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 Incoming links: Maximum laboratory deflection in relativistic elastic scattering

S^{'}

move at speed

v

in the positive

x

direction relative to

S

, with parallel axes and coincident origins at zero time. The inverse Lorentz transformation gives

d t = γ_{v} (d t^{'} + v d x^{'} / c^{2}), d x = γ_{v} (d x^{'} + v d t^{'}), d y = d y^{'}, d z = d z^{'} .

Here

γ_{v} = (1 - v^{2} / c^{2})^{- 1/2}

and

u^{'} = d x^{'} / d t^{'}

is the particle's velocity in

S^{'}

. Dividing spatial differentials by

d t

u_{x} = \frac{u _{x}^{'} + v}{1 + v u _{x}^{'} / c ^{2}}, u_{y} = \frac{u _{y}^{'}}{γ _{v} ( 1 + v u _{x}^{'} / c ^{2} )}, u_{z} = \frac{u _{z}^{'}}{γ _{v} ( 1 + v u _{x}^{'} / c ^{2} )} .

c = 1

during the algebra. Let the incident projectile have energy

E > m_{1}

p = E^{2} - m_{1}^{2}

. The target is initially at rest. The squared invariant mass, boost speed and Lorentz factor of the center-of-momentum frame are

s = m_{1}^{2} + m_{2}^{2} + 2 m_{2} E, V = \frac{p}{E + m _{2}}, Γ = \frac{E + m _{2}}{s} .

In that frame the projectile has

p_{*} = m_{2} p / s

and

E_{*} = (m_{1}^{2} + m_{2} E) / s

. An elastic collision changes the direction of its momentum, through an angle

φ

, while preserving these magnitudes. Transforming the outgoing four-momentum back to the laboratory gives

tan θ = \frac{s i n φ}{Γ ( c o s φ + B )}, B = \frac{V E _{*}}{p _{*}} = \frac{m _{1}^{2} + m _{2} E}{m _{2} ( E + m _{2} )} > 1.

The maximum occurs at

cos φ = - 1/ B

. Since

Γ^{2} (B^{2} - 1) = \frac{m _{1}^{2} - m _{2}^{2}}{m _{2}^{2}},

tan θ_{m a x} = \frac{m _{2}}{m _{1}^{2} - m _{2}^{2}}, sin θ \leq sin θ_{m a x} = \frac{m _{2}}{m _{1}} .

The printed strict inequality needs correction: equality is kinematically attainable. For example, in

c = 1

units take

m_{1} = 2

m_{2} = 1

, and incident

E = 3

. A projectile scattered through

3 0^{\circ}

with final energy

16/7

(35 /7, 15 /7)

. The target then has energy

12/7

(45 /7, - 15 /7)

; both outgoing particles satisfy their mass shells and conserve total four-momentum. This gives

sin θ = 1/2 = m_{2} / m_{1}

exactly. The inequality is strict only away from the maximizing direction.