University Quantum Field Theory I · topic UQFTI-14.06
F_1's infrared divergence and the vanishing elastic rate
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.
F₁'s infrared divergence and the vanishing elastic rate: predict a finite change
F₁'s infrared divergence is the logarithm of the photon mass left in the renormalised one-loop vertex, −(α/2π) f_IR(q²) ln(1/μ_γ²): it comes from virtual photons of vanishing momentum and is absent only at q² = 0. Use the supplied topic model Y_out = (1/(0.00232282*((2+10^lq)/sqrt(10^lq*(10^lq+4))*ln((2+10^lq+sqrt(10^lq*(10^lq+4)))/2)-1))-max(lq,0)*ln(10))/(2*ln(10)), where Y_out is ZERO CROSSING k* (decades). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm; max(a,b) selects the larger value. lq is Marker at log₁₀(−q²/m²), dimensionless. Keep k (Photon mass μ_γ = m × 10⁻ᵏ, k) = 6 fixed. For this exercise use only -1.15 ≤ lq ≤ 5.15 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 lq = 0.236 and 1.118. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlq across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ZERO CROSSING k* against lq
Predictions from the supplied F₁'s infrared divergence and the vanishing elastic rate 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): ZERO CROSSING k* / decades
Horizontal axis (x): lq: Marker at log₁₀(−q²/m²)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lq: Marker at log₁₀(−q²/m²) | ZERO CROSSING k* / decades |
|---|---|
| -1.15 | 4003.451671087192 |
| 0.4249999999999998 | 143.79939468072118 |
| 1.9999999999999996 | 24.783328813764307 |
| 3.5749999999999997 | 11.13849070323136 |
| 5.149999999999999 | 6.03445830228566 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lq separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lq = 0.236, substitute into Y_out = (1/(0.00232282*((2+10^lq)/sqrt(10^lq*(10^lq+4))*ln((2+10^lq+sqrt(10^lq*(10^lq+4)))/2)-1))-max(lq,0)*ln(10))/(2*ln(10)): Y₀ = 202.393 decades.
- At lq = 1.118, the same fixed-parameter model gives Y₁ = 53.2982 decades.
- Signed change: ΔY_out = Y₁ − Y₀ = -149.094 decades; the input change is Δlq = 0.882.
- Average sensitivity is ΔY_out/Δlq = -169.041 decades per unit input. 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 = 202.393 decades, Y₁ = 53.2982 decades; ΔY_out = -149.094 decades; average sensitivity = -169.041 decades per unit input.
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.
F₁'s infrared divergence and the vanishing elastic rate: infer an input with uncertainty
F₁'s infrared divergence is the logarithm of the photon mass left in the renormalised one-loop vertex, −(α/2π) f_IR(q²) ln(1/μ_γ²): it comes from virtual photons of vanishing momentum and is absent only at q² = 0. Use the supplied topic model Y_out = (1/(0.00232282*((2+10^lq)/sqrt(10^lq*(10^lq+4))*ln((2+10^lq+sqrt(10^lq*(10^lq+4)))/2)-1))-max(lq,0)*ln(10))/(2*ln(10)), where Y_out is ZERO CROSSING k* (decades). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm; max(a,b) selects the larger value. lq is Marker at log₁₀(−q²/m²), dimensionless. Keep k (Photon mass μ_γ = m × 10⁻ᵏ, k) = 6 fixed. For this exercise use only -1.15 ≤ lq ≤ 5.15 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 = 16.614138 decades, with a bounded reading uncertainty ±1.587 decades. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lq by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063, 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.
ZERO CROSSING k* against lq
Predictions from the supplied F₁'s infrared divergence and the vanishing elastic rate 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): ZERO CROSSING k* / decades
Horizontal axis (x): lq: Marker at log₁₀(−q²/m²)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lq: Marker at log₁₀(−q²/m²) | ZERO CROSSING k* / decades |
|---|---|
| -1.15 | 4003.451671087192 |
| 0.4249999999999998 | 143.79939468072118 |
| 1.9999999999999996 | 24.783328813764307 |
| 3.5749999999999997 | 11.13849070323136 |
| 5.149999999999999 | 6.03445830228566 |
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(lq) = 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(-1.15) = 4003.45 decades and Y_out(5.15) = 6.03446 decades. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lq_line = -1.15 + (16.6141 − 4003.45)(5.15 − -1.15)/(6.03446 − 4003.45) = 5.13333. 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 lq = 2.693; substituting back gives Y_out = 16.614138 decades.
- The interpolation error in the input is |lq_line − lq| = 2.44033. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 15.027138 decades and 18.201138 decades. Sort the roots even if the response decreases: 2.51793 ≤ lq ≤ 2.89739.
- The lower and upper input deviations are 0.175069 and 0.204386. 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 lq ≈ 2.693; compatible interval [2.51793, 2.89739]. Endpoint interpolation gives 5.13333.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.