University Physics V · unit UPV-17
Elementary Particle Physics
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
Antiparticles, charge conjugation and CPT: infer an input with uncertainty
An antiparticle is what a relativistic wave equation's negative-energy solution becomes when it is read forward in time: the particle's mass and spin, every additive charge reversed, and, by the CPT theorem, the same total lifetime. Use the supplied topic model Y_out = 197.327/max(m, 0.001), where Y_out is COMPTON LENGTH ħ/mc (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. m is Rest energy mc², measured in MeV. Keep p (Momentum pc) = 1.2 MeV fixed. For this exercise use only 0.2125 MeV ≤ m ≤ 1.7875 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 = 168.18837 fm, with a bounded reading uncertainty ±8.6693 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 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.
COMPTON LENGTH ħ/mc against m
Predictions from the supplied Antiparticles, charge conjugation and CPT 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): COMPTON LENGTH ħ/mc / fm
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | COMPTON LENGTH ħ/mc / fm |
|---|---|
| 0.2125 | 928.5976470588236 |
| 0.60625 | 325.4878350515464 |
| 1 | 197.327 |
| 1.3937499999999998 | 141.57991031390137 |
| 1.7874999999999999 | 110.39272727272729 |
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(m) = 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(0.2125) = 928.598 fm and Y_out(1.7875) = 110.393 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 0.2125 + (168.188 − 928.598)(1.7875 − 0.2125)/(110.393 − 928.598) = 1.67625 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 m = 1.17325 MeV; substituting back gives Y_out = 168.18837 fm.
- The interpolation error in the input is |m_line − m| = 0.502997 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 159.51907 fm and 176.85767 fm. Sort the roots even if the response decreases: 1.11574 MeV ≤ m ≤ 1.23701 MeV.
- The lower and upper input deviations are 0.057511 MeV and 0.063762 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 m ≈ 1.17325 MeV; compatible interval [1.11574, 1.23701] MeV. Endpoint interpolation gives 1.67625 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The lepton sector and family number: infer an input with uncertainty
The lepton sector is three copies of one left-chiral SU(2) doublet — (ν, ℓ)_L for ℓ = e, μ, τ — coupled to the W with a single universal strength and differing only in mass; family number counts the leptons of each copy separately. Use the supplied topic model Y_out = 2197000000/(max(nch,0.01)*16.817^n), where Y_out is PREDICTED TAU LIFETIME (fs). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. n is Mass exponent n in Γ ∝ mⁿ, dimensionless. Keep nch (Open channels N at the tau) = 1 fixed. For this exercise use only 2.53125 ≤ n ≤ 6.46875 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 = 1972.6924 fs, with a bounded reading uncertainty ±292.02 fs. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, 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.
PREDICTED TAU LIFETIME against n
Predictions from the supplied The lepton sector and family number 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): PREDICTED TAU LIFETIME / fs
Horizontal axis (x): n: Mass exponent n in Γ ∝ mⁿ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Mass exponent n in Γ ∝ mⁿ | PREDICTED TAU LIFETIME / fs |
|---|---|
| 2.53125 | 1734419.3300551092 |
| 3.515625 | 107784.8946543751 |
| 4.5 | 6698.255326343487 |
| 5.484375 | 416.26078088918666 |
| 6.46875 | 25.868383521454014 |
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(n) = 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(2.53125) = 1734420 fs and Y_out(6.46875) = 25.8684 fs. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 2.53125 + (1972.69 − 1734420)(6.46875 − 2.53125)/(25.8684 − 1734420) = 6.46433. 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 n = 4.93312; substituting back gives Y_out = 1972.6924 fs.
- The interpolation error in the input is |n_line − n| = 1.53121. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1680.6724 fs and 2264.7124 fs. Sort the roots even if the response decreases: 4.88421 ≤ n ≤ 4.98989.
- The lower and upper input deviations are 0.0489119 and 0.0567623. 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 n ≈ 4.93312; compatible interval [4.88421, 4.98989]. Endpoint interpolation gives 6.46433.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Quark flavours and additive quantum numbers: infer an input with uncertainty
An additive quantum number of a hadron is the eigenvalue of a diagonal, Hermitian flavour-counting operator: add one unit for each quark of that flavour, subtract one for each antiquark, and the total labels the state. Use the supplied topic model Y_out = (nu+nd+ns)/3, where Y_out is BARYON NUMBER B (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. ns is Strange quarks n_s, dimensionless. Keep nu (Up quarks n_u) = 2; nd (Down quarks n_d) = 1 fixed. For this exercise use only 0 ≤ ns ≤ 3, at the specified test settings separated by 1. 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 = 1.6666667, with a bounded reading uncertainty ±0.08. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of ns consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
BARYON NUMBER B against ns
Predictions from the supplied Quark flavours and additive quantum numbers model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BARYON NUMBER B
Horizontal axis (x): ns: Strange quarks n_s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ns: Strange quarks n_s | BARYON NUMBER B |
|---|---|
| 0 | 1 |
| 1 | 1.3333333333333333 |
| 2 | 1.6666666666666667 |
| 3 | 2 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(0) = 1 and Y_out(3) = 2. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ns_line = 0 + (1.66667 − 1)(3 − 0)/(2 − 1) = 2. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to ns = 2; only the allowed settings 2 lie inside the stated reading bound.
- The interpolation error in the input is |ns_line − ns| = 1e-7.
- The compatible allowed values span 2 to 2. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible ns: 2. Endpoint interpolation predicts 2; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.