University Quantum Field Theory I · topic UQFTI-05.04
2 -> 2 cross-sections: d sigma/d Omega = |M|^2/(64 pi^2 s), and phi^4 in microbarns
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.
2 → 2 cross-sections and φ⁴ in microbarns: predict a finite change
A 2 → 2 differential cross-section is the rate of scatters into a solid angle per target particle per unit incident flux; in the centre-of-mass frame it equals |M|² times |p_f|/|pᵢ|, divided by 64π² s. Use the supplied topic model Y_out = lam^2*389.4*sqrt(max(rs^2/4-mf^2,0))/(32*pi*rs^2*sqrt(rs^2/4-0.0625)), where Y_out is CROSS-SECTION AT MARKER (μb). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. rs is Centre-of-mass energy, measured in GeV. Keep lam (Coupling strength) = 1; mf (Final-pair mass) = 0.25 GeV fixed. For this exercise use only 1.55406 GeV ≤ rs ≤ 8.99594 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 rs = 3.19128 GeV and 4.23314 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CROSS-SECTION AT MARKER against rs
Predictions from the supplied 2 → 2 cross-sections and φ⁴ in microbarns 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): CROSS-SECTION AT MARKER / μb
Horizontal axis (x): rs: Centre-of-mass energy / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Centre-of-mass energy / GeV | CROSS-SECTION AT MARKER / μb |
|---|---|
| 1.5540625 | 1.603832960103594 |
| 3.41453125 | 0.33222622591542206 |
| 5.275 | 0.1392038248017431 |
| 7.13546875 | 0.07607659724276791 |
| 8.9959375 | 0.04786336600253621 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rs separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rs = 3.19128 GeV, substitute into Y_out = lam^2*389.4*sqrt(max(rs^2/4-mf^2,0))/(32*pi*rs^2*sqrt(rs^2/4-0.0625)): Y₀ = 0.380336 μb.
- At rs = 4.23314 GeV, the same fixed-parameter model gives Y₁ = 0.216158 μb.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.164178 μb; the input change is Δrs = 1.04186 GeV.
- Average sensitivity is ΔY_out/Δrs = -0.157581 μb 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 = 0.380336 μb, Y₁ = 0.216158 μb; ΔY_out = -0.164178 μb; average sensitivity = -0.157581 μb 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.
2 → 2 cross-sections and φ⁴ in microbarns: infer an input with uncertainty
A 2 → 2 differential cross-section is the rate of scatters into a solid angle per target particle per unit incident flux; in the centre-of-mass frame it equals |M|² times |p_f|/|pᵢ|, divided by 64π² s. Use the supplied topic model Y_out = lam^2*389.4*sqrt(max(rs^2/4-mf^2,0))/(32*pi*rs^2*sqrt(rs^2/4-0.0625)), where Y_out is CROSS-SECTION AT MARKER (μb). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. rs is Centre-of-mass energy, measured in GeV. Keep lam (Coupling strength) = 1; mf (Final-pair mass) = 0.25 GeV fixed. For this exercise use only 1.55406 GeV ≤ rs ≤ 8.99594 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 = 0.10431513 μb, with a bounded reading uncertainty ±0.0084678 μb. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rs by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000744188 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.
CROSS-SECTION AT MARKER against rs
Predictions from the supplied 2 → 2 cross-sections and φ⁴ in microbarns 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): CROSS-SECTION AT MARKER / μb
Horizontal axis (x): rs: Centre-of-mass energy / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Centre-of-mass energy / GeV | CROSS-SECTION AT MARKER / μb |
|---|---|
| 1.5540625 | 1.603832960103594 |
| 3.41453125 | 0.33222622591542206 |
| 5.275 | 0.1392038248017431 |
| 7.13546875 | 0.07607659724276791 |
| 8.9959375 | 0.04786336600253621 |
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(rs) = 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(1.55406) = 1.60383 μb and Y_out(8.99594) = 0.0478634 μb. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rs_line = 1.55406 + (0.104315 − 1.60383)(8.99594 − 1.55406)/(0.0478634 − 1.60383) = 8.72594 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 rs = 6.09361 GeV; substituting back gives Y_out = 0.10431513 μb.
- The interpolation error in the input is |rs_line − rs| = 2.63233 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.09584733 μb and 0.11278293 μb. Sort the roots even if the response decreases: 5.86039 GeV ≤ rs ≤ 6.35709 GeV.
- The lower and upper input deviations are 0.233218 GeV and 0.263479 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 rs ≈ 6.09361 GeV; compatible interval [5.86039, 6.35709] GeV. Endpoint interpolation gives 8.72594 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.