Irodov 1.370: Half a Percent Short of Light – Amit & Deepali Sinha Foundation

Problem Irodov 1.370

A proton travels in a straight line with momentum $p = 10.0$ GeV $/c$ , that is, its momentum is such that $pc = 10.0$ GeV, where $c$ is the speed of light. Find by what percentage its speed $v$ falls short of $c$ , that is, evaluate the fraction $(c-v)/c$ and express it in percent.

The energy-momentum right triangle: the legs are pc and the rest energy m_0c^2, the hypotenuse the total energy E. The speed is read off the same triangle, v/c being the cosine of θ, adjacent leg over hypotenuse. — The energy-momentum right triangle: the legs are $pc$ and the rest energy $m_0c^2$ , the hypotenuse the total energy $E$ . The speed is read off the same triangle, $v/c$ being the cosine of $\theta$ , adjacent leg over hypotenuse.

Given

$p = 10.0$ GeV $/c$
$m_pc^2 = 938$ MeV
$c = 3.00\times10^8$ m/s

Find

The shortfall

(c-v)/c

, in percent

Solution

Step 1Total energy from the given momentum

The momentum is given and the proton’s rest energy is a tabulated constant, so the quantity to fix first is the total energy. Momentum and total energy are tied together by one relation, obtained by eliminating the Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2}$ between $p = \gamma m_0v$ and $E = \gamma m_0c^2$ .

\begin{align*}E^2-p^2c^2 &= \gamma^2m_0^2c^4-\gamma^2m_0^2v^2c^2 \\ &= \gamma^2m_0^2c^4\left(1-\frac{v^2}{c^2}\right) \\ &= m_0^2c^4\end{align*}

the last step because $\gamma^2$ and $1-v^2/c^2$ are reciprocals. Solving for $E$ gives the hypotenuse of the triangle above.

E = \sqrt{(pc)^2+\left(m_0c^2\right)^2}

(1)

Step 2Speed as the ratio of momentum to energy

Dividing the momentum by the total energy cancels both $\gamma$ and $m_0$ , so the ratio is pure kinematics and gives the speed directly.

\frac{pc}{E} = \frac{\gamma m_0vc}{\gamma m_0c^2} = \frac{v}{c}

Substituting the total energy (1) and dividing numerator and denominator by $pc$ ,

\begin{align*}\frac{v}{c} &= \frac{pc}{\sqrt{(pc)^2+\left(m_0c^2\right)^2}} \\ &= \frac{1}{\sqrt{1+\left(m_0c^2/pc\right)^2}}\end{align*}

The ratio inside the root loses one power of $c$ from each side:

\frac{m_0c^2}{pc} = \frac{m_0c}{p}

so the speed depends on the momentum through that single dimensionless group.

\frac{v}{c} = \left[1+\left(\frac{m_0c}{p}\right)^2\right]^{-1/2}

(2)

The speed of against momentum in units of m_0c, approaching the asymptote v = c from below. The marked point is this problem's proton at p/m_0c = 10.7; its gap below the asymptote is the answer, too small to see here. — The speed of (2) against momentum in units of $m_0c$ , approaching the asymptote $v = c$ from below. The marked point is this problem’s proton at $p/m_0c = 10.7$ ; its gap below the asymptote is the answer, too small to see here.

Step 3The shortfall behind light

What the problem asks for is one minus the speed ratio (2). The numbers go in as energies, since $m_0c/p$ is the rest energy divided by $pc$ , both of which are known in MeV.

\frac{m_0c}{p} = \frac{m_pc^2}{pc} = \frac{938\ \text{MeV}}{10\,000\ \text{MeV}} = 0.0938

Squaring, adding one and taking the reciprocal square root,

\begin{align*}\left(\frac{m_0c}{p}\right)^2 &= 0.008798 \\ \left[1+0.008798\right]^{-1/2} &= \frac{1}{1.004390} \\ &= 0.995630\end{align*}

Subtracting this from unity gives the fractional shortfall.

\frac{c-v}{c} = 1-\left[1+\left(\frac{m_0c}{p}\right)^2\right]^{-1/2} = 0.44\%

(3)

Check

Because $(m_0c/p)^2 = 0.0088$ is small against unity, the binomial expansion $(1+x)^{-1/2} \approx 1-x/2$ applies and predicts a shortfall of $\frac{1}{2}(0.0088) = 0.0044$ , that is $0.44\%$ , agreeing with the exact evaluation above to the two figures quoted. The speed itself, $0.99563$ of $c$ , comes to $v = 2.99\times10^8$ m/s, slower than light, as any massive particle’s speed must be.

Remark

In the laboratory frame in which the momentum is $10.0$ GeV $/c$ , the Lorentz factor is the total energy over the rest energy, $\gamma = E/m_0c^2$ , which for $E = 10.04$ GeV comes to $10.7$ ; whenever $\gamma \gg 1$ the shortfall behind light is close to $1/(2\gamma^2)$ , which here is $0.0044$ as well. The useful reading is that the gap closes as the inverse square of the momentum: a proton of $100$ GeV $/c$ would sit not ten but a hundred times closer to $c$ , at about $0.0044\%$ . This is why accelerator physicists quote energy or momentum rather than speed once a beam is relativistic; the speed stops carrying information long before the momentum does.