Irodov 1.387: When the Other Two Move Together – Amit & Deepali Sinha Foundation

Problem Irodov 1.387

A particle of rest mass $m_0$ sits at rest and breaks up into three particles, of rest masses $m_1$ , $m_2$ and $m_3$ . Nothing fixes the directions in which the three fly apart or how the released energy is shared between them, so the total energy carried by the first particle differs from one disintegration to another. Working in the frame in which the parent particle was at rest, find the largest total energy $E_{1\,\mathrm{max}}$ that the particle of rest mass $m_1$ can have, in terms of $m_0$ , $m_1$ , $m_2$ , $m_3$ and $c$ .

The disintegration in the parent's rest frame, drawn to one scale. Before, one particle of rest mass m_0 sits at rest; after, three particles leave with momenta p_1, p_2 and p_3 summing to zero, shown at equal magnitudes 120 degrees apart. — The disintegration in the parent’s rest frame, drawn to one scale. Before, one particle of rest mass $m_0$ sits at rest; after, three particles leave with momenta $\vect p_1$ , $\vect p_2$ and $\vect p_3$ summing to zero, shown at equal magnitudes $120$ degrees apart.

Given

parent rest mass $m_0$ , at rest
fragments $m_1$ , $m_2$ , $m_3$
momenta sum to zero

Find

the greatest possible total energy

E_{1\,\mathrm{max}}

of the fragment

m_1

Solution

Step 1Treat the other two fragments as one system

Whatever particle 1 does, the other two must between them balance its momentum and make up the rest of the energy, so it costs nothing to treat them as a single system and give that system one total energy $E_{23}$ and one total momentum $\vect p_{23}$ . In the parent’s rest frame the two conservation laws read

E_1+E_{23} = m_0c^2, \qquad \vect p_1+\vect p_{23} = \vect 0 ,

so the two momenta have one common magnitude, which we call $p$ . Being a system, the pair also has an invariant mass $M$ , defined exactly as any particle’s is, through the combination that is the same in every frame:

E_{23}^2-p^2c^2 = M^2c^4

(1)

Step 2Express the first particle’s energy through that mass

Particle 1 obeys (1) with its own rest mass. Writing both, with $E_{23}$ replaced from the energy balance,

E_1^2-p^2c^2 = m_1^2c^4 , \qquad \left(m_0c^2-E_1\right)^2-p^2c^2 = M^2c^4 .

Subtract the first from the second. The unknown $p$ leaves with it, and so does $E_1^2$ :

\begin{aligned} \left(m_0c^2-E_1\right)^2-E_1^2 &= M^2c^4-m_1^2c^4 \\[2pt] m_0^2c^4-2m_0c^2E_1+E_1^2-E_1^2 &= M^2c^4-m_1^2c^4 \\[2pt] m_0^2c^4+m_1^2c^4-M^2c^4 &= 2m_0c^2E_1 \end{aligned}

Dividing by $2m_0c^2$ leaves the first particle’s energy as a function of one variable only:

E_1 = \frac{m_0^2+m_1^2-M^2}{2m_0}\,c^2

(2)

Step 3The smallest the pair’s invariant mass can be

Since (2) decreases as $M$ grows, the largest $E_1$ goes with the smallest $M$ . Write out the pair’s invariant mass from the two members’ energies and momenta, using the single-particle relation on each:

\begin{aligned} M^2c^4 &= \left(E_2+E_3\right)^2 -\left(\vect p_2+\vect p_3\right)^2c^2 \\[2pt] &= E_2^2-p_2^2c^2+E_3^2-p_3^2c^2 +2\left(E_2E_3-\vect p_2\cdot\vect p_3\,c^2\right) \\[2pt] &= m_2^2c^4+m_3^2c^4 +2\left(E_2E_3-\vect p_2\cdot\vect p_3\,c^2\right) \end{aligned}

The dot product is at most the product of the magnitudes, so the bracket is smallest when the two momenta are parallel, and what remains is bounded by an identity: for any two particles,

\begin{aligned} \left(E_2E_3-m_2m_3c^4\right)^2-\left(p_2p_3c^2\right)^2 &= \left(E_2E_3-m_2m_3c^4\right)^2 -\left(E_2^2-m_2^2c^4\right)\left(E_3^2-m_3^2c^4\right) \\[2pt] &= c^4\left(m_3E_2-m_2E_3\right)^2 , \end{aligned}

which is never negative. Both of the quantities squared on the left are themselves non-negative, since $E_2\ge m_2c^2$ and $E_3\ge m_3c^2$ , so the difference of their squares gives $E_2E_3-m_2m_3c^4 \ge p_2p_3c^2$ . The bracket therefore never falls below $m_2m_3c^4$ , and the pair’s invariant mass obeys

M^2c^4 \ge m_2^2c^4+m_3^2c^4+2m_2m_3c^4 = \left(m_2+m_3\right)^2c^4 .

Equality needs both conditions at once: the momenta parallel, and $m_3E_2 = m_2E_3$ , which together say the two fragments have the same velocity.

M_{\mathrm{min}} = m_2+m_3

(3)

Step 4Put the minimum mass into the energy

The configuration that maximizes E_1: the two remaining fragments travel together as one body of rest mass m_2+m_3. The single arrow above the touching pair is their combined momentum, equal and opposite to p_1. — The configuration that maximizes $E_1$ : the two remaining fragments travel together as one body of rest mass $m_2+m_3$ . The single arrow above the touching pair is their combined momentum, equal and opposite to $\vect p_1$ .

The maximum of (2) is reached at that minimum, with the two remaining fragments traveling as one body, as drawn above. Substituting (3):

E_{1\,\mathrm{max}} = \frac{m_0^2+m_1^2-\left(m_2+m_3\right)^2}{2m_0}\,c^2

(4)

Check

Test (4) at the threshold of the decay, where the parent has only just enough mass to make the three fragments, $m_0 = m_1+m_2+m_3$ . Then $m_2+m_3 = m_0-m_1$ , and the numerator is $m_0^2+m_1^2-\left(m_0-m_1\right)^2 = 2m_0m_1$ , so $E_{1\,\mathrm{max}} = m_1c^2$ : at threshold every fragment must come out at rest, and the most the first one can have is its own rest energy, with nothing over. That is exactly right, and it also confirms the sign convention, since any heavier parent gives $E_{1\,\mathrm{max}} > m_1c^2$ . Dimensionally the numerator is a mass squared and the denominator a mass, so the quotient times $c^2$ is an energy.

Remark

Reading (2) the other way round is just as useful. The pair’s invariant mass $M$ can be anywhere from $m_2+m_3$ up to $m_0-m_1$ , the latter being the case in which particle 1 is left at rest, and $E_1$ falls steadily from (4) down to $m_1c^2$ as $M$ crosses that range. So in the parent’s rest frame a three body decay gives particle 1 a continuous spectrum of energies, whereas a two body decay fixes $M$ at the single remaining rest mass and gives one sharp line. That contrast is how the neutrino was first inferred: a continuous electron spectrum in beta decay means a third body is leaving with it. Equation (4) is also the familiar two body decay energy with $m_2+m_3$ standing in for one daughter’s mass, which is the sense in which the extremal configuration is a two body decay in disguise.