University Quantum Field Theory I · topic UQFTI-10.03
Bhabha scattering e+ e- -> e+ e-: s and t channels interfering
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.
Bhabha scattering: s and t channels interfering: predict a finite change
Bhabha scattering is elastic electron-positron scattering, e⁺e⁻ → e⁺e⁻, whose tree-level amplitude is the t-channel photon-exchange diagram minus the s-channel annihilation diagram. Use the supplied topic model Y_out = (4+(1+cosd(theta))^2)/(1-cosd(theta))^2, where Y_out is EXCHANGE (s²+u²)/t² (α²/(2s)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Scattering angle θ, measured in °. Keep eta (Relative sign η of the s diagram) = -1 fixed. For this exercise use only 45.9375 ° ≤ theta ≤ 164.063 ° 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 theta = 71.925 ° and 88.4625 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtheta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EXCHANGE (s²+u²)/t² against theta
Predictions from the supplied Bhabha scattering: s and t channels interfering 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): EXCHANGE (s²+u²)/t² / α²/(2s)
Horizontal axis (x): theta: Scattering angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Scattering angle θ / ° | EXCHANGE (s²+u²)/t² / α²/(2s) |
|---|---|
| 45.9375 | 74.11472969437779 |
| 75.46875 | 9.916923843485383 |
| 105 | 2.870930405038196 |
| 134.53125 | 1.412797434581187 |
| 164.0625 | 1.0399594102515757 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of theta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At theta = 71.925 °, substitute into Y_out = (4+(1+cosd(theta))^2)/(1-cosd(theta))^2: Y₀ = 12.0166 α²/(2s).
- At theta = 88.4625 °, the same fixed-parameter model gives Y₁ = 5.33693 α²/(2s).
- Signed change: ΔY_out = Y₁ − Y₀ = -6.67971 α²/(2s); the input change is Δtheta = 16.5375 °.
- Average sensitivity is ΔY_out/Δtheta = -0.403913 α²/(2s) per °. 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 = 12.0166 α²/(2s), Y₁ = 5.33693 α²/(2s); ΔY_out = -6.67971 α²/(2s); average sensitivity = -0.403913 α²/(2s) per °.
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.
Bhabha scattering: s and t channels interfering: infer an input with uncertainty
Bhabha scattering is elastic electron-positron scattering, e⁺e⁻ → e⁺e⁻, whose tree-level amplitude is the t-channel photon-exchange diagram minus the s-channel annihilation diagram. Use the supplied topic model Y_out = (4+(1+cosd(theta))^2)/(1-cosd(theta))^2, where Y_out is EXCHANGE (s²+u²)/t² (α²/(2s)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Scattering angle θ, measured in °. Keep eta (Relative sign η of the s diagram) = -1 fixed. For this exercise use only 45.9375 ° ≤ theta ≤ 164.063 ° 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 = 1.9830646 α²/(2s), with a bounded reading uncertainty ±0.14147 α²/(2s). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0118125 °, 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.
EXCHANGE (s²+u²)/t² against theta
Predictions from the supplied Bhabha scattering: s and t channels interfering 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): EXCHANGE (s²+u²)/t² / α²/(2s)
Horizontal axis (x): theta: Scattering angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Scattering angle θ / ° | EXCHANGE (s²+u²)/t² / α²/(2s) |
|---|---|
| 45.9375 | 74.11472969437779 |
| 75.46875 | 9.916923843485383 |
| 105 | 2.870930405038196 |
| 134.53125 | 1.412797434581187 |
| 164.0625 | 1.0399594102515757 |
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(theta) = 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(45.9375) = 74.1147 α²/(2s) and Y_out(164.063) = 1.03996 α²/(2s). The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 45.9375 + (1.98306 − 74.1147)(164.063 − 45.9375)/(1.03996 − 74.1147) = 162.538 °. 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 theta = 117.994 °; substituting back gives Y_out = 1.9830646 α²/(2s).
- The interpolation error in the input is |theta_line − theta| = 44.5442 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8415946 α²/(2s) and 2.1245346 α²/(2s). Sort the roots even if the response decreases: 115.296 ° ≤ theta ≤ 121.083 °.
- The lower and upper input deviations are 2.69817 ° and 3.08895 °. 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 theta ≈ 117.994 °; compatible interval [115.296, 121.083] °. Endpoint interpolation gives 162.538 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.