University Quantum Field Theory I · topic UQFTI-13.01
Why a momentum cutoff fails: a photon mass from the regulator
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.
Why a cutoff fails: a photon mass from the regulator: predict a finite change
A sharp momentum cutoff regulates the photon self-energy by discarding loop momenta with |l_E| above Λ; because that ball is not invariant under shifts of l, it adds a non-transverse g^μν Λ² term that acts as a photon mass. Use the supplied topic model Y_out = (2/d)*10^(d*nlog)/(10^(2*nlog)+1), where Y_out is LEFTOVER AT Λ (mₑ^(d−2)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nlog is Cutoff log₁₀(Λ/mₑ), dimensionless. Keep d (Dimension d) = 4 fixed. For this exercise use only 0.31875 ≤ nlog ≤ 2.68125 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 nlog = 0.8385 and 1.16925. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δnlog across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LEFTOVER AT Λ against nlog
Predictions from the supplied Why a cutoff fails: a photon mass from the regulator 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): LEFTOVER AT Λ / mₑ^(d−2)
Horizontal axis (x): nlog: Cutoff log₁₀(Λ/mₑ)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nlog: Cutoff log₁₀(Λ/mₑ) | LEFTOVER AT Λ / mₑ^(d−2) |
|---|---|
| 0.31875 | 1.7636824774054516 |
| 0.9093749999999999 | 32.447203809535196 |
| 1.5 | 499.5004995004995 |
| 2.0906249999999997 | 7589.119549184469 |
| 2.68125 | 115204.14880496229 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of nlog separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At nlog = 0.8385, substitute into Y_out = (2/d)*10^(d*nlog)/(10^(2*nlog)+1): Y₀ = 23.2771 mₑ^(d−2).
- At nlog = 1.16925, the same fixed-parameter model gives Y₁ = 108.513 mₑ^(d−2).
- Signed change: ΔY_out = Y₁ − Y₀ = 85.2361 mₑ^(d−2); the input change is Δnlog = 0.33075.
- Average sensitivity is ΔY_out/Δnlog = 257.706 mₑ^(d−2) 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 = 23.2771 mₑ^(d−2), Y₁ = 108.513 mₑ^(d−2); ΔY_out = 85.2361 mₑ^(d−2); average sensitivity = 257.706 mₑ^(d−2) 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.
Why a cutoff fails: a photon mass from the regulator: infer an input with uncertainty
A sharp momentum cutoff regulates the photon self-energy by discarding loop momenta with |l_E| above Λ; because that ball is not invariant under shifts of l, it adds a non-transverse g^μν Λ² term that acts as a photon mass. Use the supplied topic model Y_out = (2/d)*10^(d*nlog)/(10^(2*nlog)+1), where Y_out is LEFTOVER AT Λ (mₑ^(d−2)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nlog is Cutoff log₁₀(Λ/mₑ), dimensionless. Keep d (Dimension d) = 4 fixed. For this exercise use only 0.31875 ≤ nlog ≤ 2.68125 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 = 1654.203 mₑ^(d−2), with a bounded reading uncertainty ±247.87 mₑ^(d−2). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown nlog by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, 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.
LEFTOVER AT Λ against nlog
Predictions from the supplied Why a cutoff fails: a photon mass from the regulator 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): LEFTOVER AT Λ / mₑ^(d−2)
Horizontal axis (x): nlog: Cutoff log₁₀(Λ/mₑ)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nlog: Cutoff log₁₀(Λ/mₑ) | LEFTOVER AT Λ / mₑ^(d−2) |
|---|---|
| 0.31875 | 1.7636824774054516 |
| 0.9093749999999999 | 32.447203809535196 |
| 1.5 | 499.5004995004995 |
| 2.0906249999999997 | 7589.119549184469 |
| 2.68125 | 115204.14880496229 |
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(nlog) = 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(0.31875) = 1.76368 mₑ^(d−2) and Y_out(2.68125) = 115204 mₑ^(d−2). The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives nlog_line = 0.31875 + (1654.2 − 1.76368)(2.68125 − 0.31875)/(115204 − 1.76368) = 0.352637. 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 nlog = 1.75988; substituting back gives Y_out = 1654.203 mₑ^(d−2).
- The interpolation error in the input is |nlog_line − nlog| = 1.40724. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1406.333 mₑ^(d−2) and 1902.073 mₑ^(d−2). Sort the roots even if the response decreases: 1.72464 ≤ nlog ≤ 1.79019.
- The lower and upper input deviations are 0.0352388 and 0.0303106. 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 nlog ≈ 1.75988; compatible interval [1.72464, 1.79019]. Endpoint interpolation gives 0.352637.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.