University Quantum Field Theory I · topic UQFTI-14.04
g - 2 against experiment: the measured a_e and the series in alpha/pi
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.
g − 2 against experiment: a_e and the series in α/π: predict a finite change
The electron's anomaly a_e = (g − 2)/2 = F₂(0) is predicted as a series in α/π with computed coefficients, plus small terms from loops of muons, tau leptons and hadrons and from weak bosons, and is measured in a Penning trap as the ratio ν_a/ν_c of two frequencies. Use the supplied topic model Y_out = (0.0084498*(206-166.25)+0.06762*(6.737-c5)+1.6927*(1-kh))/hypot(u,0.0929), where Y_out is RUBIDIUM RESIDUAL (σ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). u is Trap uncertainty on a_e, measured in × 10⁻¹². Keep c5 (Five-loop coefficient C₅) = 6.74; kh (Hadronic term ÷ 1.693 × 10⁻¹²) = 1 fixed. For this exercise use only 0.0978125 × 10⁻¹² ≤ u ≤ 0.452187 × 10⁻¹² 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 u = 0.175775 × 10⁻¹² and 0.225387 × 10⁻¹². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δu across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RUBIDIUM RESIDUAL against u
Predictions from the supplied g − 2 against experiment: a_e and the series in α/π 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): RUBIDIUM RESIDUAL / σ
Horizontal axis (x): u: Trap uncertainty on a_e / × 10⁻¹²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Trap uncertainty on a_e / × 10⁻¹² | RUBIDIUM RESIDUAL / σ |
|---|---|
| 0.0978125 | 2.4883587383891195 |
| 0.18640625 | 1.6117137962203707 |
| 0.275 | 1.1564378189646878 |
| 0.36359374999999994 | 0.8944834763417981 |
| 0.45218749999999996 | 0.7271523955226936 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of u separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At u = 0.175775 × 10⁻¹², substitute into Y_out = (0.0084498*(206-166.25)+0.06762*(6.737-c5)+1.6927*(1-kh))/hypot(u,0.0929): Y₀ = 1.68839 σ.
- At u = 0.225387 × 10⁻¹², the same fixed-parameter model gives Y₁ = 1.37695 σ.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.311439 σ; the input change is Δu = 0.0496125 × 10⁻¹².
- Average sensitivity is ΔY_out/Δu = -6.27743 σ per × 10⁻¹². 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 = 1.68839 σ, Y₁ = 1.37695 σ; ΔY_out = -0.311439 σ; average sensitivity = -6.27743 σ per × 10⁻¹².
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.
g − 2 against experiment: a_e and the series in α/π: infer an input with uncertainty
The electron's anomaly a_e = (g − 2)/2 = F₂(0) is predicted as a series in α/π with computed coefficients, plus small terms from loops of muons, tau leptons and hadrons and from weak bosons, and is measured in a Penning trap as the ratio ν_a/ν_c of two frequencies. Use the supplied topic model Y_out = (0.0084498*(206-166.25)+0.06762*(6.737-c5)+1.6927*(1-kh))/hypot(u,0.0929), where Y_out is RUBIDIUM RESIDUAL (σ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). u is Trap uncertainty on a_e, measured in × 10⁻¹². Keep c5 (Five-loop coefficient C₅) = 6.74; kh (Hadronic term ÷ 1.693 × 10⁻¹²) = 1 fixed. For this exercise use only 0.0978125 × 10⁻¹² ≤ u ≤ 0.452187 × 10⁻¹² 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 = 1.0251659 σ, with a bounded reading uncertainty ±0.044702 σ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000354375 × 10⁻¹², 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.
RUBIDIUM RESIDUAL against u
Predictions from the supplied g − 2 against experiment: a_e and the series in α/π 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): RUBIDIUM RESIDUAL / σ
Horizontal axis (x): u: Trap uncertainty on a_e / × 10⁻¹²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Trap uncertainty on a_e / × 10⁻¹² | RUBIDIUM RESIDUAL / σ |
|---|---|
| 0.0978125 | 2.4883587383891195 |
| 0.18640625 | 1.6117137962203707 |
| 0.275 | 1.1564378189646878 |
| 0.36359374999999994 | 0.8944834763417981 |
| 0.45218749999999996 | 0.7271523955226936 |
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(u) = 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(0.0978125) = 2.48836 σ and Y_out(0.452187) = 0.727152 σ. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives u_line = 0.0978125 + (1.02517 − 2.48836)(0.452187 − 0.0978125)/(0.727152 − 2.48836) = 0.392224 × 10⁻¹². 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 u = 0.313981 × 10⁻¹²; substituting back gives Y_out = 1.0251659 σ.
- The interpolation error in the input is |u_line − u| = 0.0782425 × 10⁻¹². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.9804639 σ and 1.0698679 σ. Sort the roots even if the response decreases: 0.299686 × 10⁻¹² ≤ u ≤ 0.32952 × 10⁻¹².
- The lower and upper input deviations are 0.0142948 × 10⁻¹² and 0.0155389 × 10⁻¹². 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 u ≈ 0.313981 × 10⁻¹²; compatible interval [0.299686, 0.32952] × 10⁻¹². Endpoint interpolation gives 0.392224 × 10⁻¹².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.