Irodov 1.371: The Excess Factor Is Just Gamma – Amit & Deepali Sinha Foundation

Problem Irodov 1.371

A particle of rest mass $m_0$ moves in a straight line with speed $v$ . Its relativistic momentum at that speed is $\eta = 2$ times as large as the Newtonian momentum $m_0v$ that the same particle moving at the same speed would be assigned in classical mechanics. Find the speed $v$ .

Given

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

Find

The speed

v

at which the two momenta stand in the ratio

\eta

Solution

Step 1The ratio of the two momenta

Both momenta are built from the same rest mass and the same speed, so writing them side by side and dividing shows at once what the ratio can depend on. With $\beta = v/c$ , the relativistic momentum and the Newtonian one are

p = \frac{m_0v}{\sqrt{1-\beta^2}}, \qquad p_{\text{N}} = m_0v .

Dividing the first by the second cancels both $m_0$ and $v$ :

\begin{align*}\frac{p}{p_{\text{N}}} &= \frac{m_0v}{m_0v\sqrt{1-\beta^2}} \\ &= \frac{1}{\sqrt{1-\beta^2}}\end{align*}

which is the Lorentz factor $\gamma$ and nothing else. The problem’s condition is therefore a condition on $\gamma$ alone.

\frac{1}{\sqrt{1-\beta^2}} = \eta

(1)

Step 2Solve the condition for the speed

Equation (1) is inverted by squaring both sides and isolating $\beta^2$ .

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

The positive root of that fraction is $\beta$ , and the speed is $c$ times it, so the whole of the algebra is now behind us and only the given excess factor is left to put in.

Step 3Put the number in

The root of the numerator and the reciprocal of $\eta$ are the two pieces (2) leaves to be evaluated, and at $\eta = 2$ both are elementary.

\sqrt{\eta^2-1} = \sqrt{3}, \qquad \frac{c}{\eta} = \frac{c}{2}

Multiplying the two and evaluating with $c = 3.00\times10^8$ m/s gives the speed.

v = \frac{c}{\eta}\sqrt{\eta^2-1} = \frac{\sqrt{3}}{2}\,c = 2.6\times10^8\ \text{m/s}

(3)

Momentum against speed, in units of m_0c. The gold curve is the relativistic γβ, the royal one the Newtonian β. At the marked abscissa the gold ordinate is 1.73 and the royal one 0.87, exactly twice as high. — Momentum against speed, in units of $m_0c$ . The gold curve is the relativistic $\gamma\beta$ , the royal one the Newtonian $\beta$ . At the marked abscissa the gold ordinate is $1.73$ and the royal one $0.87$ , exactly twice as high.

Check

Run the answer back through the condition. At $\beta = \sqrt3/2$ the square is $\beta^2 = 3/4$ , so $1-\beta^2 = 1/4$ and its square root is $1/2$ ; the reciprocal is $2$ , which is the required $\eta$ . The formula also behaves at both ends of its range: $\eta \to 1$ gives $v \to 0$ , where relativistic and Newtonian momenta agree, and $\eta \to \infty$ gives $v \to c$ from below, so any excess factor whatever is reachable and none of them needs a superluminal speed.

Remark

The excess factor is exactly the Lorentz factor $\gamma$ measured in the frame in which the particle moves at $v$ , which pins the same speed to several other statements about the particle in that frame. Its total energy $\gamma m_0c^2$ is $\eta$ times its rest energy, so at $\eta = 2$ the kinetic energy $(\gamma-1)m_0c^2$ has just risen to equal the rest energy $m_0c^2$ ; and a clock carried along with the particle, timed against the clocks of that same frame, runs slow by the same factor of two. One number, $\gamma = 2$ , fixes all three, and $v = \sqrt3\,c/2$ is the speed at which it happens.