10A (Dynamics and Relativity)

P = (P^{0}, P) = (γm c, γm u), γ = (1 - ∣ u ∣^{2} / c^{2})^{- 1/2} .

Thus

P^{0} = m c + \frac{m ∣ u ∣ ^{2}}{2 c} + \frac{3 m ∣ u ∣ ^{4}}{8 c ^{3}} + O (∣ u ∣^{6} / c^{5}) .

c

identifies the first two terms as the rest energy

m c^{2}

and the Newtonian kinetic energy

m ∣ u ∣^{2} /2

For the fixed-target reaction, the initial invariant is

(P_{1} + P_{2})^{2} = 2 m^{2} c^{2} (1 + γ) .

At threshold the four final particles are at rest in their centre-of-momentum frame, giving invariant

(4 m c)^{2}

. Hence

γ = 7

and the least laboratory speed is

u = c 1 - \frac{1}{49} = \frac{4 3}{7} c .

b

P = \sum_{i} Q_{i}

P \cdot P = M^{2} c^{2}

. Therefore

M = \frac{1}{c} (i \sum Q_{i}) \cdot (j \sum Q_{j}) .

In the parent rest frame,

M c^{2} = \sum_{i} E_{i}

E_{i} \geq m_{i} c^{2}

. Thus

M \geq i = 1 \sum N m_{i} .

B

. Two-body decay kinematics gives the massless-particle energies

E_{1} = \frac{( m _{A}^{2} - m _{B}^{2} ) c ^{2}}{2 m _{B}}, E_{2} = \frac{( m _{B}^{2} - m _{C}^{2} ) c ^{2}}{2 m _{B}} .

θ

is the angle between their momenta, then

(Q_{1} + Q_{2})^{2} = 2 Q_{1} \cdot Q_{2} = \frac{2 E _{1} E _{2}}{c ^{2}} (1 - cos θ) .

Since

0 \leq 1 - cos θ \leq 2

0 \leq (Q_{1} + Q_{2})^{2} \leq \frac{( m _{A}^{2} - m _{B}^{2} ) ( m _{B}^{2} - m _{C}^{2} ) c ^{2}}{m _{B}^{2}} .

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