The Quark Model — Question 4

Section 44-7 · Physics pages 7418–7419

Question 4 A sea pair changes abundance without changing valence quantum numbers Begin with proton valence content uud. Add two u¯u pairs and one s¯s pair. Ignore gluons for this counting exercise. uud u¯u u¯u s¯s Schematic; labels, not drawing lengths, specify the data. (a) Find total numbers of quarks and antiquarks, net up flavor and net strangeness. (b) Calculate B and Q of the entire component and compare with the original proton.
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Question 4 - Solution A sea pair changes abundance without changing valence quantum numbers uud u¯u u¯u s¯s (a) Method and calculation Nq = 3 + 3 = 6, N¯q = 3, Nu −N¯u = 4 −2 = 2 . S = −(Ns −N¯s) = −(1 −1) = 0 . (b) Interpretation and further calculation B = (6 −3)/3 = 1 . Q/e = (2/3)(4 −2) −(1/3)(1) −(1/3)(1 −1) = 1 . Physical check Only flavor differences enter the additive charges. The number of constituents in a quantum field component need not remain three.

Question 4

A sea pair changes abundance without changing valence quantum numbers

Begin with proton valence content uud. Add two uu‾u\bar u pairs and one ss‾s\bar s pair. Ignore gluons for this counting exercise.

Diagram shown in the original SVG worksheet.

  1. Find total numbers of quarks and antiquarks, net up flavor and net strangeness.

  2. Calculate B and Q of the entire component and compare with the original proton.

Question 4 - Solution

A sea pair changes abundance without changing valence quantum numbers

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Nq=3+3=6,Nq‾=3,Nu−Nu‾=4−2=2.\boxed{N_q=3+3=6,\quad N_{\bar q}=3,\quad N_u-N_{\bar u}=4-2=2}.S=−(Ns−Ns‾)=−(1−1)=0.\boxed{S=-(N_s-N_{\bar s})=-(1-1)=0}.

(b) Interpretation and further calculation

B=(6−3)/3=1.\boxed{B=(6-3)/3=1}.Q/e=(2/3)(4−2)−(1/3)(1)−(1/3)(1−1)=1.\boxed{Q/e=(2/3)(4-2)-(1/3)(1)-(1/3)(1-1)=1}.

Physical check

Only flavor differences enter the additive charges. The number of constituents in a quantum field component need not remain three.