Irodov 1.384: The Frame Where Momentum Vanishes – Amit & Deepali Sinha Foundation

Problem Irodov 1.384

In the laboratory frame a neutron of rest mass $m_0$ carries kinetic energy $T = 2m_0c^2$ and runs into a second neutron of the same rest mass that is at rest there. Take the neutron rest energy to be $m_0c^2 = 940$ MeV. Consider also the center of inertia frame of the two neutrons, the frame in which their total momentum is zero. Determine:

the combined kinetic energy $\widetilde{T}$ of the two neutrons in the center of inertia frame, and the momentum $\widetilde{p}$ of each neutron in that frame
the speed $V$ of the center of inertia relative to the laboratory frame.

For any system of particles, the quantity $E^2 - p^2c^2$ built from the total energy $E$ and the total momentum $p$ has the same value in every inertial reference frame.

The same pair of neutrons in the two frames compared, drawn to one scale. In the laboratory panel one neutron carries kinetic energy T toward a stationary partner; in the center of inertia panel the two approach with equal and opposite momenta p. — The same pair of neutrons in the two frames compared, drawn to one scale. In the laboratory panel one neutron carries kinetic energy $T$ toward a stationary partner; in the center of inertia panel the two approach with equal and opposite momenta $\widetilde{p}$ .

Given

rest mass $m_0$ each
$m_0c^2 = 940$ MeV
incident $T = 2m_0c^2$
target at rest

Find

(a)

\widetilde{T}

and

\widetilde{p}

in the center of inertia frame, (b) the speed

V

of that frame

Solution

Step 1The system’s total energy and momentum in the laboratory

Everything that follows is built from two laboratory-frame numbers: the total energy of the pair and its total momentum. The incident neutron’s total energy is its rest energy plus its kinetic energy, and the stationary one contributes its rest energy alone.

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

(1)

Only the incident neutron carries momentum, so the system’s momentum is its momentum, and that follows from the single-particle relation $E_1^2 = p^2c^2+m_0^2c^4$ applied to it:

\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 \end{aligned}

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

(2)

The energy-momentum triangle of the two-neutron system, drawn at T = 2m_0c^2. The hypotenuse is the laboratory energy E, the legs are pc and the center of inertia energy E; the angle θ at the foot satisfies θ = pc/E. — The energy-momentum triangle of the two-neutron system, drawn at $T = 2m_0c^2$ . The hypotenuse is the laboratory energy $E$ , the legs are $pc$ and the center of inertia energy $\widetilde{E}$ ; the angle $\theta$ at the foot satisfies $\cos\theta = pc/E$ .

Step 2The combined kinetic energy in the center of inertia frame

The center of inertia frame is defined by the total momentum vanishing there, so the invariant evaluated in that frame is simply the square of the total energy $\widetilde{E}$ measured there. Equating the two evaluations and substituting (1) and (2),

\begin{aligned} \widetilde{E}^2 &= 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}

In that frame the two neutrons are still two neutrons, so their total energy is their combined kinetic energy plus the two rest energies:

\widetilde{E} = \widetilde{T}+2m_0c^2 = \sqrt{2m_0c^2\left(T+2m_0c^2\right)} .

Take the factor $2m_0c^2$ out of the root to display the result as that rest energy times a pure number:

\widetilde{T}+2m_0c^2 = 2m_0c^2\sqrt{\frac{T+2m_0c^2}{2m_0c^2}} = 2m_0c^2\sqrt{1+\frac{T}{2m_0c^2}} .

Subtracting the two rest energies, and putting $T = 2m_0c^2$ with $m_0c^2 = 940$ MeV, so that the root is $\sqrt{2}$ :

\widetilde{T} = 2m_0c^2\left(\sqrt{1+\frac{T}{2m_0c^2}}-1\right) = 779\ \text{MeV}

(3)

Step 3The momentum of each neutron in that frame

In the center of inertia frame the two neutrons are identical and their momenta cancel, so they share the total energy equally and each carries half of it. Applying the single-particle relation to one of them,

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

Take the square root. One power of $c$ comes out of the root and cancels the $c$ on the left:

\widetilde{p}c = c\sqrt{\frac{m_0T}{2}}

With $T = 2m_0c^2$ the product under the root is $\left(m_0c\right)^2$ , so the momentum in that frame is one rest energy divided by $c$ :

\widetilde{p} = \sqrt{\dfrac{m_0T}{2}} = 940\ \text{MeV}/c

(4)

Step 4The speed of the center of inertia

The center of inertia frame moves relative to the laboratory at exactly the velocity that carries the system’s total momentum, and for any relativistic system total momentum, total energy and that velocity are tied by

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

so $V/c$ is the ratio $pc/E$ , which the triangle above draws as $\cos\theta$ . Substituting (1) and (2) and collecting both factors under one root,

\begin{aligned} \frac{V}{c} &= \frac{pc}{E} = \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, and putting $T = 2m_0c^2$ , so that the ratio under the root is $1/2$ :

V = c\sqrt{\dfrac{T}{T+2m_0c^2}} = 2.12\times10^{8}\ \text{m/s}

(5)

Check

The three answers must be consistent with one another in the center of inertia frame, where each neutron has energy $\sqrt{\widetilde{p}^{\,2}c^2+m_0^2c^4}$ . With $\widetilde{p}c = 940$ MeV and $m_0c^2 = 940$ MeV that energy is $940\sqrt{2} = 1329$ MeV each, so the two together carry $2659$ MeV, of which $2\times940 = 1880$ MeV is rest energy and $779$ MeV is kinetic, reproducing (3). The same total is $\sqrt{E^2-p^2c^2}$ from the laboratory numbers, since there $E = 3760$ MeV and $pc = \sqrt{1880\times3760} = 2659$ MeV. The low-energy limit also behaves: expanding (3) for $T\ll m_0c^2$ gives $\widetilde{T}\approx T/2$ , which is the Newtonian result that in a collision of equal masses only half the incident kinetic energy is available in the center of mass frame.

Remark

The quantity $\widetilde{T}$ is what a collision has to spend: it is the energy available in the center of inertia frame for making new particles, since the motion of that frame as a whole cannot be spent on anything. Its dependence on the incident energy is the discouraging part. For $T$ large compared with $m_0c^2$ (3) reduces to $\widetilde{T}\approx\sqrt{2m_0c^2T}$ , so quadrupling the accelerator’s energy only doubles what is available, while a head-on arrangement, in which the laboratory frame is already the center of inertia frame, would deliver the whole of it. That square root is the entire argument for building colliding beam machines rather than fixed target ones. Note also where the rest masses went: $E$ and $p$ are additive over the two neutrons and the rest masses are not, so the system’s own invariant mass $\widetilde{E}/c^2$ exceeds $2m_0$ by $\widetilde{T}/c^2$ , and that surplus is available even though each neutron’s own rest mass is untouched.