of the speed of light. The tolerance in energy is far tighter than the tolerance in speed because $\tau$ enters the error linearly while $\beta$ enters it quadratically. In laboratory currency the same bound reads $T < 6.8$ keV for an electron and $T < 12.5$ MeV for a proton, so a thermal or chemical energy scale is nowhere near the boundary, while an electron accelerated through some $20$ kV, as in an old television tube, is already past it and needs the relativistic formula for its speed.