University Quantum Field Theory I · topic UQFTI-13.07
The Uehling potential: vacuum polarisation in position space
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.
The Uehling potential: vacuum polarisation in position space: predict a finite change
The Uehling potential is the static potential of a point charge Ze with the one-loop electron vacuum polarisation included: Coulomb's law with the measured α, plus a sum of Yukawa potentials, one for each virtual-pair mass, felt only within about 193 fm. Use the supplied topic model Y_out = 110.791/10^lr, where Y_out is EQUIVALENT Q = 0.561/r (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lr is Probe radius log₁₀(r/fm), dimensionless. Keep lm (Heaviest pair kept log₁₀(M/MeV)) = 4.1 fixed. For this exercise use only -0.575 ≤ lr ≤ 2.575 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 lr = 0.118 and 0.559. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EQUIVALENT Q = 0.561/r against lr
Predictions from the supplied The Uehling potential: vacuum polarisation in position space 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): EQUIVALENT Q = 0.561/r / MeV
Horizontal axis (x): lr: Probe radius log₁₀(r/fm)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lr: Probe radius log₁₀(r/fm) | EQUIVALENT Q = 0.561/r / MeV |
|---|---|
| -0.575 | 416.3940185852102 |
| 0.2124999999999999 | 67.9210642680708 |
| 0.9999999999999998 | 11.079100000000006 |
| 1.7874999999999999 | 1.807192777862615 |
| 2.5749999999999993 | 0.2947843901001704 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lr separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lr = 0.118, substitute into Y_out = 110.791/10^lr: Y₀ = 84.4315 MeV.
- At lr = 0.559, the same fixed-parameter model gives Y₁ = 30.5847 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = -53.8468 MeV; the input change is Δlr = 0.441.
- Average sensitivity is ΔY_out/Δlr = -122.102 MeV 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 = 84.4315 MeV, Y₁ = 30.5847 MeV; ΔY_out = -53.8468 MeV; average sensitivity = -122.102 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.
The Uehling potential: vacuum polarisation in position space: infer an input with uncertainty
The Uehling potential is the static potential of a point charge Ze with the one-loop electron vacuum polarisation included: Coulomb's law with the measured α, plus a sum of Yukawa potentials, one for each virtual-pair mass, felt only within about 193 fm. Use the supplied topic model Y_out = 110.791/10^lr, where Y_out is EQUIVALENT Q = 0.561/r (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lr is Probe radius log₁₀(r/fm), dimensionless. Keep lm (Heaviest pair kept log₁₀(M/MeV)) = 4.1 fixed. For this exercise use only -0.575 ≤ lr ≤ 2.575 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 = 4.9888964 MeV, with a bounded reading uncertainty ±0.70412 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lr by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315, 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.
EQUIVALENT Q = 0.561/r against lr
Predictions from the supplied The Uehling potential: vacuum polarisation in position space 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): EQUIVALENT Q = 0.561/r / MeV
Horizontal axis (x): lr: Probe radius log₁₀(r/fm)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lr: Probe radius log₁₀(r/fm) | EQUIVALENT Q = 0.561/r / MeV |
|---|---|
| -0.575 | 416.3940185852102 |
| 0.2124999999999999 | 67.9210642680708 |
| 0.9999999999999998 | 11.079100000000006 |
| 1.7874999999999999 | 1.807192777862615 |
| 2.5749999999999993 | 0.2947843901001704 |
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(lr) = 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(-0.575) = 416.394 MeV and Y_out(2.575) = 0.294784 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lr_line = -0.575 + (4.9889 − 416.394)(2.575 − -0.575)/(0.294784 − 416.394) = 2.53946. 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 lr = 1.3465; substituting back gives Y_out = 4.9888964 MeV.
- The interpolation error in the input is |lr_line − lr| = 1.19296. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.2847764 MeV and 5.6930164 MeV. Sort the roots even if the response decreases: 1.28916 ≤ lr ≤ 1.41258.
- The lower and upper input deviations are 0.0573379 and 0.0660763. 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 lr ≈ 1.3465; compatible interval [1.28916, 1.41258]. Endpoint interpolation gives 2.53946.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.