Irodov 1.368: Inertia from the Square Root's Edge – Amit & Deepali Sinha Foundation

Problem Irodov 1.368

A particle travels with a speed $v$ that falls short of the speed of light by $0.010$ percent of $c$ , so that $\left(c-v\right)/c = 1.0\times10^{-4}$ . Write $\beta = v/c$ , and take the relativistic mass in the usual convention,

m = \frac{m_0}{\sqrt{1-\beta^2}},

with $m_0$ the particle’s rest mass. Find the factor $m/m_0$ by which the relativistic mass exceeds the rest mass.

Given

$\left(c-v\right)/c = 0.010\,\%$
$\beta=v/c$
$m = m_0/\sqrt{1-\beta^2}$

Find

the ratio

m/m_0

Solution

Step 1Write the ratio as a function of the gap below

c

The convention supplies the ratio directly, but in the form $1-\beta^2$ it hides what the data actually give, which is the difference between $\beta$ and $1$ . Factoring the difference of two squares splits it into a factor that is tiny and a factor that is close to $2$ .

\frac{m}{m_0} = \frac{1}{\sqrt{1-\beta^2}}

\frac{m}{m_0} = \frac{1}{\sqrt{\left(1-\beta\right)\left(1+\beta\right)}}

(1)

The ratio m/m_0 against β, climbing from 1 to about 11. The dotted gold line at β = 1 is the vertical asymptote. This problem's β=0.9999 lies past the right-hand end of the curve and above the top of this frame. — The ratio $m/m_0$ against $\beta$ , climbing from $1$ to about $11$ . The dotted gold line at $\beta = 1$ is the vertical asymptote. This problem’s $\beta=0.9999$ lies past the right-hand end of the curve and above the top of this frame.

Step 2Read the data as the small factor

The statement fixes exactly that small factor. A speed differing from $c$ by $0.010$ percent means the shortfall $c-v$ is $1.0\times10^{-4}$ of $c$ , and dividing that difference by $c$ turns it into $1-\beta$ .

\frac{c-v}{c} = 1-\frac{v}{c} = 1-\beta = 1.0\times10^{-4}

The other factor is then fixed too, since $\beta$ is now known.

\begin{gathered} 1-\beta = 1.0\times10^{-4}, \\ 1+\beta = 2-\left(1-\beta\right) = 1.9999 \end{gathered}

(2)

Step 3Evaluate the ratio

Substituting (2) into (1) needs no approximation at all, but the second factor differs from $2$ only in its fifth figure, so replacing it by $2$ costs nothing at the precision of the data and leaves the compact form that is worth remembering.

\frac{m}{m_0} = \frac{1}{\sqrt{\left(1-\beta\right) \Bigl(2-\left(1-\beta\right)\Bigr)}}

Inside the second bracket the subtraction moves $2$ by one part in twenty thousand, so it may be dropped.

\frac{m}{m_0} \approx \frac{1}{\sqrt{2\left(1-\beta\right)}} = \frac{1}{\sqrt{2.0\times10^{-4}}} \approx 70.7

(3)

The same ratio against the gap 1-β in units of 10^-4, shrinking to the left as the particle speeds up. The marked point is this problem's particle: gap 1.0·10^-4, ratio 70.7. — The same ratio against the gap $1-\beta$ in units of $10^{-4}$ , shrinking to the left as the particle speeds up. The marked point is this problem’s particle: gap $1.0\times10^{-4}$ , ratio $70.7$ .

Check

Keeping the exact second factor instead gives $1/\sqrt{1.0\times10^{-4}\times1.9999} = 70.712$ , against the $70.711$ that the approximation gives, so dropping $1-\beta$ beside $2$ changes nothing until the fifth figure. The ratio is a pure number, as it must be, being a mass over a mass. And at the other end of the range the formula behaves: with $\beta \to 0$ the boxed factorisation (1) gives $m/m_0 \to 1$ , the rest mass itself, which is the classical limit.

Remark

Because the ratio goes as the inverse square root of the gap, speed is a very poor way to describe a fast particle. Dropping the gap from $10^{-4}$ to $10^{-7}$ , a change invisible in any statement of $v$ , raises the ratio from $70.7$ to $2240$ . The ratio is the Lorentz factor $\gamma$ for this particle in the frame where its speed is $v$ , so the same number multiplies its lifetime as that frame measures it. It is also the total energy in units of the rest energy, since $E = mc^2$ : an electron at this speed carries $70.7\times0.511 \approx 36$ MeV, of which all but $0.5$ MeV is kinetic. One caution about the word mass: $m = \gamma m_0$ does not restore Newton’s second law in general. It gives the right answer for a force applied perpendicular to the velocity, where the speed and hence $\gamma$ do not change, but along the velocity the momentum $\gamma m_0 v$ carries the $\gamma$ dependence too and the effective inertia is $\gamma^3m_0$ , which for this particle is $3.5\times10^5$ times $m_0$ .