Irodov 1.386: The Price of a Stationary Target – Amit & Deepali Sinha Foundation

Problem Irodov 1.386

Two protons, each of rest mass $m_0$ , travel toward each other along one line, each carrying kinetic energy $T = 25.0$ GeV; their total kinetic energy is $2T$ . Separately, a single proton of kinetic energy $T'$ strikes a proton of the same rest mass that is at rest. Find the value of $T'$ at which the combined kinetic energy of the two protons in the frame of their center of inertia is equal to that total $2T$ of the first arrangement. Take the proton rest energy to be $m_0c^2 = 938$ MeV.

The two arrangements, drawn to one scale. On the left, two protons approach with equal kinetic energies T; on the right, one proton of kinetic energy T' runs into a proton at rest. All four protons have rest mass m_0. — The two arrangements, drawn to one scale. On the left, two protons approach with equal kinetic energies $T$ ; on the right, one proton of kinetic energy $T'$ runs into a proton at rest. All four protons have rest mass $m_0$ .

Given

$m_0c^2 = 938$ MeV
each beam $T = 25.0$ GeV
target at rest

Find

the incident kinetic energy

T'

Solution

Step 1What the head-on arrangement makes available

The two approaching protons have equal rest masses and equal kinetic energies, so their momenta are equal in magnitude and opposite in direction and the total momentum of the pair is already zero. The frame in which the two are drawn is therefore itself the center of inertia frame, and the combined kinetic energy there is simply the sum of the two individual ones:

\widetilde{T}_{\text{goal}} = 2T = 50.0\ \text{GeV}

(1)

Step 2What a stationary target makes available

For the second arrangement the center of inertia frame is not the laboratory frame, so the combined kinetic energy there has to be reached through the invariant $E^2-p^2c^2$ . In the laboratory the two rest energies and the incident kinetic energy add, and only the incident proton carries momentum:

E = T'+2m_0c^2, \qquad p^2c^2 = \left(T'+m_0c^2\right)^2-\left(m_0c^2\right)^2 = T'\left(T'+2m_0c^2\right) .

In the center of inertia frame the total momentum vanishes, so the invariant is the square of the total energy $\widetilde{E}$ there. Evaluating it in both frames and subtracting,

\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}

The total energy in that frame is the combined kinetic energy plus the two rest energies, $\widetilde{E} = \widetilde{T}+2m_0c^2$ , so the last line is what the stationary target delivers:

\left(\widetilde{T}+2m_0c^2\right)^2 = 2m_0c^2\left(T'+2m_0c^2\right)

(2)

Step 3Match the two and solve for the incident energy

Setting $\widetilde{T}$ in (2) equal to the target value (1) makes the left-hand side a perfect square in $T+m_0c^2$ , and the two terms in $m_0^2c^4$ then cancel:

\begin{aligned} \left(2T+2m_0c^2\right)^2 &= 2m_0c^2\left(T'+2m_0c^2\right) \\[2pt] 4\left(T+m_0c^2\right)^2 &= 2m_0c^2T'+4m_0^2c^4 \\[2pt] 4T^2+8Tm_0c^2+4m_0^2c^4 &= 2m_0c^2T'+4m_0^2c^4 \\[2pt] 4T^2+8Tm_0c^2 &= 2m_0c^2T' \end{aligned}

Divide by $2m_0c^2$ and take the common factor $2T$ out of the numerator. With $T = 25.0$ GeV and $m_0c^2 = 0.938$ GeV, the bracket is $26.876$ GeV:

T' = \dfrac{2T\left(T+2m_0c^2\right)}{m_0c^2} = 1.43\times10^{3}\ \text{GeV}

(3)

The curve : the incident energy T' a stationary target demands, against the per-beam energy T it must match, over 0 <= T <= 25.0 GeV at m_0c^2 = 0.938 GeV. The marked point is T' = 1.43·10^3 GeV at T = 25.0 GeV. — The curve (3): the incident energy $T'$ a stationary target demands, against the per-beam energy $T$ it must match, over $0 \le T \le 25.0$ GeV at $m_0c^2 = 0.938$ GeV. The marked point is $T' = 1.43\times10^3$ GeV at $T = 25.0$ GeV.

Check

Feed (3) back into (2) and see whether $50.0$ GeV comes out. With $T' = 1432.6$ GeV and $2m_0c^2 = 1.876$ GeV,

\widetilde{T}+2m_0c^2 = \sqrt{1.876\times\left(1432.6+1.876\right)} = \sqrt{2691.1} = 51.88\ \text{GeV},

so $\widetilde{T} = 51.88-1.88 = 50.0$ GeV, which is $2T$ exactly. The non-relativistic limit is the second test: for $T\ll m_0c^2$ (3) reduces to $T' \approx 4T$ , and that is right for slow protons, where only half the incident kinetic energy is available in the center of mass frame, so $T'/2 = 2T$ .

Remark

The formula is quadratic in $T$ , and at $25$ GeV the quadratic term is doing almost all the work: $2T^2/m_0c^2$ alone is $1333$ GeV of the $1433$ GeV total. That is the arithmetic behind every colliding beam machine built since the 1960s. Read the other way round, a fixed target machine of energy $T'$ is worth only about $\sqrt{m_0c^2T'/2}$ per beam in a collider, so the 1.43 TeV demanded here buys what two 25 GeV beams give for free. The underlying reason is the one named for the head-on beams: in the laboratory frame a fixed target experiment must conserve a momentum that is not zero, and the kinetic energy tied up in the center of inertia’s own motion, $\left(\gamma-1\right)$ times the system’s invariant rest energy, is unavailable for anything at all. The scaling has a limit worth stating: the square law holds for $T$ well above $m_0c^2$ , and near $T = 0$ the honest statement is the linear one, $T'\to4T$ .