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Vectors and Scalars — Question 9

Section 3-1 · Physics pages 328–329

Question 9 Compatible uncertainty regions do not prove equal vectors Two instruments observe what is claimed to be the same nonzero displacement in one plane. Both report magnitude m and direction θ, where θ is measured counterclockwise from laboratory east. Their hard bounds are: Instrument I allows m ∈[4.8, 5.2] m and θ ∈[28◦, 32◦]. Instrument II allows m ∈[5.1, 5.5] m and θ ∈[31◦, 35◦]. Every pair inside each instrument’s magnitude–angle rectangle is allowed. These are hard bounds with no probability distribution. None of the angle intervals crosses the 0◦/360◦boundary. m (m) θ I II 4.8 5.1 5.2 5.5 28◦ 31◦ 32◦ 35◦ Hard-bound rectangles in magnitude–angle space. Every pair inside each rectangle is allowed. (a) Find the full set of magnitude–direction pairs consistent with both reports. Exhibit one possible common displacement vector. (b) Do the overlapping bounds prove the observations were of equal vectors? Distinguish compatibility of a common vector from equality established independently. (c) In a separate case, replace II’s direction interval by [40◦, 42◦], keeping all magnitude intervals unchanged. Could the reports still describe one common vector within their bounds? (d) Would changing both instruments to the same new angle-zero direction repair that second case? Why is taking the average of the reported central values not a substitute for checking the hard bounds?
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Question 9 - Solution Compatible uncertainty regions do not prove equal vectors m (m) θ I II 4.8 5.1 5.2 5.5 28◦ 31◦ 32◦ 35◦ (a) Intersect both kinds of constraint together. [4.8, 5.2] ∩[5.1, 5.5] = [5.1, 5.2], [28, 32] ∩[31, 35] = [31, 32] nonempty The complete common set is the Cartesian product of the intersected magnitude and angular intervals. For example, 5.15 m at 31.5◦is feasible; the common vector is not unique. (b) Compatibility is an existence statement. (m, θ)I = (4.9 m, 29◦), (m, θ)II = (5.4 m, 34◦) are distinct allowed vectors Overlap establishes that a common vector is possible, not that both instruments observed the same event. The two displayed distinct choices satisfy their separate bounds. (c) Equal or overlapping magnitudes alone are insufficient. [28◦, 32◦] ∩[40◦, 42◦] = ∅⇒no common vector Disjoint direction intervals rule out a common vector even though the magnitude intervals overlap. At least one bound or shared-measurement assumption must fail. (d) A common relabeling preserves disagreement. (SI −ϕ) ∩(SII −ϕ) = (SI ∩SII) −ϕ; (30◦+ 41◦)/2 = 35.5◦ A common angular relabeling preserves intersections, with angles treated modulo 360◦. In the second case, averaging interval centers gives an angle outside both bounds. No probability model licenses replacing a hard-bound test with averaging.

Question 9

Compatible uncertainty regions do not prove equal vectors

Two instruments observe what is claimed to be the same nonzero displacement in one plane. Both report magnitude mm and direction θ\theta, where θ\theta is measured counterclockwise from laboratory east. Their hard bounds are: Instrument I allows m∈[4.8,5.2]m\in[4.8,5.2] m and θ∈[28∘,32∘]\theta\in[28^\circ,32^\circ]. Instrument II allows m∈[5.1,5.5]m\in[5.1,5.5] m and θ∈[31∘,35∘]\theta\in[31^\circ,35^\circ]. Every pair inside each instrument’s magnitude–angle rectangle is allowed. These are hard bounds with no probability distribution. None of the angle intervals crosses the 0∘/360∘0^\circ/360^\circ boundary.

Diagram shown in the original SVG worksheet.

  1. Find the full set of magnitude–direction pairs consistent with both reports. Exhibit one possible common displacement vector.

  2. Do the overlapping bounds prove the observations were of equal vectors? Distinguish compatibility of a common vector from equality established independently.

  3. In a separate case, replace II’s direction interval by [40∘,42∘][40^\circ,42^\circ], keeping all magnitude intervals unchanged. Could the reports still describe one common vector within their bounds?

  4. Would changing both instruments to the same new angle-zero direction repair that second case? Why is taking the average of the reported central values not a substitute for checking the hard bounds?

Question 9 - Solution

Compatible uncertainty regions do not prove equal vectors

Diagram shown in the original SVG worksheet.

(a) Intersect both kinds of constraint together.

[4.8,5.2]∩[5.1,5.5]=[5.1,5.2],[28,32]∩[31,35]=[31,32]nonempty[4.8,5.2]\cap[5.1,5.5]=[5.1,5.2],\quad[28,32]\cap[31,35]=[31,32]\quad\boxed{\text{nonempty}} The complete common set is the Cartesian product of the intersected magnitude and angular intervals. For example, 5.15 m at 31.5∘31.5^\circ is feasible; the common vector is not unique.

(b) Compatibility is an existence statement.

(m,θ)I=(4.9m,29∘),(m,θ)II=(5.4m,34∘)are distinct allowed vectors(m,\theta)_I=(4.9\,\mathrm m,29^\circ),\quad(m,\theta)_{II}=(5.4\,\mathrm m,34^\circ)\quad\text{are distinct allowed vectors} Overlap establishes that a common vector is possible, not that both instruments observed the same event. The two displayed distinct choices satisfy their separate bounds.

(c) Equal or overlapping magnitudes alone are insufficient.

[28∘,32∘]∩[40∘,42∘]=⌀⇒no common vector[28^\circ,32^\circ]\cap[40^\circ,42^\circ]=\varnothing\Rightarrow\text{no common vector} Disjoint direction intervals rule out a common vector even though the magnitude intervals overlap. At least one bound or shared-measurement assumption must fail.

(d) A common relabeling preserves disagreement.

(SI−ϕ)∩(SII−ϕ)=(SI∩SII)−ϕ;(30∘+41∘)/2=35.5∘(S_I-\phi)\cap(S_{II}-\phi)=(S_I\cap S_{II})-\phi;\quad(30^\circ+41^\circ)/2=35.5^\circ A common angular relabeling preserves intersections, with angles treated modulo 360∘360^\circ. In the second case, averaging interval centers gives an angle outside both bounds. No probability model licenses replacing a hard-bound test with averaging.

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