University Quantum Field Theory I · topic UQFTI-10.07
Thomson's limit, (8 pi/3) r_e^2 = 0.665 barn, and the high-energy fall-off
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.
Thomson's limit and the high-energy Compton fall: predict a finite change
The Thomson limit is the ω ≪ m end of Compton scattering, where recoil vanishes and dσ/dΩ = (rₑ²/2)(1 + cos²θ), (8π/3)rₑ² in all with rₑ = α/m; for ω ≫ m the total falls instead as (πrₑ²m/ω)[ln(2ω/m) + ½]. Use the supplied topic model Y_out = 0.511*10^lw, where Y_out is PHOTON ENERGY ω (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lw is Photon energy log₁₀(ω/m), dimensionless. Keep th (Scattering angle θ) = 90 ° fixed. For this exercise use only -1.46875 ≤ lw ≤ 2.46875 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 lw = -0.6025 and -0.05125. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PHOTON ENERGY ω against lw
Predictions from the supplied Thomson's limit and the high-energy Compton fall 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): PHOTON ENERGY ω / MeV
Horizontal axis (x): lw: Photon energy log₁₀(ω/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lw: Photon energy log₁₀(ω/m) | PHOTON ENERGY ω / MeV |
|---|---|
| -1.46875 | 0.017364844560896476 |
| -0.484375 | 0.16751199083620574 |
| 0.5 | 1.615923884346042 |
| 1.484375 | 15.5881975192645 |
| 2.46875 | 150.3733587042943 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lw separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lw = -0.6025, substitute into Y_out = 0.511*10^lw: Y₀ = 0.127621 MeV.
- At lw = -0.05125, the same fixed-parameter model gives Y₁ = 0.45412 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.3265 MeV; the input change is Δlw = 0.55125.
- Average sensitivity is ΔY_out/Δlw = 0.59229 MeV per unit input. 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 = 0.127621 MeV, Y₁ = 0.45412 MeV; ΔY_out = 0.3265 MeV; average sensitivity = 0.59229 MeV 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.
Thomson's limit and the high-energy Compton fall: infer an input with uncertainty
The Thomson limit is the ω ≪ m end of Compton scattering, where recoil vanishes and dσ/dΩ = (rₑ²/2)(1 + cos²θ), (8π/3)rₑ² in all with rₑ = α/m; for ω ≫ m the total falls instead as (πrₑ²m/ω)[ln(2ω/m) + ½]. Use the supplied topic model Y_out = 0.511*10^lw, where Y_out is PHOTON ENERGY ω (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lw is Photon energy log₁₀(ω/m), dimensionless. Keep th (Scattering angle θ) = 90 ° fixed. For this exercise use only -1.46875 ≤ lw ≤ 2.46875 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 = 4.3807241 MeV, with a bounded reading uncertainty ±0.6545 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, 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.
PHOTON ENERGY ω against lw
Predictions from the supplied Thomson's limit and the high-energy Compton fall 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): PHOTON ENERGY ω / MeV
Horizontal axis (x): lw: Photon energy log₁₀(ω/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lw: Photon energy log₁₀(ω/m) | PHOTON ENERGY ω / MeV |
|---|---|
| -1.46875 | 0.017364844560896476 |
| -0.484375 | 0.16751199083620574 |
| 0.5 | 1.615923884346042 |
| 1.484375 | 15.5881975192645 |
| 2.46875 | 150.3733587042943 |
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(lw) = 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(-1.46875) = 0.0173648 MeV and Y_out(2.46875) = 150.373 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lw_line = -1.46875 + (4.38072 − 0.0173648)(2.46875 − -1.46875)/(150.373 − 0.0173648) = -1.35448. 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 lw = 0.933125; substituting back gives Y_out = 4.3807241 MeV.
- The interpolation error in the input is |lw_line − lw| = 2.28761. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.7262241 MeV and 5.0352241 MeV. Sort the roots even if the response decreases: 0.862848 ≤ lw ≤ 0.993598.
- The lower and upper input deviations are 0.0702769 and 0.0604729. 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 lw ≈ 0.933125; compatible interval [0.862848, 0.993598]. Endpoint interpolation gives -1.35448.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.