University Quantum Field Theory I · topic UQFTI-14.08
Bloch-Nordsieck: virtual and real infrared divergences cancel in a measured 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.
Bloch-Nordsieck: virtual and real divergences cancel: predict a finite change
The Bloch-Nordsieck cancellation: the infrared divergence of the elastic vertex correction and that of emitting photons too soft to detect are equal and opposite, so the rate recorded by a detector of energy resolution E_l is finite and depends on E_l, not on the photon mass. Use the supplied topic model Y_out = (0.46456*(ln((q/1.022)+sqrt(1+(q/1.022)^2))*(1+2*(q/1.022)^2)/((q/1.022)*sqrt(1+(q/1.022)^2))-1)*(ln(el)-2.302585*lgmu)), where Y_out is REAL, k BELOW E_l (%). 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. el is Detector resolution E_l, measured in MeV. Keep lgmu (Photon mass log₁₀(μ_γ / MeV)) = -6; q (Momentum transfer √(−q²)) = 500 MeV fixed. For this exercise use only 1.50937 MeV ≤ el ≤ 8.99062 MeV 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 el = 3.15525 MeV and 4.20262 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δel across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
REAL, k BELOW E_l against el
Predictions from the supplied Bloch-Nordsieck: virtual and real divergences cancel 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): REAL, k BELOW E_l / %
Horizontal axis (x): el: Detector resolution E_l / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| el: Detector resolution E_l / MeV | REAL, k BELOW E_l / % |
|---|---|
| 1.509375 | 84.4150691961651 |
| 3.3796874999999997 | 89.19787345663215 |
| 5.25 | 91.81118920132403 |
| 7.120312499999999 | 93.61922228880847 |
| 8.990625 | 95.0030632631905 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of el separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At el = 3.15525 MeV, substitute into Y_out = (0.46456*(ln((q/1.022)+sqrt(1+(q/1.022)^2))*(1+2*(q/1.022)^2)/((q/1.022)*sqrt(1+(q/1.022)^2))-1)*(ln(el)-2.302585*lgmu)): Y₀ = 88.7902 %.
- At el = 4.20262 MeV, the same fixed-parameter model gives Y₁ = 90.4909 %.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.70075 %; the input change is Δel = 1.04737 MeV.
- Average sensitivity is ΔY_out/Δel = 1.62382 % per MeV. Its positive sign means that this response increases 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 = 88.7902 %, Y₁ = 90.4909 %; ΔY_out = 1.70075 %; average sensitivity = 1.62382 % per MeV.
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.
Bloch-Nordsieck: virtual and real divergences cancel: infer an input with uncertainty
The Bloch-Nordsieck cancellation: the infrared divergence of the elastic vertex correction and that of emitting photons too soft to detect are equal and opposite, so the rate recorded by a detector of energy resolution E_l is finite and depends on E_l, not on the photon mass. Use the supplied topic model Y_out = (0.46456*(ln((q/1.022)+sqrt(1+(q/1.022)^2))*(1+2*(q/1.022)^2)/((q/1.022)*sqrt(1+(q/1.022)^2))-1)*(ln(el)-2.302585*lgmu)), where Y_out is REAL, k BELOW E_l (%). 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. el is Detector resolution E_l, measured in MeV. Keep lgmu (Photon mass log₁₀(μ_γ / MeV)) = -6; q (Momentum transfer √(−q²)) = 500 MeV fixed. For this exercise use only 1.50937 MeV ≤ el ≤ 8.99062 MeV 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 = 92.675171 %, with a bounded reading uncertainty ±0.34918 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown el by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 MeV, 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.
REAL, k BELOW E_l against el
Predictions from the supplied Bloch-Nordsieck: virtual and real divergences cancel 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): REAL, k BELOW E_l / %
Horizontal axis (x): el: Detector resolution E_l / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| el: Detector resolution E_l / MeV | REAL, k BELOW E_l / % |
|---|---|
| 1.509375 | 84.4150691961651 |
| 3.3796874999999997 | 89.19787345663215 |
| 5.25 | 91.81118920132403 |
| 7.120312499999999 | 93.61922228880847 |
| 8.990625 | 95.0030632631905 |
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(el) = 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(1.50937) = 84.4151 % and Y_out(8.99062) = 95.0031 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives el_line = 1.50937 + (92.6752 − 84.4151)(8.99062 − 1.50937)/(95.0031 − 84.4151) = 7.34579 MeV. 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 el = 6.07294 MeV; substituting back gives Y_out = 92.675171 %.
- The interpolation error in the input is |el_line − el| = 1.27285 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 92.325991 % and 93.024351 %. Sort the roots even if the response decreases: 5.72586 MeV ≤ el ≤ 6.44106 MeV.
- The lower and upper input deviations are 0.347081 MeV and 0.36812 MeV. 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 el ≈ 6.07294 MeV; compatible interval [5.72586, 6.44106] MeV. Endpoint interpolation gives 7.34579 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.