Boosted Angle Formula

Claim \[ \Delta R \sim \frac{2m}{p_T} \]

Proof: \[ E^2 = p^2 + m^2 \] \[ m^2 = (E_1^2 + E_2^2)^2 - (\vec{p}_1^2 + \vec{p}_1^2 )^2 \] Assume particles 1 and 2 are massless: \[ = (p_1^2 + p_2^2 + 2 p_1 p_2) - (|p_1|^2 + |p_2|^2 + 2|p_1||p_2|\cos(\Delta\phi)) \] \[ = 2|p_1||p_2|(1- \cos(\Delta\phi) \] \[ = 2|p_1||p_2|(1- (1 - (\frac{\Delta R}{2})^2) \] \[ = |p_1||p_2|(\Delta R)^2 \]

Assume \[ |p_1| \sim |p_2| \sim p_C/2\] Then \[\Delta R^2 \sim \frac{4 m^2}{p_T^2} \] or \[\Delta R \sim \frac{2 m}{p_T} \]