University Physics IV · topic UPIV-14.09
Neutrino mass and what the Standard Model leaves out
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.
Neutrino mass and open questions: predict a finite change
Neutrino oscillation is the periodic change of a travelling neutrino's flavour, caused by flavour states being superpositions of mass states that accumulate different quantum phases — so it can happen only if the masses are non-zero and unequal. Use the supplied topic model Y_out = 1 - sind(2*t)^2*sin(0.001267*d*295/max(en,0.1))^2, where Y_out is P AT L = 295 km (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; sind/cosd/tand use degrees; unmarked trigonometric functions use radians. en is Beam energy E, measured in GeV. Keep d (Splitting Δm²) = 2.5 ×10⁻³ eV²; t (Mixing angle θ) = 45 ° fixed. For this exercise use only 0.67625 GeV ≤ en ≤ 2.72375 GeV 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 en = 1.1267 GeV and 1.41335 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
P AT L = 295 km against en
Predictions from the supplied Neutrino mass and open questions 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): P AT L = 295 km
Horizontal axis (x): en: Beam energy E / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Beam energy E / GeV | P AT L = 295 km |
|---|---|
| 0.67625 | 0.0353125725397605 |
| 1.1881249999999999 | 0.49893840617295393 |
| 1.7 | 0.7271059972935712 |
| 2.211875 | 0.8319010410454286 |
| 2.72375 | 0.8868543267844172 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 1.1267 GeV, substitute into Y_out = 1 - sind(2*t)^2*sin(0.001267*d*295/max(en,0.1))^2: Y₀ = 0.456119.
- At en = 1.41335 GeV, the same fixed-parameter model gives Y₁ = 0.62299.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.166871; the input change is Δen = 0.28665 GeV.
- Average sensitivity is ΔY_out/Δen = 0.582141 1 per GeV. Its positive sign means that this response increases 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 = 0.456119, Y₁ = 0.62299; ΔY_out = 0.166871; average sensitivity = 0.582141 1 per GeV.
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.
Neutrino mass and open questions: infer an input with uncertainty
Neutrino oscillation is the periodic change of a travelling neutrino's flavour, caused by flavour states being superpositions of mass states that accumulate different quantum phases — so it can happen only if the masses are non-zero and unequal. Use the supplied topic model Y_out = 1 - sind(2*t)^2*sin(0.001267*d*295/max(en,0.1))^2, where Y_out is P AT L = 295 km (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; sind/cosd/tand use degrees; unmarked trigonometric functions use radians. en is Beam energy E, measured in GeV. Keep d (Splitting Δm²) = 2.5 ×10⁻³ eV²; t (Mixing angle θ) = 45 ° fixed. For this exercise use only 0.67625 GeV ≤ en ≤ 2.72375 GeV 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 = 0.78235904, with a bounded reading uncertainty ±0.015674. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00020475 GeV, 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.
P AT L = 295 km against en
Predictions from the supplied Neutrino mass and open questions 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): P AT L = 295 km
Horizontal axis (x): en: Beam energy E / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Beam energy E / GeV | P AT L = 295 km |
|---|---|
| 0.67625 | 0.0353125725397605 |
| 1.1881249999999999 | 0.49893840617295393 |
| 1.7 | 0.7271059972935712 |
| 2.211875 | 0.8319010410454286 |
| 2.72375 | 0.8868543267844172 |
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(en) = 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(0.67625) = 0.0353126 and Y_out(2.72375) = 0.886854. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives en_line = 0.67625 + (0.782359 − 0.0353126)(2.72375 − 0.67625)/(0.886854 − 0.0353126) = 2.4725 GeV. 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 en = 1.92523 GeV; substituting back gives Y_out = 0.78235904.
- The interpolation error in the input is |en_line − en| = 0.54727 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.76668504 and 0.79803304. Sort the roots even if the response decreases: 1.85359 GeV ≤ en ≤ 2.00474 GeV.
- The lower and upper input deviations are 0.0716303 GeV and 0.0795141 GeV. 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 en ≈ 1.92523 GeV; compatible interval [1.85359, 2.00474] GeV. Endpoint interpolation gives 2.4725 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.