 Solution

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

y = 0

, both fractions equal

x = 0.0402

. Reading the original plot near this value gives representative momentum densities

U = x u_{v} ≃ 0.42, D = x d_{v} ≃ 0.24, 0.05 x Σ ≃ 0.08.

These are approximate graphical readings, not a precision PDF fit. Undo the sea-curve scale and divide by eight flavours/antiflavours:

S_{x} = x S = \frac{x Σ}{8} ≃ \frac{0.08}{0.05 \times 8} = 0.20.

All products have the same

1/ x^{2}

factor, which cancels in the fractions. The three relative weights are

w_{VV} = U D = 0.1008, w_{V S} = S_{x} (U + D) = 0.1320, w_{SS} = 2 S_{x}^{2} = 0.0800.

Their sum is

0.3128

f_{VV} ≃ 32%, f_{V S} ≃ 42%, f_{SS} ≃ 26%.

Reading the central curves more finely changes these percentages by about a percentage point, well within what the plot supports. Omitting its factor

0.05

would substantially underestimate the sea contribution.

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