 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2018/2/b/solution

At leading order, neglect the transverse momentum of the incoming partons and of the produced on-shell boson. In the hadron centre-of-mass frame,

p_{+} = \frac{s}{2} (x_{+}, 0, 0, x_{+}), p_{-} = \frac{s}{2} (x_{-}, 0, 0, - x_{-}) .

Then

m_{W}^{2} = (p_{+} + p_{-})^{2} = x_{+} x_{-} s

, and

y = \frac{1}{2} ln \frac{x _{+}}{x _{-}} .

Solving these product and ratio equations gives

x_{+} = τ e^{y}, x_{-} = τ e^{- y}, τ = \frac{80.4}{2000} = 0.0402.

Over

- 2 < y < 2

, each momentum fraction ranges between

0.00544 < x_{\pm} < 0.2970.

They are anticorrelated: increasing

y

increases

x_{+}

while decreasing

x_{-}

. Higher-order recoil changes the simple two-to-one kinematics.

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