University Physics V · topic UPV-17.01
Classifying particles: Casimir labels and the working taxonomy
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Casimir labels and the particle taxonomy: predict a finite change
A free particle is an irreducible unitary representation of the Poincaré group, labelled by the eigenvalues of its two Casimir operators — mass from P² and spin from W² — with helicity replacing spin when the mass is zero. Use the supplied topic model Y_out = m/(2*pi*max(w,0.5)), where Y_out is PHASE CYCLES M/2πΓ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Width Γ, measured in MeV. Keep m (Mass m) = 775 MeV; s (Spin s) = 1 fixed. For this exercise use only 32.7687 MeV ≤ w ≤ 268.231 MeV on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at w = 84.5705 MeV and 117.535 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PHASE CYCLES M/2πΓ against w
Predictions from the supplied Casimir labels and the particle taxonomy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PHASE CYCLES M/2πΓ
Horizontal axis (x): w: Width Γ / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Width Γ / MeV | PHASE CYCLES M/2πΓ |
|---|---|
| 32.76875 | 3.764106989012974 |
| 91.63437499999999 | 1.3460568798141406 |
| 150.5 | 0.819568643828697 |
| 209.36562499999997 | 0.5891372134094072 |
| 268.23125 | 0.4598460503622113 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 84.5705 MeV, substitute into Y_out = m/(2*pi*max(w,0.5)): Y₀ = 1.45849.
- At w = 117.535 MeV, the same fixed-parameter model gives Y₁ = 1.04943.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.409058; the input change is Δw = 32.9647 MeV.
- Average sensitivity is ΔY_out/Δw = -0.0124089 1 per MeV. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 1.45849, Y₁ = 1.04943; ΔY_out = -0.409058; average sensitivity = -0.0124089 1 per MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Casimir labels and the particle taxonomy: infer an input with uncertainty
A free particle is an irreducible unitary representation of the Poincaré group, labelled by the eigenvalues of its two Casimir operators — mass from P² and spin from W² — with helicity replacing spin when the mass is zero. Use the supplied topic model Y_out = m/(2*pi*max(w,0.5)), where Y_out is PHASE CYCLES M/2πΓ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Width Γ, measured in MeV. Keep m (Mass m) = 775 MeV; s (Spin s) = 1 fixed. For this exercise use only 32.7687 MeV ≤ w ≤ 268.231 MeV on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.69923168, with a bounded reading uncertainty ±0.035908. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0235462 MeV, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
PHASE CYCLES M/2πΓ against w
Predictions from the supplied Casimir labels and the particle taxonomy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PHASE CYCLES M/2πΓ
Horizontal axis (x): w: Width Γ / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Width Γ / MeV | PHASE CYCLES M/2πΓ |
|---|---|
| 32.76875 | 3.764106989012974 |
| 91.63437499999999 | 1.3460568798141406 |
| 150.5 | 0.819568643828697 |
| 209.36562499999997 | 0.5891372134094072 |
| 268.23125 | 0.4598460503622113 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(w) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(32.7687) = 3.76411 and Y_out(268.231) = 0.459846. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 32.7687 + (0.699232 − 3.76411)(268.231 − 32.7687)/(0.459846 − 3.76411) = 251.173 MeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives w = 176.401 MeV; substituting back gives Y_out = 0.69923168.
- The interpolation error in the input is |w_line − w| = 74.7717 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.66332368 and 0.73513968. Sort the roots even if the response decreases: 167.785 MeV ≤ w ≤ 185.95 MeV.
- The lower and upper input deviations are 8.61633 MeV and 9.54919 MeV. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer w ≈ 176.401 MeV; compatible interval [167.785, 185.95] MeV. Endpoint interpolation gives 251.173 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.