Irodov 1.385: Sticking Together Creates New Rest Mass

Problem Irodov 1.385

In the laboratory frame a particle of rest mass $m_0$ , carrying kinetic energy $T$ , runs into a second particle of the same rest mass $m_0$ that is at rest there. The two stick together and move off as a single compound particle. Find the rest mass $M_0$ of that compound particle and the speed $V$ with which it travels in the laboratory frame, both in terms of $m_0$ , $T$ and the speed of light $c$ .

The two states, laboratory frame. Before, a particle of rest mass m_0 moves at v with kinetic energy T toward an identical one at rest; after, they have merged into one body of rest mass M_0 moving at V along the same line. — The two states, laboratory frame. Before, a particle of rest mass $m_0$ moves at $\vect v$ with kinetic energy $T$ toward an identical one at rest; after, they have merged into one body of rest mass $M_0$ moving at $\vect V$ along the same line.

Given

$m_0$ for each of the two colliding particles
incident kinetic energy $T$
the target at rest in the laboratory frame
the two merge into one body

Find

the rest mass

M_0

and the speed

V

of the compound particle

Solution

Step 1The energy and the momentum carried in

Nothing leaves the system, so the total energy and the total momentum of the pair before the collision are what the compound particle must carry away. The incident particle’s total energy is its rest energy plus its kinetic energy, the target contributes its rest energy alone, and the two add:

E = \left(T+m_0c^2\right)+m_0c^2 = T+2m_0c^2

(1)

The momentum comes from the energy-momentum relation $E_1^2 = p^2c^2+m_0^2c^4$ applied to the incident particle, since the target carries none:

\begin{aligned} p^2c^2 &= \left(T+m_0c^2\right)^2-\left(m_0c^2\right)^2 \\[2pt] &= T^2+2Tm_0c^2+m_0^2c^4-m_0^2c^4 \\[2pt] &= T^2+2Tm_0c^2 \end{aligned}

p^2c^2 = T\left(T+2m_0c^2\right)

(2)

The energy-momentum triangle of the compound particle, drawn for the representative case T = m_0c^2. The hypotenuse is the total energy E, the legs are pc and the rest energy M_0c^2, and the angle θ at the foot satisfies θ = pc/E. — The energy-momentum triangle of the compound particle, drawn for the representative case $T = m_0c^2$ . The hypotenuse is the total energy $E$ , the legs are $pc$ and the rest energy $M_0c^2$ , and the angle $\theta$ at the foot satisfies $\cos\theta = pc/E$ .

Step 2The rest mass from the energy-momentum invariant

The compound particle is one body carrying the whole of $E$ and the whole of $p$ , so its rest mass follows from the same energy-momentum relation, now read backwards: the combination $E^2-p^2c^2$ is the square of the rest energy, the quantity the triangle above puts on the vertical leg. Substituting (1) and (2) into it,

\begin{aligned} M_0^2c^4 &= E^2-p^2c^2 \\[2pt] &= \left(T+2m_0c^2\right)^2-T\left(T+2m_0c^2\right) \\[2pt] &= \left(T+2m_0c^2\right) \left[\left(T+2m_0c^2\right)-T\right] \\[2pt] &= 2m_0c^2\left(T+2m_0c^2\right) \end{aligned}

Take the square root of both sides and divide by $c^2$ . That leaves the rest mass with one power of $c$ still buried inside the root:

M_0 = \frac{\sqrt{2m_0c^2\left(T+2m_0c^2\right)}}{c^2}

Bringing that factor of $c$ out of the root cancels one of the two powers of $c$ in the denominator:

M_0 = \dfrac{\sqrt{2m_0\left(T+2m_0c^2\right)}}{c}

(3)

Step 3The velocity from the ratio of momentum to energy

For any relativistic body the velocity is not an independent quantity: energy, momentum and velocity are tied together by one relation,

\vect p = \frac{E\vect V}{c^2},

so the speed in units of $c$ is the ratio $pc/E$ , which the triangle above draws as the cosine of $\theta$ . Substituting (1) and (2),

\begin{aligned} \frac{V}{c} &= \frac{pc}{E} \\[2pt] &= \frac{\sqrt{T\left(T+2m_0c^2\right)}}{T+2m_0c^2} \\[2pt] &= \sqrt{\frac{T\left(T+2m_0c^2\right)} {\left(T+2m_0c^2\right)^2}} \end{aligned}

Canceling one factor of $T+2m_0c^2$ inside the root gives the speed:

V = c\sqrt{\dfrac{T}{T+2m_0c^2}}

(4)

Check

The two answers must be consistent with each other: the compound particle’s own energy, $\gamma M_0c^2$ at the speed just found, has to reproduce (1). From the boxed $V$ ,

1-\frac{V^2}{c^2} = 1-\frac{T}{T+2m_0c^2} = \frac{2m_0c^2}{T+2m_0c^2}, \qquad \gamma = \sqrt{\frac{T+2m_0c^2}{2m_0c^2}},

so that $\gamma M_0c^2 = \sqrt{2m_0c^2\left(T+2m_0c^2\right)}\cdot \sqrt{\left(T+2m_0c^2\right)/2m_0c^2} = T+2m_0c^2$ , which is $E$ exactly. The limit $T\to0$ also behaves: the collision then never happens, and the formulas return $M_0\to2m_0$ and $V\to0$ , two particles at rest side by side.

Remark

The rest mass grew, and the amount it grew by is the kinetic energy that disappeared. Since $M_0c^2 = \sqrt{2m_0c^2(T+2m_0c^2)}$ exceeds $2m_0c^2$ for every $T>0$ , the compound is strictly heavier than its ingredients, and the excess $(M_0-2m_0)c^2$ is energy that used to be kinetic and is now locked up as rest mass. It is not all of $T$ , because momentum conservation forbids the compound from being at rest in the laboratory frame: the balance $T = (M_0-2m_0)c^2+(\gamma-1)M_0c^2$ splits the incident kinetic energy between mass made and motion kept. At low energies the split is even: expanding $M_0c^2 = 2m_0c^2\sqrt{1+T/(2m_0c^2)}$ for $T\ll m_0c^2$ gives $(M_0-2m_0)c^2\approx T/2$ , which is precisely the Newtonian result that a perfectly inelastic collision between equal masses loses half the kinetic energy. Relativity says where that half went.