An Interlude — Question 9

Section 44-2 · Physics pages 7328–7329

Question 9 Charge-to-mass comparisons constrain symmetry only with an extra assumption A trap measures the magnitude ratio R = (|q¯p|/m¯p)/(|qp|/mp) = 1 + δ, with δ = (2.0 ± 3.0) × 10−9. For a small fractional charge difference a and mass difference b, write |q¯p| = |qp|(1 + a) and m¯p = mp(1 + b). |q|/m |¯q|/ ¯m ratio R Schematic; labels, not drawing lengths, specify the data. (a) Expand R to first order and identify the constrained combination. (b) Assuming a=0, infer b and its uncertainty; explain the limitation without that assumption.
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Question 9 - Solution Charge-to-mass comparisons constrain symmetry only with an extra assumption |q|/m |¯q|/ ¯m ratio R (a) Method and calculation R = 1 + a 1 + b ≃(1 + a)(1 −b) ≃1 + a −b. a −b = (2.0 ± 3.0) × 10−9 . (b) Interpretation and further calculation a = 0 ⇒b = (−2.0 ± 3.0) × 10−9 . (a, b) →(a + η, b + η) leaves the first-order observable unchanged. Physical check This hypothetical measurement is consistent with zero difference. Measuring a ratio does not separately establish equality of charge and mass.

Question 9

Charge-to-mass comparisons constrain symmetry only with an extra assumption

A trap measures the magnitude ratio R=(|qp‾|/mp‾)/(|qp|/mp)=1+δR=(|q_{\bar p}|/m_{\bar p})/(|q_p|/m_p)=1+\delta, with δ=(2.0±3.0)×10−9\delta=(2.0\pm3.0)\times10^{-9}. For a small fractional charge difference a and mass difference b, write |qp‾|=|qp|(1+a)|q_{\bar p}|=|q_p|(1+a) and mp‾=mp(1+b)m_{\bar p}=m_p(1+b).

Diagram shown in the original SVG worksheet.

  1. Expand R to first order and identify the constrained combination.

  2. Assuming a=0, infer b and its uncertainty; explain the limitation without that assumption.

Question 9 - Solution

Charge-to-mass comparisons constrain symmetry only with an extra assumption

Diagram shown in the original SVG worksheet.

(a) Method and calculation

R=1+a1+b≃(1+a)(1−b)≃1+a−b.R=\frac{1+a}{1+b}\simeq(1+a)(1-b)\simeq1+a-b.a−b=(2.0±3.0)×10−9.\boxed{a-b=(2.0\pm3.0)\times10^{-9}}.

(b) Interpretation and further calculation

a=0⇒b=(−2.0±3.0)×10−9.a=0\Rightarrow\boxed{b=(-2.0\pm3.0)\times10^{-9}}.(a,b)→(a+η,b+η)leaves the first-order observable unchanged.\boxed{(a,b)\to(a+\eta,b+\eta)}\quad\text{leaves the first-order observable unchanged}.

Physical check

This hypothetical measurement is consistent with zero difference. Measuring a ratio does not separately establish equality of charge and mass.