University Physics IV · topic UPIV-03.09
Collisions, decays, and threshold energies
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.
Collisions, decays and thresholds: predict a finite change
The invariant mass of a system is √s/c², where s = (ΣE)² − (Σpc)² is the square of its total four-momentum: conserved through any reaction, and the same number in every inertial frame. Use the supplied topic model Y_out = max((mx*mx-2*mt*mt)/(2*mt),0), where Y_out is FIXED-TARGET E FOR M (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. mt is Beam and target rest energy m, measured in GeV. Keep eb (Beam energy per particle E) = 6.6 GeV; mx (Mass to be produced M) = 3.75 GeV fixed. For this exercise use only 0.195625 GeV ≤ mt ≤ 0.904375 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 mt = 0.35155 GeV and 0.450775 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δmt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIXED-TARGET E FOR M against mt
Predictions from the supplied Collisions, decays and thresholds 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): FIXED-TARGET E FOR M / GeV
Horizontal axis (x): mt: Beam and target rest energy m / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mt: Beam and target rest energy m / GeV | FIXED-TARGET E FOR M / GeV |
|---|---|
| 0.195625 | 35.74686701277955 |
| 0.3728125 | 18.487204264459347 |
| 0.55 | 12.234090909090908 |
| 0.7271874999999999 | 8.9419143478728 |
| 0.9043749999999999 | 6.870331288873532 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of mt separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At mt = 0.35155 GeV, substitute into Y_out = max((mx*mx-2*mt*mt)/(2*mt),0): Y₀ = 19.6492 GeV.
- At mt = 0.450775 GeV, the same fixed-parameter model gives Y₁ = 15.1474 GeV.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.5018 GeV; the input change is Δmt = 0.099225 GeV.
- Average sensitivity is ΔY_out/Δmt = -45.3696 GeV per GeV. 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 = 19.6492 GeV, Y₁ = 15.1474 GeV; ΔY_out = -4.5018 GeV; average sensitivity = -45.3696 GeV 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.
Collisions, decays and thresholds: infer an input with uncertainty
The invariant mass of a system is √s/c², where s = (ΣE)² − (Σpc)² is the square of its total four-momentum: conserved through any reaction, and the same number in every inertial frame. Use the supplied topic model Y_out = max((mx*mx-2*mt*mt)/(2*mt),0), where Y_out is FIXED-TARGET E FOR M (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. mt is Beam and target rest energy m, measured in GeV. Keep eb (Beam energy per particle E) = 6.6 GeV; mx (Mass to be produced M) = 3.75 GeV fixed. For this exercise use only 0.195625 GeV ≤ mt ≤ 0.904375 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 = 10.568964 GeV, with a bounded reading uncertainty ±0.55479 GeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000070875 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.
FIXED-TARGET E FOR M against mt
Predictions from the supplied Collisions, decays and thresholds 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): FIXED-TARGET E FOR M / GeV
Horizontal axis (x): mt: Beam and target rest energy m / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mt: Beam and target rest energy m / GeV | FIXED-TARGET E FOR M / GeV |
|---|---|
| 0.195625 | 35.74686701277955 |
| 0.3728125 | 18.487204264459347 |
| 0.55 | 12.234090909090908 |
| 0.7271874999999999 | 8.9419143478728 |
| 0.9043749999999999 | 6.870331288873532 |
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(mt) = 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(0.195625) = 35.7469 GeV and Y_out(0.904375) = 6.87033 GeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mt_line = 0.195625 + (10.569 − 35.7469)(0.904375 − 0.195625)/(6.87033 − 35.7469) = 0.813595 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 mt = 0.627963 GeV; substituting back gives Y_out = 10.568964 GeV.
- The interpolation error in the input is |mt_line − mt| = 0.185633 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.014174 GeV and 11.123754 GeV. Sort the roots even if the response decreases: 0.599756 GeV ≤ mt ≤ 0.658791 GeV.
- The lower and upper input deviations are 0.0282061 GeV and 0.0308282 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 mt ≈ 0.627963 GeV; compatible interval [0.599756, 0.658791] GeV. Endpoint interpolation gives 0.813595 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.